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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
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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.
04/19/23 3
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Principle Of Step-down Chopper
V
i0
V 0
C hopper
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.
04/19/23 5
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V d c
v 0
V
V /R
i0
Id c
t
t
tO N
T
tO F F
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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
04/19/23 7
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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
04/19/23 8
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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
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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
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2
2
Output power
But
Output power
O O O
OO
OO
O
P V I
VI
R
VP
R
dVP
R
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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
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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.
04/19/23 13
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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
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V 0
V
V
V 0
t
ttO N
tO N tO F F
tO F F
T
04/19/23 15
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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
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v 0
V
V
v 0
t
t
tO N
tO N
T
T
tO F F
tO F F
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Step-down ChopperWith R-L Load
V
i0
V 0
C hopper
R
LFW D
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 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
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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)
04/19/23 Copyright by www.noteshit.com 22
V
i0
V 0
R
L
E
+
-
04/19/23 Copyright by www.noteshit.com 23
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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
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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 Rt t
L LO
ON
V Ei t e I e
R
t t
maxurrent is O ONi t I04/19/23 25
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When Chopper is OFF
i0
R
L
E
04/19/23 26
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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
OO
O O O
O
t t
diRi L E
dt
ERI S L SI S i
St
i I
04/19/23 27
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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
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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
04/19/23 29
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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
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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
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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
EI I e e
R
I
V EI e I e
R
V e EI
R Re
04/19/23 32
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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
EI I e e
R
V e EI
R Re
I I
04/19/23 33
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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
04/19/23 34
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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
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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
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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
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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
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Where
Average source currentS
S dc
idc
I
I dI
VR
dI
04/19/23 39
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Principle Of Step-up Chopper
+
V OV
C hopper
CLOAD
DLI
+
04/19/23 40
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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, tON.
04/19/23 41
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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, 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
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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.
04/19/23 43
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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
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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
04/19/23 45
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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
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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
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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
04/19/23 48
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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.
04/19/23 49
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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.
04/19/23 50
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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
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3
Conduction period of thyristor
0.5 10 350
4600.38 msec
dcON
ON
ON
T Vt
V
t
t
04/19/23 52
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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.
04/19/23 53
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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
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6
6
Chopping frequency
1
13.33
300 10Pulse width is halved
200 10100
2ON
fT
f KHz
t s
04/19/23 55
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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
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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
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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
ONdc S ch
dc
V V R f kHz
td
TV
tV V V
T
V
04/19/23 58
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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
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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.
04/19/23 60
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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
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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
04/19/23 62
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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. 04/19/23 63
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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
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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
04/19/23 65
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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
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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
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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
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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 current04/19/23 69
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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
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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
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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
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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
101
dRT
L
RT
L
V eI
Re
eI
e
04/19/23 73
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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
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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
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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.
04/19/23 76
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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
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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 10
51
140 24.9 A
1
dRT
L
RT
L
V e EI
R Re
eI
e
eI
e
04/19/23 78
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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 10
51
140 15.1 A
1
dRT
L
RT
L
V e EI
R Re
eI
e
eI
e
04/19/23 79
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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
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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
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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
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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
04/19/23 83
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Class A Chopper
V
C hopper
FW D
+
v 0
v 0
i0
i0
LOAD
V
04/19/23 84
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• 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
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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.
04/19/23 86
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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
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Class B Chopper
V
C hopper
+
v 0
v 0
i0
i0
L
E
R
D
04/19/23 88
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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.
04/19/23 89
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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.
04/19/23 90
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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
04/19/23 Copyright by www.noteshit.com 92
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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
OO
O
LdiV Ri E
dt
i t I t
tep-down chopper
with R-L load04/19/23 93
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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 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
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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
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
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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
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• 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.
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• 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
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• 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
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G ate pu lseof C H 2
G ate pu lseof C H 1
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
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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
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• 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 .
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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 tON < tOFF .
• Hence the direction of load current is always positive but load voltage can be positive or negative.
04/19/23 103
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G ate pu lseof C H 2
G ate pu lseof C H 1
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
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G ate pu lseof C H 2
G ate pu lseof C H 1
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
04/19/23 105
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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
+
04/19/23 106
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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
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• 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.
04/19/23 109
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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.
04/19/23 110
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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.
04/19/23 111
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V
i0
v 0
C hopper
FW D
+
LOAD
04/19/23 112
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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
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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
04/19/23 114
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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
04/19/23 115
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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
04/19/23 116
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Boost Converter or
Step Up converter
Buck-Boost Converter
Buck Converter or
Step Down Converter
Simple DC-DC Converter Topologies
04/19/23 117
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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