$$$.Introduction to Power Electronics.pdf

CONTENTS

INTRODUCTION SIMPLE DIODE CIRCUITS SIMPLE SCR CIRCUITS FULLY-CONTROLLED 1-PH SCR BRIDGE RECTIFIER FULLY-CONTROLLED 3-PH SCR BRIDGE RECTIFIER SEMI-CONTROLLED RECTIFIER CIRCUITS SWITCH-MODE POWER SUPPLY

INTRODUCTION TO POWER ELECTRONICS

DEFINITION

MAIN TASK OF POWER ELECTRONICS

RECTIFICATION

DC-TO-AC CONVERSION

DC-TO-DC CONVERSION

AC-TO-AC CONVERSION

ADDITIONAL INSIGHTS INTO POWER ELECTRONICS

STRUCTURE OF THE ONLINE TEXT ON POWER ELECTRONICS

DEFINITION

Power electronics refers to control and conversion of electrical power by power semiconductor devices wherein these devices operate as switches. Advent of silicon-controlled rectifiers, abbreviated as SCRs, led to the development of a new area of application called the power electronics. Prior to the introduction of SCRs, mercury-arc rectifiers were used for controlling electrical power, but such rectifier circuits were part of industrial electronics and the scope for applications of mercury-arc rectifiers was limited. Once the SCRs were available, the application area spread to many fields such as drives, power supplies, aviation electronics, high frequency inverters and power electronics originated.

MAIN TASK OF POWER ELECTRONICS

Power electronics has applications that span the whole field of electrical power systems, with the power range of these applications extending from a few VA/Watts to several MVA / MW.

The main task of power electronics is to control and convert electrical power from one form to another. The four main forms of conversion are:

Rectification referring to conversion of ac voltage to dc voltage,

DC-to-AC conversion, DC-to DC conversion and AC-to-AC conversion.

"Electronic power converter" is the term that is used to refer to a power electronic circuit that converts voltage and current from one form to another. These converters can be classified as:

Rectifier converting an ac voltage to a dc voltage, Inverter converting a dc voltage to an ac voltage, Chopper or a switch-mode power supply that converts a dc

voltage to another dc voltage, and Cycloconverter and cycloinverter converting an ac voltage

to another ac voltage.

In addition, SCRs and other power semiconductor devices are used as static switches.

RECTIFICATION

Rectifiers can be classified as uncontrolled and controlled rectifiers, and the controlled rectifiers can be further divided into semi-controlled and fully-controlled rectifiers. Uncontrolled rectifier circuits are built with diodes, and fully-controlled

rectifier circuits are built with SCRs. Both diodes and SCRs are used in semi-controlled rectifier circuits.

There are several rectifier circuits rectifier configurations. The popular rectifier configurations are listed below.

Single-phase semi-controlled bridge rectifier, Single-phase fully-controlled bridge rectifier, Three-phase three-pulse, star-connected rectifier, Double three-phase, three-pulse star-connected rectifiers

with inter-phase transformer (IPT), Three-phase semi-controlled bridge rectifier, Three-phase fully-controlled bridge rectifier and Double three-phase fully-controlled bridge rectifiers with

IPT.

Apart from the configurations listed above, there are series-connected and 12-pulse rectifiers for delivering high power output.

Power rating of a single-phase rectifier tends to be lower than 10 kW. Three-phase bridge rectifiers are used for delivering higher power output, up to 500 kW at 500 V dc or even more. For low voltage, high current applications, a pair of three-phase, three-pulse rectifiers interconnected by an inter-phase transformer(IPT) is used. For a high current output, rectifiers with IPT are preferred to connecting devices directly in parallel. There are many applications for rectifiers. Some of them are:

Variable speed dc drives,

Battery chargers, DC power supplies and Power supply for a specific

application like electroplating

DC-TO-AC CONVERSION

The converter that changes a dc voltage to an alternating voltage is called an inverter. Earlier inverters were built with SCRs. Since the circuitry required to turn the SCR off tends to be complex, other power semiconductor devices such as bipolar junction transistors, power MOSFETs, insulated gate bipolar transistors (IGBT) and MOS-controlled thyristors (MCTs) are used nowadays. Currently only the inverters with a high power rating, such as 500 kW or higher, are likely to be built with either SCRs or gate turn-off thyristors(GTOs). There are many inverter circuits and the techniques for controlling an inverter vary in complexity. Some of the applications of an inverter are listed below:

Emergency lighting systems, AC variable speed drives, Uninterrupted power supplies, and Frequency converters.

DC-TO-DC CONVERSION

When the SCR came into use, a dc-to-dc converter circuit was called a chopper. Nowadays, an SCR is rarely used in a dc-to-dc converter. Either a power BJT or a power MOSFET is normally used in such a converter and this converter is called a switch-mode power supply. A switch-mode power supply can be of one of the types listed below:

Step-down switch-mode power supply, Step-up chopper, Fly-back converter and Resonant converter.

The typical applications for a switch-mode power supply or a chopper are:

DC drive Battery charger and DC power supply.

AC-TO-AC CONVERSION

A cycloconverter or a cycloinverter converts an ac voltage, such as the mains supply, to another ac voltage. The amplitude and the frequency of input voltage to a cycloconverter tend to be

fixed values, whereas both the amplitude and the frequency of output voltage of a cycloconverter tend to be variable. On the other hand, the circuit that converts an ac voltage to another ac voltage at the same frequency is known as an ac-chopper. A typical application of a cycloconverter is to use it for controlling the speed of an ac traction motor and most of these cycloconverters have a high power output, of the order a few megawatts and SCRs are used in these circuits. In contrast, low cost, low power cycloconverters for low power ac motors are also in use and many of these circuit tend to use triacs in place of SCRs. Unlike an SCR which conducts in only one direction, a triac is capable of conducting in either direction and like an SCR, it is also a three terminal device. It may be noted that the use of a cycloconverter is not as common as that of an inverter and a cycloinverter is rarely used.

ADDITIONAL INSIGHTS INTO POWER ELECTRONICS

There are several striking features of power electronics, the foremost among them being the extensive use of inductors and capacitors. In many applications of power electronics, an inductor may carry a high current at a high frequency. The implications of operating an inductor in this manner are quite a few, such as necessitating the use of litz wire in place of single-stranded or multi-stranded copper wire at frequencies above 50

kHz, using a proper core to limit the losses in the core, and shielding the inductor properly so that the fringing that occurs at the air-gaps in the magnetic path does not lead to electromagnetic interference. Usually the capacitors used in a power electronic application are also stressed. It is typical for a capacitor to be operated at a high frequency with current surges passing through it periodically. This means that the current rating of the capacitor at the operating frequency should be checked before its use. In addition, it may be preferable if the capacitor has self-healing property. Hence an inductor or a capacitor has to be selected or designed with care, taking into account the operating conditions, before its use in a power electronic circuit.

In many power electronic circuits, diodes play a crucial role. A normal power diode is usually designed to be operated at 400 Hz or less. Many of the inverter and switch-mode power supply circuits operate at a much higher frequency and these circuits need diodes that turn ON and OFF fast. In addition, it is also desired that the turning-off process of a diode should not create undesirable electrical transients in the circuit. Since there are several types of diodes available, selection of a proper diode is very important for reliable operation of a circuit.

Analysis of power electronic circuits tends to be quite complicated, because these circuits rarely operate in steady-state. Traditionally steady-state response refers to the state of a circuit characterized by either a dc response or a sinusoidal response. Most of the power electronic circuits have a periodic response, but this response is not usually sinusoidal. Typically,

the repetitive or the periodic response contains both a steady-state part due to the forcing function and a transient part due to the poles of the network. Since the responses are nonsinusoidal, harmonic analysis is often necessary. In order to obtain the time response, it may be necessary to resort to the use of a computer program.

Power electronics is a subject of interdisciplinary nature. To design and build control circuitry of a power electronic application, one needs knowledge of several areas, which are listed below.

Design of analogue and digital electronic circuits, to build the control circuitry.

Microcontrollers and digital signal processors for use in sophisticated applications.

Many power electronic circuits have an electrical machine as their load. In ac variable speed drive, it may be a reluctance motor, an induction motor or a synchronous motor. In a dc variable speed drive, it is usually a dc shunt motor.

In a circuit such as an inverter, a transformer may be connected at its output and the transformer may have to operate with a nonsinusoidal waveform at its input.

A pulse transformer with a ferrite core is used commonly to transfer the gate signal to the power semiconductor device. A ferrite-cored transformer with a relatively higher power output is also used in an application such as a high frequency inverter.

Many power electronic systems are operated with negative

feedback. A linear controller such as a PI controller is used in relatively simple applications, whereas a controller based on digital or state-variable feedback techniques is used in more sophisticated applications.

Computer simulation is often necessary to optimize the design of a power electronic system. In order to simulate, knowledge of software package such as MATLAB and the know-how to model nonlinear systems may be necessary.

The study of power electronics is an exciting and a challenging experience. The scope for applying power electronics is growing at a fast pace. New devices keep coming into the market, sustaining development work in power electronics.

SIMPLE DIODE CIRCUITS

This chapter describes two simple diode circuits:

A SINGLE DIODE CIRCUIT A DIODE CIRCUIT WITH A FREE-WHEELING

DIODE

A SINGLE DIODE CIRCUIT

CIRCUIT OPERATIONMATHEMATICAL ANALYSISINTERACTIVE SIMULATIONPSPICE SIMULATIONMATLAB SIMULATIONMATHCAD SIMULATIONSUMMARY

This page describes a single diode circuit. Most of the power electronic applications operate at a relative high voltage and in such cases, the voltage drop across the power diode tends to be small. It is quite often justifiable to use the ideal diode model. An ideal diode has zero conduction drop when it is forward-biased and has zero current when it is reverse-biased. The explanation and the analysis presented below is based on the ideal diode model.

CIRCUIT OPERATION

A circuit with a single diode and an RL load is shown above. The source vs is an alternating

sinusoidal source. If vs = E * sin (wt), vs is positive when 0 < wt < π, and vs is negative when π <

wt <2π. When vs starts becoming positive, the diode starts conducting and the source keeps the

diode in conduction till wt reaches π radians. At that instant defined by wt = π radians, the current through the circuit is not zero and there is some energy stored in the inductor. The voltage across an inductor is positive when the current through it is increasing and it becomes negative when the current through it tends to fall. When the voltage across the inductor is negative, it is in such a direction as to forward-bias the diode. The polarity of voltage across the inductor is as shown in the sketches shown below.

When vs changes from a positive to a negative value, there is current through the load at the

instant wt = π radians and the diode continues to conduct till the energy stored in the inductor becomes zero. After that the current tends to flow in the reverse direction and the diode blocks conduction. The entire applied voltage now appears across the diode.

MATHEMATICAL ANALYSIS

An expression for the current through the diode can be obtained as shown below. It is assumed that the current flows for 0 < wt < β, where β > π . When the diode conducts, the driving function for the differential equation is the sinusoidal function defining the source voltage. During the period defined by β < wt < 2π, the diode blocks current and acts as an open switch. For this period, there is no equation defining the behaviour of the circuit. For 0 < wt < β , the equation (1) defined below applies.

Given a linear differential equation, the solution is found out in two parts. The homogeneous equation is defined by equation (2). It is preferable to express the equation in terms of the angle θ instead of 't'. Since θ = wt, we get that dθ = w.dt. Then equation (2) then gets converted to equation (3). Equation (4) shown above is the solution to this homogeneous equation and is called the complementary integral.

The value of constant A in the complimentary solution is to be evaluated later.

The particular solution is the steady-state response and equation (5) expresses the particular solution. The steady-state response is the current that would flow in steady-state in a circuit that contains only the source, the resistor and the inductor shown in the circuit above, the only element missing being the diode. This response can be obtained using the differential equation or the Laplace transform or the ac sinusoidal circuit analysis. The total solution is the sum of both the complimentary and the particular solution and it is shown as equation (6). The value of A is obtained using the initial condition. Since the diode starts conducting at wt = 0 and the current starts building up from zero, i(0) = 0. The value of A is expressed by equation (7).

Once the value of A is known, the expression for current is known. After evaluating A, current can be evaluated at different values of wt, starting from wt = π. As wt increases, the current would keep decreasing. For some value of wt, say β , the current would be zero. If wt > β , the current would evaluate to a negative value. Since the diode blocks current in the reverse direction, the diode stops conducting when wt reaches β. Then an expression for the average output voltage can be obtained.

Since the average voltage across the inductor has to be zero, the average voltage across the resistor and the average voltage at the cathode of the diode are the same. This average value can be obtained as shown in equation (8).

INTERACTIVE SIMULATION

The operation of the circuit can be simulated as shown below. In order to simulate, the solution for current is presented in the following form, where τ = (wL)/R. Then

Again it is preferable to normalize. Here E is set to unity and E/R is also set to unity. Then

vs = sin (wt).

vo = i for 0 < wt < β,

vL = vs - i for 0 < wt < β .

To solve the expression, all we need to know is then the ratio τ. The applet shown below simulates this circuit. You have to key-in the ratio τ and then click on the button next to it. Do not key-in a NaN.

The next page presents the same circuit with an additional diode.

PSPICE SIMULATION

For simulation using Pspice, the circuit used is shown below. Here the nodes are numbered. The ac source is connected between nodes 1 and 0. The diode is connected between nodes 1 and 2 and the inductor links nodes 2 and 3. The resistor is connected from 3 to the reference node, that is, node 0.

The Pspice program is presented below.

* First Chapter: Half-wave Rectifier with RL Load* A problem to find the diode currentVIN 1 0 SIN(0 340V 50Hz)D1 1 2 DNAMEL1 2 3 31.8MHR1 3 0 10.MODEL DNAME D(IS=10N N=1 BV=1200 IBV=10E-3 VJ=0.6).TRAN 10US 60.0MS 20.0MS 10US.PROBE.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV).END

The diode is described using the MODEL statement. The TRAN statement simulates the transient operation for a period of 60 ms at an interval of 10 µs. The OPTIONS statement sets limits for tolerances. The output can be

viewed on the screen because of the PROBE statement. A snapshot of output is presented below.

MATLAB SIMULATION

The Matlab program used is re-produced below.

% Program to simulate the half-wave rectifier circuit% Enter the peak voltage, frequency, inductance L in mH and resistor R disp('Typical value for peak voltage is 340 V')peakV=input('Enter Peak voltage in Volts>');disp('Typical value for line frequency is 50 Hz')freq=input('Enter line frequency in Hz>');disp('Typical value for Load inductance is 31.8 mH')L=input('Enter Load inductance in mH>');disp('Typical value for Load Resistance is 10.0 Ohms')R=input('Enter Load Resistance in Ohms>');

w=2.0*pi*freq;X=w*L/1000.0;if (X<0.001) X=0.001; end;Z=sqrt(R*R+X*X);

loadAng=atan(X/R);A=peakV/Z*sin(loadAng);tauInv=R/X;

for n=1:360; theta=n/180.0*pi; X(n)=n; cur=peakV/Z*sin(theta-loadAng)+A*exp(-tauInv*theta); if (cur>0.0) Vind(n)=peakV*sin(theta)-R*cur; iLoad(n)=cur; Vout(n)=peakV*sin(theta); else Vind(n)=0; iLoad(n)=0; Vout(n)=0; end;end;

plot(X,iLoad)title('The diode current')xlabel('degrees')ylabel('Amps') gridpause

plot(X,Vout)title('Voltage at cathode')xlabel('degrees')ylabel('Volts') gridpause

plot(X,Vind)title('Inductor Voltage')xlabel('degrees')ylabel('Volts') grid

The plots obtained for the typical values mentioned are shown below.

It can be seen from the waveform of voltage across the inductor is that the area above the x-axis at 0 V is equal to its area below the x-axis. It can be seen that the matlab program is relatively simple.

MATHCAD SIMULATION

The simulation of this circuit is in a file called halfrec1.mcd. You can download it by clicking on the image below. You need MathCad program to open this file. If you open this file with MathCad program, you can change the parameters and see how the waveforms of output voltage, diode current and voltage across the inductor.

Alternatively, you can view the mathcad file by clicking on the image below. Then you see the file in HTML format, but you cannot change the parameters.

This page has described the circuit of a half-wave rectifier. It has been simulated using different programs. The next page is on a half-wave rectifier with a free-wheeling diode.

A DIODE CIRCUIT WITH A FREE-WHEELING DIODE

CIRCUIT DIAGRAMCIRCUIT OPERATIONMATHEMATICAL_ANALYSISSIMULATION

PSPICE SIMULATIONMATLAB SIMULATIONMATHCAD SIMULATIONSUMMARY

CIRCUIT DIAGRAM

The circuit shown above differs from the circuit described in the previous page, which had only diode D1. This circuit

has another diode, marked D2 in the circuit shown above. This diode is called the free-wheeling diode. The circuit

operation is described next. The explanation is based on the assumption that the reader knows how the circuit without a free-wheeling diode operates.

CIRCUIT OPERATION Let the source voltage vs be defined to be E*sin (wt). The source voltage is positive when 0 < wt < π radians and it is

negative when π < wt < 2π radians. When vs is positive, diode D1 conducts and the voltage vc is positive. This in turn

leads to diode D2 being reverse-biased during this period. During π < wt < 2π, the voltage vc would be negative if

diode D1 tends to conduct. This means that D2 would be forward-biased and would conduct. When diode D2 conducts,

the voltage vc would be zero volts, assuming that the diode drop is negligible. Additionally when diode D2 conducts,

diode D1 remains reverse-biased, because the voltage vs is negative.

When the current through the inductor tends to fall, it starts acting as a source. When the inductor acts as a source, its voltage tends to forward bias diode D2 if the source voltage vs is negative and forward bias diode D1 if the source

voltage vs is positive. Even when the source voltage vs is positive, the inductor current would tend to fall if the source

voltage is less than the voltage drop across the load resistor.During the negative half-cycle of source voltage, diode D1 blocks conduction and diode D2 is forced to conduct. Since

diode D2 allows the inductor current circulate through L, R and D2, diode D2 is called the free-wheeling diode. We can

say that the current free-wheels through D2.

MATHEMATICAL ANALYSIS An expression for the current through the load can be obtained as shown below. It can be assumed that the load current flows all the time. In other words, the load current is continuous. When diode D1 conducts, the driving function for the

differential equation is the sinusoidal function defining the source voltage. During the period defined by π < wt < 2π,

diode D1 blocks current and acts as an open switch. On the other hand, diode D2 conducts during this period, the driving

function can be set to be zero volts. For 0 < wt < π , the equation (1) shown below applies.For the negative half-cycle of the source, equation (2) applies. As in the previous case, the solution is obtained in two parts. The expressions for the complementary integral and the particular integral are the same. The expression for the complementary integral is presented by equation (3). The particular solution to the equation (1) is the steady-state response and is presented as equation (4). The total solution is the sum of both the complimentary and the particular solution. For 0 < θ < π, where wt = θ, the total solution is presented as equation (5).

The difference in solution is in how the constant A in complementary integral is evaluated. In the case of the circuit without free-wheeling diode, i(0) = 0, since the current starts building up from zero at the start of every positive half-cycle. On the other hand, the current-flow is continuous when the circuit contains a free-wheeling diode also. Since the input to the RL circuit is a periodic half-sinusoid function, we expect that the response of the circuit should also be periodic. That means, the current through the load is periodic. It means that i(0) = i(2π).

Since the current through the load free-wheels during π < θ < 2π , we get equation (6). We use ( θ - π ) for the elapsed period in radians instead of θ itself, since the free-wheeling action starts at θ = π . From the total solution, we can get i(π) from equation (7) by substituing θ = π. To obtain A, the following steps are necessary. From the total solution, obtain an expression for i(0) by substituting 0 for θ. Also obtain an expression for i(π) by substituting π for θ in equation (7). Using this expression for i(π) in equation (6), obtain i(2π) by letting θ = 2π . Since i(0) = i(2π), we can obtain A from equation (8). In equation (8), the terms containing constant A are grouped on the left-hand side of equation and the other terms on the right-hand side.

SIMULATION

The operation of the circuit can be simulated as shown below. During 0 < θ < π , the expression for current is presented as equation (9). During π < θ < 2π , the expression for current is shown as equation (10).

The voltage across the inductor is obtained to be

vL(θ ) = vs(θ) - R*i(θ) , for 0 < θ < π and

= - R*i(θ) , for π < θ < 2π .

vs(θ) = E*sin (θ).

It is preferable to normalize vs with respect to E and the current with respect to E/R and then the only value required to

be known for solving for the current is τ.

PSPICE SIMULATION

The circuit that is used for Pspice simulation is shown below. The nodes are numbered and the components have been labeled.

A Pspice program to simulate the circuit shown above is presented now.

* Half-wave Rectifier with free-wheeling diode and with RL Load* A problem to find the diode currentVIN 1 0 SIN(0 340V 50Hz)D1 1 2 DNAMEL1 2 3 31.8MHR1 3 0 10D2 0 2 DNAME.MODEL DNAME D(IS=10N N=1 BV=1200 IBV=10E-3 VJ=0.6).TRAN 10US 60.0MS 20.0MS 10US.PROBE.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV).END

The waveforms obtained are presented now. Since the program specifies that the waveforms be displayed from the second cycle, there is no output for the first 20 ms. The waveform of voltage at the cathode of both diodes is shown below.

The waveform of current through diode D1 is presented next.

The waveform of current through diode D2 is shown below.

The waveform of current through load resistor is shown below. It is the sum of both diode currents.

The waveform of voltage across the inductor is shown below.

The advantage with Pspice is the simplicity of the program. In addition, the devices used are also simulated using the spice models.

MATLAB SIMULATION

A matlab program for simulating the half-wave rectifier with a free-wheeling diode is presented below.

% Program to simulate the half-wave rectifier circuit% The circuit has a free-wheeling diode% Enter the peak voltage, frequency, inductance L in mH and resistor R disp('Typical value for peak voltage is 340 V')peakV=input('Enter Peak voltage in Volts>');disp('Typical value for line frequency is 50 Hz')freq=input('Enter line frequency in Hz>');disp('Typical value for Load inductance is 31.8 mH')L=input('Enter Load inductance in mH>');disp('Typical value for Load Resistance is 10.0 Ohms')R=input('Enter Load Resistance in Ohms>');

w=2.0*pi*freq;X=w*L/1000.0;if (X<0.001) X=0.001; end;Z=sqrt(R*R+X*X);tauInv=R/X;loadAng=atan(X/R);A1=peakV/Z*sin(loadAng);A2=peakV/Z*sin(pi-loadAng)*exp(-pi*tauInv);A3=(A1+A2)/(1-exp(-2.0*pi*tauInv));Ampavg=0;AmpRMS=0;

for n=1:360; theta=n/180.0*pi; X(n)=n; if (n<180) cur=peakV/Z*sin(theta-loadAng)+A3*exp(-tauInv*theta); Ampavg=Ampavg+cur*1/360;

AmpRMS=AmpRMS+cur*cur*1/360; else A4=peakV/Z*sin(pi-loadAng)*exp(-(theta-pi)*tauInv); cur=A4+A3*exp(-tauInv*theta); Ampavg=Ampavg+cur*1/360; AmpRMS=AmpRMS+cur*cur*1/360; end; if (n<180) Vind(n)=peakV*sin(theta)-R*cur; Vout(n)=peakV*sin(theta); diode2cur(n)=0; diode1cur(n)=cur; else Vind(n)=-R*cur; Vout(n)=0; diode2cur(n)=cur; diode1cur(n)=0; end iLoad(n)=cur;end;

plot(X,iLoad)title('The Load current')xlabel('degrees')ylabel('Amps') gridpause


plot(X,Vind)title('Inductor Voltage')xlabel('degrees')ylabel('Volts') gridpause

plot(X,diode1cur)title('Diode 1 current')xlabel('degrees')ylabel('Amps') gridpause

plot(X,diode2cur)title('Diode 2 current')xlabel('degrees')ylabel('Amps')

grid

AmpRMS=sqrt(AmpRMS);[A,message]=fopen('outhfr2.dat','w'); fprintf(A,'Avg Load Cur=\t%d\tRMS Load Cur=\t%f\n',Ampavg,AmpRMS);fclose(A)

The responses obtained for the typical specified values are shown below.

The average and rms values of load current are presented below.

Avg Load Cur= 1.082254e+001 RMS Load Cur= 13.954542

MATHCAD SIMULATION

This circuit can be simulated using MathCad as shown next. Click on DOWNLOAD to download the file containing the program and then open it using MathCad. Click on VIEW ONLY to view this file in HTML format.

SUMMARY

It has been shown how the half-wave rectifier with a free-wheeling diode can be simulated using different software packages. The next page shows how a half-wave controlled rectifier circuit operates.

SIMPLE SCR CIRCUITS

This chapter describes two SCR circuits which are:

A SINGLE SCR CIRCUIT A SINGLE-SCR CIRCUIT WITH A FREE-WHEELING

DIODE

A SINGLE SCR CIRCUIT

CIRCUIT OPERATIONMATHEMATICAL ANALYSISSIMULATIONPSPICE SIMULATIONMATLAB SIMULATION

MATHCAD SIMULATIONSUMMARY

This page describes a circuit with a single SCR. It is similar to the single diode circuit, the difference being that an SCR is used in place of the diode. Most of the power electronic applications operate at a relative high voltage and in such cases, the voltage drop across the SCR tends to be small. It is quite often justifiable to assume that the conduction drop across the SCR is zero when the circuit is analysed. It is also justifiable to assume that the current through the SCR is zero when it is not conducting. It is known that the SCR can block conduction in either direction. The explanation and the analysis presented below is based on the ideal SCR model. It is also assumed that the reader knows how an SCR operates.

CIRCUIT OPERATION

A circuit with a single SCR and an RL load is shown above. The source vs is an alternating sinusoidal source. If vs = E

* sin (wt), vs is positive when 0 < wt < π, and vs is negative when π < wt <2π. When vs starts becoming positive, the

SCR is forward-biased but remains in the blocking state till it is triggered. If the SCR is triggered at when wt = α, then α is called the firing angle. When the SCR is triggered in the forward-bias state, it starts conducting and the positive source keeps the SCR in conduction till wt reaches π radians. At that instant, the current through the circuit is not zero and there is some energy stored in the inductor at wt = π radians. The voltage across an inductor is positive when the current through it is increasing and it becomes negative when the current through the inductor tends to fall. When the voltage across the inductor is negative, it is in such a direction as to forward-bias the SCR.

There is current through the load at the instant wt = π radians and the SCR continues to conduct till the energy stored in the inductor becomes zero. After that the current tends to flow in the reverse direction and the SCR blocks conduction. The entire applied voltage now appears across the diode.


An expression for the current through the SCR can be obtained as shown below. It is assumed that the current flows for α < wt < δ, where δ > π . When the SCR conducts, the driving function for the differential equation is the sinusoidal function defining the source voltage. Outside this period, the SCR blocks current and acts as an open switch. For this period, there is no equation defining the behaviour of the circuit. For α < wt < δ , equation (1) applies. Given a linear differential equation, the solution is found out in two parts. The homogeneous equation is given by equation (2), where α is the firing angle. The value of constant A in the complimentary solution is to be evaluated later. The particular solution is the steady-state response and is diplayed as equation (3). The total solution is the sum of both the complimentary and the particular solution and is presented as equation (4). The value of A is obtained using the initial condition. Since the SCR starts conducting at wt = α and the current starts building up from zero, i(α) = 0. In the expression above τ = wL/R. Then A can be expressed as in equation (5).

Once the value of A is known, the expression for current is known. When the firing angle α and the extinction angle δ are known, the average output voltage at the cathode of the SCR can be evaluated as shown in equation (6).

The average load current can be obtained by dividing the average load voltage by the load resistance, since the average voltage across the inductor is zero.

SIMULATION

The operation of the circuit can be simulated as shown below. In order to simulate, the solution for current is presented in the following form, where τ = (wL)/R. Then

Again it is preferable to normalize. Here E is set to unity and E/R is also set to unity. Then

vs = sin (wt).

vo = vs - i for α < wt < δ, and

vL = vs - i for α < wt < δ .

To solve the expression, all we need to know is then the ratio τ. The applet shown below simulates this circuit. You have to key-in the ratio τ and then click on the button next to it. Do not key-in a NaN. Enter the ratio τ in the left text-field to the left of the click button and the firing angle in degrees in the textfield to its right.

The next page presents the same circuit with a free-wheeling diode.

PSPICE SIMULATIONThe program below presents a PSPICE program. The circuit used is shown below.

The PSPICE program shown below presents the SCR as a subcircuit. This model of the SCR has been described in the book SPICE FOR POWER ELECTRONICS AND ELECTRIC POWER (Muhammed H.Rashid, Prentice-Hall , 1993, pages 148-160).

* Half-wave Rectifier with RL Load* A problem to find the SCR currentVIN 1 0 SIN(0 340V 50Hz)XT1 1 2 5 2 SCRVP 5 2 PULSE(0 10 1667U 1N 1N 100U 20M)L1 2 3 31.8MHR1 3 0 10* Subcircuit for SCR.SUBCKT SCR 101 102 103 102S1 101 105 106 102 SMODRG 103 104 50VX 104 102 DC 0VY 105 107 DC 0DT 107 102 DMODRT 106 102 1CT 106 102 10UF1 102 106 POLY(2) VX VY 0 50 11.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0).MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0).ENDS SCR.TRAN 10US 60.0MS 20.0MS 10US.PROBE.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV).END

The waveforms obtained are presented below.

The voltage waveform at the cathode of the SCR

The load current waveform

The inductor voltage waveform

The voltage waveform acros the SCR

The voltage waveform of the pulse source used for triggering the SCR

MATLAB SIMULATIONThe Matlab program used for simulation is presented below.

% Program to simulate the half-wave controlled rectifier circuit% Enter the peak voltage, frequency, inductance L in mH and resistor R disp('Typical value for peak voltage is 340 V')peakV=input('Enter Peak voltage in Volts>');disp('Typical value for line frequency is 50 Hz')freq=input('Enter line frequency in Hz>');disp('Typical value for Load inductance is 31.8 mH')L=input('Enter Load inductance in mH>');disp('Typical value for Load Resistance is 10.0 Ohms')R=input('Enter Load Resistance in Ohms>');disp('Typical value for Firing angle is 30.0 degree')fangDeg=input('Enter Firing angle within range 0 to 180 in deg>');fangRad=fangDeg/180.0*pi;

w=2.0*pi*freq;X=w*L/1000.0;if (X<0.001) X=0.001; end;Z=sqrt(R*R+X*X);tauInv=R/X;loadAng=atan(X/R);A=peakV/Z*sin(loadAng-fangRad);

Ampavg=0;AmpRMS=0;

for n=1:360; theta=n/180.0*pi; X(n)=n; if (n<fangDeg) cur=0.0; Vind(n)=0; iLoad(n)=0; Vout(n)=0; else cur=peakV/Z*sin(theta-loadAng)+A*exp(-tauInv*(theta-fangRad)); if (cur>0) Ampavg=Ampavg+cur*1/360; AmpRMS=AmpRMS+cur*cur*1/360; Vind(n)=peakV*sin(theta)-R*cur; iLoad(n)=cur; Vout(n)=peakV*sin(theta); else Vind(n)=0; iLoad(n)=0; Vout(n)=0; end; end;end; plot(X,iLoad)title('The Load current')xlabel('degrees')ylabel('Amps') gridpause


plot(X,Vind)title('Inductor Voltage')xlabel('degrees')ylabel('Volts') grid

AmpRMS=sqrt(AmpRMS);[A,message]=fopen('hwavec1.dat','w'); fprintf(A,'Avg Load Cur=\t%d\tRMS Load Cur=\t%f\n',Ampavg,AmpRMS);fclose(A)

The waveforms obtained for the typical specified values are displayed now.

The output file containing the values of average load current and the RMS load current is presented below.


MATHCAD SIMULATIONThe MathCad program can be downloaded by clicking on the image below.

This program can also be viewed in the HTML format. Click on the image below to view this file.

SUMMARY

This page has described how the half-wave controlled rectifier circuit operates. The next page shows how the behaviour of this circuit can be changed by adding a free-wheeling diode.

A SINLGE-SCR CIRCUIT WITH A FREE-WHEELING DIODE

CIRCUIT DIAGRAMCIRCUIT OPERATIONMATHEMATICAL ANALYSISSIMULATIONPSPICE SIMULATIONMATLAB SIMULATIONMATHCAD SIMULATIONSUMMARY

CIRCUIT DIAGRAM

The circuit shown above differs from the circuit described in the previous page, which had only a single SCR. This circuit has a free-wheeling diode in addition, marked D in the circuit shown above. The circuit operation is described next. The explanation is based on the assumption that both the diode and the SCR are ideal. It means that the voltage drop across the device while in conduction is zero and the leakage current in the blocking state is zero.

CIRCUIT OPERATION

The source vs is an alternating sinusoidal source. If vs = E * sin (wt), vs is positive when 0 < wt < π, and it is negative

when π < wt <2π. When vs starts becoming positive, the SCR is forward-biased but remains in the blocking state till it

is triggered. If the SCR is triggered at when wt = α, then α is called the firing angle. When the SCR is triggered in the forward-bias state, it starts conducting and the positive source keeps the SCR in conduction till wt reaches π radians. At that instant, the current through the circuit is not zero and there is some energy stored in the inductor at wt = π radians. In the absence of the free wheeling diode, the inductor would keep the SCR in conduction for part of the negative cycle till the energy stored in it is discharged. But when a free wheeling diode is present as shown in the circuit, the current has a path that offers almost zero resistance. Hence the inductor discharges its energy during π < wt < (2π + α) through the free wheeling diode. When there is a free wheeling diode, the current through the load tends to be continuous, at least under ideal conditions. When the diode conducts, the SCR remains reverse-biased, because the voltage vs is negative.

MATHEMATICAL ANALYSISAn expression for the current through the load can be obtained as shown below. It can be assumed that the load current flows all the time. In other words, the load current is continuous. When the SCR conducts, the driving function for the

differential equation is the sinusoidal function defining the source voltage. During the period defined by π < wt < (2π + α), the SCR blocks current and acts as an open switch. On the other hand, the free wheeling diode conducts during this

period, and the driving function can be set to be zero volts. For α < wt < π , equation (1) applies whereas equation (2) applies for the rest of the cycle.

As in the previous cases, the solution is obtained in two parts. The expressions for the complementary integral and the particular integral are the same. The expression for the complementary integral is presented as equation (3). The particular solution is the steady-state response and is presented as equation (4).

The total solution is the sum of both the complimentary and the particular solution for α < wt < π .

The difference in solution is in how the constant A in complementary integral is evaluated. In the case of the circuit without free-wheeling diode, i(α) = 0, since the current starts building up from zero when the SCR is triggered during the positive half-cycle. On the other hand, the current-flow is continuous, when there is a free wheeling diode. Since the input to the RL circuit is a periodic function, we expect that the response of the circuit should also be periodic. That means, the current through the load is periodic. It means that

i(α) = i(2π + α).

Since the current through the load free-wheels during π < θ < (2π + α) , we get that,

i(θ) = i(π) *exp[-(θ − π)/τ] , where τ = wL/R.

We use ( θ - π ) for the elapsed period in radians instead of θ itself, since the free-wheeling action starts at θ = π . From the total solution, we get i(π).

i(π) = A*exp(- π/τ) + (E/Z)*sin (π - α).

To obtain A, the following steps are necessary. From the total solution, obtain an expression for i(α) by substituting α for θ. From the expression for the free-wheeling period, obtain i(2π + α) by letting θ = (2π + α) . Since i(α) = i(2π + α), we can obtain A to be:

SIMULATION

The operation of the circuit can be simulated as shown below. During 0 < θ < π , the expression for current is:i(θ) = A * exp [ -(θ − α)/τ) + (E/Z)*sin (θ − β ),where τ = wL/R, β = atan (τ), Z2 = R2 + (wL)2 = R2*(1 + τ2) and α is the firing angle.During π < θ < (2π + α) , the expression for current is:

i(θ) = A*exp - (θ − α) /τ + (E/Z)*sin (π - β)*exp - (θ − π) /τ .

The voltage across the inductor is obtained to be

vL(θ ) = vs(θ) - R*i(θ) , for α < θ < π and

= - R*i(θ) , for π < θ < (2π + α) .

vs(θ) = E*sin (θ).

It is preferable to normalize vs with respect to E and the current with respect to E/R and then the only value that is

required to be known for solving for the current is τ. The applet shown below simulates this circuit. You have to key-in the ratio τ in the text-field to the left of the button and the firing angle in degrees in the text-field to the right of the button and then click on the button next to it. Do not key-in a NaN.

PSPICE SIMULATIONThe circuit used for Pspice simulation is shown below.

The Pspice program is presented next.

* Half-wave Rectifier with a free-wheeling diode* A problem to find the SCR currentVIN 1 0 SIN(0 340V 50Hz)

XT1 1 2 5 2 SCRVP 5 2 PULSE(0 10 1667U 1N 1N 100U 20M)L1 2 3 31.8MHR1 3 0 10D2 0 2 DNAME.MODEL DNAME D(IS=10N N=1 BV=1200 IBV=10E-3 VJ=0.6)* Subcircuit for SCR.SUBCKT SCR 101 102 103 102S1 101 105 106 102 SMODRG 103 104 50VX 104 102 DC 0VY 105 107 DC 0DT 107 102 DMODRT 106 102 1CT 106 102 10UF1 102 106 POLY(2) VX VY 0 50 11.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0).MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0).ENDS SCR.TRAN 10US 60.0MS 20.0MS 10US.PROBE.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV).END

The waveform of voltage at the cathode of SCR

The waveform of current through the load

The waveform of voltage across the SCR

The waveform of current through the SCR

The waveform of current through the free-wheeling diode

The waveform of voltage across the inductor

MATLAB SIMULATIONThe Matlab program is presented below.

% Program to simulate the half-wave controlled rectifier circuit% The circuit has a free-wheeling diode% Enter the peak voltage, frequency, inductance L in mH and resistor R disp('Typical value for peak voltage is 340 V')peakV=input('Enter Peak voltage in Volts>');disp('Typical value for line frequency is 50 Hz')freq=input('Enter line frequency in Hz>');disp('Typical value for Load inductance is 31.8 mH')L=input('Enter Load inductance in mH>');disp('Typical value for Load Resistance is 10.0 Ohms')R=input('Enter Load Resistance in Ohms>');

disp('Typical value for Firing angle is 30.0 degree')fangDeg=input('Enter Firing angle within range 0 to 180 in deg>');fangRad=fangDeg/180.0*pi;

w=2.0*pi*freq;X=w*L/1000.0;if (X<0.001) X=0.001; end;Z=sqrt(R*R+X*X);tauInv=R/X;loadAng=atan(X/R);k1=exp(R*(pi+fangRad)/X);k2=exp(R*(fangRad-pi)/X);A=peakV/Z*(sin(pi-loadAng)+sin(loadAng-fangRad)*k1)/(k1-k2);

Ampavg=0;AmpRMS=0;

Cur180=peakV/Z*sin(pi-loadAng)+A*k2;

for n=1:360; theta=n/180.0*pi; X(n)=n; if (n<fangDeg) cur=Cur180*exp(-(pi+theta)*tauInv); Vind(n)=-R*cur; iLoad(n)=cur; Vout(n)=0; VSCR(n)=peakV*sin(theta); curSCR(n)=0; diodecur(n)=cur; Ampavg=Ampavg+cur*1/360; AmpRMS=AmpRMS+cur*cur*1/360; elseif ((n>=fangDeg) & (n<180)) cur=peakV/Z*sin(theta-loadAng)+A*exp(-tauInv*(theta-fangRad)); Ampavg=Ampavg+cur*1/360; AmpRMS=AmpRMS+cur*cur*1/360; Vind(n)=peakV*sin(theta)-R*cur; iLoad(n)=cur; Vout(n)=peakV*sin(theta); VSCR(n)=0; curSCR(n)=cur; diodecur(n)=0; else cur=Cur180*exp((pi-theta)*tauInv); Vind(n)=-R*cur; iLoad(n)=cur; Vout(n)=0; VSCR(n)=peakV*sin(theta); curSCR(n)=0; diodecur(n)=cur; Ampavg=Ampavg+cur*1/360; AmpRMS=AmpRMS+cur*cur*1/360; end;end;

plot(X,iLoad)title('The Load current')xlabel('degrees')ylabel('Amps') gridpause


plot(X,Vind)title('Inductor Voltage')xlabel('degrees')ylabel('Volts') gridpause

plot(X,VSCR)title('SCR Voltage')xlabel('degrees')ylabel('Volts') gridpause

plot(X,curSCR)title('SCR Current')

xlabel('degrees')ylabel('Amps') gridpause

plot(X,diodecur)title('diode Current')xlabel('degrees')ylabel('Amps') grid

AmpRMS=sqrt(AmpRMS);[C,message]=fopen('hwavec2.dat','w'); fprintf(C,'Avg Load Cur=\t%d\tRMS Load Cur=\t%f\n',Ampavg,AmpRMS);fclose(C)

The output file is re-produced below.


Next the plots obtained for the typical specified values are displayed.

This page has described the operation of a half-wave rectifier wirh a free-wheeling diode. Next we take up the study of

single-phase, full-wave, fully-controlled rectifier.

FULLY-CONTROLLED 1-PH SCR BRIDGE RECTIFIER

This chapter describes the operation of a single-phase fully-controlled SCR bridge rectifier circuit and two applications. The pages to follow contain:

OPERATION WITH A PURELY RESISIVE LOAD OPERATION WITH AN RL LOAD OPERATION WITH SOURCE INDUCTANCE OPERATION WITH AN RLC LOAD AND SOURCE

INDUCTANCE AN APPLICATION: A BATTERY CHARGER AN APPLICATION: A TWO-QUADRANT DC DRIVE

FULLY-CONTROLLED SINGLE-PHASE SCR BRIDGE RECTIFIER: RESISTIVE LOAD

CIRCUIT OPERATIONMATHEMATICAL ANALYSISPERFORMANCE PARAMETERS FOR CONVERTERSSIMULATIONPSPICE SIMULATIONMATHCAD SIMULATIONSUMMARY

This chapter describes the operation of a single-phase fully- controlled bridge rectifier circuit with a resistive load. The operation of this circuit can be understood more easily when the load is purely resistive. The analysis in this page is based on the assumption that the SCRs are ideal controlled switches. It means that when the SCRs are ON, the ON-state drops are zero. In the OFF- state, the leakage current is assumed to be zero. The main purpose of the fully-controlled bridge rectifier circuit is to provide a variable dc voltage from an ac source.

CIRCUIT OPERATION

The circuit of a single-phase fully-controlled bridge rectifier circuit is shown in the figure above. The circuit has four SCRs. It is preferable to state that the circuit has two pairs of SCRs, with S1 and S4 forming one pair and, S2 and S3

the other pair. For this circuit, vs is a sinusoidal voltage source. When it is positive, SCRs S1 and S4 can be triggered

and then current flows from vs through SCR S1, load resistor R, SCR S4 and back into the source. In the next half-

cycle, the other pair of SCRs conducts. Even though the direction of current through the source alternates from one half-cycle to the other half-cycle, the current through the load remains unidirectional.The main purpose of this circuit is to provide a variable dc output voltage, which is brought about by varying the firing angle. Let vs = E sin wt, with 0 < wt < 360o. If wt = 30o when S1 and S4 are triggered, then the firing angle is said to be

30o. In this instance, the other pair is triggered when wt = 210o.

When vs changes from a positive to a negative value, the current through the load becomes zero at the instant wt = π

radians, since the load is purely resistive and the SCRs cease to conduct. After that there is no current flow till the other pair is triggered. The conduction through the load is discontinuous.

The operation of the circuit is illustrated by animating the functioning of this circuit. Key in the firing angle in degrees and click the button. The source voltage and the output voltage waveforms are also displayed.


The analysis is relatively simple when the load is purely resistive. The aims of the analysis are:

i. To obtain the average output voltage as a function of firing angle, ii. To obtain the rms output voltage as a function of firing angle, iii. To obtain the ripple factor of output voltage, iv. To obtain the rms line current, v. To obtain the fundamental component of line current, vi. To obtain the Displacement power factor and power factor of line current and

vii. To obtain the total harmonic distortion(THD) in line current.

The average value of the output voltage is obtained as follows. Let the supply voltage be vs = E*Sin (θ ), where θ varies

from 0 to 2π radians. Since the output waveform repeats itself for every half-cycle, the average output voltage is expressed as a function of α, the firing angle, as shown in equation (1). The r.m.s. value of output voltage is obtained as shown in equation (2). The ripple factor in output voltage can defined in two ways. The definition followed in this text as follows. The maximum average output voltage occurs at a firing angle of 0o. Let it be Vom. Then the ripple factor RF(α)

is defined as shown in equation (3).

The alternate definition uses Vo,avg(α) as the denominator instead of Vom. If Vo,avg(α) is used as the denominator, then

RF(α) can tend to infinity. It is more logical to express the ripple content as a fraction of the maximum average voltage. The variation of average output voltage, rms output voltage and ripple factor with the firing angle have been shown below. The plots shown below have been normalized with respect to Vom. For example, when the firing angle is 90o, the

average output is shown to be 0.5. It means that the actual average output voltage is 0.5Vom. It can also be seen that

when the firing angle is 0o, the r.m.s. output voltage is about 1.1Vom and the ripple factor is about 0.48. The ripple factor

increases as the firing angle increases. It increases to 0.658 at a firing angle of 65o and then it falls as the firing angle increases further. At the firing angle of 65o, the r.m.s. ripple voltage in the output is 0.658Vom. For a sinusoidal source of

240 V r.m.s., the maximum r.m.s. ripple voltage works out to be 142 V.

PERFORMANCE PARAMETERS FOR CONVERTERS

The role of a rectifier circuit is to produce a dc output voltage from an ac sinusoidal source. The rectifier's output is not pure d.c. and it contains ripple superimposed on its d.c. content and the current drawn from the ac source is not sinusoidal either. It contains some fundamental component and harmonics. For the output voltage, its ripple content is the performance criterion. It would be desirable to obtain the amplitude frequency spectrum of the output of the rectifier in order to design a suitable filter circuit. However there is no mention of how to design a filter in this page.

As far as the ac source is concerned, the distortion in the source current is the performance criterion. The total harmonic distortion(THD) or the harmonic factor, the amplitude frequency spectrum of the source current, the apparent power factor and the displacement power factor have also to be computed. In addition, the crest factor is also to be computed, to facilitate the selection of the proper SCR.

Given a periodic function f(t) with a period of T, f(t) can be described by a trigonometric Fourier series, as shown in equation (4). The coefficients are defined as shown in equations (5), (6) and (7).

In the equations above, * is used to in place of the product sign and it should not be confused with the same symbol used for indicating convolution integrals. The source frequency, f, is taken as the fundamental and hence wo = 2πf. It is

preferable to express the above equation in terms of angle θ, where θ = wot. If T is the cycle period, woT = 2π fT = 2π,

since f = 1/T. Then the equations for the Fourier coefficients can be expressed as shown in equations (8), (9) and (10).

In the case of the full-wave bridge rectifier circuit, the period of the output is only half that of the input sinusoidal source and hence the output contains a dc component and even harmonics only. The source current has half-wave symmetry. A waveform defined as f(t) over a cycle is said to have half-wave symmetry if it satisfies equation (11). A waveform with

half-wave symmetry contains a fundamental component and odd harmonics only.

Let the Fourier coefficients for the output voltage be av0(α), av2n(α) and bv2n(α) , where α is the firing angle and these

coefficients are evaluated as shown in equations (12), (13) and (14). Since the output repeats itself twice for every cycle of source voltage, it contains only even harmonics. Then we obtain the amplitude of the harmonic as shown in equation (16).

Let the Fourier coefficients for the source current be acur0(α), acur2n(α) and bcur2n(α) . The line current waveform has

half-wave symmetry and contains only odd harmonics. When the SCRs are assumed to be ideal, load current iLine(θ) is

defined by equation (17).

For the case where the load is purely resistive, the r.m.s. source current can be computed from the value obtained for the output voltage, as shown in equation (22). Then the total harmonic distortion is defined by equation (23). Let the r.m.s current when the firing angle is 0o be Irms,max. Since the waveform of the source current is purely sinusoidal when the

firing angle is 0o, the crest factor can be taken to be square root of 2. The program that simulates the operation of this circuit computes the various values for a given firing angle and displays them in a suitable manner.

The displacement power factor, DPF, is the cosine of the angle by which the fundamental component of the line current lags the source voltage. Then apparent power factor can be estimated as shown in equation (24).

SIMULATION

The firing angle has to be keyed-in. Then click on Compute button. The plots for voltage have been normalized with respect to Vom and the currents with respect to Irms,max. The statistical details related to the output voltage have been

normalized with respect to Vom and those related to the source current with respect to Irms,max. The amplitude of each

harmonic in the output voltage has been normalized with respect to E which is the amplitude of the source voltage , whereas the amplitude of each harmonics in the source current has been normalized with respect to E/R.

PSPICE SIMULATION

The circuit used for Pspice simulation is shown below. All the nodes other than the datum node are connected to the datum node through a 1 MΩ resistor. A floating node or a node that tends to float can be a problem for Pspice simulation.

The program is presented below.

* Full-wave Bridge Rectifier with a resistive loadVIN 1 0 SIN(0 340V 50Hz)XT1 1 2 5 2 SCRXT2 0 2 6 2 SCRXT3 3 0 7 0 SCRXT4 3 1 8 1 SCRVP1 5 2 PULSE(0 10 1667U 1N 1N 100U 20M)VP2 6 2 PULSE(0 10 11667U 1N 1N 100U 20M)VP3 7 0 PULSE(0 10 1667U 1N 1N 100U 20M)VP4 8 1 PULSE(0 10 11667U 1N 1N 100U 20M)R1 2 3 10R2 1 0 1MEGR3 2 0 1MEGR4 3 0 1MEG

* Subcircuit for SCR.SUBCKT SCR 101 102 103 102S1 101 105 106 102 SMODRG 103 104 50VX 104 102 DC 0VY 105 107 DC 0DT 107 102 DMODRT 106 102 1CT 106 102 10UF1 102 106 POLY(2) VX VY 0 50 11.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0).MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0).ENDS SCR.TRAN 10US 60.0MS 20.0MS 10US.FOUR 50 V(2,3) I(VIN).PROBE

.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV)

.END

The waveforms obtained are presented below.

The voltage waveform across the load resistor

The voltage waveform across SCR1

The average load current

The Frequency spectrum of line current

The Frequency spectrum of output voltage

SUMMARY

This page has described the operation of a fully-controlled bridge rectifier circuit with a purely resistive load. The next page describes how the circuit behaves when the load has a reactive and a resistive component.

FULLY-CONTROLLED SINGLE-PHASE SCR BRIDGE RECTIFIER: RL LOAD

CIRCUIT OPERATIONMATHEMATICAL ANALYSISPERFORMANCE PARAMETERSSIMULATIONPSPICE SIMULATIONMATHCAD SIMULATIONMATLAB SIMULATIONSUMMARY

This chapter describes the operation of a single-phase fully- controlled bridge rectifier circuit with a load consisting of both a resistor and an inductor in series. The analysis in this page is based on the assumption that the SCRs are ideal controlled switches. The main purpose of the fully-controlled bridge rectifier circuit is to provide a variable dc voltage from an ac source.

CIRCUIT OPERATION

The circuit of a single-phase fully-controlled bridge rectifier circuit is shown in the figure above. The circuit has four SCRs. It is preferable to state that the circuit has two pairs of SCRs, with S1 and S3 forming one pair and, S2 and S4

the other pair. For this circuit, vs is a sinusoidal voltage source. When it is positive, the SCRs S1 and S3 can be

triggered and then current flows from vs through SCR S1, load inductor L, load resistor R, SCR S3 and back into the

source. In the next half-cycle, the other pair of SCRs conducts. Even though the direction of current through the source alternates from one half-cycle to the other half-cycle, the current through the load remains unidirectional.

The main purpose of this circuit is to provide a variable dc output voltage, which is brought about by varying the firing angle. Let vs = E sin wt, with 0 < wt < 360o. If wt = 30o when S1 and S3 are triggered, then the firing angle is said to be

30o. In this instance the other pair is triggered when wt= 210o.

When vs changes from a positive to a negative value, the current through the load does not fall to zero value at the

instant wt = π radians, since the load contains an inductor and the SCRs continue to conduct, with the inductor acting as a source. When the current through an inductor is falling, the voltage across it changes sign compared with the sign that occurs when its current is rising. When the current through the inductor is falling, its voltage is such that the inductor delivers power to the load resistor, feeds back some power to the ac source under certain conditions and keeps the SCRs in conduction forward-biased. If the firing angle is less than the load angle, the energy stored in the inductor is sufficient to maintain conduction till the next pair of SCRs is triggered. When the firing angle is greater than the load angle, the current through the load becomes zero and the conduction through the load becomes discontinuous. Usually the description of this circuit is based on the assumption that the load inductance is sufficiently large to keep the load current continuous and ripple-free.

The operation of the circuit is illustrated by animating the functioning of this circuit. Key in the firing angle in degrees and click the button. The source voltage, and the bridge output voltage are also displayed. It is assumed here that the load inductance is quite large. The animation is correct only if the firing angle is less than 90o. The programs under simulation section will run correctly for any firing angle.


Analysis by hand is based on the assumption that the load inductance is sufficiently large to keep the load current ripple-free. Programs written for computer simulation are not based on this assumption and they simulate the operation based on the parameter keyed-in.

When the load current is continuous, the average value of the output voltage is obtained as follows. Let the supply voltage be vs = E*Sin (θ ), where θ varies from 0 to 2π radians. Since the output waveform repeats itself for every half-

cycle, the average output voltage is expressed in equation (1) as a function of α, the firing angle.

If the load current is continuous, the r.m.s. value of output voltage is obtained as shown in equation (2).

The maximum average output voltage occurs at a firing angle of 0o. Let it be Vom. Then the ripple factor RF(a) is

defined as shown in equation (3). Equations (4) and (5) apply when the conduction is discontinuous.

The variation of average output voltage, r.m.s. output voltage and the ripple factor with the firing angle have been shown below, based on the assumption that the load inductance is large. The plots shown below have been normalized with respect to Vom. For example, when the firing angle is 60o, the average output is shown to be 0.5. It means that the actual

average output voltage is 0.5Vom. It can also be seen that when the firing angle is 0o, the r.m.s. output voltage is about

1.1Vom and the ripple factor is about 0.48. The ripple factor increases as the firing angle increases.

If the firing angle is highly retarded or if the load inductance is not sufficiently large, conduction of load current is not continuous. Let us consider the positive half-cycle. When θ equals the firing angle, the SCRs S1 and S3 are triggered. Let

the load current start from zero at θ = α and let the load current fall to zero when θ = π + β , before the next pair of SCRs is triggered. It means that β < α . Then the average output voltage is computed as shown in equation (4), whereas the r.m.s. output voltage is computed as shown in equation (5).

The value of β can be found out by iteration, as shown in earlier pages. For the case when the conduction is discontinuous, the plots of the average output voltage, the r.m.s. output voltage and the ripple factor are illustrated below. Key-in the ratio of wL/R, where L is the load inductance, R is the load resistance and w is the angular frequency of the source. In practice this ratio can vary from a low value to about 5. The load angle can be defined to be tan-1(wL/R). When the firing angle is less than the load angle, the conduction is continuous. As explained below, the load current is

discontinuous if the firing angle is greater than the load angle.

PERFORMANCE PARAMETERS

In this section, the expressions for the load current are first derived. Then an expression for the line current can be obtained. The r.m.s line current, the r.m.s. value of the fundamental of line current, the THD in line current, the frequency spectrum of line current, the frequency spectrum and the ripple content of the bridge output voltage and the frequency spectrum and the ripple content of the voltage across the load resistor are also determined.

For the case of discontinuous conduction, let the load current in each half cycle flow from θ = α till θ = π + β , where β < α. During this period, the differential equation that applies to the load current is calculated according to equation (6), where α is the firing angle, (π +β) is the extinction angle and the conduction angle within a cycle is (π + β − α ). The solution to the above equation is obtained as given shown in equation (7). In equation (7), τ represents wL/R,the ratio of load reactance to load resistance. φ is the load angle, defined to be tan-1(wL/R). When the conduction is discontinuous, the load current starts from zero value at θ = α . Hence equation (8) is obtained. When φ < α , A has a negative value and we have discontinuous conduction when φ < α . It means that the firing of the SCRs is retarded and α > φ .

When the conduction is continuous, equation (9) applies. The solution to equation (9) is presented as equation (10). Since the signal applied to the load circuit is a periodic signal, the load current is also periodic, after a transient period. It means that equation (11) is valid for continuous conduction.

For continuous conduction, φ > α and A is positive. When φ = α, the conduction is about to become continuous and the value of A should satisfy both expressions for A. It is seen that φ = α, A = 0 according to both expressions for A. When φ = α , the current drawn from the source is sinusoidal and equals (E/Z)*Sin (wt - φ), where Z is defined as shown in equation (12).

Once the load current is defined for a half-cycle, the expression for line current can be obtained. It equals the load current when SCRs S1 and S3 are conducting and it is the negative of the load current when the other pair of SCRs is conducting.

Further analysis can be carried out using Fourier series and the DPF and PF of the line current can be determined.

When the circuit is switched on initially, the load current may settle down to a periodic response after a few output cycles. The time constant of the load is L/R and a time period corresponding to five times the constant should elapse before the load current becomes periodic. For example if the load time constant is 20 ms, a time period of 100 ms should pass before the load current becomes periodic. This time period corresponds to five input voltage cycles at 50 Hz and six input cycles at 60 Hz.

SIMULATION

The plots for voltage have been normalized with respect to Vom, the maximum average output voltage and the plots for

line current with respect to Irms,max, which is 0.707E/R. To see the periodic response, key-in firing angle in degrees in the

box on the left-hand side and the ratio τ in the box on the right-hand side and then click start button. The boxes have default values filled-in.

The statistical details related to the output voltage have been normalized with respect to Vom and those related to the

source current with respect to Irms,max. The amplitude of each harmonic in the output voltage has been normalized with

respect to E which is the amplitude of the source voltage , whereas the amplitude of each harmonics in the source current

has been normalized with respect to E/R. The applet to follow displays the statistical details.

The applet below displays the transient response. As the ratio gets larger and larger, the load current requires larger period to elapse before it becomes periodic.

PSPICE SIMULATIONThe circuit used for simulation is shown below.

The program is presented below.

* Full-wave Bridge Rectifier with a resistive loadVIN 1 0 SIN(0 340V 50Hz)XT1 1 2 5 2 SCRXT2 0 2 6 2 SCRXT3 4 0 7 0 SCRXT4 4 1 8 1 SCRVP1 5 2 PULSE(0 10 1667U 1N 1N 100U 20M)VP2 6 2 PULSE(0 10 11667U 1N 1N 100U 20M)VP3 7 0 PULSE(0 10 1667U 1N 1N 100U 20M)VP4 8 1 PULSE(0 10 11667U 1N 1N 100U 20M)L1 2 3 31.8MR1 3 4 10R2 1 0 1MEGR3 2 0 1MEGR4 4 0 1MEG

* Subcircuit for SCR.SUBCKT SCR 101 102 103 102S1 101 105 106 102 SMODRG 103 104 50VX 104 102 DC 0

VY 105 107 DC 0DT 107 102 DMODRT 106 102 1CT 106 102 10UF1 102 106 POLY(2) VX VY 0 50 11.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0).MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0).ENDS SCR.TRAN 10US 60.0MS 0.0MS 10US.FOUR 50 V(2,4) I(VIN).PROBE.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV).END

The responses obtained are shown below.

The waveform of voltage at the output of bridge

The waveform of load current

The waveform of voltage across the SCR

The waveform of voltage across the inductor

The waveform of line current

The average load current

The rms line current

Fourier frequency spectrum of voltage at the output of bridge

Fourier frequency spectrum of line current

MATLAB SIMULATION

The Matlab program is presented below.

% Program to simulate the full-wave fully-controlled bridge rectifier% Simulation at a specified firing angle % Enter the peak voltage, frequency, inductance L in mH and resistor R disp('Typical value for peak voltage is 340 V')peakV=input('Enter Peak voltage in Volts>');disp('Typical value for line frequency is 50 Hz')freq=input('Enter line frequency in Hz>');disp('Typical value for Load inductance is 31.8 mH')L=input('Enter Load inductance in mH>');disp('Typical value for Load Resistance is 10.0 Ohms')R=input('Enter Load Resistance in Ohms>');disp('Typical value for Firing angle is 30.0 degree')fangDeg=input('Enter Firing angle within range 0 to 180 in deg>');fangRad=fangDeg/180.0*pi;

w=2.0*pi*freq;X=w*L/1000.0;if (X<0.001) X=0.001; end;Z=sqrt(R*R+X*X);tauInv=R/X;loadAng=atan(X/R);k1=peakV/Z;k2=2.0*k1*sin(loadAng-fangRad)/(1.0-exp(-pi*tauInv));k3=k1*sin(loadAng-fangRad);if (fangRad<loadAng) A=k2; sw=1;else

A=k3; sw=2;end;

Ampavg=0;AmpRMS=0;

for n=1:360; theta=n/180.0*pi; X(n)=n; if (sw==1); if (n<fangDeg); cur=k1*sin(pi+theta-loadAng)+A*exp(-tauInv*(pi+theta-fangRad)); vbr(n)=peakV*sin(theta+pi); vind(n)=vbr(n)-R*cur; iLoad(n)=cur; vSCR(n)=peakV*sin(theta); Ampavg=Ampavg+cur*1/360; AmpRMS=AmpRMS+cur*cur*1/360; elseif ((n>=fangDeg) & (n<(180+fangDeg))); cur=k1*sin(theta-loadAng)+A*exp(-tauInv*(theta-fangRad)); vbr(n)=peakV*sin(theta); vind(n)=vbr(n)-R*cur; iLoad(n)=cur; vSCR(n)=0; Ampavg=Ampavg+cur*1/360; AmpRMS=AmpRMS+cur*cur*1/360; else (n>=(180+fangDeg)); cur=k1*sin(theta-pi-loadAng)+A*exp(-tauInv*(theta-pi-fangRad)); vbr(n)=peakV*sin(theta-pi); vind(n)=vbr(n)-R*cur; iLoad(n)=cur; vSCR(n)=peakV*sin(theta); Ampavg=Ampavg+cur*1/360; AmpRMS=AmpRMS+cur*cur*1/360; end; else if (n<fangDeg); cur=k1*sin(pi+theta-loadAng)+A*exp(-tauInv*(pi+theta-fangRad)); if (cur>0); vbr(n)=peakV*sin(theta+pi); vind(n)=vbr(n)-R*cur; iLoad(n)=cur; vSCR(n)=peakV*sin(theta); Ampavg=Ampavg+cur*1/360; AmpRMS=AmpRMS+cur*cur*1/360; else; vbr(n)=0.0; vind(n)=0.0; iLoad(n)=0.0;

vSCR(n)=peakV*sin(theta)/2.0; end; elseif ((n>=fangDeg) & (n<(180+fangDeg))); cur=k1*sin(theta-loadAng)+A*exp(-tauInv*(theta-fangRad)); if (cur>0); vbr(n)=peakV*sin(theta); vind(n)=vbr(n)-R*cur; iLoad(n)=cur; vSCR(n)=0; Ampavg=Ampavg+cur*1/360; AmpRMS=AmpRMS+cur*cur*1/360; else; vbr(n)=0.0; vind(n)=0.0; iLoad(n)=0.0; vSCR(n)=peakV*sin(theta)/2.0; end; else (n>=(180+fangDeg)); cur=k1*sin(theta-pi-loadAng)+A*exp(-tauInv*(theta-pi-fangRad)); vbr(n)=peakV*sin(theta-pi); vind(n)=vbr(n)-R*cur; iLoad(n)=cur; vSCR(n)=peakV*sin(theta); Ampavg=Ampavg+cur*1/360; AmpRMS=AmpRMS+cur*cur*1/360; end; end;end; plot(X,iLoad)title('The Load current')xlabel('degrees')ylabel('Amps') gridpause

plot(X,vbr)title('Bridge Output volt')xlabel('degrees')ylabel('Volts') gridpause

plot(X,vind)title('Inductor Voltage')xlabel('degrees')ylabel('Volts') gridpause

plot(X,vSCR)title('SCR Voltage')xlabel('degrees')ylabel('Volts') grid

AmpRMS=sqrt(AmpRMS);[C,message]=fopen('fwrl1ph1.dat','w'); fprintf(C,'Avg Load Cur=\t%d\tRMS Load Cur=\t%f\n',Ampavg,AmpRMS);fclose(C);

The responses obtained for the typical specified values have been displayed below.

The results obtained with a firing angle of 60o have been displayed below. The other values are the same as for the previous set. When the firing angle is retarded this much, the load current is discontinuous.

SUMMARY

This page has described the operation of a fully-controlled bridge rectifier circuit with an RL load. Even though the circuit has been simulated using Pspice, Mathcad and Matlab, it is difficult to reach the level of interactive programming that can be achieved with Java or C by using these packages. Next page describes the operation of this circuit when the

source inductance is also taken into account.

OPERATION WITH SOURCE INDUCTANCE:1-PH. CONTROLLED RECTIFIER

CIRCUIT DIAGRAMCIRCUIT OPERATIONMATHEMATICAL ANALYSISSIMULATIONPSPICE SIMULATIONMATHCAD SIMULATIONSIMULATION USING CSUMMARY

CIRCUIT DIAGRAM

This chapter describes the operation of a single-phase fully- controlled bridge rectifier circuit with a load consisting of both resistance and inductance. In addition, the inductance of ac source supplying power to the circuit is taken into account.

CIRCUIT OPERATION

The presence of source inductance changes the way the circuit operates during commutation. The word commutation refers to switching conduction from one pair of SCRs to the other. Let vs = E sin wt, with 0 < wt < 360o. Let the load

inductance be large enough to maintain a steady current through the load. Let firing angle α be 30o. Let SCRs S2 and S4

be in conduction before wt < 30o. When S1 and S3 are triggered at wt = 30o, there is current through the source

inductance, flowing in the direction opposite to that marked in the circuit diagram and hence commutation of current from S2 and S4 to S1 and S3 would not occur instantaneously. The source current changes from - iL to iL due to the whole

of the source voltage being applied across the source inductance. When S1 is triggered with S4 in conduction, the current

through S1 would rise from zero to iL and the current through S4 would fall from iL to zero. Similar process occurs with

the SCRs S2 and S3. During this period, the current through S3 would rise from zero to iL and , the current through S2

would fall from iL to zero. The duration of the process of commutation is usually referred to as commutation overlap

period.

The operation of the circuit is illustrated with the help of an animation. There is no scope for setting any variable for animation, since the purpose is only to show how the circuit operates. It is assumed that the load inductance is large to keep the load current at a steady value. You can observe how the currents through the devices and the line current change during commutation overlap. The running of the program can be halted by clicking on the Stop button. It can be resumed by clicking on the Run button. By clicking on the One Step button, you can step through the process. To start or run the process for one more cycle, click on the Start or Repeat button.


BASED ON THE ASSUMPTION THAT LOAD INDUCTANCE IS INFINITE

When the load inductance is infinite, we can assume that the load current is continuous and steady without ripple. Let the firing angle be α. Let the commutation overlap period last from wt = α till wt = β.During the period, α < wt < β, the output voltage is zero because all the SCRs are in conduction. If the SCRs are ideal, the drop across an SCR in conduction is zero and hence the output voltage is zero. During β < wt < ( π + α), the output voltage equals E*Sin (wt) and then the average output voltage can be obtained as shown in equation (1).

This expression is not very useful because the value of β is required to be known. The value of α is known, since it is the firing angle, whereas the value of β is not likely to be known. Here β is the angle at which process of commutation overlap ends and the duration of commutation depends on the firing angle, the value of source reactance and the load impedance and value of β is variable and unknown. Hence it is preferable to derive an alternate expression. It is possible to derive an alternate expression for the case when the load reactance is large enough to ensure that the load current remains steady without ripple at a given firing angle. Let us assume that the load current be I and the firing angle be α. Let the line current change from - I to + I during commutation overlap when wt changes from α to β. From the waveforms shown above, the area of volt-seconds lost to output due to commutation overlap is computed as shown in equation (2).

Since this area is lost over π radians, the average value of output voltage lost due to commutation is calculated as shown in equation (3). It is known that the average output voltage with no commutation overlap is (2E/π)*Cos (α). By subtracting the voltage lost due to commutation, we can get the average output voltage taking into account the effect of commutation overlap, as shown in equation (4).

The waveforms appear as shown below. Key-in a firing angle less than 90o and then press the Start button.

ANALYSIS WITH A FINITE LOAD INDUCTANCE

With a finite load inductance, the conduction through the load can be either continuous or discontinuous. Let angle ϕ be defined as shown in equation (5). If the firing angle α is greater than ϕ, then the current through the load is discontinuous and the analysis is similar to that used for the circuit without a source inductance. When α < ϕ , then the conduction is continuous. The analysis is carried out as follows. In these equations, α is the firing angle, β is the angle at which commutation overlap ends and ϕ has been defined above. Let iL be the load current and is is the source current. Let the

source voltage vs = E *Sin (θ), where θ = wt and 0 < θ < 2π . Then the equation that is applicable during β < θ < (π + α)

can be expressed as shown in equation (6). During this period, the line current is equal to load current as defined by equation (7).

During β < θ < (π + α), the voltage that appears as output is almost the negative of the source voltage and hence equation (8) is used and the line current has the same magnitude as the load current, but its polarity is opposite to that of load current as indicated by equation (9).

There are two instances of switching in one input cycle and commutation overlap occurs immediately after triggering either pair of SCRs. The pair consisting of S1 and S3 is triggered at wt = α and commuation overlap lasts from wt = α

till wt =β and the output voltage is zero during this period as indicated by equation (10). Similarly after S2 and S4 are

triggered at wt = (π + α), the output voltage is zero from wt = (π + α) till wt = (π + β), as indicated by equation (11). During the commuation overlap, the entire input voltage is applied across the source inductance, as indicated by equations (12) and (13).

The average and rms output voltage can be obtained as shown by equations (14) and (15). The maximum output voltage that can occur is indicated by equation (16) neglecting the loss in output that may occur due to commutation overlap. Then the ripple factor of output voltage can be expressed as shown by equation (17). If the ripple factor is multiplied by Vom, the rms value of ripple content in output voltage is obtained.

The fundamental component of the source current can be determined and then the THD, DPF and apparent power factor can be determined. The programs for simulation have been based on the equations displayed above.

SIMULATION

The program in the applet to follow displays the bridge output waveform, the load current, the line current, the voltage across the load inductance, the voltage across the source inductance, the voltage across SCR S1, and the currents through

S1 and S4. Key-in the ratio wLL/RL, the ratio wLs/RL and the firing angle in degrees and then click on the Start button.

The program to follow produces the statistics for this circuit and needs the same inputs as the applet above.

The next applet displays the transient response of the load current. If the load inductance is large, the transient response may last for a few cycles before it settles down to a periodic response. The load current starts building up from zero from the instant one of the pairs of SCRs is triggered first. The response shown starts from this instant.

The next applet shows how the average bridge output voltage, the rms output voltage, and the ripple factor vary with firing angle for a given set of wLL/RL and wLs/RL. This program may take some time to run.

PSPICE SIMULATIONThe circuit used for simulation is presented below.

The Pspice program is presented first.

* Full-wave Bridge Rectifier with RL Load and source InductanceVIN 9 0 SIN(0 340V 50Hz)XT1 1 2 5 2 SCRXT2 0 2 6 2 SCRXT3 4 0 7 0 SCRXT4 4 1 8 1 SCRVP1 5 2 PULSE(0 10 1667U 1N 1N 100U 20M)VP2 6 2 PULSE(0 10 11667U 1N 1N 100U 20M)VP3 7 0 PULSE(0 10 1667U 1N 1N 100U 20M)VP4 8 1 PULSE(0 10 11667U 1N 1N 100U 20M)L1 2 3 31.8M

L2 1 9 1.6MR1 3 4 10R2 1 0 1MEGR3 2 0 1MEGR4 4 0 1MEG

* Subcircuit for SCR.SUBCKT SCR 101 102 103 102S1 101 105 106 102 SMODRG 103 104 50VX 104 102 DC 0VY 105 107 DC 0DT 107 102 DMODRT 106 102 1CT 106 102 10UF1 102 106 POLY(2) VX VY 0 50 11.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0).MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0).ENDS SCR.TRAN 10US 60.0MS 0.0MS 10US.FOUR 50 V(2,4) I(VIN).PROBE.OPTIONS(ABSTOL=1N RELTOL=.01 VNTOL=1MV).END

The waveforms obtained are now presented.The Voltage Output of the bridge

The Expanded view of Voltage Output of the bridge to show the commutation overlap

The waveform of Load Current


The expanded waveform of line current

The waveform of voltage across the line/source inductance

It can be seen that there is close agreement between the results obtained from the Applets and the Mathcad program.

SIMULATION USING C

Instead of a Matlab program, a C program is presented below. You can download this file by clicking on the image displayed below. This file, called fcb_1pha.cpp, can be complied just as a C program and no feature of C++ has been used in this program. This program produces an output file called tran_v1.csv and stores in it in the same directory in which the source file is located. This file can be opened by a spreadsheet program like EXCEL and the plots displayed can be obtained.

C Program

// Simulation of a single-phase fully-controlled bridge// rectifier circuit.

// This program produces the transient response, the periodic// response. Commuation overlap angle is also computed.// Calculations normalized. The average voltage that would// occur for diode-bridge rectifier with a resistive load is// taken to be unity. The average current through the load// resistor with a pure resistive load on the diode bridge// rectifier is taken as the unity current. The parameters to// be fed are just three: the ratio of load inductive reactance// to the load resistance, wL1/R , the ratio of source reactance// to the load resistance to load resistance wL2/R and the firing// angle in degrees.

#include <math.h>#include<stdlib.h>#include<stdio.h>#include<string.h>#include<iostream.h>

void kreateRespFile(void);void tranResponse(void);void OneCycle(void);void computeOneStep(void);void produceEntries(void);

const double pi=3.1415926;const double deg_rad=pi/180.0;const double step=pi/720.0;

double L1,L2,alpha,fang,elapseAng,cycleAng,tote_knt;double cur_load,cur_line,outVolt,vSource,vInput;double OverLapangle;int commute,toggle,yes_Entry,modeSw;FILE *fnew;char *tstr,*p;

int main(void) float ka1,kb2,kc3; printf(" \n"); printf(" Load Time Constant in radians = "); scanf("%f",&ka1); L1=(double)ka1; if (L1<0.05) L1=0.05; printf(" \n"); printf(" Line Reactance Time Constant in radians = "); scanf("%f",&kb2); L2=(double)kb2; printf(" \n"); printf(" Firing angle in degrees = "); scanf("%f",&kc3); alpha=(double)kc3;

fang=alpha*deg_rad; alpha=0.0; printf(" \n");

// Initialise though not necessary for global variables

elapseAng=0.0 ; cycleAng=0.0; cur_load=0.0; cur_line=0.0; outVolt=0.0; tote_knt=0.0; OverLapangle=0.0; commute=0; yes_Entry=0;

// Create a file for entering transient response fnew=fopen("tran_v1.csv","w+r"); tranResponse(); fclose(fnew); return 0;

void kreateRespFile(void) int n1,n2;

tstr="Angle,InputV,VoutBr,LoadCur,LineCur,IndVolt"; p=strtok(tstr,","); fprintf(fnew,p); fprintf(fnew,","); n2=6; for (n1=0;n1<(n2-1);n1++) p=strtok(NULL,","); if (n1!=(n2-2)) fprintf(fnew,p); fprintf(fnew,","); else fprintf(fnew,p); fprintf(fnew,"\n");

void tranResponse(void) double d1; int NumCycles,count; d1=(6.0*(L1+L2))/(2.0*pi)+0.5; NumCycles=(int)d1 + 1; kreateRespFile(); modeSw=1; for (count=0;count<=NumCycles;count++) OneCycle();

void OneCycle(void) int knt;

double dknt; dknt=0.0;

for (knt=0;knt<1440;knt++) cycleAng=dknt*step; vInput=pi/2.0*sin(cycleAng); if ((cycleAng<fang) || (cycleAng>(fang+pi))) if (cycleAng<fang) vSource=pi/2.0*sin(cycleAng+pi); else vSource=pi/2.0*sin(cycleAng-pi); toggle=0; else vSource=pi/2.0*sin(cycleAng); toggle=1; if (modeSw!=toggle) commute=1; if (L2<0.0005) commute=0; modeSw=toggle; computeOneStep(); dknt+=1.0;

void computeOneStep(void) double dLoadI,dLineI;

switch (commute) case 0: dLoadI=(vSource-cur_load)/(L1+L2)*step; cur_load=cur_load+dLoadI; if (cur_load<0.00001) cur_load=0.0; if (toggle==0) cur_line=-cur_load; else cur_line=cur_load; outVolt=cur_load+(vSource-cur_load)*L1/(L1+L2); break;

case 1: dLoadI=(0.0-cur_load)/L1*step; cur_load=cur_load+dLoadI; if (cur_load<0.00001) cur_load=0.0; dLineI=vInput/L2*step; cur_line=cur_line+dLineI; if (toggle==1) if (cur_line>cur_load) commute=0;

if (toggle==0) if ((cur_line+cur_load)<0.0) commute=0; outVolt=0.0; break; tote_knt+=1.0; elapseAng=tote_knt*step; alpha=tote_knt/4.0; yes_Entry+=1; if (yes_Entry==4) produceEntries(); if (yes_Entry==4) yes_Entry=0;

void produceEntries(void) double negate; gcvt(alpha,5,p); fprintf(fnew,p); fprintf(fnew,","); gcvt(vInput,5,p); fprintf(fnew,p); fprintf(fnew,","); gcvt(outVolt,5,p); fprintf(fnew,p); fprintf(fnew,","); gcvt(cur_load,5,p); fprintf(fnew,p); fprintf(fnew,","); gcvt(cur_line,5,p); fprintf(fnew,p); fprintf(fnew,","); if (toggle==0) negate = 1.0; else negate=-1.0; gcvt(vInput-negate*outVolt,5,p); fprintf(fnew,p); fprintf(fnew,"\n");

The responses obtained for L1=1.5, L2=0.2 and α = 30o are produced below.

Another C program is presented below. This program carries out the harmonic analysis. This file, called fcb_1phb.cpp, can be downloaded. It can also be compiled as a C program. It produces an output file, called harm_v1.csv.

C Program

// Simulation of a single-phase fully-controlled bridge

// rectifier circuit. // This program obtainss the periodic response first and

// then computes the ripple components in output current // and voltage and the harmonic components in the line current. // The parameters to be fed are just three: the ratio of load

// inductive reactance to the load resistance, wL1/R , the

// ratio of source reactance to the load resistance to load

// resistance wL2/R and the firing angle in degrees.

#include <math.h>

#include<stdlib.h>

#include<stdio.h>

#include<string.h>

void periodicResponse(void); void harmonicAnalysis(void); void OneCycle(void); void computeOneStep(void); void kreateRespFile(void); void produceEntries(void);

const double pi=3.1415926; const double deg_rad=pi/180.0; const double step=pi/720.0;

double L1,L2,fang,cycleAng; double cur_load,cur_line,outVolt,vSource,vInput; double OverLapangle,VoAvg,VoRMS,ILoadAvg,ILoadRMS; double ILineAvg,ILineRMS,THD,RFVolt,RFCur; int commute,toggle,modeSw; double anLoadCur[26],bnLoadCur[26],cnLoadCur[26]; double anLineCur[26],bnLineCur[26],cnLineCur[26]; double anOutVolt[26],bnOutVolt[26],cnOutVolt[26];

FILE *fnew; char *tstr,*p;

void main(void)

float ka1,kb2,kc3; printf(" \n"); printf(" Load Time Constant in radians = ");

scanf("%f",&ka1); L1=(double)ka1; if (L1<0.05) L1=0.1; printf(" \n"); printf(" Line Reactance Time Constant in radians = "); scanf("%f",&kb2); L2=(double)kb2; printf(" \n"); printf(" Firing angle in degrees = "); scanf("%f",&kc3); fang=deg_rad*(double)kc3; printf(" \n");


cycleAng=0.0; cur_load=0.0; cur_line=0.0; outVolt=0.0; OverLapangle=0.0; commute=0; modeSw=0; toggle=0;

// obtain the transient response. periodicResponse(); harmonicAnalysis(); printf("Close window by clicking on x button at top right corner \n");

void periodicResponse(void)

double startVal;

// Calculate one cycle of response and repeat till the

// response becomes periodic. do

startVal=cur_load; OneCycle(); while (fabs(startVal-cur_load)>0.0005);

void OneCycle(void)

int knt; double dknt; dknt=0.0;

for (knt=0;knt<1440;knt++)

cycleAng=dknt*step; vInput=pi/2.0*sin(cycleAng); if ((cycleAng<fang) || (cycleAng>(fang+pi)))

if (cycleAng<fang) vSource=pi/2.0*sin(cycleAng+pi); else vSource=pi/2.0*sin(cycleAng-pi);

toggle=0;

else

vSource=pi/2.0*sin(cycleAng); toggle=1;

// To record commutation overlap, the routine described below is used. // When commute is 1, overlap ensues. It is set to to 1 whenever

// a pair of SCRs is triggered at the set firing agnle. It is resset // by the compute_one_step routine when overlap ends. This can be

// checked from the values of load current and line current.

if (modeSw!=toggle)

commute=1; if (L2<0.0005) commute=0; modeSw=toggle;

computeOneStep(); dknt+=1.0;

void computeOneStep(void)

double dLoadI,dLineI;

switch (commute)

case 0: dLoadI=(vSource-cur_load)/(L1+L2)*step; cur_load=cur_load+dLoadI; if (cur_load<0.00001) cur_load=0.0; if (toggle==0) cur_line=-cur_load; else cur_line=cur_load; if (cur_load>0.0) outVolt=cur_load+(vSource-cur_load)*L1/(L1+L2); else outVolt=0.0; break;

case 1: dLoadI=(0.0-cur_load)/L1*step; cur_load=cur_load+dLoadI; if (cur_load<0.00001) cur_load=0.0; dLineI=vInput/L2*step; cur_line=cur_line+dLineI; if (toggle==1)

if (cur_line>cur_load) commute=0;

if (toggle==0)

if ((cur_line+cur_load)<0.0) commute=0;

outVolt=0.0; break;

void harmonicAnalysis(void)

int m; double theta,dblm; int knt; double dknt; dknt=0.0;

for (m=0;m<26;m++)

anLoadCur[m]=0.0; bnLoadCur[m]=0.0; cnLoadCur[m]=0.0; anLineCur[m]=0.0; bnLineCur[m]=0.0; cnLineCur[m]=0.0; anOutVolt[m]=0.0; bnOutVolt[m]=0.0; cnOutVolt[m]=0.0;

OverLapangle=0.0; ILoadAvg=0.0; ILoadRMS=0.0; RFCur=0.0; ILineAvg=0.0; ILineRMS=0.0; THD=0.0; VoAvg=0.0; VoRMS=0.0; RFVolt=0.0;

for (knt=0;knt<720;knt++)

cycleAng=dknt*step; vInput=pi/2.0*sin(cycleAng); if ((cycleAng<fang) || (cycleAng>(fang+pi)))

vSource=pi/2.0*sin(cycleAng-pi); toggle=0;

else

vSource=pi/2.0*sin(cycleAng); toggle=1;

if (modeSw!=toggle)

commute=1; if (L2<0.0005) commute=0; modeSw=toggle;

computeOneStep();

if ((toggle==1) && (commute==1)) OverLapangle=OverLapangle+step;

dblm=0.0; for (m=0;m<13;m++)

theta=2.0*dblm*cycleAng; while (theta>(2.0*pi)) theta=theta-2.0*pi; anLoadCur[2*m]=anLoadCur[2*m]+cur_load*cos(theta)*step; bnLoadCur[2*m]=bnLoadCur[2*m]+cur_load*sin(theta)*step; anOutVolt[2*m]=anOutVolt[2*m]+outVolt*cos(theta)*step; bnOutVolt[2*m]=bnOutVolt[2*m]+outVolt*sin(theta)*step; theta=theta+cycleAng; if (theta>(2.0*pi)) theta=theta-2.0*pi; anLineCur[2*m+1]=anLineCur[2*m+1]+cur_line*cos(theta)*step; bnLineCur[2*m+1]=bnLineCur[2*m+1]+cur_line*sin(theta)*step; dblm=dblm+1.0;

ILoadRMS=ILoadRMS+cur_load*cur_load*step; VoRMS=VoRMS+outVolt*outVolt*step; ILineAvg=ILineAvg+fabs(cur_line)*step; ILineRMS=ILineRMS+cur_line*cur_line*step; dknt+=1.0;

for (m=0;m<13;m++)

anLoadCur[2*m]=2.0/pi*anLoadCur[2*m]; bnLoadCur[2*m]=2.0/pi*bnLoadCur[2*m]; anOutVolt[2*m]=2.0/pi*anOutVolt[2*m]; bnOutVolt[2*m]=2.0/pi*bnOutVolt[2*m]; anLineCur[2*m+1]=2.0/pi*anLineCur[2*m+1]; bnLineCur[2*m+1]=2.0/pi*bnLineCur[2*m+1]; cnLoadCur[2*m]=anLoadCur[2*m]*anLoadCur[2*m]

+ bnLoadCur[2*m]*bnLoadCur[2*m]; cnLoadCur[2*m]=sqrt(cnLoadCur[2*m]); cnLineCur[2*m+1]=anLineCur[2*m+1]*anLineCur[2*m+1]

+ bnLineCur[2*m+1]*bnLineCur[2*m+1]; cnLineCur[2*m+1]=sqrt(cnLineCur[2*m+1]); cnOutVolt[2*m]=anOutVolt[2*m]*anOutVolt[2*m]

+ bnOutVolt[2*m]*bnOutVolt[2*m]; cnOutVolt[2*m]=sqrt(cnOutVolt[2*m]);

VoRMS=sqrt(VoRMS/pi); ILoadRMS=sqrt(ILoadRMS/pi); ILineAvg= ILineAvg/pi; ILineRMS= sqrt(ILineRMS/pi); ILoadAvg=anLoadCur[0]/2.0; VoAvg=anOutVolt[0]/2.0; THD=sqrt(ILineRMS*ILineRMS-cnLineCur[0]*cnLineCur[0]/2.0); RFVolt=sqrt(VoRMS*VoRMS-VoAvg*VoAvg); RFCur=sqrt(ILoadRMS*ILoadRMS-ILoadAvg*ILoadAvg);

fnew=fopen("harm_v1.csv","w+r"); kreateRespFile(); produceEntries(); fclose(fnew);

void kreateRespFile(void)

int n1,n2; tstr="harmNo,LoadCur,OutVolt,harmNo,LineCur"; p=strtok(tstr,","); fprintf(fnew,p); fprintf(fnew,","); n2=5; for (n1=0;n1<(n2-1);n1++)

p=strtok(NULL,","); if (n1!=(n2-2))

fprintf(fnew,p); fprintf(fnew,",");

else

fprintf(fnew,p); fprintf(fnew,"\n");

void produceEntries(void)

int m; for(m=0;m<13;m++)

itoa(2*m, p, 10); fprintf(fnew,p); fprintf(fnew,",");

gcvt(cnLoadCur[2*m],5,p); fprintf(fnew,p); fprintf(fnew,",");

gcvt(cnOutVolt[2*m],5,p); fprintf(fnew,p); fprintf(fnew,",");

itoa(2*m+1, p, 10); fprintf(fnew,p); fprintf(fnew,",");

gcvt(cnLineCur[2*m+1],5,p); fprintf(fnew,p); fprintf(fnew,"\n");

fprintf(fnew,"\n"); p="LoadIAvg"; fprintf(fnew,p); fprintf(fnew,","); gcvt(ILoadAvg,5,p); fprintf(fnew,p); fprintf(fnew,"\n"); p="LoadIRMS"; fprintf(fnew,p);

fprintf(fnew,","); gcvt(ILoadRMS,5,p); fprintf(fnew,p); fprintf(fnew,"\n"); p="RFCur"; fprintf(fnew,p); fprintf(fnew,","); gcvt(RFCur,5,p); fprintf(fnew,p); fprintf(fnew,"\n"); fprintf(fnew,"\n");

p="VoAvg"; fprintf(fnew,p); fprintf(fnew,","); gcvt(VoAvg,5,p); fprintf(fnew,p); fprintf(fnew,"\n"); p="VoRMS"; fprintf(fnew,p); fprintf(fnew,","); gcvt(VoRMS,5,p); fprintf(fnew,p); fprintf(fnew,"\n"); p="RFVolt"; fprintf(fnew,p); fprintf(fnew,","); gcvt(RFVolt,5,p); fprintf(fnew,p); fprintf(fnew,"\n"); fprintf(fnew,"\n"); p="ILineAvg"; fprintf(fnew,p); fprintf(fnew,","); gcvt(ILineAvg,5,p); fprintf(fnew,p); fprintf(fnew,"\n"); p="ILineRMS"; fprintf(fnew,p); fprintf(fnew,","); gcvt(ILineRMS,5,p); fprintf(fnew,p); fprintf(fnew,"\n"); p="THD"; fprintf(fnew,p); fprintf(fnew,","); gcvt(THD,5,p); fprintf(fnew,p); fprintf(fnew,"\n"); fprintf(fnew,"\n");

p="OverLapAngle"; fprintf(fnew,p); fprintf(fnew,","); if (OverLapangle>0.0001) OverLapangle=OverLapangle+step/2.0;

gcvt(OverLapangle,5,p); fprintf(fnew,p); fprintf(fnew,"\n");

The results obtained are presented below.

SUMMARY

This page has described how the fully-controlled bridge rectifier circuit behaves when the source has inductance. In the next page, it is decribed how the operation of this circuit changes when it has a filter capacitor.

OPERATION WITH RLC LOAD: 1-PH. CONTROLLED RECTIFIER

CIRCUIT DIAGRAMCIRCUIT OPERATIONMATHEMATICAL ANALYSISSIMULATIONPSPICE SIMULATIONSIMULATION USING C

SUMMARY

CIRCUIT DIAGRAM

The aim of this page is to illustrate the effect of a smoothing capacitor across the load resistor, in addition to a dc link inductor. When both an inductor and a capacitor are used as filter elements, the ripple in output voltage is reduced by a fair amount. The inductor reduces the ripple in the current output of the bridge rectifier, whereas the capacitor reduces further the ripple in the voltage across the load resistor.

CIRCUIT OPERATION

The operation of the circuit has been described in the earlier pages. One pair of SCRs conducts at a time, producing a bridge output voltage containing significant ripple content. Since the current through an inductor cannot change suddenly, the inductor in the dc link tries to maintain the current passing through it as a steady value, which in turn means that the ripple content of the current through it reduces. Nevertheless it does contain some ripple and this ripple in the inductor current causes some ripple in the capacitor voltage, whereas the current through the load can be more or less a steady value. The effectiveness of the filter circuit depends on the values of the inductor and the capacitor. The larger they are, the higher the reduction in the filter content is. Of particular interest is the comparison of the bridge output voltage and the capacitor voltage. The ripple in the capacitor voltage is almost out-of-phase with the ripple in the bridge output voltage. This aspect is important when the output voltage across the load resistor is to be controlled in closed

loop.


The analysis of this circuit is slightly more complex. The differential equations that describe the operation of the circuit are presented below. Because of the filter capacitor, the current in the inductor can become discontinuous for a light load, even if the firing angle is not high. Hence the differential equations are described for the two cases separately.

DISCONTINUOUS CONDUCTION

When the current through the load is discontinuous, the load current starts building up from zero value when one of the pair of SCRs is triggered and it falls to zero before the next pair of SCRs is triggered. Let the firing angle be α and let the current through the inductor become 0 when wt = π + β, where β < α. That is, there is current flow during α < wt < (π + β) SCRs S1 and S3 in conduction and during (π + α) < wt < (2π + β) SCRs S2 and S4 in conduction. This means

that ther will no current flow during (π + β) < wt < (π + α) and β < wt < α. Let the supply voltage vs be E*Sin (θ),

where θ = wt. Let the voltage across the capacitor be vC(θ) and the current through the inductor be iL. Then equations (1)

and (3) are for the periods when there would no current flow. When SCRs S1 and S3 are in conduction, both the line

current and the load current have the same magntitude and polarity and equation (2) applies. When SCRs S2 and S4 are

in conduction, the line current is the negative of the load current and equation (4) is to be used. The current through the capacitor is the difference of the dc link inductor current and the current through the load resistor, as shown in equation (5).

CONTINUOUS CONDUCTION

For continuous conduction, it is sufficient if the equations are described for half-a-cycle only. In the other half-cycle, the only difference is in the source current waveform. Since the source current has half-wave symmetry, it is sufficient if it is described over half-a-cycle. Let the firing angle be α. Let the commutation overlap angle be δ. Then it means the SCRs are triggered when θ = α or when θ = π + α, the process of commutation ends δ radians later. Let the source current be is.

Then during α < θ < ( α + δ ), the entire source voltage is applied across the source inductor, as shown in equation (6). During this period, the output voltage of the bridge is zero and the voltage across the inductor is then the voltage across the output capacitor. Since the voltage across the output capacitor tends to reverse the current through the inductor, equation (7) describes the current-voltage relationship in the inductor for this period. During ( α + δ ) < θ < (π + α) , the voltage across the source inductor and the dc link inductor is defined by equation (8). During this period, the line current and the load current have both the same magnitude and polarity, as shown by equation (9). Equation (10) defines the current-voltage relationship of the output capacitor. As shown in equation (10), capacitor current is the difference between the inductor current and the current through the load resistor.

SOLUTION

The solution of the equations for both the cases is carried out using numerical technique.

SIMULATION

In this section, the circuit is simulated. You have to key-in the firing angle in degrees, the ratio wLL/RL, the ratio wLs/RL

and the average load current fraction. The ratios are specified for the nominal load conditions, assuming that the average output voltage is at its maximum and that the average load current is at its rated value. The setting for load current fraction allows for simulation under different load conditions. A setting of 1.0 means that the simulation is for rated load condition. When the setting is 0.25, it means that the average load current set is 25% of the rated load current. The applet below generates the waveforms for the parameters keyed-in.

The parameters for the next applet are the same as those for the applet above. This applet prints out the statistics.

PSPICE SIMULATION

The pspice program used for simulation is as follows:

* Full-wave Bridge Rectifier with RL Load and source InductanceVIN 9 0 SIN(0 340V 50Hz)XT1 1 2 5 2 SCRXT2 0 2 6 2 SCRXT3 4 0 7 0 SCRXT4 4 1 8 1 SCRVP1 5 2 PULSE(0 10 1667U 1N 1N 100U 20M)VP2 6 2 PULSE(0 10 11667U 1N 1N 100U 20M)VP3 7 0 PULSE(0 10 1667U 1N 1N 100U 20M)VP4 8 1 PULSE(0 10 11667U 1N 1N 100U 20M)L1 2 3 31.8ML2 1 9 1.6MR1 3 4 10C1 3 4 318.5uR2 1 0 1MEGR3 2 0 1MEG

R4 4 0 1MEG



The waveform of bridge output voltage

The waveform of bridge output current

The waveform of voltage across the load

The waveform of filter capacitor current


The waveform of dc link inductor voltage

SIMULATION USING C

// Simulation of a single-phase fully-controlled bridge// rectifier circuit.// This program produces the transient response, the periodic// response. Commuation overlap angle is also computed.// Calculations normalized. The average voltage that would// occur for diode-bridge rectifier with a resistive load is// taken to be unity. The average current through the load// resistor with a pure resistive load on the diode bridge// rectifier is taken as the unity current. The parameters to// be fed are just three: the ratio of load inductive reactance// to the load resistance, wL1/R , the ratio of source reactance// to the load resistance to load resistance wL2/R and the firing// angle in degrees.

#include <math.h>

#include<stdlib.h>#include<stdio.h>#include<string.h>#include<iostream.h>

void kreateRespFile(void);void tranResponse(void);void OneCycle(void);void computeOneStep(void);void produceEntries(void);

const double pi=3.1415926;const double deg_rad=pi/180.0;const double step=pi/720.0;

double L1,L2,alpha,fang,elapseAng,cycleAng,tote_knt;double cur_load,cur_line,outVolt,vSource,vInput,capVolt;double OverLapangle,Cap,capCur,dcLinkCur,Res;int commute,toggle,yes_Entry,modeSw;FILE *fnew;char *tstr,*p;

int main(void) float ka1,kb2,kc3,kd4,ke5; printf(" \n"); printf(" Load Time Constant in radians = "); scanf("%f",&ka1); L1=(double)ka1; if (L1<0.05) L1=0.05; printf(" \n"); printf(" Line Reactance Time Constant in radians = "); scanf("%f",&kb2); L2=(double)kb2; printf(" \n"); printf(" Cap. Filter Time Constant in radians = "); scanf("%f",&kc3); Cap=(double)kc3; if (Cap<0.1) Cap=0.1; if (Cap>10.0) Cap=10.0; printf(" \n"); printf(" Load Resistance in p.u. = "); scanf("%f",&ke5); Res=(double)ke5; if (Res<0.1) Res=0.1; if (Res>10.0) Res=10.0; printf(" \n"); printf(" Firing angle in degrees = "); scanf("%f",&kd4); alpha=(double)kd4; fang=alpha*deg_rad; alpha=0.0;

printf(" \n");


elapseAng=0.0 ; cycleAng=0.0; capCur=0.0; capVolt=0.0; cur_load=0.0; cur_line=0.0; outVolt=0.0; tote_knt=0.0; OverLapangle=0.0; commute=0; yes_Entry=0; dcLinkCur=0.0; cur_load=0.0;

// Create a file for entering transient response fnew=fopen("tran_v3.csv","w+r"); tranResponse(); fclose(fnew); return 0;

void kreateRespFile(void) int n1,n2; tstr="Angle,InputV,VoutBr,dcLinkCur,LineCur,IndVolt,LoadVolt,capCur"; p=strtok(tstr,","); fprintf(fnew,p); fprintf(fnew,","); n2=8; for (n1=0;n1<(n2-1);n1++) p=strtok(NULL,","); if (n1!=(n2-2)) fprintf(fnew,p); fprintf(fnew,","); else fprintf(fnew,p); fprintf(fnew,"\n");

void tranResponse(void) double d1; int NumCycles,count; d1=(6.0*(L1+L2+Cap))/(2.0*pi)+0.5; NumCycles=(int)d1 + 1; kreateRespFile(); modeSw=1; for (count=0;count<=NumCycles;count++) OneCycle();

void OneCycle(void) int knt; double dknt;

dknt=0.0;

for (knt=0;knt<1440;knt++) cycleAng=dknt*step; vInput=pi/2.0*sin(cycleAng); if ((cycleAng<fang) || (cycleAng>(fang+pi))) if (cycleAng<fang) vSource=pi/2.0*sin(cycleAng+pi); else vSource=pi/2.0*sin(cycleAng-pi); toggle=0; else vSource=pi/2.0*sin(cycleAng); toggle=1; if (modeSw!=toggle)

commute=1; if (L2<0.0005) commute=0; modeSw=toggle; computeOneStep(); dknt+=1.0;

void computeOneStep(void) double dLoadI,dLineI,doutV;

switch (commute) case 0: dLoadI=(vSource-dcLinkCur)/(L1+L2)*step; doutV=(dcLinkCur-capVolt/Res)*step/Cap; dcLinkCur=dcLinkCur+dLoadI; capVolt=capVolt+doutV; if (capVolt<0.0) capVolt=0.0; if (dcLinkCur<0.00001) dcLinkCur=0.0; if (toggle==0) cur_line=-dcLinkCur; else cur_line=dcLinkCur; outVolt=cur_load+(vSource-cur_load)*L1/(L1+L2); cur_load=capVolt/Res; capCur=dcLinkCur-cur_load; break;

case 1: dLoadI=(0.0-cur_load)/L1*step; doutV=(dcLinkCur-capVolt/Res)*step/Cap; dcLinkCur=dcLinkCur+dLoadI;

capVolt=capVolt+doutV; if (capVolt<0.0) capVolt=0.0; if (dcLinkCur<0.00001) dcLinkCur=0.0; dLineI=vInput/L2*step; cur_line=cur_line+dLineI; if (toggle==1) if (cur_line>dcLinkCur) commute=0; if (toggle==0) if ((cur_line+dcLinkCur)<0.0) commute=0; if (dcLinkCur<0.00001) outVolt=capVolt; else outVolt=0.0; cur_load=capVolt/Res; capCur=dcLinkCur-cur_load; break; tote_knt+=1.0; elapseAng=tote_knt*step; alpha=tote_knt/4.0; yes_Entry+=1; if (yes_Entry==4) produceEntries(); if (yes_Entry==4) yes_Entry=0;

void produceEntries(void) double negate; gcvt(alpha,5,p); fprintf(fnew,p); fprintf(fnew,","); gcvt(vInput,5,p); fprintf(fnew,p); fprintf(fnew,","); gcvt(outVolt,5,p); fprintf(fnew,p); fprintf(fnew,","); gcvt(dcLinkCur,5,p); fprintf(fnew,p); fprintf(fnew,","); gcvt(cur_line,5,p); fprintf(fnew,p); fprintf(fnew,","); if (toggle==0) negate = 1.0; else negate=-1.0; gcvt(vInput-negate*outVolt,5,p); fprintf(fnew,p); fprintf(fnew,","); gcvt(capVolt,5,p); fprintf(fnew,p); fprintf(fnew,","); gcvt(capCur,5,p); fprintf(fnew,p);

fprintf(fnew,"\n");

The plots obtained for L1=1.0, L2=0.1, R=1.0, C=1.0 and α =30o have been displayed below.

SUMMARY

This page has described the effect of adding a filter capacitor. It can also be seen that no attempt is made to obtain the periodic response by determining the coefficients, since it is easier to use the numerical technique to obtain the solution. The next page describes an application of this circuit.

A BATTERY CHARGER

CIRCUIT DIAGRAMCIRCUIT OPERATIONMATHEMATICAL ANALYSISSIMULATION

CIRCUIT DIAGRAM

The circuit diagram of a battery charger is shown above. The ac source and the inductance in series with it can represent a transformer and its leakage inductance as viewed from its secondary. In a battery charger circuit, the current supplied by the source is usually discontinuous and the load across the battery is usually very light.

CIRCUIT OPERATION

The behaviour of a fully-controlled bridge circuit has been described in the previous pages. This page describes how the battery-charger can be controlled in closed-loop. Traditionally systems use controllers such as a PI controller or a PDF controller for feedback control. But a battery has a very large time constant and control of a system with a large time constant is not easy. Hence the controller described in this page is a rule-based controller. It can be implemented either with traditional logic gates, counters and ADCs or by using a micro-controller and some more ICs to complement its operation.

The battery is modeled mainly as a very large capacitor and a series resistor RB. The

capacitance used in the model is not as large as the value that can represent the stored AH capacity of the battery, since that would lead to a very large period for simulation. The capacitance used is large enough to make the behaviour of the system similar to that of a real battery-charger, but small enough to get the simulation performed in a reasonable time.

Firstly, a method by which the phase-angle can be controlled is explained and then the rule-based controller is described. A synchronizing voltage, usually the output from a secondary winding of a transformer with its primary connected to the mains supply, is fed to a zero-crossing detector. In the scheme described here, two logic signals are developed; logic signal A is set to high level when the synchronizing voltage crosses zero, becomes positive and stays positive, and logic signal B is set to high level when the synchronizing voltage crosses zero, becomes negative and stays negative. If the synchronizing signal is described to be E*Sin (θ), it would be preferable if A is at 1 from θ = 5o to θ = 175o and B is at 1 from θ = 185o to θ = 355o . Then a logic NOR signal, A NOR B , can be obtained. The NOR signal can be used to reset a ramp generator when the synchronizing signal crosses zero in either direction. The ramp output can then be compared with a control voltage VC and the output C of comparator can be ANDed

with either A or B. The logic signal, A AND C, can be used to generate a firing signal for SCRs S1 and S3. It can be seen that these SCRs are triggered when the source

voltage Vs is positive, provided that the synchronizing signal is in phase with the mains

supply voltage. The logic signal, B AND C, can be used to generate a firing signal for SCRs S2 and S4. It can be seen that these SCRs are triggered when the source voltage Vs

is negative.

The scheme outlined can be implemented in different ways. One of the methods that can be followed is described next.

From the mains supply, a synchronizing signal can be obtained. If a single-polarity

supply is to be used, CMOS opamps, suitable for operation with a single power supply, can be used. The resistors have to be chosen to suit the requirements. For example, let the peak of synchronizing voltage be 10 V. If the ramp is to start at 5o, then the magnitude of ramp signal would be 10 * Sin (5o), which is 0.871 V. If VCC is 12 V,

then R1 and R2 can be selected such that the voltage across R2 is 0.871 V. The values of

R3 and R4 are selected that the ratio (R3/R4) is very small, of the order of about 0.01.

The circuit to generate the ramp signal is shown above. Whenever the signal, A NOR B, is at 1, the transistor is switched ON and it discharges the capacitor. When the NOR signal becomes zero, the boot-strap integrator generates a ramp signal at its output. If the ramp signal is be called y(t), then

where t starts from zero from the instant the transistor stops conducting. The values of R and C should be chosen such that when the synchronizing voltage varies from 10 * Sin (5o) to 10 * Sin (175o), the ramp signal varies from 0 V to about 10 V. The time corresponding to 170o depends on the frequency. At 50 Hz, the corresponding time period is 9.444 ms and it is 7.87 ms at 60 Hz. When the ramp signal varies in this manner, the control voltage can be varied from 0 V to 10 V to vary the firing angle. For the boot-strap integrator also, a CMOS opamp can be used . It would be preferable to use a CMOS opamp as the comparator too. From the comparator output and the logic signals A and B, the signals that can be used to generate firing pulse for the SCRs is described next.

A firing signal for an SCR can be generated in several ways. For high power SCRs, the best technique is to use a pulse transformer with a ferrite core. But in an application such as the battery-charger with a single-phase fully-controlled bridge rectifier circuit, the rating of the SCRs is not likely to be in the range of hundreds of amperes and it may be preferable to use an opto-coupler with light-activated SCR detector.

One opto-coupled light-activated triggering IC is required for each SCR. It would be preferable to connect a resistor between the gate and the cathode of light-activated SCR. Its value can be in the region of several kilo-ohms and it depends on the IC used. The

light-activated SCR in turn triggers the main SCR as shown below. For the SCRs in the bridge rectifier circuit also, it would be preferable to connect a resistor between the gate and the cathode, and its value can be about 500 Ω. In addition, it would be necessary to connect a snubber circuit from the anode to cathode of each SCR. The values of components to be used for the snubber circuit can be normally obtained from the datasheet of the SCR. It should also be remembered that the blocking voltage of the light-activated SCR should be the same as that of the main SCR.

The source that drives the bridge circuit is usually the secondary of a transformer. The leakage inductance of the transformer as viewed from the secondary terminals acts as the source inductance for the circuit. With a light load, the bridge output tends to be discontinuous. The next section describes how the circuit operates.


The equations are normalized. The nominal voltage of the battery is taken to be 1 p.u and the nominal rms current of the secondary of the transformer is taken to be 1 p.u. For example, let the nominal voltage of the battery bank be 72 V. Let the rms secondary voltage be 100 V and its rated rms current be 10 A. Then the peak secondary voltage equals (1.414*100/72), that is 1.964 p.u. If the drop in terminal voltage of the battery bank when it delivers 10 A is 1 V, then its source impedance is set to be 1.414/72, that is 0.0196 p.u.

The load current drawn by the load resistor should be relatively low, less than 0.25 p.u. That is, 0.3 p.u corresponds to 4.24 A the corresponding load resistance is 72/4.24, that is 17 Ω.

The simulation routine is explained below. All setting are in per unit notation.

Initializing routine:

Set the value of peak secondary voltage

Set the value of fractional load

Set the source inductance. Set the peak battery current Set the maximum average battery current Set the initial battery voltage(internally set)

Set the time constant of the battery(set to 300 radians internally)

Set the source resistance of the battery

Set the firing angle to 175o (internal setting)

Reset to zero the battery current. peak charging current. Reset to zero the rms secondary current. average battery charging current. Go to pre-amble routine

Pre-amble routine:

Print the battery voltage

Print the average battery voltage

Print the peak charging current Print the rms secondary current Set the rule-based controller as follows:

if (average Battery Current =(1.1*maximum average battery Current))

increase the firing angle by 1o

else

if ((batteryVolt<=1.1) and (average Battery Current <maximum average battery Current))

decrease the firing angle by 1o

if ((batteryVolt<=1.1) and (average Battery Current =maximum average battery Current))

do no change the firing angle

if ((batteryVolt1.1) and (batteryVolt<=1.15)) do no change the firing angle

if (batteryVolt1.15) increase the firing angle by 1o

if (peakBtyCurpeakCurSet) increase the firing angle by 2o

Reset to zero peak charging current, average battery current and rms secondary Current. Go to One-Output-Cycle routine.

One-Output-cycle routine:

angle=0;

while ( angle =0 and angle <180 )

compute increments in battery voltage and source current compute increments to rms secondary current, average battery charging current set the peak battery charging current appropriately

increment angle by step size

SIMULATION

The applet below demonstrates how a rule-based battery charger would function.

Next page describes how a dc motor can be controlled in closed-loop.

A TWO-QUADRANT DC DRIVE

PERFORMANCE CHARACTERISITCSBLOCK DIAGRAMCLOSED-LOOP CONTROLSIMULATION

PERFORMANCE CHARACTERISTICS

This page describes how a separately-excited DC motor can be controlled in closed-loop with a single-phase fully-controlled rectifier supplying dc source to its armature. The operation of a DC motor is described briefly at first.

A symbolic representation of a separately-excited DC motor is shown above. The resistance of the field winding is Rf and its inductance is Lf, whereas the resistance of the armature is Ra and

its inductance is La. In the description of the motor, the armature reaction effects are ignored.

It is justifiable since the motor used has either interpoles or compensating winding to minimize the effects of armature reaction. The field current is described by equation (1). If a steady voltage Vf is applied to the field, the field current settles down to a constant value, as shown in

equation (2). When the field current is constant, the flux induced by the field winding remains constant, and usually it is held at its rated value φ. If the voltage applied to the armature is va,

then the differential equation that is to be applied to the armature circuit is shown in equation (3). In steady-state, equation (4) applies. The voltage, ea, is the back e.m.f. in volts. In a

separately-excited DC motor, the back e.m.f is proportional to the product of speed of motor w rad/s and the field φ Webers, as shown by equation (5).

In equation (5), Km is a coefficient and its value depends on the armature winding. If the

armature current in steady-state be Ia, then the power P that is supplied to the armature is EaIa.

This electric power is converted to mechanical power by the armature of the DC motor. Let the torque developed by the armature be Te, the unit for torque being Nm (Newton-metre). Then

power and torque can be related as shown in equation (6). On canceling the common term on both sides, the torque Te developed by the armature is obtained as presented in equation (7).

If the instantaneous armature current is ia, then equation (8) applies. Torque has been denoted

by Te in both equations.

The speed of the motor can be controlled by varying Va and holding Vf constant at its rated

value. Then as the voltage applied to the armature is raised, the armature current increases first. As the armature current increases, the torque developed by motor increases and hence the speed of motor increases. The drop across the armature resistance tends to be small and hence

the motor speed rises almost proportionately with the voltage applied to the armature. But there is a limit to the voltage that can be applied to the armature and that limit is the rated voltage of the armature voltage. The speed of the motor corresponding to the rated armature voltage and the rated field voltage is its rated speed. Thus the speed of a motor can be varied below its rated speed by controlling the armature voltage. It would be desirable that the motor should be able to develop as high as a torque as possible and hence the voltage rated applied to the field is held at its rated value. Applying higher than the rated voltage to either the field or the armature is not recommended. When the rated voltage is applied to the field, the flux would be near the saturation level in the poles. If a voltage higher than its rated voltage is applied to the field, the flux would saturate and there would not be any significant increase in the torque that the motor can deliver. On the other hand, this would only result in increased losses in the winding. Since the total heat which the DC motor can dissipate is fixed due to its surface area and cooling system, increased losses from the excitation system would mean that the other losses would have to reduce, implying that the armature current cannot be at its rated level and the maximum torque that the motor can deliver may reduce. Increasing the armature voltage above its rated value is not recommended because the insulation of the armature is designed for operation of the motor with the rated voltage applied to its armature. Moreover, the torque that the motor can deliver depends on the armature current and the field current. If the motor is operated continuously, the maximum armature current should not be higher than its rated value. When the armature current and the field voltage are at their rated level, the motor generates the rated torque. Hence the maximum torque the motor can deliver continuously over a long period of time is its rated torque when its speed is varied from a low value to its rated speed. Over this period, 0 < w < wr, where wr is its rated speed, the power output is given by:

The maximum torque which the motor can deliver continuously is called Te,max cont. What is

being referred to here is the maximum torque the motor can deliver, and not the actual torque the motor delivers. The actual torque the motor delivers depends on the mechanical load connected to its shaft. If the speed of the motor is to be increased beyond its rated value, the voltage applied to the armature can be held at its rated value and the field can be weakened by reducing the voltage applied to it. When the speed of the motor is in this manner, the maximum power that can be supplied to the armature is fixed, since both the voltage applied to the armature and the armature current cannot exceed the rated level over a long period. That means the maximum torque the motor can develop above the rated speed is:

The plots of Te,max cont and the maximum power Pa,max can be plotted as a function of rotor

speed asshown below. The rated values of speed, torque and power to the armature have been set equal to unity.

A separately-excited dc motor can be controlled, either by varying the voltage applied to the field winding or by varying the voltage applied to the armature. This page describes how the motor can be controlled by varying the armature voltage and it is assumed that the field is excited by a constant voltage, equaling the rated voltage of the field winding. It means that the discussion to follow assumes that the field current remains steady at its rated value.

BLOCK DIAGRAM

The block diagram of a dc drive is shown above. It does not show all details. The DC motor has not been represented in the form of a block diagram and the details of the load the motor drives have also not been shown. The block diagram functions as follows.

For the system described here, the output of the system is the speed of the motor. Hence when this system is to be controlled in closed-loop, the parameter that is to be set is what that speed should be. It is denoted to be Ωref. In order to control the speed in closed-loop, we need a

feedback signal too. It can be obtained in several ways. A digital tacho or an analogue tachogenerator can be used. It is assumed that an analogue tachogenerator is used here. It is coupled to the motor shaft and its output voltage varies linearly with its speed. Let the speed feedback signal be Ωf. This signal can be compared with the speed reference signal and the

error can be processed by a controller. The controller can be of one of several types. It can be

an integral controller, or a PI controller and PDF (pseudo-derivative feedback) controller or a PID controller or a rule-based fuzzy logic controller. Here both the controllers used are PI (proportional plus integral) controllers. A PI controller can lead to fast response and zero-error for a step input.

The PI controller for speed has as its input the error between the two signals, Ωref and Ωf. If the

speed feedback signal Ωf is lower than the reference signal Ωref , it means that the DC motor

speed is running below the set speed and it needs to be accelerated. In order to accelerate the motor, it should develop greater torque. To develop greater torque, its armature current has to increase. Hence the output of speed controller is set to function as the reference signal for armature current. It will be a voltage corresponding to armature current with an appropriate coefficient linking the two quantities. When Ωf < Ωref, the difference causes the output of speed

controller to increase. Since the output of speed-controller is set to function as the armature current reference signal, an increase in the value of speed-controller output would in turn lead to an increase in the armature current.

The rectifier circuit is made up of SCRs and the SCRs have a current rating. Hence it is necessary to ensure that the current through the SCRs remains within a safe level. Hence the output of speed controller is limited at both ends. Its maximum value corresponds to the safe level for SCRs. It is not normally the rated current of the motor and it is usually set at a value ranging from 1.5 times to 2 times the rated armature current. The reason is that the motor may have to develop more than the rated torque under transient conditions to achieve fast response. In order to ensure that the motor armature current remains within its rated value, another supervisory loop may be used. Another option is to use a circuit-breaker. The instantaneous trip action in the circuit breaker can be due to magnetic effect and the overload trip can be due to thermal action. A bi-metallic strip within the circuit-breaker expands due to temperature and would trip circuit-breaker. The lower limit on the output of speed-controller would correspond to zero current in the armature, since the motor current in this scheme cannot be in the reverse direction.

The current controller has two inputs, the reference current signal which is the output of the speed controller and a feedback signal proportional to the armature current. The feedback signal can be obtained in several ways. A current transformer can be introduced in the path of ac current from the ac supply. Another option would be to use a DC current transducer that makes use of a Hall-effect sensor or isolated opamp. The transducer used produces a voltage proportional to the current in the armature. The difference between these two signals is processed by another PI controller and its output is also limited to correspond to 0o and 180o firing angle. The output of current controller may vary between 0 V and 10 V, with 0 V corresponding to 180o firing angle and 10 V corresponding 0o firing angle. If the firing angle be α and the output of current controller VC, then

As the output voltage of current controller increases due to the difference between the reference signal and the feedback signal corresponding to armature current, the firing angle is advanced towards 0o and the average output voltage of the bridge rectifier increases. This in turn leads to increased torque generation and the motor accelerates.

If the speed reference is brought down suddenly, the current in the motor cannot be reversed and hence the motor slows down due to friction and the load. This process can be slow.

The question that can be raised is whether we need the current loop. The answer is that it improves the performance. If there is a change in the supply voltage even by a small amount, the output of the bridge circuit tends to a fall a bit for the same firing angle. The reduction in output voltage causes a large change in the armature current, with the speed remaining more or less constant. The current loop comes into action, correcting the firing angle to the required value. The time constant of the armature, due to its inductance and resistance, tends to be of the order of a few tens of ms and the mechanical time constant, due to the moment of inertia of motor and load and the friction, is of the order of a few tenths of a second. If a current controller is not used, the speed would have to change before the speed controller can come into action. Since the mechanical time constant is about at least 10 times greater, there would be a significant change in speed if there be no current controller.

Normally a filter may be necessary in the feedback circuit for speed. The tacho signal usually contains a small ripple superimposed on its dc content. The frequency of the ripple is usually dependent on the speed and the lower the speed is the lower is the frequency of this ripple. Hence the time constant of the filter may have to be set to correspond to the lowest speed at which the motor would be required to run. Usually the motor speed does not have to vary over a range larger than 0.1 p.u to 1 p.u. Since the power output varies proportionately with the speed, there is usually no justification to run the motor at an extremely low speed. The next section describes how the simulation is carried out. The routines are explained with the help of pseudo-code that can be understood by a reader with some knowledge of one of the programming languages such as C, PASCAL, BASIC, Fortran or Matlab.

CLOSED-LOOP CONTROL

This section explains how the simulation can be carried out.

Initialize_Routine:

Set and get Parameters

Initialize controller Outputs

Go to Calculation_Routine

Calculation_Routine

Execute One_Cycle_Routine

Plot the results

One_Cycle_Routine

Set angle to zero. For (angle = 0; angle <360o; angle +stepsize)

Set the SCR pair based on firing angle and angle. Go to Next_Values

Next_Values

Calculate the increments in

armature current motor speed

tacho filter output speed controller output current controller output Add the increments

Compute the firing angle

The pseudo-routine presents only the main steps.

SIMULATION

Before selecting the type of response, set the value of the selected parameter. When you select a parameter, the textfield shows the default value set inside the program. Change the parameter value if you want to and then you must click on the SET VALUE button for the change to take effect. You can go from one type of response to another after the present calculations are carried out. When you have selected a new type of response, you must click on Click to Start. If you click on Reset button, initializing routine is carried out and the motor speed is set to zero, and the other values are also reset. The program has been written assuming that the frequency of operation is 50 Hz. If the frequency is different, the parameter values should be scaled suitably.

Next page describes how a three-phase fully-controlled bridge rectifier circuit operates.

FULLY-CONTROLLED 3-PH SCR BRIDGE RECTIFIER

This chapter describes the operation of a three-phase fully-controlled SCR bridge rectifier circuit and two applications. The pages to follow contain:

OPERATION OF A 3-PHASE FULLY-CONTROLLED RECTIFIER

OPERATION WITH A RESISTIVE LOAD OPERATION WITH AN RL LOAD OPERATION WITH AN RL LOAD AND SOURCE

INDUCTANCE AN APPLICATION: A DC POWER SUPPLY AN APPLICATION: A FOUR-QUADRANT DC DRIVE

OPERATION OF A 3-PHASE FULLY-CONTROLLED RECTIFIER

CIRCUIT OPERATIONSYNCHRONIZING SIGNALSMATHEMATICAL ANALYSISSIMULATIONMATHCAD SIMULATION

CIRCUIT OPERATION

The operation of a 3-phase fully-controlled bridge rectifier circuit is described in this page. A three-phase fully-controlled bridge rectifier can be constructed using six SCRs as shown below.

The three-phase bridge rectifier circuit has three-legs, each phase connected to one of the three phase voltages. Alternatively, it can be seen that the bridge circuit has two halves, the positive half consisting of the SCRs S1, S3 and S5 and the negative half consisting of the SCRs S2, S4 and S6. At any time when there

is current flow, one SCR from each half conducts. If the phase sequence of the source be RYB, the SCRs are triggered in the sequence S1, S2 , S3 , S4, S5 , S6 and S1 and so on.

The operation of the circuit is first explained with the assumption that diodes are used in place of the SCRs. The three-phase voltages vary as shown below.

Let the three-phase voltages be defined as shown below.

It can be seen that the R-phase voltage is the highest of the three-phase voltages when θ is in the range from 30o to 150o. It can also be seen that Y-phase voltage is the highest of the three-phase voltages when θ is in the range from 150o to 270o and that B-phase voltage is the highest of the three-phase voltages when θ is in the range from 270o to 390o or 30o in the next cycle. We also find that R-phase voltage is the lowest of the three-phase voltages when θ is in the range from 210o to 330o. It can also be seen that Y-phase voltage is the lowest of the three-phase voltages when θ is in the range from 330o to 450o or 90o in the next cycle, and that B-phase voltage is the lowest when θ is in the range from 90o to 210o. If diodes are used, diode D1 in place of S1 would conduct from 30o to 150o, diode D3

would conduct from 150o to 270o and

diode D5 from 270o to 390o or 30o in the next cycle. In the same way, diode D4 would conduct from 210o

to 330o, diode D6 from 330o to 450o or 90o in the next cycle, and diode D2

would conduct from 90o to

210o. The positive rail of output voltage of the bridge is connected to the topmost segments of the envelope of three-phase voltages and the negative rail of the output voltage to the lowest segments of the envelope.

At any instant barring the change-over periods when current flow gets transferred from diode to another, only one of the following pairs conducts at any time.

Period, range of θ Diode Pair in conduction

30o to 90o D1 and D6





330o to 360o and 0o to 30o D5 and D6

If SCRs are used, their conduction can be delayed by choosing the desired firing angle. When the SCRs are fired at 0o firing angle, the output of the bridge rectifier would be the same as that of the circuit with diodes. For instance, it is seen that D1 starts conducting only after θ = 30o. In fact, it can start conducting

only after θ = 30o , since it is reverse-biased before θ = 30o. The bias across D1 becomes zero when θ = 30o

and diode D1 starts getting forward-biased only after θ =30o. When vR(θ) = E*Sin (θ), diode D1 is reverse-

biased before θ = 30o and it is forward-biased when θ > 30o. When firing angle to SCRs is zero degree, S1

is triggered when θ = 30o. This means that if a synchronizing signal is needed for triggering S1, that signal

voltage would lag vR(θ) by 30o and if the firing angle is α, SCR S1 is triggered when θ = α + 30o. Given

that the conduction is continuous, the following table presents the SCR pair in conduction at any instant.

Period, range of θ SCR Pair in conduction

α + 30o to α + 90o S1 and S6

α + 90o to α + 150o S1 and S2

α + 150o to α + 210o S2 and S3

α + 210o to α + 270o S3 and S4

α + 270o to α + 330o S4 and S5

α + 330o to α + 360o and α + 0o to α + 30o S5 and S6

The operation of the bridge-rectifier is illustrated with the help of an applet that follows this line. You can set the firing angle in the range 0o < firing angle < 180o and the instantaneous angle. The applet displays the SCR pair in conduction at the chosen instant. The current flow path is shown in red colour in the circuit diagram. The instantaneous angle can be either set in its text-field or varied by dragging the scroll-bar button. The rotating phasor diagram is quite useful to illustrate how the circuit operates. Once the firing angle is set, the phasor position for firing angle is fixed. Then as the instantaneous angle changes, the pair that conducts is connected to the thick orange arcs. One way to visualize is to imagine two brushes which are 120o wide and the device in the phase connected to the brush conducts. The brush that has "Firing angle " written beside it acts as the brush connected to the positive rail and the other acts as if it is

connected to the negative rail. This diagram illustrates how the rectifier circuit acts as a commutator and converts ac to dc. The output voltage is specified with the amplitude of phase voltage being assigned unity value.

SYNCHRONIZING SIGNALS

To vary the output voltage, it is necessary to vary the firing angle. In order to vary the firing angle, one commonly used technique is to establish a synchronizing signal for each SCR. It has been seen that zero degree firing angle occurs 30o degrees after the zero-crossing of the respective phase voltage. If the synchronizing signal is to be a sinusoidal signal, it should lag the respective phase by 30o and then the circuitry needed to generate a firing signal can be similar to that described for single-phase. Instead of a single such circuit for a single phase rectifier, we would need three such circuits.

When the 3-phase source supply connected to the rectifier is star-connected, the line voltages and the phase voltages have a 30o phase angle difference between them, as shown below.

The line voltage can also be obtained as:

This line voltage lags the R-phase voltage by30o and has an amplitude which is 1.732 times the amplitude of the phase voltage. The synchronizing signal for SCR S1 can be obtained based on vRB line voltage. The

synchronizing signals for the other SCRs can be obtained in a similar manner.

To get the synchronizing signals, three control transformers can be used, with the primaries connected in delta and the secondaries in star, as shown below.

For S1, voltage vS1 is used as the synchronizing signal. Voltage vS2 is used as the synchronizing signal for

SCR S2 and so on. The waveforms presented by the synchronizing signals are as shown below. The

waveforms do not show the effect of turns ratio, since any instantaneous value has been normalized with respect to its peak value. For example, let the primary phase voltage be 240 V and then its peak value is 339.4 V. The primary voltage is normalized with respect to 339. V. If the peak voltage of each half of secondary is 10 V, the secondary voltage are normalized with respect to 10 V.


Analysis of this three-phase controlled rectifier is in many ways similar to the analysis of single-phase bridge rectifier circuit. We are interested in output voltage and the source current. The average output voltage, the rms output voltage, the ripple content in output voltage, the total rms line current, the fundamental rms current, THD in line current, the displacement power factor and the apparent power factor are to be determined. In this section, the analysis is carried out assuming that the load current is a steady dc value.

AVERAGE OUTPUT VOLTAGE

Before getting an expression for the output voltage, it is preferable to find out how the output voltage waveform varies as the firing angle is varied. In one cycle of source voltage, six pairs conduct, each pair for 60o. This means that the period for output waveform is one-sixth of the period of line voltage. The output waveform repeats itself six times in one cycle of input voltage. The waveform of output voltage can be determined by considering one pair. It is seen that when vR(θ) = E* Sin (θ), SCR S1 and S6 conduct

when θ varies from 30o + α to 90o + α , where α is the firing angle. Then

The waveform of output can be plotted for different firing angles. The applet below takes in the firing angle as an input and plots the output. The peak line-to-line voltage is marked as 'U' and the applet starts with the instant an SCR is fired and displays the output waveform for one input cycle period.

The average output voltage of the bridge circuit is calculated as follows, with a change in variable, where θ = α + 60o.

In the expression above, U is the peak line-to-line voltage, whereas E is the amplitude of phase voltage of 3-phase supply.

RMS OUTPUT VOLTAGE

The rms output voltage is calculated as follows:

The ripple factor of the output voltage is then:

The applet below displays the average output voltage, the rms output voltage and the ripple factor for the case of continuous conduction through the load.

It is seen that the average output voltage is negative when firing angle exceeds 90o. It means that power flow is from the dc side to the ac source. When the firing angle is kept in the region 0o < α < 90o, this circuit is said to be operating in the rectifier region. When the firing angle is kept in the region 90o < α < 180o, this circuit is said to be operating in the inverter region. When the circuit operates in the rectifier region, the net power flow is from the ac source to the dc link. In the inverter region, the net power flow is in the reverse direction. To operate in the inverter region, it is necessary to have a dc source present in the dc link which can provide the power that is fed back to the ac source.

RMS LINE CURRENT

The rms line current is relatively easy to find out if the dc current is ripple-free and steady. The load current is ripple-free if the inductance in the dc link is relatively large. To maintain load current at any firing angle, it is necessary that the dc link should contain a voltage source. Given that the resistance of the load circuit is zero, the voltage source should equal the average output voltage of the bridge circuit. The waveforms shown below are based on the assumption that these conditions are met. It has been shown that if vR(θ) = E*Sin (θ), SCR S1 conducts when θ varies from α + 30o to α + 90o and that SCR S4 conducts

when θ varies from α + 210o to α + 270o. If the amplitude of dc load current is assigned to be unity, the line current waveform is then a rectangular pulse, remaining at + 1 from α + 30o to α + 150o, at - 1 from α + 210o to α + 330o, and zero elsewhere. The amplitude of the fundamental in line current is then 3.464/π ( which evaluates to nearly 0.78) and the amplitude of other odd harmonics is 3.464/nπ, where n is the odd harmonic number. When the dc load current is steady and has a magnitude of unity, the rms line current is obtained as shown in equation (5). The rms value of the fundamental is obtained as shown in equation (6). Equation (6) is based on how trigonometric Fourier coefficients are defined for waveforms with quarter-wave symmetry. When the line current is a rectangular and symmetric, the phase current is the same as the line current and the fundamental component of the phase current lags the phase voltage by an angle equal to the firing angle. Hence the displacement power factor is expressed as shown by equation (7). Since the line current is not sinusoidal, the apparent power factor, usually referred to just as the power factor in most of the texts, is less than DPF and is represented by equation (8). Since the line current is not sinusoidal, the distortion component in the line current has to be computed. This component, called the THD( Total Harmonic Distortion ), is calculated as shown in equation (9).

OPERATION WITH A RESISTIVE LOAD

CIRCUIT OPERATIONMATHEMATICAL ANALYSISSIMULATIONPSPICE SIMULATIONSUMMARY

CIRCUIT OPERATION

The three-phase fully-controlled bridge rectifier with a purely resistive load is shown above. With any single-controlled rectifier circuit, the load current is unidirectional. Hence when the output voltage of the bridge tends to become negative, the load current is zero and the conduction is discontinuous, with the output voltage clamped to zero volts. The operation of this circuit has been described in detail in the previous page.


Since the load current tends to be discontinuous, two expressions for the average output voltage can be derived, one for the continuous mode and the other for the discontinuous mode. In the continuous mode, the average output voltage is as shown in equation (1). The conduction becomes discontinuous when the firing angle exceeds 60o and remains less than 120o. Then the average voltage is obtained as shown in equation (2).

In equations (1) and (2), the amplitude of phase voltage is designated as E and the amplitude of line voltage is designaed as U.

The rms voltage is computed as follows. When α < 60o, the conduction is continuous and the expression for the rms voltage is presented in equation (3), whereas equation (4) expresses the rms voltage obtained when firing angle α > 60o.

The ripple factor can be found out as defined in the previous page.

The line/phase current can be defined as follows. Let R-phase voltage be defined to be

vR(θ) = E*Sin (θ).

Then the R-phase current iR(θ) is defined as follows, when the conduction is continuous.

When the conduction is discontinuous,

From this definition of phase current, the rms line current, the rms of the fundamental in line current and its THD can be found out.

The applet below displays the average output voltage, the rms output voltage, its ripple factor, the rms line current, the fundamental rms content in line current and its THD as a function of firing angle. The peak average output voltage, (3U/π), is taken to be unity and (3U/πR) is set equal to unity, where R is the load resistor.

SIMULATION

The applet shown below displays the source voltage, the output voltage, the current in R-phase and the voltage across SCR S1. In addition, the relevant statistical details are also displayed. To run the applet, key-in the firing angle and then

click on Start Button.

PSPICE SIMULATION

The Pspice program for simulation of this three-phase rectifier circuit is presented below. The model used for the SCRs is the same as defined for the single-phase fully-controlled bridge rectifier. The three-phase brdige rectifier contains six SCRs and it is necessary to define six pulse sources, one for each SCR. The pulse sources have been defined for a firing angle of 30o and the frequency of the three-phase source is 50 Hz. At any time two SCRs need to conduct, one from the top half and another bottom half and hence only if one SCR is triggered at a time, conduction may never get established. To overcome this problem, two SCRs are triggered at the same time. For example, when SCR S2 is to be triggered, SCR

S1 is also triggered. In the same way, when SCR S3 is to be triggered, SCR S2 is also triggered and so on. In order to

effect this in program, one voltage-controlled voltage source is defined for each SCR. The dependent source defined for SCR S1 is dependent on two sources, the pulse source that defines when SCR S1 is to be triggered and the pulse source

that defines when SCR S2 is to be triggered.

* Three-phase Full-wave Fully-Controlled Bridge RectifierVA 1 0 SIN(0 340V 50Hz)VB 2 0 SIN(0 340V 50Hz 0 0 -120)VC 3 0 SIN(0 340V 50Hz 0 0 -240)XT1 1 4 11 4 SCRXT3 2 4 13 4 SCRXT5 3 4 15 4 SCRXT4 5 1 14 1 SCRXT6 5 2 16 2 SCRXT2 5 3 12 3 SCRRP 4 0 100KRN 5 0 100KR1 4 5 10

VP1 21 0 PULSE(0 10 3333.3U 1N 1N 100U 20M)VP2 22 0 PULSE(0 10 6666.7U 1N 1N 100U 20M) VP3 23 0 PULSE(0 10 10M 1N 1N 100U 20M) VP4 24 0 PULSE(0 10 13333.3U 1N 1N 100U 20M) VP5 25 0 PULSE(0 10 16666.7U 1N 1N 100U 20M) VP6 26 0 PULSE(0 10 0M 1N 1N 100U 20M) RP1 21 0 100KRP2 22 0 100KRP3 23 0 100KRP4 24 0 100KRP5 25 0 100KRP6 26 0 100KEP1 11 4 poly(2) (21,0) (22,0) 0 1 1EP2 12 3 poly(2) (22,0) (23,0) 0 1 1EP3 13 4 poly(2) (23,0) (24,0) 0 1 1EP4 14 1 poly(2) (24,0) (25,0) 0 1 1EP5 15 4 poly(2) (25,0) (26,0) 0 1 1EP6 16 2 poly(2) (26,0) (21,0) 0 1 1

* Subcircuit for SCR.SUBCKT SCR 101 102 103 102S1 101 105 106 102 SMODRG 103 104 50VX 104 102 DC 0VY 105 107 DC 0

DT 107 102 DMODRT 106 102 1CT 106 102 10UF1 102 106 POLY(2) VX VY 0 50 11.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0).MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0).ENDS SCR

.TRAN 10US 60.0MS 0.0MS 10US

.PROBE


.END

The responses obtained for a load resistance of 10Ω are presented below.

The Output Voltage Waveform

The Line Current(phase A) Waveform

The Voltage(SCR S1) Waveform

SUMMARY

This page has described the operation of the three-phase fully-controlled bridge rectifier with a resistive load. The next page describes how this rectifier functions when the load contains an inductor also.

OPERATION WITH RL LOAD

CIRCUIT OPERATIONMATHEMATICAL ANALYSISSIMULATION

PSPICE SIMULATIONMATHCAD SIMULATIONSUMMARY

CIRCUIT OPERATION

The circuit of the three-phase fully-controlled bridge rectifier circuit with an RL load and a voltage source has been shown above. The purpose of providing a dc source in the dc link is to illustrate how two-quadrant operation can take place. When the firing angle is held between 0o and 90o, the circuit operates in the rectifier region. When the firing angle is held between 90o and 180o, the circuit can operate in the inverter region if the source E is sufficiently negative. In the inverter region, the power flow is from the dc link to the ac 3-phase source. The inductor reduces the ripple in dc link current.


When the current flow in the dc link is continuous, the average output voltage can be calculated as outlined in the previous page. It is difficult to estimate what the average output would be if there is a source present in the dc link and the conduction is discontinuous. If there be no source in the dc link, the average output voltage for discontinuous conduction is expressed by equation (1). The analysis of the circuit is along the lines described for the single-phase controlled rectifier circuit. In order to get an expression for the line/phase current, it is necessary to get an expression for the load current. The expression for load current is obtained from the expression for output voltage. The output voltage is described by equation (2). The differential equation that describes the load current is expressed by equation (3). The solution is of the form expressed by equation (4). The impedance of load is Z and the load angle is φ. They are defined as shown in equation (5).

In the equation above, w is the angular frequency in radians/second corresponding to the source frequency. When the load current is continuous, equation (6) is valid. Using equation (6), equation (7) is obtained. Solving for A, we get equation (8). When the conduction is discontinuous, iL(0) = 0 and then A is evaluated as shown in equation (9).

Once the value of A is known, the load current can be found out. From the load current, an expression for the line current can be obtained. When SCR S1 is ON, the line current equals the load current. The line current is the negative

of load current when SCR S4 is ON, and it is zero when neither S1 nor S4 is ON. From the expression of the load

current, the rms value of line current, the rms value of the fundamental component of line current, the THD in line current, the harmonic spectrum of line current, the DPF and the apparent power factor can be determined as outlined in the earlier pages.

SIMULATION

The parameters to be keyed in are the firing angle, the value of source in the dc link (between 1.0 and -1.0 preferably) and the ratio of load reactance to load resistance. The load reactance is wL, where w is the angular frequency in rad/s corresponding to the ac source frequency. Click on Start Button for the program to respond.

PSPICE SIMULATION

* Three-phase Full-wave Fully-Controlled Bridge RectifierVA 1 0 SIN(0 340V 50Hz)VB 2 0 SIN(0 340V 50Hz 0 0 -120)VC 3 0 SIN(0 340V 50Hz 0 0 -240)XT1 1 4 11 4 SCRXT3 2 4 13 4 SCRXT5 3 4 15 4 SCRXT4 5 1 14 1 SCRXT6 5 2 16 2 SCRXT2 5 3 12 3 SCRRP 4 0 100KRN 5 0 100KR1 4 6 10L1 6 5 31.8M

VP1 21 0 PULSE(0 10 3333.3U 1N 1N 100U 20M)VP2 22 0 PULSE(0 10 6666.7U 1N 1N 100U 20M) VP3 23 0 PULSE(0 10 10M 1N 1N 100U 20M) VP4 24 0 PULSE(0 10 13333.3U 1N 1N 100U 20M) VP5 25 0 PULSE(0 10 16666.7U 1N 1N 100U 20M) VP6 26 0 PULSE(0 10 0M 1N 1N 100U 20M) RP1 21 0 100KRP2 22 0 100KRP3 23 0 100KRP4 24 0 100KRP5 25 0 100KRP6 26 0 100KEP1 11 4 poly(2) (21,0) (22,0) 0 1 1EP2 12 3 poly(2) (22,0) (23,0) 0 1 1EP3 13 4 poly(2) (23,0) (24,0) 0 1 1EP4 14 1 poly(2) (24,0) (25,0) 0 1 1EP5 15 4 poly(2) (25,0) (26,0) 0 1 1EP6 16 2 poly(2) (26,0) (21,0) 0 1 1

* Subcircuit for SCR.SUBCKT SCR 101 102 103 102S1 101 105 106 102 SMODRG 103 104 50VX 104 102 DC 0VY 105 107 DC 0DT 107 102 DMODRT 106 102 1CT 106 102 10UF1 102 106 POLY(2) VX VY 0 50 11.MODEL SMOD VSWITCH(RON=0.0105 ROFF=10E+5 VON=0.5 VOFF=0).MODEL DMOD D((IS=2.2E-15 BV=1200 TT=0 CJO=0).ENDS SCR

.TRAN 10US 60.0MS 0.0MS 10US

.PROBE

.FOUR 50 V(4,5) I(VA) I(L1)


.END

The results of the Fourier series analysis are presented below.

The Fourier Series Spectrum of the output voltage

The Fourier Series Spectrum of the Line Current (Phase A)

The Fourier Series Spectrum of the Load Current

It can be seen that the lowest harmonic frequency in the output voltage is at 300 Hz, the sixth harmonic, whereas there is hardly any harmonic present in the load current because of the relatively large inductance in the load circuit. On the other hand, the 5-th and the 7-th harmonics are visibly high in the line current.

CIRCUIT OPERATIONMATHEMATICAL ANALYSISSIMULATIONMATHCAD SIMULATION

SIMULATION USING CSUMMARY

CIRCUIT OPERATION

The circuit of a three-phase fully-controlled bridge rectifier with source inductance is presented above. The presence of source inductance introduces an additional mode of operation when the firing angle is less than a certain value. Let us assume that SCRS S1 and S2 are in conduction when

SCRS S3 is triggered. Then the current from the source does not transfer from S1 to S3

instantaneously, and the transfer of current, called commutation, takes a while. During this

commutation overlap, both S1 and S3 conduct in addition to S2. SCR S1 continues to conduct till

the current through S3 rises to equal the dc link current.

The effects of commutation overlap are:

i. A slight reduction in output voltage, ii. A notch in the supply voltage to the circuit during commutation overlap.

When the source has inductance, other loads connected to this source along with the controlled rectifier are supplied voltages with notches in them and some of these loads can be sensitive to these notches and they may operate improperly. Hence in order to reduce the magnitude of notches, it is mandatory in some countries for the rectifier to be provided with an inductance in series with each of its three-phase input lines. If these inductors are much larger than the source inductance, the notch voltages are absorbed by these inductances and the other loads connected to the same 3-phase source are not supplied with distorted voltages. The internal inductances connected in series with the source are sometimes referred to as 4% inductances. If the inductor is such that the voltage drop across it is about 4% of the phase voltage at rated current, it is normally sufficient to reduce the notches at the source terminals to an acceptable level.


When there is an inductor in series with each input line, it is necessary to find out its effect. We need to find out:

a. The reduction in output voltage. b. The duration of commutation overlap. c. The relationship between the firing angle and the commutation overlap.

REDUCTION IN OUTPUT VOLTAGE

Calculations by hand are carried out assuming that the dc link current remains steady without any ripple. The source voltages at its terminals and the output voltage appear as shown below, assuming that the inductances belong to the source.

It is seen that there are six notches in one input cycle. The reduction in average output voltage can be found out as follows. Let SCR S1 be in conduction and let S3 be triggered. Let the current

through the dc link be IDC. Then current throughY-phase has to rise from zero to IDC, whereas

current through R-phase has to fall from - IDC to zero. On the other hand, loop current iLOOP

marked in the sketch below has to rise from zero to IDC. This means that during commutation

current through Y-phase would rise from zero to IDC and the volt-second area the output misses out

is L2IDC, that absorbed by the inductor in the Y-phase.

From the volt-seconds lost per commutation, we can find the total volt-seconds lost in one input cycle. Since there are six commutations per cycle, the total volt-seconds lost per cycle is expressed as shown by equation (1). Dividing this area by the time corresponding to one cycle, we get the average voltage reduction in output. The time corresponding to one input cycle is 1/f, where f is the line frequency. Then the average reduction in output voltage is obtained as shown in equation (2). In equation (2), we make use of the relation that the angular frequency, w = 2πf. It is to be noted that commutation overlap occurs only when there is continuous conduction through the load and the average output voltage is expressed by equation (3). In equation (3), U is the peak line-to-line voltage and α is the firing angle.

COMMUTATION OVERLAP ANGLE

The commutation of commutation overlap depends on:

a. the firing angle, b. the dc link current and

c. the source inductance or the inductance in series with each phase.

To find out the commutation overlap, it is sufficient to analyse one commutation. Let SCR S5 be in

conduction and let SCR S1 be triggered at a firing angle of α. As seen earlier, the loop current

iLOOP builds up from zero to IDC. This means that the current in the inductance in R-phase builds

up to IDC, whereas it decreases to zero in the inductance in the B-phase. Let the commutation last

for an angle µ. Then during commutation, the voltage across the source inductance is expressed as shown in equation (4). In equation (4), the loop current is denoted as ‘i’. Since θ = wt, the equation for commutation overlap can be represented as shown in equation (5). During this interval, the loop current changes from zero to IDC. Hence equation (6) defines how current in R phase changes.

The solution of equation (6) is presented in equation (7). Equation (7) can be re-arranged and presented as shown in equation (8). Solving for µ, we obtain equation (9).

It is seen that overlap angle µ increases

a. as the firing angle moves closer to either 0o or 180o , b. as the dc link current becomes larger, and

c. as the source inductance gets larger.

The above equation has been obtained based on the assumption that the dc link current remains steady, which happens only when the dc link inductance is relatively large. In practice, it is not true and hence the above equation yields only an approximate result.

The rest of mathematical analysis follows the familiar route. The items of interest are:

a. RMS output voltage of bridge circuit, b. Average output voltage of bridge circuit, c. Ripple Factor of output voltage of the bridge, d. RMS output voltage/voltage across load resistor, e. Average output voltage (across load resistor), f. Ripple factor of output voltage (across load resistor), g. RMS line current, h. RMS value of fundamental component in line current, i. THD in line current, j. Displacement power factor, k. Apparent power factor and

l. Harmonic analysis.

In order to simulate it is necessary to have an expression for line current and load current. Let us consider one output cycle, starting from the instant SCR S1 is triggered till SCR S2 is triggered.

During this period, the R-phase voltage is defined as shown in equation (10). When SCR S1 is

triggered, we have commutation overlap till the load current is transferred from SCR S5 to SCR S1

and let the commutation overlap angle be µ. During commutation overlap, the current through SCR S1 rises from zero to load current. At the end of commutation overlap, line current is obtained from

the expression in equation (11). During the period of commuation overlap, the voltage source that is seen by the dc link circuit is described as shown in equation (12), where vOL represents the

source during overlap period.

On simplifying equation (12), we get equation (13). During this period, the source appears to have a source inductance equalling 1.5L2 . We get this value because the path through SCR S6 contains

L2 whereas the path through both S1 and S5 appears to have an equivalent inductance of value

equal to 0.5L2. The current through the load at the instant when SCR S1 is triggered can be

expressed as shown in equation (14).. The values of Z and φ in equation (14) are expressed in equation (15).

The constant A in equation (14) is to be evaluated and τ2 = tan (φ). At the end of commutation, this

current would be equal to the line current. Hence we obtain equation (16). The terms used in equation (17) are defined in equation (18).

Equation (16) is the first of the three equations we need in order to obtain a solution and this equation equates the line current to the load current at the end of commutation. Equation (17) expresses the load current at the end of commutation, using another constant B and B is also to be evaluated. Another equation can be formed as shown in equation (19), which equates the line current at the end of commuation to the load current. Unlike equation (16), equation (19) makes use of constant B. Equations (16) and (19) are two of the three equations required to solve for A, B and δ. The third equation is obtained as follows. The load current at the instant when θ = π /3 can be computed in two ways, one from equation (16) and the other from equation (19) and these two expressions can be equated to yield equation (20).

From these three equations, the three unknowns, A, B and µ, can be obtained. These equations have been used in the program written for simulation.

SIMULATION

Three applets are presented in this section. The first applet animates the circuit. The only purpose is to illustrate the sequence of operation of this circuit. It does not take in any parameter. To run it, click on the RUN button. The SINGLE STEP button allows the user to step through, the PAUSE button allows the user to stop the program and the RESET button allows the user to view the simulation once more. To view the simulation once more, click on the RESET button and then click on the RUN button. During the simulation, the user can ask the program to pause for a while,

then step through for a while and allow it to run through to the end of its cycle.

The second applet takes in these parameters:

a. the firing angle in degrees,

b. the ratio of load reactance to load resistance (load reactance evaluated at line frequency),

c. the ratio of source reactance to load resistance(preferably below 0.1 p.u), and

d. the value of dc link source.

The program allows you to view either the waveforms or the statistics. The values of load and line reactance are to be entered in per unit. For example, if the line reactance is entered as 0.05 p.u., rated load current would cause a drop of 4% of phase voltage across the line reactance.

An example is presented now to explain how the per unit values can be set. Let a 3 phase, 415 V, 50 Hz source supply power to the converter. Then the maximum average voltage that can be obtained is obtained as shown below. Let the nominal rated dc link current be 100 A. Then the nominal load resistance or the base impedance for the system is computed as shown below. It is also shown how the line reactance can be obtained, if its value in p.u. is known. Given that the current through the load is free of ripple, the rated RMS line current is obtained as illustrated below. From the total rms line current, inclusive of both the fundamental component and the harmonic components, the fundamental rms component is obtained as illustrated below.

Given that the line inductance is 1 mH, its p.u. value is obtained as shown below. Given that the dc link inductance is 10 mH, its p.u. value is computed as illustrated below. Usually the line inductance is called as the 4% reactor, implying that when the line current is at its rated value, the rms value of the fundamental component of voltage across the line reactor is 4% of the phase voltage. For example, if the rms phase voltage is 240 V, the drop across 4% reactor at rated current would be 9.6 V. When the line voltage is 415 V, the phase voltage is calculated as

shown below.The drop across the line inductor can now be stated as a fraction of the phase voltage as given below.

This means that if the drop across the line inductor is to be 4% of phase voltage, the inductance should be 0.4 mH and not 1 mH.

The third applet takes in these parameters:

a. the ratio of load reactance to load resistance (load reactance evaluated at line frequency), b. the ratio of source reactance to load resistance(preferably below 0.1 p.u), and

c. the value of dc link source.

Within the program the firing angle is varied and the plots of rms bridge output voltage, the average bridge output voltage, its RF, the rms line current, the fundamental rms component of line current, the THD in line current, the average output current, its RF and the overlap angle as a function of firing angle are displayed. The value of dc source in the dc link is assigned to be zero in this applet.

SIMULATION USING C

A C-program called, ss_resp.cpp, can be downloaded by clicking on the image below. It can be compiled as a C program and executed.

The program was run with the following values: Time constant of the Link inductor in radians: 1.00

Time constant of the Source Inductance in radians : 0.03

Firing Angle = 30o.


Another C-program called, harm3ph.cpp, can be downloaded by clicking on the image below. It can be compiled as a C program and executed.

The results obtained for the same parameters are presented below.

harmNo LoadCur OutVolt harmNo LineCur0 1.6874 1.6872 1 0.931052 2.69E-08 1.16E-08 3 2.31E-084 2.75E-08 2.94E-08 5 0.199756 0.029517 0.1743 7 0.11428 2.85E-08 2.82E-08 9 1.85E-08

10 2.85E-08 3.00E-08 11 0.08092112 0.00639 0.074642 13 0.06150914 2.87E-08 2.90E-08 15 1.82E-0816 2.86E-08 2.97E-08 17 0.04848418 0.0023099 0.040382 19 0.04023320 2.87E-08 2.91E-08 21 1.84E-0822 2.87E-08 2.94E-08 23 0.03274124 0.0009234 0.021504 25 0.02832

LoadIAvg 0.84371 LoadIRMS 0.84398 RFCur 0.021433 VoAvg 0.84362 VoRMS 0.85546 RFVolt 0.14184

ILineAvg 0.56248 ILineRMS 0.68481 THD 0.684811 OverLapAngle 0.085085

SUMMARY

This page has explained how source inductance leads to commutation overlap and how it affects the output voltage. Next page how a DC power supply can be built using the 3-phase fully-controlled bridge rectifier.

AN APPLICATION: A DC POWER SUPPLY

CIRCUIT OPERATIONMATHEMATICAL ANALYSISSIMULATION

CIRCUIT OPERATIONA three-phase fully-controlled bridge circuit is a much more suitable circuit to be used for generating a variable dc output voltage than the single-phase fully-controlled bridge circuit, on account of two reasons, which are:a. reduced ripple content in its output andb. much higher ripple frequency.Both these factors lead to an LC filter which is relatively small and economical. This page describes how such a power supply can be built and controlled.

An inductor in the dc link reduces ripple in the output current of the bridge circuit, whereas the capacitor absorbs the ripple in output voltage. The inductor has to be designed such that it does not saturate even when it carries the maximum current. This means that it should have an airgap in the path of flux. The ripple current through the capacitor can also be significant. Hence it needs to be checked from the datasheet that the capacitor chosen has the required ripple current rating. For such an

application, an electrolytic capacitor is normally chosen and its voltage rating should also be adequate.

We can have a block diagram to describe the operation of this dc power supply obtained using a three-phase fully-controlled bridge rectifier. The output voltage Vo is varied by varying the firing angle α.

The firing angle in turn is controlled by voltage VC, which is the output of a PI controller. The inputs

to the PI controller are a voltage named Vref representing the desired output voltage and the output

voltage Vo of the bridge circuit. If Vo is less than the desired output voltage, the resultant error causes

the output, VC, to increase, which in turns should advance firing angle. As the firing angle is

advanced, the output voltage of the bridge circuit increases. The next section describes how the block diagram can be analysed, leading to simulation of the system.


The simulation program is based on the pseudo-code displayed below.

Start block:

Set the values of load reactance, line reactance, capacitor, load fraction. Set the desired output voltage, PI controller parameters

Set the current firing angle to be 120o. Set base reference angle to 60o. Set Commute = 0. ( Indicates no commutation overlap exists at start. )

Go to Loop Routine. Set theta to zero.

Loop Routine

Call Compute routine. Increment theta. If (theta + base reference angle) = (next firing angle)

[

current firing angle = next firing angle.

base reference angle = next firing angle - 60o. Set Commute = 1 . ( Indicates next SCR is triggered)

]

Execute Loop Routine

Compute Routine:

If (Commute == 0)

[

compute next value of link inductor current. ]

else

[

compute next value of triggered SCR current. compute next value of link inductor current. if (SCR current link inductor current) Reset Commute to 0. ]

Compute next value of capacitor voltage. Compute next value of PI controller's output. Compute next firing angle.

The equations used in the Compute Routine are obtained as follows.

When there is commutation overlap, the output voltage behind the source inductance is expressed by equation (1). In equation (1), µ is the overlap angle and U is the amplitude of line voltage. If the output voltage be vo(θ) during this period, the differential equation for the current through the dc link

inductance is presented as equation (2). During commutation overlap, the current in the SCR just triggered on is described by equation (3). The commutation overlap ends when the current through this SCR equals the dc link inductor current. When there is no overlap, the bridge output voltage is described by equation (4).

The differential equation that describes the dc link inductor current is then described by equation (5). The differential equation for the capacitor voltage is easily obtained and it is expressed as equation (6). Next the equations relating to closed-loop control are described. Let the output of PI controller be vC(θ) and it is expressed by equation (7). In equation (7), A is a constant to be evaluated, K is the

proportional gain of the controller and T is its time constant. The above equation is represented as equation (8) , which is more convenient for use in simulation. In equation (8), Vref is the desired output

voltage. The output of the controller is normally checked to ensure that it is within the set limits. From the output of the controller, the firing angle, α can be obtained. The maximum output voltage of the controller should correspond to zero degree firing angle and the minimum to 120o firing angle. Hence we get the following equation for firing angle. This means that the range for vC(θ) is from 0 to

VCmax.

The simulation program uses the above equations and displays the results in a graphical format.

SIMULATION

The applet below can be run with the default parameters. To set any parameter, click on the arrow pointing downwards beside the Peak Source Voltage, a menu would appear. The default value of the parameter highlighted appears in the textfield for Set Value. To change the parameter, select the parameter and then click within the editable textfield for Set Value. In order to change this parameter, you must click on Set Value button. You can set the desired response to be one of three responses.

An example is presented now to explain how the per unit values can be set. Let a 3 phase, 415 V, 50 Hz source supply power to the converter. Then the maximum average voltage that can be obtained is presented in equation (10). Let the nominal rated dc link current be 100 A. Then the nominal load resistance or the base impedance for the system is assigned as shown by equation (11). Given that the current through the load is free of ripple, the rms line current is obtained according to equation (12) and this value includes both the fundamental component and the harmonic components. The fundamental rms component is obtained as illustrated by equation (13). Given that the dc link inductance is 10 mH, its p.u. value is obtained as shown by equation (14).

Given that the line inductance is 1 mH, its p.u. value is obtained as shown by equation (16). Usually the line inductance is called as the 4% reactor, implying that when the line current is at its rated value, the rms value of the fundamental component of voltage across the line reactor is 4% of the phase voltage. For example, if the rms phase voltage is 240 V, the drop across 4% reactor at rated current would be 9.6 V. When the line voltage is 415 V, the phase voltage is obtained as shown by equation (16). The drop across the line inductor can now be stated as a fraction of the phase voltage as given by equation (17).This means that if the drop across the line inductor is to be 4% of phase voltage, the inductance should be 0.4 mH and not 1 mH.

It is possible to set the load resistance to a value other than its nominal value. The nominal value of load resistance is 5.6 Ω. If the load resistance is to be 10 Ω, then set Load Fraction as shown by equation (18).

Some explanation is offered here to indicate how the values for different parameters can be set in the applet shown below. The rated dc output voltage corresponds to 1 per unit. If the rated dc output voltage is 400 V, the peak voltage of 415 V, 3 phase, 50 Hz supply is 1.467 p.u.. Let the nominal load

current through the load resistor be 100 A. Then Rnom is 4 Ω. If the dc link inductance is 10 mH, its

time constant in radians is (2πfLLink/Rnom) . Given that frequency is 50 Hz, it works out to be 0.7854

radians or p.u., since it expresses the ratio of the reactance over the base impedance. The per cunit value to be set for the line reactance is also obtained similarly. Given the value of the filter capacitance, its per unit value is obtained as (1/2πfCRnom). The applet allows the dc output voltage to be set to a

value other than unity and the load fraction can also be varied.

INTRODUCTION

CIRCUIT OPERATION

CURRENT LOOP

THE PER UNIT NOTATION

SPEED LOOP

PARAMETERS IN PER UNIT NOTATION

CURRENT CONTROLLER DESIGN

SPEED CONTROLLER DESIGN

FIELD CONTROLLER DESIGN

LOGIC FOR FOUR-QUADRANT OPERATION

SWITCH-OVER OF CONTROL FROM ARMATURE TO FIELD

SIMULATION OF THE FOUR-QUADRANT DRIVE

SUMMARY

INTRODUCTION

This page describes how a separately-excited dc shunt motor can be operated in either direction in either of the two modes, the two modes being the motoring mode and the regenerating mode. It can be seen that the motor can operate in any of the four quadrants and the armature of the dc motor in a fast four-quadrant drive is usually supplied power through a dual converter. The dual converter can be operated with either circulating current or without circulating current. If both the converters conduct at the same time, there would be circulating current and the level of circulating current is restricted by an inductor. It is possible to operate only one converter at any instant, but switching from one converter to the other would be carried out after a small delay. This page describes the operation of a dual converter operating without circulating current.

As shown in Fig. 1, the motor is operated such that it can deliver maximum torque below its base speed and maximum power above its base speed. To control the speed below its base speed, the voltage applied to the armature of motor is varied with the field voltage held at its nominal value. To control the speed above its base speed, the armature is supplied with its rated voltage and the field is weakened. It

means that an additional single-phase controlled rectifier circuit is needed for field control. Closed-loop control in the field-weakening mode tends to be difficult because of the relatively large time constant of the field.

The power circuit of the dual-converter dc drive is shown in Fig. 2.

Each converter has six SCRs. The converter that conducts for forward motoring is called the positive converter and the other converter is called the negative converter. Instead of naming the converters as

positive converter and negative converter, the names could have been forward and reverse converters. The field is also connected to a controlled-bridge in order to bring about field weakening.

The circuit shown above can be re-drawn as shown in Fig. 3. Usually an inductor is inserted in each line as shown in Fig. 3 and this inductor reduces the impact of notches on line voltages that occur during commutation overlap.

CIRCUIT OPERATION

The operation of the circuit in the circulating-current free mode is not very much different from that described in the previous pages. In order to drive the motor in the forward direction, the positive converter is controlled. To control the motor in the reverse direction, the negative converter is controlled. When the speed of motor is to be changed fast from a high value to a low value in the forward direction, the conduction has to switch from the positive converter to the negative converter. Then the direction of current flow changes in the motor and it regenerates, feeding power back to the source. When the speed is to be reduced in the reverse direction, the conduction has to switch from the negative converter to the positive converter. It is seen that conduction has to switch from one converter to the other when the direction of motor rotation is to change, so that regeneration can occur. During regeneration, the direction torque developed by motor is opposite to that of the motoring torque. Thus the regenerating torque acts as the breaking torque and the motor decelerates fast.

At the instant when the switch from one converter to the other is to occur, it would be preferable to ensure that the average output voltage of either converter is the same. Let the firing angle of the positive converter be αP, and the firing angle of the negative converter be αN . If the peak line voltage be U, then

equation (1) should apply. Equation (1) leads to equation (2). Then the sum of firing angles of the two

converters is π, as shown in equation (3).

In a dual-converter, the firing angles for the converter are changed according to equation (3). But it needs to be emphasized that only one converter operates at any instant.

When the speed of the motor is to be increased above its base speed, the voltage applied to the armature is kept at its nominal value and the phase-angle of the single phase bridge is varied such that the field current is set to a value below its nominal value. If the nominal speed of the motor is 1500 rpm, then the maximum speed at which it can run cannot exceed a certain value, say 2000 rpm. Above this speed, the rotational stresses can affect the commutator and the motor can get damaged.

Next it is shown how the operation of motor can be represented by means of a block diagram. This approach can be helpful in designing the closed-loop system.

CURRENT LOOP

Let the field excitation be assumed to remain constant at its nominal level. Let the voltage applied to armature be va volts, the back e.m.f. eb volts and rotor speed wr rad/s. The back emf is expressed by

equation (4), where Km is the coefficient relating speed of motor to its back emf. If Ra be the resistance

of armature and La its inductance, then the applied armature voltage equals the sum of the motor back

e.m.f, the drop across its armature resistance and the drop across the armature inductance, as shown in equation (5). In equation (5), va is the voltage applied to the armature and ia is the current though the

armature. The above equation can be represented in terms of Laplace transform, leading to equation (6).

The block diagram shown in Fig. 4 represents equation (6).

Given that the excitation of the motor is constant and that the effects of armature reaction are negligible either due to interpoles or series compensation winding, the torque output Me can be expressed as shown

in equation (7). If the load torque be ML N-m, the combined polar moment of inertia of motor and load

be J kg.m2 and its friction coefficient be B N-m/rad/sec, then the torque output of motor equals the expression on the right-hand side of equation (8). Equation (8) can be represented in terms of Laplace transform, as shown in equation (9), where the Laplace transform of w, the motor speed, is assigned to be Ω(s). A block diagram, as shown in Fig. 5, can now be drawn based on equations (4), (6), (7) and (9). It can be seen that unit for Km is N-m/A.

With the load torque set to zero, a transfer function linking current Ia(s) and the input voltage Va(s) can

be obtained. It is expressed in equation (10).

The reason for obtaining this transfer function is to facilitate the design of a controller for controlling the armature current. The two output parameters of interest are the torque and the speed. The armature current is selected as one of the state-variables to be controlled in closed-loop, since the torque output varies linearly with it. It is preferable that the variable to be controlled by negative feedback is a variable that reflects some energy stored in a system. Here the armature current reflects the energy stored in the inductance in the armature circuit. If the motor has a compensating winding and/or a compound winding, the inductance of this winding should be added to La. In some drives, an additional inductor is

used in series with the armature and this value should also be added to La. Let G1(s) reflect the transfer

function in equation (10) and equation (11) shows the expression for G1(s). The part of the closed-loop

system that is usually used for controlling the armature current is shown in Fig. 6.

The block diagram in Fig. 6 is now described. If the armature current is to be controlled in closed-loop, it is necessary to have a current reference signal, marked as IR(s) in Fig. 6. This signal is internally

generated, most often as the output of the controller for speed and it is shown later how that is achieved.

It is possible to use a controller other than a proportional plus gain(PI) controller. A PDF controller(a pseudo-derivative controller) or a PID-controller can be used. But a PI controller is often sufficient, since the integrating part of the PI controller leads to zero steady-state error for a step input and the

proportional gain can be adjusted to yield fast response and stability. The output of the current controller is often a voltage which sets the firing angle for the fully-controlled bridge circuit. Since the gain KB(α)

is negative, the sign of both the proportional gain KI and the integrating time-constant TI should be

negative in order to keep the loop-gain of the system represented by block diagram in Fig. 6 as negative.

A variation in the output of the current controller does not change the firing angle instantaneously since the SCRs in the bridge are triggered in a sequence at an interval of 60o on the average and there is a delay before the change in the output of the current controller has an effect on the firing angle. This delay can be classified as a transportation lag and it can be approximated by a first-order transfer function, as shown in equation (12). In equation (12), y has been used in place of sTD.

For a system with 50 Hz input source, one-sixth of a cycle is about 3.3 ms and then the delay TD can be

set to be half of that value, that is 1.67 ms. What is carried out is an approximation to facilitate the design of current controller.

As the firing angle α increases, the instant of triggering of each SCR is delayed more and more from its reference point corresponding to 0o firing angle. When the current flow in the dc link is continuous, the average output voltage of the bridge changes from its positive maximum average value to its negative maximum average value, as α is allowed to very from 0o to 180o. In order to ensure that the loop gain is negative, it is necessary that the gain due to controlled rectifier circuit is inverted. It is explained later how it can be brought about for practical realization.

Another point to be noted is that the gain of the controlled rectifier is not constant and it varies with firing angle. Let the maximum average output voltage be Vom. Then equation (13) shows how the

average output voltage at any firing angle, α, is obtained. The actual gain of the controlled rectifier is defined by equation (14). If the rated armature voltage, VRA, is assigned to be the base voltage, then the

gain of the controlled rectifier in per unit notation can be defined as in equation (15).

In equation (15), KA defines the ratio of maximum average output voltage of the bridge to the rated

armature voltage. It is seen that the gain varies and hence the controller has to be designed such that it operates properly over this range of variation.

THE PER UNIT NOTATION

It is better to design the current loop first before the outer loop design is attempted. But it is necessary to describe the per unit notation that is adopted here. Let the rated armature voltage VRA be the base

voltage. Then equation (16) is valid. As shown in equation (17), the rated armature current IRA is chosen

to be the base current.

Per Unit Value of Rated Armature Voltage = 1 (16)

Per Unit Value of Rated Armature Current = 1 (17)

Then the base impedance for the armature circuit is obtained as shown in equations(18) whereas equation (19) shows how the per unit value of armature resistance can be obtained if it is ra Ω. Given

that the inductance present in the armature circuit is La H, the voltage across it is obtained as shown in

equation (20). Equation (21) is obtained by dividing both sides of equation (20) by VRA. Equation (21)

uses symbol τa, representing the time constant of the armature circuit and it is defined by equation (22).

For a 3-phase controlled-bridge rectifier circuit, the maximum average output voltage that can be obtained at 0o firing angle is shown in equation (23). Then the amplitude of line voltage of 3-ph supply is described by equation (24). The per unit value of the peak line voltage is obtained from equation (25).

We have seen so far how voltages, currents and impedances related to armature circuit can be expressed in per unit values. Next, it is shown how the torque developed, moment of inertia J and friction coefficient B can be expressed in per unit values. Let the torque developed by motor be Me N-m. Then

when the motor is operating with the nominal or the rated flux, the torque developed by motor is defined by equation (26), where ia is the armature current. Also, let wr be the armature shaft speed in rad/s. Then

the per unit value of the torque developed is expressed as shown in equation (27), where IRA is the rated

armature current.

The per unit value of moment of inertia is obtained as follows. Let ΩR be the rated shaft speed in rad/s,

and the moment of inertia of motor and the coupled load be J kg-m2. Let the torque required to accelerate this moment of inertia be MJ. Then equation (28) can be used to relate J and MJ. Dividing both sides of equation (28) by the rated torque, we get equation (29). From equation (29), it is seen how the per unit value of the moment of inertia can be obtained. Similarly, we can get an expression for friction coefficient, as shown by equation (30).

It is necessary to state how the parameters for the current controller should be specified. The gain, KI , is

just a ratio whereas the time constant, TI , should be specified in seconds.

SPEED LOOP

Before the design of speed loop is to be attempted, the current loop should be approximated by a suitable transfer function. The block diagram in Fig. 6 can be expressed by a transfer function, say G2(s). Then

G2(s) = Ia(s)/IR(s). Using G2(s), the speed loop can be represented as shown in Fig. 7. Again a PI

controller is used and it may be necessary to provide a filter in the path of speed feedback signal. The time constants are to be specified in seconds. In per unit notation, the value of KM marked in Fig. 7

would be 1.

PARAMETERS IN PER UNIT NOTATION

At first, typical per unit values are obtained from the datasheet of a dc motor. Then it is shown how the current loop can be designed. Next, the design is verified by simulation. Finally, the speed loop design is illustrated.

In the per-unit calculations, a per unit value of 1 is assigned to the rated armature voltage, the rated motor speed and the rated armature current. From the values of the rated armature voltage and the rated motor speed, obtain Km, the coefficient for the motor. Then other per unit values can be obtained as

outlined earlier. The applet displayed below computes the per unit values given the actual values. The textfields contain default values and the applet computes and displays the per unit values when the Compute Button is clicked.

CURRENT CONTROLLER DESIGN

The transfer function G1(s) expressed as equation (11) can expressed in terms of per unit values and then

Va(s) and Ia(s) marked in Fig. 6 would be values in per unit notation. Conversion of equation (11) such

that it conforms to the per unit notation is explained below.

Both the numerator and the denominator of expression in equation (11) can be divided by BRa. The ratio

of J/B can be represented as the mechanical time constant, τm. The resulting expression for G1(s)

presented as equation (31). Then equations (32) and (33) explain how equation (31) can be converted such that it is in per unit notation.

The block diagram shown in Fig. 6 conforms to per unit notation. Here G1(s) is expressed by equation

(34) and Va(s) and Ia(s) marked in Fig. 6 are values in per unit notation. Using (34), the transfer

function, G2(s) representing the block diagram in Fig. 6 can be represented as shown below. In this case,

the controlled-rectifier is assigned to have its highest gain, KA. It is logical to do so, because the system

designed for stability at gain KA would be stable at lower gains too.

The second applet in this page finds the location of the poles and zeros of the closed-loop system in Fig. 6, given the necessary data. It is seen for a wide range of controller parameters, the zeros are located such that they cancel almost two poles. The other two poles are located away from the origin. It is seen that the design of current controller is fairly easy.

SPEED CONTROLLER DESIGN

It is necessary to design the speed controller next. To design the speed controller, it is necessary to represent the transfer function G2(s) suitably. It is found that for a wide range of values, two of the zeros

of G2(s) are located near two of the poles and the other two poles are away from the origin. Hence while

designing the speed loop, G2(s) is set equal to unity. The simulation of the drive presented later would

show whether this simplification is justified.

The design of the speed controller is carried out based on the assumption that the motor is on no load. A variable drive system tends to exhibit oscillatory behaviour under no load conditions and hence the design based on no load condition is assumed to be justified. Here the output of the speed controller is not clamped, whereas there would be limits on the output of speed controller. The output of speed controller corresponds to armature current and it is necessary to limit the peak value of speed controller in order to protect the SCRs used in the bridge. The applet presented later for simulating the drive has limits imposed on the output of the speed controller.

FIELD CONTROLLER DESIGN

The block diagram for closed-loop operation with the field controller in action turns out to be somewhat complex. The interaction that occurs within a separately-excited DC motor is first presented in Fig. 8.

The block diagram in Fig.8 is now described. The field current, marked as IF, produces magnetic flux in

the motor and the back e.m.f of the motor is then proportional to the product of the field current and the speed of the motor. This statement is based on the assumption that the field flux in the motor is not saturated and that the field flux varies linearly with the field current. If the field current is in per unit notation, where the field current corresponding to the rated current equals unity, then the back emf can be shown to be equal to Km × iF × wR, where both Km and wR are also in per unit representing the motor

coefficient and the speed of the motor. Once the back e.m.f and the applied voltage are known, the armature current can be obtained as shown in Fig. 8. From the values of armature current and field current, the torque output of motor is obtained and the speed of the motor changes as shown.

For design of field controller, the block diagram in Fig. 8 is too complex. The design is carried out using a simplified or a simplistic block diagram and the performance of the controller is evaluated using the final simulation program, which uses a model that is reasonably close to real system.

The design of field controller is based on the block diagram shown in Fig. 9.

It is easy to represent the block diagram in Fig. 9 in per unit notation. The gain of the controlled bridge for the field circuit is KF. Its value equals the ratio of the maximum rate of bridge output voltage to the

rated voltage of the field circuit and normally the value of KF is likely to be near 1.2. The delay due to

firing circuit is again approximated by TD2, and it is set equal to (1/4f), where f is the frequency of the ac

source. Then the field current is obtained in per unit value and it can be made equal to the torque, assuming that the armature circuit has comparatively a small time constant and that the armature current stays at the rated value. The friction coefficient, the mechanical time constant and the time constant of the filter in the speed feedback signal are the same signals used for design of the speed controller. The applet below can be used to design the field controller. This applet runs somewhat slowly. The poles and zeros are calculated for the block diagram shown in Fig.9, whereas the step response is obtained using the block diagram in Fig. 10.

The design of field controller is somewhat difficult because both the field circuit time constant and the mechanical time constant are relatively large.

The applet displayed below shows step response of the drive with the field controller.

LOGIC FOR FOUR-QUADRANT OPERATION

For four-quadrant operation, the scheme outlined in this page makes use of two converters, called the positive converter and the negative converter. It has also been shown that the sum of the firing angles of the two converters should be π radians. Hence the synchronizing signals for the SCRs in both converters can be obtained as shown in Fig. 11.

It is also necessary to find out when the switch from the positive converter to the negative converter or vice-versa can be made. One possible method is outlined in Fig. 12. Based on the polarity of current reference signal, a logic signal, called W can be developed. A comparator can be designed to yield an output of 1 (W =1 ) when the current reference signal is positive and an output of 0 (W =0 ) when the current reference signal is negative. Along with W, another signal can be derived based on the armature current. A signal, called Enable, can be produced such that Enable is 1 when the armature current is zero. When Enable is 1, the output of a flip-flop can be set. Output Q takes on the polarity of W signal. When both W and Q are at logic 1, the positive converter is allowed to be triggered. When both W and Q are at logic 0, the negative converter is allowed to be triggered.

The rest of the control arrangement for generating the firing signals is shown in Fig. 13. The block diagram shows that there are active limits on the firing angle. The voltage output of the bridge can be sensed and when it is at about 1.05 times the rated armature voltage, the firing angle may not be allowed to become any smaller.

The variation of firing angle towards either 0o or 180o can be blocked, avoiding further rise in the output voltage. It is better to have a provision that would allow for varying the limiting values to accommodate changes in source voltages. It would not be difficult to implement such a scheme for a micro-controller based control system.

SWITCH-OVER OF CONTROL FROM ARMATURE TO FIELD

Field control is necessary above the base speed of the motor and the field current has to vary only over a limited range, say from 0.6 pu to 1 pu. The signal that sets the active limits on the firing angle for armature control can be used also for field control. The block diagram related to field control is shown in Fig. 14.

When the active limit is not set, the firing angle of the field controller is set such that the field current remains at the nominal value. When the active limits are applied to the armature control circuit, the firing angle is varied such that the field current gets adjusted to the value required for the speed reference set.

SIMULATION OF THE FOUR-QUADRANT DRIVE

Before selecting the type of response, set the value of the selected parameter. When you select a parameter, the textfield shows the default value set inside the program. Change the parameter value if you want to and then you must click on the SET VALUE button for the change to take effect. You can go from one type of response to another after the present calculations are carried out. When you have selected a new type of response, you must click on Click to Start. If you click on Reset button, initializing routine is carried out and the motor speed is set to zero, and the other values are also reset. It is possible to see the effect of step changes in speed or load or unbalance in source or unbalance in firing circuit.

SEMI-CONTROLLED RECTIFIERS

This chapter describes two semi-controlled rectifier circuits:

HALF-CONTROLLED SINGLE-PHASE BRIDGE RECTIFIER

HALF-CONTROLLED THREE-PHASE BRIDGE RECTIFIER

CIRCUIT OPERATION


SIMULATION

PSPICE SIMULATION

MATHCAD SIMULATION

SUMMARY

CIRCUIT OPERATION

A fully-controlled rectifier circuit contains only controlled-rectifiers, whereas a semi-controlled rectifier circuit is made up of both controlled and uncontrolled rectifiers. Due to presence of diodes, free-wheeling operation takes place without allowing the bridge output voltage to become negative. In a semi-controlled rectifier, control is effected only for positve output voltage, and no control is possible when its output voltage tends to become negative since it is clamped at zero volt. This page describes the operation of a single-phase half-controlled rectifier.

A semi-controlled full-wave bridge rectifier can be configured in a few ways. They are shown below.

The circuit in Configuration 1 contains two SCRs and two diodes. When source Vin is positive, SCR S1 can be triggered

at a firing angle called α and then current flows out of the source through SCR S1 first, then through the load and returns

via diode D3. If

then SCR S1 and diode D3 conduct during α < wt < π. When π < wt < 2π, Vin is negative and SCR S2 is normally

triggered when wt = π + α. During π < wt < (π + α) , the output of the bridge circuit would have been negative if we had used a fully-controlled bridge rectifer and if the current flow was continuous. But here we have two diodes D3 and D4

instead of two SCRs. When the output of the bridge tends to becomes negative just after wt exceeds π, diode D4 tends to

get forward-biased and it starts conducting. Then diode D3 is reverse-biased and it stops conducting. During π < wt < (π

+ α) , the devices in conduction are SCR S1 and diode D4 and the output of the bridge is clamped at zero, assuming that

the on-state drops across devices in conduction is zero. During ( π + α) < wt < 2π , the devices in conduction are SCR S2

and diode D4. SCR S2 and diode D3 would conduct during 0 < wt < α .

The circuit in configuration 1 has SCRs as the devices in the top-half and diodes as the devices in the bottom-half. Instead, it it is possible to use SCRs as the devices in the bottom-half and diodes as the devices in the top-half.

It is also possible to build a semi-controlled full-wave bridge rectifier as shown by the circuit in configuration 2.

The behaviour of the circuit is the same as described earlier. In this circuit, SCR S1 and diode D3 conduct during α < wt

< π . During π < wt < (π + α) , the devices in conduction are diodes D3 and D4 and the output of the bridge is clamped at

zero. During (π + α) < wt < 2π , the devices in conduction are SCR S2 and diode D4. Diodes D3 and D4 would conduct

during 0 < wt < α .

Yet another configuration is available for semi-controlled bridge rectifier, as shown by the circuit in configuration 3.

In this circuit, SCRs S1 and S3 conduct during α < wt < π. During π < wt < (π + α) , the device in conduction is diode D

and the output of the bridge is clamped at zero. During (π + α) < wt < 2π , the devices in conduction are SCRs S2 and S4.

Diode D would conduct during 0 < wt < α .


The aim of analysis is to obtain the following values:

1. The average output voltage of the bridge as a function of firing angle.

2. The rms output voltage of the bridge as a function of firing angle.

3. The ripple factor of output voltage of the bridge as a function of firing angle.

4. The rms line current as a function of firing angle and the ratio wL/R.

5. The fundamental rms line current as a function of firing angle and the ratio wL/R.

6. The THD in line current as a function of firing angle and the ratio wL/R.

The Average Output Voltage

The ripple factor is defined then as

Next it is shown how the line current is to be analysed. An expression for load current over half-a-cycle can be obtained first. The load current during α < wt < π can be defined as follows.

where

From the expression for load current,

The load current during π < wt < (π + α) can be defined as follows.

When the load current is repetitive, we have that

That is,

and

Hence we obtain that

Once A is known, the total rms value of line current and the rms value of its fundamental component can be estimated.

Let

and

Then the rms line current given t and a is obtained as follows.

To obtain the rms value of the fundamental component of the line current, we obtain the trigonometric Fourier series coefficients of the fundamental component. The line current has half-wave symmetry and hence these coefficients are obtained as follows.

Then

We obtain the rms value of fundamental component as:

Total harmonic distortion in line current is then

SIMULATION

The applet shown below simulates this circuit. The parameters to be keyed-in are the ratio of load reactance to load resistance and the firing angle in degrees.

PSPICE SIMULATION

The semi-controlled bridge rectifier that has been simulated has four SCRs and a single free-wheeling diode. The program is presented below.

* Full-wave Bridge Rectifier with a resistive loadVIN 1 0 SIN(0 340V 50Hz)XT1 1 2 5 2 SCRXT2 0 2 6 2 SCRXT3 4 0 7 0 SCRXT4 4 1 8 1 SCRVP1 5 2 PULSE(0 10 1667U 1N 1N 100U 20M)VP2 6 2 PULSE(0 10 11667U 1N 1N 100U 20M)VP3 7 0 PULSE(0 10 1667U 1N 1N 100U 20M)VP4 8 1 PULSE(0 10 11667U 1N 1N 100U 20M)D1 4 2 DNAMEL1 2 3 31.8MR1 3 4 10R2 1 0 1MEGR3 2 0 1MEGR4 4 0 1MEG.MODEL DNAME D(IS=10N N=1 BV=1200 IBV=10E-3 VJ=0.6)


The waveforms obtained are displayed.

The waveform of bridge output voltage

The waveform of load current

The waveform of mains current

The waveform of current through the free-wheeling diode

INTRODUCTION

CIRCUIT OPERATION


SIMULATION WITH AN APPLET

MATHCAD SIMULATION

SUMMARY

INTRODUCTION

This page describes the operation of the three-phase semi-controlled rectifier. When a semi-controlled rectifier is used, the output of the bridge is controlled when it remains positive. If it tends to become negative, the output is clamped to zero volt through the free-wheeling action of the diodes.

Fig. 1

The circuit of the semi-controlled bridge rectifier is shown in Fig. 1, and this circuit contains three-

SCRs and three diodes. It is possible to configure the circuit in two more ways. For example, the top-half can contain the diodes and the bottom half the SCRs. Alternatively, six SCRs can be used as in the case of the fully-controlled bridge rectifier and an additional diode can be connected from the negative rail to the positive rail of the bridge, with the anode connected to the negative rail and the cathode to the positive rail. The operation of this circuit is different from that of the circuit displayed above.

CIRCUIT OPERATION

The operation of the semi-controlled rectifier circuit displayed in Fig. 1 is now explained. The waveform of the bridge output voltage for a firing angle of 30o is shown in Fig. 2. .It can be seen that the trace of the positive rail of output voltage is that of a controlled rectifier, since only the top-half of the bridge is controlled. The trace of the bottom rail is that of an uncontrolled bridge rectifier, since the bottom half contains only diodes.

Fig. 2

Let the 3-phase supply be defined as shown in equation (1).

Given that the firing angle is 30o, SCR S1 is triggered when wt = 60o . The conduction range of the

SCRs in the top half can be is now expressed in equation (2).

As long as the firing angle α remains less than 60o, the expression for output voltage over one output cycle can be expressed as follows.

Substituting for vR, vY, and vB from equation (1), we get that

When the firing angle α is higher than 60o, the expression for output voltage over one output cycle can be expressed as follows, if conduction through the load is continuous.

Substituting for vR, and vB from equation (1), we get that

Depending on the firing angle, the bridge output repeats itself every 120o. The applet shown below plots the output voltage, given the firing angle. It is assumed that conduction through the load impedance is continuous.

Fig. 3

If the firing angle α is less than 60o, when SCR S1 is triggered, SCR S1 and diode D6 conduct during

the period (α + 30o) ≤ wt < 90o. Figure 3 shows the path of conduction that would exist during this period. The pairs that conduct vary depending on the firing angle. Table 1 show the pairs that would conduct when the firing angle α is less than 60o, whereas Table 2 shows the pairs that would conduct when the firing angle α is greater than 60o.


The aims of anlysis are:

1. To obtain an expression for the average bridge output voltage as a function of firing angle,2. To obtain an expression for the rms bridge output voltage as a function of firing angle,3. To obtain an expression for the ripple factor of bridge output voltage as a function of firing

angle,4. To obtain an expression for the instantaneous load current as a function of firing angle,5. To obtain an expression for the instantaneous line current as a function of firing angle,6. To obtain an expression for the rms value of the fundamental of line current as a function of

firing angle,7. To obtain an expression for the rms line current as a function of firing angle,8. To obtain an expression for the THD of line current as a function of firing angle,9. To carry out harmonic analysis of line current, bridge output voltage and load current at a given

firing angle.

TABLE 1: PAIRS IN CONDUCTION WHEN α < 60o

S1 – D6 (α + 30o) ≤ wt < 90o

S1 – D2 90o ≤ wt < (α + 150o)

S3 – D2 (α + 150o) ≤ wt < 210o

S3 – D4 210o ≤ wt < (α + 270o)

S5 – D4 (α + 270o) ≤ wt < 330o

S5 – D6 330o ≤ wt < (α + 390o)

TABLE 2: PAIRS IN CONDUCTION WHEN α > 60o

S1 – D2 (α + 30o) ≤ wt < 210o

S1 – D4 210o ≤ wt < (α + 150o)

S3 – D4 (α + 150o) ≤ wt < 330o

S3 – D6 330o ≤ wt < (α + 390o)

S5 – D6 (α + 270o) ≤ wt < 450o

S5 – D2 90o ≤ wt < (α + 30o)

Average Output Voltage

When the conduction through the load is continuous, the average bridge output voltage is obtained as shown below.

Given that the firing angle is less than 60o,

If the firing angle is greater than 60o and the conduction is continuous,

The maximum output voltage occurs when α = 0o and let it be Vdm:

From equation (7) and (9), we obtain that

RMS Output Voltage

The expression for the rms output voltage is found separately for the two cases. The assumption here is that the conduction is continuous. When the firing angle is greater than 60o,

When the firing angle is less than 60o,

Ripple Factor of the Bridge Output Voltage

The ripple factor, RF(α), of the bridge output voltage can be computed as follows:

Since both VRMS(α) and VDC(α) are known, the ripple factor can be computed.

Instantaneous Load Current

An expression for the instantaneous load current as a function of firing angle can be obtained through a somewhat tedious process.

When the firing angle is less than 60o, the instantaneous bridge output voltage expressed by equation (4) is reproduced below.

The above expression can be written with the origin shifted to the instant of triggering an SCR. Then

Let the load angle be:

The expression for load current during 0o ≤ θ ≤ (60o – α) can be expressed to be:

In the expression, A1 is a constant to be evaluated. Then

The expression for load current during (60o – α) ≤ θ ≤ 120o can be expressed to be:

In equation (20), A2 is a constant to be evaluated. Then

Another expression for the load current can be obtained when the output cycle ends.

When the load current is periodic, then

From equations (18), (19), (21) , (22) and (23), we can determine A1 and A2. From equations (19) and

(21), we get that

On simplifying the above expression, we get that

From equations (18), (22) and (23), we get that

On simplifying the above expression, we get that

Substituting for A2 from equation (26),

Then A2 can be determined from equation (25).

Since both A1 and A2 are known, the expression for load current has been obtained for the case when

the firing angle is less than 60o.

When the firing angle is greater than 60o,

Then the load current can be expressed to be:

In the above equation, A3 and A4 have to be determined. The conditions we have are:

On solving for A3, we obtain that

Now an expression for the load current is known for any firing angle.

Instantaneous Load Current

An expression for the instantaneous line current as a function of firing angle can be obtained from the expression for the load current. Here an expression for current through R-phase over an input cycle is

obtained with the origin coinciding with the triggering of SCR S1. When SCR S1 is in conduction, the

line current is equal to the load current Then

When the firing angle is less than 60o,

When the firing angle is greater than 60o,

RMS Line Current

Since the line current iL(θ) is known over one input cycle, the rms line current can be obtained.

RMS Value of the Fundamental of Line Current

By performing Fouries series analysis of the line current, the rms value of the fundamental can be obtained. From the rms line current and the rms value of the fundamental of line current, THD can be computed.

SWITCH-MODE POWER SUPPLY

Power electronics deals with four forms of power conversion.

1. ac-dc conversion called rectification2. ac-ac conversion ,3. dc-ac conversion and4. dc-dc conversion.

DC-DC converters were referred to as choppers earlier, when SCRs were used. Nowadays, IGBTs and MOSFETs are the devices used for dc-dc conversion and these circuits can be classified as switch mode power supply circuits. The abbreviation or acronym for switch mode power supply is SMPS.

A switch mode power supply circuit is versatile. It can be used to:

1. step down an unregulated dc input voltage to produce a regulated dc output voltage using a circuit known as Buck Converter or Step-Down SMPS,

2. step up an unregulated dc input voltage to produce a regulated dc output voltage using a circuit known as Boost Converter or Step-Up SMPS,

3. step up or step down an unregulated dc input voltage to produce a regulated dc output voltage ,

4. invert the input dc voltage using usually a circuit such as the Cuk converter, and

5. produce multiple dc outputs using a circuit such as the fly-

back converter.

A switch mode power supply is a widely used circuit nowadays and it is used in a system such as a computer, television receiver, battery charger etc. The switching frequency is usually above 20 kHz, so that the noise produced by it is above the audio range. It is also used to provide a variable dc voltage to armature of a dc motor in a variable speed drive. It is used in a high-frequency unity-power factor circuit.

This chapter describes the basics, operation and design of switched-mode power supplies. The pages to follow contain:

STEP-DOWN /BUCK CONVERTER: IDEAL CIRCUIT STEP-DOWN /BUCK CONVERTER: PRACTICAL

CIRCUIT STEP-UP SWITCH MODE POWER SUPPLY /IDEAL

BOOST CONVERTER PRACTICAL BOOST CONVERTER AND UNITY-

POWER FACTOR RECTIFIER

INTRODUCTION

BASIC CIRCUIT OPERATION

DISCONTINUOUS OPERATION

CONTROL BY PULSE-WIDTH MODULATION

CLOSED-LOOP CONTROL

MICRO-CONTROLLER IMPLEMENTATION

IMPLEMENTATION USING A MULTIPLIER

SUMMARY

INTRODUCTION

A buck converter or step-down switch mode power supply can also be called a switch mode regulator. Popularity of a switch mode regulator is due to its fairly high efficiency and compact size and a switch mode regulator is used in place of a linear voltage regulator at relatively high output, because linear voltage regulators are inefficient . Since the power devices used in linear regulators have to dissipate a fairly large amount of power, they have to be adequately cooled, by mounting them on heatsinks and the heat is transferred from the heatsinks to the surrounding air either by natural convection or by forced-air cooling. Heatsinks and provision for cooling makes the regulator bulky and large. In applications where size and efficiency are critical, linear voltage regulators cannot be used.

A switch mode regulator overcomes the drawbacks of linear regulators. Switched power supplies are more efficient and they tend to have an efficiency of 80% or more. They can be packaged in a fraction of the size of linear regulators. Unlike linear regulators, switched power supplies can step up or step down the input voltage.

The buck converter is introduced in this page using the evolutionary approach. Let us consider the circuit in Fig. 1, containing a single pole double-throw switch.

For the circuit in Fig. 1, the output voltage equals the input voltage when the switch is in position A and it is zero when the switch is in position B. By varying the duration for which the switch is in position A and B, it can be seen that the average output voltage can be varied, but the output voltage is not pure dc. The output voltage contains an average voltage with a square-voltage superimposed on it, as shown in Fig. 2.. Usually the desired outcome is a dc voltage without any noticeable ripple content and the circuit in Fig. 1 is to be modified.

The circuit in Fig. 1 can be modified as shown in Fig. 3 by adding an inductor in series with the load resistor. An inductor reduces ripple in current passing through it and the output voltage would contain less ripple content

since the current through the load resistor is the same as that of the inductor. When the switch is in position A, the current through the inductor increases and the energy stored in the inductor increases. When the switch is in position B, the inductor acts as a source and maintains the current through the load resistor. During this period, the energy stored in the inductor decreases and its current falls. It is important to note that there is continuous conduction through the load for this circuit. If the time constant due to the inductor and load resistor is relatively large compared with the period for which the switch is in position A or B, then the rise and fall of current through inductor is more or less linear, as shown in Fig. 3.

The next step in evolutionary development of the buck converter is to add a capacitor across the load resistor and this circuit is shown in Fig. 4. A capacitor reduces the ripple content in voltage across it, whereas an inductor smoothes the current passing through it. The combined action of LC filter reduces the ripple in output to a very low level.

The circuit in Fig. 4 contains a single-pole double-throw switch. It is a difficult configuration to realize using power semiconductor devices. On the other hand, an understanding of the circuit in Fig. 4 leads to a realizable

and simple configuration. When the switch is in position A, the current through the inductor and it decreases when the switch is in position B. It is possible to have a power semiconductor switch to correspond to the switch in position A. When switch is in position B, the inductor current free-wheels through it and hence a diode can be used for free-wheeling operation. Then only the power semiconductor switch needs to be controlled, and in practice, a pulse-width modulating IC is used. The circuit that results is shown in Fig. 5.

Generally any basic switched power supply consists of five standard components:

a. a pulse-width modulating controller,b. a transistor switch,c. an inductor ,d. a capacitor ande. a diode.

Control by pulse-width modulation, usually effected by an IC, is necessary for regulating the output. The transistor switch is the heart of the switched supply and it controls the power supplied to the load. Power MOSFETs are more suited than BJTs at power outputs of the order of 50 W. Transistors chosen for use in switching power supplies must have fast switching times and should be able to withstand the voltage spikes produced by the inductor.

An inductor is used in a filter to reduce the ripple in current. This reduction occurs because current through the inductor cannot change suddenly. When the current through an inductor tends to fall, the inductor tends to maintain the current by acting as a source. Inductors used in switched supplies are usually wound on toroidal cores, often made of ferrite or powdered iron core with distributed air-gap to minimize core losses at high frequencies.

A capacitor is used in a filter to reduce ripple in voltage. Since switched power regulators are usually used in high current, high-performance power supplies, the capacitor should be chosen for minimum loss. Loss in a capacitor occurs because of its internal series resistance and inductance. Capacitors for switched regulators are chosen on the basis of effective series resistance (ESR). Solid tantalum capacitors are the best in this respect. For very high performance power supplies, sometimes it is necessary to parallel capacitors to get a low enough effective series resistance.

The diode used in a switched regulator is usually referred to as free-wheeling diode or sometimes as a catch diode. The purpose of this diode is not to rectify, but to direct current flow in the circuit and to ensure that there is always a path for the current to flow into the inductor. It is also necessary that this diode should be able to turn off relatively fast. Diodes known as the fast recovery diodes are used in these applications.

Most of the switched supplies needs a minimum load, in order to ensure that the inductor carries current always. If the current flow through the inductor is not continuous, regulation may become poorer.

The buck converter or SMPS can be controlled in two ways, known as :

1. Constant-frequency operation or pulse-width modulation control2. Variable-frequency operation or control by frequency modulation

With pulse-width modulation control, the regulation of output voltage is achieved by varying the duty cycle of the switch, keeping the frequency of operation constant. Duty cycle refers to the ratio of the period for which the power semiconductor is kept ON to the cycle period. Usually control by pulse width modulation is the preferred method since constant frequency operation leads to optimization of LC filter and the ripple content in output voltage can be controlled within the set limits. On the other hand, if the load on the converter is below a certain level, voltage regulation of output becomes a problem and in such a case, control by frequency modulation is to be preferred.

When control by frequency modulation is to be achieved, the ON period of the power semiconductor switch is kept constant and the frequency of operation is varied to effect voltage regulation. Design of LC filter is not easy in such a case.

If a micro-controller is used instead of a specific PWM IC, it is possible to switch from one mode of control to the other depending on the load conditions.

BASIC CIRCUIT OPERATION

The operation of the buck converter is explained first. This circuit can operate in any of the three states as explained below. The first state corresponds to the case when the switch is ON. In this state, the current through the inductor rises, as the source voltage would be greater than the output voltage, whereas the capacitor current may be in either direction, depending on the inductor current and the load current. When the inductor current rises, the energy stored in it increases. During this state, the inductor acquires energy.

When the switch is closed, the elements carrying current are shown in red colour in Fig. 6, whereas the diode is in gray, indicting that it is in the off state. In Fig. 6(a), the capacitor is getting charged, whereas it is discharging in Fig. 6(b).

FIG. 6(a):

FIG. 6(b):

Fig. 6.: Buck Converter : First State

The equations that govern the operation of the circuit in the first state are shown below.

The second state relates to the condition when the switch is off and the diode is ON. In this state, the inductor current free-wheels through the diode and the inductor supplies energy to the RC network at the output. The energy stored in the inductor falls in this state. In this state, the inductor discharges its energy and the capacitor current may be in either direction, depending on the inductor current and the load current Figure 7 illustrates the second state.

FIG. 7(a):

FIG. 7(b):

Fig. 7.: Buck Converter : Second State

The equations that govern the operation of the circuit in the second state are shown below.

When the switch is open, the inductor discharges its energy. When it has discharged all its energy, its current falls to zero and tends to reverse, but the diode blocks conduction in the reverse direction. In the third state, both the diode and the switch are OFF and Fig.8 illustrates the third state. During this state, the capacitor discharges its energy and the inductor is at rest, with no energy stored in it. The inductor does not acquire energy or discharge energy in this state.

FIG. 8:

Fig. 8.: Buck Converter : Third State

The equation that governs the operation of the circuit in the third state is shown below.

When the circuit receives a periodic signal, the response of the circuit also becomes periodic. Here it is assumed that the source voltage remains constant with no ripple, and the frequency of operation is kept fixed with a fixed duty cycle. If the RC time constant due to the load resistor and the filter capacitor is very large compared to the cycle period of the switching frequency, the output voltage is more or less constant, with no noticeable ripple. When both the input voltage and the output voltage are constant, the current through the inductor rises linearly when the switch is ON and it falls linearly when the switch is OFF. Under this condition, the current through the capacitor also varies linearly when it is getting charged or discharged.

The responses obtained for a particular set of parameters are displayed in Fig. 9. The values of parameters used are:

Source voltage = 100 V dc, Switching frequency = 20 kHz, L = 500 µH, C = 500 µF, R = 10 Ω, and

duty cycle = 0.5.

The value of 1 in a voltage plot in Fig. 9 corresponds to 100 V and the value of 1 in a current plot corresponds to 10 A. Figure 9 displays the responses over one cycle.

Fig.9: Periodic Response with a dc voltage Input

An expression for the average output voltage can be obtained as follows. It is assumed that there is continuous conduction in the inductor. Given that the cycle period is T, the ON-period is DT, and the source voltage is E,

In eqn. (6), D stands for the duty cycle. The same expression for output voltage can be obtained in another way. When the responses in the circuit are periodic, the inductor current is the same at the beginning and end of a cycle. That is,

Equation (7) can be expressed as follows:

When the switch is ON, vL(t) = E - Vo,avg and when the diode is conducting,

vL(t) = - Vo,avg. Therefore

On evaluation,

The change in inductor current when the switch is On can be determined as follows. Let the change in inductor current be ∆I, as shown in Fig. 10. In this figure, the change in output voltage has been exaggerated for sake of clarity. When the switch is ON, the voltage across the inductor can be expressed as:

When the output voltage remains steady at Vo,avg, the inductor current linearly during the ON period of the

switch. Then

During the ON-period, the inductor current rises from (Vo,avg/R - ∆I/2) to (Vo,avg/R + ∆I/2). That is,

The capacitor current iC is expressed as follows:

Now an assumption is made to find out the change in output voltage. It is assumed that the capacitor gets charged for half of the cycle period and gets discharged during the other half , as shown in Fig. 11. Since the current through the capacitor varies linearly, the average charging current is half of its peak value of the

triangular waveform. The peak value of its triangular waveform is shown to be ∆I/2. Hence

If a periodic signal has zero dc value over its cycle period, its average is defined based only on its positive part and hence the average capacitor current is obtained as shown above. For a capacitor

Based on the average charging current and half of the cycle period as the charging period, we get the change in output voltage ∆V as:

Using equation (12), the above equation can be expressed as:

Assuming that the ripple in output voltage is sinusoidal, the rms value of ripple content in output voltage is:

Note that the variation in output voltage is not shown to be sinusoidal in Fig. 10.. Even though the variation appears to be triangular, equation (20) gives a better approximation of the rms value of ripple content in output voltage.

Given that source voltage = 100 V dc, switching frequency = 20 kHz, L = 500 µH, C = 500 µF, R = 10 Ω, and duty cycle = 0.5, the results obtained are:

Vo,avg = 50 V,

∆I = 2.5 A,

∆V = 31.25 mV, and

Vrms,ripple = 11.05 mV.

In order that the capacitor current and the inductor current vary linearly, it is necessary that the RC time constant should be relatively large, equal to about four or five times the cycle period. When the RC time constant is much smaller than the cycle period, the responses obtained are not linear. To illustrate, Fig. 12 displays another set of responses. The only change is that a 1 µF capacitor is used in place of the 500 µF capacitor.

When the RC time constant is small, the output voltage contains noticeable ripple. In addition, the ripple in output voltage appears to be sinusoidal, justifying the equation used for finding out the rms value of the ripple

content in output voltage. Another aspect can be noticed in Fig. 12. It is that the ripple in output voltage goes through its negative half-cycle when the switch is ON. When this circuit is to be controlled by negative feedback, the feedback at the ripple frequency would become positive and closed-loop control effected without taking this aspect into account would produce a larger ripple in the output.

Fig.12: Periodic Response with a dc voltage Input

Transient Response

Fig. 13 Transient Response

Let us assume that the output voltage is zero and that there is no current through the inductor at start. If the source voltage is connected suddenly and the switch is turned ON and OFF at a fixed frequency with a preset duty cycle, the transient response of the circuit lasts for several cycles before it settles down to periodic response. The inrush current through the inductor is quite high, several times the maximum current that can flow under settled conditions. The transient response obtained over the first 10 ms with source voltage = 100 V dc, switching frequency = 20 kHz, L = 500 µH, C = 500 µF, R = 10 Ω, and duty cycle = 0.5, is shown in Fig. 13 . Even after 200 cycles, the response is still in the transient state.

Effect of ripple in input voltage

Usually the input to an SMPS happens to be an unregulated dc voltage provided by a rectifier-filter circuit. Such a filter contains significant ripple content at double the line frequency. The first applet in this page illustrates the response obtained when the input voltage contains ripple. In this program, the ripple frequency, the peak-to-peak ripple and the source voltage can be set. If the source voltage = 100 V, peak-to-peak ripple voltage = 20 V, and the ripple frequency = 100 Hz, then the input voltage falls from 110 V to 90 V in the first 7.5 ms and rises from 90 V to 110 V in the remaining 25% of the input cycle period. The input voltage falls linearly in the initially for 75% of the cycle period from (E + Vrip,pk-to-pk/2) to (E - Vrip,pk-to-pk/2) and then rises linearly during

the remainder of the input cycle period.

When the input voltage varies cyclically, the response of the circuit is periodic over its input cycle period and it is not periodic not over the period corresponding the switching frequency. In addition, there is significant overshoot in the inductor current when the input voltage is rising linearly. It can also be seen that the output voltage has nearly the same waveform as the input voltage, which is only to be expected. Since the duty cycle is kept fixed, the output voltage would tend to rise as the input voltage rises. The response obtained with a peak-to-peak ripple voltage of 20 V at 100 Hz is shown in Fig. 14. It becomes clear from the response that closed-loop control is necessary to maintain the output voltage when the input voltage has some ripple content. The closed-loop control circuit has to be designed with care, since the duty cycle has to be continually varied to maintain the output voltage at its set value.

Fig. 14: Periodic Response over One Input Cycle Period

Effect of Step Change in Load

When there is a step change in load, the circuit goes through transient response before it settles back to periodic response. Here the circuit is allowed to be a settled state with a load resistance of 10 Ω. Then the load resistance is changed to 5 Ω and the transient response that is obtained is presented in Fig. 15. In Fig. 15, '1' on the axis for corresponds to 10 A.

When the load resistor was 10 Ω, the load current would have been 5 A given that the duty cycle is 0.5 and the input voltage is 100 V. It can be seen that after the step change in load resistor, the output voltage dips first and then recovers. If the change is in the other direction from 5 Ω to 10 Ω, the output voltage rises first before falling back, as shown in Fig. 16. For this figure, '1' on the axis for corresponds to 20 A.

Effective closed-loop control would reduce the transients. The under-damped nature of inductor current response can be improved.

Fig. 15: Effect of Step Load Change: Increase in Load

DISCONTINUOUS OPERATION

When the load resistor becomes high, the buck converter circuit operates in the discontinuous mode. The inductor current falls to zero when the switch is open and remains at zero till the switch becomes ON in the next cycle. Such a mode is the desirable mode of operation when variable frequency of operation is employed for

voltage control. Additionally, discontinuous operation puts less stress on the diode and the circuit operates well. This aspect will be explained later.

The effect of load resistance on the inductor current is shown in Fig. 17. Given that the conduction is continuous, the average load current is Vo,avg/R. As the load resistance increases, the average load current

decreases, but on the other hand, the change in inductor current as defined by equation (12) remains constant. As indicated by equation (13), the inductor current remains within the range specified below.

Fig. 16: Effect of Step Load Change: Reduction in Load

When the inductor current is continuous, the value of Vo,avg is as defined by equation (10) and it is not so if the

conduction is discontinuous. After substituting for ∆I from equation (12), we get that

Hence a critical resistance, say RC can be defined that makes the circuit operate at boundary of continuous

conduction, a situation illustrated in Fig. 17. When the load resistance is higher than the conduction in inductor is discontinuous and it is continuous if the load resistance is less than RC .

It can be seen from equation that at the boundary of continuous and discontinuous conduction,

Discontinuous operation is illustrated in Fig. 18. In Fig. 18, the duty cycle for the switch is defined to D1, and

the duty cycle for the diode is defined to D2. It means that in each cycle, the switch conducts for a time interval

equal to D1T and the diode conducts for a time interval equal to D2T, where T is the cycle period. It can be seen

that for discontinuous conduction

At the boundary of continuous and discontinuous conduction, (D1 + D2) = 1. Given that the discontinuous, there

is no current in the inductor for a time interval equal to (1 - D1 -D2)T.

The average load current is the average of the inductor current in Fig. 18. From Fig. 18,

Substituting for ∆I from equation (24), we get that

Solving for D2,

When R > RC, D2 can be obtained from equation (27), assuming that D1, f and L are known. Since the average

of the inductor voltage over a cycle is zero, we obtain from Fig. 18 that,

Then

Substituting for D2 from equation (27),

The second applet shows how D2 and Vo,avg vary when R > RC.

The first applet is shown below. It has a pull-down list of items, of which the only first item is displayed. On clicking on the downward arrow, the list is displayed. If any item of the list is high-lighted, its default value is displayed in the window adjacent to the label with the caption Set Value.. If one scrolls down or up the list, the default value of the item high-lighted is displayed in the window adjacent to the label with the caption Set Value. To change the default value of an item, highlight it first and then click inside the text-field window adjacent to the label with the caption Set Value. A line cursor would appear and the value can be changed. After changing the value, click on the Set Value label and then the program recognizes the change. It can be verified that the program has made the change out by inspecting the pull-down list again.

The default values of items in pull-down list are:

Input Voltage,dc avg = 100

Input Ripple Volt., pk-pk = 0

Input Ripple Frequency = 100

Switching Frequency,kHz = 20

Inductance, microHenry = 500

Capacitor, microFarad = 500

Load Resistance, Ohms = 10

Duty Cycle= 0.5

Slow Response,0-200 = 0.

It is possible to see one of the four responses:

Periodic response over one output cycle, Transient response over one output cycle, Periodic response over one input cycle, and

Transient response over one input cycle.

Initially, the program sets '1' on the voltage axis of the plots equal to 100 V and sets '1' on the current axis to 10 A. If periodic response over one output cycle is chosen, the program assumes that the peak-to-peak ripple in input voltage is zero. The response can be slowed by varying the parameter called Slow Response. This

parameter is effective only for periodic or transient response over one output cycle.

If the load resistance is changed from 10 Ω to 5 Ω, still '1' on the current axis would correspond to 10 A. If the Reset button is clicked, the program would make changes so that '1' on the voltage axis of the plots equal to input voltage and sets '1' on the current axis to (input voltage/load resistance).

When Start/Continue button is clicked, the program responds to the visible response type and starts with the values of inductor current and the capacitor voltage it had from the previous calculation. Initially they are set to zero values. If the Reset button is clicked, the program would make them zero again.

FIRST APPLET

The second applet takes in three parameters, namely the switching frequency, the inductor value and the duty cycle and plots the duty cycle of the diode and the output voltage as a function of the load resistance varying from RC to (11 × RC).

SECOND APPLET

CONTROL BY PULSE-WIDTH MODULATION

The pulse-width modulator controls the semiconductor switch and is a complex part of a switched regulator. Nowadays, switched regulator uses a pulse-width modulator integrator circuit. The principle of control by pulse-

width modulation is illustrated in Fig. 19. The simplified functional diagram of a typical pulse-width modulator is shown in Fig.19a, whereas the waveforms in Fig. 19b explain the operation.

The pulse-width modulator circuit consists of a saw-tooth generator, an error amplifier, and a comparator. The frequency of saw-tooth generator can usually be set by choosing proper values of an RC network. The error amplifier compares the reference voltage and the feedback signal. The feedback signal is obtained using a voltage divider network across the output of the SMPS circuit. For example, let the feedback signal be Vf and

the reference voltage be Vref. Then

The output of the error amplifier is compared with the saw-tooth waveform and when this voltage is greater than the output of sawtooth generator, the output of the comparator would be at logic '1'. When the output of comparator is at logic '1', the switch in the SMPS circuit can be kept in the ON state. When the comparator is at logic '0', the switch in the SMPS circuit can be kept in the OFF state.

If the output voltage tends to be greater than that indicated by equation (32), the output voltage of the error amplifier would fall and the duration for which the output of comparator remains at logic '1' would decrease. Thus the duty cycle of the switch reduces and the output of the SMPS would fall, according to equation (10). Thus it can be seen that the negative feedback control maintains the output at the desired value. For negative feedback control, the feedback signal should be applied to the inverting input of the error amplifier.

When control by frequency modulation is desired, the ON-period is kept constant, but the frequency is varied in order to bring about regulation. Such a technique is necessary if the load on the regulator tends to become very low. It is difficult to make the ON-period below a certain time duration and when this limit is reached, control by pulse width modulation becomes impossible. Then the duty cycle is reduced by keeping the ON period fixed and increasing the cycle period. The value of minimum ON period depends on the transistor switch.

CLOSED-LOOP CONTROL

In order to regulate the output voltage, a controller is needed to be designed. For this purpose, the power circuit is first represented by a transfer function.

The passive part of the power circuit is shown in Fig. 20. Since the DC output of the SMPS is defined by equation (10), the effective input to the passive circuit in Fig. 20 can be stated to be E.D(s), whereas D(s) is the Laplace transform of the output of the error amplifier shown in Fig. 19. In the design procedure outlined below, D(s) is taken to be the output of the controller, varying between 0 and 1.

For the circuit in Fig. 20,

where

The block diagram of the SMPS with a PI controller is shown in Fig. 21. The transfer functions of the different blocks in Fig. 21 are as follows:

The PI controller has two inputs, the reference signal Vref corresponding to the desired output voltage and the

feedback signal Vfdb. The error signal, the difference, e, between Vref and Vfdb is fed to the PI controller and the

output of the PI controller is D(s), the signal that sets the duty cycle. Then

In equation (35), K is the proportional gain and T is the integrating time constant. The output of the PI controller varies between 0 and 1 and has two limits, one corresponding to the lowest duty cycle and the other corresponding to the highest duty cycle.

There is a filter in the feedback path. The filter time constant is about 4 times the cycle period corresponding to the switching frequency of the SMPS. Then the ripple at the switching frequency gets filtered out and the feedback signal is essentially a dc signal. Otherwise, the ripple in the output at the switching frequency has nearly 180o phase shift with respect to the fundamental of the input square pulse to the SMPS and then feedback at switching frequency becomes positive, which in turn leads to amplification of the ripple content. The filter avoids this problem, without seriously affecting its dynamic performance.

The transfer function of the power circuit is obtained as follows. Since the output varies linearly with the duty cycle, equation (33) can be presented as follows:

For designing the closed loop system, the poles due to the source can be ignored. Since the power circuit is very much under-damped, derivative feedback is required for stable operation. The differentiating circuit provides the feedback signal. For simulation, the derivative signal is obtained based on the capacitor current. The derivative feedback voltage is set to be kd.iC, where iC is expected to be its per unit value.

The third applet presented below allows the user to design a suitable PI controller. The parameters to be set are: switching frequency in kHz, gain of the PI controller, its integrating time-constant , the inductance, the capacitance, the load resistance, the time constant of the filter in the feedback path and the derivative coefficient.

THIRD APPLET

The feedback arrangement used for actual simulation contains additional circuitry. It is necessary to limit the inductor current since it is seen from the open loop response that the inductor current can be several times the rated current, where the rated load current is the nominal current rating of the SMPS. It is a parameter that can be set in the program.. In this program, the current limit is set at 1.5 times the rated load current, and when the inductor current exceeds the current limit, the current in excess of the current limit is amplified and added to the feedback signal. In addition, a derivative feedback signal, obtained using the capacitor current, is added to the feedback signal. Derivative feedback improves damping of the system and without it, the system is oscillatory in spite of a well-designed PI controller. While designing the PI controller using the third applet, the switching circuit is replaced by a linear amplifier and this approximation is necessary for the design of the PI controller. But in reality, the square-wave voltage input to the power circuit makes the system to be oscillatory since the power circuit is heavily under-damped. The feedback arrangement used is shown in Fig. 22.

When the derivative feedback coefficient is set at unity, the derivative feedback amounts to 10% of the rated voltage when the capacitor current equals the rated current.

The fourth applet that is presented below simulates the circuit. It has a pull-down list of items, of which the only first item is displayed. On clicking on the downward arrow, the list is displayed. If any item of the list is high-lighted, its default value is displayed in the window adjacent to the label with the caption Set Value.. If one scrolls down or up the list, the default value of the item highlighted is displayed in the window adjacent to the label with the caption Set Value. To change the default value of an item, highlight it first and then click inside the text-field window adjacent to the label with the caption Set Value. A line cursor would appear and the value can be changed. After changing the value, click on the Set Value label and then the program recognizes the change. It can be verified that the program has made the change out by inspecting the pull-down list again.

The default values of items in pull-down list are:


Input Ripple Volt., pk-pk = 0

Input Ripple Frequency = 100

Switching Frequency,kHz = 20

Inductance, microHenry = 500

Capacitor, microFarad = 500

Load Resistance, Ohms = 10

Output Voltage set at = 50

Gain of the PI controller = 5. Time-constant of the PI controller in µs =500

Time-constant of the Filter in µs =40

Derivative Feedback Coefficient = 1.0

Rated Current in Amps = 10

Slow Response,0-200 = 0.

It is possible to see one of the five responses:

Periodic response over one output cycle, Transient response over one output cycle,

Periodic response over one input cycle, Transient response over one input cycle and

Statistics

Initially, the program sets '1' on the voltage axis of the plots equal to 100 V and sets '1' on the current axis to 10 A. If periodic response over one output cycle is chosen, the program assumes that the peak-to-peak ripple in input voltage is zero. The response can be slowed by varying the parameter called Slow Response. This parameter is effective only for periodic or transient response over one output cycle.

If the load resistance is changed from 10 Ω to 5 Ω, still '1' on the current axis would correspond to rated current. If the Reset button is clicked, the program would make changes so that '1' on the voltage axis of the plots equal to input voltage and sets '1' on the current axis to (input voltage/load resistance).

When Start/Continue button is clicked, the program responds to the visible response type and starts with the values of inductor current and the capacitor voltage it had from the previous calculation. Initially they are set to zero values. If the Reset button is clicked, the program would make them zero again.

FOURTH APPLET


Nowadays the use of a micro-controller is popular. But control of a circuit switching at 20 kHz tends to be difficult, essentially due to delays involved in A/D conversion and some mathematical operations. Hence the approach to closed-loop control is a combination of rule-based logic and action of a PID controller. The delay that occurs from the instant the A/D converter samples an input signal to the instant when the corresponding digital output is available for use tends to be of the same order as the switching cycle period.

At 20 kHz, the switching cycle period is 50 µS and the delay involved due to the use of an A/D converter is about 40 µS. This means that no more than one sample can be obtained during one switching cycle period. If this sampling is done in an asynchronous manner, closed-loop operation can be difficult. For example, the inductorcurrent can vary considerably over a cycle period and asynchronous sampling of inductor current may not lead to stable closed-loop control. It is preferable to sample the inductor current at a predetermined instant of the cycle period. For example, at the start of a switching cycle, a sample and hold circuit can be used to store the instantaneous value of inductor current and the A/D converter can convert this value. In the fifth applet displayed below, the A/D converter delay is set to be equal to one switching cycle period. When that is so, the sampling occurs at the beginning of each cycle. The inductor current and the capacitor voltage are sampled once in two cycles, since conversion period for each sample equals one switching cycle period.

If the sampling is to be done at a faster rate, the sampling has to be asynchronous and then an analogue filter has to be used to reduce the variations in the feedback signal corresponding to inductor current before it is used as the input to the A/D converter. Alternatively, fast A/D converters such as

flash A/D converter with the sampling period of the order of 1 µs can be used.

The scheme suggested for controlling the SMPS with a micro-controller is outlined with the help of a pseudo-code presented below.

Initialize:

Output Voltage Count = 0 // Output of A/D converter Inductor Current Count =0 // Output of A/D converter Old Current Count = 0

// Previous current count required for derivative feedback

swCycleCount = 100 // Set the period of switching frequency as a number of micro-controller clock frequency.

atod_delay =100 // Set the A/D converter's conversion period and computing delay as a number of // micro-controller clock frequency.

lowCount= 5; highCount=95; dutyCycle=lowCount; // Set the limits for duty cycle and set the duty cycle at its lowest value

on_off=true; // boolean value indicating switch is ON when it is true

rampCount=0; // Used with dutyCycle and swCycleCount for setting on_off to either true or false value

atod_Count=0; feedback Voltage=0 integrator Output=0 //Output of an integrator used in feedback control Go to Main Loop

Main Loop:

Call Parameters Subroutine Increment ramp Count If (rampCount<dutyCycle) on_off = true; Else on_off = false; If (rampCount == swCycleCount) on_off = true rampCount=0;

Increment atod_Count Increment rampCount If (atod_Count>atod_delay) call NextValues subroutine Return to Main Loop

Parameters Subroutine: Begin Desired Output Voltage=50.0 // numerical setting using thumb-wheel switches/KeyPad Current Limit = 125% // Set in software, using KeyPad or thumb-wheel switches

kd=20 // Derivative feedback coefficient in Ohms. // Feedback in Volt equals (kd × Inductor Current).

dkI=0.1 // Integrating Coefficient effective when output voltage is within // a close band of ± 5% of source voltage from the desired output voltage.

End subroutine

NextValues Subroutine Begin atod_Count=0;

Inductor Current Count = output of A/D output Ouput Voltage Count = output of A/D output if (Inductor Current Count <=Current Limit) feedback Voltage= Ouput VoltageCoun +kd*( Inductor Current Count - Old Current Count); if (Desired Output Voltage>(feedback Voltage+5%)) dutyCycle++; if (Desired Output Voltage<(feedback Voltage-5%)) dutyCycle--; if (Desired Output Voltage>(feedback Voltage-5%)) AND (Desired Output Voltage<(feedback Voltage+5 integrator output+= dkI*(Desired output Voltage-Ouput Voltage Count) dutyCycle+= integrator output; if (dutyCycle<=lowCount) dutyCycle=lowCount if (dutyCycle>=lowCount) dutyCycle=highCount

else integrator output =0; else dutyCycle = dutyCycle-2; if (dutyCycle<=lowCount) dutyCycle=lowCount; End Subroutine

The fifth applet presented below simulates the operation of the switch mode step-down power supply controlled by a micro-controller. The applet has default values for the parameters listed below.


Input Ripple Volt., pk-pk = 0; Input Ripple Frequency = 100

Inductance, microHenry = 500; Capacitor, microFarad = 500 ; Load Resistance, Ohms = 10; Desired Output Voltage = 50; Rated Current, Amp = 10; Micro's Clock Freq, MHz = 2 ; Switching Period:Clock Cycles = 100; A/D Delay :Clock Cycles = 100; Derivative FeedBack Coef. =20; Integrating Coefficient = 0.1;

This applet has the same structure as the fourth applet and is hence not described any further.

FIFTH APPLET


Implementation using a multiplier is an attractive option since this scheme can deliver nearly desired output voltage in spite of fluctuations in the source voltage. The scheme to be used is shown in Fig. 23.

The inputs to the multiplier are the source voltage or a part thereof and the output of the opamp in the feedback loop. Since the output of the multiplier should equal Vref representing the desired output voltage, the output of the opamp reflects the duty cycle. This signal can be compared with the output of a sawtooth generator and a pulse to turn on the switch can be obtained. If Vref can be modified to include the effects of drop in the switch, the output of the SMPS can equal the set value. The sixth applet simulates the SMPS controlled in this manner.

SUMMARY

This page has described the operation of a switch mode step-down power supply circuit. For description in this page, it has been assumed that the components used in the power circuit are ideal, whereas they are not ideal in reality. The next page goes into details of the practical limitations of the power circuit components.

INTRODUCTION

DIODE REVERSE RECOVERY CURRENT

TURN-ON TRANSIENT IN A DIODE

TRANSIENT PROCESSES IN A MOSFET

LOSSES IN ENERGY STORAGE ELEMENTS

CONTROL BY PWM

CURRENT MODE CONTROL: IDEAL CIRCUIT

SUMMARY

INTRODUCTION

The previous page on the step-down switch-mode power supply circuit was described and analyzed assuming that the components used are ideal. The real diode and the MOSFET have some limitations and they have to be taken into account for designing a practical power supply. First, the characteristics of the diode are outlined first.

DIODE REVERSE RECOVERY CURRENT

Even if a fast-recovery or an ultra-fast recovery diode is used for free-wheeling operation, the reverse-recovery characteristic of the diode imposes some constraints. The circuit that can be used to test the reverse recovery characteristic of a diode is shown in Fig. 1. An applet, named the first applet, simulates the reverse recovery property of the diode and is displayed below.

The operation of the test circuit is explained at first. The time constant due to load inductance and load resistance should be several times the time corresponding to the switching frequency. Then the current through the remains more or less steady and its value is given by:

When the MOSFET is ON, the load current flows through the transistor. When it is turned off, the current flows through the snubber circuit initially and then the free-wheeling diode, marked as DUT (device under test). When the MOSFET is switched ON again, the source tends to reverse the current through the diode. The current through the diode falls rapidly and reverses before it becomes zero. The reverse recovery has been simulated using the model of the diode shown in Fig. 1.

The reverse-recovery transient process depends on the diode itself and it also depends on the junction temperature of the diode, the forward current prior to being reverse biased, the rate of fall forward current and the source voltage that applies to the reverse bias to the diode. As the value of any one of these parameters rises with the other parameters remaining unchanged, the reverse recovery transient process becomes worse, reflected by an increase in the peak reverse

current. The reverse recovery turn-off period starts once the diode current becomes negative and lasts till the reverse diode current increases in ampltidue first and then decays to about 10% of its reverse peak current. It represents the time that has to elapse before the diode recovers ability to block reverse voltage.

The manufacture specifies the reverse recovery transient period and the maximum reverse current for a set defined by its forward current, the rate of fall of forward current and the junction temperature. For many diodes, the maximum rate of fall forward current is specified to be 100 A/µs. This means the test inductance L should set be such that

If reverse recovery transient is to be minimized, it would be preferable to select L such that

where the rated current of diode is Irated,diode and diF/dt is the permissible ,axi,u, rate of fall of

diode current. It is difficult to select an inductor based on trr, its reverse recovery transient

period, because the inductor chosen based on the diF/dt rating of the diode is more

conservative. For example, the data sheet for a diode with 30 A rating may have its diF/dt

rating to be 100 A/µs and its turn-off time at 100o C of junction temperature may be specified to be 150 ns. If the diode is carrying 30 A when it is suddenly reverse-biased, a time interval of 300 ns is required for its current to fall to zero and then a further 150 ns

would need to elapse before the diode is off. It is seen why the inductor is to be selected based on the diF/dt rating of the diode. It is better that the inductor restricts the rate of fall of diode

current to a value lower than its diF/dt rating in order to restrict the reverse-recovery current.

The applet simulates the reverse recovery transient. It is an approximation of the behaviour of the diode. No accuracy for this model is claimed, but it does show the dependence of recovery transient on the parameters mentioned. The model developed is a mathematical model, based purely on the turn-off time, load current, rated current, junction temperature, and the maximum rate of fall of current of the diode. No physical model has been developed, as is the common practice. The process that occurs within a diode during turn-off and turn-on is a complex one and it is difficult to develop a physical model that is stable as well as realistic. On the other hand, the mathematical model is easy to develop. When the MOSFET is turned on, the current through the diode falls, its rate of fall restricted by the inductor. In the process, diode current

reduces and becomes negative. The peak reverse current that can occur is calculated based on the load current, the rate of fall of diode current,and the junction temperature. Once this peak reverse current is reached, the diode voltage builds up, depending on the reverse voltage rating, the junction temperature and the turn-off time and the diode current can be calculated once the diode voltage is determined.

The pull-down choice menu contains rated current of the diode, its turn-off period, its turn-on period, load current, source voltage, test inductance in nH, test resistance in Ohms and the junction temperature as its items. To change the value of a menu-item, pull down the menu, highlight the item, change its value in the adjacent text-field and then click on Set Value button. When the Run/More button is clicked, the program displays the diode current and the voltage across the diode for a period corresponding to 12 times the turn-off period. If the diode had not turned off by then, click again on the Run/More button, and the process that occurs for the next 12 times the turn-off period would be displayed. Whenever the rate of fall of forward current increases due to either lower test inductance or higher source voltage, the total period required for the diode to turn-off decreases, but there would a significant increase in the peak reverse current. In the plots shown by the applet, unit on the y-axis represents the rated current for the top plot and it represents the rated reverse voltage for the bottom plot.

APPLET FOR REVERSE RECOVERY TRANSIENT IN A DIODE

TURN-ON TRANSIENT IN A DIODE

The turn-on transient in a diode is not as predominant as its turn-off process is, but nonetheless it affects the performance of the circuit in which it is placed. The circuit for simulating the turn-on transient process is shown in Fig.2 and the diode model used is the same as shown in Fig. 1.

In a practical circuit, it is often difficult to notice the turn-on transient process, for two reasons. Firstly, the turn-on process lasts only for a brief period. Secondly, the transient process of some other device can often mask the turn-on process of a diode. For example, the turn-off delay of the MOSFET can mask the turn-on process of the diode. The simulation, shown in the second applet, makes the assumption that the transistor is ideal and simulates the turn-on process in an approximate manner. It is necessary to have a snubber capacitor across the FET as shown in Fig. 2. Again a mathematical model is used. When the FET is turned off, the snubber capacitor voltage builds up to the source voltage due to load current charging it. After that, the diode current rises at a predetermined rate, depending on the rated current, the turn-on period and the junction temperature. After the diode current becomes equal to the load current, the diode voltage exponentially decays, based on its turn-on period. It is assumed that the diode current remains constant.

In some circuits, such as the flyback converter to be described in one of the pages to follow, the turn-on delay of the diode is crucial. The only practical remedy to address this problem is to connect an RC snubber circuit across the diode.

APPLET FOR TURN-ON TRANSIENT IN A DIODE

TRANSIENT PROCESSES IN A MOSFET

For a MOSFET, three transient processes can be identified. The first is the turn-on transient process associated with the MOSFET and the second is its turn-off process and the third is the turn-off transient process associated with the body-diode of the MOSFET. The turn-on transient of the body diode is relatively fast and can be ignored.

For a fast MOSFET, the turn-on transient process is characterized by two time periods, one is the turn-on time delay and the second is the rise time. During the turn-on time delay, the gate-to-source voltage builds up to its threshold value and during the rise time, the device current rises to about 90% of its final value. There is a further delay before the drain-to-source voltage becomes equal to its conduction drop. The turn-on delay time can be reduced to some extent by a stiff gate drive signal. When the MOSFET is turned off, there is a delay period corresponding to the period in which the gate voltage reduces to its threshold level. It is followed by the cross-over period that includes a delay period and a fall period. During the delay period, the drain-to-source voltage rises from its conduction value to its blocking value. After this delay period has elapsed, the fall period follows during which the current through the MOSFET decreases. For a fast MOSFET, the total turn-on transient lasts for about 50 ns, whereas the turn-off process lasts for about 100 ns.

The turn-on characteristic of a MOSFET is shown below. The voltage and the current associated with the device can change in this manner and the datasheets display a similar turn-on characteristic. When the MOSFET is placed in a circuit, the rise in current and the fall in voltage get modified due to the influence of the external components.

The turn-off characteristic of a MOSFET is shown below. The voltage and the current associated with the device can change in this manner and the datasheets display a similar turn-off characteristic. When the MOSFET is placed in a circuit, the fall in current and the rise in voltage get modified due to the influence of the external components.

The body diode is comparatively slow. It turns on quite fast, but its turn-off process is quite slow, of the order of 500 ns. Hence when the MOSFET is to be operated at high frequency, it is preferable to use an external diode. In such a case, an additional diode may be required, that has to be connected in series with the MOSFET.

A circuit that can be used for simulating the transient processes in a MOSFET is shown in Fig. 3. Here Q1 is turned on and off, whereas Q2 remains off. When Q1 is turned off, the load current is diverted through diode D2, the body diode of Q2 . When Q1 is turned on, both the turn-on transient of Q1 and the reverse recovery transient of D2 occur. The simulation that is displayed as applet 3 is again quite approximate. The purpose is to illustrate how MOSFET functions as a switch. An air-cored inductor, labeled L3 in Fig.3, is necessary to be used to reduce the reverse recovery current of the diode. The time constant due L3 and R3 should be much less compared with the cycle period corresponding to switching frequency. The turn-on process of the MOSFET is quite slow, due to the slow turn-off of the body diode D2. In applications where the MOSFET has to be switched on and off at higher frequency of the order of 20 kHz and above, it is the practice to bypass the body diode by an external diode. In this application, such a technique is unnecessary because the body diode does not have to conduct at all.

When the applet for simulating the dynamic characteristics of a MOSFET is run, the waveforms produced may appear to be bizarre. When the applet is run with the default values to simulate the turn-on process of the MOSFET, the waveforms produced are presented below.

The logic behind the waveforms is as follows. When the MOSFET is turned on by applying a gate pulse, theere is a turn-on delay. During this period, the voltage across the FET remains equal to the source voltage and there is no current through the FET. Then the rise period follows, during which the FET current rises and the FET voltage is determined by the external circuit. Here when the current the FET rises, inductor L3 absorbs the voltage and the voltage across the FET is zero. The current through the FET equals the load current well before the rise period elapses and hence the voltage across the FET becomes equal to the source voltage once the rate of rise FET current is zero. At the end of the rise period, the body diode of Q2 contains still a large number of charge carriers and it is predisposed to conduct in the reverse direction. As the reverse recovery current increases, the FET voltage falls in some piecewise linear fashion. There is some behind the simulation, even though its accuracy is questionable.

Now the waveforms obtained for the turn-off process are explained.

When the MOSFET is turned off, there is a storage delay period during which both the current and the voltage of the FET do not change. After that, the FET voltage rises, with its current remaining unchanged. Then the FET current falls. As the current falls, the voltage across L3 changes polarity and the FET voltage is the sum of the source voltage and the voltage across the inductor L3.

The losses in a FET are due to three factors:

1. Conduction losses due to on-state voltage,2. Turn-on losses and3. Turn-off losses.

The fourth applet simulates the step-down SMPS with PWM control in the open loop. The aim of this simulation is to estimate the losses in the switch and the diode and to show how the modifications to the power circuit lead to reliable operation, whether the conduction is continuous or discontinuous.

LOSSES IN ENERGY STORAGE ELEMENTS

The non-ideal properties of filter capacitor at the output and the inductor affect the performance of the circuit. The inductor is not lossless because of the winding resistance and the core loss. The model of the inductor can be changed to a series network containing a resistor reflecting its losses and an inductor. Due to the losses in the inductor, there is a slight reduction in the output voltage and also a drop in the efficiency of the circuit. Given a duty cycle and a steady input voltage, the output voltage is has been obtained in the previous page as:

The average inductor current is the same as the average load current. Then the average inductor current IL is:

If the internal resistance of the inductor is Rind, the drop in output voltage is approximately:

The effect of the ESR of capacitor is to increase the ripple content in output. From the previous page, the change in capactor current from its higheest value to its lowest value is:

The worst-case increase in peak-to-peak ripple in output voltage can be obtained by just adding the ripple due to ESR with the previously obtained value. If ESR of capavitor be Rcap, then

The actual peak-to-peak ripple would be less than the value stated above.

CONTROL BY PWM

It is necessary to modify the step-down SMPS if it is to operate at relatively high voltage in the continuous conduction mode. The problem with continuous conduction occurs when the MOSFET is turned on with the diode still in conduction. To explain this aspect, Fig. 4 is presented. When the MOSFET is turned on, the diode can act as a short circuit till it recovers. To overcome this problem, the circuit in Fig. 4 is to be modified.

The modifed circuit is presented in Fig. 5. This circuit, when designed properly, would well whether the conduction is continuous or discontinuous. This circuit contains a few additional components. The components added are an additional diode marked as D1 and an RC dissipating circuit for D1. Diode D3, inductor L2 and capacitor C2 are the same components, marked as D, L and C respectively in Fig. 4. The operation of the circuit in Fig. 5 is explained now. The circuit in Fig. 5 does not show the snubber circuit required for the diodes D1 and D2 and possibly MOSFET. Some MOSFETs do not need a snubber circuit, whereas a diode needs a snubber circuit. There is a separate page on design of snubber circuit.

The conduction path that exists when the MOSFET is ON is shown in red colour in Fig. 6. The current flow is through the MOSFET and the inductors.

When the MOSFET is turned off, the set of components in conduction varies. In mode 1 following immediately after the turn-off of the MOSFET, diode D1, inductor L1, capacitor C1, resistor R1 and components L2 C2 and RL are in conduction. The RC circuit in series with D1 conducts, the time constant associated with C1 and R1 being very small compared with the on-off periods of the MOSFET. During this mode, inductor L1 discharges its energy to R1 and C1. The components in conduction in mode 1 are shown in Fig. 7.

When mode 1 is over after inductor L1 has discharged its energy, mode 2 follows. Here only D2 continues to conduct, but L1 and D1 are not in conduction. This mode is illustrated in Fig. 8. During this phase, C1 would discharge any energy it may have acquired into R1. After the end of the period corresponding to the switching frequency, the MOSFET is turned ON again and the circuit reverts to the state shown in Fig. 9.

The circuit lasts in the state shown in Fig. 9 till the current through L1 becomes equal to that through L2. After that, the circuit reverts to the state shown in Fig. 6.

When the MOSFET is switched on, current through L1 rises gradually and current through D2 falls gradually. If inductor L1 is sufficiently large, there would be hardly any reverse recovery transient due to diode D2. Typically L1 should be such that the rate of rise current is less than 50% of the maximum rate of fall specified for diode D2. Inductor L1 can even be an air-core inductor, whereas inductor L2 has a ferrite-core with an air gap.

The complete power circuit along with the required snubber circuits is shwon in Fig. 10. The design of power circuit is described in a separate page. The snubber circuit for the MOSFET may not be required if a MOSFET that does not need a snubber is chosen.

The fourth applet presented below simulates the circuit in Fig. 5. It is assumed that the turn-on delay of diodes is negligible.The pull-down menu contains two items that are not shown explicitly in Fig.5. They are the equivalent-series resistance of capacitor C2 and the internal resistance of inductor L2. In practice, even with a fixed duty cycle, a fixed switching frequency and a steady-input voltage, the ripple content in output tends to be higher than the calculated value, mainly due to the ESR of C2. The default value of ESR of C2 is set to be zero. To see its effect, the type of response should be set as Statistics. Then when the program is run, the peak-to-peak ripple in output voltage is displayed. The program has to be run a few times in this mode before the peak-to-peak ripple in output settles down to its periodic value. For example, when ESR of C2 is increased to 0.1 Ω , it can be seen that the peak-to-peak ripple that results is much higher.

The internal resistance of L2 reflects the winding resistance and the core loss and the losses in inductor L2 tend to reduce the efficiency of this converter. The default value of internal resistance has been set to be zero. Its realistic value can be calculated by assuming a quality factor of about 100 at the switching frequency. For the value of inductor L2, an appropriate value of internal resistance of L2 can be set to be about 0.5 Ω. To see its effect, the type of response should be set as Statistics in the program.

When the type of response is Statistics, the program should be run a few times before the results become repetitive. Even then, the sum of output power and all the losses may not add upto the input power, due to errors in modelling and the presence of energy storage elements. The loss calculation of diode D1 in particular is quite incorrect, because the turn-on losses of a diode have been ignored. Even though the average current of D1 turns out to be small, its turn-

on losses are quite significant and it is advisable to use the same diode selected to be used as D2. These diodes should be fast-switching diodes with a low reverse period of about 50 ns. It is likely that there will be some time delay before the selected response is displayed. In the meantime, the message that computing is going on will be flashed on the screen.

CURRENT MODE CONTROL: IDEAL CIRCUIT

Current mode control of a SMPS is a popular technique built into several PWM integrated-circuits nowadays. This topic should have been covered in the previous page, but has been held over and presented now.

The block diagram of the current-mode control is similar to that described the previous page on the ideal step-down SMPS circuit. The PI controller illustrated in Fig. 22 of the previous page can used be as it is. In the previous page, closed-loop control was effected by comparing the output of the PI controller with a ramp signal (Fig. 19). For current-mode control, the output of the PI controller should be compared with the signal reflecting the current though the inductor L2, with this signal suitably scaled. This signal is compared with the PI controller output and when the signal corresponding to inductor current tends to exceed the PI controller output, the MOSFET is turned off for the rest of the output cycle. At the start of each cycle, the MOSFET should be turned on.

With just this type of control, the pulse-width tends to fluctuate from one cycle to the next, leading to oscillations in output. Ideally duty cycle should be equal to the ratio of desired output voltage to the source voltage, but the intersection of the PI controller output and the signal reflecting inductor current may not always occur after a time lapse from the start of a cycle that corresponds to the desired duty cycle. To overcome this problem, a component proportional to the time elapsed from the instant the cycle starts can be added to the signal corresponding to the inductor current and this sum can be compared with the output of the PI controller. Let us say that the maximum of the PI controller output be 10 V and let the signal corresponding to inductor current be 10 V when the inductor current is at nominal rated value. Let also a ramp voltage be generated such that it rises from 0 V to 10 V from the start to the end of a cycle. Then a fraction of the ramp voltage can be added to the signal reflecting current, the fraction being equal to the duty cycle which in turn is the ratio of desired output voltage to the source voltage. When the source voltage tends to vary over an input cycle, this fraction computed as the ratio of desired output voltage to the source voltage automatically gets adjusted and the correction to the signal reflecting the inductor current becomes the right adjustment.

INTRODUCTION

ANALYSIS OF THE IDEAL CIRCUIT



CLOSED-LOOP CONTROL USING PWM


CURRENT-MODE CONTROL


SUMMARY

INTRODUCTION

The boost converter, also known as the step-up converter, is another switching converter that has the same components as the buck converter, but this converter produces an output voltage greater than the source. The ideal boost converter has the five basic components, namely a power semiconductor switch, a diode, an inductor, a capacitor and a PWM controller. The placement of the inductor, the switch and the diode in the boost converter is different from that of the buck converter. The basic circuit of the boost converter is shown in Fig. 1.

The operation of the circuit is explained now. The essential control mechanism of the circuit in Fig. 1 is turning the power semiconductor switch on and off. When the switch is ON, the current through the inductor increases and the energy stored in the inductor builds up. When the switch is off, current through the inductor continues to flow via the diode D, the RC network and back to the source. The inductor is discharging its energy and the polarity of inductor voltage is such that its terminal connected to the diode is positive with respect to its other terminal connected to the source. It can be seen then the capacitor voltage has to be higher than the source voltage and hence this converter is known as the boost converter. It can be seen that the inductor acts like a pump, receiving energy when the switch is closed and transferring it to the RC network when the switch is open.

When the switch is closed, the diode does not conduct and the capacitor sustains the output voltage. The circuit can be split into two parts, as shown in Fig. 2. As long as the RC time constant is very much larger than the on-period of the switch, the output voltage would remain more or less constant.

When the switch is open, the equivalent circuit that is applicable is shown in Fig. 3. There is a single connected circuit in this case.

ANALYSIS OF THE IDEAL CIRCUIT

Analysis of the circuit is carried out based on the following assumptions. The circuit is ideal. It means when the switch is ON, the drop across it is zero and the current through it is zero when it is open. The diode has zero voltages drop in the conducting state and zero current in the reverse-bias mode. The time delays in switching on and off the switch and the diode are assumed to be negligible. The inductor and the capacitor are assumed to be lossless.

1. The responses in the circuit are periodic. It means especially that the inductor current is periodic. Its value at the start and end of a switching cycle is the same. The net increase in inductor current over a cycle is zero. If it is non-zero, it would mean that the average inductor current should either be gradually increasing or decreasing and then the inductor current is in a transient state and has not become periodic.

2. It is assumed that the switch is made ON and OFF at a fixed frequency and let the period corresponding to the switching frequency be T. Given that the duty cycle is D, the switch is on for a period equal to DT, and the switch is off for a time interval equal to (1 - D)T.

3. The inductor current is continuous and is greater than zero.4. The capacitor is relatively large. The RC time constant is so large, that the

changes in capacitor voltage when the switch is ON or OFF can be neglected for calculating the change in inductor current and the average output voltage. The average output voltage is assumed to remain steady, excepting when the change in output voltage is calculated.

5. The source voltage VS remains constant.

Inductor Current with Switch Closed

When the switch is closed, the equivalent circuit that is applicable is shown in Fig. 2. The source voltage is applied across the inductor and the rate of rise of inductor current is dependent on the source voltage VS and inductance L. The differential

equation describing this condition is:

If the source voltage remains constant, the rate of rise of inductor current is positive and remains fixed, so long as the inductor is not saturated. Then equation (1) can be expressed as :

The switch remains ON for a time interval of DT in one switching cycle and hence DT can be used for ∆t. The net increase in inductor current when the switch is ON can be obtained from equation (2) to be:

Inductor Current with Switch Open

When the switch is open, the circuit that is applicable is shown in Fig. 3. Now the voltage across the inductor is:

Given that the output voltage is larger than the source voltage, the voltage across the inductor is negative and the rate of rise of inductor current, described by equation (5), is negative. Hence if the switch is held OFF for a time interval equal to (1 - D)T, the change in inductor current can be computed as shown in equation (6)

.

The change in inductor current reflected by equation (6) is a negative value, since Vo

> VS. Since the net change in inductor current over a cycle period is zero when the

response iL(t) is periodic, the sum of changes in inductor current expressed by (4)

and (6) should be zero. That is,

On simplifying equation (7), we get that

It has been stated that when iL(t) is periodic, the net change in inductor current over

a cycle is zero. Since change in inductor current is related to its volt-seconds, the net volt-seconds of the inductor has to be zero. The expression for the net volt-seconds can be obtained from equation (7) and it can be seen that the numerator of equation (7) should be zero. That is,

The value of D varies such that 0 < D < 1 and it can be seen from equation (8) that output voltage is greater than the source voltage, and hence this circuit is called the boost converter. The output voltage has its lowest value when D = 0 and then the output voltage equals the source voltage. When D approaches unity, output voltage tends to infinity. Usually D is varied such that 0.1 < D < 0.9 .

The waveforms of inductor voltage and inductor current are shown in Fig. 4. These waveforms are drawn assuming that both the output and the source voltage remain steady. These waveforms illustrate how the inductor voltage is related to its current.

Output Voltage Ripple with Switch Closed

In this sub-section, the change in output voltage is calculated. It needs to be emphasized that the peak-to-peak ripple in output voltage is quite small for a well-designed circuit. For the inductor, the net change in inductor current over a cycle is zero when iL(t) is periodic. For the capacitor, the net change in capacitor voltage

over a cycle is zero when it is periodic. When the switch is closed, the equivalent circuit in Fig. 2 shows that the boost converter is split into two sub-circuits, with the loop currents decoupled from each other. When the switch is closed, the output voltage is sustained by the capacitor. During this period, the capacitor discharges part of its stored energy and it re-acquires this energy when the switch is open. When the switch is open, part of the inductor current charges the capacitor since the inductor current usually remains larger than the current through the load resistor. From Fig. 2,

When current through a capacitor charges it up, its rate of rise of capacitor voltage is positive since the capacitor voltage is increasing. When the switch is open, the capacitor is discharging its energy with its voltage falling and the current through the capacitor is then a negative value. The output voltage remains positive and hence the output current is positive and it is the negative of the capacitor current, as can be seen from Fig. 2. Since the change in output voltage is quite small, it can be assumed that the load current remains constant at its average value and equation (10) can be now expressed as:

When the capacitor current is constant, its voltage changes linearly with time. Here the period for which the switch is closed is DT and the DT can be used in place of ∆t. The peak-to-peak ripple in output voltage expressed as ∆vo and it is then

expressed as:

Equation (12) yields the value of the peak-to-peak ripple in output voltage. In equation (12), 1/f replaces T since T is the reciprocal of switching frequency.

Figure 5 shows how the capacitor current and voltage vary over a cycle. The ripple in output voltage is exaggerated in Fig. 5, whereas in practice it would be much smaller. If the output voltage is drawn to scale, the ripple in output voltage would not be noticeable.

Expression for Average Inductor Current

The average inductor current can be found out by equating the power drawn from the source to the power delivered to the load resistor. Again the ripple in output voltage is ignored and it is assumed justifiably that the output voltage remains steady at its average value. Power Po absorbed by load resistor is then:

It can be seen from the circuit in Fig. 1 that the current drawn from the source flows through the inductor. Hence the average value of inductor current is also the average value of source current. Let the average inductor current be IL. Then power PS

supplied by the source is then:

After equating equations (13) and (14), we get the average inductor current as:

Since load current Io is:

Using equations (8) and (16), equation (15) can be re-presented as:

Since 0 < D < 1, it can be seen from equation (17) that IL > Io.


The analysis thus far is based on the assumption that the current through the inductor is continuous. The inductor current varies over a cycle, varying between a minimum value and a maximum value. The minimum and maximum values can be expressed in terms of its mean value and its change as expressed in equation (3). That is,

and

It is shown in Fig. 6 how the maximum and the minimum inductor current can be obtained. It is also shown that as the load resistor becomes greater, the average inductor current reduces, but the peak-to-peak ripple in inductor current does not change. It has to be so and expression for ∆IL in equation (3) does not indicate any

term reflecting the load resistor.

For continuous conduction,

At the boundary of continuous and discontinuous conduction,

Another expression for IL is now obtained. Substituting for Vo in equation (15) the

expression in equation (8), we obtain that

Substituting for IL from the equation above and for ∆iL from equation (3), equation

(18) becomes:

and

From equations (23) and (24), the condition for continuous conduction is:

Equation (25) can be interpreted as follows, assuming that only one of the four parameters is varied at a given time with the other three parameters remaining unchanged.

The circuit tends to become discontinuous,

i. if the switching frequency f is decreased, orii. if the duty cycle D is reduced, or

iii. if the load resistance increases, oriv. if the inductance used has lower value.


When the conduction is discontinuous, the voltage across the inductor is zero for part of the cycle since there is no current through the inductor. Let D1T be the time

for which the switch is ON in one cycle and let D2T be the period for which the

diode conducts. Since the conduction is discontinuous,

An expression for the output voltage can be obtained in terms of source voltage, duty cycle D1 of the switch and duty cycle D2 of the diode. Since the net change in

inductor current is over a cycle, the net volt-seconds area associated with the inductor is zero. The waveforms relevant to the inductor when the conduction is discontinuous are shown in Fig. 7. From Fig. 7,

On simplifying, an expression for Vo can be obtained. Then

The value of D1, the duty cycle of the switch, is usually known, but the period for

which the diode conducts is an unknown quantity depending on the other circuit parameters. The value of D2 can be determined in several ways. Here it is

determined using the power balance between the input and output. When the circuit is ideal, the input power equals output power. Let the average source current be IS

and the average output current be Io. Then

Using equation (28), we get that

The average source current be IS can be obtained from Fig. 7. The average source

current is the same as the average inductor current. Let the peak inductor current be ∆IL and the period for which this current flows is (D1T + D2T). This period is the

base of the triangle that defines the inductor current. The average inductor current is obtained as the area of this triangle divided by the cycle period. We have that

Equating equations(30) and (31),

From equation (3),

Substituting for ∆IL from equation (33) in equation (32), we get that

Equation (34) can be re-written as:

Solving for D2,

Equation (36) states how D2 varies as a function of R, D1 , f and L. Once D2 is

known, Vo can be obtained from equation (28).

It is possible to get an expression for Vo as a function of R, D1 , f and L. For this, we

equate the average load current with the average diode current. The average output current can be obtained from the average output voltage and the load resistor. The average diode current is:

Using the expression for ∆IL from equation (33), and replacing the L.H.S. by the

average load current,

Hence we obtain that

By substituting for D2 from equation (36) in the above equation, we can get an

expression for Vo/VS. Alternatively, equation (28) can be re-written as:

Using the expression for D2 from equation (39) in equation (40),

That is,

Solving for the ratio of output to source voltage and taking the positive root of the expression on the R.H.S. of equation (42),

Equation (43) states how (Vo/VS) varies as a function of R, D1 , f and L

Two applets are presented below, the first applet simulates the behaviour of the ideal circuit in open loop, whereas the second applet is about the discontinuous mode of operation.

The first applet presents four types of responses. When the circuit is switched on at a fixed duty cycle with no energy stored in either the inductor or capacitor initially, the transient inductor current happens to be large. If the input voltage has any ripple content, its effect can be seen by selecting either the periodic response or the transient response over one input cycle.

The second applet displays two sets of curves. The first set illustrates how the ratio of output voltage to input voltage varies as a function of the ratio of load resistance to critical resistance for different duty cycles of the switch, where the critical resistance is calculated from equation (25). The critical resistance at a given duty cycle can be stated to be:

The second set of curves illustrates how the duty cycle of the diode varies as a function of the ratio of load resistance to critical resistance for different duty cycles of the switch. The values of critical resistance at various values of duty cycle are

also displayed. The parameters to be set initially are the frequency of operation and the value of inductor.

CLOSED-LOOP CONTROL USING PWM

Under this section, the closed-loop control of the ideal circuit is considered. For both the buck converter and the boost converter, the output voltage increases as the duty cycle of the switch increases and hence the same PWM circuit can be used. The block diagram is also similar, but the transfer function is only slightly different. The block diagram is shown in Fig. 8.

A more detailed block diagram is shown in Fig. 9.

The transfer function of the circuit in open loop is obtained as outlined below by using an approximate method. Even though it is difficult to justify the approximation, it is presented. The output voltage is expressed by equation (8). Let the quiescent duty cycle be assumed to be D and let the small increment to duty cycle be d. Then equation (8) can be presented as:

If d<< (1-D), then the above equation can be approximated using the binomial expansion and retaining only the first two terms. That is,

The change in output, denoted as ∆Vo, can be expressed to be

The transfer function can then be obtained as:

The design of the controller is based on the block diagram in Fig. 10. The third applet has been developed based on this block diagram. The difference between the inductor current and the load current is the capacitor current and the current through

the capacitor is proportional to the derivative of the capacitor voltage and the derivative feedback is obtained as shown. . The third applet presented below displays the location of poles and zeros for the selected parameters. If any poles happen to be in the right-half side of s-plane, the program may not work

For simulation of the power circuit using PWM, closed loop scheme used is shown in Fig. 9. As long as the inductor current is less than the maximum current, the feedback signal is the sum of the output voltage of the filter circuit and a derivative signal obtained from the inductor current and the load current. When the inductor voltage exceeds the set maximum value, the feedback signal contains an additional element proportional to the amount by which the inductor current exceeds the maximum current. The output of the PI controller determines the duty cycle.

The simulation of the power circuit with closed-loop control and pulse-width modulation mode is presented in the fourth applet. One of the five responses can be selected and the program interface is similar to many of the preceding applets. For all the applets to follow, transient responses are displayed without any delay, whereas there can be quite a delay before the periodic responses or the statistical details are displayed.


It is possible to use a multiplier for boost converter also. Using equation (8), an expression for the duty cycle can be obtained to be:

The above equation is valid if the conduction is continuous. The above equation can be realized using a multiplier and a few opamps as shown in Fig. 11. It needs to be mentioned that components should be chosen suitably, reflecting the required scaling. For example, if the multiplier yields an output of 10 V when both its inputs are at 10 V, an input of 10 V should correspond to the maximum input voltage. If the maximum duty cycle is set to be 0.9 and it corresponds to 9 V, then an input of 9 V to the opamp in the feedback path should correspond to the maximum desired output voltage. In this case, the ramp signal with which the output of this circuit is

Power Electronic Circuits

Section 1 - Introduction to Power Electronics

ObjectivesHistory and Applications of Power Electronics

Power Semiconductor Devices

Ratings and Characteristics of Power DevicesControl Characteristics of Power DevicesClassification of Power Semiconductor Switching Devices

Types of Power Electronic Circuits

Design of Power Electronic EquipmentPeripheral Effects

Section 2 - Power Semiconductor Diodes

ObjectivesIntroduction

The Junction Diode

V-I Diode Characteristics

Reverse Recovery Characteristics

Power Diodes Types

Effects of Forward and Reverse Recovery Time

Series Connected Diodes

Parallel Connected Diodes

SPICE Diode Model

Section 3 - Diode Circuits and Rectifiers


Freewheeling DiodesLaplace Transform used to solve equations - Revision

Performance Parameters of Rectifiers

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Single Phase Half-Wave Rectifier (R Load) Example 3.1

Single Phase Half-Wave Rectifier (RL Load) Example 3.2

Single Phase Full Wave Rectifiers - Center Tapped Transformer Example 3.3

Bridge RectifierSolving Exact and Non-Exact First Order Linear Differential Equations - Revision

Three Phase Bridge Rectifier Example 3.4

Rectifier Filter Circuits Example 3.5 Example 3.6 Example 3.7 Example 3.8

Effects of Source Inductance

Section 4 - Thyristors


Thyristor Characteristics

Two - Transistor Model of Thyristor

Thyristor Turn-on

di/dt Protectiondv/dt Protection

Example 4.1

Thyristor Turn-Off

Thyristor Types

Series Operation of Thyristors Example 4.2

Parallel Operation of Thyristors

Thyristor firing circuits

Section 5 - Controlled Rectifiers

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Single Phase Half-Wave Thyristor Converter With R Load Example 5.1

Single Phase Semiconverters

Single-Phase Semiconverter With RL Load

Single-Phase Full Converter

Single-Phase Full Converter with RL load

Three-Phase Half-Wave Converter

Three-Phase Semiconverter

Three-Phase Full Converter

Fourier Transform - Revision

Section 6 - AC Voltage Controllers


On-off Control

Single-Phase Unidirectional Controller

Single-Phase Bidirectional Controller With R Load

Single-Phase Bidirectional Controller With RL Load

Single-Phase Transformer Tap Changers

Three-Phase Full-Wave Controller



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Objectives

Learn the scope and application of power electronics

Become familiar with different types of power semiconductor devices, their characteristics and use

Understand the fundamental techniques of power conversion

History & Application of Power Electronics

History of Power Electronic Devices

Power electronics began with the introduction of the mercury arc rectifier in 1900. This was followed by the first electronic revolution which began in 1948 with the invention of the silicon transistor.

The second electronic revolution began in 1958 with the development of the thyristor. This caused the beginning of a new era for power electronics, since many power semiconductor devices and power conversion techniques were introduced using thyristors.

Next, was the microelectronics revolution which gave the ability to process a huge amount of data in a very short time. The power electronics revolution which merges power electronics and microelectronics provides the ability to control large amounts of power in a very efficient manner.

Definition of Power Electronics

Power Electronics may be defined as the application of solid-state electronics for the control and conversion of electric power. Power Electronics is based on the switching of power semiconductor devices whose power handling capabilities and switching speeds have improved tremendously over the years.

Power Electronics is presently playing an important role in modern technology and is used in a variety of high power products e.g. Motor controls, heat controls, light controls and power supplies.








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Power Semiconductor Devices

Power semiconductor devices can be broken up into five different groups:

1. Power Diodes

2. Thyristors

3. Power Bipolar Junction Transistors (BJT)

4. Power MOSFETs

5. Insulated Gate Bipolar Transistors (IGBT) and Static Induction Transistors (SITs)

Power Diodes

Diode Operation

A diode is a two terminal device consisting of an anode and a cathode. The diode conducts when its anode voltage is more positive than that of the cathode. If the cathode voltage is more positive than its anode voltage, the diode is said to be in the blocking mode.

Diode Types

There are three types of power diodes:

1. General Purpose

2. High-Speed (or fast recovery) - used for high frequency switching of power converters

3. Schottky - have low on state voltage and very small recovery time, typically nanoseconds

Thyristors

Thyristor Operation

A thyristor is a three terminal device consisting of an anode, a cathode and a gate. It is physically made up of four layers of alternate p-type and n-type silicon semiconductor. The terminals connected to the ending p-type and the n-type layers




are the anode and cathode respictively. This configuration will give three p-n junctions. When the anode is held more positive than the cathode, two of the p-n junctions are foward biased, offering very little resistance, and one is reverse biased, offering high resistance.

When a small current is passed through the gate to cathode circuit, and the anode is at a higher potential than the cathode, the thyristor conducts current from anode to cathode. In other words when triggered the thyristor has aproximately the same characteristics as a single diode. Once the thyristor has been turned on, the gate circuit looses control of the thyristor and the forward voltage drop across the device is very small in the region of 0.5 to 2V.

Once on, the device loses control over the anode current, and the only way to turn it off is to reduce the anode current below some value referred to as the holding value. This can be acheived in one of two ways; by making the anode potential equal or less than the cathode potential, due to the sinusoidal nature of an ac voltage which is called line commutation or by the use of an auxilixry as in the case of forced-commutation.

Thyristors can be subdivided into eight groups:

1. Forced-commutated thyristors

2. Line-commutated thyristors

3. Gate turn off thyristors (GTO) - self turned-off device.

4. Reverse conducting thyristors (RCT) - can be considered as a thyristor with an inverse parallel diode, used for high speed switching.

5. Static induction thyristors (SITH) - self turned-off device.

6. Gate assisted turn-off thyristors (GATT) - used for high speed switching.

7. Light activated silicon-controlled rectifier (LASCR)

8. MOS-controlled thyristors (MCTs)

Triacs can be considered as two thyristors connected in inverse parallel and having only one gate terminal. The current flow through the triac can be controlled in either direction. They are used in simple heat, light and motor controls

GTOs and SITHs are self turned-off devices. They are turned on by the application of a short positive gate pulse and turned-off by a short negative gate pulse and do not require a commutation circuit

Because of the nature of the construction of a thyristor, there exists some capacitance between the anode and the gate. If a sharply rising voltage is applied to the thyristor, the associated inrush of charge can switch on te thyristor. These surges can be the result of switching in circuits, and can be accomadated for by providing RC circuits for diverting this surge.

All p-n junctions have a leakage current which increases with increasing temperature. If the temperatture of operation of a thyristor were allowed to rise too much, the leakage current could rise enough to turn on thr thyristor. A possible use of this feature is

in the manufacture of a switching system, which needs to be turned on exceeding certain temperatures.

Bipolar Power Transistors

These are three terminal devices consisting of emitter, base and collector which is operated as a switch in the common emitter configuration. These devices are turned-on when the base-emitter junction is forward biased with the base current sufficiently large to drive the device into saturation. Under these conditions, the collector-emitter voltage drops in a range of 0.5 to 1.5V. If the base-emitter junction is reversed biased the device switches to the off or non-conducting state.

Power MOSFETs

These are three terminal devices consisting of gate, source and drain. It is a fast switching device with a high input impedance and has applications in high switching frequency circuits. MOSFETs are available in both N-channel and P-channel types. To turn-on an N-channel device, the drain is made positive with respect to the source and a small positive voltage is applied to the gate with respect to the source. The device is turned-off when the gate voltage is removed

Power MOSFETs possesses faster switching speeds than power BJTs.

IGBTs

The insulated gate bipolar transistor (IGBT) is a three terminal device consisting of gate, emitter and collector. It combines the low on-state voltage drop characteristics of the BJT with the excellent switching characteristics and high input impedance of the MOSFET. They are available in current and voltage ratings much higher than those found in MOSFETs.

To turn-on the N-channel IGBT the collector must be at a positive potential with respect to the emitter and a positive gate potential will turn-on the device. The removal of this positive gate voltage would turn-off the device






Ratings & Characteristics of Power Devices

The power ratings of some semiconductors are shown in table 1.1, while table 1.2 displays the characteristics and symbols of some power devices

Table 1.1 - Ratings of Power Semiconductor Devices




Table 1.2 - Characteristics & Symbols of Some Power Devices

Figure 1.1 Output voltage and control characteristics of a few power switching devices

Control Characteristics of Power Devices

Power semiconductor devices can be operated as switches by applying control signals to the gate of thyristors, base of power transistors to the gate of power MOSFETs and to the gate of IGBTs. The required output voltage is obtained by varying the conduction time of these devices. Figure 1.1 shows the output voltage and control characteristics of a few power switching devices.





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Classification of Power Semiconductor Switching Devices

Power semiconductor devices can be classified as follows:

1. Uncontrolled turn on and off (Diode)

2. Controlled turn on and uncontrolled turn off (SCR)

3. Controlled turn on and off characteristics (BJT, MOSFET, GTO, SITH, IGBT, SIT MCT)

4. Continuous gate signal requirement (BJT, MOSFET, IGBT, SIT)

5. Pulse gate requirements (SCR, GTO, MCT)

6. Bipolar voltage-withstanding capability (SCR, GTO)

7. Unipolar voltage-withstanding capability (BJT, MOSFET, GTO, IGBT, MCT)

8. Bidirectional current capability (TRIAC, RCT)

9. Unidirectional current capability (SCR, GTO, BJT, MOSFET, MCT, IGBT, SITH, SIT, Diode)

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Types of Power Electronic Circuits

The switching characteristics of power devices permit the control and conversion of electric power from one form to another. These converters are called static power converters and consist of a matrix of switches. Using a combination of these devices allows us to create circuit configurations that allow us to convert between a.c. and d.c. signals.

The resulting power electronic circuits are classified into six types:

1. diode rectifiers

2. ac-dc converters (controlled rectifiers)

3. ac-ac converters (ac voltage controllers)

4. dc-dc converters (dc choppers)

5. dc-ac converters (inverters)

6. static switches

Diode Rectifiers

A diode rectifier circuit converts ac voltage into a fixed dc voltage. Figure 1.2 shows a diode rectifier circuit.




Figure 1.2 Single Phase Diode Rectifier Circuit

Since diodes have a much greater conductivity in one direction than the other, when connected in series with an alternating supply and load, they will produce a direct component of the current during one half cycle of the supply. This property is exploited in the main application of diodes - as rectifiers.

In figure 1.2, the load is connected to the center tap of a transformer. The center tap is such that during the positive phase of the input, the ouput of the secondary is divided equally between the two halves. During this phase, the diode D1 is conducting while D2 is non-conducting. So across the load we will get a positive half cycle.

During the negative half cycle, the diode D2 is conducting while the diode D1 is not, and again we will get a positive half cycle across the load. In this way each diode conducts on alternate cycles, passing current through the load in the same direction.

Ac-dc Converters

An ac-dc converter with two naturally commutated thyristors are shown in figure 1.3.

Figure 1.3 Single Phase ac-dc Thyristor Converter

The circuit shown is similar to the diode rectifier but with thyristors used in the place of diodes. Recall that thyristors need to be fired to turn on. If each thyristor is fired alternately the same effect would be obtained as with the diode rectifier. The average output voltage is controlled by varying the conduction time or firing delay angle α of the thyristors. These converters are also called controlled rectifiers.

Thyristors need to be fired to be turned on, and each one is fired at the same point in the appropriate half cycle. This pont, or delay into the cycle, is termed the firing angle.

Ac-ac Converters

These converters ore used to produce a variable ac output voltage from a fixed ac source. Figure 1.4 shows a single phase converter using a TRIAC. The output voltage is controlled by varying the conduction time of firing/delay angle α of the TRIAC. These converters are also known as ac voltage controllers.

Figure 1.4 Single Phase ac-ac TRIAC Converter

Dc-dc Converters

A dc-dc converter is also known as a chopper or switching regulator. A transistor chopper is shown in figure 1.5. The average output voltage is obtained by controlling the conduction time t of transistor Q1. If T is the chopping period, then t1 = δT. δ is called the duty cycle of the chopper.

Figure 1.5 Dc-dc Converter

Dc-ac Converters

A dc-ac converter is also known as an inverter. A single phase transistor inverter is shown in figure 1.6. We need to switch the transistors on and off in pairs, which is why we did not choose thyristors for this circuit. Transistors M1 and M2 conduct for one-half cycle while M3 and M4 conduct for the other half cycle.

When transistors M1 and M2 are on the voltage VO appears across the load. When the other pair of transistors is on the voltage appears across the load, but in the opposite direction thereby producing an alternating output voltage waveform. The frequency of the output voltage is controlled by varying the conduction time of the transistors.

Figure 1.6 Single Phase dc-ac Converter

Between the time when one pair of transistors is on and the other pair is on, there is a period during which both are off, and the output across the load is zero.

Static Switches

Power devices can be used to operate as static switches or contactors, When the supply to these switches is an ac supply, they are called ac static switches and when the supply is dc, they are called dc static switches.





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Design of Power Electronic Equipment

Power electronic equipment design can be divided into four main areas:

1. Design of power circuits

2. Protection of power devices

3. Determination of control strategy

4. Design of logic and gating circuits

Peripheral Effects

Power converters operate on the basis of switching power semiconductor devices on and off. This switching action of converters introduce current and voltage harmonics into:

1. The supply system

2. The output of converters

The problems caused by these harmonics are:

1. Distortion of the output voltage

2. Distortion of the supply voltage

3. Interference with communication and signaling circuits

4. Reduction of input power factor

The following methods can be used to solve or reduce harmonic problems caused by power converters

1. Use of input and output filters on power converters

2. Choice of control strategy used

3. Grounded shielding

Last updated June 27th 2001








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Objectives

Identify different types of power semiconductor diodes

Understand the importance of forward and reverse recovery times

Describe the methods of connecting diodes in series and in parallel for applications that require higher voltages and currents

Introduction

Function of power semiconductor diodes in power electronic circuits:

1. Switches in rectifiers

2. Freewheeling in switching regulators

3. Charge reversal of capacitor & energy transfer between components

4. Voltage isolation

5. Energy feedback from load to power source

6. Trapped energy recovery








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Junction Diode

Recall that a n-type semiconductor is one in which there are free electrons and the same number of fixed positive ions. Also recall that a p-type semiconductor is one in which there are fixed negative ions and the same number of free moving holes.

Consider a crystal, one half of which has p-type impurity added, and the other half has n-type. Initially the p-type semiconductor has free moving holes, and the n-type has free moving electrons, hence each region is initially neutral. Because of random movement in nature and due to the difference in concentration of carriers, some holes will diffuse across to the n-type material. Likewise some electrons will diffuse across to the p-type material.

The movement of electrons and holes is only restricted to the area immediately around the junction. Becasue of this some point will be reached when the area of the p-type semiconductor closest to the junction will have a build-up of electrons repelling the movement of any more electrons. Similarly, the area of the n-type semiconductor closest to the junction will have a build up of holes repelling the movement of more holes. These positive and negative charges are concentrated near the junction and therefore form a potential barrier between the two regions.

Figure 2.1 pn-junction and diode symbol

Forward Biased Diode

Consider a potential difference applied across the diode, the positive of the supply connected to the p-type material, and the negative connected to the n-type material. Once this potential difference is greater than that created by the concentration of holes and electrons at the junction, the orientation of the magnetic field will produce a drift of holes towards the n-type conductor and electrons towards the p-type conductor. At the junction, free electrons and holes will combine. For each combination, at the p-type terminal, an electron is freed and flows to the supply creating a free hole which moves towards the junction. Similarly at the n-type terminal an electron enters the region from the supply and moves towards the center junction.

With this in mind a diode is said to be forward biased or conducting when the anode potential is positive with respect to the cathode. In this state, the diode has




a small forward voltage drop across it. The magnitude of this voltage drop depends on: -> the manufacturing process and -> the junction temperature.

Reverse Biased Diode

When the cathode potential is positive with respect to the anode, the diode is said to be reverse biased. In such a configuration, the holes will be attracted to the negative electrode, and the electrons will be attracted towards the positive electrode. This will create an area at the junction void of free holes or electrons. This area is called the depletion layer in which there are no charge carriers to facilitate current flow.

In practice, due to thermal agitation, some carriers build up sufficient velocity to jump the gap. This causes a small reverse or leakage current. Since this current is due to the effects of heat, the higher the temperature, the greater this leagake current will be.

In addition, the magnitude of the reverse current increases in magnitude with reverse voltage until the avalanche or zener voltage is reached.





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V-I Diode Characteristics

The steady-state V-I characteristics of a diode is shown in figure 2.2(a). In most cases a diode can be considered to behave as an ideal diode with characteristics shown in figure 2.2(b).

Figure 2.2 V-I characteristics of a diode

The practical characteristics shown in figure 2.2(a) can be expressed by the Shockley diode equation, which is given by:

..............(2.1)

whereID = current through the diode in amperes

VD = diode voltage with anode positive w.r.t cathode in volts

IS = leakage (or reverse saturation) current, usually in the range 10-6 to 10-

15A

n = empirical constant known as emission coefficient or ideality factor which varies between 1 and 2 and depends on material used and physical construction

VT = thermal voltage constant and is given by:

..............(2.2)




whereq = electron charge given as 1.6022 * 10-19 coulomb (C)

T = absolute temperature in kelvin K = 273 + ° C

k = Boltzmann's constant given as 1.3806 * 10-3 J/K

At a specified temperature, the leakagae current, Is is a constant for a specified diode. In addition at a specified temperature, VT can be calculated and is also a constant. Therefore for a given diode at a given temperature:

..............(2.1.a)

Examination of the diode characteristics in figure 2.2(a) reveals three distinct regions:

1. Forward-biased region, where VD > 0

2. Reversed-biased region, where VD < 0

3. Breakdown region, where VD < -VZK

Forward-biased Region

In this region, VD > 0. The diode current ID is small if VD is less than the threshold voltage or cut-in voltage or turn-on voltage written as VTD. This voltage is small and is usually in the range 0.5V to 0.7V. The diode conducts fully if VD is higher than VTD.

The diode equation can be simplified if VD > 0.1 volts

For VD = 0.1 V, n = 1 and VT = 25.8 mV, equation 2.1 can be used to obtain the corresponding value of diode current ID.

with 2.1% error.

For VD > 0.1 volts which is usually the case,

ID >> IS

Hence the diode equation can be approximated to within 2.1% error to

..............(2.3)

Reverse-biased Region

In this region, VD < 0 volts and if |VD| >> VT, or in other words if the magnitude of the diode voltage is much greater than the thershold voltage (which is generally the case), the exponential term in equation 2.1 becomes very small compared to unity and the diode current ID can be written as:

..............(2.4)

Equation 2.4 indicates that the diode current in the reverse direction is constant and equal to IS.

Breakdown Region

When the reverse voltage exceeds the breakdown voltage VBR, the diode is said to be in the breakdown region. In this region, the reverse current increases rapidly for small increases in reverse voltage beyond VBR. Diode operation in the breakdown region is not destructive, provided that the power dissipation is within a safe level as specified by the manufacturer.

It is generally advisable to implement circuitry which will limit the reverse current in the breakdown region. This is because in this region, although it may be a non-destructive region, it should be noted that the slightest change in VD would cause a large change in ID which could damage the device.





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Reverse Recovery Characteristics

The current in a forward-biased junction diode is made up of Majority carriers and Minority carriers. Once there is a forward current, there will be free minority carriers. A forward conducting diode whose forward current has been reduced to zero, will continue to conduct for some small time after due to minority carriers stored in the pn-junction and carriers stored in the bulk semiconductor material.

The forward current in a diode goes to zero if the diode goes from foward biased to reverse biased, or in other words V goes from +ve to -ve. According to the characteristics of a diode, ignoring the leakage current, when reverse biased there should be no reverse current once the reverse voltage does not exceed in magnitude to the breakdown voltage. However, in practice, the diode does exhibit a reverse characteristic for a short space of time due to the free carriers. These minority carriers require some finite time, the reverse recovery time, to recombine with opposite charges in order to be neutralized. This time is called the reverse recovery time.

Two reverse recovery characteristics exists. They are:

1. Soft recovery

2. Abrupt recovery

as shown in Figure 2.3

Figure 2.3 Reverse recovery characteristics

trr = reverse recovery time, measured as the time between the initial zero crossing of the diode current to the time when this current reaches 25% of the peak reverse current.

IRR = maximum reverse current

ta = time between zero crossing and the maximum reverse current and it is due to the charge stored in the depletion region of the junction




tb = time between maximum reverse current IRR and 25% of the of the maximum reverse current IRR and is due to charge stored in the bulk semiconductor material

The reverse recovery time is measured from the initial zero crossing from forward conduction to reverse blocking condition of the diode current to 25% of the maximum reverse current IRR. Its magnitude depends on:

1. junction temperature

2. rate of fall of forward current

3. forward current prior to commutation

From the graph it can be seen that,

..............(2.5)

..............(2.6)

= softness factor (SF)

Reverse Recovery Charge

This is the amount of charge carriers that flow across the diode in the reverse direction due to changeover from forward conduction to reverse blocking condition. Its value is determined from the area enclosed by the path of the reverse recovery current (Recall ∆ Q = ∆ I ∆ t). That is

..............(2.7)

..............(2.8)

From equations 2.6 and 2.8 we get

..............(2.9)

If tb is negligible in comparison to ta which is usually the case, then

Hence equation 2.9 becomes

..............(2.10)

and

..............(2.11)

Ideally diodes should not have a reverse recovery time, and it is possible to construct such a diode. However, the manufacturing cost of such a diode would be quite high for such a feature which in most cases has minor consequences.





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Power Diodes Types

Power diodes can be classified into three categories depending on the recovery characteristics. These types are:

1. Standard or general-purpose diodes

2. Fast-recovery diodes

3. Schottky diodes

General Purpose Diodes

These diodes have a generally high reverse recovery time, typically around 25 µs. Used in low speed applications e.g. rectifiers and converters with frequencies up to 1 kHz, where the recovery time is not critical.

Fast-recovery Diodes

These diodes have a low recovery time, typically around 5 µs. They are used in applications where the recovery time is of critical importance such as dc-dc and dc-ac converters.

Schottky Diodes

The barrier potential is accomplished with a contact between a metal and a semiconductor. This barrier simulates the behavior of a pn-junction

Recovery charge of this diode is much less than the equivalent pn-junction diode. It is due only to junction capacitance and is independent to reverse di/dt

It has a relatively low forward voltage drop

The leakage current is higher than that of a pn-junction diode

They are mainly used in high current low voltage power supplies





Effects of Forward and Reverse Recovery Time

Consider the chopper circuit shown below without a di/dt limiting inductor. This circuit would be used to display the effects of forward and reverse recovery time.

Figure 2.4 Chopper circuit without di/dt limiting inductor

The switch SW is turned on at time t = 0 and remains on long enough for a steady load current to flow. This load current is given by

..............(2.12)

Under these conditions, the freewheeling diode Dm is reverse-biased. Say switch SW is opened at timet = t1. Inductor L will now discharge its stored energy through diode Dm.

Switch SW is turned on again at time t = t2. Now at time t2, diode D1 would conduct almost instantaneously, and diode Dm goes instantaneously from foward baised to reverse biased. As discussed earlier, there would be free carriers which




would result in a reverse current. This reverse current, IRR, is directly proportional to di/dt. And since the fall of foward current of Dm would be very high, this reverse current would also be high, as seen from equation 2.13.

..............(2.13)

Since diode Dm would have free carriers, it would conduct in the reverse direction. This would cause it to behave as a short in the circuit. This short would cause the rate of rise in the foward current od D1 to be very high and consequently increate the reverse current of Dm. This run-away effect could damage the diodes.

This problem can be overcome by connecting a current limiting inductor in series with diode D1.





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Series-Connected Diodes

In high-voltage applications one commercially available diode is unable to meet the required voltage ratings of a circuit. Connecting diodes in series increases the reverse blocking capabilities of the diodes. Each diode must carry the same leakage current, and have the same blocking voltage. However, in reality even two diodes of the same part number will not have the same characteristics due to tolerances in the production process. This is shown for two diodes of the same part number connected in series in Figure 2.5, along with their v-i charactieristics. This gives rise to problems when diodes are connected in series, since the blocking voltages will differ slightly.

Figure 2.5 Two Series Connected Diodes in Reverse Bias

From the graphs it can be seen that in the forward-biased condition, both diodes conduct the same amount of current and the forward voltage drop for each diode would be almost equal. In the reversed-biased condition, however, where each diode has to carry the same leakage current, the blocking voltage would differ significantly as shown in figure 2.5b.

This problem is solved by forcing equal voltage sharing by connecting a resistor across each diode as shown in figure 2.6.




Figure 2.6 Series-Connected Diodes with Steady-State Voltage Sharing

Due to the equal voltage sharing the leakage current of each diode would be different as shown in figure 2.6b.

The relationship between the resistors for equal voltage sharing is developed below

In this arrangement, the total leakage current must be shared by a diode and a resistor. Hence

..............(2.14)

but we need to get,

..............(2.15)

Now we know that,

and

Hence using equation 2.14 under conditions of equal voltage sharing yields

..............(2.16)

Hence the relationship between the two resistors for equal voltage sharing is given by equation 2.16.





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Parallel-Connected Diodes

In high power applications, diodes are connected in parallel to increase the current carrying capability in order to meet circuit requirements. In parallel operation of diodes, current sharing depends on the magnitude of their forward voltage drops. Uniform current sharing can be achieved either by the use of equal inductances or by connecting current sharing resistors, the later option may not be practical due to power losses incurred by the resistive components. Selecting diodes with equal forward voltage drops would minimise the unequal sharing of current.

Figure 2.7 Parallel-Connected Diodes

For steady-state current sharing, the circuit of figure 2.7a with series resistors are used.

Dynamic current sharing is achieved with the use of coupled inductors as indicated by figure 2.7b. If the current through diode D1 rises, then the voltage across inductor L1 (Recall VL = L di/dt) increases, causing a voltage of opposite polarity to be induced across inductor L2. This causes a low impedance path for current flow through diode D2, and more current is shifted through this diode.

The disadvantage of using current sharing devices under dynamic conditions is that the inductors would generate voltage spikes and would be expensive and bulky.









SPICE Diode Model

The SPICE model for a diode is shown in figure 2.8 below.

Figure 2.8 SPICE Model for Diode




The diode current ID depends on the diode voltage VD and is represented by a current source.

Resistance RS is the bulk resistance of the semiconductor material and depends on the amount of doping.

Capacitance Cd is a nonlinear function of the diode voltage vd and is represented by

where qd is the depletion layer charge.





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Section 3 - Diode Circuits And Rectifiers

Objectives

Understand the importance of freewheeling diodes in power electronic circuits.

Examine the harmonic distortion of voltage and current on the load and supply caused by diode rectifiers.

Examine the performance parameters of rectifiers.

Select the ratings of diodes for rectifier circuits.

Select the parameters of simple input and output filters for diode circuits.

Introduction

Diodes are widely used in power electronics circuits for the conversion of electric power and power processing. Ac-dc converters, usually called rectifiers, utilize diodes and provide a fixed output dc voltage. The diodes considered in the application circuits to follow would be considered to be ideal, i.e. having negligible reverse recovery time and forward voltage drop.








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Freewheeling Diodes

Say we have a circuit with an inductive load in it. From the instant that electrical power is supplied to the circuit (by the closing of a switch say), the inductive load will accumulate stored energy. If an attempt is made to open the switch this energy will arc across the contacts of the switch, and could cause damage to the circuit components. Freewheeling diodes are placed across inductive loads to provide a path for the release of energy stored in the load when the load voltage drops to zero.

Figure 3.1 shows a diode circuit with a freewheeling diode. Diode Dm is the freewheeling diode. The circuit operation is divided into two modes. Mode 1, which begins when the switch is closed, and Mode 2, which begins when the switch is opened.

Mode 1

Mode 1 begins when the switch is closed at time t = 0. The equivalent circuit for this mode is shown in figure 3.1b.

Examination of the equivalent circuit for Mode 1, using voltage simple analysis, reveals the following:

Now we know that for an inductor

Therefore

..............(3.1)




Figure 3.1 Circuit With Freewheeling Diode

We need to obtain an equation for the current in the circuit. This can be obtained by taking the Laplace transform of the equation, making I(s) the subject of the formulae, and then taking the inverse Laplace.

Recall The Laplace Transform of a function F(t) is denoted by LF(t) and is defined by

......(L.1)

The Integral above is a function of the parameter s; therefore we call that function F(s) such that F(s) = LF(t)

An important property of the Laplace function is its linearity, similar to the Differential function. If F1(t) and F2(t) have Laplace transforms and if c1 and c2 are arbitrary constants,

......(L.2)

An important Theorem (that will not be proven here) in working in the Laplace plane is as follows:If F(t) is continuous for t≥ 0 and is also of exponential order as t → ∞ , and if F´(t) is of "Class A"*, then

LF´(t) = sLF(t) - F(0) ......(L.3)

Where a function of Class A is one that is

a. Sectionally Continuous over every finite interval in the range t ≥ 0

b. Of exponential order as t → ∞

We can also say, from L.3, that

LF′′(t) = sLF′(t) - F′(0) LF′′(t) = s[sLF(t) - F(0)] - F′(0)

F′′(s) = s²F(s) - sF(0) - F′(0) ......(L.4)

and the process can be repeated as many times as we wish for more derivatives.

Let us assume that the initial conditions are zero, i.e i(t=0)=0. The Laplace transform can be applied to equation 3.1, using L.4 yielding

..............(3.2)

Solving for I1(s) yields

..............(3.3)

where

..............(3.4)

The inverse transform of equation (3.3) in the time domain yields,

..............(3.5)

When the switch is opened at time t = t1 at the end of this mode, the current becomes

..............(3.6)

If t1 is sufficiently long such that the exponential term in equation 3.6 becomes negligible, then the steady-state current is given by

..............(3.7)

Mode 2

This mode begins when the switch is opened and the load current starts to flow through the freewheeling diode Dm. The voltage equation for this mode is given by

..............(3.8)

with initial conditions

Applying the Laplace transform to equation 3.8 yields

and solving for I2 yields

..............(3.9)

where

Applying the inverse Laplace transform to equation 3.9 yields

..............(3.10)

Equation 3.10 is the freewheeling current which decays exponentially to zero at time t = t2.





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Performance Parameters of Rectifiers

The performance of rectifiers are evaluated using the following parameters:

1. The average value of output (load) voltage given as Vdc

2. The average value of output (load) current given by Idc

3. The output dc power given by Pdc = VdcIdc

4. The rms value of output voltage given as Vrms

5. The rms value of output current given as Irms

6. The output ac power given by Pac = VrmsIrms

7. The efficiency or rectification ratio of a rectifier given by η =

8. The output voltage consists of two components an ac component and a dc component. The effective or (rms) value of the ac component of output voltage is given by

9. The form factor which is a measure of the shape of the output voltage is given by

10. The ripple factor which is a measure of the ripple content is given by

By substituting the equation for the effective value of the qac component of the output voltgae into the ripple factor equation, we can express the ripple factor as

11. The transformer utilization factor is defined as

where Vsand Isare the rms voltage and rms current of the transformer




secondary respectively.

Figure 3.2 - Waveforms of Input Voltage & Current

Consider the waveforms in figure 3.2

vsis the sinusoidal input voltage

isis the instantaneous input current

is1is the fundamental component of is

12. The displacement angle φ is the angle between fundamental components of input current and voltage

13. The displacement factor (DF) or Displacement Power Factor (DPF) is defined as

14. The harmonic factor (HF) also known as total harmonic distortion (THD) is a measure of the distortion of a waveform. The harmonic factor of the input current is given as

Where both currents are recorded as rms values

15. The crest factor is a comparison of the peak input current to its rms value. It is given as

For a pure sinusoidal input current and voltage, power factor is defined as the cosine of the load angle φ, i.e.

where the voltages and currents are stated in rms values and

where

For a rectifier circuit, the input power factor is given by

Power factor in rectifier circuits is related to harmonic voltages and currents given in the diagrams

where

Hence,

An ideal rectifier should have η = 100%, Vac = 0, RF = 0, TUF = 1, HF = THD = 0, and PF = PDF = 1.





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Single Phase Half-Wave Rectifier ( R Load)

A rectifier is a circuit that converts an ac voltage to a dc voltage. A single-phase half-wave rectifier, shown in figure 3.3, is the simplest type, and not commonly used in many applications. It is however, useful in understanding its operation as they can be applied even to the complex rectifier circuits.

Figure 3.3 - Single Phase Half-Wave Rectifier

During the positive half cycle of input voltage, diode D1 conducts and the input voltage appears across the load. During the negative half cycle of input voltage, the diode is in the reverse biased condition, so it does not conduct and the output voltage is zero. A point to note is that when in foward baising, there is a voltage drop, VTD, across the diode equal to the turn on voltage. This voltage is small, usually 0.5 to 0.7 Volts, and is generally assumed to be zero volts.

Disadvantages of this rectifier:




1. Dc output voltage is discontinuous and contains harmonics.

2. Input current is not sinusoidal

The performance of this half wave rectifier with resistive load is examined in Example 3.1






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Section 3 - Diode Circuits And RectifiersExample 3.1

Consider the circuit below with a purely resistive load R. Determine

a. the efficiency

b. the ripple factor

c. the transformer utilization factor

d. the peak inverse voltage (PIV) of diode D1

e. form factor

f. the crest factor of the input current

Solution

Now for time interval ½T ≤ t ≤ T, vL(t) = 0 hence

But and

∴




For the output voltage from the diode rectifier,

∴ The rms value of the output voltage is given by

a. Efficiency,

Pdc = VdcIdc and Pac = VrmsIrms

∴

b. Form factor

c. Ripple factor

d. Transformer utilization factor

RMS value of transformer secondary voltage Vs is given by

The rms value of the transformer secondary current is the same as the rms value of the load current

Note:VsIs= Volt-ampere rating (VA) of the transformer

e. The peak inverse voltage is the peak reverse voltage seen by the diode and is equal to Vm.

f. Crest factor

These results reveal

a. the rectifier has a high ripple factor of 121%

b. a low efficiency of 40.4%

c. a poor TUF of 0.286

d. the transformer must carry a dc current which may result in dc saturation problems of the transformer core.





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Single Phase Half-Wave Rectifier ( RL Load)

The half wave rectifier with an inductive load (RL) is shown in figure 3.4.

Figure 3.4 - Half-Wave Rectifier With RL Load

The operation of the circuit is as follows:

As in the case of a resistive load, the diode turns on when its anode is positive w.r.t its cathode, and the foward voltage is greater than the threshold voltage. Assuming a turn-on voltage of zero volts, the voltage across the load is the same as the positive half cycle of the ac source.

During the interval 0 to π/2

The source voltage vs increases from zero to its positive maximum, while the voltage across the inductor vL opposes the change of current through the load. It must be noted that the current through an inductor cannot change instantaneously, hence the current gradually increases until it reaches its maximum value. The current does not reach its peak when the voltage is at its maximum, which is consistent with the fact that the current through an inductor lags the voltage across it. During this time, energy is transferred from the ac source and is stored in the




magnetic field of the inductor.

For the interval π/2 and π

The source voltage decreases from its positive maximum to zero. The induced voltage in the iductor reverses polarity and oposes the associated decrease in current, thereby aiding the diode foward current. Therefore, the current starts decreasing gradually at a delayed time, becoming zero when all the energy stored by then inductor is released to the circuit. Again this is consistent with the fact that current lags voltage in an inductive circuit. Hence, even after the source voltage has dropped past zero volts, there is still load current, which exists a little more than half a cycle.

For the interval greater than π

At π, the source voltage reverses and starts to increase to its negative maximum. However, the voltage induced across the inductor is still positive and will sustain forward conduction of the diode until this induced voltage decreases to zero. When this induced voltage falls to zero, the diode will now be reversed biased, but would have conducted forward current for an angle θ, where θ = π + σ. σ is the extended angle of current conduction due to the energy stored in the magnetic field being returned to the source.

The instantaneous supply voltage vs is given by

vs = vR + vL

From figure 3.4, for angles less than ωt2 the inductor is storing energy in its magnetic field from the source and the inductor voltage would be such as to oppose the growth of current and the supply voltage. For angles greater than ωt2 the inductor voltage would have reversed and would aid the supply voltage to prevent the fall of current. Hence, the average inductor voltage is zero.

The average output voltage is given by

The average load current is given by

The addition of a freewheeling diode

The average dc voltage varies proportionately to [1 - cos(π + σ)]. This can be made to be a maximum, thereby increasing the average dc voltage, by making cos(π + σ) a maximum. The maximum value that this can take is given by cos(π + σ) = 1, which can be obtained if σ = 0. We can make σ = 0 with the addition of a freewheeling diode given by Dm as shown with the dotted line.

When the supply voltage goes to zero, the current from D1 is transferred across to diode Dm. This is called commutation of diodes. The result is the charge in the inductor will be used to keep diode Dm on, instead of previously forcing D1 to remain in its foward state. This would reduce the value of the extended angle of conduction of the diode D1, σ to zero.

Intuitively we can see that if the value of the inductance is high, it will store more charge and therefore be able to keep diode Dm on for a longer time. In other words, if the circuit has a long time constant given by

then the inductor would be able to keep diode Dm on for the entire duration of the negative half-cycle, and by so doing, maintain a continuous load current.

Example 3.2






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Consider the circuit shown below, where the battery voltage E = 12V and its capacity is 100 W-h. The average charging current should be Idc = 5A. The primary input voltage is Vp = 120 V, 60 Hz and the transformer has a turns ratio of n = 2:1.

Calculate

a. the conduction angle σ of the diode

b. the current-limiting resistor R

c. the power rating PR

d. the charging time, h, in hours

e. the rectifier efficiency η and

f. the peak inverse voltage PIV of the diode

The waveforms of the voltage and load current are shown below.

Solution

E = 12 V




Vp = 120 V

Vm = √2 Vs = √2(60) = 84.85 V

a.

for Vs > E the diode D1 conducts.

The angle α when the diode starts to conduct is found from:

Vm sin α = E

α = 8.13°

β = 180° - α = 180° - 8.13° = 171.87°

conduction angle δ= β- α= 171.87° - 8.13°

δ = 163.74°

b. The average charging surrent Idc is given by

β = π - α

R = 4.26 Ω

c. The rms battery current Irms is given by

= 67.4

Irms = √67.4 = 8.2A

Power rating of resistor = = 286.4 W

d. Power delivered to the battery = Pdc

Pdc = IdcE = 5 x 12 = 60 W

But WH rating of battery = 100 WH

∴ 100 WH = 60 x h

e. Rectifier efficiency η is given by

f. The PIV of the diode is

PIV = Vm + E = 84.85 + 12 = 96.85





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Ronald De FourSection 3 - Diode Circuits And Rectifiers

Single Phase Full Wave Rectifiers

Two types of full wave rectifiers exist. They are:

1. That formed by using a center-tapped transformer and two diodes and

2. That formed by using a transformer and four diodes also known as a bridge rectifier.

Full-Wave Rectifier With Center-Tapped Transformer (R Load)

The circuit and associated waveforms are shown in figure 3.5. Each half of the transformer with its associated diode acts as a half-wave rectifier. On the positive half cycle of supply voltage, diode D1 is forward biased while diode D2 is reversed biased. On the negative half cycle of supply voltage, diode D2 is forward biased while diode D1 is reversed biased.

Figure 3.5 - Full-Wave Rectifier With Center-Tapped Transformer

Each diode conducts on alternate half cycles of supply voltage producing a full-wave output voltage across the load. Dc saturation of the transformer core does not exist here since there is no dc current flowing through the transformer.





The peak inverse voltage of the diodes is 2Vm

Comparison Between Half-Wave & Full-Wave Center-Tapped Rectifiers R load

Table 3.1

Performance Parameter Half-Wave Rectifier (R Load)

Full-Wave Center-Tapped Transformer Rectifier (R

Load)

Efficiency (η) 40.5% 81%

Form Factor (FF) 157% 111%

Ripple Factor (RF) 121% 48.2%

Transformer Utilization Factor (TUF) 28.6% 57.32%

Peak Inverse Voltage (PIV) Vm 2Vm

Crest Factor (CF) 2 1.414

Table 3.1 shows that the full-wave center-tapped rectifier has improved performance over the half-wave rectifier.

Example 3.3






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The rectifier shown in the figure below has a purely resistive load. Determine

a. the efficiency

b. the form factor

c. the ripple factor

d. the transformer utilization factor

e. the peak inverse voltage of the diode D1 and

f. the crest factor of the input current

Solution



The rms value of the output voltage is given by

a.




b. Form factor

c. Ripple factor

d. Transformer utilization factor

Rms value of transformer secondary voltage = vs

This is the value of the transformer secondary voltage for ½ of the secondary winding.

Rms value of transformer secondary voltage associated with this ½ winding secondary voltage is the same as the (rms) load current for this ½ winding.

= rms value of load current for this ½ winding

The Volt-Ampere rating of the transformer

e. PIV = 2Vm

f.





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Single Phase Full Wave Rectifiers

Full-Wave Bridge Rectifier R Load

As mentioned earlier, one can also implement a single-phase full-wave rectifier using four diodes. The diagram of the full-wave bridge rectifier and associated waveforms are shown in figure 3.6.

On the positive half cycle of transformer secondary supply voltage, diodes D1 and D2 conduct, supplying this voltage to the load. On the negative half cycle of supply voltage, diodes D3 and D4 conduct supplying this voltage to the load.

It can be seen from the waveforms that the peak inverse voltage of the diodes is only Vm The average output voltage is the same as that for the center-tapped transformer full-wave rectifier.

Figure 3.6 - Full-Wave Bridge Rectifier With R load and Associated Waveforms

Single Phase Full-Wave Bridge Rectifier With RL Load

With a resistive load, the load current is identical in shape to the output voltage. Most loads are inductive and the load current with these loads depends on the value of load resistance and load inductance.

Figure 3.7 is a circuit of a bridge rectifier with an inductive load and a battery.




Figure 3.7 - Bridge Rectifier With RL Load & Battery

The input voltage is given by:

and the load current can be obtained from the solution of the differential equation

.............(1)

Recall The above equation is a non-exact first order linear differential equation, hence by definition it must be of the form

A(x) dy/dx + B(x)y = C(x) .............(LE.1)

If an equation is not exact, it is natural to make it exact by the introduction of an appropriate factor, which is then called an integrating factor. By dividing each member of equation LE.1 by A(x) we obtain

dy/dx + P(x)y = Q(x) .............(LE.2)

For the moment, let us assume that there exists for equation LE.2 a positive integrating factor v(x) > 0, a function of x alone. Then

v(x)[dy/dx + P(x)y] = v(x)Q(x) .............(LE.3)

must be an exact equation. But equation LE.3 can be easily put into the form

[v(x)P(x)y - v(x)Q(x)]dx + v(x)dy = 0 .............(LE.4)

which is of the form of an exact first order differential equation.

Mdx + Ndy = 0 .............(LE.5)

where M = vPy - vQ and N = v, and v, P, and Q are functions of x alone.

Recall Say we have an equation of the form

M(x,y)dx + N(x,y)dy = 0 .............(E.1)

in which seperation of variables is not possible. Suppose, also, that a function f(x,y) can be found such that it has for its differential the expression Mdx + Ndy = 0; that is

dF = Mdx + Ndy .............(E.2)

Then certainly

F(x,y) = c .............(E.3)

following from the fact that dF = 0. Now we know that if an equation E.1 is exact then a function F exists which will satisfy E.2. But from calculus

.............(E.4)

so

.............(E.5)

These two equations lead to

.............(E.6)

therefore

.............(E.7)

Therefore if equation LE.5 is to be exact it follows from the requirement of equation E.7 that v must satisfy the equation

vP = dv/dx .............(LE.6)

Solving LE.6 we can readily get v

.............(LE.7)

In other words, we have shown that if equation LE.2 has a positive integrating factor independant of y, then that factor must be given by equation LE.7. Let us multiply equation LE.2 by the integrating factor, obtaining

.............(LE.8)

Now the left member of LE.8 is the derivative of the product

.............(LE.9)

and the right side is a function of x only. Hence equation LE.8 is exact, which is what we wanted to show. Now let us proceed to solve and obtain a general solution for y. Integrating equation LE.8 with respect to x is done as shown below

.............(LE.10)

Solving for y yields a general solution as shown in equation LE.11

(where r = Q, and h = ).

.............(LE.11)

Now applying this solution of a first order linear differential equation to equation (1) which is

.............(1)

which can be rewritten as

.............(16)

where

yields the solution of equation (2) as

.............(3)

.............(4)

Now

where

Therefore

and

.............(5)

where Z is the load impedance given by

Continuous Load Current

Examination of the load current waveform for this circuit reveals the following:

1. at ωt = π, iL = I1

2. at ωt = 0, iL = I1

hence the value of the constant C in equation (5) and the value of I1 can be determined by imposing these conditions in the load current equation.

Imposing the condition at ωt = π, iL = I1

on the load current equation

yields

.............(6)

Substituting for C in equation (5) yields

.............(7)

Under steady-state conditions

Applying this condition to equation (7) yields

where

.............(8)

Now substituting for I1 in equation (7) yields

.............(9)

for and

Equation (9) is the equation for the continuous load current of a bridge rectifier with an inductive load and a battery in series with the load.

The rms diode current for diodes D1 and D2 is given by

.............(10)

and the rms output current can be obtained by combining the rms diode current of diodes D1 and D2 with that of diodes D3 and D4 and is given by

.............(11)

The average diode current is given by

.............(12)

Discontinuous Load Current

Under these conditions, the load current flows only for the interval given by

The diode starts to conduct at ωt = α where α is given by

At ωt = α, and the general solution of the load current equation which is given by

yields

Substituting for C in the load current equation yields

At ωt = β, the load current falls to zero and .

Substituting this condition in the load current equation yields

The value of β can be determined from this transcendental equation by an iterative method of solution. Starting with β = 0, and increasing its value by a very small amount until the left hand side of the equation becomes zero.

Knowing α, β and the load current equation, the rms current through diodes D1 and D2 is obtained from

and the average diode current from


Three-Phase Bridge Rectifier

The diagram for a three-phase bridge rectifier is shown in figure 3.8. together with the supply voltage, output voltage and diode current waveforms.

Figure 3.8 - Three-Phase Bridge Rectifier & Associated Waveforms




The diodes are numbered in order of conduction sequences where the conduction sequence for the diodes is 12, 23, 34, 45, 56 and 61. In one cycle, each pair of diodes conducts for 60° and each diode conducts for 120°. The pair of diodes connected to the supply lines with the highest instantaneous line-to-line voltage will conduct. Since there are six commutations taking place in one cycle this rectifier is also called a six-pulse converter.

Three-Phase Bridge Rectifier R Load

Taking the line voltage waveform as a cosine wave, since each pair of diodes is on for 60° then they begin conduction at -30° and end at +30°.

If Vm is the peak phase voltage of the transformer secondary winding, and there are six pulses in one cycle, then the average output voltage is given by

The rms output voltage is given by

The peak current through the diode is given by Im where

The rms diode current is obtained by integrating over 60° and multiplying the integral by 2 since each diode conducts for 60° in a cycle. This current is given by

From the line current waveform, it is clear that the current in line-a flows for four sixty degrees in one cycle. Hence, the rms value of transformer secondary current is given by

Example 3.4

Performance Parameters of Half-Wave, Full-Wave Center-Tapped & 3-Phase Bridge Rectifiers R load

Table 3.2

PerformanceParameter

Single PhaseHalf-

Wave Rectifier(R Load)

Single PhaseFull-Wave Center-

Tapped Transformer

Rectifier(R Load)

Three-Phase BridgeRectifier

(R Load)

Efficiency (η) 40.5% 81% 99.83%

Form Factor (FF) 157% 111% 100.08%

Ripple Factor (RF) 121% 48.2% 4%

Transformer Utilization Factor

(TUF)28.6% 57.32% 95.42%

Peak Inverse Voltage (PIV)

Vm 2Vm √3Vm

Crest Factor (CF) 2 1.414

From the above table, it is clear that the three-phase bridge rectifier is far superior with respect to performance parameters in comparison to the single phase half wave and single phase full wave center-tapped rectifiers.

Three-Phase Bridge Rectifier RL Load

The instantaneous output voltage of this rectifier is given by

for .............(1)

The load current can be obtained from the solution of the equation

.............(2)

The solution is of the form


.............(3)


and

Applying the condition at

, IL = I1

to equation (3) the constant C can be determined. Application of this condition yields

Substituting for C in equation (3) yields

Under steady-state conditions

Applying this condition to the above equation gives the value of I1 as

Substituting for I1 in the equation of instantaneous load current yields

for

Since each diode conducts for 120°, the rms diode current is twice the rms current over the 60° interval in a cycle. This is given by

The rms output current is obtained by combining the rms current of each diode over a cycle. This is given by

The average diode current is given by





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A three phase rectifier has a purely resistive load of R. Determine

a. the efficiency

b. the form factor

c. the ripple factor

d. the transformer utilization factor

e. the peak inverse voltage of each diode and

f. the peak current through a diode.

The rectifier delivers Idc = 60 A at an output voltage of Vdc = 280.7V and the source frequency is 60 Hz.

Solution

a. Efficiency,

NowPdc = VdcIdc

where Vdc= average output voltage Idc= average output (load) current

Pac = VrmsIrms

where Vrms= rms value of output voltage Irms= rms value of output (load) current

Since

where Vm= peak voltage of transformer secondary




then

and sinceVrms = 1.6554Vm

then

b. Form factor

c. Ripple factor

Now, the output voltage is comprised of two componentsa. a dc value b. an ac or ripple component


d.

where Vs = rms voltage of transformer secondary Is = rms current of transformer secondary

Vs = 0.707VmIs = current in one line of transformer = 0.7804Imwhere Im = peak secondary line current

But

∴ VA rating of transformer = 3VsIs

e. Vm = peak line to neutral voltageBut Vdc = 1.654Vm = 280.7 V

PIV= peak value of secondary line to line voltage = 169.7√3 = 293.9 V

f. The average diode current Idc is given by

If the average load current is Idcand each diode is on for 120° of a cycle of 360°then average diode current = average load current






Rectifier Filter Circuits

It has been observed that the output of rectifiers contain harmonics.

DC filters are used to smooth out the dc output voltage of rectifiers.

DC filters are usually L type, C type or LC type, these are shown in figure 3.9.

Due to the rectification action, the input current of a rectifier contain harmonics.

An ac filter of LC type is used to reduce the harmonic content of the supply current. An ac filter is shown in figure 3.9.

The examples given below would examine the steps involved when designing ac and dc filters for rectifier circuits.

Figure 3.9 - AC & DC Filters

Example 3.5

Example 3.6

Example 3.7

Example 3.8








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A single phase bridge rectifier is supplied from a 120 V 60 Hz source. The load resistance isR = 500 ohms.

a. Design a C filter so that the ripple factor of the output voltage is less than 5%

b. With the value of the capacitor C in part (a), calculate the average load voltage Vdc

Solution

i. Charging of capacitorWhen the instantaneous voltage vs is higher than the capacitor voltage vc, diodes D1 and D2 or D3 and D4 conduct supply voltage thereby charging the capacitor.

ii. Discharging of capacitorWhen the instantaneous supply voltage vs falls below the instantaneous capacitor voltage vc diodes D1 and D2 or D3 and D4 are reverse biased and the capacitor discharges through the load resistor.

iii. The capacitor voltage varies between Vcmin and Vcmax and this is known as the output ripple voltage.

iv. The output voltage waveforms and equivalent circuit on charging and discharging are shown below.




At t = t1, the capacitor is charged at the peak supply voltage Vm.

The equation for discharge of the capacitor is given by

differentiating we get

Integrating bith sides yields

Let ek = K

Applying the initial condition of discharge:

at t = 0, vc = Vm and

Hence the output or capacitor voltage during the discharge period is given by

The peak to peak ripple voltage Vr(pp) is given by

Vr(pp) = vL(t=0) - vL(t=t2)

Now e-x ≈ 1 - x

For large R, t1 « t2

hence

but

The average load voltage Vdc is given by

The rms output ripple voltage Vac can be approximated to

Ripple factor RF is given by

b.


Reduction of harmonics in output voltage with the use of an LC filter.

An LC filter shown below is used to reduc the ripple content of the output voltage for a single phase full wave rectifier. The load resistance is R = 40 ohms, load inductance L = 10 mH and source frequency is 60 Hz.

a. Determine the values of Le and Ce so that the ripple factor of the output voltage is 10%.

Solution

To make it easier for the nth harmonic ripple current to pass through the capacitor, the load impedance must be 10 times greater than the capacitor reactance, ie.,

The rms value of the nth harmonic component appearing in the output is obtained using

where Vn = rms value of the nth harmonic

The total rms ripple voltage due to all harmonics is given by

The output voltage of a bridge rectifier can be described by a power series given by




where

∴ bn = 0

The computation can be simplified by only considering the dominant harmonics. This can be seen as the second harmonic whose rms value is given by

Ce can be obtained using

∴ Le = 30.83 mH





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This example examines an input ac filter design in order to keep the harmonic line current within a specified value.

An LC filter shown below is used to reduce the input current harmonics in a single phase full wave rectifier given below. The load current is ripple free and its average value is Ia If the supply frequency is f = 60 Hz determine

a. the resonant frequency of the filter so that the total input harmonic current is reduced to 1% of the fundamental component.

b. the HF of the input current

c. and input PF in that filter.




Solution

a. The equivalent circuit of the nth harmonic current is shown above where:

Isn = rms value of the nth harmonic current appearing in the supplyIn = rms value of the nth harmonic current produced by the rectifier switching action

NOTE: It is Isn that must be reduced to a specified value relative to the fundamental current so that:

a. the supply would have a small harmonic content andb. the remainder of the harmonic current would flow through the capacitor

and diodes

The potential difference across Li and Ci due to the current source In(ω) is given by

The total rms harmonic current flowing in the supply is given by Inwhere

The harmonic factor r is given by

Now the input current expressed in a fourier series is given by i1(t) where

where

The rms value of the fundamental current is given by Is1where

and the rms value of the nth harmonic is given by Inwhere

If we only consider the third harmonic

Now the filter circuit would be in resonance when

NOTE: At this frequency 30.72 Hz the filter would ensure that the total input harmonic current is reduced to 1% of the fundamental component.

b. Harmonic factorThe rms value of the input current is given by

HF = 48.43 %

c. Power factorFrom the waveform, the displacement angle φ is zero,hence

displacement factor DF = cos φ = cos 0 = 1

PF = 0.9 lagging





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Effects of Source Inductance

The effects of source inductance were ignored in the previous derivations of output voltages produced by rectifiers.

The effect of source inductance is to reduce the rectifier output voltage.

This effect can be observed with the use of the three-phase bridge rectifier and associated waveforms in figure 3.10.

Figure 3.10 - Three-phase Bridge Rectifier With Source Inductance

The effect of the source inductances are as follows:

In this circuit, the diodes with the most positive voltage will conduct.




Circuit operation at the point where ωt = π will now be considered.

Just before ωt = π, Vac is more positive than the other two line voltages and as a result, diodes D1 and D2 supply load current.

At the point just after ωt = π, line voltage Vbc is the most positive and diode D1 is turning off while diode D3 is turning on to supply load current along with diode D2.

But at ωt = π, line voltages Vac and Vbc are both equal, and load current is still flowing through diode D1.

Due to the source inductance L1, the load current flowing through diode D1 cannot fall to zero immediately for load current to be transferred to diode D3.

As the current through diode D1 given as id1 decreases, a voltage +vL1 is induced across inductor L1. The resulting output voltage is given by vL, where

vL = vac + vL1

At the same time, id3 the current through diode D3 increases from zero, inducing an equal voltage-vL2 across inductor L2. The resulting output voltage becomes

vL = vbc - vL2

The result is that the anode voltages of diodes D1 and D3 are equal and both diodes conduct for a specified period called the commutation or overlap angle µ.

The transfer of current from one diode to another is called commutation and the reactance corresponding to the inductance causing the overlap is called the commutating reactance.

The effect of this overlap is to reduce the average output voltage of the rectifier. The reduction in output voltage is determined using the following method:

The voltage across the inductor L2 is given by

Assuming a linear rise of current from zero to Idc , we can write

This process of transferring current from one diode to another occurs six times for a three-phase bridge rectifier, hence the average reduction in voltage due to the presence of commutating inductances is given by Vx where

If all the inductances are equal then





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Objectives

Understand the different types of thyristors and their characteristics.

Examine the methods of turning on and turning off thyristors.

Demonstrate the techniques for di/dt and dv/dt protection of thyristors.

Identify the conditions and techniques for series and parallel operation of thyristors.

Examine thyristor firing circuits.

Introduction

The thyristor is one of the most important power semiconductor devices.

They are operated as bistable switches, operating from nonconducting to conducting state.








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Thyristor Characteristics

A thyristor is a four layer pnpn semiconductor device consisting of three pn junctions. It has three terminals: an anode a cathode and a gate. Figure 4.1 shows the thyristor symbol and a sectional view of the three pn junctions.

Figure 4.1 Thyristor Symbol & pn Junctions

When the anode voltage is made positive with respect to the cathode, junctions J1 and J3 are forward biased and junction J2 is reverse biased. The thyristor is said to be in the forward blocking or off-state condition. A small leakage current flows from anode to cathode and is called the off-state current.

If the anode voltage VAK is increased to a sufficiently large value, the reverse biased junction J2 would breakdown. This is known as avalanche breakdown and the corresponding voltage is called the forward breakdown voltage VBO. Since the other two junctions J1 and J3 are already forward biased, there will be free movement of carriers across all three junctions. This results in a large forward current. The device is now said to be in a conducting or on-state. The voltage drop across the device in the on-state is due to the ohmic drop in the four layers and is very small (in the region of 1 V). In the on-state the anode current is limited by an external impedance or resistance as shown in figure 4.2(a).

V-I Characteristics of Thyristor




Figure 4.2 shows the V-I characteristics and the circuit used to obtain these characteristics.

Figure 4.2 Thyristor Circuit & V-I Characteristics

The important points on this characteristic are :

Latching Current IL

This is the minimum anode current required to maintain the thyristor in the on-state immediately after a thyristor has been turned on and the gate signal has been removed.

If a gate current greater than the threshold gate current is applied until the anode current is greater than the latching current IL then the thyristor will be turned on or triggered.

Holding Current IH

This is the minimum anode current required to maintain the thyristor in the on-state.

To turn off a thyristor, the forward anode current must be reduced below its holding current for a sufficient time for mobile charge carriers to vacate the junction. If the anode current is not maintained below IH for long enough, the thyristor will not have returned to the fully blocking state by the time the anode-to-cathode voltage rises again. It might then return to the conducting state without an externally-applied gate current.

Reverse Current IR

When the cathode voltage is positive with respect to the anode, the junction J2 is forward biased but junctions J1 and J3 are reverse biased. The thyristor is said to be in the reverse blocking state and a reverse leakage current known as reverse current IR will flow through the device.

Forward Breakover Voltage VBO

If the forward voltage VAK is increased beyond VBO , the thyristor can be turned on. But such a turn-on could be destructive. In practice the forward voltage is maintained below VBO and the thyristor is turned on by applying a positive gate signal between gate and cathode.

Once the thyristor is turned on by a gate signal and its anode current is greater than the holding current, the device continues to conduct due to positive feedback even if the gate signal is removed. This is because the thyristor is a latching device and it has been latched to the on-state.





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Two - Transistor Model of Thyristor

This model is used to demonstrate the regenerative or latching action due to positive feedback in the thyristor. A thyristor can be considered as two complementary transistors. One being pnp and the other npn. The two-transistor model is shown in figure 4.3 below.

Figure 4.3 Two-Transistor Model of Thyristor

The collector current IC of a transistor is related to the emitter current IE and the leakage current of the collector base junction ICBO as

.............(1)

The emitter current of transistor Q1 is the anode current IA of the thyristor and collector current IC1 is given by

.............(2)

where α1 and ICBO1 are the current gain and leakage current respectively for transistor Q1.

Similarly, the collector current for transistor Q2 is IC2 where

.............(3)




where α2 and ICBO2 are the current gain and leakage current respectively for transistor Q2.

Combining the two collector currents IC1 and IC2 yields

.............(4)

When a gate current IG is applied to the thyristor

.............(5)

Solving for anode current IA in equation 5 yields

.............(6)

The current gain α1 varies with emitter current IE1 which is equal to IA; and α2 varies with emitter current IE2 which is equal to Ik.

A typical variation of current gain α with emitter current IE is shown in figure 4.4.

Figure 4.4 Typical Variation of Current Gain With Emitter Current

If the gate current IG is increased from zero to some positive value, this will increase the anode current IA as shown by equation 6. An increase of IA which is an increase of IE1 would increase α1 as shown in figure 4.4 and also α2 since

. The increase in values of both α1 and α2 would further increase the

value of anode current IA which is a regenerative or positive feedback effect.

If α1 and α2 approach unity, the denominator of equation 6 approaches zero and a large value of anode current is produced causing the thyristor to turn on as a result of the application of a small gate current.

The capacitance of the pn junctions are shown in figure 4.5 below.

Figure 4.5 Two-transistor Transient Model of Thyristor

Under transient conditions, the capacitances of the pn junctions influence the characteristics of the thyristor.

If a thyristor is in the blocking state and a rapidly rising voltage is applied to the device, high currents would flow through the junction capacitors. The current through capacitor Cj2 can be expressed as

where

Cj2 = capacitance of junction j2

Vj2 = voltage of junction j2

qj2 = charge in junction j2

If the rate of rise of voltage dv/dt is large, then ij2 would be large, which would result in increased leakage currents ICBO1 and ICBO2. High enough values of ICBO1 and ICBO2 may cause α1 and α2 to approach unity, resulting in undesirable turn on of the thyristor.

It must be noted that a large current through the junction capacitors may cause damage to the device.





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Thyristor Turn-on

A thyristor is turned on by increasing the anode current. This can be accomplished in the following ways.

Thermals

If the temperature of a thyristor is high, there will be an increase in the number of electron-hole pairs. This would increase the leakage current. This increase in leakage current causes the anode current to increase and as a result causes α1 and α2 to increase. Due to the regenerative action, the sum α1 + α2 may tend to unity and the thyristor may be turned on. This type of turn-on may cause thermal runaway and should be avoided.

Light

If light is allowed to strike the junction of a thyristor, the electron-hole pairs will increase and this may cause the thyristor to be turned on. This is the principle of operation of light activated thyristors.

High Voltage

If the forward anode to cathode voltage VAK is increased beyond the forward breakdown voltage VBO , high enough leakage currents will flow, causing regenerative turn-on. This type of turn-on is destructive and should be avoided.

dv/dt

From equation 6, if the rate or rise of the anode to cathode voltage is high, (for example, when there is a voltage spike), the charging current of the capacitive junctions may be high enough to turn on the thyristor. A high value of charging current may cause damage to the thyristor and must be avoided. Hence, thyristors must be protected against high dv/dt and must be operated within the manufacturer's dv/dt specifications.

Gate Current

The injection of gate current into a forward biased thyristor would turn-on the device. As the gate current is increased, the forward voltage required to turn-on the device decreases. This is shown in figure 4.6.




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Figure 4.6 Effects of Gate Current on Forward Blocking Voltage

Figure 4.7 shows the waveform of anode current as a function of time, following the application of a gate signal. The following time delays can be described from the waveforms:

Figure 4.7 Thyristor Turn-on Characteristics

Turn-on Time ton

The turn-on time ton is defined as the time interval between 10% of steady-state gate current and 90% of steady-state thyristor on-state current.

Delay Time td

The delay time td is defined as the time interval between 10% of gate current and 10% of thyristor on-state current.

Rise Time tr

The rise time is defined as the time required for the anode current to rise from 10% of the on-state current to 90% of the on-state current.

Note:

Gate Control Circuit Design

Consideration must be given to the following points when designing gate control circuits.

The gate signal should be removed after the thyristor has been turned on. A continuous gate signal will increase the power loss in the gate junction.

No gate signal should be applied when the thyristor is reversed biased. If a gate signal is applied under these conditions, the thyristor may fail due to an increased leakage current.

The width of the gate pulse must be greater than the time required for the anode current to rise to the holding current. In practice, the gate pulse width is made wider than the turn-on time of the thyristor.





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Di/dt Protection

A minimum time is required for the thyristor to spread the current conduction uniformly throughout the junctions. If this time is not allotted and the rate of rise of anode current is very high compared to the spreading velocity at turn-on, then this could lead to localised "hot-spot" heating and the device may fail as a result of excessive heating.

Protection against di/dt is necessary and an example is shown in figure 4.8 below. The circuit analysis is as follows:

Figure 4.8

For an inductive load, when thyristor T1 is turned off, free-wheeling diode Dm conducts load current. If thyristor T1 is fired when diode Dm is still conducting, di/dt can be very high. In order to reduce the high di/dt a series inductor Ls is added to the circuit as shown. The forward di/dt is given as

dv/dt Protection

As seen earlier, a high dv/dt may cause damage to a thyristor. In order to protect a thyristor from high dv/dt, the circuits shown in figure 4.9 below could be used.




Figure 4.9 dv/dt Protection

If the switch S1 in figure 4.9(a) is closed at time t = 0, a step voltage will be applied across thyristor T1 and dv/dt may be high enough to turn on the thyristor. dv/dt can be limited by connecting capacitor Cs across the thyristor as shown in figure 4.9(b). Since

then

and the rate of rise of voltage is limited by the value of the capacitor used. In order to limit the capacitor discharge current when the thyristor is turned on, a resistor Rs is inserted in series with the capacitor as shown in figure 4.9(c). This resistor capacitor arrangement is known as a snubber circuit.

For figure 4.9(c), when switch S1 is closed at time t = 0, the voltage across the capacitor is given by

and this charging capacitor voltage is seen by the thyristor anode to cathode terminals as VAK. This is depicted by the waveform of figure 4.9(d). The rate of rise

of voltage across the thyristor can be represented by

where 0.632VS is one time constant.

The value of the snubber time constant RSCS can be found for a known dv/dt. And for a known discharge current ITD , the value of resistor RS can be found using

It is sometimes necessary to use one resistor for dv/dt and another for limiting the discharge current of the snubber capacitor. This arrangement is shown in figure 4.9(e). In this circuit, R1 and CS are used for dv/dt protection, while R1 + R2 is used for limiting the capacitor discharge current.

The load can also be placed in series with the snubber components as shown in figure 4.9(f).

Example 4.1






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Section 4 - ThyristorsExample 4.1

The input voltage to the circuit shown below is Vs = 200 V with load resistance R = 5 ohms. The load and stray inductances are negligible and the thyristor is operated at a frequency of 2 kHz. If the required dv/dt is 100 V/µs and the discharge current is to be limited to 100 A, determine

a. the values of Rs and Cs

b. the snubber losses and

c. the power rating if the snubber resistor.

Solution

ITD = 100 A

R = 5 Ω

Vs = 200 V

(a) Using Kirchoffs voltage law

at t = 0, vc = 0

differentiating




Therefore integrating each side yields

Recall that 102 = 100Therefore log10 100 = 2So if we let i =





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Thyristor Turn-Off

A thyristor which is in the on-state can be turned off or commutated by reducing the anode current to a level below the holding current and keeping the anode current below this level for a sufficiently long time so that the excess carriers in the four layers are swept out or recombined.

Thyristors can either be line commutated or forced commutated. Figure 4.10 displays the turn-off characteristics of a line commutated thyristor.

The two outer pn junctions exhibit torn-off characteristics similar to that of a diode. The reverse recovery time trr and peak reverse recovery current IRR are given by

trr = ta + tb

For this line commutated thyristor, a reverse voltage appears across the thyristor immediately after the forward current goes through the zero value. This reverse voltage will accelerate the turn-off process by sweeping out excess carriers in junctions J1 and J3.

Figure 4.10 Turn-Off Characteristics of Line Commutated Thyristor

The inner junction J2 will require a time known as the recombination time trc to recombine the excess carriers and the negative reverse voltage reduces this recombination time.

Turn-off Time tq

This is the sum of the reverse recovery time trr and the recombination time trc. It is defined as the time interval between the instant when the on-state




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current has decreased to zero and the instant when the thyristor is capable of withstanding forward voltage without turning on. It depends on the peak value of on-state current and the instantaneous on-state voltage.

Reverse Recovery Charge QRR

This is defined as the amount of charge which has to be recovered during the turn-off process. Its magnitude is determined by the area enclosed by the path of the reverse recovery current and depends on:

1. The rate of fall of on-state current and

2. The peak value of on-state current before turn-off





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Thyristor Types

Thyristors are manufactured almost exclusively by diffusion. Thyristors can be classified into nine categories depending on the physical construction, turn-on time and turn-off time. These categories are:

1. Phase Control Thyristors (SCRs)

2. Fast Switching Thyristors (SCRs)

3. Gate Turn-off Thyristors (GTOs)

4. Bidirectional Triode Thyristors (TRIACs)

5. Reverse Conducting Thyristors (RCTs)

6. Static Induction Thyristors (SITHs)

7. Light Activated Silicon Controlled Rectifiers (LASCRs)

8. FET Controlled Thyristors (FET-CTHs)

9. MOS Controlled Thyristors (MCTs)

Phase Control Thyristors

This type of thyristor operates at line frequency and is turned-off by natural commutation. The turn-off time is of the order of 50 to 100 µs. Since the thyristor is made of silicon and it is a controlled device, it is called a silicon control rectifier (SCR). This type is used in low speed switching applications and is also known as a converter thyristor. Modern thyristors use an amplifying gate circuit to turn-on the main thyristor. In figure 4.11 an auxiliary thyristor TA is gated on by an external gate signal. The amplified output of this auxiliary thyristor is used to supply gate signal to the main thyristor TM.

Figure 4.11 Amplifying Gate Thyristor

Fast Switching Thyristor

This type of thyristor is used in high speed switching applications and are forced




commutated. They have fast turn-off times in the range of 5 to 50 µs. The on-state voltage drop varies approximately as an inverse function of the turn-off time tq. This type is known as an inverter thyristor.

Gate Turn-off Thyristor

A gate turn-off thyristor is turned on by applying a positive gate signal and turned off with the application of a negative gate signal. GTOs have several advantages over SCRs. These advantages are:

1. Elimination of commutating components resulting in a reduction in cost, weight and volume.

2. Reduction in acoustic and electromagnetic noise due to the elimination of commutation chokes.

3. Faster turn-off permitting high switching frequency applications.

4. Improved efficiency of converters.

GTOs, when used in low power applications have several advantages over bipolar transistors. These advantages are:

1. Higher blocking voltage capability.

2. A high ratio of peak controllable current to average current.

3. A high ratio of peak surge current to average current.

4. A high on-state gain i.e. anode current to gate current.

5. A pulsed gate signal of shorter duration.

Controllable Peak On-state Current ITGQ

This is defined as the peak value of on-state current which can be turned off by gate control.

Bi-directional Triode Thyristor

The bi-directional thyristor or TRIAC conducts current in both directions and is normally used in ac phase control circuits. It can be considered as two SCRs connected in anti-parallel with a common gate connection. The TRIAC symbol and V-I characteristics are shown in figure 4.12.

Figure 4.12 TRIAC Characteristics

A bi-directional device cannot have an anode and a cathode, hence the terminals are labeled gate, MT1 and MT2. If terminal MT1 is positive with respect to MT2, the triac can be turned on by applying a positive gate signal between gate and MT1. If terminal MT2 is negative with respect to MT1, the triac can be turned on by applying a negative gate signal between gate and MT1. It is not necessary to have both polarities of gate signals to turn on a thyristor. If operated in quadrant 1, it needs a positive gate voltage and current, while if operated in quadrant III it needs a negative gate voltage and current.

Reverse Conducting Thyristor

An RCT is a thyristor with a built in anti-parallel diode as shown in figure 4.13.

Figure 4.13 Reverse Conducting Thyristor

This thyristor is used in chopper and inverter circuits to allow reverse current due to inductive loads to flow through the antiparallel diode. The reverse blocking voltage of an RCT is very low typically in the range of 30 to 40 volts.

Static Induction Thyristor

A SITH is a minority carrier device. It is turned on by applying a positive gate voltage and turned off by applying a negative gate voltage. It has fast switching speed in the range of 1 to 6 µs and high dv/dt and di/dt capabilities.

Light Activated Silicon Controlled Thyristor

This device is turned on when light is incident on the silicon wafer. Electron-hole pairs which are produced by the incident light trigger the device when an electric field is applied. LASCRs are used in high voltage high current applications like HVDC transmission. These devices offer complete electrical isolation between the light triggering source and the switching device. LASCRs cannot be turned off by the gate.

FET Controlled Thyristor

The FET controlled thyristor consists of a FET in parallel with a thyristor as shown in figure 4.14. The thyristor is triggered when a gate voltage is applied to the MOSFET. These devices cannot be turned off by gate control.

Figure 4.14 FET Controlled Thyristor

MOS Controlled Thyristor

These devices combines the features of a regenerative four layer thyristor and a MOS gate structure. The equivalent circuit and symbol for this device is shown in figure 4.15. The features of a MCT are:

1. Low forward voltage drop during conduction.

2. Fast turn-on and turn-off times.

3. Low switching losses.

4. Low reverse voltage blocking capability.

5. High gate input impedance.

This type is used in low speed switching applications and is also known as a converter thyristor. Modern thyristors use an amplifying gate circuit to turn-on the main thyristor. In figure 4.11 an auxiliary thyristor TA is gated on by an external gate signal. The amplified output of this auxiliary thyristor is used to supply gare signal to the main thyristor TM.

Figure 4.15 MOS Controlled Thyristor





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Series Operation of Thyristors

Thyristors are connected in series to improve their overall voltage rating. The characteristics of thyristors of the same type are not the same, as shown in figure 4.16 and hence auxiliary components must be added to thyristors connected in series to ensure proper operation.

Figure 4.16 Off-State Characteristics of Two Thyristors of Same Type

In figure 4.16 it is clearly seen that for the same off-state current the off-state voltages differ. Voltage sharing networks are required for both reverse and off-state conditions. Resistors placed in parallel with the thyristors are used to accomplish voltage sharing between thyristors placed in series. The voltage sharing resistors for n thyristors in series are shown in figure 4.17.

Figure 4.17 Three Series Connected Thyristors

For equal voltage sharing, the off-state currents differ as shown in figure 4.18.




Figure 4.18 Forward Leakage Currents For Equal Voltage Sharing

Let the off-state current of thyristor T1 be represented by ID1. For ns thyristors in the string, and the other ns - 1 thyristors having the same off-state current, then

ID2 = ID3 = IDn

and

ID1 < ID2

Note: Since thyristor T1 has the least off-state current, then this thyristor will share the highest voltage because the voltage drop across the resistor R in parallel with this thyristor will be larger than that across the resistor R

for the other thyristors.

If I1 is the current through resistor R which is connected across thyristor T1, then the current through the other resistors are equal and given by

ID2 = ID3 = IDn

The off-state current spread is given by ∆ID where

∆ID = ID2 - ID1

∆ID = (IT - I2) - (IT - I1)

∆ID = I1 - I2

therefore

I2 = I1 - ∆ID

The voltage drop across thyristor T1 is given by VD1 where,

VD1 = RI1

Using Kirchhoff's voltage law where supply voltage is given by Vs yields

Vs = VD1 + (ns - 1)I2R = VD1 + (ns - 1)(I1 - ∆ID)R

Vs = VD1 + (ns - 1)I1R - (ns - 1)∆IDR

Vs = nsVD1 - (ns - 1)∆IDR

Solving the above equation for VD1 the voltage across thyristor T1 yields

The voltage across the thyristor T1 will be a maximum when the off-state current spread ∆ID is maximum. The worse case steady-state voltage across the thyristor T1 is seen when no current flows through thyristor T1. Under these conditions,

ID1 = 0

and

∆ID = ID2

The voltage across thyristor T1 is now given as VDS(max), where

During turn-off the differences in forward leakage currents causes differences in stored charge which in turn causes differences in reverse voltage sharing. This is shown in figure 4.19. The thyristor with the least recovery charge or with the lowest reverse recovery time will see the highest transient voltage. The junctions capacitances which control the transient voltage distribution will be inadequate for this process and an external capacitor C1 should be used as shown in figure 4.17. Resistor R1 limits the capacitor discharge current. The same components R1 and C1 are used for both transient voltage sharing and dv/dt protection.

Figure 4.19 Reverse Recovery Time & Voltage Sharing

The voltage difference between thyristors T1 and the other thyristors is given by

where Q1 is the charge stored by thyristor T1 and Q2 is the charge stored by the other thyristors such that

Q2 = Q3 = Qn

and

Q1 < Q2

The transient voltage across thyristor T1 is given by

The worse case transient voltage occurs when Q1 = 0, hence ∆Q = Q2 and is given by

The transient voltage derating factor which is normally used to increase the reliability of the string is given by

Example 4.2






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Parallel Operation of Thyristors

When the load current exceeds the rating of a single thyristor, thyristors are connected in parallel to increase the overall current capability. Contrary to what might be expected, the load current is not shared equally between the thyristors, because thyristors are not perfectly matched. Figure 4.20 shows the V-I characteristics of two thyristors T1 and T2 which are connected in parallel.

Figure 4.20 V-I Characteristics of Two Parallel Thyristors

The thyristor carrying the higher current would dissipate more power which in turn will increase the junction temperature and hence decrease the internal resistance. This in turn will increase its current sharing capcity and maybe damage the thyristor. This process is termed thermal runawy and is not common only to thyristors. Thermal runaway may be prevented by using one common heat sink to ensure that both thyristors are operating at the same temperature. Equal current sharing could be accomplished with the use of a small resistor or inductor in series with each thyristor as shown in figure 4.21.

Figure 4.21 Current Sharing of Thyristors in Parallel

When a resistor is used to produce equal current sharing, the losses in the series resistor is very high and may be unacceptable. When magnetically coupled inductors are used for the purpose of current sharing, if thyristor T1 current increases, a voltage of opposite polarity to that of the coil in series with T1 will be induced in the coil in series with thyristor T2. The polarity of this voltage is as such to increase the anode potential of thyristor T2, thereby increasing the current flow through this thyristor.





Thyristor Firing Circuits

In thyristor converters high ac voltages exists between anode and cathode of the thyristor, while low voltage level pulses are placed between gate and cathode. Isolation is necessary between the gate-cathode circuit and the anode-cathode circuit. This isolation is accomplished with the use of:

1. Optocouplers and 2. Pulse transformers.

In the case of optocoupler isolation, the low voltage gate drive circuit is optically isolated from the high voltage anode-cathode circuit as shown in figure 4.22.

Figure 4.22 Optical Isolation Using Photo SCR

In the above circuit, the gate drive circuit is connected to the light emitting diode D1 via a current limiting resistor R1. Pulses sent to the light emitting diode D1 turns on the photo SCR T1 which in turn triggers the power thyristor TL. Hence the gate drive circuit is optically isolated from the output circuit. Instead of using an optocoupler, a pulse transformer could be used to magnetically isolate the gate drive circuit from the anode-cathode circuit.








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Objectives

To examine the principles of operation of single-phase and three-phase controlled rectifiers to convert a fixed ac voltage into a variable dc voltage.

Describe the application of thyristors in controlled rectification

Analyse and evaluate the performance of controlled rectifiers.

Examine the problems of poor power factor and harmonic distortion of supply current due to controlled rectifiers.

Determine the filtering requirements for controlled rectifiers.

Introduction

As was observed earlier, diode rectifiers produce a fixed output voltage. In order to control the output voltage of a rectifier, phase control thyristors must be used instead of diodes.

The output voltages of these rectifiers are varied by varying the delay or firing angle of the thyristor.

Phase controlled thyristors are turned on by the application of a short pulse to the gate and they are turned off by the process of natural or line commutation.

Phase controlled rectifiers are also called ac-dc converters and are widely used in industrial applications.

There are two types of phase control converters:

1. Single phase converters.

2. Three-phase converters

Each type of converter can be divided into three categories:

1. Semiconverter

2. Full converter

3. Dual converter

A semiconverter is a one quadrant converter, having one polarity of voltage and current.

A full converter is a two quadrant device whose output voltage polarity can be either positive or negative but whose output current has one polarity.

A dual converter is a four quadrant device whose output voltage and current




can be of either positive of negative polarity.





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Electrical Home


Single Phase Half-Wave Thyristor Converter With R Load

Figure 5.1 below shows a single phase half-wave thyristor converter with a resistive load. For the positive half cycle of input voltage, the thyristor T1 is forward biased

and when the thyristor is fired at ωt = α it conducts and the input voltage appears across the load. When the input voltage goes negative at ωt = π, the thyristor is reversed biased and it is turned off. The delay angle α, is defined as the time the input voltage starts to go positive to the time the thyristor is fired.

Figure 5.1 - Single Phase Thyristor Converter With R Load

The average output voltage Vdc is given by




The output voltage Vdc can be varied from Vm/π to zero as the firing angle α varies

from zero to π.


Example 5.1






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Section 5 - Controlled RectifiersExample 5.1

The converter shown below has a purely resistive load R and a delay angle .

Determine

a. the rectification efficiency b. the form factor (FF) c. the ripple factor (RF) d. the transformer utilization factor (TUF) e. and the peak inverse voltage of the thyristor

Solution

delay angle,

Vdc = 0.1592 Vm




a.

b. Form Factor (FF) =

c. Ripple Factor (RF)

d. Vs= rms voltage of transformer secondary winding

Is= rms value of transformer secondary current = rms value of load current

e. Peak Inverse Voltage (PIV) = Vm





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Single Phase Semiconverters

Figure 5.2 shows a single phase semiconverter with a highly inductive load, such that the load current is assumed continuous and ripple free.

Figure 5.2 - Single-Phase Semiconverter

Click the stop/play button to stop or play the animation or click the step button to step forward through the animated sequence above.

During the positive half cycle of input voltage, thyristor T1 is forward biased. During the interval 0 ≤ ωt ≤ α, no supply current can flow through the thyristors and the load current freewheels through the freewheeling diode Dm. When T1 is fired at ωt = α, the load is connected to the input supply via T1 and D2 for the period α ≤ ωt ≤ π.

For the period π ≤ ωt ≤ (π + α), the input voltage is negative and the thyristor T1 and diode D2 will be reversed biased. Under these conditions, freewheeling diode




Dm is forward biased and provides a path for the flow of load current. During the negative half cycle of input voltage, thyristor T2 is forward biased and when a firing pulse occurs at ωt = π + α, diode Dm will be reversed biased and the load will be connected to the supply via thyristor T2 and diode D1.

The average output voltage is found from

The output voltage can be varied from a maximum of 2Vm/π to a minimum of zero as the firing angle α varies from zero to π.






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Single-Phase Semiconverter With RL Load

The load current of a semiconverter is dependent on the load resistance, inductance and the battery voltage E in series with the load. The load current of a semiconductor is made up of two components:

1. That component in the range 0 ≤ ωt ≤ α when the freewheeling diode is forward biased and load current is provided by the energy stored in the magnetic field.

2. The component in the range α ≤ ωt ≤ π when load current is provided by the supply.

Load Current For Interval 0 ≤ ωt ≤ α

In this time range, the freewheeling diode Dm is forward biased and load current is supplied by the energy stored in the magnetic field of the inductor. The load current is given by iL1 and the voltage equation for the circuit is given by

.............(1)

.............(2)

where α is given as

Equation (2) is of the form

whose solution is given as

where

Therefore the solution for equation 2 can be written as




Applying the initial conditions yields

Hence,

At the end of this interval i.e. when ωt = α, the load current is given by IL1, where

Load Current For Interval α ≤ ωt ≤ π

During this interval the thyristor conducts and supply voltage is connected to the load via thyristor T1 and diode D2.The load current for this interval is given by iL2 and the voltage equation can be written as

Solving the above equation for the load current iL2 yields

Now

where

Therefore


The constant C can be determined from the initial conditions

Therefore the load current is given by

When thyristor T1 is turned off at ωt = π, the load current iL2(t) goes back to its original value IL0 where,

The rms current flowing through one thyristor is given by

The average current flowing through one thyristor is given by

The rms value of load current is given by







Single-Phase Full Converter

Figure 5.3 shows a single-phase full converter with a highly inductive load so that the load current is continuous and ripple free. Thyristors T1 and T2 are forward biased during the positive half cycle of supply voltage. When these two thyristors are fired simultaneously at ωt = α, the load is connected to the supply via thyristors T1 and T2. Thyristors T1 and T2 will continue to conduct beyond ωt = π as a result of the presence of an inductive load. During the negative half cycle, thyristors T1 and T2 are forward biased and at ωt = π + α, these thyristors are fired into conduction causing thyristors T1 and T2 to be reversed biased and be turned off due to line commutation. For the period α ≤ ωt ≤ π, the input voltage and current are positive and power flows from supply to load and the converter is said to be in rectification mode.

Figure 5.3 - Single-Phase Full Converter

Click the stop/play button to stop or play the animation or click the step button to step through the animated sequence above.

For the period π ≤ ωt ≤ π + α, the input voltage is negative, input current is positive and power flows from the load to the supply. The converter is said to be operated




in the inversion mode. This converter supplies two quadrant operation since the output voltage can either be positive or negative depending on the value of the firing angle.

The average output voltage is given as

The output voltage can be varied from 2Vm/π to -2Vm/π when a varies from 0 to π.






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Single-Phase Full Converter with RL load

The output load current of the converter comprises of two components per cycle. One component flows when thyristors T1 and T2 are fired and connects the supply voltage to the load and the other component flows when thyristors T3 and T4 are turned on again connecting supply voltage to the load. Since both components of current are identical, only one component will be studied. For the interval α ≤ ωt ≤ (α + π)

the load current is given by

The solution for iL takes the form

at ωt = α, iL = IL0 hence

Hence the load current during this interval is given by

The current magnitude at the end of the first conponent of load current is the same as that at the beginning of the second component of load current, that is at ωt = α + α, iL = IL0 and the current IL0 can be obtained by substituting this condition in the above equation which yields

The value of firing angle a at which current IL0 = 0 can be obtained for known values of the parameters in the above equation, using an iterative method.

The rms value of thyristor current is given by




The rms output current is given by

The average current of one thyristor is given by

The average output current is given by

Idc = IA + IA = 2IA





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Three-Phase Half-Wave Converter

Three-phase converters provide higher average output voltage than their single phase counterparts. In addition, the output voltage ripple frequency is higher in three-phase converters than in single phase converters. Figure 5.4 shows a three-phase half wave converter and the associated waveforms.

Thyristor T1 is fired at ωt = π/6 + α since the phase voltage van is the most positive phase voltage for the interval π/6 ≤ ωt ≤ 5π/6. Thyristor T2 is fired at ωt = 5π/6 + α, since vbn is the most positive phase voltage for the next 120°. When thyristor T2 is fired on, T1 will be turned off due to the line to line voltage Vab is now negative. Thyristor T3 is fired at ωt = 3π/2 + α. When this happens, thyristor T2 will be turned off.




Figure 5.4 Three-Phase Half-Wave Converter

Click the stop/play button to stop or play the animation or click the step button to step forward through the animated sequence above.

In figure 5.4 the load current is continuous since the load is highly inductive. For a purely resistive load and firing angle α > π/6, the load current would be discontinuous and each thyristor would be commutated when the polarity of the phase voltage is reversed. The frequency of the output ripple voltage is 3fs, where fs is the frequency of the supply. This converter is not normally implemented in practical circuits because the supply current contain dc components.

For a continuous load current, the average output voltage is given by:

The rms output voltage is obtained from

For a resistive load and α ≥ π/6

And the rms output voltage for a resistive load is given by





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Three-Phase Semiconverter

Figure 5.5 Three-Phase Semiconverter with Highly Inductive Load

Figure 5.6 Three-Phase Semiconverter for α < π/3

Figures 5.5 and 5.6 show a three-phase semiconverter with a highly inductive load.

For delay angle α ≥ π/3

The delay angle α can be varied between 0 and π. For the period π/6 ≤ ωt ≤ 7π/6, thyristor T1 is forward biased and if it is fired at ωt = π/6 + α, thyristor T1 and diode D1 conducts the supply line voltage vac across the load. At ωt = 7π/6, the line voltage vac starts to go negative. This causes freewheeling diode Dm to conduct load current causing thyristor T1 and diode D1 to turn off. Each thyristor conducts for a duration of π - α which is less than 2π/3.

For delay angle α ≤ π/3

Under these conditions, each thyristor conducts along with two diodes (one diode and a thyristor at any one time) for the interval 2π/3.

Output Voltage of Semiconverter

The three line to neutral voltages are given by:

The line to line voltages are given by:




where Vm is the peak phase voltage.

For delay angle α ≥ π/3

Under these conditions the output voltage is discontinuous and the average output is given by

The maximum output voltage occurs when α = 0 and is given by


For delay angle α ≤ π/3

Under these conditions the output voltage is continuous and the output voltage is given by






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Three-Phase Full Converter

Figure 5.7 Three-Phase Full Converter

A diagram of a three-phase full converter with a highly inductive load is shown in figure 5.7 together with the associated voltage and current waveforms. This converter provides two quadrant operation and thyristors are fired at an interval of π/3 degrees. Since thyristors are fired every 60°, the frequency of the output ripple voltage is six times the frequency of the supply voltage. At ωt = π/6 + α, thyristor T6 is already conducting and thyristor T1 is turned on. For the interval π/6 ≤ ωt ≤ π/2 thyristors T1 and T6 conduct, and line to line voltage vab appears across the load. At ωt = π/2 + α, thyristor T2 is turned on and thyristor T6 is turned off due to natural commutation. This occurs because when thyristor T2 is turned on the line to line voltage across thyristor T6 is the positive voltage vbc from cathode to anode which reverse biases thyristor T6. During the interval (π/2 + α) ≤ ωt ≤ (5π/6 + α), thyristors T1 and T2 conduct and line to line voltage appears across the load. The firing sequence of the thyristors is: 12, 23, 34, 45, 56 and 61.

Determination of average output voltage and rms value of output voltage.

The line to neutral voltages are defined as:

The corresponding line to line voltages are:





The maximum output voltage is obtained when α = 0 and is given by

The rms value of output voltage is given by:





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Fourier Transform - Revision

Introduction

Periodic phenomena occur quite frequently in Electrical Engineering, indeed the source of electrical power is a periodic signal so it stands to reason that manipulation of this a.c. source will result in mostly periodic signals. It is an important practical problem to represent the corresponding periodic functions in terms of simple periodic functions such as sine and cosine.

Fourier series are series of sine and cosine terms and arise in the important task of representing complex periodic functions. The theory of Fourier series is quite involved, but the application of these series is simple.

Recall A function f(x) is called periodic if it is defined for all real x and if there is some positive number p such that

f(x + p) = f(x) for all x ................F.1

This number p is called the period of f(x). The graph of such a function is obtained by periodic repetition of its graph in any interval of length p.

Note that f = c = const is also a periodic function in the sense of the definition, because it satisfies (F.1) for every positive p.

From (F.1) we have

f(x + 2p) = f[(x + p) + p] = f(x + p)

= f(x)

So we can say

f(x + np) = f(x) for all x ................F.2

Furthermore, if f(x) and g(x) have period p, then the function

h(x) = af(x) = bg(x) (a,b = const)

also has the period p.

The Fourier Coefficients

The Fourier Series, as already mentioned allows us to represent a periodic function as a series of sine and cosine terms. Say we have such a periodic funtion f(x). Then we can reprensent f(x) as shown below.

f(x) = a0 + a1cosx + b1sinx + a2cos2x + b2sin2x + .....

Grouping the sine and cosine terms yields,

................F.3


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That is we assume that the series converges. We see that each term of the series has the period 2π. Hence, if the series converges, its sum will be a function of period 2π.

Given such a function we want to find the coefficients a0, an and bn of the corresponding series (F.3).

We determine a0. Integrating on both sides of (F.3) from -π to π, we get

Hence our first result is

................F.4

We now determine a1, a2,... by a similar procedure. We multiply (F.3) by cos mx, where m is any

fixed positive integer, and integrate from -π to π.

................F.5

Recall

The first integral in (F.5) is zero. By applying (1) and (2) to (F.5), we obtain

Integration shows that all the terms on the right of equations (F.6) and (F.7) are zero, except for the last term in (F.6) which equals π when n = m. Since in (F.5) this term is multiplied by am, the right

side in (F.5) equals amπ. Our second result is (if we write n in place of m).

................F.8

We finally determine b1,b2,..., in (F.3). If we multiply (F.3) by sin mx, where m is any fixed positive integer, and then integrate from integrate from -π to π.

................F.9

The first integral in is zero. The next integral is of the kind considered before, and is zero for all n = 1,2,... For the last integral, using (3), we obtain

................F.10

The last term is zero. The first term is also zero, except at n = m when it is equal to π. Since in (F.9) this term is multiplied by bm, the right side in (F.9) equals bmπ. Our final result is (if we write n in place of m).

................F.10

Summary

So in summary, if we have a function f(x) of period 2π, we can represent it as a sum of sin and cosine terms given by the fourier series

where

n = 1,2,...

n = 1,2,...





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Objectives

Introduction

An ac voltage controller is defined as a circuit consisting of thyristors between the supply and the load for the purpose of varying the rms value of ac load voltage.

AC voltage controllers are used in:

a. Industrial heating

b. No-load transformer tap changing

c. Light controls and

d. Speed control of induction motors

Two types of control strategy are used to control power flow between supply and load:

a. On-off control

b. Phase angle control

In on-off control, thyristors connect the load to the ac source for a few cycles of input voltage and then disconnect it for another few cycles.

In phase angle control, thyristors connect the load to the ac source for a portion of each cycle of input voltage.

AC voltage controllers are classified as follows:

1. Single phase controllers

a. Unidirectional or half wave control

b. Bidirectional or full-wave control

2. Three-Phase Controllers

a. Unidirectional or half wave control

b. Bidirectional or full-wave control





On-off Control

A single-phase full-wave controller is shown in figure 6.1 to explain the principle of on-off control.

Figure 6.1 - On-off Control

The thyristor switch which consists of thyristors T1 and T2 connect the ac supply to the load for a time tn where tn consists of an integral number of cycles.

It must be noted that the thyristors are turned on at the zero-voltage crossing of input voltage.

This type of control is applied in applications which have a high mechanical inertia and high thermal time constant e.g. industrial heating and speed control of motors.

The harmonics generated by the switching action is reduced due to zero-voltage and zero-current switching.

If a sinusoidal voltage of the form

is connected to the load for n cycles and disconnected for m cycles the rms output voltage is given by

where

and k is called the duty cycle

The input power factor is given by




The following must be noted when using an on-off ac voltage controller:

1. The power factor and output voltage varies with the square root of the duty cycle.

2. If T is the period of the input voltage, then (m+n)T is the period of the on-off control. (m+n)T must be less than the mechanical or thermal time constant of the load.

3. If m and n are in days, then the square root of k would give erroneous results if used to compute the power factor.





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Phase Control

Single-Phase Unidirectional Controller

A single-phase unidirectional phase control ac voltage controller is shown in figure 6.2 below. The power flow to the load is controlled by delaying the firing angle of thyristor T1 over the positive half cycle of voltage waveform.

Figure 6.2 - Single-Phase Half-Wave Controller


The average value of output voltage is given by

It should be noted that as the delay angle is varied between 0 and π, the rms output voltage varies between Vm/√2 and Vm/2 , while the average output voltage varies between 0 and -Vm/π.

The output voltage and input current are asymmetrical and contain a dc component. This may produce a saturation problem if an input transformer exists.





Phase Control

Single-Phase Bidirectional Controller With R Load

A single-phase full-wave rectifier with resistive load is shown in figure 6.3 below.

Figure 6.3 - Single-Phase Full-Wave Controller

The problem of dc input current could be prevented with the use of the bidirectional controller.

Thyristor T1 controls power flow during the positive half-cycle of input voltage, while thyristor T2 provides control during the negative half cycle of input voltage by varying the delay angle.

It must be noted that the firing pulses of thyristors T1 and T2 are kept 180 degrees apart.

If the delay angle of thyristors T1 and T2 are equal and given by α, then the rms output voltage is given by





Phase Control

Single-Phase Bidirectional Controller With RL Load

Figure 6.4 shows a single-phase full-wave controller with an inductive load.

Figure 6.4 Single-Phase Full-Wave Controller with RL Load

If thyristor T1 is fired on the positive half cycle and carries the load current, then,

at ωt = π, when the input voltage starts to go negative, the load current does not fall to zero due to the load inductance.

The thyristor T1 will continue to conduct until its current falls to zero at ωt = β.

The conduction angle of thyristor T1 is given by δ, where

δ = β - α

The angle δ depends on

1. delay angle α 2. power factor angle of load θ.

The instantaneous thyristor current can be found from the equation

The solution of the above equation is of the form

where load impedance

and

The constant A can be determined from the initial condition:At ωt = α, i1 = 0 hence A can be given as




therefore the current can be given as

When the thyristor T1 is turned off at ωt = β, the current i1 falls to zero. Substituting this condition in the above equation yields

from which the angle β which is known as the extinction angle can be determined by an iterative method.


The rms thyristor current is given by

and the rms output current is obtained by combining the rms current of each thyristor and given as Io where

For resistive loads, the gating signals for a controller could be short pulses.

For inductive loads, short pulses should not be used as gating signals. This can be explained as follows:

When thyristor T2 is fired at ωt = π + α, thyristor T1 is still conducting due to load

inductance. When thyristor T1 current falls to zero at time ωt = β, the gate pulses to thyristor T2 has already ceased and hence thyristor T2 will not turn on. This can

be corrected by using continuous gating signals of duration π - α. The disadvantage of continuous gating signal is an increase in switching losses and hence a larger isolation transformer is required for the gate circuit.


Single-Phase Transformer Tap Changers

The diagram of a single phase transformer tap changer is shown in figure 6.5 below.

Figure 6.5 - Single-Phase Transformer Tap Changer

The waveforms for the transformer tap changer is shown in figure 6.6.

Figure 6.6 Waveforms of Single-Phase Transformer Tap Changer

Three-Phase Half-Wave Controllers

The circuit diagram of a three-phase half-wave controller with a star connected resistive load is shown in figure 6.7 below.

The current flow to the load is controlled by thyristors T1, T3 and T5 while the three diodes provide the return current path. The thyristor firing sequence is T1, T3, T5 and current flow through the controller is only possible if at least one thyristor is conducting.

For rms input phase voltage of Vs, the instantaneous input phase voltages are given by

and the input line voltages are given by

The input and output waveforms for α = 60o and α = 150o are shown in figure 6.8.

Three conduction ranges would be described for this controller.




For 0° ≤ α ≤ 90°

For this range, two or three devices could conduct simultaneously and the possible combinations are:

a. two thyristors and one diode b. one thyristor and one diode c. one thyristor and two diodes

If three devices conduct, then normal three-phase operation is in effect and the output phase voltage is the same as the input phase voltage i.e.

van = vAN = √2 Vs sin ωt

If two devices conduct at the same time, current flows only through two lines and the third line can be considered to be open circuited and the line to line voltage would appear across the load. The output phase voltage would be one half the line voltage. Example if c is open circuited, then

Figure 6.8 Waveforms of Three-Phase Half-Wave Controller & Y-Connected Load

The waveforms of an output phase voltage can be drawn directly from the input phase and line voltages by noting the output phase voltage corresponds to the input phase voltage if three devices are conducting, the output phase voltage corresponds to one-half the input line voltage if two devices are conducting and the output phase voltage corresponds to zero if all thyristors are turned off.

For this range of firing angle, one thyristor is conducting at any time and the return path is shared by either one or two diodes.

The rms output phase voltage is given by

Output phase voltage van is made up of the following parts:

1. At ωt = α, thyristor T1 is fired and line voltage vAB supplies load current via thyristor T1 and diode D6 line voltage vAC supplies load current via thyristor T1 and diode D2

Hence, three devices are conducting and the output phase voltage van is equal to the input phase voltage vAN.

These three devices conduct until line voltage vBC becomes more positive than line voltage vAB and this reverse biases diode D6, resulting in only line voltage vAC supplying load current thereafter.

The three devices conduct for the interval ωt = α to ωt = 2π/3, hence the output phase voltage van for this interval is given by the input phase voltage VAN:

2. At ωt = 2π/3 diode D6 commutates and:

line voltage vAC supplies load current via thyristor T1 and diode D2. Hence, only two devices are conducting and the output phase voltage van is given as one-half the line voltage vAC.

These two devices conduct until thyristor T3 is fired at ωt = 2π/3 + α for the vBC waveform which reverse biases thyristor T1 and causes it to

commutate. Now the output sees vAC for the interval ωt = 2π/3 to ωt =

2π/3 + α. If the vAC waveform starts from zero on the ωt axis, then the

output sees vAC from ωt = 2π/3 to ωt = 2π/3 + α, where the vAC waveform is given by:

vAC = √6Vssinωt

The output phase voltage van for this interval is given by:

3. At ωt = 2π/3 + α thyristor T3 fires and:

line voltage vBA supplies load current via thyristor T3 and diode D4 line voltage vBC supplies load current via thyristor T3 and diode D2.


These three devices conduct until line voltage vCA becomes more positive than line voltage vBC and this reverse biases diode D2, resulting in only line voltage vBA supplying load current thereafter. The three devices

conduct for the interval ωt = 2π/3 + α to ωt = 4π/3, hence the output phase voltage van for this interval is given by the input phase voltage vAN:

4. At ωt = 4π/3 diode D2 commutates and:

line voltage vBA supplies load current via thyristor T3 and diode D4. Hence, only two devices are conducting and the output phase voltage van is given as one-half the line voltage vBA.

These two devices conduct until thyristor T5 is fired at ωt = 4π/3 + α for the vCA waveform which reverse biases thyristor T3 and causes it to

commutate. Now the output sees vBA for the interval ωt = 4π/3 to ωt =

4π/3 + α. If the vBA waveform starts from zero on the ωt axis, then the

output sees vBA from ωt = 3π/2 to ωt = 3π/2 + α, where the vBA waveform is given by:

vBA = √6Vssinωt

The output phase voltage van for this interval is given by:

5. At ωt = 4π/3 + α thyristor T5 fires and:

line voltage vCA supplies load current via thyristor T5 and diode D4 line voltage vCB supplies load current via thyristor T5 and diode D6.


These three devices conduct until line voltage vAB becomes more positive than line voltage vCA and this reverse biases diode D4, resulting in only line voltage vCB supplying load current thereafter. The three devices

conduct for the interval ωt = 4π/3 + α to ωt = 6π/3, hence the output phase voltage van for this interval is given by the input phase voltage vAN:

The full rms output phase voltage over an interval of 2π is given by:

For 90° ≤ α ≤ 120°

When the thyristor anode to cathode voltage goes negative, the thyristor will commutate, hence the range of the line voltages will change as follows:

1. vAC will be placed across the load for the interval ωt = π/2 and ωt = π since the thyristor in question will commutate when its anode voltage goes negative.

2. vBA will be placed across the load for the interval ωt = 3π/2 and ωt = 2π, since the thyristor in question will commutate when its anode voltage goes negative.

Hence the rms output voltage for this interval is given by:

For 120° ≤ α ≤ 210°

In this interval, only one thyristor and one diode conduct at the same time since:

1. with α>=120o, a return diode is reversed biased and only one line voltage supply appears across the load.

2. The supply voltage vAC is connected to the load for the interval ωt = α -

π/6 and π, since for ωt >= π, vAC is negative and the thyristor in question commutates.

3. Line voltage vBA will be placed across the load for the interval ωt = 3π/2 -

2π/3 + α and ωt = 2π, since the thyristor in question will commutate when its anode voltage goes negative.

Hence the rms output voltage for this interval is given by:





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Three-Phase Full-Wave Controller

The diagram of a three-phase full-wave controller is shown in figure 6.9.

Figure 6.9 Three-Phase Bidirectional Controller

The unidirectional controller contain dc input current and high harmonic content due to the asymmetrical nature of the output voltage waveform.

The firing sequence of thyristors is T1, T2, T3, T4, T5and T6.

For rms input phase voltage of Vs, the instantaneous input phase voltages are given by

and the input line voltages are given by

The waveforms of output phase voltage and input voltages are shown in figure 6.10.

Figure 6.10 Waveforms of 3-Phase Bidirectional Controller

For the range 0° ≤ α ≤ 60°




Two and three thyristors conduct at one time, and the rms output phase voltage is given by:


Two thyristors conduct at one time, and the rms output phase voltage is given by:


Two thyristors conduct at one time, and the rms output phase voltage is given by:





Basic Circuit Theory

Basic Definitions

Electron: an indivisible particle of negative charge. The amount of charge is measured in coulombs (C). The magnitude of the charge associated with an electron is 1.602x10-l9 C.

Current: charge in motion (electrons). Current is measured in units of amperes, or more simply amp.

Voltage: an electric potential difference that causes electron flow. It is also called electromotive force (EMF). An analogy often used to describe current and voltage is water in a pipe. Current is analogous to the flow of water, while voltage is analogous to the pressure.

Conductor: a material that allows a continuous current to pass through it under the action of a fixed voltage. An example of a good conductor is copper or aluminum which is used in homes and offices for all electrical connections.

Insulator: the opposite of a conductor, it does not allow a continuous current to pass though it under the action of a fixed voltage. An example of an insulator is the plastic on electrical cords. Using our water analogy, a conductor can be envisioned as the region inside a pipe, while an insulator can be envisioned as the actual material of the pipe which contains the water flow.

Switch: used to control the flow of electrons, or current as it is commonly called. Ideally, a switch turns on or off instantly, and has no voltage across it while it is conducting. In our water analogy, an ideal switch would cut the flow immediately, from completely on to completely off in an instant.

Common Passive Circuit Elements

All circuit elements can be separated into two groups: active and passive. The electrical definition is very similar to the common definition: active circuit elements are capable of delivering power, while passive elements are capable of receiving, and possibly storing, power. In our water analogy, a pump would be an active element. A narrow section of pipe that restricts the flow, a tank, and a water wheel would all be examples of passive elements.

Resistors: circuit elements that literally "resist" current flow. Voltage is higher on the end of the resistor that sees the current first. Figure 1 shows two schematic representations of a resistor. In our water analogy, a resistor would be a narrow section of pipe that restricts the flow.

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Figure 1. Schematic representations of a resistor

The on-resistance (RDS(on)) of our HEXFET® power MOSFETs is usually one of two parameters critical to the designer. The other is breakdown voltage (V(BR)DSS) or how much voltage the device can block when it is off. On-resistance is merely the resistance from drain to source of the power MOSFET in the "on" state. In the "off" state, the resistance is extremely high, but instead of RDS(off), we measure it as leakage current, or IDSS.

Capacitors: circuit elements that store electrons. In many instances, they are used as a rechargeable battery, providing a stable voltage reference far from the input power point. They have many different uses in electrical circuits in addition to simply storing electrons. There are many different types of capacitors, including aluminum electrolytic, tantalum electrolytic, ceramic disk, mica, polycarbonate, polypropylene, and polystyrene.

Two important considerations in the selection of capacitors are equivalent series inductance (ESL), and equivalent series resistance (ESR). Ideally, these two parameters should be as close to zero as possible, especially as frequency increases. The capacitors above are mentioned approximately in order of decreasing ESL and ESR. Aluminum electrolytic capacitors have extremely high capacitive values, but also high ESL and ESR. This makes them good for dc applications, such as the capacitors on the output of a bridge rectifier, to provide the dc supply to the rest of the circuit. Polypropylene and polystyrene capacitors have very low capacitive values, but also extremely low ESR and ESL values making them good for extremely high frequency applications.

Stray capacitance exists in all circuits to some extent. While usually to ground, it can occur between any two points with different potentials. All semiconductor devices have capacitance between their external terminals, and are specified on the data sheets. Figure 2 shows several different schematic representations of capacitors. In our water analogy, a capacitor would be a tank storing water for later use.

Figure 2.Schematic Representations of Capacitors

Stray capacitance is also responsible for electro-static discharge (ESD). ESD is responsible for the shock you receive in the winter after walking across a carpeted room and touching the doorknob. ESD is particularly dangerous to MOS-gated semiconductors. The amount of static required to cause damage is so small, that a person can damage a device without knowing it. This is why anyone who handles MOS-gated semiconductors must follow strict ESD prevention procedures. Following proper procedures is essential as devices can be damaged, reducing their lifetime, with no perceivable effects at the time of damage.

Figure 3. Schematic representations of inductors

Inductors: circuit elements that resist change. If, after a period of current flow, an attempt is made to interrupt the current flow, the inductor will continue to force current. Figure 3 shows the schematic representations of two different inductors. In our water analogy, an inductor would be a water wheel - it is difficult to start spinning, but once it is spinning, it is difficult to stop.

Figure 4.Toroidal Inductor

Inductors are typically manufactured by winding wire in a toroidal (donut) shape shown in Figure 4. If the inductor is wound around a non-ferromagnetic material such as plastic, ceramic, cardboard, or merely air, the inductance per unit volume is considerably less than if the inductor is wound on a ferromagnetic core. The upper inductor in Figure 4 depicts an air-cored inductor, while the lower inductor depicts a ferromagnetic cored inductor. Ferromagnetic refers to magnetic materials, whose characteristics greatly vary.

Figure 5. B-H Characteristics for a Magnetic Material.

Figure S shows the B-H characteristics for a ferromagnetic material where B is the magnetic flux density, and H is the magnetic field. Operation follows the line, in the direction indicated by the arrow. Although the explanation of this figure is beyond the scope of this module, some important concepts can be observed without a thorough understanding of the plot. During operation, the operating point slides along the curve in the direction of the arrows. If the positive magnetic flux density (B) is not offset by an equal negative magnetic flux density, the operation curve will slowly creep up, until the material saturates (magnetic flux density (B) is at a maximum and cannot further increase).

At saturation the inductance drops to the value of an equivalent air-cored inductor, and the current through it is merely limited by the core's internal resistance which is usually quite low. This is seen at the top of the above curve where the lines flatten, and further increases in flux density (B) are not allowed. Saturation can be caused by one of two mechanisms. First, if the magnetic material is underdesigned, and the flux generated by the current in the winding is greater than the core can handle, the material will saturate. In the above figure, this would place the operating point at the top of the B-H curve.

The second method applies if the magnetic material is not allowed to reset between consecutive pulses. Sufficient time between pulses is necessary to allow the energy stored in the magnetic element to go to zero, or reset. If the design does not allow this to occur, the flux in the magnetic element will build up, or staircase, with each consecutive pulse until the device saturates. This results in a large current which usually destroys the semiconductors in its path. This phenomenon also affects transformers which are merely special cases of the inductor.

Figure 6. Schematic Representation of a Transformer

The final circuit element is the transformer. Figure 6 shows the schematic representation of a transformer. A transformer could be thought of as a ferromagnetic-cored inductor with two or more sets of wires wound on it. Saturation is also a problem in transformers. Thus transformers and inductors are sometimes lumped together and simply called magnetics.

Transformers are most commonly used for one of two purposes. The first is isolation, which is typically needed between two sections of a system which have different ground levels. The second is to change voltage levels. A familiar example is the large ac adapter wall plug supplied with most portable equipment for home use. The adaptor box contains a transformer which steps the voltage down from the line voltage, usually to around 12V, which is then further conditioned by two diodes, and finally supplied to the equipment.

Leakage inductance is a critical parameter for transformers, generators, and motors. Leakage inductance is the difference between the self-inductance and the mutual inductance of the primary and secondary windings. Its value is typically quite small, but very important in determining the characteristics and operation of the circuit. It is of particular interest as the switching device may be asked to dissipate the energy stored in the leakage inductance. The leakage inductance contributes to a turn-off voltage spike seen by the switching device. If the energy and/or voltage is sufficient, a snubber may need to be added to the circuit to protect the switching device from damage due to this spike. IR specifies the amount of energy HEXFET® power MOSFETs can dissipate in this mode and are tested as shown in Figure 7, the unclamped inductive test circuit.

Figure 7. Unclamped Inductive Test Circuit

Basic Electrical Definitions

Power is defined as current multiplied by voltage:

P=Vx I

where P is the power measured in watts (W) (also joules per second), V is the steady state voltage measured in volts (V), and I is the steady state current measured in amps (A).

Energy is defined as current multiplied by voltage, multiplied by time:

E=IxVxT

where "E" is the energy measured in joules (also watt-seconds), "V" is the instantaneous voltage measured in volts, "i" is the instantaneous current measured in amps, and "T" is the time period measured in seconds.

To calculate power, given energy and frequency, multiply energy by the frequency. For example, if an IGBT has a total switching energy loss of 1.4mJ under a given set of operating conditions, and is operated at 20kHz, the total power loss due to switching will be 28W.

E (1.4mJ) x f (20kHz) = P (28W)

ac versus dc

Direct current (dc) has a constant magnitude. In contrast, alternating current (ac) has a magnitude dependent on time. it follows a sinusoidal waveform, shown below. ac is generated by moving a copper winding through a magnetic field. This causes a voltage to be developed on the winding. Generators in the United States operate at 60Hz, but many places in the world, 50Hz is the standard. Hz is the abbreviation for Hertz, which is the unit of measure for frequency. Frequency is only defined for regular waveforms that repeat indefinitely. Frequency is how many times per second the same position on the waveform occurs. Thus, in the figure below, sixty peaks will pass in one second if the frequency is 60Hz. T is the period, while 1/T is the frequency.

Figure 8. 60Hz Sine Wave.

Nearly all current starts off as ac, which is generated through an electromechanical process, and is then converted to dc. It is difficult to generate dc directly, as it requires either a dynamo or a chemical reaction such as the one within in a solar cell which converts sunlight into dc voltage. In applications where dc is present, there is usually a nearby ac source. For example, in your automobile the battery that drives the lights, all the electronics, and all the motors are typically 12 volts dc. This battery is charged by the alternator which is basically a small generator driven by the engine. A three-phase diode bridge is responsible for converting the ac output of the alternator to be compatible with the dc battery.

The last important concept is the role of frequency on magnetics. It is beyond the scope of this training module to explain why, but as the operating frequency of a circuit increases, the physical size of the magnetics (remember this means both inductors and transformers) shrinks. This is one of the reasons designers are constantly increasing the frequency of their designs. In the power supply world, one of the benchmarks of a design is how many watts per cubic inch the power supply delivers. One way to substantially increase this number is by moving to higher frequency, and hence, physically smaller magnetic components. The tradeoff of higher frequency operation is increased switching losses in the semiconductor devices, whether it be a diode, IGBT, or power MOSFET.

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Basic Semiconductor Theory

Semiconductor is an appropriate name for the device because it perfectly describes the material from which it's made -- not quite a conductor, and not quite an insulator. To produce a useful device, semiconductor material is processed into MOSFET, IGBT, SCR, or diode devices, etc.

All IR semiconductor products originate from silicon wafers. Silicon is one of the most common elements on earth. It is the basis of sand and glass in the form of silicon dioxide (SiO2). After being refined, silicon is supplied as amorphous silicon which means that the atoms are randomly arranged in the material. Under the proper conditions, silicon can be manufactured into epitaxial chunks which basically means a single crystal. Most gems are examples of single crystals. Diamonds, for example, are merely carbon atoms arranged in a particular 3-D or lattice structure shown in Figure 9. We make use of silicon in a similar lattice form to manufacture semiconductors.

Figure 9. Silicon Lattice Structure.

A pure silicon lattice is a good insulator because the atoms are arranged so that all electrons are bonded to a silicon nucleus. In order to change the resistivity of the silicon, it is necessary to introduce impurities, which changes the number of electrons in the lattice structure. This is called doping, and may result in extra electrons (called n-type), or missing electrons (called p-type) in the lattice. Typical dopants include boron and phosphorous.

Missing electrons are also called holes. There is no physical basis for this nomenclature, but it is easy to understand how it came about. Assume that you have six paper cups and five balls. Line the cups in a row, and put balls in the five right cups. Now move the ball in the second cup to the first, the ball in the third to the second cup and so on. It appears that the empty cup is moving to the right when in reality, the balls are merely shifting to the left. This movement is possible because of the different number of cups and balls. The significance of this is that electrons can move approximately two times faster than holes, and as such, n-type material is much preferred to

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p-type material.

Semiconductor devices can also be divided into two groups based on how current is conducted through the material. A device in which the current is conducted by the charges dominant in the lattice is called a majority carrier device (e.g., electrons in n-type material, or holes in p-type material. If the current is conducted by charges not dominant in the lattice, the resulting device is called a minority carrier device (e.g., electrons in p-type material, or holes in retype material). This very important distinction has a large bearing on the device's operation, especially the recovery characteristics. A fish tank can be used to illustrate how current is conducted through silicon. If you put an air hose onto the bottom of the fish tank, it takes some time for the air to bubble out of the tank. This is analogous to a minority carrier device (current carrier is different from the bulk media). On the other hand, if you put a water hose onto the bottom of the tank, it doesn't take any time for the water to reach equilibrium. This is analogous to a majority carrier device (current carrier is the same as the bulk media).

Basic Semiconductor Processing

Wafers are sliced from a single silicon crystal which has to be "grown." This is done by melting silicon in a crucible. Pure silicon occurs in two forms - either as a single crystal, or as a collection of atoms with no particular arrangement, called polysilicon. A "seed" or a small silicon crystal is inserted into the crucible holding the molten polysilicon. As the seed is slowly drawn out, the molten silicon aligns with the crystal lattice in the seed. As it cools, the molten silicon expands on this crystal lattice forming an ingot as shown in Figure 10. The entire ingot is drawn out as a single crystal made up of many silicon atoms. This ingot is then sliced into thin wafers, and each wafer is polished to a mirror-like finish. The mirror-like finish of the silicon wafer needs to have a pattern etched into it to make a useful circuit, or circuit element (discrete).

Figure 10. Monocrystalline Silicon Ingot.

Figure 11 depicts a mask used to transfer the desired pattern onto the silicon wafer. The mask pattern (either positive or negative) is then projected onto the wafer by one of several different methods. A particular device, or design, requires a number of different masks - this collection is known as a mask set. The fewer masks in the mask set. the lower the processing costs.

Figure 11. Mask.

The mask contains one image repeated numerous times. The masks are used in the photolithography process to transfer the patterns to the silicon wafer. A photo-sensitive material is applied to the wafer, and the mask exposes certain areas of the wafer. This causes a chemical change in the photo-sensitive material. A chemical is then used to etch portions away, leaving a pattern on the wafer. This is repeated a number of times, with some or all of the following intermediate steps occurring between mask steps.

Diffusion is the process by which dopants are added to the wafer. By using the appropriate mask, a certain pattern is diffused into the wafer. Diffusion is usually followed by a drive-in step by heating the wafer for a particular amount of time. Controlling the time and temperature defines the profile diffused into the wafer.

The diffusion process must be tightly controlled. With advances in IC processing, a method called ion implantation has been developed. Instead of controlling the time and temperature of a diffusion furnance, ion implantation makes use of an extremely high voltage electron gun which accelerates the dopants, and "shoots" them into the wafer. By adjusting the high voltage, the implant depth is controlled. It is possible to get very precise profiles by using this method.

The "wires" of the integrated circuit world are constructed using either metal, or polysilicon, which may be heavily doped to reduce its resistance. Either of these materials can be used to cover the entire surface, or to make tightly controlled patterns via the aforementioned processes. Sometimes problems happen with "step coverage" of the metal as shown in Figure 12., when the metal has to cover what looks like a single stair step. The metal thickness tends to thin at the outer most portion of the step, and can lead to failure if the metal becomes discontinuous across this step. Silicon dioxide is commonly used to insulate the metal from contacting other layers.

Figure 12. Metal Step Coverage.


Device Cross Section

The following sections show the cross sectional views and describes the operation of six different devices. More detailed information is available in each device's specific Training Module. The goal of this section is to compare and contrast the various devices: pn diode, Schottky diode, SCR, MOSFET, IGBT, and Control IC.

pn Diode

As shown in Figure 13, the top metal is the anode, while the bottom is the cathode. The action occurs at the interface, called the junction, between the implanted p-type and n-type materials. When a positive voltage is applied between the anode and cathode, current will flow through the diode, provided the voltage is greater than "a diode drop" which, for standard pn diodes, is usually around 0.7V. As the forward current (IF) increases, the voltage drop (VF) will also increase. However, most of the voltage drop is the initial 0.7V drop which occurs when any amount of current flows through the diode.

Figure 13. PN Diode

If a negative voltage is applied across the pn junction (anode to cathode), the device exhibits very high resistance to current flow, and the small amount of current that does conduct is called the leakage current (IRM).

When a diode is conducting in the forward direction and is asked to block in the reverse direction, it

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"forgets" it is a diode for a period of time and allows current to conduct. After this short period of time, called the reverse recovery time, or trr, the diode "remembers" it is a diode and begins blocking current. However, during this recovery time, a large current conducts through the diode, called reverse recovery current, or Irr. The shape of the waveforms during this period are critical to the operation of the rest of the circuit, which is why IR has developed the HEXFRED® diode, an ultra-fast, but ultra-soft diode unlike snappier diodes from our competitors that usually cause excessive voltage ringing in the circuit. As temperature increases, the forward voltage decreases, while the reverse recovery current and charge increase.

Schottky Diode

As shown in Figure 14, the Schottky diode is very similar to a standard pn diode, but instead of having an implanted p-layer, the action occurs at the interface between the barrier metal and the silicon. The guard rings are used to make the device's reverse breakdown characteristics more rugged. Since both metal and the silicon are n-type materials, the conduction occurs through majority carriers only, with no minority carrier injection, storage, or recombination . This explains the Schottky diode's lack of reverse recovery, making it ideal for high frequency applications.

Figure 14. Schottky Diode

The barrier metal also is responsible for the Schottky diode's low forward voltage drop, making it ideal for use in low voltage systems. Of course, the tradeoff is the reverse leakage current, which is many times that seen in pn-junction diodes. In some applications, and especially during burn-in, this leakage current may cause the device to exceed its rated junction temperature. It needs to be included in any junction temperature calculations. As temperature increases, the forward drop decreases, while the reverse leakage current greatly increases.

Figure 15. SCR

SCR

The SCR (silicon controlled rectifier), or thyristor, is one of the original high power semiconductor switching technologies. As shown in Figure 15, the SCR is a four layer device, npnp from top to bottom (or cathode to anode). It is a latching device; once it is turned on, or "fired," it remains on until the current is removed. For this reason, its primary application is phase-control of ac signals. Figure 16 shows that by controlling where on the cycle the SCR is turned on, the output power level is controlled. SCRs designed for these line frequency (50-60 Hz) applications are called phase control SCRs.

Figure 16. Phase Control of ac Waveform

The second family of SCRs is the inverter type. These are used in pulsed power applications involving higher frequencies. The main difference between the two families is the turn-off time (tq). A device's tq is measured as the time required for the device to be in the "off" state before voltage is reapplied. Inverter SCRs typically have a tq of less than 30 microseconds (ms). Similar phase control SCRs have tq ratings of several hundred ms.

Like with the pn diode, as the temperature increases, the voltage drop decreases. Perhaps more importantly, as temperature increases, the current required to fire the SCR decreases. At low temperatures, the gate triggering circuitry must supply enough current to ensure the device fires, while at high temperatures, the SCR is susceptible to spurious firing due to noise. The device's gate triggering circuitry must ensure this does not happen.

Figure 17. HEXFET power MOSFET

MOSFET

As shown in Figure 17 above, the HEXFET® power MOSFET is named for the hexagonal shape of its individual cells. Current flows from the source metallization down through the device, and out through the drain contact. Vertical current flow is the reason the HEXFET is also called a vertical MOSFET. Nearly all power MOSFETs on the market employ this vertical structure.

The MOSFET (Metal Oxide Semiconductor Field Effect Transistor) is used primarily in medium-power circuits where switching speed is critical. This power device is extremely easy to drive as it requires voltage, not current on the gate. And due to its wide acceptance, the MOSFET market is growing at a rapid pace.

It is important to select the proper voltage MOSFET as the RDS(on) increases exponentially with increasing breakdown voltage. Also, as the device heats up due to power dissipation, its RDS(on) increases. Thus, in most applications, the 25 degC RDS(on) rating is not accurate. The actual RDS(on) rating could be twice as high. This is not always a negative attribute. It's what allows power MOSFETs to easily be used in parallel.

IGBT

The IGBT (Insulated Gate Bipolar Transistor) represents the union of a power MOSFET and a power bipolar (BJT) transistor, incorporating the best features of each. But while the MOSFET can be used in applications exceeding 1 MHz, the fastest IGBTs are limited to only a fraction of that. Therefore, the only real drawback of the IGBT is its switching speed. Yet the conduction characteristics of the IGBT really outshine those of the power MOSFET, especially at voltages greater than about 200V. If you have a midfrequency, high voltage design, look at IGBTs.

Increasing the operating temperature of an IGBT causes its switching losses to increase substantially, thus decreasing the maximum operating frequency. The conduction voltage of IGBTs is not strongly affected by temperature, and can even decrease with temperature at certain current densities. This causes concern for designers with regards to the IGBT's parallelibility. Still, IR believes that by following a few simple guidelines, IGBTs can be successfully paralleled. (See Design Tip 94-6 for more information on parallel operation of IGBTs.)

Control IC

In 1987, IR introduced the IR2110, a half-bridge, high voltage, MOS Gate Driver, or Control IC . Since that time, many variations on the IR2110 have been introduced. These devices are unique in that they simplify the drive circuitry required to drive a high-side, n-channel MOSFET at voltages up to 600V. These devices allow designers to reduce the parts count and design time for the drive circuitry. Figure 18 shows a cross-section of the die. One of the most successful devices in this family is the IR2151, a self-oscillating, half-bridge driver designed primarily for electronic ballast applications.

Future 18. Control IC

Modes of Operation

The switching devices (MOSFET, IGBT, and SCR) can be operated in one of several modes. The SCR is unique in this group in that it is a self-commutating device, which means the user is required to turn on, or "fire" the SCR. But at the zero crossing, the SCR switches off. This makes the SCR extremely useful in ac applications, or where the current decays to zero at the point when the SCR should be turned off. One limitation of the SCR is that it cannot easily be used to switch dc loads.

By far the largest number of IR devices are used in hard switching as depicted in Figure 19. The switching waveforms on data sheets are for hard switching. The conditions are very difficult on the switching device, i.e., high current must be switched off to high voltage. Considerable power is dissipated in the switching interval due to these conditions.

Figure 19. Hard Switching Waveforms

Pulse-Width Modulation (PWM) is a special case of hard switching. In many applications, it is desirable to replicate a sine wave as shown in Figure 20. One way to accomplish this is to approximate the sine wave with narrow square pulses of varying duty cycle. After this waveform is smoothed, typically by an inductor, it appears very similar to the desired sine wave. The frequency of the desired sine wave is called the carrier frequency, while the frequency at which the switch operates is called the modulation frequency. To make the replicated waveform closely match the desired waveform, the modulation frequency is usually at least ten times the carrier frequency.

Some applications employ resonant mode switching. In these applications, the current and/or voltage is a sine wave as opposed to the square waves common in PWM techniques. Operation in the resonant mode has lower switching losses, and is used with devices that have high switching losses, or to push operating frequency higher.

Figure 20. Sine Wave Generation through PWM.

Some applications operate in the linear mode. As applied to switching devices, linear mode means that discrete changes in the control signal result in proportional discrete changes in the output. When a circuit is operated in the linear mode, the switching element limits the current in the circuit, while normally the circuit itself, rather than the switching element, limits the current. This limitation leads to high power dissipation in the switching device.

Parallel operation of semiconductors requires extra effort on the part of the design engineer. When operating semiconductors in parallel, the critical parameter is the temperature coefficient. The temperature coefficient reflects how the semiconductor's voltage drop responds to changes in temperature. In general, semiconductors have a negative temperature coefficient, with the one notable exception of the power MOSFET in that its temperature coefficient is positive. A negative temperature coefficient means that the voltage drop across the semiconductor decreases as the temperature increases. This causes problems such as current hogging, thermal runaway, and hotshots. As the device heats up due to normal power dissipation, the voltage drop decreases. This allows more current to flow, generates more heat, and further reduces the voltage drop, creating a regenerative effect. When paralleling any IR device (other than the MOSFET) special design considerations are necessary to prevent

potential current sharing problems.

The following documents provide more information on this subject: IR Design Tip 94-6A, "Parallel Operation of IGBTs"; IR Application Note AN-990, "Application Characterization of IGBTs"; and "Paralleling of Power MOSFETs for Higher Power Output," by James B. Forsythe (PowerCon '81).

Packaging

Many IR devices are available either in plastic or hermetically sealed metal packages. In the past, the deciding which package to use was divided between commercial versus military. Today the division is not as clear. Military does not automatically mean hermetic, nor does hermetic automatically mean more reliable. However, most space applications still require hermetic packages. In applications where the device is exposed to high temperature, and high humidity, a hermetically packaged device will improve its reliability. The TO-3, once the power transistor package, is no longer competitive with the new package styles. The TO-3 is difficult to heat sink, and must be isolated externally. New packages, such as the TO-254 (M-Pak) are tab mounted, and isolated, making assembly much easier.

Most heat sinkable plastic packages have a metal tab connected to the heat sink by the user. The die is mounted directly on this tab, usually the positive terminal of the device. In some applications, it is desireable to have the heat sink grounded, while in other applications it is easier to insulate the heat sink from the rest of the system. In cases with a grounded heat sink, it is necessary to isolate the device from the heat sink. For this purpose, IR manufactures special versions of the TO-220 and TO-247 packages called Full-Paks that have a very thin plastic coating on the exposed metal of the device. The plastic provides up to 2500Vrms isolation voltage, while being thin enough to only moderately increase the thermal resistance of the mounting system. As an alternative to buying these Full-Pak devices from IR, the user can isolate the back of the package from the heat sink using a thermally conductive, electrically isolating material as shown in Figure 22. Some common types are mica, pressed ceramic wafer, polyimide, and elastomeric insulators, with the latter gaining more and more popularity. The screw hole in the TO-247ac package is already isolated, so only the back of the package needs to be isolated. The screw hole of the TO-220AB is not isolated, so the user must isolate it. One way is by using a nylon shoulder washer with a standard steel screw. The steel screw is typically 4-40 whereas a 6-32 screw is normally used when isolation is of no concern.

The extremely broad R product line is packaged in anything from a tiny,3-lead surface mount package (barely visible to the naked eye) to a huge "hockey puk" package (greater than 4" in diameter!). IR application note AN-995 discusses IR's various surface mount packages, and how to mount them. Most high power products (SCRs, and diodes in both discrete and module packages) are simple to mount. These devices are typically large, and connnected using large bolts, or are stud mounted. Remember, however, not to exceed the torque or force specified on the data sheet. For the hockey puk packages, a suitable mounting clamp must be used as shown in Figure 22 below. In fact, the puk will appear open-circuited if pressure is not applied since this is a compression bonded device.

Figure 22. Isolation System for Standard TO-220AB and TO-247AC

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Figure 23. Mounting Hockey Puk SCRs and Diodes










Mounting Techniques

Most mounting questions concern the TO-220AB and TO-247AC package styles used to house diodes, MOSFETs, and IGBTs, and the three methods used to make electrical connections to the die within these packages: wire bonding, soldering, and compression mounting. Each method has its own advantages and disadvantages.

Figure 23. Wire Bonding

Wire bonding, shown in Figure 23, uses a small diameter (typically <=20 mils) wire that is ultrasonically bonded (melted) at each connection point. Advantages: wire bonding is quick and easy, and can be completely automated. Disadvantages: increased voltage drop due to the small wires, low fusing current, expensive equipment required.

Solder mounting shown in Figure 24 below is typically used in smaller diodes (< 10A), mostly the familar axial-leaded diode.

Figure 24. Solder Mounting

Some smaller IR Schottky diodes (SMB and SMC) also use this technique. Advantages: both the voltage drop, and fusing current are improved. For example, the 30BQ015 Schottky diode is rated at 3 amps and 15 volts, but the surge rating is 650 amps because the die is soldered directly to the leadframe. Disadvantage: both sides of the die must be solderable.

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Figure 25. Compression Bonding

Compression bonding (Figure 25 above) is used in devices where power cycling capability is required, typically high power diodes and SCRs. Advantage: in these applications, compression bonding makes a much better connection because there is no fixed (soldered) interface to fatigue, and the fusing current is very high. Disadvantage: compression bonding is more expensive, and requires physically rugged die. All IR hockey puks are compression bonded. Some stud mounted diodes and SCRs, and some diode and SCR modules are also compression bonded.

Topical Applications

International Rectifier manufactures components used in the efficient control of energy. High efficiency is a major requirement because of today's rising energy costs and the need for battery-powered systems to be able to run extended periods of time.

In Figure 26, the line power or raw energy source may be an electric utility, automobile alternator (ac), battery (dc), or a power supply. IR components take this raw power and condition it into more useful energy. Some examples are switching off unneeded circuits in portable electronics, variable ac to control motors, and variable dc to control motors, electronic lamp ballasts, etc. All IR components fit into the basic power conversion function sub-blocks:

Figure 26. The Basic Power Conversion Functions

These blocks are similar to the USDA's Basic Food Groups wherein each meal should include one item from each group. Similarly, designs will often need one or more items from each group. The following sections discuss each block in detail, the associated IR product, and how to use them.

Input Rectification

Input rectification is the first stage in most electronic devices using ac power. The design uses four diodes/rectifiers arranged in a bridge configuration for single-phase inputs, or six diodes arranged in a bridge for three-phase inputs. Standard recovery diodes are specified since the speed of the diode is not important. Standard recovery diodes have excellent forward voltage drop, and have lower relative costs than other families of diodes.

Figure 27. Bridge Rectifier

A bridge rectifier (both single- and three-phase shown in Fig. 27 above) converts the ac waveform on the left to the dc level on the right. IR sells both single- and three-phase bridges in plastic isolated-base modules. Diode bridges can also comprise discrete diodes, or doubler diode modules (two diodes in series). The choice depends on the desired mounting method, and the current requirement.

In the above example, the dc voltage will equal the peak of the ac input voltage due to the capacitor on the output of the bridge rectifier. Since ac voltage is measured in volts-rms, peak voltage is equal to 1.414 times the rms voltage. Select the rectifiers based on this peak voltage, allowing extra margin for high line conditions.

SCRs may be used to limit current, or output voltage. In some cases, designers require SCRs in their input bridges. Phase control SCRs are used in smart bridge configurations, so called because they can be used to limit inrush current.

A problem with passive rectification is that current is only drawn from the line when the line voltage exceeds the capacitor voltage. This results in a current waveform shown on the left in the above diagram. The mismatch between the current and voltage waveforms is called power factor, and results in inefficient utilization of the power source. Additionally, some government regulations require new designs to meet a specific power factor.

Typical Applications

A boost converter (shown below) is the most common circuit used to improve the power factor. As shown on the left, the voltage and current waveforms are very similar. This results in near-perfect power factoring.

Figure 28. Boost Converter

Figure 29. DC Chopper

DC Chopper

The dc output voltage is in the 400 to 500 volt range, and calls for a 450 to 600 volt rating on both the power switch and the diode. IR has focused on this application with its low gate charge HEXFET® power MOSFETs and HEXFRED® ultrafast diodes. This configuration is one most commonly used in power electronics circuits. Depending on the application, a transformer may be used to provide voltage isolation or to change voltage levels according to the turns ratio between the primary and secondary sides. This single switch, single ended configuration is also known as a forward converter in the power supply world. In power supply applications, the inductance and the freewheeling diode are on the secondary side of the transformer.

Much emphasis is placed on the selection of the switch (either a MOSFET or IGBT), but the selection of the diode is equally important. The diode characteristics affect the operation of the switch itself. For low voltage applications, a Schottky diode is ideal. For higher voltage applications, IR's HEXFRED is an excellent choice.

Figure 30. Half Bridge

Half Bridge

The half-bridge is a higher power version of the previous circuit. It is the workhorse of the power electronics industry, finding use in power supplies, motor controls, and lighting ballasts. In a power supply, the inductor is actually a transformer. In a motor drive, the inductor is the motor windings. In a lighting ballast, the inductor is in series with the lamp, which is in parallel with the lower capacitor.

The greatest design problem with the half-bridge configuration is driving the upper switch, which can be either an IGBT or a HEXFET. To properly drive a MOS-gated transistor, the gate voltage must be greater than the emitter/source voltage by approximately 15V. When the emitter/source terminal is connected to a fixed voltage reference (i.e., ground in the case of the bottom switch), this is a simple task. However, the emitter/source of the high-side switch swings between ground (when the lower switch is conducting), and nearly the positive rail, which requires the gate voltage to be above the positive rail. This is typically a problem, since the positive rail

is usually the highest voltage available in the system.

Several methods are used to solve the problem of driving the upper switch (for a list, see AN978A), all of which are relatively complex, and each has drawbacks. IR produces a range of devices to solve this problem: the IR2100 Series Control IC drivers. They use a technique called bootstrapping to generate the gate drive signal for the upper switch. IR manufactures a line of these devices that focus on the electronic ballast market, the IR215x. They include both control circuitry and gate drive circuitry--all on one chip.

In even higher power circuits, the two capacitors on the right side of the circuit may be replaced with a half-bridge identical to the one on the left. This is called a full bridge. In addition to higher power applications, full bridges are also used in reversible dc motor drives. The configuration allows voltage to be applied to the load in both directions.

Typical Applications

Figure 31. Three-Phase Bridge

The three-phase bridge can be thought of as three half-bridges. Three-phase outputs are mainly used for motor drives or ac inverters. The IR2130 Control IC has been designed specifically for this application.

Figure 32. Push-Pull Configuration

The push-pull configuration is used in power supply and low power UPS systems.

Figure 33. Two Transistor Forward Converter

The two transistor forward converter topology is commonly used in power supply and switched reluctance (SR) motor designs. Its advantage is that the voltage requirement of the switch and freewheeling diode is half that of the one transistor forward topology, often used in off-line power supplies requiring the greater capabilitiesof 800 volt HEXFET® power MOSFETs. The increased benefits and lower cost outweigh the complexity of the design. Typical applications include low power converters with high switching frequencies.

Figure 34. High Side dc Switch

Placing the switch on the high side offers protection against the most common short circuit, a short to the chassis (ground). If this occurs, the load is connected to ground on both ends, so the ground is unenergized. However, if the switch and load were reversed, a short to ground would energize the load.

IR manufactures the lR62xx series of high side intelligent switches. They offer overcurrent, overtemperature, and ESD protection. And because these devices accept ground-referenced logic-level signals as control, the problem of driving a high-side switch is addressed.

In some applications, high-side intelligent switches can be used to replace electromagnetic relays. However, applications that require voltage isolation must use IR microelectronic relays. These solid state devices cost more than equivalent mechanical devices, but feature superior reliability. Additionally, microelectronic relays are available to control ac loads. This line has found application in telecommunications, instrumentation, and process control.










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