Chopper-2002.pdf

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Power Electronics andDrives (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.

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DC-DC Converter (Chopper)

• DEFINITION: Converting the unregulatedDC 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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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 highat high current due

to:

T Lo R I P 2=

T Lo R I V =

+

−

V o

R L

+ V CE

− I L

MODEL OF LINEAR

REGULATOR

R T

EQUIVALENT

CIRCUIT

V s

R L

+ V CE

−

I L

V s

V o

+

−

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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 theswitch.

– Power is 100%

transferred from

source to load.

• Switching regulator

is the basis of all

DC-DC converters

+

−

V o

R L

+ V CE

− I L

MODEL OF LINEAR REGULATOR

EQUIVALENT CIRCUIT

V s

R L

I L

V s

V o

+

−

(ON)

closed

(OFF)

open

(ON)

closed

DT T

OUTPUT VOLTAGE

V o

SWITCH

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5

Buck (step-down) converter

V d

L

D C R L

S

+

V o

−

V o

+

CIRCUIT OF BUCK CONVERTER

CIRCUIT WHEN SWITCH IS CLOSED

CIRCUIT WHEN SWITCH IS OPENED

V o

+

−

−

i L

V d D R L

S

V d D R L

S

+ −v L

+ v L −

i L

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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

od L V V v −=

∫=⇒

=

dt v L

i

dt di Lv

L L

L L

1

V d V D

+ v L -

C R L

+

−

V o

V d −V

o

−V o

closed

openedclosed

opened

t

DT T t

i Lmin

i Lmax

I L

v L

i L

i L

+

−

S

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7

Operation when switch turned

off (opened)

• Because of

inductive energy

storage, i L

continues to flow.

• Diode is forward

biased

• Current now flows

through the diode

and

o L V v −=

V d

+ v L -

C R L

+

−

V o

V d −V

o

−V o

closed

openedclosed

opened

t

DT T t

i Lmin

i Lmax

I L

v L

i L

i L

S

D

(1-D)T

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8

Analysis for switch closed

( ) DT L

V V i

LV V

DT i

t i

dt di

i

i

LV V

dt di

dt

di L

V V v

od closed L

od L L L

L

L

od L

L

od L

⋅

−=∆

−=∆=∆∆=

−=⇒

=

−=

FigureFrom

linearly.increasemustTherefore

constant.tive- posiais of vative

-derithesince: Note

oltage,inductor vThe

I L

i L max

DT T

i L

V d − V

o

v L

t

t

i L min

closed

∆i L

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9

Analysis for switch opened

( ) T D L

V i

L

V

T D

i

t

i

dt

di

ii

LV

dt di

dt

di L

V v

oopened L

o L L L

L

L

o L

L

o L

)1(

)1(

FigureFrom

linearly.decrease

mustconstant,tive-negaais of vative

-derithesince: Note

opened,switchFor

−⋅

−=∆

−=

−

∆=

∆

∆=

−=⇒

=

−=

I L

i L max

DT T

i L

V d − V

o

v L

t

t

i L min

opened

∆i L

(1− D)T

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10

Steady-state operation

( ) ( )

d o

so

sod

opened Lclosed L

L

L

DV V

T D L

V DT

L

V V

ii

i

i

=⇒

=−⋅

−−⋅

−

=∆+∆

0)1(

0

:i.ezero,is periodoneoverof changetheisThatcycle.nexttheof begining

at thesametheiscycleswitchingof end at thethatrequiresoperationstate-Steady

i L

Unstable current

Decaying current

Steady-state current

t

t

t

i L

i L

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11

−−=

∆−=

−+=

−+=∆+=

==⇒=

Lf

D

RV

i I I

Lf

D

RV

T D LV

RV i I I

R

V I I

o L

L

o

oo L L

o R L

2

)1(1

2

:currentMinimum

2

)1(1

)1(21

2

:currentMaximum

R incurrentAveragecurrentinductorAverage

min

max

L

Average, Maximum and

Minimum inductor current

I L

I max

I min

i L

∆i L

t

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12

Continuous current operationi L

I max

I min

t 0

min

min

min

min

be bechosenis Normallyoperation.of modecontinousensure

current toinductorminimumtheisThis2

)1(

02

)1(1

,0 operation,continuousFor

2

)1(1

2

analysis, previousFrom

L L

R f

D L L

Lf

D

RV

I

Lf

D

RV

i I I

o

o L

L

>>

⋅−

=≥⇒

≥

−−⇒

≥

−−=

∆−=

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13

Output voltage ripple

2

2

8

)1(

factor,rippletheSo,

8

)1(

8

8222

1

:formulaareatriangleusefigure,From

LCf

D

V

V r

LCf

D

C

iT V

iT iT Q

C

QV V C QCV Q

iii

o

o

Lo

L L

ooo

R Lc

−=

∆=

−=

∆=∆∴

∆=

∆

=∆

∆=∆⇒∆=∆⇒=

+=

i Ri L

LiC

i L

i L=I R

imax

imin

0

0V o

V o / R

+

−

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14

Design procedures for Buck

Vd

(input

spec.)

SWITCH

f = ?

D = ?

TYPE ?

D

L

Lmin

= ?

L = 10Lmin

C

ripple ?

R L

Po

= ?

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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15

Design procedures for Buck

• Determine Lmin. Increase Lmin by about 10times 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 I p/3, where I p is the

peak capacitor current given by ∆i L/2

• Wire size consideration:

– Normally rated in RMS. But i L is known as

peak. RMS value for i L is given as:

22

,3

2

∆+= L L RMS L

i I I

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16

Examples of Buck converter

• A buck converter is supplied from a 50V batterysource. 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 ripplemust 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

V d

L D

C

R L

S

V d

LD

CR

LS

V d

LD

C R LS

+ v L−

+

V o

−

+ v L

-

V o

+

CIRCUIT OF BOOST CONVERTER



V o

+

−

−

i L

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Boost analysis:switch closed

V d

L

D

CS

+ v L−

i L

+

vo

−

( ) L

DT V i

LV

dt di

DT

i

t

i

dt

di

L

V

dt

didt

di L

V v

d closed L

d L

L L L

d L

L

d L

=∆

=⇒

∆=

∆∆

=

=⇒

=

=

DT T

i L

v L

CLOSED

t

t

V d

V d − V

o

∆i L

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19

Switch opened

( ) ( )

L

DT V V i

L

V V

dt

di

T D

it

i

dt

di

L

V V

dt

didt

di L

V V v

od opened L

od L

L

L L

od L

L

od L

)1(

)1(

−−=∆⇒

−=⇒

−∆

=∆∆

=

−=⇒

=−=

DT T

( 1-D )T

i L

v L

OPENED

t

t

V d

V d − V

o

∆i L

V d

D

CS

+ v L

-

i L

+

vo

-

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20

Steady-state operation

( ) ( )

( )

D

V

V

L

T DV V

L

DT V

ii

d

o

od d

opened L

closed L

−=⇒

=−−

−

=∆+∆

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 fromthe inductor. Hence output is large.

– Output voltage is maintained constant by

virtue of large C.

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21

Average, Maximum,

Minimum inductor current

L

DT V

R D

V i I I

L

DT V

R D

V i I I

R D

V

I

R D

V

R

D

V

I V

R

V I V

d d L L

d d L L

d L

d

d

Ld

od d

2)1(

2

2)1(

2

currentinductorminMax,

)1(

currentinductorAverage

)1(

)1(

power Output power Input

2min

2max

2

2

2

2

2

−−

=∆

−=

+

−

=∆

+=

−=

−=

−=

=

=

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22

L and C values

( )

( )

RCf

D

V

V r

RCf

DV

RCf

DT V V

V C DT R

V Q

f

R D D

TR D D L

L

DT V

R D

V

I

o

o

oo

o

oo

d d

=∆

=

==∆

∆=

=∆

−=

−=

≥−−

≥

factor Ripple

2

12

1

02)1(

0

operation,continousFor

2

2

min

2

min

DT T

imax

imin

imin

imax

ic

i D

i L

V d

v L

∆Q

V d −V

o

I o=V

o / R

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23

Examples

• The boost converter has the following

parameters: Vd=20V, D=0.6, R=12.5ohm,

L=65uH, C=200uF, f s=40KHz. Determine

(a) output voltage, (b) average, maximum

and minimum inductor current, (c) outputvoltage 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

V d L

D

CR

L

S

+

V o

−

V o

+

CIRCUIT OF BUCK-BOOST CONVERTER



V o

+

−

−

i LV d v L

+

−

i LV d v L

+

−

D

DS

S

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25

Buck-boost analysis

DT T

imax

imin

imin

imax

ic

i D

i L

V d

v L

∆Q

V d −V

o

I o=V

o / R

L

T DV i

L

V

T D

i

t

i

L

V

dt

didt di LV v

L

DT V i

L

V

DT

i

t

i

L

V

dt

didt

di

LVd v

oopened L

o L L

o L

Lo L

d closed L

d L L

d L

L L

)1()(

)1(

openedSwitch

)(

closedSwitch

−=∆

=−∆

=∆∆

=⇒

==

=∆

=∆

=∆

∆

=⇒==

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26

Output voltage

−−=⇒

=−

+

D

DV

L

T DV

L

DT V

s

od

1V

0)1(

:operationstateSteady

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

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27

Average inductor current

2

2

2

2

)1(

,forngSubstituti

:ascurrentinductoraverage torelatediscurrentsourceaverageBut

i.e.source, bythesupplied powerequalmustload by the absorbed power

converter,in theloss powernoAssuming

D R

DV

DV

P

RDV

V I

V

D I V R

V

D I I

I V RV

P P

d

d

o

d

o L

o

Ld o

L s

sd o

so

−===⇒

=⇒

=

=

=

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28

L and C values

RCf

D

V

V r

RCf

DV

RC

DT V V

V C DT R

V

f

R D L

L

DT V

D R

D

L

DT V

D R

DV i I I

L

DT V

D R

DV i I I

o

o

ooo

oo

d

d d L L

d d L L

=∆

=

==∆

∆=

=∆

−=⇒

=+−

−−

=∆

−=

+−

=∆+=

Q

ripple,tageOutput vol

2

)1(

02)1(

V

current,continuousFor

2)1(2

2)1(2

current,inductorminandMax

2

min

2d

2min

2max

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Control of DC-DC converter

using pulse width modulation-

PWM

Comparator

Vcontrol

Sawtooth

Waveform

Vo(desired)

Vo (actual)

+

-

Switch control

signal

Sawtooth

Waveform

Vcontrol 1

Switch

control

signalton 2

T

Vcontrol 2

ton 1

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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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Linear and switched mode

power supplies block diagram

Basic Block diagram of linear power supply

Base/gate

Drive

Error

Amp.

Line

Input

φφ 3/1 50/60 HzIsolation

Transformer

Rectifier +

Vd

-

Vce=Vd-Vo

C E

B

Vo

Vref

RL

+Vo

+

Vo

-

Basic Block diagram of SMPS

EMI

FILTER

RECTIFIER

AND

FILTER

High

Frequency

rectifier

and

filter

Base/

gate

drive

PWM

Controller

error

Amp

Vo

Vref

DC

Regulated

DC-DC CONVERSITION + ISOLATION

DCUnregulated

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High frequency transformer

:Models

;

iprelationshoutput-inputBasic voltagevarying-meup/down tistepii)

isolationelectricaloutput-Input i)

:functionBasic

1

2

2

1

2

1

2

1

N

N

i

i

N

N

v

v==

V 1

V2

+

−

+

−

i1 i2 N 1 N 2

Ideal model

V 1

V 2

+

−

+

−

i1 i2 N 1 N 2

Model used for

most PE application L

m

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33

Flyback Converter

V s

+

−V

o

V s

V s

V s

N 1

N 2i

1

i2

+

-

v2v1

+

-

i LM

i D

+ -v D

iC

i R

C R

vSW

+ −

+

−

V o

iS

N 1

N 2

+

-

0

0

v1

v1=V

s

i LM

i s=i

LM

+=

2

1

N

N V V V o s sw

+ −

−

2

11

N

N V v o

v1

+

−i LM

N 1

N 2

v2= -V S

+−

+

−V o

i D

L M

vSW

V o

+

−

Flyback converter circuit

Model with magnetising

inductance

Switch closed

Voltage and current

conditions when switch

opened

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34

Flyback waveforms

DT T

i Lm

DT

T

t

t

i s

DT

T

t

t

t

i D

iC

T

v1

-V(N 1 /N

2 )

V o/ R

V s

DT

DT T

∆i LM

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Analysis: switched closed

( )

00

Therefore,

0

er,transformtheof sideloadOn the

2

1

1

2

1

2

1

212

1

==

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36

Analysis: switch opened

( )

( )

( ) ( )

( )

−

=⇒

=

−+⇒

=∆+∆

−−=∆⇒

−=

−

∆=

∆=

−==

−=

=⇒

−=

−=

2

10

2

10

2

10

2

10

2

1

01

2

10

2

121

022

101

)1(

01

0

operation,state-steadyFor

)1(

1

;

N

N

D

DV V

N

N

L

T DV

L

DT V

ii

N

N

L

T DV i

N

N

L

V

T D

i

dt

i

dt

di

N

N

V vdt

di

L

N

N V

N

N vv

V v N

N

V v

d

mm

d

opened Lclosed L

mopenm L

m

m Lm Lm L

m L

m

mm

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37

Output voltage

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

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

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

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38

Average inductor current

( )

( )

−

=

−=

=⇒

=

==

=

=

1

20

2

1

22

20

20

20

0

)1()1(

:aswrittenalsoiscurrentinductoraverageThe

forsolvingandSubstitute

:as torelatedis

N

N

R D

V

N

N

R D

DV I

DRV

V I

RV D I V

I

D I T

DT I I

I I

R

V I V

P P

d L

d L

Ld

L

L L

s

L s

sd

s

m

m

m

m

m

m

m

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39

Max, Min inductor current,

Lmin, C values

( )

RCf

D

V

V r

N

N

f

R DV L

f L

DV

L

DT V

N

N

R D

DV

I

L

DT V

N

N

R D

DV i

I I

L

DT V

N

N

R D

DV i I I

d m

m

d

m

d d

L

m

d d L

L L

m

d d L L L

m

m

mm

m

mm

=∆

=

−=

==

−

=

−

−=

∆

−=

+

−=

∆+=

0

0

2

2

12

min

2

1

22

min

2

1

2

2min,

2

1

22max,

converter, boostsimilar toisncalculatiorippleThe

2

)1(

22)1(

0,operation,continuosFor

2)1(2

2)1(2

8/20/2019 Chopper-2002.pdf

40/41


Zainal Salam

40

Full-bridge converter

SW2SW4

V S

N S

N S

+

−v x C R

+

−V

o

+

−

v p

SW1,SW2

SW3,SW4 DT T

2

T DT

T +

2V P V

S

-V S

V x

P

S S

N

N V

DT T T T

SW3 L

x

SW1

2 DT +

2

8/20/2019 Chopper-2002.pdf

41/41

Full bridge: basic operation

• Switch “pair”: [S1 & S2];[S3 & S4].

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

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

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

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

• Output Voltage D N

N V V

p

s so ⋅

= 2

Documents

Chopper-2002.pdf