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Contents
Manual for K-Notes ................................................................................. 2
Power Semi-Conductor Devices .............................................................. 3
Phase Controlled converter .................................................................. 10
Chopper ................................................................................................ 15
Inverters ................................................................................................ 21
AC - AC Converters ................................................................................ 26
© 2014 Kreatryx. All Rights Reserved.
3
Power Semi-Conductor Devices
Properties of ideal switch
1. Conduction state , ON ONV 0, I
2. Blocking state , OFF OFFV 0, V
3. Ideal switch can change its state instantaneously ON OFFT 0 , T 0
4. No power loss while switching.
5. Stable under all operating conditions.
Classification of switches
1. Uncontrolled switch (Passive switch)
Switching state cannot be controlled by any control signal E.g. Diode
2. Semi-controlled switch
Only one switching state can be controlled by an external control signal. E.g. SCR
3. Fully controlled switch
If both switching states can be controlled by switchable control signal. E.g. BJT, MOSFET.
Other Classification
1. Unipolar switch
The switch can block only one polarity of voltage when it is in OFF state.
2. Bipolar switch
This switch can block both polarity of voltage when it is in blocking state.
3. Unidirectional switch
This switch can carry current in only one direction when it is in conduction state.
4. Bidirectional switch
This switch can carry current in both the directions when it is in conduction state.
4
Ideal characteristics of power semiconductor switches
Device Characteristic
Diode
BJT
MOSFET
IGBT
SCR
GTO
TRIAC
5
Power loss in a switch
1) The average power has in a switch is given by
PT
o
1vidt
T
Where v = instantaneous voltage
i = instantaneous current
2) If the device is modeled as a resistance, as in case of a MOSFET
2 2
rms ON rms ONP I R V R
3) If the device is modeled as a voltage source.
avgP V I
Silicon Controlled Rectifier
In forward blocking mode, 1 3J , J are forward biased and 2J is reverse biased.
In forward conduction mode, 2J breakdown, 1 3J , J are forward biased.
In reverse blocking mode, 1 3J , J are reverse biased & 2J is forward biased.
Latching Current
This is the minimum value of anode current above which SCR turns ON. This is related to
minimum gate pulse width requirement for SCR.
Holding current
Minimum value of anode current below which SCR turns OFF.
6
Slope of characteristics =
di
dt
If a rrt t
Area under the curve = RQ
R RM rr
1Q I t
2
RM rrdiI t
dt
2
R rr
1 diQ tdt2
Device & Circuit Turn-off time
Device turn off time, q rr grt t t
rrt = reverse recovery time
grt = gate recovery time
Circuit turn-off time ct is the time period for which communication circuit applies reverse
voltage across SCR after anode current becomes zero.
For successful communication, c qt t
Turn-ON methods of SCR
1) Forward voltage triggering
If AK BOV V , then 2J breakdown & SCR conducts. This can damage the SCR.
2) dV
dt Triggering
c j
dvI C
dt , if
dv
dt is high, charging current increase and SCR conducts when c latchingI I .
3) Light Triggering
If light is incident on 2J , charge carriers are generated and 2J starts conducting.
4) Thermal Triggering
When temperature is increased then charge carriers are generated & SCR conducts.
5) Gate Triggering
By applying gate pulse in SCR, BOV is lowered and SCR can easily conduct.
7
Static V-I characteristics of SCR
Communication of thyristor
Communication is defined as process of turning OFF the thyristor.
Types of Commutations:
1. Natural or line communication
In this case nature of supply supports the commutation.
E.g. Rectifier, AC voltage controllers, Step-down cyclo-converters.
2. Forced Commutation
1) Class A commutation
Circuit should be under-damped.
2 4LR
C for damped oscillations.
Ringing frequency, 2
2r
1 R
LC 4L
Thyristor conducts for a period of =
r
8
2) Class-B commutation or current commutation
a) oTM peakI I
b) s PTA peak
CI V I
L
c) Time required to turn OFF MT after AT ON
1 o
p
ILC LC sin
I
d) Conduction time of AT LC
e) RCM
o
CVt
I = circuit turn off time
Where 1 o
R S
p
IV V cos sin
I
Other Implementation
1 o
CM
p
It 2sin LC
I
Rest all parameters remains same.
3) Class-C commutation or Impulse commutation
S S
T1 peak1 2
V 2VI
R R
S S
T2 peak2 1
V 2VI
R R
C1 1
t R ln2
C2 2
t R ln2
9
Class-D commutation or voltage commutation
o STM peak
CI I V
L
oTA peakI I
MON minT for T LC
sCM
o
CVt
I
Conduction time of sA CM
o
2CVT 2t
I
S
O ON CMavg
VV T 2t
T , T = Switching internal
Thermal Protection of SCR
jc = Thermal resistance b/w J & C
CS = Thermal resistance b/w C & S
SA = Thermal resistance b/w S & A
Unit of 0C / w
In electrical circuit representation
jAT = Temperature difference b/w J & A
10
Phase Controlled converter Form factor
or
o
VFF
V
orV : rms value of output voltage.
oV : Average value of output voltage.
Ripple Factor
RF = 2FF 1
Distortion factor
01
or
VDF
V
01V : rms value of fundamental components of oV
orV : rms value of output voltage.
Total harmonic Distortion
2
1THD 1
DF
Single phase half wave uncontrolled rectifier
R – load RL – Load L – Load
OV mV
mV
1 cos2
0
OI mV
R mV
1 cos2 R
mV
L
ϒ 2
OI max 2
,2
= Extinction angle, Angle at which ω goes to zero.
If a free-wheeling diode is connected across the load (RL) that behaves as R-load as output
voltage goes to zero after t when FD conducts.
11
Single phase half wave controlled rectifier
i) R – load
m
O avg
VV 1 cos
2
m
O avg
VI 1 cos
2 R
2
m
or
V sin2V
4 2
Input power factor =
2
or or
S S S
V R V
V I V
mS
V V
2
α = firing angle
ii) R – L load
m
o avg
VV cos cos
2
m
o avg
VI cos cos
2 R
m
or
V 1V sin2 sin2
22
Circuit turn off time,
c
2t
Single phase full – wave rectifier
12
1
full converter
1
Semi converter
OV
m2V
cos
mV
1 cos
S1I
o
2 2I
O
2 2I cos
2
SI oI
O
I
DF 2 2
2 2cos
2
DPF cos cos2
IPF
2 2cos
21 cos
DPF: Displacement power factor = S S1cos angle b w V & I
S1I = fundamental components of SI
IPF: Input power factor
IPF = DPF x DF
DF: Distortion factor
In case of continuous conductions, outgoing thyristors stop conduction before incoming
thyristor start
Load 1
Full converter
1
Semi – converter
R – load
m
o
VV 1 cos
m
o
VV 1 cos
R – L load
m
o
VV cos cos
m
o
VV 1 cos
RLE – load
mo
VV cos cos E
o m
1V V 1 cos E
13
Three phase half wave controlled rectifier
ml
o
3VV cos
2
mlV : Peak value of line voltage
12
or mp
1 3 3V V cos2
2 8
mpV : Peak value of phase voltage
Three phase full wave rectifier
14
3
Full converter
3
Semi converter
oV
ml3V
cos
ml3V
1 cos2
orV
ml
1 3 3V cos2
2 4
Expression varies for 0 060 & 60
For 060 , it becomes 3-pulse converter.
S1I O
6I
O
6I cos
2
SI O
2I
3
O
I
DF 3
6cos
2
DPF cosα cos2
IPF
3cos
26cos
2x
S1I : Fundamental rms value of source current
SI : rms value of source current
Effect of source inductance
Assuming source inductance equal to SL .
Due to source inductance, there is an overlap b/w incoming and outgoing thyristor, given by
overlap angle .
For 2-pulse converter
Sm
O O
L2VV cos I
m
O
VV cos cos
Displacement power factor =
cos2
15
For 6 – pulse converter
Sm
O O
3 L3VV cos I
m
O
3VV cos cos
2
Displacement power factor =
cos2
Chopper
Buck Converter
When CH is ON o t DT
Voltage across inductor L S OV V V
When CH is OFF (DT < t < T)
Voltage across inductor L OV V
Applying volt-sec balance across inductor
S O OV V DT V T DT 0
S O OV V D V 1 D 0
O SV DV
D = duty cycle = ONT
T
Where T = switching period = 1f
f = switching frequency
16
Average output voltage = SDV
rms output voltage = SDV
Average source current = ODI
Average current of FD = O1 D I
Ripple in output current
When CH is ON 0 t DT
L S O SV V V 1 D V
During this period, since voltage is positive current increase from minimum value to maximum
value.
max mini I I
t DT 0 DT
S
iL 1 D V
DT
SD 1 D V
ifL
This formula gives approximate value of output ripple current for maximum ripple, D = 0.5
S
max
Vi
4fL
Lmax O
II I
2
Lmin O
II I
2
Critical Inductance (LC)
Value of inductance at which inductor voltage waveform is just discontinuous.
c
1 D RL
2f
17
Critical Capacitance (CC)
Value of capacitance at which capacitor voltage waveform is just discontinuous.
C
1C
8fR
Step-up chopper (Boost converter)
L S
when CHis ON 0 t DT , V V
L S O
when CHis OFF DT t T , V V V
Applying volt-sec balance across inductor
S S OV DT V V 1 D T 0
S
O
VV
1 D
Since D < 1, O SV V
C O
when CHis ON 0 t DT , I I
C L O
when CHis OFF DT t T , I I I
O L O
Applying Ampere sec balance across capacitor
I DT I I 1 D T 0
OL
II
1 D
Ripple in inductor current
min max
When CHis ON 0 t DT , current increase from I to I
S SL
S L
V DT DViL V i
DT L fL
18
Ripple in output voltage
C O
when CHis ON , I I
C
O
VC. I
DT
O
O C
I DTV V
C
-ve sign indicates voltage decrease
O
O
I DTV
C
Critical Inductance (Lc)
LL
II
2
C
D 1 D RL
2f
Critical Capacitance (Cc)
O
O
VV
2
C
DC
2fR
If inductor also has an internal resistance, then
O S 2
1 DV V
r 1 DR
r = internal resistance of inductor
R = load resistance
19
Buck-Boost Converter
When CH is ON (O < t < DT)
L SV V
C OI I
When CH is OFF (DT < t < T)
L OV V
C L OI I I
Applying volt-sec balance across inductor
S O
V DT V 1 D T 0
S
O
DVV
1 D
Applying Ampere-sec balance across inductor
O L OI DT I I 1 D T 0
O
L
II
1 D
2
O SL
V DVI
R 1 D R 1 D
Ripple in inductor current
When CH is ON (O < t < DT)
Inductor current increase from min maxI to I
L
S
IL V
DT
SL
DVI
fL
20
Ripple in output voltage
When CH is ON (O < t < DT)
Capacitor discharge & voltage decrease from max min
V to V
O
O
C VI
DT
OO
DIV
fC
Critical inductance (Lc)
LL
II
2
2
C
R 1 DL
2f
Critical capacitance (Cc)
O
O
VV
2
O
C
S
I 1 D TC
2V
If internal resistance (r) of inductor is also considered then
O S2
D 1 DV V
r 1 DR
R = load resistance
21
Inverters Inverters circuits will convert DC power to AC power at required voltage & required frequency.
Classification
1) Voltage source Inverter
Input source is a voltage source.
Switching device is bidirectional & unipolar.
Load voltage depends on source voltage & load current depends on load parameters.
2) Current source Inverters
Input source is a current source.
Switching device is bidirectional & bipolar
Load voltage depends on source current & load voltage on load parameters.
Single phase half bridge VSI
When 1S is ON, O OV 0, I 0
When 2S is ON, O OV 0, I 0
When 1
D is ON, O O
V 0, I 0
When 2D is ON, O OV 0, I 0
The output voltage is a square wave of amplitude dcV
2
The fourier series of output voltage is given by
O
dc
n 1,3,5
2VV sin n t
n
rms value of fundamental components is given by
dc
or1 dc
2V 1 2V V
2
rms value of output voltage dc
or
VV
2
Distortion Factor(DF) =
or1
or
V 2 2
V
% Total Harmonic Distortion 2
1THD 1
DF= 48.43%
22
If load power factor is lagging, then it requires forced commutation.
If load power factor is leading, then natural commutation occurs.
Single phase Full Bridge VSI
When 1S , 2S conduct O OV 0, I 0
When1
D , 2D conduct, O O
V 0, I 0
When3
S ,4
S conduct, O O
V 0, I 0
When3 4
D ,D conduct, O OV 0, I 0
The output voltage is a square wave of amplitude dc
V
The fourier series of output voltage is given by
O
dc
n 1,3,5
4VV sin n t
n
rms value of fundamental components is given by
or1 dc
2V V
rms value of output voltage or dc
V V
Distortion Factor(DF) =
or1
or
V 2 2
V
% Total Harmonic Distortion 2
1THD 1
DF= 48.43%
Three phase full bridge VSI
23
1800conduction mode
In this mode, each switch will conduct for a period of 0180 and phase displacement between
any two poles is 0120
Phase voltage
ph dcrms
2V V
3
dcRN
n 6k 1
2VV sin n t
n
R1V = rms value of fundamental component of RNV
dcR1
2VV
Distortion factor,
R1
ph,rms
V 3DF
V
2
1THD 1 100 31%
DF
Line voltage
L L dcrms
2V V
3
dcRY
n 6k 1
4V nV sin sin n t3 6n
RY 1V = rms value of fundamental component of RYV =
RY 1
6V
Distortion factor = 3
In each phase, each switch conducts for 0180 out of 0360
o, rms
r.rms
II
2
dc dc2V V
3R3R 2
, Where R = load resistance
24
Voltage Total RMS Fundamental RMS
Phase dc
2V
3 dc
2V
Line dc
2V
3 dc
6 V
This conversion from total rms to fundamental rms can be performed by multiplication of
3 DF .
This conversion from phase to line voltage can be performed by multiplication of 3 .
1200conduction mode
For each thyristor, conduction angle is 0120 & last 060 for commutation.
Phase Voltage
dcph rms
VV
6
dcRN
n 6k 1
2VV sin n sin n t n
3 6n
R1 dc
6V V
Distortion factor,
3DF
THD = 31%
Line Voltage
dcL RMS
VV
2
dcRY
n 6k 1
3VV sin n t
3n
RY dc1
3V V
2
25
Distortion factor, = 3 ; THD 31%
In each, phase each switch conducts for 0120 out of 0360
o, rms
T, rms
II
3 dcV
2R
R = load resistance
Voltage Total RMS Fundamental RMS
Phase dcV
6
dc
6V
Line dcV
2 dc
3V
2
The conversion factor remain same as in 0180 conduction mode.
In both 0 0120 & 180 conduction mode both phase & line voltages are free from even & triplen
harmonics.
Voltage control using PWM techniques
1) Single PWM techniques
In this case, width of positive & negative cycle is not but rather equal to 2d.
O
S
n 1,3,5
4VV sin n sin nd sin n t
2n
To eliminate nth harmonics
Sin (nd) = 0
dn
Pulse width, 2 4 62d , , ,...................n n n
but 2d
To eliminate 3rd harmonics
23d ; d ; 2d3 3
So pulse width of 0120 is required.
26
2) Multiple PWM techniques
Here a single pulse of ‘2d’ width is divided into ‘n’ pulses each of width2d
n.
c
r
fn
2f
fc = carrier signal frequency
fr = reference signal frequency
AC - AC Converters
These circuits control AC power. They are of 2 types:
1) AC voltage regulator
2) Cyclo-converter
AC voltage regulator
These transfer AC power from 1 circuit to another by controlling output voltage & fixed
frequency.
Single phase half wave ACVR
m
O avg
VV cos 1
2
m
O avg
VI cos 1
2 R
1
2m
O rms
V 1V 2 sin2
22
1
2or
sr
V 1 1pf 2 sin2
V 22
27
Single phase fully controlled ACVR
o avg
V 0
1
2m
o rms
V 1V sin2
22
If R – L load is used, then in steady state OI lags OV by an angle
1 wLtan
R
If r , then above formulas remain valid & output voltage is controllable by controlling α.
If r , output voltage is not controllable & or srV V
So, range of firing angle is 0180
Integral cycle control (ON/OFF) control
If in fully controlled ACVR, thyristors conduct for m cycle & are OFF for n cycle then
1
2
O srrms
mV V
m n
For R – load,
1
2or
sr
V mpf
V m n
28
m
T1 avg
V mI
R m n
1
2m
T1 rms
V mI
2R m n
R = load resistance ; mV is maximum value of SV
1
Contents Manual for Kuestion ........................................................................... 2
Type 1: Power Semi-Conductor Devices ............................................. 3
Type 2: Phase Controlled Converters .................................................. 6
Type 3: Chopper ................................................................................ 10
Type 4: Inverters ............................................................................... 14
Type 5: DC & AC Drives ..................................................................... 17
Type 6: Commutation Circuit ............................................................ 20
Answer Key ....................................................................................... 23
3
Type 1: Power Semi-Conductor Devices For Concept, refer to Power Electronics K-Notes, Power Semi-Conductor Devices Sample Problem 1:
An SCR having a turn ON times of 5 μsec, latching current of 50A and holding current of 40
mA is triggered by a short duration pulse and is used in the circuit shown in figure. The
minimum pulse width required to turn the SCR ON will be
(A) 251 μsec
(B) 150 μsec
(C) 100 μsec
(D) 5 μsec
Solution: (B) is correct option
In this given circuit minimum gate pulse
width time= Time required by ia rise up to iL
2 3
40t
1
100i 20mA
5 10
100i 1 e
20
Sample Problem 2:
L − C circuit is used to commutate a thyristor, which is initially carrying a current of 5A as
shown in the figure below. The initial current through the inductor is zero, while the initial
capacitor voltage is 100 V. The values of L and C are 1 mH and 10 µF respectively. The switch
is closed at t = 0. If the forward drop is negligible, the time taken for the device to turn off is
(A) 52 μs
(B) 156 μs
(C) 312 μs
(D) 26 μs
40t
1 2
40t
40t
anode currrent I=I +I 0.02 5 1 e
0.05=0.02+5 1 e
0.03 1 e
5
T=150 s
4
Solution: (A) is correct option
Current through the LC circuit is
4c 0
Li(t) V sin t 10sin10 t
C
The device will be turned off only when current through it is zero. Commutation circuit has to
supply 5A current in opposite direction to the thyristor current.
i.e. i(t)=5 A
4So, 10sin10 t 5 A t=52.3 sec
Unsolved Problems:
Q.1 The latching current of thyristor is 4mA. The minimum width of gate pulse required to
turn on thyristor in inductive circuit is 4 ms. If the value of inductance is increased, then the
width of the gate pulse would be
(A) 6 s (B) 2 s (C) 4 s (D) None
Q.2 An SCR during it’s turn on process has the following data:
Anode voltage: 300 V 0V
Anode current: 0A 100A
During turn on time of 8s, the anode current and anode voltage very linearly. If triggering
frequency is 100 Hz, the average power loss in thyristor is
(A) 8 W (B) 16 W (C) 48 W (D) 4 W
Q.3 An LC combination circuit with thyristor in the circuit is operated with no external voltage,
but with an initial current of 250 A in inductor. The circuit parameters are L= 10 H, C = 50F.
Then the current extinguishes in the circuit for a time of
(A) 70.25s (B) 35.12s (C) 140.5s (D) 17.56 s
Q.4 A thyristor switching circuit is shown in figure. The thyristor is rated for a latching current
of 10 mA and is switched on within 5 s. Design the value of resistance R?
(A) 100 k
(B) 20 k
(C) 30 k
(D) 40 k
5
Q.5 The value of ‘L’ required to limit the di
dt to 20 A/S for the circuit shown in figure is
(A) 8.7H
(B) 17.2H
(C) 14.7H
(D) 12.3H
Q.6 A thyristor has an i2 dt rating of 15 Amp2.S and is being used to supply the circuit shown
from a 120 V. AC supply when a fault occurs, short circuiting the 10 resistor to earth. What
is the shortest fault clearance time to be achieved, if damage to the thyristor is to be
prevented?
(A) 1.04 ms
(B) 1.37 ms
(C) 0.96 ms
(D) 2.03 ms
Q.7 The junction capacitance of thyristor used in the circuit is 15 PF. The limiting value of
charging current to turn on thyristor is 5 mA and the critical value dv/dt is 200 V/s. The value
of capacitance Cs so that the thyristor will not be turned on due to dv/dt is---------
(A) 12 PF
(B) 7 PF
(C) 15 PF
(D) 20 PF
Q.8 A string of thyristor is connected in series to withstand a dc voltage of Vs = 10 KV. The
maximum leakage current and recovery charge difference of 100 mA and 150 C respectively
consider a derating factor of 20% and maximum voltage across each SCR is 1000 volts, the
transient voltage capacitance and steady state voltage sharing resistance are?
(A) 0.6 f, 25 K (B) 0.6 PF, 25 (C) 0.3 F, 25 K (D) 0.6 F, 25
6
Q.9 An SCR can withstand maximum junction temperature of 3900 K with ambient temperature
of 3450 K. Its thermal resistance from junction to ambient is 1.50C/W, the maximum internal
power dissipation allowed will be
(A) 30 W (B) 45 W (C) 60 W (D) 90 W
Q.10 For a thyristor, maximum junction temperature is 0125 C . The thermal resistance for the
thyristor sink combination are jC 0.16 and 0
CS 0.08 C / W . For a heat sink temperature
of 070 C , In case the heat sink temperature is brought down to 060 C by forced cooling, find
the percentage increase in the device rating.
(A) 8.00% (B) 9.71%
(C) 9.00% (D) 8.71%
Q.11 For the circuit shown in figure, dV
dt rating of SCR is 350 V / s and its junction
capacitance, is 20 pF. Switch S is closed at t=0. Calculate to value of sC so that SCR T is not
turn ON due to dV
dt
(A)0.0357 F
(B)0.0257 F
(C)2.5 F
(D)3.5 F
Type 2: Phase Controlled Converters For Concept, refer to Power Electronics K-Notes, Phase Controlled Converters Common Mistake: In some questions where line commutated converter works as an inverter you need to reverse the direction of current before calculating terminal voltage.
Sample Problem 3:
A single-phase bridge converter is used to charge a battery of 200 V having an internal
resistance of 0.2 Ω as shown in figure. The SCRs are triggered by a constant dc signal. If SCR2
gets open circuited, what will be the average charging current?
(A) 23.8 A (B) 15 A
(C) 11.9 A (D) 3.54 A
7
Solution: (C) is correct option
In this circuitry if SCR gets open circuited, than circuit behaves like a half wave rectifier.
So Iavg=Average value of current
1
1
avg m
avg m 1 1
1I (V sin t E)d
2 R
1I 2V cos E( 2 )
2 R
1
1
m
1 0
avg 1 12
Esin ( )
V
200 sin ( ) 38 0.66 rad
230 2
1I 2 230 cos 200( 2 )
2 2
0
avg2 11.9A
1I 2 230 cos38 200( 2 0.66)
2 2
Sample Problem 4:
Consider a phase-controlled converter shown in the figure. The thyristor is fired at an angle α
in every positive half cycle of the input voltage. If the peak value of the instantaneous output
voltage equals 230 V, the firing angle α is close to
(A) 450
(B) 1350
(C) 900
(D) 83.60
Solution: (B) is correct option
We know that rms
V 230V
So, m
2 VV 230
0 0
0
0 0
peak m
145 or 135
2
peak value of output is 230 V(given)
if <90 then the peak value of the out put will be 230 2
should be >90 i.e
sin
. =135
V V 230
sin
8
Sample Problem 5:
A single phase fully controlled converter bridge is used for electrical braking of a separately
excited dc motor. The dc motor load is represented by an equivalent circuit as shown in the
figure. Assume that the load inductance is sufficient to ensure continuous and ripple free
load current. The firing angle of the bridge for a load current of I0 = 10 A will be
(A) 440
(B) 510
(C) 1290
(D) 1360
Solution: (C) is correct option
Here for continuous conduction mode, by Kirchoff’s voltage law, average load current
a a
1
V 150V 2I 150 0 I
2
I 10 A
So, V=-130 V
m
0
2Vcos =-130
2 2 230cos 130 129
Unsolved Problems:
Q.1 A 1- diode rectifier is feeding a capacitor load. The supply is 230V, 50Hz and capacitor
is 1 F. The conduction time and final voltage across capacitor would be
(A) 1800, 230V (B) V2230 ,1800 (C) 900, 230V (D) 2230 ,90 0
Q.2 A single phase full bridge rectifier with freewheeling diode feeds an inductive load. The
load resistance is 15.53 and it has a large inductance to make current ripple free. The supply
voltage is 230V, 50Hz. For a firing angle delay of 60o, the average value of thyristor current is
(A) 3.33 A (B) 6.66 A (C) 10 A (D) 5 A
Q.3 A single phase half wave rectifier is feeding a resistive load with firing angle .2
Then
the form factor of the output voltage waveform will be
(A) 1.11 (B) 2.22 (C) 1 (D) 1.32
9
Q.4 A 1- full wave rectifier consists of 3 diodes and 1 thyristor feeding a resistive load of
10 as shown in figure. The average value of output voltage for firing angle delay of 600 is
(A) 181V
(B) 77.65 V
(C) 155V
(D) 210V
Q.5 A semi converter employs 02 thyristors and two diodes. Due to malfunction one of the
thyristor is working as a diode. The converter is feeding from an a.c. supply of 200 sin314 t
and gives power to resistive load of 100 . Then average output current is ………. for firing
angle of = 60.
(A) π
3.5 (B)
π
2.5 (C)
π
1.5 (D)
π
3
Q.6 A single phase full converter feeding RLE load has the following data source voltage:
230 V, 50 Hz; R 2.5 , E = 100 V, Firing angle = 30. If load inductance is large enough, so that
the output current is ripple free. Then the load power factor will be
(A)0.78 lag (B) 0.38 lag (C) 0.5 lag (D) unity
Q.7 A single phase semi converter is operated from a 50 Hz, 240 V ac source. If a resistive load
of 10 is connected at the dc terminals of the converter and the average output voltage is
25% of the maximum possible average output voltage, calculate the rms load current
(A) 5.4 A (B) 24 A (C) 10.6 A (D) 8.5 A
Q.8 A single phase fully controlled bridge circuit shown in fig. and the firing angle is
maintained at /3. If SCR3 (T3) is damaged and gets open circuited, Average direct current (Id)
output is
(A) 1 A
(B) 2 A
(C) 4 A
(D) 5 A
Q.9 A single phase fully controlled bridge converter supplies an inductive load (R-L load).
Assuming that the output current is virtually constant at 10 A. Firing angle is maintained at
/3 radians and the supply voltage is 230 V (RMS). Reactive power input and input power
factor are
(A) 1.793 kVAR and 0.45 lag (B) 1.234 kVAR and 0.5 lag
(C) 1.563 kVAR and 0.866 lag (D) 1.372 kVAR and 0.866 lag
10
Q.10 A three phase fully controlled bridge converter is connected to 440 v (line) supply, having
a reactance of 0.3 /phase and resistance of 0.05 /phase. The converter is working in the
inversion mode at a firing advance angle of 35. Assume load current is constant at 70 A. The
no load output voltage is
(A) 459 V (B) 486 V (C) 372 V (D) 356 V
Q.11 Three-phase fully controlled bridge converter is connected to supply voltage of 400 V
(line), source inductance is 3 mH. The load current on D.C side is constant at 15 A. Load
consists of a D.C source voltage of 400 V having an external resistance of 0.5 , overlap angle
is?
(A) 3.87 (B) 2.62 (C) 4.38 (D) 5.6
Q.12 A three phase fully controlled converter charges a battery from a three phase supply of
400 V (line), 50 Hz supply. The battery e.m.f is 200 V and its internal resistance is 1.5. Battery
charging current is constant at 20A, due to the inductance connected in series with the battery.
If the power flows from D.C. source to A.C., firing angle of the converter for the same current
is
(A) 107 (B) 116.43 (C) 127.72 (D) 132.3
Q.13 A 230V, 50Hz one pulse SCR converter is triggered at a firing angle of 040 and the load
current extinguishes at an angle of 0210 . Find the circuit turn off time & average load voltage
for R 5 , L 2mH and E 110V
(A)C
0
t 9.432ms
V 142.532V
(B)
C
0
t 9.11ms
V 141.16V
(C)C
0
t 8.126ms
V 112.532V
(D)
C
0
t 9.432ms
V 231.63V
Type 3: Chopper For Concept, refer to Power Electronics K-Notes, Chopper
Point to Remember:
For approximate ripple in Inductor Current we can directly use the concept of Voltage across the inductor so no need to memorize the formulas for maximum and minimum current in chopper. Sample Problem 6:
In the figure shown below, the chopper feeds a resistive load from a battery source. MOSFET
Q is switched at 250 kHz, with duty ratio of 0.4. All elements of the circuit are assumed to be
ideal. The average source current in Amps in steady-state is
11
(A) 3/2
(B) 5/3
(C) 5/2
(D) 15/4
Solution: (C) is correct option
Here, the average current through the capacitor will be zero. (since, it is a boost converter).
We consider the two cases :
Case I : When MOSFET is ON
ic1 =- i0 (i0 is output current)
(since, diode will be in cut off mode)
Case II : When MOSFET is OFF
Diode will be forward biased and so
ic1 = Is - i0 (Is is source current)
Therefore, average current through capacitor
c1 c2
c,avg
0 0 0 s 0
i II
2
DT ( i ) (1 D)T (I i ) 0 (D is duty ratio)
2
Solving the equation, we get
0
s
iI
(1 D)
Since, the output load current can be given as
s
0
0
0
s
V 12V (1 D) 0.6i 1 AR R 20
i 1 5I
(1 D) 0.6 3
Sample Problem 7:
In the circuit shown in the figure, the switch is operated at a duty cycle of 0.5. A large capacitor
is connected across the load. The inductor current is assumed to be continuous. The average
voltage across the load and the average current through the diode will respectively be
12
(A) 10 V, 2 A
(B) 10 V, 8 A
(C) 40 V 2 A
(D) 40 V, 8 A
Solution: (C) is correct option
when switch S is open I0 = IL = 4 A,Vs = 20 V
when switch S is closed ID = 0,V0 = 0 V
Duty cycle = 0.5 so average voltage is =Vs/(1-D)
Average current=(0+4)/2 A
Average Voltage=20/(1-0.5) = 40 V
Unsolved Problems:
Q.1 A step down chopper feeding an inductive load and current is ripple free in the output
circuit. The ratio of average currents through freewheeling diode and chopper is ____ (Take
duty cycle =0.8)
(A) 0.25 (B) 4 (C) 2 (D) 0.5
Q.2 A step up chopper is required to deliver a voltage of 330 V from 110 V dc source. If the
conduction time of chopper is 200 S, then turn off time of chopper is
(A) 100 s (B) 300 s (C) 200 s (D) 400 s
Q.3 Type-A chopper supplies power to a resistive load of 10 from 200 V dc source. The
output voltage is ON for 4 ms duration and OFF for 6 ms duration. The rms value of
fundamental component of output voltage is
(A) 80 V (B) 126.5 V (C) 90 V (D) 85.6 V
Q.4 A step down d.c chopper feeding a resistive load of R = 20 and input voltage =220 V.
When the chopper remains ON, it’s voltage drop is 1.5 V and chopping frequency is 10 KHz.
What is the chopper efficiency?
(A) 99.3 % (B) 89% (C) 99% (D) 98.3%
Q.5 A DC chopper circuit is connected to a 120 V DC source supplies to inductive load having
50 mH in series with a resistance of 6. A Freewheeling diode is placed across the load. The
load current varies between the limits of 12 A and 16 A. Time ratio (Ton / Toff) of chopper is
(A) 2.122 (B) 1.222 (C) 2.333 (D) 1.732
13
Q.6 A simple D.C chopper is operating at a frequency of 4 kHz from a 100 VDC source to
supply a load resistance of 6. The load time constant is 6 ms, average load voltage is 62 V.
Assume load current variations are linear. Magnitude of ripple current is
(A) 0.127 A (B) 0.164 A (C) 0.237 A (D) 0.282 A
Q.7 The chopper shown in figure operates at a frequently of 100 Hz with an “ON” time of 4ms.
The average value of load current is 25A, with a peak to peak ripple of 6A. Average value of
source and diode currents are
(A) 10A, 8A
(B) 12.5A,12.5A
(C)13.7A/7.4A
(D) 9A, 7A
Q.8 The step up chopper shown in figure operates at a frequency of 1kHz. The source voltage
is 100v D.C and the load resistance is 4 .If the average value of load current is 200 A, the
average value of the switch current is
(A) 200A
(B) 300A
(C) 100A
(D) 400A
Q.9 A battery with it’s terminal voltage of 200 volts is supplied with power from type-A
chopper ckt. The output voltage of the chopper consists of rectangular pulses of 2 ms duration
in an overall cycle time of 5 ms. The average, rms values of output voltage, Ripple voltage and
ripple factors are respectively -------
(A) 80 V, 126.49 V, 149.66 V, 1.87 (B) 80 V, 126.49 V, 97.98 V, 1.87
(C) 80 V, 126.49 V, 149.66 V, 1.581 (D) 80 V, 126.49 V, 97.98 V, 1.2247
Q.10 A voltage commutated chopper has the following parameters: Vs = 220 V, load circuit
parameters = 0.5, 2 mH, 40 V commutation circuit parameters L = 20 H, C = 50 F
Ton = 800 s, T = 2000 s. For a constant load current of 80 Amps, the effective on period, peak
current through main thyristor are?
(A) 525 s, 427.85 A (B) 525 s, 160 A
(C) 1075 s, 80 A (D) 1075 s, 427.85 A
14
Q.11 In figure, the ideal switch “S” is switched on and off with a switching frequency f = 10KHz.
The circuit is operated in steady state at the boundary of continuous and discontinuous
conduction, so that the inductor current ‘i’ is as shown. Find the ON time of the switch
(A)60 s (B) 72 s (C) 83.37 s (D) 74.34 s
Q.12 A D.C Chopper is used for Regenerative breaking of a separately excited D.C motor. The
D.C supply voltage is 400 V. The motor has ra = 0.2, Km = 1.2 V-S/Rad. The average armature
current during regenerative breaking is kept constant at 300 A with negligible ripple, For a
duty cycle of 0.55 for a Chopper. Power returned to the supply is
(A) 54 KW (B) 48KW (C) 58KW (D) 42KW
Q.13 In a boost converter, the duty ratio is adjust to regulate the output voltage 0V at 48 V.
the input voltage varies in a wide range from 12 to 36 V. the maximum output power is 120W.
for stability reasons, it is required that the converter always operate in a discontinuous current
conduction mode. The switching frequency is 50 kHz. Assuming ideal components and C as
very large, Calculate the maximum value of L that can be used.
(A)9 H (B)12 H
(C)20 H (D) 9 H
Type 4: Inverters For Concept, refer to Power Electronics K-Notes, Inverters
Point to Remember:
The understanding of basic operation of Inverter is a must to tackle some non-trivial
problems other than that remember the basic formulas of Inverters.
Sample Problem 8:
A single-phase voltages source inverter is controlled in a single pulse-width modulated
mode with a pulse width of 1500 in each half cycle. Total harmonic distortion is defined as
2 2
rms 1
1
V VTHD 100
V
15
where V1 is the rms value of the fundamental component of the output voltage. The
THD of output ac voltage waveform is
(A) 65.65% (B) 48.42% (C) 31.83% (D) 30.49%
Solution: (C) is correct option
Given that, total harmonic distortion
2 2
rms 1
1
V VTHD 100
V
Pulse width is 1500
Here
rms s s
150V V 0.91V
180
0s1 rms(fundamental) s
0.4VV V sin75 0.8696V
2
2 2
s s
2
s
0.91V 0.87VTHD 31.9%
0.87V
Sample Problem 9:
An inverter has a periodic output voltage with the output wave form as shown in Figure. When
the conduction angle α = 1200, the rms fundamental component of the output voltage is
(A) 0.78 V
(B) 1.10 V
(C) 0.90 V
(D) 1.27 V
Solution: (A) is correct option
Output voltage
s
0n 1,3,5
4VV sinnd sinn t sinn
n 2
s
rms(fundamental)
0 0 0
RMS value of fndamental component
4VV sind 1
2
120 ,2d 120 d 60
0s
rms(fundamental)
s
4VV sin60
2
0.78V 0.78V
16
Unsolved Problems:
Q.1 A single phase full bridge inverter is supplying power to RLC load with R=3 , and
12XL The bridge operates with a periodicity of 0.2 ms. The circuit turn off time required
for transistors is 24 s. The suitable value of c that results in to load commutation would be
(A) 3 F (B) 2.5 F (C) 4 F (D) 2 F
Q.2 In single pulse modulation of PWM inverters, the pulse width is 80. For an input voltage
of 330 V dc, the r.m.s value of output voltage is
(A) 110 V (B) 220 V (C) 330 V (D) 200 V
Q.3 A 1 - full bridge inverter is supplied from 230 V d .c and load is series RLC with R = 1 ,
L = 2 and 1 / = 1.5 . The output voltage is controlled by single pulse modulation and
pulse width is 1200. The r.m.s values of fundamental and third harmonic components of current
are
(A) 50 A, 0 A (B) 160 A, 55 A (C) 82.3 A, 31.4 A (D) 160 A, 0 A
Q.4 A single phase-bridge inverter delivers power to a series connected RLC load with R = 3
and XL = 12. The periodic time is 0.2 ms. Thyristor turn off time is 1.2 s. Circuit turn off time
is 1.5 tq (where tq is the each SCR turn off time). Assume that load current contains only
fundamental component. The value of ‘C’ should be in the load circuit in order to obtain the
load commutation of SCR is
(A) 1.37F (B) 2.23 F (C) 1.87 F (D) 2.62 F
Q.5 A 3-phase inverter is fed from a 600 V source. For a star connected resistive load
15 / phase, the rms load current for 120 conduction is.
(A) 32.66A (B) 16.33A (C) 8.16A (D) 12.33A
Q.6 Output voltage of inverter is controlled by one of the PWM technique as shown in fig is
multiple pulse modulation (two pulse modulation)
Peak value of fundamental voltage component is
(A) 278 V (B) 343 V (C) 247 V (D) 175 V
17
Q.7 A single phase full bridge inverter has RMS value of the fundamental component of output
voltage, with single pulse modulation, equal to 110v, and DC source voltage is 220v. If we take
the two symmetrically speed pulses per half cycle, RMS value of 3rd harmonic of the two pulse
waveform is
(A) 54.3v (B) 37.4v (C) 29.6v (D) 41.8v
Q.8 The output voltage from inverter is controlled by a single pulse modulation. The output
voltage waveform is shown in figure. The peak value of 5th harmonic is
(A)0.9 V
(B)1.273 V
(C)0.11 V
(D)0 V
Q.9 A D.C Chopper Circuit connected to a 100 V DC source supplies an inductive load having
40mH in series with a resistance of 5. A freewheeling diode is placed across the load. The
load current varies between the limits of 10 A and 12 A. The time ratio of the Chopper is?
(A) 0.55 (B) 1.22 (C) 0.67 (D) 0.73
Q.10 A single phase full bridge inverter controls the power in a resistive load. The nominal
value of input dc voltage is Vs = 220 V and a uniform pulse width modulation with five pulse
per half cycle is used. For the required control, the width of each pulse is 025 . Determine the
r.m.s. value of the load.
(A)200.8 V (B)183.33 V
(C)166.67 V (D)150 V
Type 5: DC & AC Drives
Point to Remember:
For AC and DC drives remembering the basic formulas of Machines and Power Electronics is
sufficient.
Sample Problem 10:
A single-phase half-controlled rectifier is driving a separately excited dc motor. The dc motor
has a back emf constant of 0.5 V/rpm. The armature current is 5 A without any ripple. The
armature resistance is 2 Ω. The converter is working from a 230 V, single-phase ac source with
a firing angle of 300. Under this operating condition, the speed of the motor will be
(A) 339 rpm (B) 359 rpm (C) 366 rpm (D) 386 rpm
18
Solution: (C) is correct option
E.m.f. constant Kb=0.5 V/rpm
For separately excited d.c. motor
Ea= KbN and
0 a a
E V I R
0mV 230 2
E (1 cos ) 5 2 (1 cos30 ) 10
b
E 183.20
E 183.20N 366.4r.p.m.
K 0.5
Unsolved Problems:
Q.1 A separately excited DC motor is fed from two single phase semi converters, one in the
armature circuit and other in the field circuit. Field current is constant at 2A, motor armature
resistance is 0.8 and motor constant is K=0.5 V.S/A. rad. AC voltage is 230V, 50HZ. For a
ripple free armature current and speed of 1500rp.m, firing angle of 300. The torque developed
by the motor would be
(A) 45.12Nm (B) 90.24Nm (C) 22.66Nm (D) 10.24Nm
Q.2 A 3–, 440 V, 50 Hz ac mains fed thyristor bridge is feeding a 440 V, ac, 15 kw, 1500 rpm
separately excited dc motor with a ripple free continuous current in the dc link under all
operating conditions neglecting the losses, the motor is running at 3/4 rated speed. The rms
value of line current is
(A) 32.5 A (B) 37.5 A (C) 25 A (D) 40 A
Q.3 A Separately excited d.c. motor is fed from the single phase semi converters, one in the
armature circuit and the other in field circuit. Field current is constant at 2A, motor armature
resistance is 0.8 and motor constant is K = 0.5 V.S/A-rad. AC voltage is 230 V, 50 Hz. For a
ripple free armature current and speed of 1500 r.p.m. Motor torque is
(A) 45 Nm (B) 22.5 Nm (C) 90 Nm (D) None
Q.4 A small separately excited DC motor is supplied via a half controlled, single phase bridge
rectifier. The supply is 240v, the thyristors are triggered at 1100, and the armature current
continuous for 500 beyond the voltage zero. The motor torque characteristic is 1.0Nm/A and
its armature resistance is 6. Motor speed at a torque of 1.8Nm is (neglect all converter losses)
(A) 864rpm (B) 778rpm (C) 932rpm (D) 884rpm
Q.5 A D.C. motor with armature resistance of 0.2 is fed from a step down chopper in the
continuous mode and operates at some known speed and known excitation current. The
motor current rises from 6A to 10A in the ON period of the chopper and drop from 10A to 6A
19
in the off period of the chopper of the same circuit. Both rise and fall of the current may be
assumed linear. Calculate the average power loss in the machine armature
(A) 27.2 W (B) 46.67 W (C) 13.07 W (D) 11.02 W
Q.6 A separately excited D.C. motor has the following name plate data 220V, 100A, 2200 R.P.M.
The armature resistance is 0.1 and inductance is 5m.H. The motor is fed by a chopper, which
is operating from a D.C. supply of 250V. The chopper is operated at a duty cycle of 80%.
Determine the speed at which the motor can be operated at rated torque.
(A) 1200 R.P.M (B) 400 R.P.M (C) 1900 R.P.M (D) 2090 R.P.M
Q.7 The speed of the separately excited D.C motor is controlled by a Chopper. The D.C supply
voltage is 100 V, armature circuit resistance is Ra=0.4, armature circuit inductance La= 10 mH
and the motor constant is Ka = 0.05v/R.P.M. The motor drives a constant torque load
requiring an average current of 30A. Assume that the motor current is continuous. The range
of duty cycle is
(A) 1/44<α<1 (B) 0.12< α <1 (C) 0.3< α <1 (D) 0.27< α <1
Q.8 A single phase ac regulator fed from 50 Hz system supplies a load having resistance and
inductance 4.0 ohms and 12.73 mH respectively. The control range of firing angle is ___________
(A) 0 00 180 (B) 0 090 180
(C) 0 045 180 (D) 0 00 45
Q.9 In the following circuit, The RMS value of load current in amps by assuming 090 is
____________ (in amp)
(A)10
(B)20
(C)30
(D)40
Q.10 A 200 V, 1000 rpm, 10 A separately excited DC motor is controlled in constant torque
mode by using a buck-boost converter. Armature resistance sR 1 . If the motor is to be run
at 700 rpm. What is the duty cycle required (Input to the converter is 220 V)
(A)0.216 (B)0.316
(C)0.416 (D)0.516
20
Type 6: Commutation Circuit For Concept, refer to Power Electronics K-Notes, Power Semi-Conductor Devices.
Point to Remember:
Try to understand the basic operation of Commutation Circuits as sometimes some non-
standard case may come where you need to apply the principle of operation. Remember
Standard cases direct formulas as well.
Sample Problem 11:
Figure shows a chopper. The device S1 is the main switching device. S2 is the auxiliary commutation device. S1 is rated for 400 V, 60 A. S2 is rated for 400 V, 30 A. The load current
is 20 A. The main device operates with a duty ratio of 0.5. The peak current through S1 is
(A) 10 A
(B) 20 A
(C) 30 A
(D) 40 A
Solution: (D) is correct option
When S1 is triggered, tank CKT is formed by C(2µF) and L(200µH).
The current through the circuit is
sc 0
0
Vi sin t
L
Peak capacitor current
scp s
0
V CI V
L L
Therefore peak current through S1 is
6
s
cp 0
6
C 2 10 V 20 200 20
L 200 1
I I
0
I
40 A
Unsolved Problems:
Q.1 A voltage commutated chopper shown in figure employs a thyristor of device turn off time
100 s. Factor of safety is 2. The chopping frequency is 500Hz.
The average value of output voltage is
21
(A) 88 V
(B) 110 V
(C) 131 V
(D) cannot be evaluated
Q.2 A voltage commutated chopper is shown in figure. The chopper operates at frequency of
400 Hz and 50 % duty ratio. The load current remains constant 10A. Assuming the input
voltage to be 200 V and device to be ideal. Maximum value of current through capacitor is
(A) 8.94 A
(B) 9 A
(C) 10 A
(D) 9.5 A
Q.3 A resonant pulse commutation circuit is shown in fig. For this circuit C = 16 F and
L = 4H. Initial voltage across capacitor is 250 V. Load current is constant at 360 A. Voltage
across main thyristor (VT) when it gets commutated is
(A) 173.5 V
(B) 137.8 V
(C) 152.3 V
(D) 167.4 V
Q.4 A voltage commutated circuit is shown in fig. For this circuit VS = 250 V, L = 24 H and
C = 52 F. Load current is constant at 132 A. Circuit turn off time for auxiliary thyristor (TA) is
(A) 26.7 s (B) 33.3 s
(C) 44.4 s (D) 55.5 s
22
Q.5 For the circuit shown in figure, the value of resistance “R” is 10, VDC = 100v and thyristor
turn “OFF” time is 50 sec. The value of C is required for commutation of thyristor is
(A) 8.4F
(B)5.7 F
(C)7.2 F
(D)6.3 F
Q.6 The circuit of figure employing a resonant pulse commutation has C=20 F and L=5 H.
Initial voltage across capacitor Vs is 230V. For a constant load current of 300 A, the circuit turn
of time for the main thyristor?
(A) 31.46 s
(B) 15.70 s
(C) 11.62 s
(D) 23.24 s
Q.7 In the circuit shown below, load draws a constant current of OI
Pick the condition of C for reliable communication of 1T
L=1 mH
(A)C 0.1 F Io=10A
(B)C 0.1 F Vs=100 V
(C)C 0.3 F
(D)C 0.3 F
Q.8 An impulse –commutated chopper feeds inductive load requiring a constant current of
260 A. The source voltage is 220 V dc and the chopping frequency is 400 Hz Turn – off time
for main thyristor is 18 s . Peak current through main thyristor is limited to 1.8 times the
23
constant load current. Taking a factor of safety 2 for the main thyristor, calculate the value of
commutating components C and L.
(A)C 41.545 F, L 46.596 H (B)C 34.214 F, L 44.72 H
(C)C 42.545 F, L 47.596 H (D)C 36.712 F, L 37.16 H
Answer Key 1 2 3 4 5 6 7 8 9 10 11 12 13
Type 1 A D B B C A B D A D A
Type 2 D A B A A A C B A B C C A
Type 3 A A D A C B B A D D A A D
Type 4 D B D D B D C C B B
Type 5 A A A A C C B C B C
Type 6 B C A D C C B C
Type 7
Type 8
