chopper.pdf

Power Electronics and Drives (Version 3-2003)

Dr. Zainal Salam, UTM-JB

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 fromfeedback 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:

Lceo

TLo

IVP

RIP

=

=

or

2

+

Vo

RL

+ VCEce IL

LINEAR REGULATOR

RT

EQUIVALENTCIRCUIT

Vin

RL

+ Vce

IL

VinV

o

+

ceino VVV =



4

Switching Regulator

Transistor is operated in switched-mode:

Switch closed: Fully on (saturated)

Switch opened: Fully off (cut-off)

When switch is open, no current flow in it

When switch is closed no voltage drop across it.

Since P=V.I, no lossesoccurs in the switch.

Power is 100% transferred from source to load.

Power loss is zero (for ideal switch):

Switching regulator is the basis of all DC-DC converters

+

Vo

RL

+ Vce

IL

SWITCHING REGULATOR

EQUIVALENT CIRCUIT

Vin

RL

IL

VinV

o

+

(ON)closed

(OFF)open

(ON)closed

DT T

OUTPUT VOLTAGE

Vo

SWITCH

Vin



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

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

Switch turned off (opened)

Because of inductive energy storage, iL continues to flow.

Diode is forward biased

Current now flows (freewheeling) through the diode.

The inductor voltage can be derived as:

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

( )

( ) TDLVi

LV

TDi

ti

dtdi

LV

dtdi

dtdiLVv

DTL

VVi

LVV

DTi

ti

dtdi

ii

LVV

dtdi

dtdiLVVv

oopenedL

oLLL

oL

LoL

odclosedL

odLLL

L

L

odL

LodL

)1(

)1(

opened,switch For

Figure From linearly. increased

must hereforeconstant.T positive a is of Derivative

:(on) closed isswitch When the

=

=

=

=

=

==

=

=

=

=

=

==

IL

iL max

DT T

iL

Vd Vo

vL

t

t

iL min

closed

iL



9

Steady-state operation

( ) ( )

do

so

sod

openedLclosedL

L

L

DVV

TDLVDT

LVV

iii

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



10

minmax

min

max

L

:ripplecurrent Inductor 2

)1(12

:current Minimum2

)1(1

)1(21

2

:current Maximum

Rin current Average currentinductor Average

IIi

LfD

RViII

LfD

RV

TDL

VR

ViII

RVII

L

oL

L

o

ooLL

oRL

=

=

=

+=

+=

+=

==

=

Average, Maximum and Minimum Inductor Current

IL

Imax

Imin

iL

iL

t



11

Continuous Current Mode (CCM)

min

min

min

min

be bchosen 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

ILf

DR

ViII

o

oL

L

>>

=

=

=

iL

Imax

Imin

t0



12

Output voltage ripple

size.capacitor Increasing 3)sizeinductor Increasing 2)frequency switching Increasing 1)

:by reduced becan Ripple :Note8

)1(factor, ripple theSo,

8)1(

8

peak) to-(Peak voltageRipple8

2221

:formula area triangleUse

:as witten becan charge The

:currentCapacitor KCL,

2

2

LCfD

VV

r

LCfD

CiTV

iT

iTQ

CQV

VCQCVQ

iii

o

o

Lo

L

L

oo

o

RLc

=

=

=

=

=

=

=

==

+=

iL

iL=IR

imax

imin

0

0

Vo

+

Vo/R

iRL iL

iC

iC



13

Basic design procedures

Vd(inputspec.)

SWITCH

f = ?D = ?TYPE ?

D

L

Lmin= ?

L = 10Lmin

Cripple ?

RLP

o = ?

Io = ?

Calculate D to obtain required output voltage.

Select a particular switching frequency (f) and device preferably f>20KHz for negligible acoustic noise higher fs results in smaller L and C. But results in higher losses.

Reduced efficiency, larger heat sink. Possible devices: MOSFET, IGBT and BJT. Low power MOSFET can

reach MHz range. Calculate Lmin. Choose L>>10 Lmin 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. ECAPs can be used

Wire size consideration: Normally rated in RMS. But iL is known as peak. RMS value

for iL is given as:2

2, 3

2

+= LLRMSLiII



14

Examples 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.



15

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



16

Boost analysis:switch closed

( )LDTVi

LV

dtdi

DTi

ti

dtdi

LV

dtdi

dtdiL

Vv

dclosedL

dL

LLL

dL

L

dL

=

=

=

=

=

=

=

DT T

iL

vL CLOSED

t

t

Vd

Vd Vo

iL

Vd

L D

CS

+ vL

iL

+v

o



17

Switch opened

( ) ( )L

DTVVi

LVV

dtdi

TDi

ti

dtdi

LVV

dtdi

dtdiL

VVv

odopenedL

odL

L

LL

odL

L

odL

)1(

)1(

=

=

=

=

=

=

=

DT T

( 1-D )T

iL

vL OPENED

t

t

Vd

Vd Vo

iL

Vd

D

CS

+ vL

-

iL

+v

o

-



18

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.



19

Average, Maximum, Minimum Inductor Current

LDTV

RDViII

LDTV

RDViII

RDVI

RDV

RD

V

IV

RVIV

ddLL

ddLL

dL

dd

Ld

odd

2)1(

2

:currentinductor Minimum2)1(

2

:currentinductor Maximum)1(

:currentinductor Average)1(

)1(

powerOutput powerInput

2min

2max

2

2

2

2

2

=

=

+

=

+=

=

=

=

=

=



20

L and C values

( )

( )

RCfD

VV

r

RCfDV

RCfDTVV

VCDTR

VQ

fRDD

TRDDL

LDTV

RDV

I

o

o

ooo

oo

dd

=

=

==

=

=

=

=

factor Ripple

21

21

02)1(

0CCM,For

2

2min

2

min

Imax

Imin

Imin

Imax

ic

iD

iL

VdvL

VdVo

Io=V

o / R

DT T

Q



21

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.



22

Buck-Boost converter

Vd L

D

C RL

S+

Vo

Vo

+

CIRCUIT OF BUCK-BOOST CONVERTER



Vo

+

iLVd vL

+

iLVd vL

+

D

DS

S



23

Buck-boost analysis

DT T

Imin

Imax

ic

iD

iL

VdvL

Q

VdVo

Io=V

o / R

Imax

Imin

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

=

=

=

=

==

=

=

=

=

==



24

Output voltage

NOTE: Output of a buck-boost converter either be higher or lower than input. If D>0.5, output is higher than input If D



25

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

ddo

do

L

o

Ldo

Ls

sdo

so

===

=

=

=

=



26

L and C values

RCfD

VV

r

RCfDV

RCDTVV

VCDTR

V

fRDL

LDTV

DR

D

LDTV

DRDViII

LDTV

DRDViII

o

o

ooo

oo

d

ddLL

ddLL

=

=

==

=

=

=

=+

=

=

+

=

+=

Q

ripple, tageOutput vol2

)1(

02)1(

V

CCMFor 2)1(2

2)1(2

current,inductor min andMax

2min

2d

2min

2max



27

Converters in CCM: Summary

Vd

L

DC RL

S

+

Vo

fRDL

LCfD

VV

DVV

o

o

do

2)1(

81

Buck

min

2

=

=

=

fRDDL

RCfD

VV

DVV

o

o

do

2)1(

11

Boost

2min

=

=

=

fRDL

RCfD

VV

DD

VV

o

o

do

2)1(

1

BoostBuck

2min

=

=

=

Vd

L

D C RL

S+V

o

Vd

L D

CRL

S+

Vo



28

Control of DC-DC converter:pulse width modulation (PWM)

ComparatorV

control

SawtoothWaveform

Vo (desired)

Vo (actual)

+

-

Switch control signal

SawtoothWaveform

Vcontrol 1

Switchcontrolsignalton 2

T

Vcontrol 2

ton 1



29

Isolated DC-DC Converter

Isolated DC-DC requires isolation transformer Two types: Linear and Switched-mode

Advantages of switched mode over linear power supply-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)



30

Linear and SMPS block diagramBasic Block diagram of linear power supply

Basic Block diagram of SMPS

Base/gateDrive

ErrorAmp.

LineInput

3/150/60 HzIsolation

Transformer

Rectifier/Filter

+

Vd-

Vce=Vd-VoC E

B

Vo

Vref

RL

+Vo

+

Vo

-

DC UnregulatedDC Regulated

EMIFILTER

RECTIFIERAND

FILTER

HighFrequency

rectifierandfilter

Base/gatedrive

PWMController

errorAmp

Vo

Vref

DCRegulated

DC-DC CONVERSITION ANDISOLATIONDC

Unregulated



31

High frequency transformer

: Models

;

iprelationshoutput -input Basic voltage varying- meup/down ti step 2)

isolation electricaloutput -Input 1):function Basic

1

2

2

1

2

1

2

1

NN

ii

NN

v

v

=

=

V1 V2

+

+

i1 i2N1 N2

Ideal model

V1 V2

+

+

i1 i2N1 N2

Model used formost PE application

Lm



32

Flyback Converter

Vd

+

Vo

Vd

N1 N2i1

i2

+

v2v1

+

iLM

iD

+ vDiC iR

C R

vSW+

+

iS

LM

Vo

Flyback converter circuit

Model with magnetisinginductance



33

Operation: switch closed

VdV

o

N2+

0

0

v1

v1=Vs

iLM

is=iLM

+

N1

+

v2

( )

0 and 0Therefore,

off turneddiode i.e. ,0

er, transform theof side load On the

12

1

2

1

2

1

212

1

==



34

Switch opened

Vs

+

v1

+

iLM

N1 N2

v2= VS+

+

Vo

iD

vSW

( )( )

+=

=

=

=

=

==

=

=

=

=

2

1

2

1

21

2

11

2

1

2

121

2

2

11

:switch theacross Voltage

)1(1

But

NNVVv

NN

LTDVi

NN

LV

TDi

dti

dtdi

NNV

dtdi

Lv

NNV

NN

vv

VvNNVv

odSW

m

oopenmL

m

omLmLmL

omL

m

o

o

o



35

Output voltage

( ) ( )( )

=

=

=+

1

2

21

1

01

0operation, state-steadyFor

NN

DDVV

NN

LTDV

LDTV

ii

do

m

o

m

d

openedLclosedL mm

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.



36

Flyback waveforms

iLm

is

t

t

iD

iC

v1

-V(N1 /N2)

Vo/ R

Vs

DT T

iLM( )

( )

=

=

=

=

==

=

=

1

20

2

1

22

20

20

20

0

)1(

)1(

:asn can writte

for Solving

:as torelated is

NN

RDV

NN

RDDVI

IDRV

VI

RVDIV

I

DITDTI

I

IIR

VIV

PP

dL

L

dL

Ld

L

LL

s

Ls

sd

s

m

m

m

m

m

m

m

m



37

Max, Min inductor current

( )

RCfD

VV

r

NN

fRDVL

fLDV

LDTV

NN

RDDV

I

LDTV

NN

RDDV

iII

LDTV

NN

RDDV

iII

dm

m

dm

dd

L

m

dd

LLL

m

dd

LLL

m

m

mm

m

mm

=

=

=

==

=

=

=

+

=

+=

00

2

2

12

min

2

1

22

min

2

1

22

min,

2

1

22

max,

boost, similar to isn calculatio Ripple2

)1(

22)1(

0, CCM,For 2)1(

2

2)1(

2



38

ExampleThe Flyback converter has these specifications:DC input voltage: 40VOutput voltage: 25VDuty cycle: 0.5Rated load: 62.5WMax peak-peak inductor current ripple:25% of the average inductor current.Maximum peak-peak output voltage: 0.1VSwitching frequency: 75kHz

Based on the abovementioned specifications, determinea) Transformer turns ratiob) Value of magnetizing inductor Lm. c) Maximum and minimum inductor current.d) Value of capacitor C.



39

Full-bridge converter

SW2SW4

SW3

VS

NS

NS

+

vx C

Lx

R+

Vo

+

vp

SW1

SW1,SW2

SW3,SW4 DT T

2T DTT +

2VPVS

-VSV

x

P

SS N

NV

DT2T

DTT +2

T



40

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 (C ) acts like the buck converter circuit.

Output Voltage

DNNVV

p

sso

= 2

Documents

chopper.pdf