DC Choppers VIT

8/8/2019 DC Choppers VIT

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Power Electronics 22

Introduc

tion

Chopper is a static device.

A variable dc voltage is obtained from aconstant dc voltage source.

Also known as dc-to-dc converter. Widely used for motor control.

Also used in regenerative braking.

Thyristor converter offers greaterefficiency, faster response, lowermaintenance, smaller size and smoothcontrol.


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Choppers are of Two Types

y Step-down choppers.

y Step-up choppers.

y In step down chopper outputvoltage is less than input voltage.

y In step up chopper output voltage

is more than input voltage.


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A step-down chopper with resistiveload.

The thyristor in the circuit acts as aswitch.

When thyristor is ON, supplyvoltage appears across the load

When thyristor is OFF, the voltageacross the load will be zero.


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Vdc

v0

V

V/R

i0

Idc

t

t

tON

T

tOFF


7/89


verage value of output or load voltage.

verage value of output or load current.

Time interval for which SCR conducts.

Time interval for which SCRis OFF.

Period ofswitching

dc

dc

ON

OFF

ON OFF

V A

I A

t

t

T t t

!

!

!

!

! !or chopping period.

1Freq. of chopperswitching or chopping freq.f

T! !


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Average Output Voltage

.

duty cycle

ONdc

ON OFF

ONdc

ON

tV V

t t

tV V V d

T

tbut d t

!

! !

! !


9/89


2

0

Average Output Current

RMS value of output voltage

1 ON

dcdc

ONdc

t

O o

VI

R

tV VI d R T R

V dtT

!

! !

!


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2

0

2

But during ,

Therefore RMS output voltage

1

.

.

ON

ON o

t

O

ON

O ON

O

t V

V V dt T

tV

V t VT T

V d V

!

!

! !

!


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2

2

Outputpo er

But

Outputpo er

O O O

OO

OO

O

P V I

VI

VP

RdV

PR

!

!

@

!

!


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Effective input resistance of chopper

The output voltage can be varied by

varying the duty cycle.

i

dc

i

VR

I

RR

d

!

!


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Methods Of Control

The output dc voltage can be variedby the following methods.

Pulse width modulation control orconstant frequency operation.

Variable frequency control.


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Pulse Width Modulation

tON is varied keeping choppingfrequency f& chopping period T

constant. Output voltage is varied by varying

the ON time tON


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V0

V

V

V0

t

ttON

tON tOFF

tOFF

T


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Variable Frequency Control

Chopping frequency fis variedkeeping either tON or tOFF constant.

To obtain full output voltage range,frequency has to be varied over awide range.

This method produces harmonics in

the output and for large tOFF loadcurrent may become discontinuous


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v0

v0

t

t

tON

tON

T

T

tOFF

tOFF


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Step-down Chopper

With R-L Load

V

i

V0

Chopp r

LFWD

E


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When c

hopper is ON, supply is connectedacross load.

Current flows from supply to load.

When chopper is OFF, load current

continues to flow in the same directionthrough FWD due to energy stored ininductor L.


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Load current can be continuous ordiscontinuous depending on the values of

Land duty cycle d For a continuous current operation, load

current varies between two limits Imaxand I

min When current becomes equal to Imax the

chopper is turned-off and it is turned-onwhen current reduces to Imin.


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Outputvoltage

Outputcurrent

v0

V

i0

Imax

Imin

t

t

tON

T

tOFF

Continuouscurrent

Outputcurrent

t

Discontinuouscurrent

i0


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Expressions ForLoad Current

iO For Continuous Current Operation

WhenChopperIs ON (0e te tON)


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V

i0

V0

R

L

E

+

-


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min

min

Taking Laplace Transform

. 0

t 0, initial current 0

O

O

O O O

O

O

di

Vi

R L Edt

V E

RI S L S I S iS S

t i I

IV EI SRR

SLS SLL

!

! -

! !

!


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min

Taking Inverse aplace Transform

1

This expression is valid for0 ,

i.e., during the period chopperis ON.

Atthe instantthe chopperisturned off,

load c

R Rt t

L L

O

O N

V Ei t e I e

R

t t

! - e e

maxurrentis O O Ni t I!


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When Chopperis OFF

i0

R

L

E


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max

When Chopperis OFF 0

0

Talking aplace 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

E RI S L SI S i

S

t

i I

e e

!

! -

!

!


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max

max

Taking Inverse aplace Transform

1

O

R Rt tL L

O

I EI S

R RS LS S

L L

Ei t I e eR

@ !

! -


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min

The expression is vali for0 ,

i.e., during the period chopperisOFF

tthe instantthe chopperisturned ON or at

the end ofthe offperiod, the load currentis

OFF

O OFF

t t

i t I

e e

!


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min

max

max

max min

min

From equation

1

At ,

To Find &

1

R Rt t

L L

O

ON O

dRT dRT

L L

V Ei t e I eR

t t dT i t I

V E I e I e

I I

R

! -

! ! !

@ !

-


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max

min

From equation

1

At ,1

R R

t tL LO

OFF ON O

OFF

Ei t I e eR

t t T t i t It t d T

! -

! ! !! !


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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

dRTL

RT

L

E

I I e eR

I

V E I e I eR

V e EI

R Re

@ ! -

!

-

!

-


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max

1 1

min max

min

max min

ubstituting for in equation

1

e get,

1

1

is kno n asthe steadystate ripple.

d RT d RT

L L

dRT

L

RTL

I

E I I e e

R

V e EI

R Re

I I

! -

!

-


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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

( !

!

!


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min max

min

max minmin

Assuming load current varies linearlyfrom to instantaneous

load currentis given by

. 0O ON

O

I I

I ti I for t t dT

dT

I Ii I t

dT

(! e e

!


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2

0

0

2

max min

min

0

2

min max min2 2max minmin

0

RMS value ofload current

1

1

21

dT

O RMS

dT

O RMS

dT

O RMS

I i dtdT

I I t I I dt

dT dT

I I I t I I I I t dtdT dT dT

!

!

-

! -


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12 2

max min2

min min max min

2

0

0

2

max minmin

0

RMS value of output current

3

RMS chopper current

1

1

O RMS

dT

CH

dT

CH

I I I I I I I

I i dtT

I I I I t dt

T dT

!

-

!

!

-


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12 2

max min2

min min max min3

Effective input resistance is

CH

CH O RMS

i

S

I I I d I I I I

I d I

VR

I

!

-

!

!


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WhereAverage source currentS

S dc

i

dc

I

I dI

VR

dI

!

!

@ !


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Principle Of Step-up Chopper

+

VOV

Chopper

CLOA

D

DLI

+


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

The values ofL and Care chosendepending upon the requirement of

output voltage and current. When the chopper is ON, the inductor L is

connected across the supply.

The inductor current Irises and the

inductor stores energy during the ONtime of the chopper, tON.


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When the chopper is off, the inductorcurrent I is forced to flow through thediode D and load for a period, tOFF.

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

Therefore voltage across load is given by

. .,O OdI

V V L i e V Vdt

! "


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A large capacitor C connectedacross 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.


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44

Expression For OutputVoltageAssume the average inductor currentto be

during ON and OFF time of Chopper.

Voltage acrossinductor

Therefore energystored in inductor

= . .

Where

When Chopper

period of chopper.

is ON

ON

ON

I

L V

V I t

t ON

!

!


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45

(energyissupplied byinductorto load)

Voltage across

Energysupplied byinductor

where period of Chopper.

Neg

When Chopper

lecting losses, energystored in inductor

is OFF

O

O OFF

OFF

L V V

L V V It

t OFF

L

!

!

!

= energysupplied byinductorL


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46

? A

WhereT = Chopping period orperiod

ofswitching.

ON O OFF

ON OFF

O

OFF

O

ON

VIt V V It

V t tV

t

TV V

T t

@ !

!

!


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47

1

1

1

1

Where duty cyle

ON OFF

OON

O

ON

T t t

V Vt

T

V Vd

td

T

!

!

@ !

! !


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48

For variation ofduty cycle ' ' in the

range of 0 1 the output voltage

ill varyin the range

O

O

d

d V

V V

g


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49

Performance Parameters The thyristor requires a certain minimum time to turn

ONand turnOFF.

Duty cycle dcan be varied only between a min. &

max. value, limiting the min. and max. value ofthe

output voltage.

Ripple inthe load current depends inversely onthe

chopping frequency, f.

To reduce the load ripple current, frequency shouldbe as high as possible.


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50

Problem

AChopper circuitisoperatingonTRCata frequencyof 2 kHzona

460V

supply.If theloadvoltageis350 volts, calculate the conductionperiodof the thyristorineach cycle.


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51

3

460V, 350V, f 2 kHz

1Chopping period

10.5 sec

2 10

Output voltage

dc

ONdc

V V

T

f

T m

tV V

T

!

!

! !v

!


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52

3

Conductionperiod ofthyristor

0.5 10 350

4600.38msec

dcON

ON

ON

T Vt

V

t

t

v!

v v!

!


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53

Problem

Input to thestep up chopperis 200 V.Theoutputrequiredis 600 V.If the

conducting timeof thyristoris 200 Qsec.Compute

Chopping frequency,

If thepulse widthishalved for

constant frequencyofoperation, findthenewoutput voltage.


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54

6

200 , 200 , 600

600 200200 10

Solving for

300

ON dc

dc

ON

V V t s V V

TV V

T tT

T

T

T s

Q

Q

! ! !

!

! v

!


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55

6

6

Chopping frequency

1

1 3.33300 10

Pulse i th is halved

200 10100

2ON

fT

f KHz

t sQ

!

! !v

v@ ! !


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56

6

6

Frequencyis constant

3.33

1300

Output voltage

300 10200 300Volts

300 100 10

ON

f KHz

T sf

TV

T t

Q

@ !

! !

@

v! !


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57

Problem

Adc chopperhasaresistiveloadof 20;

andinput voltageVS = 220V. Whenchopperis ON, its voltagedropis 1.5voltsand chopping frequencyis 10 kHz.If theduty cycleis 80%, determine the

averageoutput voltageand the chopperon time.


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58

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 kHzt

dT

V

tV V VT

V

! ! ; !

! !

!

! !


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59

3

3

3

3

ChopperONtime,

1Chopping period,

10.1 10 secs 100 secs10 10

ChopperONtime,

0.80 0.1 10

0.08 10 80 secs

ON

ON

ON

ON

t dT

Tf

T

t dT

t

t

!

!

! ! v !v

!

! v v

! v !


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60

Problem

Inadc chopper, theaverageloadcurrentis 30 Amps, choppingfrequencyis 250 Hz, supply voltageis 110 volts.Calculate the ONandOFFperiodsof the chopperif theloadresistanceis 2 ohms.


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61

3

30 , 250 , 110 , 21 1

Chopping period, 4 10 4 msecs250

30 20.545

110

dc

dcdc dc

dc

dc

I Ampsf Hz V V R

Tf

V I V dV R

dVI

RI R

dV

! ! ! ! ;

! ! ! v !

! !

@ !

v! ! !


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62

3

3 3

3

Chopper ON period,

0.545 4 10 2.18msecsChopper OFF period,

4 10 2.18 10

1.82 10 1.82msec

ON

OFF ON

OFF

OFF

t dT

t T t

t

t

! ! v v !

!

! v v

! v !


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63

Adc chopperin figure hasaresistiveloadof R = 10; andinput voltageof V

= 200 V. When chopperis ON, its voltagedropis 2 Vand the chopping frequencyis

1 kHz.If theduty cycleis 60%,determine

Averageoutput voltage

RMS valueofoutput voltage Effectiveinputresistanceof chopper

Chopperefficiency.


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64

V

iChopper

+

v0

200 , 10 , 2

0.60, 1 .

chV V R Chopper voltage drop V V

d f kHz

! ! ; !

! !


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65

? A


0.60 200 2 118.8VoltsRMS value ofoutput voltage

0.6 200 2 153.37 Volts

dc ch

dc

O ch

O

V d V V

V

V d V V V

!

! !

!

! !


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66

22

0

0 0

Effective input resistance of chopperis

118.811.88Amps

10200

16.8311.88

Outputpoweris

1 1

i

S dc

dcdc

i

S dc

dT dT

ch

O

V VRI I

VI

RV V

RI I

V V P dt dt

T R T R

! !

! ! !

! ! ! ! ;

! !


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67

? A

2

2

0

0

0.6 200 22352.24 watts

10

Inputpower,

1

1

ch

O

O

dT

i O

dT

ch

O

d V V

P R

P

P Vi dtT

V V V P dt

T R

!

! !

!

!


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68

? A0.6 200 200 22376 watts

10

Chopper efficiency,

100

2352.24100 99%

2376

ch

O

O

O

i

dV V V P

R

P

P

P

L

L

!

v ! !

! v

! v !


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69

Problem A chopper is supplying an inductive loadwith a

free-wheelingdiode. The load inductance is 5 H

and resistance is 10;.. The inputvoltage to the

chopper is 200 volts andthe chopper isoperating ata frequency of1000 Hz. Ifthe

ON/OFFtime ratio is 2:3. Calculate

Maximum andminimum values of loadcurrent in one cycle ofchopper operation.

Average loadcurrent


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70

5 , 10 , 1000 ,

200 , : 2 : 3

Chopping period,

1 11 msecs1000

2

32

3

ON OFF

ON

OFF

ON OFF

L H R f Hz

V V t t

T f

t

t

t t

! ! ; !

! !

! ! !

!

!


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71

3

2

3

53

3

53

1 10 0.6 msec5

ON OFF

OFF OFF

OFF

OFF

T t t

T t t

T t

t T

T

!

!

!

!

! v v !


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72

3

3

3

max

1 0.6 10 0.4 msec

Duty cycle,

0.4 100.41 10

Maximum value of load currentis given by

1

1

ON OFF

ON

ON

dRT

L

RT

L

t T t

t

td T

V e EI

R Re

!

! v !

v! ! !v

!

-


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73

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

200 110

1

dRT

L

RT

L

V e

I Re

eI

e

v v v

v v

@ ! -

! -


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74

3

3

0.8 10

max 2 10

max

min

120

1

8.0047A

Minimum value ofload current ith E 0

is given by

1

1

dRT

L

RT

L

eI

e

I

V eIR

e

v

v

! -

!

!

-


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75

3

3

0.4 10 1 10

5

min 10 1 10

5

max min

200 17.995A10

1

Average load current

2

8.0047 7.995 8A2

dc

dc

eI

e

I II

I

v v v

v v

! ! -

!

! }


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76

Problem A chopper feedingon RLloadisshownin

figure, withV= 200 V, R = 5;, L= 5mH, f= 1 kHz, d= 0.5and E= 0 V.Calculate

Maximumandminimum valuesofloadcurrent.

Average valueofloadcurrent.

RMSloadcurrent. Effectiveinputresistanceasseen bysource.

RMS choppercurrent.


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77

3

3

V 200V, R 5 , L 5mH,

f 1kHz, d 0.5, E 0Chopping period is

1 11 10 secs

1 10

T

f

;

! ! ! v

v i0

v0

C er

R

LFW

E

+


78/89


78

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

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 currentis given by

1

1

200 10

51

140 24.9A

1

dRTL

RT

L

V e EI

R Re

eI

e

eI

e

v v v

v

v v

v

!

-

! -

! ! -


79/89


79


3

3

3

3

min

0.5 5 1 10

5 10

min 5 1 10

5 10

0.5

min 1

Minimum value ofload currentis 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

v v vv

v v

v

!

-

! -

! !

-


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80


1 2

12 2

max min2

min min max min

Average value ofload currentis

2

for linear variation ofcurrents

24.9 15.1 20 A2

RMS load currentis given by

3

dc

dc

O RMS

I II

I

I I I I I I I

!

@ ! !

!

-


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81


12 2

2

1

2

24.9 15.115.1 15.1 24.9 15.1

3

96.04228.01 147.98 20.2A

3

RMS chopper currentis given by

0.5 20.2 14.28A

O RMS

O RMS

ch O RMS

I

I

I d I

!

-

! ! -

! ! v !


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82

Effective input resistance is

= Average source current

0.5 20 10 A

Therefore effective input resistance is

20020

10

i

S

S

S dc

S

i

S

VRI

I

I dI

I

VR

I

!

!

! v !

! ! ! ;


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83

Classification Of Choppers

Choppers are classified as

Class A Chopper

Class B Chopper

Class CChopper

Class D Chopper

Class E Chopper


84/89


84

Class A Chopper

V

Chopper

FWD

+

v0

v0

i0

i0

LOAD

V


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85

Class B Chopper

V

Chopper

+

v0

v0

i0

i0

L

E

R

D


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86

Class C Chopper

V

e

+

v0

D1

D2

CH2

CH1

v0i0

i0

L

E


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87

Class D Chopper

V+ v0

D2

D1 C 2

C 1

v0

i0

L ER i0


88/89


88

Class E Chopper

V

v

0

i0L ER

CH2 H2 4

1H1 H3

+


89/89

Four Quadrant Operationv0

i0

CH CH ON

CH Conducts1 4

4 2

2 Conducts

CH Conducts4 2

CH CH ON

CH Conducts3 2

2 4

CH Conducts

Conducts2 4

1 4

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