Principle Of Step-up Chopper - Dronacharyagn.dronacharya.info/.../unit-3/ied_Unit3_Lecture2.pdf•In...

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

Principle Of Step-up Chopper

+

VOV

Chopper

CLOAD

DLI

+


• 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.


• 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


• 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.


Expression For Output Voltage

Assume 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


(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


Where

T = Chopping period or period

of switching.

ON O OFF

ON OFF

O

OFF

O

ON

VIt V V It

V t tV

t

TV V

T t


1

1

1

1

Where duty cyle

ON OFF

OON

O

ON

T t t

V Vt

T

V Vd

td

T


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


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.


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.


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


3

Conduction period of thyristor

0.5 10 350

460

0.38 msec

dcON

ON

ON

T Vt

V

t

t


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.


6

200 , 200 , 600

600 200200 10

Solving for

300

ON dc

dc

ON

V V t s V V

TV V

T t

T

T

T

T s


6

6

Chopping frequency

1

13.33

300 10

Pulse width is halved

200 10100

2ON

fT

f KHz

t s


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


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.


220 , 20 , 10

0.80

= Voltage drop across chopper = 1.5 volts

Average output voltage

0.80 220 1.5 174.8 Volts

S

ON

ch

ONdc S ch

dc

V V R f kHz

td

T

V

tV V V

T

V


3

3

3

3

Chopper ON time,

1Chopping period,

10.1 10 secs 100 μsecs

10 10

Chopper ON time,

0.80 0.1 10

0.08 10 80 μsecs

ON

ON

ON

ON

t dT

Tf

T

t dT

t

t


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.


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

R

dVI

R

I Rd

V


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


• 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.


V

i0

Chopper

+

R v0

200 , 10 , 2

0.60, 1 .

chV V R Chopper voltage drop V V

d f kHz


Average output voltage

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


22

0

0 0

Effective input resistance of chopper is

118.811.88 Amps

10

20016.83

11.88

Output power is

1 1

i

S dc

dcdc

i

S dc

dT dT

ch

O

V VR

I I

VI

R

V VR

I I

V VvP dt dt

T R T R


2

2

0

0

0.6 200 22352.24 watts

10

Input power,

1

1

ch

O

O

dT

i O

dT

ch

O

d V VP

R

P

P Vi dtT

V V VP dt

T R


0.6 200 200 22376 watts

10

Chopper efficiency,

100

2352.24100 99%

2376

ch

O

O

O

i

dV V VP

R

P

P

P


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 current


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


3

2

3

5

3

3

5

31 10 0.6 msec

5

ON OFF

OFF OFF

OFF

OFF

T t t

T t t

T t

t T

T


3

3

3

max

1 0.6 10 0.4 msec

Duty cycle,

0.4 100.4

1 10

Maximum 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


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


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


3

3

0.4 10 1 10

5

min 10 1 10

5

max min

200 17.995 A

101

Average load current

2

8.0047 7.9958 A

2

dc

dc

eI

e

I II

I


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.


3

3

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

v0

Chopper

R

LFWD

E

+


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


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


1 2

12 2

max min2

min min max min

Average value of load current is

2

for linear variation of currents

24.9 15.120 A

2

RMS load current is given by

3

dc

dc

O RMS

I II

I

I II I I I I


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


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

VR

I

I

I dI

I

VR

I

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Principle Of Step-up Chopper - Dronacharyagn.dronacharya.info/.../unit-3/ied_Unit3_Lecture2.pdf•In...

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