Choppers

Power Electronics and Drives (Version 2) Dr. Zainal Salam

1

Chapter 3DC to DC CONVERTER

(CHOPPER) Basic non-isolated DC-DC converter

topologies: Buck, Boost, Buck-Boost, Cuk in CCM and DCM mode

Non-ideal effects on converter performance

Isolated DC-DC converters, switched-mode power supply

Control of DC-DC converters High frequency transformer and inductor

design 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


3

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 RIP2=

TLo RIV =

+

VoRL

+ VCE IL

MODEL OF LINEARREGULATOR

RT

EQUIVALENTCIRCUIT

Vs

RL

+ VCE IL

VsVo

+


4

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


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


6

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

VdVo

Vo

closedopened

closedopened

t

DT Tt

iLmin

iLmaxIL

vL

iL

iL

+

S


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

VdVo

Vo

closedopened

closedopened

t

DT Tt

iLmin

iLmaxIL

vL

iL

iL

S

D

(1-D)T


8

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


9

Analysis for switch opened

( ) TDLVi

LV

TDi

ti

dtdi

ii

LV

dtdi

dtdiLVv

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


10

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


11

==

+=

+=+=

===

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


12

Continuous current operationiL

Imax

Imin t0

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

>>

=

==


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

==

==

=

=

===+=

iRiLL

iC

iLiL=IR

imax

imin

0

0Vo

Vo/R+


14

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.


15

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

+= LLRMSL iII


16

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.


17

Boost (step-up) converter

Vd

L D

C

RL

S

Vd

L D

CRLS

Vd

LD

C RLS

+ vL

+

Vo

+ vL -

Vo

+

CIRCUIT OF BOOST CONVERTER



Vo

+

iL


18

Boost analysis:switch closed

Vd

L D

CS

+ vL

iL

+vo

( )LDTVi

LV

dtdi

DTi

ti

dtdi

LV

dtdi

dtdiLVv

dclosedL

dL

LLL

dL

LdL

=

=

==

==

=

DT T

iL

vL

CLOSED

t

t

Vd

Vd Vo

iL


19

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-


20

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.


21

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

==

+=+=

=

=

=

==


22

Continuous Current Mode (CCM)

( )( )

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

VdvL

Q

VdVo

Io=Vo / R


23

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.


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


25

Buck-boost analysis

DT T

imax

imin

imin

imax

ic

iD

iL

VdvL

Q

VdVo

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

=

==

===

===

===


26

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


27

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

===

=

=

=

=


28

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


29

Cuk Converter

CIRCUIT OF CUK CONVERTER

Vd

L2

D C2 RLS+

Vo

L1 C1

iL1 iL2+ vc1-+

vc2

Vd

L2

DC2 RL

+

Vo

L1

C1iL1 iL2

+

vc2S

Vd

L2

DC2 RLS

+

Vo

L1

C1iL1 iL2

+

vc2


21 LC ii =

11 LC ii =


30

Cuk analysis: from capacitor current point of view

iC1

OPENED

t

iL2

iL1

CLOSED

( )

( )

( )[ ] ( )[ ]

)1(

0)1(ng,Substituti

0)1(zero, iscurrent average theoperation, periodicFor

i.e. source, by the suppliedpower the toequal is load by the absorbedpower The

:is C1in current The on. diode theforce L2 and L1in current theopened, isswitch When the

:is C1in current theand off is diodeclosed, isswitch When the

KVL,by computed is C1 across voltageaverage The

2

112

11

12

11

21

1

DD

II

TDIDTI

TDiDTi

IVIV

ii

ii

VVV

L

LLL

openCclosedC

LsLo

LopenC

LclosedC

odC

==+

=+

=

=

=

=


31

Cuk analysis

fLDV

LDTVi

orLV

DTi

dtdi

Lvv

fCLD

VV

DD

VV

VV

II

IVIV

dd

d

dL

o

o

s

o

s

o

L

LLsLo

L

L

L

11

1

11

222

2

112

1

1

1

closed, isswitch when DT interval In time8

1Hence, converter.buck theasion configurat same the

in are adnR) C2 (L2, stageoutput that theNote

)1(

:as written becan tageoutput vol Combining,

i.e. source, by the suppliedpower the toequal is load by the absorbedpower The

==

===

=

=

==


32

Cuk design parameters

fRDL

DfRDL

fLDV

LDTVi

or

dtdiLv

VVVVVvvVvv

dd

LL

doodocLoLc

L

2)1(

2)1(

operation,current continuousFor

)(0

closed, isswitch when DT interval in time L2,For

min,2

min,1

2

2

22

222

1221

=

=

==

==+=+=

++=


33

Cuk analysis from inductor current point of view

decrease to causes This .n larger tha is since negative iswhich

:as written becan oltageinductor v The L1. and Vdfromenergy by diode he through tcharged C1isCapacitor

diode. he through tflow and off, isswitch When the

and n larger tha is seen that becan alsoIt

). ofpolarity the(Note ,

zero. are and thatassumed becan it state,steady In

11

11

21

01

1

21

LdC

CdL

LL

dC

oodC

LL

iVV

VVv

ii

VVV

VVVV

V V

+=

+=

Vd

L2

DC2 RLS

+

Vo

L1

C1iL1 iL2

+

vc2

SWITCH IS OPENED

Closed

Open

t

DT T

t

vL1

iL1

vd

-vo

iL1

Closed

Open

t

DT T

t

vd

-vo

IL1

vL2

iL2

iL1


34

Cuk analysis

Vd

L2

DC2 RL

+

Vo

L1

C1iL1 iL2

+

vc2S


21 LC ii =

increase. to causing L1, energy to feedsinput The

increases. reL2.Therefonoutput theenergy to ngtransferriswitch, he through tdischarged C1capacitor Since

switch. he through tflow andcurrent inductor The diode. thebiased-reverse on, isswitch When the

decrease. to causesh which whic

side,output on theSimilarly

1

21

221

1

L

LoC

LLC

Ldo

i

iVV

iiV

iVV

>

=

Closed

Open

t

DT T

t

vL1

iL1

vd

-vo

iL1

Closed

Open

t

DT T

t

vd

-vo

IL1

vL2

iL2


35

Cuk Analysis

output theofpolarity theNote

1

:Combining

10)1)(()( :L2

1

10)1)(( :L1

L2, and L1 across voltages theof integral theEquating

1

1

1

1

DD

VV

VD

VTDVDTVV

VD

VDVVDTV

d

o

oc

ooc

dc

cdd

=

==+

==+


36

Converters in CCM: Summary

Vd

L2

D C2 RLS+Vo

L1 C1

iL1 iL2+ vc1-+

vc2

Vd

L D

CRL

S+Vo

Vd

L

DC RL

S

+Vo

Vd

L

D C RL

S+Vo f

RDLLCf

DVVDVV

Buck

do

do

2)1(

81

min

2

=

==

fRDDL

RCfDVVD

VVBoost

do

do

2)1(

11

2

min=

==

fRDL

RCfDVV

DDVV

BoostBuck

do

do

2)1(

1

2

min=

==

fRDL

DfRDL

LCfDVV

DDVV

Cuk

do

do

2)1(

2)1(81

1

2

2

1

2

=

=

==


37

Buck in discontinuous current mode (DCM)

Vd VD

+ vL -

C RL

+

Vo

VdVo

Vo

closedopened

closedopened

t

tDT

is

vL

iL

iL

+

S

Imax

Imax

BUCK CONVERTER

D1T

T

( )

DTI

DTi

ti

LVV

dtdi

VVv

TDDI

TDIDTIT

I

RVII

DDD

VV

DVDVVTDVDTVV

LL

odL

odL

L

oRL

d

oood

ood

max

1max

1maxmax

1

11

inductor, across Voltage21

21

211

figure, From

zero) iscurrent capacitor average (because

current resistor equalscurrentinductor Average

)(0)(

zero, isoltageinductor v Average

==

=

=

+=

+=

==

+=

==


38

Buck in DCM

( ) ( )

++=

+=

+=

=+

=+

=+

=

==

RTLDD

DVDD

DVV

RTLDD

D

D

RTLDDD

RVDD

LTDVDDI

LTDVDT

LVViI

DVVI

ddo

oo

oosL

os

82

Hence,

2

8,for Solving

02gives,Which

21

21

,Substitute

,)( using and for Solving

21

2

1

1

12

1

11

1max

1max

max


39

Example

example. thisof parameters for the ratioduty and voltage

output ebetween th iprelationsh theshows below Figure

.97.13

DCMin circuit 0.64,0.29 i.e. D),-1( Since

29.020

)10)(10)(200(84.04.021

2

8:by calculated becan

D-1 current, usdiscontinoFor

tage,output vol theDetermine b):ousdiscontinu iscurrent inductor that theShow a)

4.0,10,100,20,200,24converter,buck For the

1

1

62

2

1

11

VDD

DVV

D

K

RTLDD

D

DD

V

DKHzfuFCRuHLVV

do

o

d

=

+=


40

Boost Converter in DCM

Vd

Vd-Vo

closedopened

closedopened

t

tDT

iD

vL

iLImax

Imax

BOOST CONVERTER

D1T

T

Vd

L D

CRL

S

++=

=

=

=

=

==

=

=

+==

=+

LRTD

VV

LRTD

VV

VV

RDTL

VVD

DR

VDLDTVI

LDTViI

DITDIT

I

DDD

VV

DVDVVTDVVDTV

d

o

d

o

d

o

d

o

odD

dL

D

d

oood

odd

2

22

1

1

1

max

1max1max

11

1

21121

02

ng,Substituti

2,for Solving

21

closed, isswitch en thecurrent whinductor

in change theas same theis21

211

:iscurrent diode Average

)(0)(

zero, isoltageinductor v Average


41

Non-ideal effects: switch/diode voltage drop

case. ideal for the than less isWhich

)1(

Solving,

,0)1)(()(period, switching the

for zero isinductor theacross voltageaverage The

diode. theacross voltage theis where

(off),open switch During

switch conducting theacross voltage theis where

(on), closedswitch During

DVV

DVDVDVV

DVVDVVVV

VVVv

VVVVv

do

DQdo

DdQodL

DDdL

QQodL

==

=+=

=

=

converterBuck :ExampleLVQ

VD VoVd

+

+

+_

_

_


42

Inductor (winding) resistance

converterBoost :Example

VdS

+ vL -

iL

+vo-

rL

LLod

LLLoLd

LD

LLDoLd

rLos

rIDVV

rIDIVIV

DII

rIIVIVPPP

+=

+=

=

+=+=

)1(becomes,Which

)1(ng,Substituti

)1(

current, diode (DC) average theBut,

i.e. source, by the suppliedpower equalmust (rs), resitanceinductor theand load by the absorbedPower

2

2


43

Inductor resistance

decreases.converter boost of efficiency theincreases, ratioduty theAs

)1(1

1

)1(

,for ngSubstituti

:Efficiencyresistanceinductor for account factor to correction a includes

but converter boost idealfor similar isequation output The)1(

1

1)1(

Solving,

)1()1(

Hence,)1()1(

But,

22

2

2

22

2

2

DRr

rDRVRV

RVI

rIRVRV

PPP

DRrD

VV

DVDR

rVV

DRV

DII

LL

oo

oL

LLo

o

losso

o

Ld

o

oLo

d

oDd

+=

+=

+=+=

+

=

+=

==


44

Other non-idealities

Capacitors Equivalent Series Resistor (ESR) Producing ripple greater than ideal capacitor Output C must be chosen on the basis of ESR

and not only capacitance value.

Switching losses


45

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)


46

Linear and switched mode power supplies block diagram

Basic Block diagram of linear power supply

Base/gateDrive

ErrorAmp.

LineInput 3/1 50/60 Hz

IsolationTransformer

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


47

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 ==

V1 V2

+

+

i1 i2N1 N2

Ideal model

V1 V2

+

+

i1 i2N1 N2

Model used formost PE application

Lm


48

Flyback Converter

Vs

+

Vo

Vs

Vs

Vs

N1 N2i1

i2

+

-v2v1

+

-

iLM

iD

+ -vDiC

iR

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


49

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


50

Analysis: switched closed

( )

00

Therefore,

0

er, transform theof side load On the

21

1

2

1

2

1

212

1

==


51

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


52

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.


53

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


54

Max, Min inductor current, Lmin, C values

( )

RCfD

VVr

NN

fRDVL

fLDV

LDTV

NN

RDDV

I

LDTV

NN

RDDViII

LDTV

NN

RDDViII

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


55

Full-bridge converter

SW2SW4

VS

NS

NS

+

vx C R+

Vo

+

vp

SW1,SW2

SW3,SW4 DT T

2T DTT +

2VPVS

-VSVx

P

SS N

NV

DT T T T

SW3Lx

SW1

2DT+

2


56

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


57

Control of DC-DC Converter

Comparator

Vcontrol

SawtoothWaveform

Vo (desired)

Vo (actual)

+

-

Switch control signal

SawtoothWaveform

Vcontrol 1

Switchcontrolsignalton 2

T

Vcontrol 2

ton 1

Chapter 3DC to DC CONVERTER(CHOPPER)DC-DC Converter (Chopper)Linear regulatorSwitching RegulatorBuck (step-down) converterCircuit operation when switch is turned on (closed)Operation when switch turned off (opened)Analysis for switch closedAnalysis for switch openedSteady-state operationAverage, Maximum and Minimum inductor currentContinuous current operationOutput voltage rippleDesign procedures for BuckDesign procedures for BuckExamples of Buck converterBoost (step-up) converterBoost analysis:switch closedSwitch openedSteady-state operationAverage, Maximum, Minimum inductor currentContinuous Current Mode (CCM)ExamplesBuck-Boost converterBuck-boost analysisOutput voltageAverage inductor currentL and C valuesCuk ConverterCuk analysis: from capacitor current point of viewCuk analysisCuk design parametersCuk analysis from inductor current point of viewCuk analysisCuk AnalysisConverters in CCM: SummaryBuck in discontinuous current mode (DCM)Buck in DCMExampleBoost Converter in DCMNon-ideal effects: switch/diode voltage dropInductor (winding) resistanceInductor resistanceOther non-idealitiesSwitch-mode power supply (SMPS)Linear and switched mode power supplies block diagramHigh frequency transformerFlyback ConverterFlyback waveformsAnalysis: switched closedAnalysis: switch openedOutput voltageAverage inductor currentMax, Min inductor current, Lmin, C valuesFull-bridge converterFull bridge: basic operationControl of DC-DC Converter

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