UNIT III DC Choppers 7/3/2015Copyright by .

UNIT III

DC Choppers

04/19/23 Copyright by www.noteshit.com 1

Copyright by www.noteshit.com

Introduction• Chopper is a static device.• A variable dc voltage is obtained from a

constant dc voltage source.• Also known as dc-to-dc converter.• Widely used for motor control.• Also used in regenerative braking.• Thyristor converter offers greater

efficiency, faster response, lower maintenance, smaller size and smooth control.

04/19/23 2


Choppers are of Two Types

Step-down choppers. Step-up choppers. In step down chopper output voltage is

less than input voltage. In step up chopper output voltage is

more than input voltage.

04/19/23 3


Principle Of Step-down Chopper

V

i0

V 0

C hopper

R

+

04/19/23 4


• A step-down chopper with resistive load.

• The thyristor in the circuit acts as a switch.

• When thyristor is ON, supply voltage appears across the load

• When thyristor is OFF, the voltage across the load will be zero.

04/19/23 5


V d c

v 0

V

V /R

i0

Id c

t

t

tO N

T

tO F F

04/19/23 6


verage value of output or load voltage.

verage value of output or load current.

Time interval for which SCR conducts.

Time interval for which SCR is OFF.

Period of switching

dc

dc

ON

OFF

ON OFF

V A

I A

t

t

T t t

or chopping period.

1 Freq. of chopper switching or chopping freq.f

T

04/19/23 7


Average Output Voltage

.

duty cycle

ONdc

ON OFF

ONdc

ON

tV V

t t

tV V V d

T

tbut d

t

04/19/23 8


2

0

Average Output Current

RMS value of output voltage

1 ON

dcdc

ONdc

t

O o

VI

RtV V

I dR T R

V v dtT

04/19/23 9


2

0

2

But during ,

Therefore RMS output voltage

1

.

.

ON

ON o

t

O

ONO ON

O

t v V

V V dtT

tVV t V

T T

V d V

04/19/23 10


2

2

Output power

But

Output power

O O O

OO

OO

O

P V I

VI

R

VP

R

dVP

R

04/19/23 11


Effective input resistance of chopper

The output voltage can be varied by

varying the duty cycle.

idc

i

VR

I

RR

d

04/19/23 12


Methods Of Control

• The output dc voltage can be varied by the following methods.

– Pulse width modulation control or constant frequency operation.

– Variable frequency control.

04/19/23 13


Pulse Width Modulation

• tON is varied keeping chopping frequency ‘f’ & chopping period ‘T’ constant.

• Output voltage is varied by varying the ON time tON

04/19/23 14


V 0

V

V

V 0

t

ttO N

tO N tO F F

tO F F

T

04/19/23 15


Variable Frequency Control

• Chopping frequency ‘f’ is varied keeping either tON or tOFF constant.

• To obtain full output voltage range, frequency has to be varied over a wide range.

• This method produces harmonics in the output and for large tOFF load current may become discontinuous

04/19/23 16


v 0

V

V

v 0

t

t

tO N

tO N

T

T

tO F F

tO F F

04/19/23 17


Step-down ChopperWith R-L Load

V

i0

V 0

C hopper

R

LFW D

E

+

04/19/23 18


• When chopper is ON, supply is connected across load.

• Current flows from supply to load.

• When chopper is OFF, load current continues to flow in the same direction through FWD due to energy stored in inductor ‘L’.

04/19/23 19


• Load current can be continuous or discontinuous depending on the values of ‘L’ and duty cycle ‘d’

• For a continuous current operation, load current varies between two limits Imax and Imin

• When current becomes equal to Imax the chopper is turned-off and it is turned-on when current reduces to Imin.

04/19/23 20


O utpu tvo ltage

O utpu tcurrent

v 0

V

i0

Im a x

Im in

t

t

tO N

T

tO F F

C ontinuouscurrent

O utpu tcurrent

t

D iscon tinuouscurrent

i0

04/19/23 21

Expressions For Load Current

iO For Continuous Current Operation When

Chopper Is ON (0 t tON)


V

i0

V 0

R

L

E

+

-



min

min

Taking Laplace Transform

. 0

At 0, initial current 0

OO

O O O

O

O

diV i R L E

dt

V ERI S L S I S i

S S

t i I

IV EI S

RR SLS SLL

04/19/23 24


min

Taking Inverse Laplace Transform

1

This expression is valid for 0 ,

i.e., during the period chopper is ON.

At the instant the chopper is turned off,

load c

R Rt t

L LO

ON

V Ei t e I e

R

t t

maxurrent is O ONi t I04/19/23 25


When Chopper is OFF

i0

R

L

E

04/19/23 26


max

When Chopper is OFF 0

0

Talking Laplace transform

0 0

Redefining time origin we have at 0,

initial current 0

OFF

OO

O O O

O

t t

diRi L E

dt

ERI S L SI S i

St

i I

04/19/23 27


max

max

Taking Inverse Laplace Transform

1

O

R Rt t

L LO

I EI S

R RS LS SL L

Ei t I e e

R

04/19/23 28


min

The expression is valid for 0 ,

i.e., during the period chopper is OFF

At the instant the chopper is turned ON or at

the end of the off period, the load current is

OFF

O OFF

t t

i t I

04/19/23 29


min

max

max

max min

min

From equation

1

At ,

To Find &

1

R Rt t

L LO

ON O

dRT dRT

L L

V Ei t e I e

R

t t dT i t I

V EI e I e

I I

R

04/19/23 30


max

min

From equation

1

At ,

1

R Rt t

L LO

OFF ON O

OFF

Ei t I e e

R

t t T t i t I

t t d T

04/19/23 31


1 1

min max

min

max min

max

1

Substituting for in equation

1

we get,

1

1

d RT d RT

L L

dRT dRT

L L

dRT

L

RT

L

EI I e e

R

I

V EI e I e

R

V e EI

R Re

04/19/23 32


max

1 1

min max

min

max min

Substituting for in equation

1

we get,

1

1

is known as the steady state ripple.

d RT d RT

L L

dRT

L

RT

L

I

EI I e e

R

V e EI

R Re

I I

04/19/23 33


max min

max min

Therefore peak-to-peak ripple current

Average output voltage

.

Average output current

2

dc

dc approx

I I I

V d V

I II

04/19/23 34


min max

min

max minmin

Assuming load current varies linearly

from to instantaneous

load current is given by

. 0O ON

O

I I

I ti I for t t dT

dTI I

i I tdT

04/19/23 35


20

0

2

max minmin

0

2min max min2 2max min

min

0

RMS value of load current

1

1

21

dT

O RMS

dT

O RMS

dT

O RMS

I i dtdT

I I tI I dt

dT dT

I I I tI II I t dt

dT dT dT

04/19/23 36


12 2

max min2min min max min

20

0

2

max minmin

0

RMS value of output current

3

RMS chopper current

1

1

O RMS

dT

CH

dT

CH

I II I I I I

I i dtT

I II I t dt

T dT

04/19/23 37


12 2

max min2min min max min3

Effective input resistance is

CH

CH O RMS

iS

I II d I I I I

I d I

VR

I

04/19/23 38


Where

Average source currentS

S dc

idc

I

I dI

VR

dI

04/19/23 39


Principle Of Step-up Chopper

+

V OV

C hopper

CLOAD

DLI

+

04/19/23 40


• Step-up chopper is used to obtain a load voltage higher than the input voltage V.

• The values of L and C are chosen depending upon the requirement of output voltage and current.

• When the chopper is ON, the inductor L is connected across the supply.

• The inductor current ‘I’ rises and the inductor stores energy during the ON time of the chopper, tON.

04/19/23 41


• When the chopper is off, the inductor current I is forced to flow through the diode D and load for a period, tOFF.

• The current tends to decrease resulting in reversing the polarity of induced EMF in L.

• Therefore voltage across load is given by

. ., O O

dIV V L i e V V

dt

04/19/23 42


• A large capacitor ‘C’ connected across the load, will provide a continuous output voltage .

• Diode D prevents any current flow from capacitor to the source.

• Step up choppers are used for regenerative braking of dc motors.

04/19/23 43


Expression For Output VoltageAssume the average inductor current to be

during ON and OFF time of Chopper.

Voltage across inductor

Therefore energy stored in inductor

= . .

Where

When Chopper

period of chopper.

is ON

ON

ON

I

L V

V I t

t ON

04/19/23 44


(energy is supplied by inductor to load)

Voltage across

Energy supplied by inductor

where period of Chopper.

Neg

When Chopper

lecting losses, energy stored in inductor

is OFF

O

O OFF

OFF

L V V

L V V It

t OFF

L

= energy supplied by inductor L

04/19/23 45


Where

T = Chopping period or period

of switching.

ON O OFF

ON OFFO

OFF

OON

VIt V V It

V t tV

t

TV V

T t

04/19/23 46


1

1

1

1

Where duty cyle

ON OFF

OON

O

ON

T t t

V Vt

T

V Vd

td

T

04/19/23 47


For variation of duty cycle ' ' in the

range of 0 1 the output voltage

will vary in the range O

O

d

d V

V V

04/19/23 48


Performance Parameters• The thyristor requires a certain minimum time to

turn ON and turn OFF.

• Duty cycle d can be varied only between a min. & max. value, limiting the min. and max. value of the output voltage.

• Ripple in the load current depends inversely on the chopping frequency, f.

• To reduce the load ripple current, frequency should be as high as possible.

04/19/23 49


Problem

• A Chopper circuit is operating on TRC at a frequency of 2 kHz on a 460 V supply. If the load voltage is 350 volts, calculate the conduction period of the thyristor in each cycle.

04/19/23 50


3

460 V, = 350 V, f = 2 kHz

1Chopping period

10.5 sec

2 10

Output voltage

dc

ONdc

V V

Tf

T m

tV V

T

04/19/23 51


3

Conduction period of thyristor

0.5 10 350

4600.38 msec

dcON

ON

ON

T Vt

V

t

t

04/19/23 52


Problem

• Input to the step up chopper is 200 V. The output required is 600 V. If the conducting time of thyristor is 200 sec. Compute

– Chopping frequency,

– If the pulse width is halved for constant frequency of operation, find the new output voltage.

04/19/23 53


6

200 , 200 , 600

600 200200 10

Solving for

300

ON dc

dcON

V V t s V V

TV V

T t

T

T

T

T s

04/19/23 54


6

6

Chopping frequency

1

13.33

300 10Pulse width is halved

200 10100

2ON

fT

f KHz

t s

04/19/23 55


6

6

Frequency is constant

3.33

1300

Output voltage =

300 10200 300 Volts

300 100 10

ON

f KHz

T sf

TV

T t

04/19/23 56


Problem

• A dc chopper has a resistive load of 20 and input voltage VS = 220V. When chopper is ON, its voltage drop is 1.5 volts and chopping frequency is 10 kHz. If the duty cycle is 80%, determine the average output voltage and the chopper on time.

04/19/23 57


220 , 20 , 10

0.80

= Voltage drop across chopper = 1.5 volts


0.80 220 1.5 174.8 Volts

S

ON

ch

ONdc S ch

dc

V V R f kHz

td

TV

tV V V

T

V

04/19/23 58


33

3

3

Chopper ON time,

1Chopping period,

10.1 10 secs 100 μsecs

10 10Chopper ON time,

0.80 0.1 10

0.08 10 80 μsecs

ON

ON

ON

ON

t dT

Tf

T

t dT

t

t

04/19/23 59


Problem

• In a dc chopper, the average load current is 30 Amps, chopping frequency is 250 Hz, supply voltage is 110 volts. Calculate the ON and OFF periods of the chopper if the load resistance is 2 ohms.

04/19/23 60


3

30 , 250 , 110 , 2

1 1Chopping period, 4 10 4 msecs

250

&

30 20.545

110

dc

dcdc dc

dc

dc

I Amps f Hz V V R

Tf

VI V dV

RdV

IR

I Rd

V

04/19/23 61


3

3 3

3

Chopper ON period,

0.545 4 10 2.18 msecs

Chopper OFF period,

4 10 2.18 10

1.82 10 1.82 msec

ON

OFF ON

OFF

OFF

t dT

t T t

t

t

04/19/23 62


• A dc chopper in figure has a resistive load of R = 10 and input voltage of V = 200 V. When chopper is ON, its voltage drop is 2 V and the chopping frequency is 1 kHz. If the duty cycle is 60%, determine

– Average output voltage

– RMS value of output voltage

– Effective input resistance of chopper

– Chopper efficiency. 04/19/23 63


V

i0

C hopper

+

R v 0

200 , 10 , 2

0.60, 1 .chV V R Chopper voltage drop V V

d f kHz

04/19/23 64



0.60 200 2 118.8 Volts

RMS value of output voltage

0.6 200 2 153.37 Volts

dc ch

dc

O ch

O

V d V V

V

V d V V

V

04/19/23 65


220

0 0

Effective input resistance of chopper is

118.811.88 Amps

10200

16.8311.88

Output power is

1 1

iS dc

dcdc

iS dc

dT dTch

O

V VR

I I

VI

RV V

RI I

V VvP dt dt

T R T R

04/19/23 66


2

2

0

0

0.6 200 22352.24 watts

10Input power,

1

1

chO

O

dT

i O

dTch

O

d V VP

R

P

P Vi dtT

V V VP dt

T R

04/19/23 67


0.6 200 200 22376 watts

10Chopper efficiency,

100

2352.24100 99%

2376

chO

O

O

i

dV V VP

R

P

P

P

04/19/23 68


Problem• A chopper is supplying an inductive load with a

free-wheeling diode. The load inductance is 5 H and resistance is 10.. The input voltage to the chopper is 200 volts and the chopper is operating at a frequency of 1000 Hz. If the ON/OFF time ratio is 2:3. Calculate – Maximum and minimum values of load current

in one cycle of chopper operation.– Average load current04/19/23 69


5 , 10 , 1000 ,

200 , : 2 : 3

Chopping period,

1 11 msecs

1000

2

3

2

3

ON OFF

ON

OFF

ON OFF

L H R f Hz

V V t t

Tf

t

t

t t

04/19/23 70


3

2

35

33

53

1 10 0.6 msec5

ON OFF

OFF OFF

OFF

OFF

T t t

T t t

T t

t T

T

04/19/23 71


3

3

3

max

1 0.6 10 0.4 msec

Duty cycle,

0.4 100.4

1 10Maximum value of load current is given by

1

1

ON OFF

ON

ON

dRT

L

RT

L

t T t

t

td

T

V e EI

R Re

04/19/23 72


3

3

max

0.4 10 1 10

5

max 10 1 10

5

Since there is no voltage source in

the load circuit, E = 0

1

1

200 1

101

dRT

L

RT

L

V eI

Re

eI

e

04/19/23 73


3

3

0.8 10

max 2 10

max

min

120

1

8.0047A

Minimum value of load current with E = 0

is given by

1

1

dRT

L

RT

L

eI

e

I

V eI

Re

04/19/23 74


3

3

0.4 10 1 10

5

min 10 1 10

5

max min

200 17.995 A

101

Average load current

28.0047 7.995

8 A2

dc

dc

eI

e

I II

I

04/19/23 75


Problem• A chopper feeding on RL load is shown in figure,

with V = 200 V, R = 5, L = 5 mH, f = 1 kHz, d = 0.5 and E = 0 V. Calculate – Maximum and minimum values of load

current.– Average value of load current.– RMS load current.– Effective input resistance as seen by source.– RMS chopper current.

04/19/23 76


33

V = 200 V, R = 5 , L = 5 mH,

f = 1kHz, d = 0.5, E = 0

Chopping period is

1 11 10 secs

1 10T

f

i0

v 0

C hopper

R

LFW D

E

+

04/19/23 77


3

3

3

3

max

0.5 5 1 10

5 10

max 5 1 10

5 10

0.5

max 1

Maximum value of load current is given by

1

1

200 10

51

140 24.9 A

1

dRT

L

RT

L

V e EI

R Re

eI

e

eI

e

04/19/23 78


3

3

3

3

min

0.5 5 1 10

5 10

min 5 1 10

5 10

0.5

min 1

Minimum value of load current is given by

1

1

200 10

51

140 15.1 A

1

dRT

L

RT

L

V e EI

R Re

eI

e

eI

e

04/19/23 79


1 2

12 2

max min2min min max min

Average value of load current is

2for linear variation of currents

24.9 15.120 A

2RMS load current is given by

3

dc

dc

O RMS

I II

I

I II I I I I

04/19/23 80


12 2

2

1

2

24.9 15.115.1 15.1 24.9 15.1

3

96.04228.01 147.98 20.2 A

3

RMS chopper current is given by

0.5 20.2 14.28 A

O RMS

O RMS

ch O RMS

I

I

I d I

04/19/23 81


Effective input resistance is

= Average source current

0.5 20 10 A

Therefore effective input resistance is

20020

10

iS

S

S dc

S

iS

VR

I

I

I dI

I

VR

I

04/19/23 82


Classification Of Choppers

• Choppers are classified as – Class A Chopper– Class B Chopper– Class C Chopper– Class D Chopper– Class E Chopper

04/19/23 83


Class A Chopper

V

C hopper

FW D

+

v 0

v 0

i0

i0

LOAD

V

04/19/23 84


• When chopper is ON, supply voltage V is connected across the load.

• When chopper is OFF, vO = 0 and the load current continues to flow in the same direction through the FWD.

• The average values of output voltage and current are always positive.

• Class A Chopper is a first quadrant chopper .

04/19/23 85


• Class A Chopper is a step-down chopper in which power always flows form source to load.

• It is used to control the speed of dc motor.

• The output current equations obtained in step down chopper with R-L load can be used to study the performance of Class A Chopper.

04/19/23 86


O utpu t cu rren t

Thyristo rgate pu lse

O utpu t vo ltage

ig

i0

v 0

t

t

ttO N

T

C H O N

FW D C onduc ts

04/19/23 87


Class B Chopper

V

C hopper

+

v 0

v 0

i0

i0

L

E

R

D

04/19/23 88


• When chopper is ON, E drives a current through L and R in a direction opposite to that shown in figure.

• During the ON period of the chopper, the inductance L stores energy.

• When Chopper is OFF, diode D conducts, and part of the energy stored in inductor L is returned to the supply.

04/19/23 89


• Average output voltage is positive.• Average output current is negative. • Therefore Class B Chopper operates in

second quadrant.• In this chopper, power flows from load to

source.• Class B Chopper is used for regenerative

braking of dc motor.• Class B Chopper is a step-up chopper.

04/19/23 90


O utpu t cu rren t

D con d u cts C h o pp e r

con d u cts

Thyristo rgate pu lse

O utpu t vo ltage

ig

i0

v 0

t

t

t

Im in

Im ax

T

tO NtO F F

04/19/23 91

Expression for Output Current



min

For the initial condition i.e.,

During the interval diode 'D' conduc

at 0

The solution of the ab

ts

voltage equation

ove equation is obtained

along similar lines as in s

is given by

OO

O

LdiV Ri E

dt

i t I t

tep-down chopper

with R-L load04/19/23 93


min

max

max min

During the interval chopper is ON voltage

equation is g

1 0

At

1

0

iven by

OFF OFF

R Rt t

L LO OFF

OFF O

R Rt t

L L

OO

V Ei t e I e t t

R

t t i t I

V EI e I e

R

LdiRi E

dt

04/19/23 94


max

max

min

min max

Redefining the time origin, at 0

The solution for the stated initial condition is

1 0

At

1ON ON

O

R Rt t

L LO ON

ON O

R Rt t

L L

t i t I

Ei t I e e t t

R

t t i t I

EI I e e

R

04/19/23 95


Class C Chopper

V

C hopper

+

v 0

D 1

D 2C H 2

C H 1

v 0i0

i0

L

E

R

04/19/23 96


• Class C Chopper is a combination of Class A and Class B Choppers.

• For first quadrant operation, CH1 is ON or D2 conducts.

• For second quadrant operation, CH2 is ON or D1 conducts.

• When CH1 is ON, the load current is positive.• The output voltage is equal to ‘V’ & the load

receives power from the source. • When CH1 is turned OFF, energy stored in

inductance L forces current to flow through the diode D2 and the output voltage is zero.

04/19/23 97


• Current continues to flow in positive direction.• When CH2 is triggered, the voltage E forces

current to flow in opposite direction through L and CH2 .

• The output voltage is zero.• On turning OFF CH2 , the energy stored in the

inductance drives current through diode D1 and the supply

• Output voltage is V, the input current becomes negative and power flows from load to source.

04/19/23 98


• Average output voltage is positive• Average output current can take both

positive and negative values.• Choppers CH1 & CH2 should not be turned

ON simultaneously as it would result in short circuiting the supply.

• Class C Chopper can be used both for dc motor control and regenerative braking of dc motor.

• Class C Chopper can be used as a step-up or step-down chopper.

04/19/23 99


G ate pu lseof C H 2


O utput cu rren t

O utpu t vo ltage

ig 1

ig 2

i0

V 0

t

t

t

t

D 1 D 1D 2 D 2C H 1 C H 2 C H 1 C H 2

O N O N O N O N

04/19/23 100


Class D Chopper

V+ v 0

D 2

D 1 C H 2

C H 1

v 0

i0

L ER i0

04/19/23 101


• Class D is a two quadrant chopper.

• When both CH1 and CH2 are triggered simultaneously, the output voltage vO = V and output current flows through the load.

• When CH1 and CH2 are turned OFF, the load current continues to flow in the same direction through load, D1 and D2 , due to the energy stored in the inductor L.

• Output voltage vO = - V .

04/19/23 102


• Average load voltage is positive if chopper ON time is more than the OFF time

• Average output voltage becomes negative if tON < tOFF .

• Hence the direction of load current is always positive but load voltage can be positive or negative.

04/19/23 103




O utpu t curren t

O utpu t vo ltage

Average v 0

ig 1

ig 2

i0

v 0

V

t

t

t

t

C H ,C HO N1 2 D 1,D 2 Conducting

04/19/23 104




O utpu t curren t

O utpu t vo ltage

Average v 0

ig 1

ig 2

i0

v 0

V

t

t

t

t

C HC H

1

2

D , D1 2

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Class E Chopper

V

v 0

i0L ER

C H 2 C H 4D 2 D 4

D 1 D 3C H 1 C H 3

+

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Four Quadrant Operationv 0

i0

C H - C H O NC H - D C onduc ts

1 4

4 2

D D2 3 - C onductsC H - D C onduc ts4 2

C H - C H O NC H - D C onduc ts

3 2

2 4

C H - D C onduc tsD - D C onducts

2 4

1 4

04/19/23 107


• Class E is a four quadrant chopper

• When CH1 and CH4 are triggered, output current iO flows in positive direction through CH1 and CH4, and with output voltage vO = V.

• This gives the first quadrant operation.

• When both CH1 and CH4 are OFF, the energy stored in the inductor L drives iO through D2 and D3 in the same direction, but output voltage vO = -V.

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• Therefore the chopper operates in the fourth quadrant.

• When CH2 and CH3 are triggered, the load current iO flows in opposite direction & output voltage vO = -V.

• Since both iO and vO are negative, the chopper operates in third quadrant.

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• When both CH2 and CH3 are OFF, the load current iO continues to flow in the same direction D1 and D4 and the output voltage vO = V.

• Therefore the chopper operates in second quadrant as vO is positive but iO is negative.

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Problem• For the first quadrant chopper shown in figure,

express the following variables as functions of V, R and duty cycle ‘d’ in case load is resistive.– Average output voltage and current– Output current at the instant of commutation– Average and RMS free wheeling diode current.– RMS value of output voltage– RMS and average thyristor currents.

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V

i0

v 0

C hopper

FW D

+

LOAD

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Average output voltage,

Average output current,

The thyristor is commutated at the instant

output current at the instant of commutation is

since V is the output v

ONdc

dcdc

ON

tV V dV

T

V dVI

R Rt t

V

R

oltage at that instant.04/19/23 113


20

0

Free wheeling diode (FWD) will never

conduct in a resistive load.

Average & RMS free wheeling diode

currents are zero.

1

But during

ONt

O RMS

O ON

V v dtT

v V t

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2

0

2

1

Where duty cycle,

ONt

O RMS

ONO RMS

O RMS

ON

V V dtT

tV V

T

V dV

td

T

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RMS value of thyristor current

= RMS value of load current

Average value of thyristor current

= Average value of load current

O RMSV

R

dV

R

dV

R

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Boost Converter or

Step Up converter

Buck-Boost Converter

Buck Converter or

Step Down Converter

Simple DC-DC Converter Topologies

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SMPS benefits

– Very wide input voltage range.• For example: most personal computer power supplies are

SMPSs - accepting AC input 90V to 250V.– Lower Quiescent Current than linear regulators– Less heat than an equivalent linear regulator.

• Much Lower Green House Gas emissions– Overall Smaller geometry components are used – Lighter Weight– Lower running cost - Lower total cost of ownership (TCO).– Battery operated devices - longer lifetime.

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SMPS disadvantages

– Significant Output Ripple• May need a post filter to decrease ripple• May need a secondary linear low drop out

regulator to ensure damaging voltage transients keep away from voltage sensitive elements - electronics.

• An SMPS May add too much cost. – How much is too much?

04/19/23 119

UNIT III DC Choppers 7/3/2015Copyright by .

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