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Power Electronics by Prof. M. MadhusudhanRao

1

1Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

DC Choppers

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2

2 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

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

sie and smooth control.

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3


Choppers are of Two Types

• !tep-down choppers.

• !tep-up choppers.• "n step down chopper output voltage is lessthan input voltage.

• "n step up chopper output voltage is morethan input voltage.

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%


# d c

v $

#

# * %

i $

" d c

t

t

t ' (

T

t ' ) )

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& Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

verage value of output or load voltage.

verage value of output or load current.

Time interval for which !C% conducts.

Time interval for which !C% is ')).

+eriod of switching

dc

dc

ON

OFF

ON OFF

V A

I A

t

t

T t t

=

=

==

= + =or chopping period.

)re. of chopper switching or chopping fre. f

T = =

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'! Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Average 'utput #oltage

.

duty cycle

ON dc

ON OFF

ON dc

ON

t V V

t t

t V V V d

T

t but d t

= ÷+

= = ÷

= = ÷

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("Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

$

Average 'utput Current

%/! value of output voltage

ON

dcdc

ON dc

t

O o

V I R

t V V

I d R T R

V v dt T

=

= = ÷

= ∫

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1)1# Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

$

0ut during ,

Therefore %/! output voltage

.

.

ON

ON o

t

O

ON

O ON

O

t v V

V V dt T

t V

V t V T T

V d V

=

=

= ==

∫

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'utput power

0ut

'utput power

O O O

OO

OO

O

P V I

V I

R

V P

RdV

P R

=

=

∴=

=

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1ffective input resistance of chopper

The output voltage can be varied by

varying the duty cycle.

i

dc

i

V R

I

R R

d

=

=

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Methods !f Control

• The output dc voltage can be varied by the

following methods. 2 +ulse width modulation control or constant

freuency operation.

2 #ariable freuency control.

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Pulse *idth Modulation

• t ON is varied keeping chopping freuency 3 f’ 4

chopping period 3T5 constant.• 'utput voltage is varied by varying the '(

time t ON

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1$15 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

# $

#

#

# $

t

tt ' (

t ' ( t ' ) )

t ' ) )

T

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1%16 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

+ariable ,re-uency Control

• Chopping freuency ‘f’ is varied keepingeither t ON or t OFF constant.

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

• This method produces harmonics in the output

and for large t OFF load current may becomediscontinuous

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1&1 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

v $

#

#

v $

t

t

t ' (

t ' (

T

T

t ' ) )

t ' ) )

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1'1! Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

"tep#down Chopper

*ith R# oad

#

i $

# $

C h o p p e r

%

6) W 7

1

&

−

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1(1"Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

• When chopper is '(, supply is connected

across load.

• Current flows from supply to load.

• When chopper is ')), load current continuesto flow in the same direction through )W7 due

to energy stored in inductor ‘L’ .

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• 6oad current can be continuous or

discontinuous depending on the values of ‘L’

and duty cycle ‘d’ • )or a continuous current operation, load current

varies between two limits I max and I min

• When current becomes eual to I max the chopperis turned-off and it is turned-on when current

reduces to I min.

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' u t p u tv o l t a g e

' u t p u tc u r r e n t

v $

#

i $

" m a 8

" m i n

t

t

t ' (

T

t ' ) )

C o n t i n u o u sc u r r e n t

' u t p u tc u r r e n t

t

7 i s c o n t i n u o u sc u r r e n t

i$

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E/pressions ,or

oad Current

i O ,or Continuous Current !peration*hen

Chopper Is !0 (0 ≤ t ≤ t ON )

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#

i $

# $

%

6

1

&

-

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( ) ( ) ( )( )

( )

min

min

Taking 6aplace Transform

. $

At $, initial current $

OO

O O O

O

O

diV i R L E

dt

V E

RI S L S I S iS S

t i I

I V E I S R R

S LS S L L

−

−

= + +

= + − + = =

−= + ++ ÷

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( ) min

Taking "nverse 6aplace Transform

This e8pression is valid for $ ,

i.e., during the period chopper is '(.

At the instant the chopper is turned off,

load c

R Rt t L L

O

ON

V E i t e I e

R

t t

− − ÷ ÷ −= − +

≤ ≤

( ) ma8urrent is O ON i t I =

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*hen Chopper is !,,i $

%

6

1

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( )

( ) ( ) ( )

( ) ma8

When Chopper is ')) $

$

Talking 6aplace transform

$ $

%edefining time origin we have at $,

initial current $

OFF

OO

O O O

O

t t

di Ri L E

dt

E RI S L SI S i

S

t

i I

−

−

≤ ≤

= + +

= + − +

==

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( )

( )

ma8

ma8

Taking "nverse 6aplace Transform

O

R Rt t L L

O

I E I S R R

S LS S

L L

E i t I e e R

− −

∴ = − + + ÷

= − −

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( ) min

The e8pression is valid for $ ,

i.e., during the period chopper is '))

At the instant the chopper is turned '( or at

the end of the off period, the load current is

OFF

O OFF

t t

i t I

≤ ≤

=

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( )

( )

min

ma8

ma8

ma8 min

min

)rom euation

At ,

To )ind 4

R Rt t

L L

O

ON O

dRT dRT

L L

V E

i t e I e R

t t dT i t I

V E I e I e

I I

R

− − ÷ ÷

− −

−

= − + = = =

−∴ = − +

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( )

( )( )

ma8

min

)rom euation

At ,

R Rt t

L LO

OFF ON O

OFF

E i t I e e R

t t T t i t I t t d T

− −

= − −

= = − == = −

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( ) ( )

min ma8

min

ma8 min

ma8

!ubstituting for in euation

we get,

d RT d RT

L L

dRT dRT

L L

dRT

L

RT

L

E I I e e

R

I

V E I e I e R

V e E I

R Re

− −− −

− −

−

−

∴ = − −

−= − +

− = − −

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( ) ( )

( )

ma8

min ma8

min

ma8 min

!ubstituting for in euation

we get,

is known as the steady state ripple.

d RT d RT

L L

dRT

L

RT L

I

E I I e e

R

V e E I

R Re

I I

− −− − = − −

− = −

− −

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( )

ma8 min

ma8 min

Therefore peak-to-peak ripple current

Average output voltage

.

Average output current

dc

dc appox

I I I

V d V

I I I

∆ = −

=

+=

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( )( )

min ma8

min

ma8 minmin

Assuming load current varies linearly

from to instantaneous

load current is given by

. $O ON

O

I I

I t i I fo t t dT

dT

I I i I t dT

∆= + ≤ ≤

− = + ÷

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( )

( )( )

( )

( )

$

$

ma8 min

min

$

min ma8 min ma8 min

min

$

%/! value of load current

dT

O R!S

dT

O R!S

dT

O R!S

I i dt dT

I I t I I dt

dT dT

I I I t I I

I I t dt dT dT dT

=

− = +

−− = + + ÷

∫

∫

∫

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( )

( )( )

ma8 min

min min ma8 min

$

$

ma8 minmin

$

%/! value of output current

9

%/! chopper current

O R!S

dT

"#

dT

"#

I I I I I I I

I i dt T

I I I I t dt

T dT

−= + + −

=

− = + ÷

∫

∫

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( )( )

( )

ma8 min

min min ma8 min9

1ffective input resistance is

"#

"# O R!S

i

S

I I I d I I I I

I d I

V R

I

−= + + −

=

=

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Where

Average source currentS

S dc

i

dc

I

I dI

V R

dI

=

=

∴ =

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)4# Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Principle !f "tep#up Chopper

&

−

# '#

C h o p p e r

C6'A

7

76"

& −

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• !tep-up chopper is used to obtain a load voltage

higher than the input voltage V .• The values of L and " are chosen depending

upon the reuirement 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 inductorstores energy during the ON time of the

chopper, t ON .

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• When the chopper is off, the inductor current I is forced to flow through the diode % and load

for a period, t OFF .

• The current tends to decrease resulting inreversing the polarity of induced 1/) in L.

• Therefore voltage across load is given by

. .,O OdI

V V L i e V V dt

= + >

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• A large capacitor 3C5 connected across the load,

will provide a continuous output voltage .

• 7iode % prevents any current flow from

capacitor to the source.

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

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E/pression ,or !utput +oltae

Assume the average inductor current to be

during '( and ')) time of Chopper.

#oltage across inductor

Therefore energy stored in inductor

: . .

Where

When Chopper

period of chopper.

is '(

ON

ON

I

L V

V I t

t ON

=

=

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$45 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

( )

;energy is supplied by inductor to load<

#oltage across

1nergy supplied by inductor

where period of Chopper.

(eg

When Chopper

lecting losses, energy stored in inductor

is '))

O

O OFF

OFF

L V V

L V V It

t OFF

L

= −

= −=

: energy supplied by inductor L

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%46 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

( )

[ ]

WhereT : Chopping period or period

of switching.

ON O OFF

ON OFF

O

OFF

O

ON

VIt V V It

V t t V

t

T V V

T t

∴ = −

+=

= ÷−

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&4 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Where duty cyle

ON OFF

OON

O

ON

T t t

V V t

T

V V d

t d

T

= +

÷= ÷

÷− ∴ = ÷−

= =

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'4! Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

)or variation of duty cycle = = in the

range of $ the output voltage

will vary in the range

O

O

d

d V

V V < <

< < ∞

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(4"Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Perforance Paraeters

The thyristor re-uires a certain iniu tieto turn ON and turn OFF .

Duty cycle d can be 4aried only between a in.

5 a/. 4alue6 liitin the in. and a/. 4alueof the output 4oltae.

Ripple in the load current depends in4ersely on

the choppin fre-uency6 f .

To reduce the load ripple current6 fre-uency

should be as hih as possible.

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$)5#

Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Proble• A "&oppe cicuit i' opeatin( on TR" at a

fe)uenc* of + ,#- on a /0 V 'upp1*. If t&e

1oad vo1ta(e i' 230 vo1t'$ ca1cu1ate t&e

conduction peiod of t&e t&*i'to in eac&

c*c1e.

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$151


9

>?$ #, : 9@$ #, f : k

Chopping period

$.@ sec

$

'utput voltage

dc

ON dc

V V

T

f

T m

t V V

T

−

=

=

= =×

= ÷

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Power Electronics by Prof. M. MadhusudhanRao $353


Proble• Input to t&e 'tep up c&oppe i' +00 V. T&e

output e)uied i' /00 V. If t&e conductin( time

of t&*i'to i' +00 µ 'ec. "ompute 2 "&oppin( fe)uenc*$

2 If t&e pu1'e 4idt& i' &a1ved fo con'tant

fe)uenc* of opeation$ find t&e ne4 outputvo1ta(e.

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Power Electronics by Prof. M. MadhusudhanRao $54


?

$$ , $$ , ?$$

?$$ $$$$ $

!olving for

9$$

ON dc

dc

ON

V V t ' V V

T V V

T t T

T

T

T '

µ

µ

−

= = =

= ÷

− = ÷− ×

=

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Power Electronics by Prof. M. MadhusudhanRao $$55


?

?

Chopping freuency

9.999$$ $

+ulse width is halved

$$ $$$

ON

f T

f 5#-

t ' µ

−

−

=

= =×

×∴ = =

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Power Electronics by Prof. M. MadhusudhanRao $%56


( )

?

?

)reuency is constant

9.99

9$$

'utput voltage :

9$$ $$$ 9$$ #olts

9$$ $$ $

ON

f 5#-

T ' f

T V

T t

µ

−

−

∴ =

= =

∴ ÷−

×= = ÷ ÷−

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Power Electronics by Prof. M. MadhusudhanRao $&5


Proble

• A dc c&oppe &a' a e'i'tive 1oad of +0Ω and

input vo1ta(e V S 6 ++0V. 7&en c&oppe i' ON$it' vo1ta(e dop i' 8.3 vo1t' and c&oppin(

fe)uenc* i' 80 ,#-. If t&e dut* c*c1e i' 90:$

detemine t&e avea(e output vo1ta(e and t&ec&oppe on time.

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Power Electronics by Prof. M. MadhusudhanRao $'5!


( )

( )

$ , $ , $

$.B$

: #oltage drop across chopper : .@ voltsAverage output voltage

$.B$ $ .@ >.B #olts

S

ON

c&

ON dc S c&

dc

V V R f ,#-

t d

T

V

t

V V V T

V

= = Ω =

= =

= − ÷ = − =

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Power Electronics by Prof. M. MadhusudhanRao $(5"


9

9

9

9

Chopper '( time,

Chopping period,

$. $ secs $$ Dsecs$ $

Chopper '( time,

$.B$ $. $

$.$B $ B$ Dsecs

ON

ON

ON

ON

t dT

T f

T

t dT

t

t

−

−

−

=

=

= = × =×

=

= × ×

= × =

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Power Electronics by Prof. M. MadhusudhanRao %)6#


Proble• In a dc c&oppe$ t&e avea(e 1oad cuent i' 20

Amp'$ c&oppin( fe)uenc* i' +30 #-$ 'upp1*

vo1ta(e i' 880 vo1t'. "a1cu1ate t&e ON and

OFF peiod' of t&e c&oppe if t&e 1oad

e'i'tance i' + o&m'.

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Power Electronics by Prof. M. MadhusudhanRao %161


9

9$ , @$ , $ ,

Chopping period, > $ > msecs

@$

4

9$ $.@>@

$

dc

dcdc dc

dc

dc

I Amp' f #- V V R

T f

V I V dV R

dV I

R I R

d V

−

= = = = Ω

= = = × =

= =

∴ =

×= = =

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9

9 9

9

Chopper '( period,

$.@>@ > $ .B msecs

Chopper ')) period,

> $ .B $

.B $ .B msec

ON

OFF ON

OFF

OFF

t dT

t T t

t

t

−

− −

−

= = × × =

= −

= × − ×= × =

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• A dc c&oppe in fi(ue &a' a e'i'tive 1oad of R6 80Ω and input vo1ta(e of V 6 +00 V. 7&en

c&oppe i' ON$ it' vo1ta(e dop i' + V and t&e

c&oppin( fe)uenc* i' 8 ,#-. If t&e dut* c*c1e

i' /0:$ detemine

2 Avea(e output vo1ta(e

2 R!S va1ue of output vo1ta(e 2 Effective input e'i'tance of c&oppe

2 "&oppe efficienc*.

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#

i $C h o p p e r

&

−

% v $

$$ , $ ,

$.?$, .

c&V V R "&oppe vo1ta(e dop V V

d f ,#-

= = Ω == =

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Power Electronics by Prof. M. MadhusudhanRao %$65


( )

[ ]

( )( )

Average output voltage

$.?$ $$ B.B #olts%/! value of output voltage

$.? $$ @9.9 #olts

dc c&

dc

O c&

O

V d V V

V

V d V V

V

= −

= − =

= −= − =

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Power Electronics by Prof. M. MadhusudhanRao %%66


( )

$

$ $

1ffective input resistance of chopper is

B.B.BB Amps

$$$

?.B9.BB

'utput power is

i

S dc

dcdc

i

S dc

dT dT

c&

O

V V R I I

V I

RV V

R I I

V V v P dt dt

T R T R

= =

= = =

= = = = Ω

−= =∫ ∫

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%&6 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

( )

[ ]

( )

$

$

$.? $$ 9@.> watts

$

"nput power,

c&

O

O

dT

i O

dT

c&

O

d V V P

R

P

P Vi dt T

V V V P dt

T R

−=

−= =

=

−=

∫

∫

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%'6! Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

( )

[ ]$.? $$ $$ 9? watts

$

Chopper efficiency,

$$

9@.>$$ EEF

9?

c&

O

O

O

i

dV V V P

R

P

P

P η

η

−=

× −= =

= ×

= × =

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%(6"Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Proble

A chopper is supplying an inductive load with a

free-wheeling diode. The load inductance is 5

and resistance is !0 Ω .. The input voltage to the

chopper is "00 volts and the chopper is operatingat a fre#uency of !000 $. %f the ON&OFF ti'e

ratio is ". *alculate

7 +a,i'u' and 'ini'u' values of load currentin one cycle of chopper operation.

7 Average load current

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&)# Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

@ , $ , $$$ ,

$$ , G G 9Chopping period,

msecs$$$

9

9

ON OFF

ON

OFF

ON OFF

L # R f #-

V V t t

T f

t

t

t t

= = Ω =

= =

= = =

=

=

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&11Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

9

9

@

9

9

@9

$ $.? msec@

ON OFF

OFF OFF

OFF

OFF

T t t

T t t

T t

t T

T −

= +

= +

=

=

= × × =

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&22 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

( )9

9

9

ma8

$.? $ $.> msec7uty cycle,

$.> $

$.> $

/a8imum value of load current is given by

ON OFF

ON

ON

dRT

L

RT

L

t T t

t

t

d T

V e E I

R Re

−

−

−

−

−

= −

= − × =

×= = =×

− = − −

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9

9

ma8

$.> $ $

@

ma8 $ $

@

!ince there is no voltage source in

the load circuit, 1 : $

$$ $

dRT

L

RT L

V e I

R e

e I

e

−

−

−

−

× × ×−

× ×−

− ∴ =

−

− = −

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Power Electronics by Prof. M. Madhusudhan

Rao


9

9

$.B $

ma8 $

ma8

min

$

B.$$>A

/inimum value of load current with 1 : $is given by

dRT

L

RT

L

e

I e

I

V e I R

e

−

−

− ×

− ×

−

= − =

− = −

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Rao

&$5 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

9

9

$.> $ $

@

min $ $

@

ma8 min

$$ .EE@ A$

Average load current

B.$$> .EE@B A

dc

dc

e I

e

I I I

I

−

−

× × ×

× ×

− = = −

+=

+= ≈

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Rao

&%6 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Proble

• A c&oppe feedin( on RL 1oad i' '&o4n in fi(ue$ 4it& V 6 +00 V$ R 6 3Ω $ L 6 3 m#$

f 6 8 ,#-$ d 6 0.3 and E 6 0 V. "a1cu1ate

2 !aximum and minimum va1ue' of 1oadcuent.

2 Avea(e va1ue of 1oad cuent.

2 R!S 1oad cuent. 2 Effective input e'i'tance a' 'een b* 'ouce.

2 R!S c&oppe cuent.

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Rao

&& Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

9

9

# : $$ #, % : @ , 6 : @ m,

f : k, d : $.@, 1 : $

Chopping period is

$ secs

$

T

f

−

Ω

= = = ×

× i $

v $

C h o p p e r

%

6) W 7

1

&

−

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Rao

&'! Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

9

9

9

9

ma8

$.@ @ $

@ $

ma8 @ $

@ $

$.@

ma8

/a8imum value of load current is given by

$$ $

@

>$ >.E A

dRT

L

RT

L

V e E I

R Re

e I

e

e I

e

−

−

−

−

−

× × ×−×

× ×−

×

−

−

− = − −

− = − − −

= = −

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Rao

&("Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

9

9

9

9

min

$.@ @ $@ $

min @ $

@ $

$.@

min

/inimum value of load current is given by

$$ $

@

>$ @. A

dRT

L

RT

L

V e E I R R

e

e I

e

e I

e

−

−

−

−

× × ××

× ××

− = − −

− = − −

−= = −

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Rao

')!# Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

( )

( )( )

ma8 min

min min ma8 min

Average value of load current is

for linear variation of currents

>.E @. $ A

%/! load current is given by

9

dc

dc

O R!S

I I I

I

I I I I I I I

+=

+∴ = =

−= + + −

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Rao

'1!1Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

( )

( )( )

( )

( )

>.E @.@. @. >.E @.

9

E?.$>B.$ >.EB $. A

9

%/! chopper current is given by

$.@ $. >.B A

O R!S

O R!S

c& O R!S

I

I

I d I

−= + + −

= + + =

= = × =

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Rao

'2!2 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

1ffective input resistance is

: Average source current

$.@ $ $ A

Therefore effective input resistance is

$$$

$

i

S

S

S dc

S

i

S

V R I

I

I dI

I

V R

I

=

== × =

= = = Ω

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Rao

'3!3Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Classification !f Choppers• Choppers are classified as

2 Class A Chopper

2 Class 0 Chopper

2 Class C Chopper

2 Class 7 Chopper

2 Class 1 Chopper

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Rao

'!4Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Class 8 Chopper

#

C h o p p e r

) W 7

&

−

v $

v $

i $

i $

6'A7

#

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Rao

'$!5 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

• When chopper is ON$ supply voltage V isconnected across the load.

• When chopper is ')), v' : $ and the loadcurrent continues to flow in the same directionthrough the )W7.

• The average values of output voltage andcurrent are always positive.

• "1a'' A "&oppe is a first uadrant chopper .

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Rao

'%!6 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

• "1a'' A "&oppe is a step-down chopper in

which power always flows form source to load.

• "t is used to control the speed of dc motor.

• The output current euations obtained in step

down chopper with R;L load can be used to

study the performance of "1a'' A "&oppe.

i

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Rao

'&! Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

' u t p u t c u r r e n t

T h y r i s t o rg a t e p u l s e

' u t p u t v o l t a g e

i g

i $

v $

t

t

tt ' (

T

C ' (

) W 7 C o n d u c t s

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Rao

''!! Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Class 9 Chopper

#

C h o p p e r

&

−

v $

v $

− i $

i $

6

1

%

7

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Rao

'(!"Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

• When chopper is '(, E drives a currentthrough L and % in a direction opposite to that

shown in figure.• 7uring the '( period of the chopper, the

inductance L stores energy.

•When Chopper is ')), diode % conducts, and part of the energy stored in inductor L isreturned to the supply.

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Rao

()"# Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

• Average output voltage is positive.• Average output current is negative.

• Therefore C1a'' < "&oppe operates in seconduadrant.

• "n this chopper, power flows from load to source.

• "1a'' < "&oppe is used for regenerative brakingof dc motor.

• "1a'' < "&oppe is a step-up chopper.

T h y r i s t o ri g

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Rao

(1"1Prof. T.K. Anantha Kumar, E&E Dept., MSRIT


7c o n d u c t s C h o p p e r

c o n d u c t s

T h y r i s t o rg a t e p u l s e


i $

v $

t

t

t

" m i n

" m a 8

T

t ' (t ' ) )

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Rao

(2"2 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

E/pression for !utput Current

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Rao

(3"3Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

( ) min

)or the initial condition i.e.,

7uring the interval diode =7= conduc

at $

The solution of the ab

ts

voltage euation

ove euation is obtainedalong similar lines as in s

is given by

OO

O

LdiV Ri E

dt

i t I t

= + +

= =

tep-down chopper

with %-6 load

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Rao

("4Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

( )

( ) ( )

min

ma8

ma8 min

7uring the interval chopper is '( voltage

euation is g

$

At

$

iven by

OFF OFF

R Rt t

L LO OFF

OFF O

R Rt t

L L

OO

V E i t e I e t t

Rt t i t I

V E I e I e R

Ldi Ri E

dt

− −

− −

−∴ = − + <

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Rao

($"5 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

( )

( )

( )

ma8

ma8

min

min ma8

%edefining the time origin, at $

The solution for the stated initial condition is

$

At

ON ON

O

R Rt t

L LO ON

ON O

R Rt t

L L

t i t I

E i t I e e t t R

t t i t I

E I I e e

R

− −

− −

= =

= − − <

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Rao

(%"6 Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

Class C Chopper

#

C h o p p e r

&

−

v $

7

7 .C .

C

v $i $

i $

6

1

%

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Rao

(&" Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

• "1a'' " "&oppe is a combination of "1a'' A and

"1a'' < "&oppe'.• )or first uadrant operation, "# 8 is '( or %+

conducts.

• )or second uadrant operation, "# + is '( or %8

conducts.• When "# 8 is '(, the load current is positive.

• The output voltage is eual to ‘V’ 4 the loadreceives power from the source.

• When "# 8 is turned ')), energy stored ininductance L forces current to flow through thediode %+ and the output voltage is ero.

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Rao

('"! Prof. T.K. Anantha Kumar, E&E Dept., MSRIT

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

current to flow in opposite direction through 6and "# + .

• The output voltage is ero.

• 'n turning ')) "# + , the energy stored in the

inductance drives current through diode %8 and

the supply• 'utput voltage is V , the input current becomes

negative and power flows from load to source.

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Rao

((""


• Average output voltage is positive

• Average output current can take both positiveand negative values.

• Choppers "# 8 = "# + should not be turned

'( simultaneously as it would result in shortcircuiting the supply.

• "1a'' " "&oppe can be used both for dc motor

control and regenerative braking of dc motor.• "1a'' " "&oppe can be used as a step-up orstep-down chopper.

H a t e p u l s eo f C

i g

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Rao

1))1##


H a t e p u l s eo f C .

o f C



i g .

i $

# $

t

t

t

t

7 , 7 ,7 . 7 .C , C . C , C .' ( ' ( ' ( ' (

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Rao

1)11#1


Class D Chopper

#& −v $

7 .

7 C .

C

v $

i $

6 1% i $

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Rao

1)21#2


• Class 7 is a two uadrant chopper.• When both "# 8 and "# + are triggered

simultaneously, the output voltage vO 6 V and

output current flows through the load.• When "# 8 and "# + are turned ')), the load

current continues to flow in the same direction

through load, %8 and %+ , due to the energystored in the inductor 6.

• 'utput voltage vO 6 ; V .

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Rao

1)31#3


• Average load voltage is positive if chopper'( time is more than the ')) time

• Average output voltage becomes negative if

t ON > t OFF .• ence the direction of load current is always

positive but load voltage can be positive or

negative.


i g

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Rao

1)1#4





A v e r a g e v $

i g .

i $

v $

#

t

t

t

t

C , C

' ( . 7 , 7 . C o n d u c t i n g


i g

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Rao

1)$1#5





A v e r a g e v $

i g .

i $

v $

#

t

t

t

t

C

C

.

7 , 7 .

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Rao

1)%1#6


Class E Chopper

#

v $

i $6 1%

C . C >7 . 7 >

7 7 9C C 9

& −

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Rao

1)&1#


,our :uadrant !perationv

$

i $

C - C ' (

C - 7 C o n d u c t s >

> .

7 7. 9- C o n d u c t s

C - 7 C o n d u c t s> .

C - C ' (C - 7 C o n d u c t s

9 .

. >

C - 7 C o n d u c t s

7 - 7 C o n d u c t s. >

>

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Rao

1)'1#!


• Class 1 is a four uadrant chopper • When "# 8 and "# . are triggered, output

current iO flows in positive direction through

"# 8 and "# ., and with output voltage vO 6 V .• This gives the first uadrant operation.

• When both "# 8 and "# . are ')), the energy

stored in the inductor 6 drives iO through %+ and %2 in the same direction, but output

voltage vO 6 ;V .

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Rao

1)(1#"


• Therefore the chopper operates in the

fourth uadrant.• When "# + and "# 2 are triggered, the load

current iO flows in opposite direction 4 output

voltage vO 6 ;V .• !ince both iO and vO are negative, the chopper

operates in third uadrant.

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Rao

11)11#


• When both "# + and "# 2 are ')), the load

current iO continues to flow in the same

direction %8 and %. and the output voltage

vO 6 V .

• Therefore the chopper operates in second

uadrant as vO is positive but iO is negative.

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Rao

111111


Effect !f "ource 5

oad Inductance

• The source inductance should be as small as

possible to limit the transient voltage.• Also source inductance may cause

commutation problem for the chopper.

• Isually an input filter is used to overcome the problem of source inductance.

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Rao

112112


• The load ripple current is inversely

proportional to load inductance and chopping

freuency.• +eak load current depends on load inductance.

• To limit the load ripple current, a smoothing

inductor is connected in series with the load.

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Rao

113113


Proble

For the first #uadrant chopper shown in figuree,press the following variales as functions of /

and duty cycle 1d2 in case load is resistive.

7 Average output voltage and current 7 Output current at the instant of co''utation

7 Average and +3 free wheeling diode current.

7 +3 value of output voltage 7 +3 and average thyristor currents.

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Rao

11114


#

i $

v $

C h o p p e r

) W 7

&

−

6'A7

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Rao

11$115


Average output voltage,

Average output current,

The thyristor is commutated at the instant

output current at the instant of commutation is

since # is the output v

ON dc

dc

dc

ON

t V V dV T

V dV

I R R

t t

V

R

= = ÷

= ==

∴oltage at that instant.

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Rao

11%116


( )

$$

)ree wheeling diode ;)W7< will never

conduct in a resistive load.

Average 4 %/! free wheeling diode

currents are ero.

0ut during

ON t

O R!S

O ON

V v dt T

v V t

∴

=

=∫

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Rao

11&11


( )

( )

( )

$

Where duty cycle,

ON t

O R!S

ON

O R!S

O R!S

ON

V V dt T

t

V V T

V dV

t d

T

=

= ÷

=

=

∫

%/! value of thyristor current

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Rao

11'11!


( )

%/! value of thyristor current

: %/! value of load current

Average value of thyristor current

: Average value of load current

O R!S V

R

dV R

dV

R

=

=

=

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Rao

11(11"


Ipulse

Coutated Chopper • "mpulse commutated choppers are widely used in

high power circuits where load fluctuation is not

large.• This chopper is also known as

2 +arallel capacitor turn-off chopper

2 #oltage commutated chopper 2 Classical chopper.

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Rao

12)12#


6'A7

6

C

" 6

6 !

# !

&

J

&

J

T .

T

7

a

b

i C

i T

v '

&

J

) W 7

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Rao

121121


• To start the circuit, capacitor ‘"’ is initially

charged with polarity ;with plate 3a5 positive< bytriggering the thyristor T +.

• Capacitor ‘"’ gets charged through V S $ "$ T + and

load.

• As the charging current decays to ero thyristorT + will be turned-off.

• With capacitor charged with plate 3a5 positive the

circuit is ready for operation.• Assume that the load current remains constant

during the commutation process.

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122122


• )or convenience the chopper operation isdivided into five modes.

2 /ode-

2 /ode-

2 /ode-9

2 /ode->

2 /ode-@

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Rao

123123


Mode#1 !peration

6'A7

6

C

" 6

6 !

# !

&

J

&

J

T

7

# C i C

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Rao

12124


• Thyristor T 8 is fired at t 6 0.

• The supply voltage comes across the load.

• 6oad current I L flows through T 8 and load.

• At the same time capacitor discharges through T 8 $ %8 $ L8 $ 4 ‘"’ and the capacitor reverses its voltage.

• This reverse voltage on capacitor is held constant by

diode %8.

C i 7i h C

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Rao

12$125


( )

( )

Capacitor 7ischarge Current

sin

Where

4 Capacitor #oltagecos

"

"

" i t V t

L

L"

V t V t

ω

ω

ω

=

=

=

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12%126


Mode#2 !peration

6'A7

C

6 !

# !&

J&

J

T .

# C

" 6

" 6

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12&12


• Thyristor T + is now fired to commutate thyristor T 8.

• When T + is '( capacitor voltage reverse biases T 8

and turns if off.

• The capacitor discharges through the load from ?V

to 0.

• 7ischarge time is known as circuit turn-off time.

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Rao

12'12!


C

Circuit turn-off time is given by

Where is load current.t depends on load current, it must be designed

for the worst case condition which occur at the

ma8imum value of load current and mini

" "

L

L

V " t

I

I

×=

mum

value of capacitor voltage.

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12(12"


• Capacitor recharges back to the supply voltage

;with plate 3a5 positive

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13)13#


• At the end of /ode- capacitor has recharged

to V S and the free wheeling diode starts

conducting.

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131131


Mode#3 !peration

6'A

7

C

6!

# !

& J

&

J

T .# !

) W 7

" 6

" 6

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132132


• F7% starts conducting and the load current

decays.

• The energy stored in source inductance LS istransferred to capacitor.

• ence capacitor charges to a voltage higher

than supply voltage, T naturally turns off.

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133133


( )

The instantaneous capacitor voltage is

sin

Where

S " S L S

S

S

LV t V I t

"

L "

ω

ω

= +

=

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13134


Mode# !peration

6'A7

C

6 !

# !

& J

&

J

7

6 ) W 7

" 6# C

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Rao

13$135


• Capacitor has been overcharged i.e. its voltage

is above supply voltage.

• Capacitor starts discharging in reversedirection.

• ence capacitor current becomes negative.• The capacitor discharges through LS $ V S $ F7%$

%8 and L.

• When this current reduces to ero %8 will stopconducting and the capacitor voltage will besame as the supply voltage

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13%136


Mode#$ !peration

• 0oth thyristors are offand the load current

flows through the )W7.

• This mode will end oncethyristor T 8 is fired.

6'A7

" 6

) W 7

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Rao

13&13


C a p a c i t o r C u r r e n t" 6

t

t

" p C u r r e n t t h r o u g h T

i c

$

I pi T

$

" 6

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Disad4antaes

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13(13"


Disad4antaes

• A starting circuit is reuired and the starting circuitshould be such that it triggers thyristor T + first.

• 6oad voltage Kumps to almost twice the supply

voltage when the commutation is initiated.• The discharging and charging time of commutation

capacitor are dependent on the load current and

this limits high freuency operation, especially at

low load current.

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• Chopper cannot be tested without connecting

load.

• Thyristor T 8 has to carry load current as wellas resonant current resulting in increasing its

peak current rating.

Documents

2433774.ppt