ECE 320 Energy Conversion and Power Electronics

Dr. Tim Hogan

Chapter 8: Power Electronics (Textbook Chapter 10, and Sections 11.2, 11.3, and reserve book: Power Electronics)

Chapter Objectives

As we saw in the last chapter, control over the torque and speed of the motor can be gained through voltage and frequency control to the motor. This can be accomplished by converting the input AC source power to a DC source (rectifying it), then filtering it to reduce harmonics, and finally converting it back to an AC source having the desired frequency and amplitude (inverter).

8.1 Line Controlled Rectifiers We start with a description of how to draw power from a 1-phase or 3-phase system to provide DC to

a load. The characteristics of the systems here include that the devices used will turn themselves off (commutate) and that the systems draw reactive power from the loads.

8.1.1 One-Phase and Three-Phase Circuits with Diodes If the source is 1-phase, a diode is used and the load is purely resistive, as shown in Figure 1 then it is

a relatively simple configuration. When the source voltage is positive, the current flows through the diode and the voltage of the source equals the voltage of the load.

Rvd

+

i

vs

+

vdiode+

i

vs, v

d

vs, v

diode

vdiode t

Figure 1. Simple circuit with diode and resistive load. If the load includes an inductance and a source (such as a battery we wish to charge), as in Figure 2,

then the diode will continue to conduct even when the load voltage becomes negative as long as the current is maintained. This comes from the characteristics of the inductor:

( ) ( ) 0013

0

3=−=∫ itidtv

L

tL (8.1)

Thus, the shaded area A in Figure 2 must equal the shaded area B.

vd

+

i

vs

+

vdiode+

L

+ vL

Ed

+

00

0

00

00

0

00

0s 5ms 10ms 15ms 20ms 25ms0V

0V

0V

0s 5ms 10ms 15ms 20ms 25ms0V

0V

0V

0V

0V

0V

0V

0V

0V

i

Ed

t

vdiode

t

t

vL

t1 t2 t3

vs

A

B

Figure 2. Simple circuit with diode and inductive load with voltage source.

8.1.2 One-Phase Full Wave Rectifier More common is a single phase diode bridge rectifier such as shown in Figure 3. The load can be

modeled with one of two extremes: either as a constant current source, representing the case of a large

inductance that keeps the current through it almost constant, or as a resistor, representing the case of minimum line inductance. We will study the first case with AC and DC side current and voltage waveforms shown in Figure 4 for the ideal case of Ls = 0.

vs

+ L

+

vdCd

i

s

s

id

Figure 3. One-phase full wave rectifier.

0 5 10 15 20 250

0

0

0 5 10 15 20 250

0

0

vs

is

Idt

0

0

0

0

0

0

vd

id = Id

t

Figure 4. Waveforms for a one-phase full wave rectifier with inductive load.

If we analyze these waveforms, the output voltage will have a DC component, Vdo (where the

subscript o represents that this is the ideal case with Ls = 0):

ssdo VVV 9.022≈=

π (8.2)

where Vs is the RMS value of the input AC voltage. On the other hand the RMS value of the output voltage will be

ds VV = (8.3)

containing components of higher frequency. Similarly on the AC side the current is not sinusoidal, rather it changes abruptly between Id and –Id.

dds III 9.0221 ≈=

π (8.4)

and again, the RMS values are the same

sd II = (8.5)

The total harmonic distortion, THD, can then be found as:

%43.481

21

2≈

−=

s

ssI

IITHD (8.6)

It is important to note here that if the source has some inductance (and it usually does), then commutation will be delayed after the voltage reaches zero, until the current has dropped to zero as shown in Figure 5. This will lead to a decrease of the output DC voltage below what is expected from (8.2)

vs

+ L

+

vd

i

s

s

id

I d

16 20 24 28 32 36 400V

0V

0V

16 20 24 28 32 36 400V

0V

0V

vd

t

0V

0V

0V

0V

0V

0V

AA

vs

vL

t

is

t

Figure 5. One-phase full wave rectifier with inductive load and source inductance.

8.1.3 Three-Phase Full Diode Rectifiers The circuit of Figure 3 can be modified to handle three phases, by using 6 diodes as shown in Figure 6. Figure 7 shows the AC side currents and DC side voltage for the case of high load inductance. Similar analysis as before shows that on the DC side, the voltage is:

lllldo VVV −− == 35.123π

(8.7)

From Figure 6, we see that on the AC side, the RMS current, Is is:

dds III 816.032

== (8.8)

while the fundamental current, i.e. the current at power frequency is:

dds III 78.0611 ==

π (8.9)

Again, inductance on the AC side will delay commutation, causing a voltage loss, i.e. the DC voltage will be less than that predicted by equation (8.7).

a+

L

i

s

a

id

R

+

vdCdb+

c+

n

Figure 6. Three-phase full-wave rectifier with diodes.

0s 10ms 20ms 30ms 40ms0V

0V

0V

0s 10ms 20ms 30ms 40ms0V

0V

0V

0s 10ms 20ms 30ms 40ms0V

0V

0V

vd

Vdo

van vbn vcn

t

0A0A

AA

ia

D1

D4

120º

120º t

60º

ib

0A0A

D2

D5 t

ic

D3

D6 t

Figure 7. Waveforms of a three-phase full-wave rectifier with diodes and inductive load.

8.1.4 Controlled Rectifiers with Tyristors Thyristors give us the ability to vary the DC voltage. Remember that to make a thyristor start conducting, the thyristor must be forward biased and a gate pulse provided to its gate. Also, to turn off the thyristor the current through it must reverse direction for a short period of time, trr, and return to zero.

8.1.5 One-Phase Controlled Rectifiers Fig. shows the same 1-phase bridge we have already studied, now with thyristors instead of diodes, and fig shows the output voltage and input current waveforms. In this figure α is the delay angle, corresponding to the time we delay triggering the thyristors after they became forward biased.

vs

+

+

vd

is

I d

T1 T3

T4 T2

Figure 8. One-phase full wave converter with thyristors.

Thyristors 1 and 2 are triggered together and of course so are 3 and 4. Each pair of thyristors is turned off immediately (or shortly) after the other pair is turned on by gating. Analysis similar to that for diode circuits will give:

( ) ( )ααπ

cos9.0cos22ssdo VVV == (8.10)

and the relation for the currents is the same

( ) ( )ααπ

cos9.0cos221 dds III == (8.11)

It should also be pointed out that in Figure 9 the current waveform on the AC side is offset in time with respect to the corresponding voltage by the same angle α, hence so is the fundamental of the current, resulting in a lagging power factor.

vd

is

ωt

α = 0 vs

1 6 2 0 2 4 2 8 321 4 3 500 V

0 V

00 V

1 6 2 0 2 4 2 8 321 4 3 500 V

0 V

00 V

vd

ωt

-100

0

100

-100

0

100

α > 0 vs

α

is

ωt

vs

Figure 9. Waveforms of one-phase full wave converter with thyristors.

On the DC side, only the DC component of the voltage carries power, since there is no harmonic

content in the current. On the AC side the power is carried only by the fundamental, since there are no harmonics in the voltage.

( ) ddss IVIVP == αcos1 (8.12)

8.1.5.1 Inverter Mode If the current on the DC side is sustained even if the voltage reverses polarity, then power will be

transferred from the DC to the AC side. The voltage on the DC side can reverse polarity when the delay angle exceeds 90º, as long as the current is maintained. This can only happen when the load voltage is as shown in Figure 10, e.g. a battery.

vs

+

+

vd

is

T1 T3

T4 T2

L

Ed+

id

Figure 10. Operation of a one-phase controlled converter as an inverter.

8.1.6 Three-Phase Controlled Converters

As with diodes, only 6 thyristors are needed to accommodate three phases. Figure 11 shows the schematic of the system, and Figure 12 shows the output voltage waveforms.

a+ia

id

b+

c+

n

+

vd

I dia

ia

T1 T3 T5

T2T6T4

Figure 11. Schematic of a three-phase full-wave converter based on thyristors.

15ms 20ms 25ms 30ms 35ms00

0

00

15ms 20ms 25ms 30ms 35ms00

0

00

15ms 20ms 25ms 30ms 35ms00

0

00

vdVdo

van vbn vcnia

ωt

α = 0

16ms 20ms 24ms 28ms 32ms16ms 20ms 24ms 28ms 32mss 16ms 20ms 24ms 28ms 32ms16ms 20ms 24ms 28ms 32mss 16ms 20ms 24ms 28ms 32ms16ms 20ms 24ms 28ms 32ms16ms 20ms 24ms 28ms 32mss

vd

Vdαvan vbn vcnia

ωt

α > 0

Figure 12. Waveforms of a three-phase full-wave converter based on thyristors. The delay angle α is again measured from the point that a thyristor becomes forward biased, but in this case the point is at the intersection of the voltage waveforms of two different phases. The voltage on the DC side is then (the subscript o here again meaning Ls = 0):

( ) ( )ααπ

cos35.1cos23lllldo VVV −− == (8.13)

while the power for both the AC and DC side is

( ) ( )αα cos3cos35.1 1slldllddo IVIVIVP ⋅⋅=== −− (8.14)

which leads to:

ds II 78.01 = (8.15)

and the relationship between Vdo and Vdα in Figure 12 is:

( )αα cosdod VV = (8.16)

Again, if the delay angle α is extended beyond 90º, the converter transfers power from the DC side to the AC side, becoming an inverter. We should keep in mind, though that even in this case the converter is drawing reactive power from the AC side.

8.1.7 Notes • For both 1-phase and 3-phase controlled rectifiers, a delay in α creates a phase displacement of the

phase current with respect to the phase voltage, equal to α. The cosine of this angle is the power factor of the fundamental harmonic.

• For both motor and generator modes the controlled rectifier absorbs reactive power from the three-phase AC system, although it can either absorb or produce real power. It also needs the power line to commutate the thyristors. This means that inverter operation is possible only with the rectifier is connected to a power line.

• When a DC motor or a battery is connected to the terminals of a controlled rectifier and α becomes greater than 90º, the terminal DC voltage changes polarity, but the direction of the current stays the same. This means that in order for the rectifier to draw power from the battery or a motor that operates as a generator turning in the same direction, the terminals have to be switched.

8.2 Inverters Here we study systems that can convert DC to AC through the use of devices that can be turned on and off such as GTOs, BJTs, IGBTs, and MOSFETs, which allows the transfer of power from the DC source to any AC load, and gives considerable control over the resulting AC signal. The general block diagram of the complete system is shown in Figure 13.

Rectifier

AC Motor

60 (Hz) AC

Filter Capacitor

Switch-mode inverter

+ Vd _

Figure 13. Typical variable voltage and variable frequency system. 8.2.1 One-Phase Inverter

Figure 14 shows the operation of one leg of an inverter regardless of the number of phases. To illustrate the point better, the input DC voltage is divided into two equal parts. When the upper switch,

S1, is closed while S2 is open, the output voltage VAo will be 2dV+ , and when the lower switch, S2, is

closed while S1 is open, the output voltage will be 2dV−

vAo +

vAN

+

S1

S2Vd

+Vd2

Vd2

Figure 14. One leg of an inverter.

To control the output waveform the switches can be controlled by pulsed width modulation, PWM,

where the time each switch is closed can be determined by the difference between a control waveform, and a carrier (or triangular) waveform as shown in Figure 15. When the control wave is greater than the triangular wave, S1 is closed, and S2 is open. When the control wave is less than the triangular wave, S1 is open, and S2 is closed. In this way, the width of the output is modulated (hence the name).

T im e

16 m s 1 8 ms 2 0 m s 2 2m s 2 4 m s 26 m s 2 8 ms 3 0 m s 3 2m s 34 m sV 1 (V t r i ) V 1( V c o nt )

- 1 . 0 V

- 0 . 5 V

0 . 0 V

0 . 5 V

1 . 0 V

Figure 15. Control (red sinewave) and carrier or triangular waveform (green) determine when the switches are closed. For the times when the control wave is greater than the triangular wave, S1 is closed, and S2 is open. When the control wave is less than the triangular wave, S1 is open, and S2 is closed.

Time

16ms 18ms 20ms 22ms 24ms 26ms 28ms 30ms 32ms 34msV(3)- V(2)

-60V

-40V

-20V

-0V

20V

40V

60V

Figure 16. Output voltage VAo corresponding to the control and carrier waves shown in Figure 15.

The frequency of the triangular wave is fc, and the frequency of the control wave is fo. We define the ratio of these as the frequency modulation index, mf.

o

cf f

fm = (8.17)

Likewise, the amplitude modulation index, ma, is defined as the ratio of the control voltage to the triangular wave voltage.

triangular

controlVVma = (8.18)

A one phase, bull wave inverter is shown in Figure 17. It has four controlled switches, each with an

antiparallel diode.

S1

S2

S4

S3AoVd

+Vd2

Vd2

B

v =AB v -Ao vBo

+

Figure 17. One-phase full wave inverter. The diagonal switches operate together such that S1 and S3 open and close together, and S2 and S4

open and close together. The output will oscillate between 2dV+ and 2

dV− . If the Fourier

transformation of the pulse width modulated square wave shown in Figure 18 is taken, the amplitude of the fundamental will be a linear function of the amplitude index Vo = maVd/2 as long as ma ≤ 1. Then the RMS value of the output voltage will be:

dada

o VmVmV 353.0221 == (8.19)

Time

16ms 18ms 20ms 22ms 24ms 26m s 28m s 30 ms 32 ms 34m sV(2)- V(3)

-12 0V

-8 0V

-4 0V

- 0V

4 0V

8 0V

12 0V

Figure 18. Output voltage VAB for the one-phase, full wave inverter of Figure 17.

When ma increases beyond 1, the output voltage increases also, but not linearly with ma. The output

amplitude can reach a peak value of dVπ4 when the reference signal becomes infinite and the output is a

square wave. Under this condition, the RMS value is:

dd

o VVV 45.02

221 ==

π (8.20)

Equating the power of the DC side with that of the AC side gives

pfIVIVP oodod ⋅== 11 (8.21)

Thus for normal operation:

pfImI oado ⋅= 1353.0 (8.22)

and in the limit for a square wave:

pfII odo ⋅= 145.0 (8.23)

8.2.2 Three-Phase Inverter For three-phase loads it makes more sense to use a three-phase inverter shown in Fig., rather than using three one-phase inverters.

dV+

S1

S4

S3

S6

A C

S2

S5

B

Figure 19. Three-phase, full wave inverter.

The basic PWM scheme for a three-phase inverter has one common carrier and three separate control

waveforms. If the waveforms we want to achieve are sinusoidal the frequency modulation index, mf, is low we use a synchronized carrier signal with mf an integer multiple of 3.

T i m e

1 6 m s 1 7 m s 1 8 m s 1 9 m s 2 0 m s 2 1 m s 2 2 m s 2 3 m s 2 4 m sV 1 ( V c o n t 1 ) V 1 ( V c o n t 2 ) V 1 ( V c o n t 3 ) V 1 ( V t r i )

- 1 . 0 V

- 0 . 5 V

0 . 0 V

0 . 5 V

1 . 0 V

(a)

Time

16m s 17 ms 1 8ms 19ms 20ms 21ms 22ms 23ms 24msV(2)

0V

40V

80V

110V

(b)

Time

16m s 17 ms 1 8ms 19ms 20ms 21ms 22ms 23ms 24msV(3)

0V

40V

80V

110V

(c)

Time

16ms 17ms 18ms 19ms 20ms 21ms 22ms 23ms 24msV(2)- V(3)

-120V

-80V

-40V

-0V

40V

80V

120V

(d) Figure 20. Three-phase, full wave inverter showing (a) control and triangular wave, (b) phase A line-to-

neutral, (c) phase B line-to-neutral, and (d) line-to-line signal from phase A to B.

Time

1 7.00ms 17 .04ms 17. 08ms 17. 12ms 17.1 6ms 17.20 ms 17.24m s 17.28ms 17.32ms 17.36ms 17.40m sV(2)

0V

40V

80V

110V

(a)

Time

1 7.00ms 17 .04ms 17. 08ms 17. 12ms 17.1 6ms 17.20 ms 17.24m s 17.28ms 17.32ms 17.36ms 17.40m sV(3)

0V

40V

80V

110V

(b)

Tim e

17.00ms 17.04ms 1 7.08ms 17.12ms 17. 16ms 17.20ms 17.24 ms 17.28ms 17.32ms 17 .36ms 17.40msV(2)- V( 3)

-120V

-80V

-40V

-0V

40V

80V

120V

(c)

Figure 21. Shorter time span for the (a) phase A line-to-neutral, (b) phase B line-to-neutral, and (c) line-to-

line signal from phase A to B of the outputs in Figure 20.

The PSPICE netlist for this three phase PWM circuit is given below: 3 phase FULL-BRIDGE INVERTER ************ OUTPUT IS V(2,3) **************** **********INPUT PARAMETERS **************** .PARAM Vsource = 100 ; DC input to inverter .PARAM Fo = 200 ; fundamental frequency .PARAM Mf = 21 ; carrier, multiple of Fo .PARAM Ma = .8 ; amplitude ratio .PARAM Fc = Mf*Fo ; carrier frequency VS 1 0 DC Vsource ; dc source ******* VOLTAGE-CONTROLLED SWITCHES ***** S1 1 2 40 30 SWITCH S2 1 3 50 30 SWITCH S3 1 4 60 30 SWITCH S4 2 0 30 40 SWITCH S5 3 0 30 50 SWITCH S6 4 0 30 60 SWITCH ************** FEEDBACK DIODES ************* D1 2 1 DMOD D2 3 1 DMOD D3 4 1 DMOD D4 0 2 DMOD D5 0 3 DMOD D6 0 4 DMOD ******************** LOAD ****************** R1 2 5 10 ; load between nodes 2 and 0 L1 5 0 60MH R2 3 6 10 ; load between nodes 3 and 0 L2 6 0 60MH R3 4 7 10 ; load between nodes 4 and 0 L3 7 0 60MH *************** TRIANGLE CARRIER ************** Vtri 30 0 PULSE (1 -1 0 1/(2*Fc) 1/(2*Fc) 1ns 1/Fc) ******************** REFERENCE ***************** Vcont1 40 0 SIN(0 Ma Fo 0 0 -90/Mf) Vcont2 50 0 SIN(0 Ma Fo 0 0 -90/Mf - 120) Vcont3 60 0 SIN(0 Ma Fo 0 0 -90/Mf - 240) ************ MODELS AND COMMANDS ************* .MODEL SWITCH VSWITCH(RON=0.001 VON=.007 VOFF=-.007) .MODEL DMOD D ; default diode .PROBE .TRAN 0.5MS 33.33MS 16MS 0.1MS .FOUR 200 25 I(R1) ; Fourier transform .OPTIONS NOPAGE ITL5=0 .END

As long as ma is less than 1, the RMS value of the fundamental of the output voltage is a linear function of it:

dadall VmVmV 612.022

31

≈=− (8.24)

In the limit, when the control voltage becomes infinite, the RMS value of the fundamental of the output is then:

dd

ll VVV 78.02

423

1≈=− π

(8.25)

Again, equating the power on the DC and AC sides we obtain:

pfIVIVP olldod ⋅== − 113 (8.26)

or for normal PWM operation:

pfImI oado ⋅= 106.1 (8.27)

and in the limit for the square wave:

pfII odo ⋅= 135.1 (8.28)

Finally, there are other ways to control the operation of an inverter. If it is not he output voltage

waveform we want to control, but rather the current, we can either impose a fast controller on the voltage waveform, driven by the error between the current signal and the reference, or we can apply a hysteresis band controller, shown for one leg of the inverter in Figure 22.

vAo +

vAN

+

S1

S2Vd

+Vd2

Vd2

Ti me

16 ms 18 ms 20 ms 22 ms 24 ms 26 ms 28 ms 30 ms 32 ms 34msI( R)

-4 .0A

-2 .0A

0A

2 .0A

4 .0A

iεΣ Switch-Mode

Inverter

iε

iA

*Ai

+ _

Comparator tolerance

band

Figure 22. Current control with hysteresis band.

8.2.3 Inverter Notes

• With a sine-triangle PWM the harmonics of the output voltages are of frequency around nfn, where n is an integer and fn is the frequency of the carrier (triangle) waveform. The higher this frequency is the easier to filter out these harmonics. On the other hand, increasing the switching frequency also increases proportionally the switching losses. For 6-step operation of a 3-phase inverter the harmonics are even, except the tripled ones, i.e. they are of order 5, 7, 11, 13, 17 etc.

• When the load of an inverter is inductive the current in each phase remains positive after the voltage in that phase became negative, i.e. after the top switch has been turned off. The current then flows through the antiparallel diode of the bottom switch, returning power to the DC link. The same

happens of course when the bottom switch is turned off and the current flows through the antiparallel diode of the top switch.

8.2.4 Example 1

A three-phase controlled rectifier is supplying a DC motor that has k = 1 (V·s) and R = 1 (Ω). The rectifier is fed from a 208 (Vl-l) source.

Filter 208 (V)

Figure 23. Figure for example problem 8.2.4.

a) Calculate the maximum no-load speed of the DC motor:

Without a load the current is zero, therefore:

ωω kIRkV =+= The maximum speed is then found by the maximum DC voltage

maxmax ωkV =

The maximum DC voltage is provided by the controlled rectifier for α = 0.

(V) 8.28135.1max == −llVV Therefore,

(rad/s) 8.280max =ω

b) The motor now is producing torque of 20 (N·m). What is the maximum speed the motor can achieve? Now that there is load torque, there is also current:

(A) 20=⇒= IkIT Then

(rad/s) 8.2601

1208.280=

⋅−=

−=

kIRVω

c) For the case in b) calculate the total RMS current of the fundamental and the power factor at the AC side

At the maximum, the fundamental of the AC current is:

(A) 6.1578.01 =≈ ds II The power factor is then 1.

d) If the motor is now connected as a generator with a counter torque of 20 (N·m) at 1500 (rpm). What should be the delay angle and AC current?

For a DC generator:

(V) 08.137112

60215001 =⋅−⋅⋅−=−=πωω R

kTkIRkV

Since this is a generator, the voltage is negative for the inverter.

( )αcos2081.35(V) 08.137 ⋅⋅=− º22.119=α

8.2.5 Example 2 For the system shown in Figure 24, the AC source is constant, and the load voltage is 150 (Vl-l),

20 (A), 52 (Hz), 0.85 pf lagging.

Filter 208 (V)

α

6-step inverte

Figure 24. Figure for example problem 8.2.5.

a) Calculate the voltage on the DC side and the DC component of the current. For the 6-step inverter

(V) 19278.01, =⇒=− ddll VVV

(A) 23192

85.02015035.13 00 =⋅⋅⋅

=⇒=⋅= − dddlll IIVpfIVP

b) Calculate the AC source side RMS and fundamental current and power factor.

For a 3-phase rectifier

( ) ( ) ( ) 685.0coscos20835.1192cos35.1 =⇒⋅⋅=⇒= − αααlld VV (A) 94.1778.01 == ds II

( ) lagging 685.0cos == αpf

8.3 DC-DC Conversion DC to DC conversion is often associated with stabilizing the output while the input varies, however

the converse is also required in some applications, which is to produce a variable DC from a fixed or variable source. The issues of selecting component parameters and calculating the performance of the system is the focus of this section. Since these converters are switched mode systems, they are often referred to as choppers.

8.3.1 Step-Down or Buck Converters The basic circuit of this converter is shown in Figure 25 connected to a purely resistive load. If we

remove the low pass filter shown and the diode, the output voltage vo(t) is equal to the input voltage Vd when the switch is closed, and zero when the switch is open. The average output, Vo, is then:

ds

onT

t

td

so V

TtdtdtV

TV

s

on

on=⎥

⎦

⎤⎢⎣

⎡+= ∫∫ 01

0 (8.29)

L

R(load)

vo

+

io

C

Low-pass filter

o= V

iL

+ vL

+

Vd

vo

oVdV

t0

sT = 1sf

ontofft

Figure 25. Topology of the buck chopper.

The ratio of ton/Ts = D, the duty ratio.

The low pass filter attenuates the high frequencies (multiples of the switching frequency) and leaves almost only the DC component. The energy stored in the filter inductor (or the load inductor) has to be absorbed somewhere other than the switch, hence the diode, which conducts when the switch is open. Here we study this converter in the continuous mode of operation such that the current through the inductor never becomes zero. As the switch opens and closes the circuit assumes one of the topologies of Figure 26.

L

R

+

C oV

iL

+ vL

+

Vd

L

R

+

C oV

iL

+ vL

t

vL

A

B

t

i L

0

ontofft

i L = Io

sT = 1sf

Vd - Vo

- Vo

Figure 26. Operation of the buck chopper.

We will use the fact that the average voltage across the inductor is zero, and assume a perfect filter

such that the voltage across the inductor is (Vd – Vo) during ton, and –Vo for the remainder of the cycle.

(8.30) ( ) ( ) 00

=−+−= ∫∫s

on

on T

to

todo dtVdtVVV

( ) ( ) 0=−−− onsoonod tTVtVV (8.31)

DTt

VV

s

on

d

o == (8.32)

Also using the fact that the input and output powers are the same gives:

oodd IVIV = (8.33)

DVV

II

d

o

o

d == (8.34)

We analyzed this under the assumption of continuous mode of operation. In the discontinuous mode, the output DC voltage is less than what is given here, and the chopper is less easy to control. At the boundary between continuous and discontinuous modes, the inductor current reaches zero for one instant every cycle, as shown in Figure 27.

0

sT = 1sf

ontofft

vo

t

i L

i L, avg= IL

Vd - Vo

- Vo

i L

sT= 8L

0 0.5 1.0D

IL,max

Vd

IL

Figure 27. Operation of the buck converter at the boundary of continuous conduction.

From this figure we can see that at this operating point, the average inductor current is IL = ½ iL, and:

( ) ( )ods

odonL VVL

DTVVtI −=−=22

1 (8.35)

Since the average inductor current is the average output current (the average capacitor current is zero), equation (8.35) defines the minimum load current that will sustain continuous condition.

Finally, a consideration is the output voltage ripple. We assume the ripple current is absorbed by the capacitor, i.e. the voltage ripple is small. The ripple voltage is then due to the deviation from the average of the inductor current as show in fig. Under these conditions:

222

11 sLo

TILC

QV Δ=

Δ=Δ (8.36)

where

( ) so

L TDL

VI −=Δ 1 (8.37)

( )DLCT

VV s

o

o −=Δ 1

81 2

(8.38)

t

vL

A

B

t

i L

0

ontofft

i L = Io

sT = 1sf

Vd - Vo

- Vo

oΔV

2

vo

t0

Vo

Figure 28. Analysis of the output voltage ripple of the buck converter.

Another way to view this is to define the switching frequency fs = 1/Ts and use the corner frequency of the filter ( )LCfcorn π21= :

( )22

12 ⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

Δ

s

corn

o

of

fDVV π (8.39)

8.3.2 Step-Up or Boost Converter Here the output voltage is always higher than the input. The schematic is shown in Figure 29.

L

R

+

C oV

iL

+ vL

+

Vd

Figure 29. Schematic diagram of a boost converter. Based on the condition of the switch, there are two possible topologies as shown in Figure 30. Again,

the way to calculate the relationship between input and output voltage we take the average current of the inductor to be zero, and the output power equal to the input power giving:

( )( ) 0=−−+ onsodond tTVVtV (8.40)

DV

V

d

o−

=⇒1

1 (8.41)

DII

d

o −=⇒ 1 (8.42)

L

R

+

C oV

iL

+ vL+

Vd

L

R

+

C oV

iL

+ vL

t

vL

A

B

t

i L

0

ontofft

i L = Io

sT = 1sf

Vd - Vo

- Vo

+

Vd

Figure 30. Two circuit topologies of the boost converter.

To determine the values of inductance and capacitance we study the boundary of continuous

conduction like before and the output voltage ripple. At the boundary of the continuous conduction, the geometry of the current waveform gives:

( )212

DDLVTI os

o −= (8.43)

The output current must exceed this value for continuous conduction. Using the ripple analysis as shown in Figure 31 we find:

RCDT

VV s

o

o =Δ (8.44)

It is important to note that the operation of a boost converter depends on parasitic components, especially for duty cycle approaching unity. These components will limit the output voltage to levels well below those given by equation (8.41).

tΔQ

ΔQ

i D

i D= Io

oΔV

vo

t0

Vo

DTs (1-D)Ts

Figure 31. Calculating the output voltage ripple for a boost inverter.

8.3.3 Buck-Boost Converter This converter has a schematic shown in fig. and can provide output voltage that can be lower or higher than the input voltage.

Again the operation of the converter can be analyzed using the two topologies resulting from the operation of the switch as shown in fig.

By equating the integral of the inductor voltage to zero we get:

( )( ) 01 =−−+ sosd TDVDTV (8.45)

D

DVV

d

o−

=⇒1

(8.46)

At the boundary between continuous and discontinuous conduction we find

( )212

DLVTI os

o −= (8.47)

The output ripple, as calculated from Fig. is

RCDT

VV s

o

o =Δ (8.48)

L R

+

C oV

iD

+

vL

+

Vd

io

iL

Figure 32. Basic Buck-Boost converter.

ss

t

vL

A

B

t

i L

0

IL = (I + I )o

sT = 1sf

- Vo

Vd

L R

+

C oV

iD

+

vL

+

Vd

io

iL

L R

+

C oV

+

vL

+

Vd

io

iL

DT (1-D)T

d

Figure 33. Operation of a Buck-Boost chopper.

8.3 Example

The input of a step down converter varies from 30 (V) to 40 (V) and the output voltage is to be constant at 20 (V), with output power varying between 100 (W) and 200 (W). The switch is operating at 10 (kHz). What is the inductor needed to keep the inductor current continuous? What is then the filter capacitor needed to keep the output ripple below 2%?

The duty cycle will vary between D1 = 20/30 = 0.667 and D2 = 20/40 = 0.5. The load current will

range between Io1 = 100/20 = 5 (A), and Io2 = 200/20 = 10 (A). The minimum current needed to keep the inductor current continuous is:

( )ods

o VVL

DTI −=2min,

Since the constant is the output voltage, Vo, and the minimum load current must be greater than Io,min, we can express it as a function of Vo and make it less than or equal to 5 (A), or:

( ) ( ) ( )DLTVVV

LDTI so

ods

o −=−=≥ 122

A5 min,

Ts = 1/10(kHz), Vo = 20 (V), and the maximum value is achieved for D = 0.5, leading to Lmin = 50 (µH). For the ripple, the highest will occur at 1-D = 0.5, thus:

(Hz) 9001010

5.02

02.02

3

2=⇒⎟⎟

⎠

⎞⎜⎜⎝

⎛

×= corn

corn ffπ

(Hz) 90010502

16

=×

⇒− Cπ

C = 625 (µF)