3

Power Electronics and Drives (Version 2): Dr.

Zainal Salam

1

Chapter 3

DC to DC CONVERTER(CHOPPER)

• General• Buck converter• Boost converter• Buck-Boost converter• Switched-mode power supply• Bridge converter• Notes on electromagnetic compatibility

(EMC) and solutions.

DC-DC Converter (Chopper)

• DEFINITION: Converting the unregulated DC input to a controlled DC output with a desired voltage level.

• General block diagram:

LOAD

Vcontrol(derived from

feedback circuit)

DC supply(from rectifier-filter, battery,fuel cell etc.)

DC output

• APPLICATIONS: – Switched-mode power supply (SMPS), DC

motor control, battery chargers


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Linear regulator• Transistor is

operated in linear (active) mode.

• Output voltage

• The transistor can be conveniently modelled by an equivalent variable resistor, as shown.

• Power loss is high at high current due to:

TLo RIP 2=

TLo RIV =

+

−

VoRL

+ VCE − IL

MODEL OF LINEARREGULATOR

RT

EQUIVALENTCIRCUIT

Vs

RL

+ VCE −IL

VsVo

+

−


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Switching Regulator• Power loss is zero

(for ideal switch):– when switch is

open, no current flow in it,

– when switch is closed no voltage drop across it.

– Since power is a product of voltage and current, no losses occurs in the switch.

– Power is 100% transferred from source to load.

• Switching regulator is the basis of all DC-DC converters

+

−

Vo

RL

+ VCE − IL

MODEL OF LINEARREGULATOR

EQUIVALENT CIRCUIT

Vs

RL

IL

VsVo

+

−

(ON)closed

(OFF)open

(ON)closed

DT T

OUTPUT VOLTAGE

Vo

SWITCH


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5

Buck (step-down) converter

Vd

L

D C RL

S+

Vo

−

Vo

+

CIRCUIT OF BUCK CONVERTER

CIRCUIT WHEN SWITCH IS CLOSED

CIRCUIT WHEN SWITCH IS OPENED

Vo

+

−

−

iL

Vd D RL

S

Vd D RL

S

+ −vL

+ vL −

iL


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Circuit operation when switch is turned on (closed)

• Diode is reversed biased. Switch conducts inductor current

• This results in positive inductor voltage, i.e:

• It causes linear increase in the inductor current

odL VVv −=

∫=⇒

=

dtvL

i

dtdiLv

LL

LL

1

Vd VD

+ vL -

C RL

+

−

Vo

Vd−Vo

−Vo

closedopened

closedopened

t

DT Tt

iLmin

iLmax

IL

vL

iL

iL

+

−

S


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7

Operation when switch turned off (opened)

• Because of inductive energy storage, iL

continues to flow.

• Diode is forward biased

• Current now flows through the diode and

oL Vv −=

Vd

+ vL -

C RL

+

−

Vo

Vd−Vo

−Vo

closedopened

closedopened

t

DT Tt

iLmin

iLmax

IL

vL

iL

iL

S

D

(1-D)T


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Analysis for switch closed

( ) DTL

VVi

LVV

DTi

ti

dtdi

i

i

LVV

dtdi

dtdiL

VVv

odclosedL

odLLL

L

L

odL

LodL

⋅

−

=∆

−=

∆=

∆∆

=

−=⇒

=

−=

Figure From

linearly. increase must Therefore

constant. tive-posi a is of vative

-deri thesince :Note

oltage,inductor v The

IL

iL max

DT T

iL

Vd− Vo

vL

t

t

iL min

closed

∆iL


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Analysis for switch opened

( ) TDLVi

LV

TDi

ti

dtdi

ii

LV

dtdi

dtdiL

Vv

oopenedL

oLLL

LL

oL

LoL

)1(

)1(

Figure From

linearly. decreasemust constant, tive

-nega a is of vative-deri thesince :Note

opened,switch For

−⋅

−=∆

−=

−∆

=∆∆

=

−=⇒

=

−=

IL

iL max

DT T

iL

Vd− Vo

vL

t

t

iL min

opened

∆iL

(1− D)T


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Steady-state operation

( ) ( )

do

so

sod

openedLclosedL

L

L

DVV

TDLVDT

LVV

ii

i

i

=⇒

=−⋅

−−⋅

−

=∆+∆

0)1(

0

:i.e zero, is period oneover of change theisThat cycle.next theof begining

at the same theis cycle switching of endat the that requiresoperation state-Steady

iL Unstable current

Decaying current

Steady-state current

t

t

t

iL

iL


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−−=

∆−=

−+=

−+=∆+=

==⇒

=

LfD

RViII

LfD

RV

TDL

VR

ViII

RVII

oL

L

o

ooLL

oRL

2)1(1

2

:current Minimum

2)1(1

)1(21

2

:current Maximum

Rin current Average currentinductor Average

min

max

L

Average, Maximum and Minimum inductor current

IL

Imax

Imin

iL

∆iL

t


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Continuous current operationiL

Imax

Imint

0

min

min

min

min

be bechosen is Normally operation. of mode continous ensure

current toinductor minimum theis This2

)1(

02

)1(1

,0 operation, continuousFor

2)1(1

2

analysis, previous From

LL

RfDLL

LfD

RV

I

LfD

RViII

o

oL

L

>>

⋅−

=≥⇒

≥

−−⇒

≥

−−=

∆−=


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13

Output voltage ripple

2

2

8)1(

factor, ripple theSo,8

)1(8

82221

:formula area triangleuse figure, From

LCfD

VVr

LCfD

CiTV

iTiTQ

CQVVCQCVQ

iii

o

o

Lo

LL

ooo

RLc

−=

∆=

−=

∆=∆∴

∆=

∆

=∆

∆=∆⇒∆=∆⇒=

+=

iRiL

LiC

iL

iL=IR

imax

imin

0

0Vo

Vo/R+

−


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Design procedures for Buck

Vd(inputspec.)

SWITCH

f = ?D = ?TYPE ?

D

L

Lmin= ?L = 10Lmin

Cripple ?

RLPo = ?Io = ?

• Calculate D to obtain required output voltage.

• Select a particular switching frequency:– preferably >20KHz for negligible acoustic

noise– higher fs results in smaller L, but higher device

losses. Thus lowering efficiency and larger heat sink. Also C is reduced.

– Possible devices: MOSFET, IGBT and BJT. Low power MOSFET can reach MHz range.


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Design procedures for Buck

• Determine Lmin. Increase Lmin by about 10 times to ensure full continuos mode.

• Calculate C for ripple factor requirement.

• Capacitor ratings:– must withstand peak output voltage– must carry required RMS current. Note RMS

current for triangular w/f is Ip/3, where Ip is the peak capacitor current given by ∆iL/2

• Wire size consideration:– Normally rated in RMS. But iL is known as

peak. RMS value for iL is given as:

22

, 32

∆+= L

LRMSLiII


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Examples of Buck converter

• A buck converter is supplied from a 50V battery source. Given L=400uH, C=100uF, R=20 Ohm, f=20KHz and D=0.4. Calculate: (a) output voltage (b) maximum and minimum inductor current, (c) output voltage ripple.

• A buck converter has an input voltage of 50V and output of 25V. The switching frequency is 10KHz. The power output is 125W. (a) Determine the duty cycle, (b) value of L to limit the peak inductor current to 6.25A, (c) value of capacitance to limit the output voltage ripple factor to 0.5%.

• Design a buck converter such that the output voltage is 28V when the input is 48V. The load is 8Ohm. Design the converter such that it will be in continuous current mode. The output voltage ripple must not be more than 0.5%. Specify the frequency and the values of each component. Suggest the power switch also.


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17

Boost (step-up) converter

Vd

L D

C

RL

S

Vd

L D

CRL

S

Vd

LD

C RLS

+ vL −

+

Vo

−

+ vL -

Vo

+

CIRCUIT OF BOOST CONVERTER



Vo

+

−

−

iL


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Boost analysis:switch closed

Vd

L D

CS

+ vL −

iL

+vo−

( )LDTVi

LV

dtdi

DTi

ti

dtdi

LV

dtdi

dtdiL

Vv

dclosedL

dL

LLL

dL

LdL

=∆

=⇒

∆=

∆∆

=

=⇒

=

=

DT T

iL

vL

CLOSED

t

t

Vd

Vd− Vo

∆iL


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Switch opened

( ) ( )L

DTVViL

VVdtdi

TDi

ti

dtdi

LVV

dtdi

dtdiL

VVv

odopenedL

odL

L

LL

odL

LodL

)1(

)1(

−−=∆⇒

−=⇒

−∆

=∆∆=

−=⇒

=

−=

DT T

( 1-D )T

iL

vL

OPENED

t

t

Vd

Vd− Vo

∆iL

Vd

D

CS

+ vL -

iL

+vo-


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Steady-state operation( ) ( )

( )

DVV

LTDVV

LDTV

ii

do

odd

openedLclosedL

−=⇒

=−−

−

=∆+∆

1

0)1(

0

• Boost converter produces output voltage that is greater or equal to the input voltage.

• Alternative explanation:– when switch is closed, diode is reversed. Thus

output is isolated. The input supplies energy to inductor.

– When switch is opened, the output stage receives energy from the input as well as from the inductor. Hence output is large.

– Output voltage is maintained constant by virtue of large C.


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Average, Maximum, Minimum inductor current

LDTV

RDViII

LDTV

RDViII

RDVI

RDV

RD

V

IV

RVIV

ddLL

ddLL

dL

d

d

Ld

odd

2)1(

2

2)1(

2

currentinductor min Max,

)1(

currentinductor Average

)1()1(

powerOutput powerInput

2min

2max

2

2

2

2

2

−−

=∆

−=

+−

=∆

+=

−=

−=

−

=

=

=


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22

L and C values

( )

( )

RCfD

VVr

RCfDV

RCfDTVV

VCDTR

VQ

fRDD

TRDDL

LDTV

RDV

I

o

o

ooo

oo

dd

=∆

=

==∆

∆=

=∆

−=

−=

≥−−

≥

factor Ripple

21

21

02)1(

0

operation, continousFor

2

2

min

2

min

��

DT T

imax

imin

imin

imax

ic

iD

iL

Vd

vL

∆Q

Vd−Vo

Io=Vo / R


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Examples

• The boost converter has the following parameters: Vd=20V, D=0.6, R=12.5ohm, L=65uH, C=200uF, fs=40KHz. Determine (a) output voltage, (b) average, maximum and minimum inductor current, (c) output voltage ripple.

• Design a boost converter to provide an output voltage of 36V from a 24V source. The load is 50W. The voltage ripple factor must be less than 0.5%. Specify the duty cycle ratio, switching frequency, inductor and capacitor size, and power device.


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24

Buck-Boost converter

Vd L

D

C RL

S+

Vo

−

Vo

+

CIRCUIT OF BUCK-BOOST CONVERTER



Vo

+

−

−

iLVd vL

+

−

iLVd vL

+

−

D

DS

S


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Buck-boost analysis

��

DT T

imax

imin

imin

imax

ic

iD

iL

Vd

vL

∆Q

Vd−Vo

Io=Vo / R

LTDVi

LV

TDi

ti

LV

dtdi

dtdiLVv

LDTVi

LV

DTi

ti

LV

dtdi

dtdiLVdv

oopenedL

oLL

oL

LoL

dclosedL

dLL

dL

LL

)1()(

)1(

openedSwitch

)(

closedSwitch

−=∆

=−∆

=∆∆

=⇒

==

=∆

=∆

=∆∆

=⇒

==


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Output voltage

−

−=⇒

=−

+

DDV

LTDV

LDTV

s

od

1V

0)1(

:operation stateSteady

o

• NOTE: Output of a buck-boost converter either be higher or lower than the source voltage.– If D>0.5, output is higher– If D<0.5, output is lower

• Output voltage is always negative• Note that output is never directly • connected to load. Energy is stored in

inductor when switch is closed and transferred to load when switch is opened.


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Average inductor current

2

2

2

2

)1(

,for ngSubstituti

:ascurrent inductor average torelated iscurrent source averageBut

i.e. source,by thesuppliedpower equalmust load by the absorbedpower

converter, in the losspower no Assuming

DRDV

DVP

RDVVI

V

DIVR

V

DII

IVR

V

PP

d

d

o

d

oL

o

Ldo

Ls

sdo

so

−===⇒

=⇒

=

=

=


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L and C values

RCfD

VVr

RCfDV

RCDTVV

VCDTR

V

fRDL

LDTV

DRD

LDTV

DRDViII

LDTV

DRDViII

o

o

ooo

oo

d

ddLL

ddLL

=∆

=

==∆

∆=

=∆

−=⇒

=+−

−−

=∆

−=

+−

=∆

+=

Q

ripple, tageOutput vol

2)1(

02)1(

Vcurrent, continuousFor

2)1(2

2)1(2

current,inductor min andMax

2

min

2d

2min

2max


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Control of DC-DC converterusing pulse width modulation-

PWM

Comparator

Vcontrol

SawtoothWaveform

Vo (desired)

Vo (actual)

+

-

Switch control signal

SawtoothWaveform

Vcontrol 1

Switchcontrolsignalton 2

T

Vcontrol 2

ton 1


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30

Switch-mode power supply (SMPS)

• Advantages over linear power-Efficient (70-95%)-Weight and size reduction

• Disadvantages -Complex design-EMI problems

• However above certain ratings,SMPS is the only feasible choice

• Types of SMPS-Flyback-forward-Push-pull-Bridge (half and full)


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31

Linear and switched mode power supplies block diagram

Basic Block diagram of linear power supply

Base/gateDrive

ErrorAmp.

LineInput

φφ 3/1 50/60 HzIsolation

Transformer

Rectifier+

Vd

-

Vce=Vd-VoC E

B

Vo

Vref

RL

+Vo

+

Vo

-

Basic Block diagram of SMPS

EMIFILTER

RECTIFIERAND

FILTER

HighFrequency

rectifierandfilter

Base/gatedrive

PWMController

errorAmp

Vo

Vref

DCRegulated

DC-DC CONVERSITION + ISOLATION

DCUnregulated


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32

High frequency transformer

: Models

;

iprelationshoutput -input Basic voltage varying- meup/down ti step ii)

isolation electricaloutput -Input i)

:function Basic

1

2

2

1

2

1

2

1NN

ii

NN

vv

==

V1V2

+

−

+

−

i1 i2N1 N2

Ideal model

V1V2

+

−

+

−

i1 i2N1 N2

Model used formost PE application

Lm


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33

Flyback Converter

Vs

+

−Vo

Vs

Vs

Vs

N1 N2i1

i2

+

-v2v1

+

-

iLM

iD

+ -vD

iCiR

C R

vSW+ −

+

−

Vo

iS

N1 N2

+

-

0

0

v1

v1=Vs

iLM

is=iLM

+=

2

1NNVVV ossw

+ −

−

2

11 N

NVv o

v1

+

−iLM

N1 N2

v2= -VS

+

− +

−Vo

iD

LM

vSW

Vo

+

−

Flyback converter circuit

Model with magnetisinginductance

Switch closed

Voltage and currentconditions when switchopened


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34

Flyback waveforms

DT T

iLm

DT

T

t

t

is

DT

T

t

t

t

iD

iC

T

v1

-V(N1/N2)

Vo/ R

Vs

DT

DT T

∆iLM


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Analysis: switched closed

( )

00

Therefore,

0

er, transform theof side load On the

21

1

2

1

2

1

212

1

==

<

−−=

=

=

=∆⇒

=∆

=∆

=

==

ii

NNVVv

NNV

NNvv

LDTVi

LV

DTiL

dtiL

dtdiL

dtdiLVv

doD

d

m

dclosedL

m

dmmm

Lmmd

m


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36

Analysis: switch opened

( )

( )

( ) ( )

( )

−

=⇒

=

−+⇒

=∆+∆

−−=∆⇒

−=

−∆

=∆

=

−==

−=

=⇒

−=

−=

2

10

2

10

2

10

2

10

2

101

2

10

2

121

022

101

)1(

01

0

operation, state-steadyFor

)1(

1

;

NN

DDVV

NN

LTDV

LDTV

ii

NN

LTDVi

NN

LV

TDi

dti

dtdi

NNVv

dtdi

L

NNV

NNvv

VvNNVv

d

mm

d

openedLclosedL

mopenmL

m

mLmLmL

mLm

mm


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Output voltage

• Input output relationship is similar to buck-boost converter.

• Output can be greater of less than input,depending upon D.

• Additional term, i.e. transformer ratio is present.


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38

Average inductor current

( )

( )

−

=

−

=

=⇒

=

==

=

=

1

202

1

22

20

20

20

0

)1()1(

:as written also iscurrent inductor average The

for solving and Substitute

:as torelated is

NN

RDV

NN

RDDVI

DRVVI

RVDIV

I

DITDTI

I

II

RVIV

PP

dL

dL

Ld

L

LL

s

Ls

sd

s

m

m

m

m

mm

m


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39

Max, Min inductor current, Lmin, C values

( )

RCfD

VVr

NN

fRDVL

fLDV

LDTV

NN

RDDV

I

LDTV

NN

RDDVi

II

LDTV

NN

RDDVi

II

dm

m

d

m

dd

L

m

ddLLL

m

ddLLL

m

mmm

mmm

=∆

=

−=

==

−

=

−

−

=∆

−=

+

−

=∆

+=

0

0

2

2

12

min

2

1

22

min

2

1

22min,

2

1

22max,

converter,boost similar to isn calculatio ripple The

2)1(

22)1(

0, operation, continuosFor

2)1(2

2)1(2


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40

Full-bridge converter

SW2SW4

VS

NS

NS

+

−vx C R

+

−Vo

+

−

vp

SW1,SW2

SW3,SW4 DT T

2T DTT

+2VP

VS

-VS

Vx

P

SS N

NV

DT T T T

SW3Lx

SW1

2DT+

2


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41

Full bridge: basic operation

• Switch “pair”: [S1 & S2];[S3 & S4].

• Each switch pair turn on at a time as shown. The other pair is off.

• “AC voltage” is developed across the primary. Then transferred to secondary via high frequency transformers.

• On secondary side, diode pair is “high frequency full wave rectification”.

• The choke (L) and © acts like the “buck converter” circuit.

• Output Voltage DNNVV

p

sso ⋅

= 2

3

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