EE 435 SCR Notes3

Fig.3.1 Half wave single phase controlled rectifier.

EE 435- Electric Drives Dr. Ali M. Eltamaly Chapter 3

Thyristor Converters or Controlled Converters 3.1 Introduction The controlled rectifier circuit is divided into three main circuits:-

(1) Power Circuit This is the circuit contains voltage source, load and switches as diodes, thyristors or IGBTs.

(2) Control Circuit This circuit is the circuit, which contains the logic of the firing of switches that may, contains amplifiers, logic gates and sensors.

(3) Triggering circuit This circuit lies between the control circuit and power thyristors. Sometimes this circuit called switch drivers circuit. This circuit contains buffers, opt coupler or pulse transformers. The main purpose of this circuit is to separate between the power circuit and control circuit.

The method of switching off the thyristor is known as Thyristor commutation. The thyristor can be turned off by reducing its forward current below its holding current or by applying a reverse voltage across it. The commutation of thyristor is classified into two types:-

1- Natural Commutation If the input voltage is AC, the thyristor current passes through a natural zero, and a reverse voltage appear across the thyristor, which in turn automatically turned off the device due to the natural behavior of AC voltage source. This is known as natural commutation or line commutation. This type of commutation is applied in AC voltage controller rectifiers and cycloconverters.

2- Forced Commutation In DC thyristor circuits, if the input voltage is DC, the forward current of the thyristor is forced to zero by an additional circuit called commutation circuit to turn off the thyristor. This technique is called forced commutation. Normally this method for turning off the thyristor is applied in choppers.

There are many thyristor circuits we can not present all of them. In the following items we are going to present and analyze the most famous thyristor circuits.

3.2 Half Wave Single Phase Controlled Rectifier 3.2.1 Half Wave Single Phase Controlled Rectifier With Resistive Load The circuit with single SCR 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 analyzed. 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 are based on the ideal SCR model. All simulation carried out by using PSIM computer simulation program.

A circuit with a single SCR and resistive load is shown in Fig.3.1. The source vs is an alternating sinusoidal

Chapter Three 30

source. If ( )tVv ms ωsin= , vs is positive when 0 < ω t < π, and vs is negative when π < ω t <2π. When vs starts become positive, the SCR is forward-biased but remains in the blocking state till it is triggered. If the SCR is triggered at ω t = 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 ω t reaches π radians. At that instant, the current through the circuit is 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 SCR. Various voltages and currents waveforms of the half-wave controlled rectifier with resistive load are shown in Fig.3.2 for α=40o. FFT components for load voltage and current of half wave single phase controlled rectifier with resistive load at α=40o are shown in Fig.3.3. It is clear from Fig.3.3 that the supply current containes DC component and all other harmonic components which makes the supply current highly distorted. For this reason, this converter does not have acceptable practical applecations.

Fig.3.2 Various voltages and currents waveforms for half wave single-phase controlled rectifier with

resistive load at α=40o.

Fig.3.3 FFT components for load voltage and current of half wave single phase controlled rectifier at α =40o.

The average voltage, dcV , across the resistive load can be obtained by considering the waveform

shown in Fig.3.2. )cos1(2

))cos(cos(2

)sin(21 α

παπ

πωω

π

π

α

+=+−== ∫ mmmdc

VVtdtVV (3.1)

SCR Rectifier or Controlled Rectifier 31

The maximum output voltage and can be acheaved when 0=α which is the same as diode case which obtained before in (2.12). π/mdm VV = (3.2) The normalized output voltage is the DC voltage devideded by maximum DC voltage, dmV which can be obtained as shown in equation (3.3). )cos1(5.0/ α+== mdcn VVV (3.3)

The rms value of the output voltage is shown in the following equation:-

( )

+−== ∫ 2

)2sin(12

)sin(21 2 ααπ

πωω

π

π

α

mmrms

VtdtVV (3.3)

The rms value of the transformer secondery current and load is: RVI rmss /= (3.4)

Example 1 In the rectifier shown in Fig.3.1 it has a load of R=15 Ω and, Vs=220 sin 314 t and unity transformer ratio. If it is required to obtain an average output voltage of 70% of the maximum possible output voltage, calculate:- (a) The firing angle, α, (b) The efficiency, (c) Ripple factor (d) Peak inverse voltage (PIV) of the thyristor Solution: (a) Vdm is the maximum output voltage and can be acheaved when 0=α , The normalized output voltage is shown in equation (3.3) which is required to be 70%. Then,

7.0)cos1(5.0 =+== αdm

dcn V

VV . Then, α=66.42o =1.15925 rad.

(b) 220=mV V , VVVV mdmdc 02.49*7.0*7.0 ===

π, A

RVI dc

dc 268.315

02.49===

+−=

2)2sin(1

2ααπ

πm

rmsVV at α=66.42o, Vrms=95.1217V. Then, Irms=95.122/15=6.34145A

VVV mS 56.155

2== , AII rmsS 34145.6==

Then, the rectification efficiency is: %56.2634145.6*121.95268.3*02.49

**

====rmsrms

dcdc

ac

dc

IVIV

PPη

(b) 94.12202.49

121.95====

π

dc

rmsVVFF

(c) 6624.1194.11 22 =−=−== FFVVRF

dc

ac , (d) The PIV is mV

3.2.2 Half Wave Single Phase Controlled Rectifier With RL Load A circuit with single SCR and RL load is shown in Fig.3.4. The source vs is an alternating sinusoidal source. If vs = Vm sin (ω t), vs is positive when 0 < ω t < π, and vs is negative when π <ω t <2π. When vs starts become positive, the SCR is forward-biased but remains in the blocking state till it is triggered. If the SCR is triggered when ω t = α, then it starts conducting and the positive source keeps the SCR in conduction till ω t

Fig.3.4 Half wave single phase controlled rectifier with RL load.

Chapter Three 32

reaches π radians. At that instant, the current through the circuit is not zero and there is some energy stored in the inductor at ω t = π radians. The voltage across the 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 ω t = 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. Fig.3.5 shows the output voltage, resistor, inductor voltages and thyristor voltage drop waveforms.

Fig.3.5 Various voltages and currents waveforms for half wave single phase controlled rectifier with RL

load.

3.3 Single-Phase Full Wave Controlled Rectifier 3.3.1 Single-Phase Center Tap Controlled Rectifier With Resistive Load Center tap controlled rectifier is shown in Fig.3.8. When the upper half of the transformer secondary is positive and thyristor T1 is triggered, T1 will conduct and the current flows through the load from point a to point b. When the lower half of the transformer secondary is positive and thyristor T2 is triggered, T2 will conduct and the current flows through the load from point a to point b. So, each half of input wave a unidirectional voltage (from a to b ) is applied across the load. Various voltages and currents waveforms for center tap controlled rectifier with resistive load are shown in Fig.3.9 and Fig.3.10.

Fig.3.8 Center tap controlled rectifier with resistive load.

a b


Fig.3.9 The output voltgae and thyristor T1 reverse voltage wavforms along with the supply voltage

wavform.

Fig.3.10 Load current and thyristors currents for Center tap controlled rectifier with resistive load.

The average voltage, Vdc, across the resistive load is given by:

)cos1())cos(cos()sin(1 απ

αππ

ωωπ

π

α

+=−−== ∫ mmmdc

VVtdtVV (3.27)

Vdm is the maximum output voltage and can be acheaved when α=0 in the above equation. The

normalized output voltage is: )cos1(5.0 α+==dm

dcn V

VV (3.28)

Chapter Three 34

From the wavfrom of the output voltage shown in Fug.3.9 the rms output voltage can be obtained as

following: ( )2

)2sin(2

)sin(1 2 ααππ

ωωπ

π

α

+−== ∫ mmrms

VtdtVV (3.29)

Example 4 The rectifier shown in Fig.3.8 has load of R=15 Ω and, Vs=220 sin 314 t and unity transformer ratio. If it is required to obtain an average output voltage of 70 % of the maximum possible output voltage, calculate:- (a) The delay angle α, (b) The efficiency, (c) The ripple factor (d) The peak inverse voltage (PIV) of the thyristor. Solution : (a) Vdm is the maximum output voltage and can be acheaved when α=0, the normalized output voltage is shown in equation (3.28) which is required to be 70%. Then:

7.0)cos1(5.0 =+== αdm

dcn V

VV , then, α=66.42o

(b) 220=mV , then, VVVV mdmdc 04.982*7.0*7.0 ===

π

AR

VI dcdc 536.6

1504.98

=== , 2

)2sin(2

ααππ

+−= mrms

VV

at α=66.42o Vrms=134.638 V. Then, Irms=134.638/15=8.976 A VVV mS 56.1552/ == , AII rmsS 347.62/ ==

Then, The rectification efficiency is: %04.53976.8*638.134

536.6*04.98**

====rmsrms

dcdc

ac

dc

IVIV

PPη

(c) 3733.104.98638.134

===dc

rmsVVFF and, 9413.013733.11 22 =−=−== FF

VVRF

dc

ac

(d) The PIV is 2 Vm 3.3.2 Single-Phase Fully Controlled Rectifier Bridge With Resistive Load This section describes the operation of a single-phase fully-controlled bridge rectifier circuit with resistive load. The operation of this circuit can be understood more easily when the load is pure resistance. The main purpose of the fully controlled bridge rectifier circuit is to provide a variable DC voltage from an AC source.

The circuit of a single-phase fully controlled bridge rectifier circuit is shown in Fig.3.11. The circuit has four SCRs. For this circuit, vs is a sinusoidal voltage source. When the supply voltage is positive, SCRs T1 and T2 triggered then current flows from vs through SCR T1, load resistor R (from up to down), SCR T2 and back into the source. In the next half-cycle, the other pair of SCRs T3 and T4 conducts when get pulse on their gates. Then current flows from vs through SCR T3, load resistor R (from up to down), SCR T4 and back into the source. Even though the direction of current through the source alternates from one

Fig.3.11 Single-phase fully controlled rectifier bridge with resistive load.


half-cycle to the other half-cycle, the current through the load remains unidirectional (from up to down).

Fig.3.12 Various voltages and currents waveforms for converter shown in Fig.3.11 with resistive load.

Fig.3.13 FFT components of the output voltage and supply current for converter shown in Fig.3.11.

The main purpose of this circuit is to provide a controllable DC output voltage, which is brought about by varying the firing angle, α. Let vs = Vm sin ω t, with 0 < ω t < 360o. If ω t = 30o when T1 and T2 are triggered, then the firing angle α is said to be 30o. In this instance, the other pair is triggered when ω t = 30+180=210o. When vs changes from positive to negative value, the current through the load becomes zero at the instant ω t = π radians, since the load is purely resistive. After that there is no current flow till the other is triggered. The conduction through the load is discontinuous. The average value of the output voltage is obtained as follows.:-

Let the supply voltage be vs = Vm*Sin (ω t), where ω t varies from 0 to 2π radians. Since the output waveform repeats itself every half-cycle, the average output voltage is expressed as a function of α, as shown in equation (3.27).

( )[ ] )cos1()cos(cos)sin(1 απ

αππ

ωωπ

π

α

+=−−−== ∫ mmmdc

VVtdtVV (3.27)

Vdm is the maximum output voltage and can be acheaved when α=0, The normalized output voltage is: )cos1(5.0/ α+== dmdcn VVV (3.28) The rms value of output voltage is obtained as shown in equation (3.29).

Chapter Three 36

( )2

)2sin(2

)sin(1 2 ααππ

ωωπ

π

α

+−== ∫ mmrms

VtdtVV (3.29)

Example 5 The rectifier shown in Fig.3.11 has load of R=15 Ω and, Vs=220 sin 314 t and unity transformer ratio. If it is required to obtain an average output voltage of 70% of the maximum possible output voltage, calculate:- (a) The delay angle α, (b) The efficiency, (c) Ripple factor of output voltage(d) The peak inverse voltage (PIV) of one thyristor. Solution: (a) Vdm is the maximum output voltage and can be acheaved when α=0,

The normalized output voltage is shown in equation (3.31) which is required to be 70%.

Then, 7.0)cos1(5.0 =+== αdm

dcn V

VV , then, α=66.42o

(b) 220=mV , then, VVVV mdmdc 04.982*7.0*7.0 ===

π, A

RVI dc

dc 536.615

04.98===

2)2sin(

2ααπ

π+−= m

rmsVV

At α=66.42o Vrms=134.638 V. Then, Irms=134.638/15=8.976 A

VVV mS 56.155

2==

The rms value of the transformer secondery current is: AII rmsS 976.8==

Then, The rectification efficiency is %04.53976.8*638.134

536.6*04.98**

=====rmsrms

dcdc

ac

dc

IVIV

PPη

(c) 3733.104.98638.134

===dc

rmsVVFF , 9413.013733.11 22 =−=−== FF

VVRF

dc

ac

(d) The PIV is Vm

3.3.3 Full Wave Fully Controlled Rectifier With RL Load In Continuous Conduction Mode The circuit of a single-phase fully controlled bridge rectifier circuit is shown in Fig.3.14. The

main purpose of this circuit is to provide a variable DC output voltage, which is brought about by varying the firing angle. The circuit has four SCRs. For this circuit, vs is a sinusoidal voltage source. When it is positive, the SCRs T1 and T2 triggered then current flows from +ve point of voltage source, vs through SCR T1, load inductor L, load resistor R (from up to down), SCR T2 and back into the –ve point of voltage source. In the next half-cycle the current flows from -ve point of voltage source, vs through SCR T3, load resistor R, load inductor L (from up to down), SCR T4 and back into the +ve point of voltage source. 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 (from up to down). Fig.3.15 shows various voltages and currents waveforms for the converter shown in Fig.3.14. Fig.3.16 shows the FFT components of load voltage and supply current. Fig.3.14 Full wave fully controlled rectifier with RL load.


Fig.3.15 Various voltages and currents waveforms for the converter shown in Fig.3.14 in continuous

conduction mode.

Fig.3.16 FFT components of load voltage and supply current in continuous conduction mode.

Let vs = Vm sin ω t, with 0 < ω t < 360o. If ω t = 30o when T1 and T2 are triggered, then the

firing angle is said to be 30o. In this instance the other pair is triggered when ω t= 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 ω t = 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.

Chapter Three 38

Since the output waveform repeats itself every half-cycle, the average output voltage is expressed in equation (3.33) as a function of α, the firing angle. The maximum average output voltage occurs at a firing angle of 0o as shown in equation (3.34). The rms value of output voltage is obtained as shown in equation (3.35).

απ

ωωπ

απ

α

cos2)sin(1 mmdc

VtdtVV == ∫+

(3.33)

The normalized output voltage is αcos/ == dmdcn VVV (3.34) The rms value of output voltage is obtained as shown in equation (3.35).

( )2

)2cos(1(2

)sin(1 2 mmmrms

VtdtVtdtVV =−== ∫∫++ απ

α

απ

α

ωωπ

ωωπ

(3.35)

Example 6 The rectifier shown in Fig.3.14 has pure DC load current of 50 A and, Vs=220 sin 314 t and unity transformer ratio. If it is required to obtain an average output voltage of 70% of the maximum possible output voltage, calculate:- (a) The delay angle α, (b) The efficiency, (c) Ripple factor (d) The peak inverse voltage (PIV) of the thyristor and (e) Input displacement factor. Solution: (a) Vdm is the maximum output voltage and can be acheaved when α=0. The normalized

output voltage is shown in equation (3.30) which is required to be 70%. Then, 7.0cos === αdm

dcn V

VV ,

then, α=45.5731o= 0.7954 (b) 220=mV , VVVV mdmdc 04.98/2*7.0*7.0 === π , 2/mrms VV = At α=45.5731o Vrms=155.563 V. Then, Irms=50 A, VVV mS 56.1552/ == The rms value of the transformer secondery current is: AII rmsS 50==

Then, The rectification efficiency is %02.6350*563.155

50*04.98**

====rmsrms

dcdc

ac

dcIVIV

PP

η

(c) 587.104.98563.155

===dc

rmsVVFF , 23195.113733.11 22 =−=−== FF

VVRF

dc

ac

(d) The PIV is Vm 3.3.5 Single Phase Full Wave Fully Controlled Rectifier With Source Inductance: Full wave fully controlled rectifier with source inductance is shown in Fig.3.19. The presence of source inductance changes the way the circuit operates during commutation time. Let vs = Vm sin wt, with 0 < ω t < 360o. Let the load inductance be large enough to maintain a steady current through the load. Let firing angle α be 30o. Let SCRs T3 and T4 be in conduction before ω t < 30o. When T1 and T2 are triggered at ω t = 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 T3 and T4 to T1 and T2 would not occur instantaneously. The source current changes from dcI− to dcI due to the whole of the source voltage being applied across the source inductance. When T1 is triggered with T3 in conduction, the current through T1 would rise from zero to dcI and the current through T3 would fall from dcI to zero. Similar process occurs with the SCRs T2 and T4. During this period, the current through T2 would rise from zero to dcI and, the current through T4 would fall from dcI to zero.


Fig.3.19 single phase full wave fully controlled rectifier with source inductance

Various voltages and currents waveforms of converter shown in Fig.3.19 are shown in Fig.3.20 and Fig.3.21. You can observe how the currents through the devices and the line current change during commutation overlap.

Fig.3.20 Output voltage, thyristors current along with supply voltage waveform of a single phase full

wave fully controlled rectifier with source inductance.

Fig.3.21 Output voltage, supply current along with supply voltage waveform of a single phase full wave

fully controlled rectifier with source inductance.

Chapter Three 40

( ) αω

α −

−= −

m

os

VILu 2coscos 1 (3.42)

osos

rd IfLILV 42

4==

πω (3.47)

The DC voltage with source inductance taking into account can be calculated as following:

osm

rdceinducsourcewithoutdcactualdc IfLVVVV 4cos2tan −=−= α

π (3.48)

The rms value of supply current is the same as obtained before in single phase full bridge

diode rectifier in (2.64)

−=

322 2 uII o

sπ

π (3.49)

The Fourier transform of supply current is the same as obtained for single phase full bridge diode rectifier in (2.66) and the fundamental component of supply current 1sI is shown in (2.68)

as following: 2

sin*28

1u

uII o

S π= (3.50)

The power factor of this rectifier is shown in the following:

+=

2cos. 1 u

IIfp

s

s α (3.51)

3.3.6 Inverter Mode Of Operation The thyristor converters can also operate in an inverter mode, where dV has a negative value, and hence the power flows from the do side to the ac side. The easiest way to understandd the inverter mode of operation is to assume that the DC side of the converter can be replaced by a current source of a constant amplitude dI , as shown in Fig.3.25. For a delay angle a greater than 90° but less than 180°, the voltage and current waveforms are shown in Fig.3.26. The average value of dv is negative, given by (3.48), where 90° < α < 180°. Therefore, the average power

ddd IVP *= is negative, that is, it flows from the DC to the AC side. On the AC side, 11 cosφSsac IVP = is

also negative because o90>φ .

Fig.3.26 Waveform output from single phase inverter assuming DC load current.

Fig.3.25 Single phase SCR inverter.


There are several points worth noting here. This inverter mode of operation is possible since there is a source of energy on the DC side. On the ac side, the ac voltage source facilitates the commutation of current from one pair of thyristors to another. The power flows into this AC source.

Generally, the DC current source is not a realistic DC side representation of systems where such a mode of operation may be encountered. Fig.3.27 shows a voltage source dE on the DC side that may represent a battery, a photovoltaic source, or a DC voltage produced by a wind-electric system. It may also be encountered in a four-quadrant DC motor supplied by a back-to-back connected thyristor converter.

An assumption of a very large value of dL allows us to assume di to be a constant DC, and hence the waveforms of Fig.3.28 also apply to the circuit of Fig.3.27. Since the average voltage across dL is zero,

dsdodd ILVVE ωπ

α 2cos −== (3.55)

Fig.3.27 SCR inverter with a DC voltage source.

Fig.3.28 dV versus dI in SCR inverter with a DC voltage source.

Chapter Three 42

Fig.3.29 Voltage across a thyristor in the inverter mode. Inverter startup

For startup of the inverter in Fig.3.25, the delay angle α is initially made sufficiently large (e.g.,165o) so that di is discontinuous as shown in Fig.3.30. Then, α is decreased by the controller such that the desired dI and dP are obtained.

Fig.3.30 Waveforms of single phase SCR inverter at startup.

3.5 Three Phase Half Wave Controlled Rectifier 3.5.1 Three Phase Half Wave Controlled Rectifier with Resistive Load Fig.3.31 shows the circuit of a three-phase half wave controlled rectifier, the control circuit of this rectifier has to ensure that the three gate pulses for three thyristor are displaced 120o relative to each other’s. Each thyristor will conduct for 120o. A thyristor can be fired to conduct when its anode voltage is positive with respect to its cathode voltage. The maximum output voltage occurred when α=0 which is the same as diode case. This rectfier has continuous load current and voltage in case of α ≤ 30. However, the load voltage and current will be discontinuous in case of α > 30.

In case of α ≤ 30, various voltages and currents of the converter shown in Fig.3.31 are shown in Fig.3.32. Fig.3.33 shows FFT components of load voltage, secondary current and primary current. As we see the load voltage contains high third harmonics and all other triplex harmonics. Also secondary current contains DC component, which saturate the transformer core. The saturation of the transformer core is the main drawback of this system. Also the primary current is highly distorted but without a DC component. The average output voltage and current are shown in equation (3.57) and (3.58) respectively. The rms output voltage and current are shown in equation (3.59) and (3.60) respectively.

ααπ

ααπ

ωωπ

απ

απ

cos675.0cos23cos827.0cos

233sin

23 6/5

6/LLLLm

mmdc VVVVtdtVV ===== ∫

+

+

(3.57)

ααπ

cos*827.0cos**2

33R

VR

VI mmdc == (3.58)

( ) απ

ωωπ

απ

απ

2cos8

3613sin

23 6/5

6/

2 +== ∫+

+mmrms VtdtVV (3.59)

Fig.3.31 Three phase half wave controlled rectifier.


απ

2cos8

3613

+=RVI m

rms (3.60)

Then the thyristor rms current is equal to secondery current and can be obtaiend as follows:

απ

2cos8

361

3+===

RVIII mrms

Sr (3.61)

The PIV of the diodes is mLL VV 32 = (3.62)

Fig.3.32 Voltages and currents waveforms for rectifier shown in Fig.3.31 at α ≤ 30.

Fig.3.33 FFT components of load voltage, secondary current and supply current for the converter shown in Fig.3.31

for α ≤ 30.

Chapter Three 44

In case of α > 30, various voltages and currents of the rectifier shown in Fig.3.31 are shown in Fig.3.34. Fig.3.35 shows FFT components of load voltage, secondary current and primary current. As we can see the load voltage and current equal zero in some regions (i.e. discontinuous load current). The average output voltage and current are shown in equation (3.63) and (3.64) respectively. The rms output voltage and current are shown in equation (3.65) and (3.66) respectively. The average output voltage is :-

++=

++== ∫

+

απαππ

ωωπ

π

απ 6cos14775.0

6cos1

23sin

23

6/m

mmdc VVtdtVV (3.63)

++= απ

π 6cos1

23

RVI m

dc (3.64)

( ) )23/sin(81

42453sin

23

6/

2 απππ

αωωπ

π

απ

++−== ∫+

mmrms VtdtVV (3.65)

)23/sin(81

42453

απππ

α++−=

RVI m

rms (3.66)

Then the diode rms current can be obtaiend as follows:

)23/sin(81

4245

3απ

ππα

++−===RVIII mrms

Sr (3.67)

The PIV of the diodes is mLL VV 32 = (3.68)


Fig.3.34 Various voltages and currents waveforms for converter shown in Fig.3.22 for α > 30.

Fig.3.35 FFT components of load voltage, secondary current and supply current for the converter shown

in Fig.3.22 for α > 30.

Example 7 Three-phase half-wave controlled rectfier is connected to 380 V three phase supply via delta-way 380/460V transformer. The load of the rectfier is pure resistance of 5Ω . The delay angle o25=α . Calculate: The rectfication effeciency (b) PIV of thyristors

Solution: From (3.57) the DC value of the output voltage can be obtained as following:

VVV LLdc 5.28125cos46023cos

23

===π

απ

Then; AR

VI dcdc 3.56

55.281

===

From (3.59) we can calculate rmsV as following:

απ

απ

2cos8

361*22cos

83

613 +=+= LLmrms VVV

Chapter Three 46

Then, ( ) VVrms 8.29825*2cos8

361*460*2 =+=

π

Then AR

VI rmsrms 76.59

58.298

===

Then, the rectfication effeciency is, %75.88100* ==rmsrms

dcdcIVIVη

VVPIV LL 54.650460*22 === Example 8 Solve the previous example (evample 7) if the firing angle o60=α Slution: From (3.63) the DC value of the output voltage can be obtained as following:

VVV mdc 33.179

36cos1

2

460*323

6cos1

23

=

++

=

++=

πππ

αππ

Then; AR

VI dcdc 87.35

533.179

===

From (3.65) we can calculate rmsV as following:

VVV mrms 230)3/23/sin(81

43/

245*460*2)23/sin(

81

42453 =++−=++−= ππ

πππαπ

ππα

Then AR

VI rmsrms 46

5230

===

Then, the rectfication effeciency can be calculated as following

%79.60100* ==rmsrms

dcdcIVIVη and VVPIV LL 54.650460*22 ===

3.5 Three Phase Half Wave Controlled Rectifier With DC Load Current

The Three Phase Half Wave Controlled Rectifier With DC Load Current is shown in Fig.3.36, the load voltage will reverse its direction only if α > 30. However if α < 30 the load voltage will be positive all the time. Then in case of α > 30 the load voltage will be negative till the next thyristor in the sequence gets triggering pulse. Also each thyristor will conduct for 120o if the load current is continuous as shown in Fig.3.37. Fig.3.38 shows the FFT components of load voltage, secondary current and supply current for the converter shown in Fig.3.36 for α > 30 and pure DC current load.

Fig.3.36 Three Phase Half Wave Controlled Rectifier With DC Load Current


Fig.3.37 Various voltages and currents waveforms for the converter shown in Fig.3.36 for α > 30 and

pure DC current load.

Fig.3.38 FFT components of load voltage, secondary current and supply current for the converter shown

in Fig.3.36 for α > 30 and pure DC current load.

As explained before the secondary current of transformer contains DC component. Also the source current is highly distorted which make this system has less practical significance. The THD of the supply current can be obtained by the aid of Fourier analysis as shown in the following:-

If we move y-axis of supply current to be as shown in Fig.3.33, then the waveform can be represented as odd function. So, an=0 and bn can be obtained as the following:-

−== ∫ 3

2cos12)sin(2 3/2

0

ππ

ωωπ

π nn

ItdtnIb dcdcn for n=1,2,3,4,…

Then, 23*2

nIb dc

n π= for n=1,2,4,5,7,8,10,….. (3.69)

t=0

Chapter Three 48

And 0=nb For n=3,6,9,12,….. (3.70) Then the source current waveform can be expressed as the following equation

+++++= ......7sin

715sin

514sin

412sin

21sin3)( tttttIti dc

p ωωωωωπ

ω (3.71)

The resultant waveform shown in equation (3.61) agrees with the result from simulation (Fig.3.38). The THD of source current can be obtained by two different methods. The first method is shown below:-

21

21

2

p

pp

I

IITHD

−= Where, dcp II *

32

= (3.72)

The rms of the fundamental component of supply current can be obtained from equation (3.71) and it will be as shown in equation (3.74)

π23

1dc

pII = (3.74)

Substitute equations (3.73) and (3.74) into equation (3.72), then,

%68

29

29

32

22

22

2

=

−

=dc

dcdc

I

IITHD

π

π (3.75)

Another method to determine the THD of supply current is shown in the following:- Substitute from equation (3.71) into equation (3.72) we get the following equation:-

%68....141

131

111

101

81

71

51

41

21 222222222

≅+

+

+

+

+

+

+

+

+

=THD (3.76)

The supply current THD is very high and it is not acceptable by any electric utility system. In case of full wave three-phase converter, the THD in supply current becomes much better than half wave (THD=35%) but still this value of THD is not acceptable.

Example 9 Three phase half wave controlled rectfier is connected to 380 V three phase supply via delta-way 380/460V transformer. The load of the rectfier draws 100 A pure DC current. The delay angle, o30=α . Calculate: (a) THD of primary current. (b) Input power factor. Solution: The voltage ratio of delta-way transformer is 380/460V. Then, the peak value of

primary current is A05.121380460*100 = . Then, AI rmsP 84.98

32*05.121, == .

1PI can be obtained from equation (372) where AII dcP 74.81

205.121*3

23

1 ===ππ

.

Then, ( ) %98.67100*174.8184.98100*1

22

1

, =−

=−

=

P

rmsPI I

ITHD

P

The input power factor can be calculated as following:

LaggingI

IfPrmsP

P 414.066

cos*84.9874.81

6cos*.

,

1 =

+=

+=

πππα


3.7 Three Phase Full Wave Fully Controlled Rectifier Bridge 3.7.1 Three Phase Full Wave Fully Controlled Rectifier With Resistive Load

Three-phase full wave controlled rectifier shown in Fig.3.42. As we can see in this figure the thyristors has labels T1, T2,……,T6. The label of each thyristor is chosen to be identical to triggering sequence where thyristors are triggered in the sequence of T1, T2,……,T6 which is clear from the thyristors currents shown in Fig.3.43.

Fig.3.42 Three-phase full wave controlled rectifier.

Fig.3.43 Thyristors currents of three-phase full wave controlled rectifier.

The operation of the circuit explained here depending on the understanding of the reader the three phase diode bridge rectifier. The Three-phase voltages vary with time as shown in the following equations:

)120(sin)120(sin

)(sin

+=−=

=

tVvtVvtVv

mc

mb

ma

ωωω

It can be seen from Fig.3.44 that the voltage av is the highest positive voltage of the three

phase voltage when tω is in the range ot 15030 << ω . So, the thyristor T1 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.3.44 the firing angle 40=α as an example. So, T1 takes a pulse at

ot 70403030 =+=+= αω as shown in Fig.3.44. Also, it is clear from Fig3.38 that thyristor T1

or any other thyristor remains on for o120 .

Chapter Three 50

Fig.3.44 Phase voltages and thyristors currents of three-phase full wave controlled rectifier at o40=α .

It can be seen from Fig.3.44 that the voltage bv is the highest positive voltage of the three

phase voltage when tω is in the range of ot 270150 << ω . So, the thyristor T3 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.3.44, the firing angle 40=α as an example. So, T3 takes a pulse at

ot 19040150150 =+=+= αω . It can be seen from Fig.3.44 that the voltage cv is the highest positive voltage of the three

phase voltage when tω is in the range ot 390270 << ω . So, the thyristor T5 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.3.44, the firing angle 40=α as an example. So, T3 takes a pulse at

ot 310270 =+= αω . It can be seen from Fig.3.44 that the voltage av is the highest negative voltage of the three

phase voltage when tω is in the range ot 330210 << ω . So, the thyristor T4 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.338, the firing angle 40=α as an example. So, T4 takes a pulse at

ot 25040210210 =+=+= αω . It can be seen from Fig.3.44 that the voltage bv is the highest negative voltage of the three

phase voltage when tω is in the range ot 450330 << ω or ot 90330 << ω in the next period of supply voltage waveform. So, the thyristor T6 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.3.44, the firing angle

40=α as an example. So, T6 takes a pulse at ot 370330 =+= αω . It can be seen from Fig.3.44 that the voltage cv is the highest negative voltage of the three

phase voltage when tω is in the range ot 21090 << ω . So, the thyristor T2 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.3.44, the firing angle 40=α as an example. So, T2 takes a pulse at ot 13090 =+= αω .


From the above explanation we can conclude that there is two thyristor in conduction at any time during the period of supply voltage. It is also clear that the two thyristors in conduction one in the upper half (T1, T3, or, T5) which become forward bias at highest positive voltage connected to its anode and another one in the lower half (T2, T4, or, T6) which become forward bias at highest negative voltage connected to its cathode. So the load is connected at any time between the highest positive phase voltage and the highest negative phase voltage. So, the load voltage equal the highest line to line voltage at any time which is clear from Fig.3.45. The following table summarizes the above explanation.

Period, range of wt SCR Pair in conduction α + 30o to α + 90o T1 and T6 α + 90o to α + 150o T1 and T2 α + 150o to α + 210o T2 and T3 α + 210o to α + 270o T3 and T4 α + 270o to α + 330o T4 and T5 α + 330o to α + 360o and α + 0o to α + 30o T5 and T6

Fig.3.45 Output voltage along with three phase line to line voltages of rectifier in Fig.3.42 at o40=α .

The line current waveform is very easy to obtain it by applying kerchief's current law at the terminals of any phase. As an example 41 TTa III −= which is clear from Fig.3.42. The input current of this rectifier for ( )60,40 ≤= αα is shown in Fig.3.46. Fast Fourier transform (FFT) of output voltage and supply current are shown in Fig.3.47.

Analysis of this three-phase controlled rectifier is in many ways similar to the analysis of single-phase bridge controlled rectifier circuit. 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 is pure resistance.

απ

ωπωπ

απ

απ

cos33

)6

sin(33 2/

6/

mmdc

VtdtVV =+= ∫

+

+

(3.81)

The maximum average output voltage for delay angle α=0 is

Chapter Three 52

πm

dmVV 33

= (3.82)

The normalized average output voltage is as shown in (3.83)

αcos==dm

dcn V

VV (3.83)

The rms value of the output voltage is found from the following equation:

+=

+= ∫

+

+

απ

ωπωπ

απ

απ

2cos4

33213)

6sin(33 2/

6/

2

mmrms VtdtVV (3.84)

Fig.3.46 The input current of this rectifier of rectifier in Fig.3.42 at ( )60,40 ≤= αα .

Fig.3.46 FFT components of output voltage and supply current of rectifier in Fig.3.42 at

( )60,40 ≤= αα .


In the converter shown in Fig.3.42 the output voltage will be continuous only and only if o60≤α . If o60>α the output voltage, and phase current will be as shown in Fig.3.47.

Fig.3.47 Output voltage along with three phase line to line voltages of rectifier in Fig.3.42 at o75=α . The average and rms values of output voltage is shown in the following equation:

( )[ ]αππ

ωπωπ

π

απ

++=+= ∫+

3/cos133)6

sin(33 6/5

6/

mmdc

VtdtVV (3.85)

The maximum average output voltage for delay angle α=0 is: π

mdm

VV 33= (3.86)

The normalized average output voltage is ( )[ ]απ ++== 3/cos1dm

dcn V

VV (3.87)

The rms value of the output voltage is found from the following equation:

+−−=

+= ∫

+6

2cos24313)

6sin(33 6/5

6/

2 πααπ

ωπωπ

π

απmmrms VtdtVV (3.88)

Example 10 Three-phase full-wave controlled rectifier is connected to 380 V, 50 Hz supply to feed a load of 10Ω pure resistance. If it is required to get 400 V DC output voltage, calculate the following: (a) The firing angle, α (b) The rectfication effeciency (c) PIV of the thyristors. Solution: From (3.81) the average voltage is :

VR

VV mdc 400cos380*

32*33cos33

=== ααπ

.

Then o79.38=α , AR

VI dcdc 40

10400

===

From (3.84) the rms value of the output voltage is:

( )

+=

+= 79.38*2cos

433

21*380*

32*32cos

433

213

πα

πmrms VV

Then, VVrms 412.412= Then, AR

VI rmsrms 24.41

10412.412

===

Then, %07.94100*24.41*4.412

40*400100***

===rmsrms

dcdcIVIVη

The PIV= 3 Vm=537.4V

Chapter Three 54

Example 11 Solve the previous example if the required dc voltage is 150V. Solution: From (3.81) the average voltage is :

VVV mdc 150cos

380*32*33

cos33=== α

πα

π. Then, o73=α

It is not acceptable result because the above equation valid only for 60≤α . Then we have to use the (3.85) to get dcV as following:

( )[ ] VVdc 1503/cos1380*

32*33

=++== αππ

. Then, o05.75=α

Then AR

VI dcdc 15

10150

===

From (3.88) the rms value of the output voltage is:

( )

+−−=

+−−= 3005.75*2cos

180*05.75*2

431*380*

32*3

62cos2

4313 π

ππαα

πmrms VV

Then, VVrms 075.198= , Then, AR

VI rmsrms 8075.19

10075.198

===

Then, %35.57100*81.19*075.198

15*150100***

===rmsrms

dcdcIVIVη

The PIV= 3 Vm=537.4V

3.7.1 Three Phase Full Wave Fully Controlled Rectifier With pure DC Load Current Three-phase full wave-fully controlled rectifier with pure DC load current is shown in Fig.3.48. Fig.3.49 shows various currents and voltage of the converter shown in Fig.3.48 when the delay angle is less than 60o. As we see in Fig.3.49, the load voltage is only positive and there is no negative period in the output waveform. Fig.3.50 shows FFT components of output voltage of rectifier shown in Fig.3.48 for o60<α .

Fig.3.48 Three phase full wave fully controlled rectifier with pure dc load current

In case of the firing angle is greater than o60 , the output voltage contains negaive portion as shown in Fig.3.51. Fig.3.52 shows FFT components of output voltage of rectifier shown in Fig.3.48 for o60>α . The average and rms voltage is the same as in equations (3.81) and (3.84) respectively. The line current of this rectifier is the same as line current of three-phase full-wave diode bridge rectifier typically except the phase shift between the phase voltage and phase current is zero in case of diode bridge but it is α in case of three-phase full-wave controlled rectifier


with pure DC load current as shown in Fig.3.53. So, the input power factor of three-phase full-wave diode bridge rectifier with pure DC load current is:

αcos1

s

sIIrPowerFacto = (3.89)

Fig.3.49 Output voltage and supply current waveforms along with three phase line voltages for the rectifier

shown in Fig.3.48 for α < 60o with pure DC current load.

Fig.3.50 FFT components of SCR, secondary, primary currents respectively of rectifier shown in

Fig.3.48.

Chapter Three 56

Fig.3.51 Output voltage and supply current waveforms along with three phase line voltages for the

rectifier shown in Fig.3.48 for α > 60o with pure DC current load.

Fig.3.52 FFT components of SCR, secondary, primary currents respectively of rectifier shown in Fig.3.48

for α > 60o.

Fig.3.53 Phase a voltage, current and fundamental components of phase a of three phase full bridge fully

controlled rectifier with pure DC current load and 60>α .


In case of three-phase full-wave controlled rectifier with pure DC load and source inductance the waveform of output voltage and line current and their FFT components are shown in Fig.3.54 and Fig.3.55 respectively. The output voltage reduction due to the source inductance is the same as obtained before in Three-phase diode bridge rectifier. But, the commutation time will differ than the commutation time obtained in case of Three-phase diode bridge rectifier. It is left to the reader to determine the commutation angle u in case of three-phase full-wave diode bridge rectifier with pure DC load and source inductance. The Fourier transform of line current and THD will be the same as obtained before in Three-phase diode bridge rectifier with pure DC load and source inductance which explained in the previous chapter.

Fig.3.54 Output voltage and supply current of rectifier shown n Fig.3.48 with pure DC load and source

inductance the waveforms.

Fig.3.55 FFT components of output voltage of rectifier shown in Fig.3.48 for α > 60o and there is a

source inductance.

( ) αωα −

−= −

LL

oSV

ILu 2coscos 1 (3.99)

ord fLIV 6= (3.104) The DC voltage without source inductance tacking into account can be calculated as following:

Chapter Three 58

osLLrdceinducsourcewithoutdcactualdc IfLVVVV 6cos23tan −=−= α

π (3.105)

Then

−=

632 2 uII o

Sπ

π (3.106)

=

2sin62

1u

uII o

S π (3.110)

The power factor can be calculated from the following equation:

+

−

=

+

−

=

+=

2cos

63

2sin*32

2cos

632

2sin62

2cos

21 u

uu

uu

uI

uu

Iu

IIpf

o

o

S

S αππ

απ

π

πα (3.111)

Note, if we approximate the source current to be trapezoidal as shown in Fig.3.58n, the

displacement power factor will be as shown in (3.111) is

+

2cos uα . Another expression for the

displacement power factor, by equating the AC side and DC side powers [ ] as shown in the following derivation: From (3.98) and (3.105) we can get the following equation:

( )( )uVVV LLLLdc +−−= ααπ

απ

coscos23cos23

( )( )[ ]uVV LLdc +−−=∴ αααπ

coscoscos22

23 ( )[ ]uVV LLdc ++=∴ ααπ

coscos2

23 (3.112)

Then the DC power output from the rectifier is odcdc IVP = . Then,

( )[ ]uIVP oLLdc ++= ααπ

coscos*2

23 (3.113)

On the AC side, the AC power is: 11 cos3 φSLLac IVP = (3.114) Substitute from (3.110) into (3.114) we get the following equation:

11 cos2

sin26cos2

sin2

343 φπ

φπ

=

=

uu

IVuu

IVP oLLoLLac (3.115)

By equating (3.113) and (3.115) we get the following: ( )[ ]

++=

2sin*4

coscoscos 1 uuu ααφ (3.116)

3.7.2 Inverter Mode of Operation Once again, to understand the inverter mode of operation, we will assume that the do side of the converter can be represented by a current source of a constant amplitude dI , as shown in Fig.3.61. For a delay angle a greater than 90° but less than 180°, the voltage and current waveforms are shown in Fig.3.62a. The average value of dV is negative according to (3.81). On the ac side, the negative power implies that the phase angle 1φ , between sv and si , is greater than 90°, as shown in Fig.3.62b.


Fig.3.61 Three phase SCR inverter with a DC current.

Fig.3.62 Waveforms in the inverter shown in Fig.3.56.

In a practical circuit shown in Fig.3.63, the operating point for a given dE and α can be obtained from the characteristics shown in Fig.3.64.

Similar to the discussion in connection with single-phase converters, the extinction angle ( )uo −−= αγ 180 must be greater than the thyristor turn-off interval qtω in the waveforms of

Fig.3.54, where 5v is the voltage across thyristor 5.

Fig.3.63 Three phase SCR inverter with a DC voltage source.

Chapter Three 60

Fig.3.64 dV versus dI of Three phase SCR inverter with a DC voltage source.

Inverter Startup As discussed for start up of a single-phase inverter, the delay angle α in the three-phase inverter of Fig.3.63 is initially made sufficiently large (e.g., 165°) so that id is discontinuous. Then, α is decreased by the controller such that the desired dI and dP are obtained.

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EE 435 SCR Notes3