UNIT III

DC Choppers
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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.
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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.
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Principle Of
Step-down Chopper
V
i
0
V
0
Chopper
R
+

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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.
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V
dc
v
0
V
V/R
i
0
I
dc
t
t
t
ON
T
t
OFF
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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
= =
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Average Output Voltage
.
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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2
0
Average Output Current
RMS value of output voltage
1
ON
dc
dc
ON
dc
t
O o
V
I
R
t V V
I d
R T R
V v dt
T
=
| |
= =
|
\
=

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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 V dt
T
t V
V t V
T T
V d V
=
=
= =
=

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2
2
Output power
But
Output power
O O O
O
O
O
O
O
P V I
V
I
R
V
P
R
dV
P
R
=
=

=
=
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Effective 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 Of Control
The output dc voltage can be varied by the
following methods.
Pulse width modulation control or
constant frequency operation.
Variable frequency control.
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Pulse Width Modulation
t
ON
is varied keeping chopping frequency
f & chopping period T constant.
Output voltage is varied by varying the ON
time t
ON

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V
0
V
V
V
0
t
t
t
ON
t
ON
t
OFF
t
OFF
T
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Variable Frequency Control
Chopping frequency f is varied keeping
either t
ON
or t
OFF
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 t
OFF
load current may
become discontinuous
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v
0
V
V
v
0
t
t
t
ON
t
ON
T
T
t
OFF
t
OFF
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Step-down Chopper
With R-L Load
V
i
0
V
0
Chopper
R
L
FWD
E
+

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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.
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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 I
max
and
I
min
When current becomes equal to I
max
the
chopper is turned-off and it is turned-on
when current reduces to I
min.

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Output
voltage
Output
current
v
0
V
i
0
I
max
I
min
t
t
t
ON
T
t
OFF
Continuous
current
Output
current
t
Discontinuous
current
i
0
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Expressions For
Load Current
i
O
For Continuous Current
Operation When
Chopper Is ON (0 t t
ON
)
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V
i
0
V
0
R
L
E
+
-
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( ) ( )
( )
( )
( )
min
min
Taking Laplace Transform
. 0
At 0, initial current 0
O
O
O O O
O
O
di
V i R L E
dt
V E
RI S L S I S i
S S
t i I
I V E
I S
R
R
S
LS S
L
L

= + +
(
= + +

= =

= +
| |
+
+
|
\
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( )
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 R
t t
L L
O
ON
V E
i t e I e
R
t t
| | | |

| |
\ \
(

= +
(
(


( )
max
urrent is
O ON
i t I =
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When Chopper is OFF
i
0
R
L
E
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( )
( ) ( )
( )
( )
max
When Chopper is OFF 0
0
Talking Laplace transform
0 0
Redefining time origin we have at 0,
initial current 0
OFF
O
O
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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( )
( )
max
max
Taking Inverse Laplace Transform
1
O
R R
t 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 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

=
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( )
( )
min
max
max
max min
min
From equation
1
At ,
To Find &
1
R R
t 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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( )
( )
( )
max
min
From equation
1
At ,
1
R R
t t
L L
O
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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( ) ( )
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
E
I I e e
R
I
V E
I e I e
R
V e E
I
R R
e

(
=
(
(

(

= +
(

(

(
=
(

(

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( ) ( )
( )
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
E
I I e e
R
V e E
I
R R
e
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 I
I
=
=
+
=
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( )
( )
min max
min
max min
min
Assuming load current varies linearly
from to instantaneous
load current is given by
.
0
O ON
O
I I
I t
i I for t t dT
dT
I I
i I t
dT

= +

| |
= +
|
\
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( )
( )
( )
( )
( )
2
0
0
2
max min
min
0
2
min max min 2 2
max min
min
0
RMS value of load current
1
1
2
1
dT
O RMS
dT
O RMS
dT
O RMS
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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( )
( )
( )
1
2
2
max min 2
min min max min
2
0
0
2
max min
min
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 dt
T
I I
I I t dt
T dT
(

= + +
(
(

=
(
| |
= +
|
(
\

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( )
( )
( )
1
2
2
max min 2
min min max min
3
Effective input resistance is
CH
CH
O RMS
i
S
I I
I d I I I I
I d I
V
R
I
(

= + +
(
(

=
=
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Where
Average source current
S
S dc
i
dc
I
I dI
V
R
dI
=
=
=
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Principle Of Step-up Chopper
+

V
O
V
Chopper
C
L
O
A
D
D
L I
+

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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, t
ON
.
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When the chopper is off, the inductor
current I is forced to flow through the diode
D and load for a period, t
OFF
.
The current tends to decrease resulting in
reversing the polarity of induced EMF in L.
Therefore voltage across load is given by
. .,
O O
dI
V V L i e V V
dt
= + >
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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.
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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
=
=
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( )
(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
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( )
[ ]
Where
T = 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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1
1
1
1
Where duty cyle
ON OFF
O
ON
O
ON
T t t
V V
t
T
V V
d
t
d
T
= +
| |
|
=
|
|

\
| |
=
|

\
= =
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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
< <
< <
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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.
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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.
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3
460 V, = 350 V, f = 2 kHz
1
Chopping period
1
0.5 sec
2 10
Output voltage
dc
ON
dc
V V
T
f
T m
t
V V
T

=
=
= =

| |
=
|
\
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3
Conduction period of thyristor
0.5 10 350
460
0.38 msec
dc
ON
ON
ON
T V
t
V
t
t

=

=
=
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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.
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6
200 , 200 , 600
600 200
200 10
Solving for
300
ON dc
dc
ON
V V t s V V
T
V V
T t
T
T
T
T s

= = =
| |
=
|

\
| |
=
|

\
=
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6
6
Chopping frequency
1
1
3.33
300 10
Pulse width is halved
200 10
100
2
ON
f
T
f KHz
t s

=
= =

= =
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( )
6
6
Frequency is constant
3.33
1
300
Output voltage =
300 10
200 300 Volts
300 100 10
ON
f KHz
T s
f
T
V
T t

=
= =
| |

\
| |

= =
|
|

\
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Problem
A dc chopper has a resistive load of 20
and input voltage V
S
= 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.
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( )
( )
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
ON
dc S ch
dc
V V R f kHz
t
d
T
V
t
V V V
T
V
= = =
= =
| |
=
|
\
= =
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3
3
3
3
Chopper ON time,
1
Chopping period,
1
0.1 10 secs 100 secs
10 10
Chopper ON time,
0.80 0.1 10
0.08 10 80 secs
ON
ON
ON
ON
t dT
T
f
T
t dT
t
t

=
=
= = =

=
=
= =
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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.
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3
30 , 250 , 110 , 2
1 1
Chopping period, 4 10 4 msecs
250
&

30 2
0.545
110
dc
dc
dc dc
dc
dc
I Amps f Hz V V R
T
f
V
I V dV
R
dV
I
R
I R
d
V

= = = =
= = = =
= =
=

= = =
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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

= = =
=
=
= =
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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.
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V
i
0
Chopper
+

R
v
0
200 , 10 , 2
0.60, 1 .
ch
V V R Chopper voltage drop V V
d f kHz
= = =
= =
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( )
[ ]
( )
( )
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
=
= =
=
= =
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( )
2
2
0
0 0
Effective input resistance of chopper is
118.8
11.88 Amps
10
200
16.83
11.88
Output power is
1 1
i
S dc
dc
dc
i
S dc
dT dT
ch
O
V V
R
I I
V
I
R
V V
R
I I
V V
v
P dt dt
T R T R
= =
= = =
= = = =

= =

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( )
[ ]
( )
2
2
0
0
0.6 200 2
2352.24 watts
10
Input power,
1
1
ch
O
O
dT
i O
dT
ch
O
d V V
P
R
P
P Vi dt
T
V V V
P dt
T R

= =
=

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( )
[ ]
0.6 200 200 2
2376 watts
10
Chopper efficiency,
100
2352.24
100 99%
2376
ch
O
O
O
i
dV V V
P
R
P
P
P

=

= =
=
= =
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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
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5 , 10 , 1000 ,
200 , : 2: 3
Chopping period,
1 1
1 msecs
1000
2
3
2
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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3
2
3
5
3
3
5
3
1 10 0.6 msec
5
ON OFF
OFF OFF
OFF
OFF
T t t
T t t
T t
t T
T

= +
= +
=
=
= =
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3
3
3
max
1 0.6 10 0.4 msec
Duty cycle,
0.4 10
0.4
1 10
Maximum value of load current is given by
1
1
ON OFF
ON
ON
dRT
L
RT
L
t T t
t
t
d
T
V e E
I
R R
e

=
= =

= = =

(
=
(

(

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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
10
1
dRT
L
RT
L
V e
I
R
e
e
I
e

(
=
(

(

(

(
=
(
(

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3
3
0.8 10
max
2 10
max
min
1
20
1
8.0047A
Minimum value of load current with E = 0
is given by
1
1
dRT
L
RT
L
e
I
e
I
V e
I
R
e

=
(

(

=
(

(
=
(

(

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3
3
0.4 10 1 10
5
min
10 1 10
5
max min
200 1
7.995 A
10
1
Average load current
2
8.0047 7.995
8 A
2
dc
dc
e
I
e
I I
I
I

(
= =
(
(

+
=
+
=
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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.
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3
3
V = 200 V, R = 5 , L = 5 mH,
f = 1kHz, d = 0.5, E = 0
Chopping period is
1 1
1 10 secs
1 10
T
f

= = =

i
0
v
0
Chopper
R
L
FWD
E
+

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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 1
0
5
1
1
40 24.9 A
1
dRT
L
RT
L
V e E
I
R R
e
e
I
e
e
I
e

(
=
(

(

(

(
=
(
(

= =
(

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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 1
0
5
1
1
40 15.1 A
1
dRT
L
RT
L
V e E
I
R R
e
e
I
e
e
I
e

(
=
(

(

(

(
=
(
(

= =
(

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( )
( )
1 2
1
2
2
max min 2
min min max min
Average value of load current is
2
for linear variation of currents
24.9 15.1
20 A
2
RMS load current is given by
3
dc
dc
O RMS
I I
I
I
I I
I I I I I
+
=
+
= =
(

= + +
(
(

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( )
( )
( )
( )
1
2
2
2
1
2
24.9 15.1
15.1 15.1 24.9 15.1
3
96.04
228.01 147.98 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
(

= + +
(
(

(
= + + =
(

= = =
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Effective input resistance is
= Average source current
0.5 20 10 A
Therefore effective input resistance is
200
20
10
i
S
S
S dc
S
i
S
V
R
I
I
I dI
I
V
R
I
=
=
= =
= = =
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Classification Of Choppers
Choppers are classified as
Class A Chopper
Class B Chopper
Class C Chopper
Class D Chopper
Class E Chopper
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Class A Chopper
V
Chopper
FWD
+

v
0
v
0
i
0
i
0
L
O
A
D
V
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When chopper is ON, supply voltage V is
connected across the load.
When chopper is OFF, v
O
= 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 .
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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.
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Output current
Thyristor
gate pulse
Output voltage
i
g
i
0
v
0
t
t
t
t
ON
T
CH ON
FWD Conducts
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Class B Chopper
V
Chopper
+

v
0
v
0
i
0
i
0
L
E
R
D
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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.
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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.
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Output current
D
conducts
Chopper
conducts
Thyristor
gate pulse
Output voltage
i
g
i
0
v
0
t
t
t
I
min
I
max
T
t
ON
t
OFF
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Expression for Output Current
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( )
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
O
O
O
Ldi
V Ri E
dt
i t I t
= + +
= =
tep-down chopper
with R-L load
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( )
( )
( )
min
max
max min
During the interval chopper is ON voltage
equation is g
1 0
At
1
0
iven by
OFF OFF
R R
t t
L L
O OFF
OFF
O
R R
t t
L L
O
O
V E
i t e I e t t
R
t t i t I
V E
I e I e
R
Ldi
Ri E
dt


| |

= + < <
|
\
= =
| |

= +
|
\
= + +
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( )
( )
( )
max
max
min
min max
Redefining the time origin, at 0
The solution for the stated initial condition is
1 0
At
1
ON ON
O
R R
t t
L L
O ON
ON O
R R
t 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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Class C Chopper
V
Chopper
+

v
0
D
1
D
2
CH
2
CH
1
v
0
i
0
i
0
L
E
R
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Class C Chopper is a combination of Class A
and Class B Choppers.
For first quadrant operation, CH
1
is ON or D
2

conducts.
For second quadrant operation, CH
2
is ON or D
1

conducts.
When CH
1
is ON, the load current is positive.
The output voltage is equal to V & the load
receives power from the source.
When CH
1
is turned OFF, energy stored in
inductance L forces current to flow through the
diode D
2
and the output voltage is zero.
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Current continues to flow in positive direction.
When CH
2
is triggered, the voltage E forces
current to flow in opposite direction through L
and CH
2
.
The output voltage is zero.
On turning OFF CH
2
, the energy stored in the
inductance drives current through diode D
1
and
the supply
Output voltage is V, the input current becomes
negative and power flows from load to source.
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Average output voltage is positive
Average output current can take both
positive and negative values.
Choppers CH
1
& CH
2
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.
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Gate pulse
of CH
2
Gate pulse
of CH
1
Output current
Output voltage
i
g1
i
g2
i
0
V
0
t
t
t
t
D
1
D
1
D
2
D
2
CH
1
CH
2
CH
1
CH
2
ON ON ON ON
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Class D Chopper
V
+
v
0
D
2
D
1
CH
2
CH
1
v
0
i
0
L
E
R
i
0
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Class D is a two quadrant chopper.
When both CH
1
and CH
2
are triggered
simultaneously, the output voltage v
O
= V
and output current flows through the load.
When CH
1
and CH
2
are turned OFF, the
load current continues to flow in the same
direction through load, D
1
and D
2
, due to
the energy stored in the inductor L.
Output voltage v
O
= - V .
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Average load voltage is positive if chopper
ON time is more than the OFF time
Average output voltage becomes negative
if t
ON
< t
OFF
.
Hence the direction of load current is
always positive but load voltage can be
positive or negative.
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Gate pulse
of CH
2
Gate pulse
of CH
1
Output current
Output voltage
Average v
0
i
g1
i
g2
i
0
v
0
V
t
t
t
t
CH ,CH
ON
1 2
D1,D2 Conducting
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Gate pulse
of CH
2
Gate pulse
of CH
1
Output current
Output voltage
Average v
0
i
g1
i
g2
i
0
v
0
V
t
t
t
t
CH
CH
1
2
D , D
1 2
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Class E Chopper
V
v
0
i
0
L
E
R
CH
2
CH
4
D
2
D
4
D
1
D
3
CH
1
CH
3
+
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Four Quadrant Operation
v
0
i
0
CH - CH ON
CH - D Conducts
1 4
4 2
D D
2 3
- Conducts
CH - D Conducts
4 2
CH - CH ON
CH - D Conducts
3 2
2 4
CH - D Conducts
D - D Conducts
2 4
1 4
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Class E is a four quadrant chopper
When CH
1
and CH
4
are triggered, output
current i
O
flows in positive direction
through CH
1
and CH
4
, and with output
voltage v
O
= V.
This gives the first quadrant operation.
When both CH
1
and CH
4
are OFF, the
energy stored in the inductor L drives i
O

through D
2
and D
3
in the same direction,
but output voltage v
O
= -V.
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Therefore the chopper operates in the
fourth quadrant.
When CH
2
and CH
3
are triggered, the load
current i
O
flows in opposite direction &
output voltage v
O
= -V.
Since both i
O
and v
O
are negative, the
chopper operates in third quadrant.
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When both CH
2
and CH
3
are OFF, the
load current i
O
continues to flow in the
same direction D
1
and

D
4
and the output
voltage v
O
= V.
Therefore the chopper operates in second
quadrant as v
O
is positive but i
O
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
i
0
v
0
Chopper
FWD
+

L
O
A
D
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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
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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( )
( )
( )
2
0
2
1
Where duty cycle,
ON
t
O RMS
ON
O RMS
O RMS
ON
V V dt
T
t
V V
T
V dV
t
d
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 RMS
V
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?

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