http://electrical-mentor.blogspot.in/
POWER ELECTRONICS
Question Bank
by
Shankar
Version: PEQBTNC06
Conventional, Objective and Interview questions in
Power Electronics for GATE |IES | All PSUs
Version Code: PEQBTNC06
http://electrical-mentor.blogspot.in/
Page 2 of 120
Copyright © Reserved 2014
Page 3 of 120
Version Code: PEQBTNC06
PREFACE
I would like to present this Question bank on Power electronics to my student community
at free of cost.
I have prepared both conventional and objective questions in the subject of Power
Electronics from various sources and knowledge gained from my teaching experience over a
span of 7 years. The content of this Question bank is mainly useful for GATE and
Engineering Service (ESE) aspirants to gain in depth analysis into the subject. As previous
GATE and ESE papers are available in various modes, I have not repeated those questions
here.
It is expected that the reader must have basic knowledge in the area of Power Electronics
and its applications at under graduate level before solving this booklet.
This booklet contains the following sections:
Conventional Questions: By solving these questions, the reader can enhance his/her basic
concepts in Power electronics and can establish the link between other branches of
electrical engineering. By solving these types of questions, I am sure your confidence levels
in the subject will increase which is the key thing for success in any competitive exam and
in career as well. I have provided answers for around 90% of questions and remaining 10%
is left as open for the readers so that they can sharpen their knowledge. I will address these
questions in the next release of this booklet based on response and will provide some more
open questions in subsequent releases
Objective Questions: In the present trend, every exam is based on Objective questions.
After solving the conventional questions, the reader can test his/her understanding in the
concepts by taking 4 practice tests based on objective questions
Interview Questions: These questions are collected from various interviews like M.Tech
admissions in IITs, OCES & DGFS interviews in BARC etc from student community itself.
In fact, these questions are not my creation and collected from various students. If you
attend any interview, you can also share your experience for the benefit of your next
generation
And then I have given practical approach for compensator design for PE
converter
After solving this booklet, I am expecting you can face any exam, or interview very
confidently especially in the field of Power Electronics.
With initial thoughts in my mind, this booklet came out. I am planning to update this
booklet based on feedback received and will revise in regular intervals and need basis
Finally I would like to express my sincere thanks to Mr Saida (my student) for his valuable
suggestions and efforts in the drafting corrections
If you have any suggestions for further development of this booklet, if you find any
mistakes or corrections required, please feel free to write an email to
electrical.mentor@gmail.com by referring version code
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 4 of 120
Version Code: PEQBTNC06
INDEX
Sl.
No
0
1
2
3
4
5
6
7
8
10
11
12
Description
Preface
Conventional Questions
Objective Questions - Practice Test 1
Objective Questions - Practice Test 2
Objective Questions - Practice Test 3
Objective Questions - Practice Test 4
Interview Questions
Answers for Conventional Questions
Answers for Objective Questions
Compensator Design
Useful units for Electrical Engineering
Useful Mathematical Formulae
http://electrical-mentor.blogspot.in/
Page
No
3
5
35
47
60
68
76
89
92
93
114
116
Copyright © Reserved 2014
Page 5 of 120
Version Code: PEQBTNC06
Power Electronics
Conventional Questions
Q1.
In a power electronics laboratory, an experiment is conducted to find circuit
component value and its circuit diagram is shown in Fig. The voltage and current
waveforms for periodic time of 20 ms are captured from oscilloscope are also shown
below. Find out what could be the circuit element and its value
vi
10V
i1(t)
ii
1A
t
Vi(t)
−10V
Ts
2
Fig for Q1
Q2.
In a power electronics laboratory, an experiment is conducted to find circuit
component value and its circuit diagram is shown in Fig. The voltage and current
waveforms for periodic time of 10 ms are captured from oscilloscope are also shown
below. Find out what could be the circuit element and its value
Ii
v(t)
10A
vi
1V
Ii(t)
t
Ts
2
Fig for Q2
Q3.
In a power electronics laboratory, the impedance Z(s) diagram (bode plot) for a pure
inductor is captured using network analyzer as shown in Fig
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 6 of 120
Version Code: PEQBTNC06
dBΩ
Z(s)
6 dB
0 dB
ω1
−6 dB
ω2
log10(ω)
ω0
Fig for Q3
(a) If ω0 = 100 rad/s, find the value of inductance
(b) If ω1 = 50 rad/s and ω2 = 200 rad/s then find the value of Z (s) in dB and Ω at ω =
1000 rad/s
Q4.
In a power electronics laboratory, the impedance Z (s) diagram (bode plot) for a pure
capacitor is captured using network analyzer as shown in Fig
dBΩ
Z(s)
20 dB
0 dB
ω1
ω0
ω2
log10(ω)
−20 dB
Fig for Q4
(a) If C = 10µF then find the values of ω0, ω1 and ω2 in rad/s (These frequencies are
in decade fashion)
(b) Find the frequency in rad/s when Z (s) = 2 Ω
Q5.
The current through and the voltage across a power semi conductor switch is shown
in Fig.
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 7 of 120
Version Code: PEQBTNC06
20A
Fig for Q5
Evaluate,
(a) The average current and the RMS current rating of the device.
(b) The conduction loss in the device
Q6.
The approximate wave shape of a capacitor current in a commutation circuit is
shown in Fig. The capacitor has an equivalent series resistance (ESR) of 20 mΩ.
Fig for Q6
Evaluate the power dissipation in the capacitor
Q7.
In an inverter, the current through the active device is measured and found to be as
shown in Fig. The switching frequency may be considered very high compared to the
fundamental frequency of the output current.
Fig for Q7
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 8 of 120
Version Code: PEQBTNC06
Determine,
(a) The average and RMS current rating of the switch.
(b) If the power device is a power transistor with a Vce drop of 1.2 V, evaluate the
conduction loss
Q8.
The SCR is used in an application carrying half sinusoidal current of period 1 ms
and a peak of 100 A as shown in Fig. The SCR may be modeled during conduction to
have a constant voltage drop of 1.1 V and a dynamic resistance of 8 mΩ. Calculate
the average conduction loss in the device for this application
Fig for Q8
Q9.
The periodic current through a power-switching device in a switching converter
application is shown in Fig.
Fig for Q9
(a) Evaluate the average current through the device.
(b) Evaluate the RMS current through the device.
(c) A BJT with a device drop of 1.2 V and a MOSFET with an of 150 mΩ are
considered for this application. Evaluate the conduction loss in the device in either
case.
Q10.
A power diode (ideal in blocking and switching) shown in Fig, is capable of
dissipating 75 W. For square wave operation, it is rated for peak current of 100 A
and 135 A at duty ratios 0.5 and 0.33 respectively.
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 9 of 120
Version Code: PEQBTNC06
Fig for Q10
(a) Evaluate the ON state model of the diode (This procedure is known as piecewise modeling of semiconductor device).
(b) The above diode while dissipating 40W at an ambient temperature of 350 C, is
running with a case temperature of 750 C and 1250 C respectively. Evaluate the
thermal resistances of the device
Q11.
The diode (20ETS08) is a 20 A, 800 V rectifier diode. It has a voltage drop of 0.8 V at
2 A and 1.2 V at 30 A.
(a) Find a piece-wise linear model for this diode consisting of a cut-in voltage and
dynamic resistance.
(b) With this piece-wise model evaluate its conduction loss for a 30 A peak half sine
wave of current.
Q12.
A power-switching device is rated for 600 V and 30 A. The device has an on state
voltage drop of 1.5 V to 2.4 V for conduction current in the range of 15 to 30 A. The
device has a leakage current of 5 mA while blocking 600 V.
Evaluate
(a) The maximum conduction loss,
(b) The maximum blocking loss, and
(c) The ratio of the conduction and blocking loss with maximum possible power that
may be controlled by this switch and make your comment on the result.
Q13.
A composite switch used in a power converter is shown in Fig. The periodic current
through the switch is also shown.
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 10 of 120
Version Code: PEQBTNC06
Fig for Q13
Evaluate,
(a) The average current and RMS current through the composite switch.
(b) The power loss in the MOSFET and the diode of the composite switch.
Q14.
A power MOSFET has an Rds(on) of 50 mΩ. The device carries a current as shown in
Fig. Consider the switching process to be ideal and evaluate the conduction loss in
the device. (Explore if you can simplify the evaluation of RMS value by applying
superposition).
Fig for Q14
Q15.
A power-switching device is ideal in conduction and blocking (0 V during conduction
and 0 A in blocking). It is used in a circuit with switching voltages and currents as
shown. The switching waveforms under resistive loading and inductive loading are
shown in Fig. The switching times tr and tf are 100 ns and 200 ns respectively.
Evaluate,
(a) The switch-on and switch-off energy loss (in joule) for resistive loading
(b) The switch-on and switch-off energy loss (in joule) for inductive loading
(c) The resistive and inductive switching losses in watt for a switching frequency of
100 kHz.
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 11 of 120
Version Code: PEQBTNC06
Fig for Q15
Q16.
The current through and the voltage across a switching device is given in Fig.
Evaluate the approximate switch-off and switch-on energy loss in the device.
Fig for Q16
Q17.
A disc type Thyristor is shown with its cooling arrangement in Fig. The device is
operating with a steady power dissipation of 200 W.
Fig for Q17
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 12 of 120
Version Code: PEQBTNC06
Various thermal resistances are defined as below:
= 0.3 C/W; = 0.3 C/W; = 0.05 C/W; = 0.05 C/W;
= 0.5 C/W; = 0.4 C/W;
Evaluate the steady state temperature rise of the junction
Q18.
A composite switch (Q1 and Q2 in parallel) carrying a load current of 10 A is shown
in Fig. The switches may be considered ideal in switching. The on-state resistances
of the devices Q1 and Q2 are respectively 0.8 Ω and 0.2 Ω. The devices are mounted
on a common heat sink held at a temperature of 800 C.
Fig for Q18
Evaluate,
(a) RMS values of I1 and I2
(b) The average power dissipation (P1 and P2) in Q1 and Q2.
(c) The junction temperatures of Q1 and Q2 (Note: RJC1 and RJC2 is thermal
resistances from Junction to case of Q1 and Q2).
Q19.
Fig for Q19
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 13 of 120
Version Code: PEQBTNC06
The voltage across a capacitor used for a power electronic application is shown in
Fig. The capacitance value is 2.5 µF. The capacitor has an equivalent series
resistance (ESR) of 10 mΩ. The dielectric of the capacitor has a thermal resistance
of 0.2 0C/W to the ambient.
(a) Sketch the current waveform through the capacitor for one cycle
(b) Evaluate the losses in the capacitor
(c) Evaluate the temperature rise in the dielectric of the capacitor
Q20.
A power electronic capacitor is specified to have the following values. Capacitance =
10 µF; ESR = 30 mΩ; ESL = 75 nH; Sketch the impedance of the capacitor as a
function of frequency in the dBΩ vs log ω. Determine the range of frequency for
which the capacitor may be idealized to be a pure capacitance of 10 µF
Q21.
The current through a diode is shown in Fig. Consider the following data for
waveform analysis.
t1 = 100 µs, t2 = 350 µs, t3 = 500 µs, f = 250 Hz, fs = 5 kHz, Im = 450 A and Ia = 150 A
Determine, (a) Average diode current and (b) RMS diode current
Im
Ia
i1=Imsinωst
i2
t1
T=
t2
t3
T
t
1
f
Fig for Q21
*Q22.
Visit a manufacturer's website, identify a controlled power switching device (BJT,
or MOSFET, or IGBT etc) of rating > 10A and > 600V. Download the datasheet and
fill in the following.
(a) Manufacturer (b) Device and Type No (c) On-state voltage (V) (d) ON-state
current (A) (e) Transient switching times (s) (f) Maximum junction temperature (K)
(g) Recommended drive conditions (?) (h) Conduction loss at rated current (W) (i)
Blocking loss at rated voltage (W) (j) Switching energy loss (J).
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 14 of 120
Version Code: PEQBTNC06
Q23.
The magnetic circuit of a coupled inductor is shown in Fig. The magnetic material of
the core may be assumed to be ideal.
N1= 100 T; N2 = 200 T; Ag1= Ag2 = 40mm2; Ag = 80mm2; lg1 = 1mm; lg2 = 2mm;
lg = 1.5mm
Fig for Q23
Evaluate the inductances L1; L2; L12; L21
*Q24.
The following figures (a, b, and c) show three magnetic circuits with an exciting
winding on each having 100 turns. The core in (c) is obtained by assembling
together one each of cores shown in (a) and (b). The magnetic material for the core
may be considered to have very large permeability with saturation flux density of
0.2 T.
(a) Evaluate the expression for flux linkages (Nϕ) for cores (a) and (b) as a function
of the exciting current ia and ib.
(b) Plot the characteristics Nϕ vs i for the cores (a) and (b).
(c) From the above plot Nϕ vs i for the composite core (c).
(d) Comment on the inductance of the circuit (c).
Fig for Q24
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 15 of 120
Version Code: PEQBTNC06
*Q25.
0
A
0
(a)
π
A
2π
0
(b)
0
µ µ
2 2
µ µ
2 2
0
A
(c)
A
µ µ
2 2
0
0
µ µ
2 2
(d)
A
0
A
ωt
(e)
A
A
0
ωt
(f)
t
DT
T
Fig for Q25
(a) For the waveforms shown in Fig, calculate their average value, RMS values of
the fundamental and the harmonic frequency components
(b) For the waveforms shown in Fig, consider A = 100 and µ = 600 where applicable.
Calculate their total RMS values
(c) For the waveforms shown from a to d in Fig shown, calculate the ratio of (i) the
fundamental frequency component to the total RMS value and (ii) the distortion
component to the total RMS value
(d) For the waveforms shown from e to f in Fig shown, calculate the ratio of the
average value to the total RMS value (form factor)
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 16 of 120
Version Code: PEQBTNC06
(e) For some common rectifiers, the line currents may be like the waveforms shown
in a to b of Fig 2 with µ = 600. The need for power per phase is the same in the two
cases i.e, the RMS value of the fundamental component the line currents are 100 A
in both cases.
(i) Calculate the amplitude and the RMS value for waveform a in Fig shown
(ii) Calculate the amplitude and the RMS value for waveform b in Fig shown
(iii) Comment on the above answers
Q26.
An inductive load connected to a 120 V, 60 Hz ac source draws 1 kW at a power
factor of 0.8. Calculate the capacitance required in parallel with the load in order to
bring the combined pf to be 0.95 lag
Q27.
A 110 V/220 V, 60 Hz single phase 1 kVA transformer has a leakage reactance of 4
%. Calculate its total leakage inductance referred to (a) the 110 V side and (b) 220 V
side
*Q28.
An input voltage of a repetitive waveform is filtered and the applied across the load
resistance as shown in Fig. Consider the system to be in steady state. It is given
that L = 5 µH and Pload = 250 W
vi
iload
+
+
+
iL
vi
v0
ic
−
15V
R
−
(load)
(Fig 3)
t
0
4 µs
6 µs
Fig for Q28
(a) Calculate the average output voltage V0
(b) Assume that C ∞ so that vo (t) = V0. Calculate Iload and the RMS value of the
capacitor current ic
(c) In part (b), plot vo and iL
*Q29.
The voltage v across load and current i into the positive polarity are as follows (ω1 ≠
ω3 )
= + √2 !"# + √2 "$%# + √2& !"#& 'V
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 17 of 120
Version Code: PEQBTNC06
$ = ) + √2) !"# + √2)& !"#& − +& 'A
Calculate the following:
(a) The average power P supplied to the load
(b) The RMS value of $and (c) The power factor at which the load is operating
Q30.
A single phase half wave diode rectifier is designed to supply dc output voltage of
200 V and load resistance of 10 Ω. Calculate the average and RMS current ratings
of diode, PIV of diode and transformer for this circuit arrangement
Q31.
(a) In the circuit shown in Fig, The PMMC ammeter reads 10 A. Find the
inductance value. Also find volt meters reading if they are PMMC type
V1
A
220V
∼
V2
L
50Hz
Fig for Q31
(b) If all the meters in part (a) are replaced with MI type instruments, then find the
meter readings
Q32.
(a) In the circuit shown in Fig, Ideal PMMC instruments are placed. Find voltmeter
readings
V1
A
220V
∼
1µF
C
V2
50Hz
Fig for Q32
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
Page 18 of 120
(b) In case voltmeter 2 in part (a) is replaced by MI type, then find its reading
Q33.
A battery is to be charged by a single phase half wave diode rectifier. The supply
voltage is 30 V, 50 Hz and the battery emf is constant at 6 V. Determine,
(a) The resistance to be inserted in series with the battery to limit the charging
current to 4 A. Take a voltage drop of 1 V across the diode when it is ON
(b) PIV of diode
(c) In case battery capacity is 100 W.h, find the charging time in hours
Q34.
Find the time required to deliver a charge of 200 A.h through a single phase half
wave diode rectifier with an output current of 100 A (RMS) and with sinusoidal
input voltage. Assume diode conduction over a half cycle.
*Q35.
(a) A dc battery is to be charged through a resistor R from a single phase half wave
uncontrolled rectifier. For an ac source voltage of 230 V 50 Hz, find the value of
average charging current and supply power factor for R = 8 Ω and E = 150 V
(b) In case, if diode is replaced by SCR and fired continuously through a constant dc
signal, the repeat part (a)
(c) In case, SCR in part (b) is triggered after 1 ms from its forward bias point. Then
repeat part (b)
(d) Comment on all the calculations
Q36.
Fig for Q36
(a) For the single phase half wave rectifier shown find out the PIV rating of diode
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 19 of 120
Version Code: PEQBTNC06
(b) Will the required PIV rating change if a inductor is placed between the diode
and capacitor
(c) What will be the required VRRM rating if the capacitor is removed? Assume a
resistive load.
(d) The source of the single phase rectifier circuit has an internal resistance of 2 Ω.
Find out the required Non repetitive peak surge current rating of the diode. Also
2
find the i t rating of the protective fuse to be connected in series with the diode.
*Q37.
A single phase midpoint converter is shown in Fig, where we assume the
transformer is to be ideal and the dc side load to be represented by a current stiff
load. Calculate the VA rating of the transformer as a ratio of the average power
supplied to the load.
n:1:1
D1
Id
Vp=Vmsinωt
∼
D2
Fig for Q37
Q38.
A single bridge consists of one SCR and three diodes operating with a firing angle of
450. Find the average load current and power delivered to the load in case the load
consists of R = 8.356 Ω, L = 8 mH and E = 100 V. Assume the load current is
constant in the entire working range
Q39.
A single phase full converter feeds power to RLE load with R = 10 Ω, L = 6 mH and
E = 60 V. The ac source voltage is 220 V, 50 Hz. In case one of the four SCRs gets
open circuited due to fault, find the average value of load current by assuming the
load current as continuous and firing angle is 450.
Q40.
A three phase half wave phase controlled rectifier delivers power to a resistive load
of 10 Ω. Input to the rectifier is 400 V, 50 Hz three phase ac supply. Find power
delivered to the load for a firing angle of (a) 150 and (b) 600
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
Page 20 of 120
Q41.
A three phase half wave phase controlled rectifier is operated from a 3-ph 230 V, 50
Hz supply with load resistance of 10 Ω. An average output voltage of 50 % of the
maximum possible output voltage is required. Determine, (a) Firing angle of the
converter (b) Average and RMS values of load current
Q42.
A three phase half wave phase controlled rectifier is fed from a 3-ph, 400 V 50 Hz
source and is connected to load taking a constant current of 30 A. SCRs are having
a voltage drop of 1.9 V during their conduction. Calculate,
(a) Average value of load voltage for a firing angle of 300 and 600
(b) Average and RMS current ratings of SCRs as well as PIV of SCRs
(c) Power loss in each SCR
(d) In case, if freewheeling diode (FD) is connected across load, find the average
value of output voltage, average and RMS value of FD current for firing angles of
300 and 600
Q43.
A three phase half wave phase controlled rectifier is operating from a 3-ph, 400 V 50
Hz and delivers power to the armature of a dc motor with negligible resistance and
large inductor in the dc bus. The source transformer has DY-11 connection with
unity phase turns ratio. Back emf of the motor is 300 V. Find the firing angle of the
rectifier
Q44.
A three phase fully controlled rectifier is delivers a ripple free load current of 10 A
with a firing angle of 300. The average output voltage is 400 V. Find active and
reactive power input to the bridge and input power factor of the converter
Q45.
A battery consists of R = 5 Ω and E = 150 V is charging through a three phase half
wave phase controlled rectifier. Input voltage to the converter is 230 V (RMS) from
any line to neutral and firing angle for SCRs is 300. Find average current flowing
through the battery
Q46.
A three phase full converter is fed from 230 V, 50 Hz supply having source
inductance of 4 mH per phase. The load current is 10 A and ripple free
(a) Calculate the voltage drop in dc output voltage due to source inductance
(b) If dc output voltage is 210 V, find firing angle and overlap period
(c) In case, the bridge is made to operate as a line commutated inverter with dc
voltage of 210 V, find firing angle for the same load current
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 21 of 120
Version Code: PEQBTNC06
Q47.
A three phase half wave diode rectifier delivers power to an inductive load which
takes ripple free current of 100 A. The source voltage to the bridge is 3-ph 440 V, 50
Hz. Determine,
(a) The average and RMS current ratings of diode
(b) PIV of diode
(c) RMS value of source current
*Q48.
A battery with a nominal voltage of 200 V and internal resistance of 10 mΩ has to
be charged at constant current of 20 A from a 3-phase 220 V, 50 Hz AC power
supply. Which of the following converter circuit will give better performance in
terms of Distortion factor in source current, fundamental power factor, and total
input power factor? (i) 3-ϕ Full converter (ii) 3-ϕ Semi converter
Q49.
(a) For the same average DC output voltage of 100 V, calculate the PIV of SCR for
the following configurations (Consider α = 00)
(i) 1-ϕ full wave center tap converter (midpoint converter)
(ii) 1-ϕ full converter
(iii) 1-ϕ semi converter
(iv) 3-ϕ half wave converter
(v) 3-ϕ full converter
(vi) 3-ϕ semi converter
(b) From the above calculations, which configuration is having maximum and
minimum PIV rating for SCR
Q50.
Fig for Q50
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 22 of 120
Version Code: PEQBTNC06
Two six pulse converters, used for a bipolar HVDC transmission system (shown in
figure) are rated at 1000 MW, ±200 kV. Evaluate, the RMS current and peak
reverse voltage ratings for each of the SCRs
Q51.
A buck converter has an input voltage of 16 V. The required average output voltage
is 8 V and peak to peak ripple in output voltage is 10 mV. The switching frequency
of the converter is 25 kHz. If the peak to peak ripple in inductor current is limited
to 0.7 A. Determine, (a) Duty cycle ratio (b) Filter inductance (c) Filter capacitance
Q52.
The input voltage to a boost converter is 8 V. The required average output voltage is
16 V and the average output load current is 0.5 A. The switching frequency of the
converter is 30 kHz. If L = 160 µH and C = 380 µF, calculate, (a) Duty cycle ratio (b)
The peak to peak ripple in inductor current (c) The peak current of the switch (d)
The ripple voltage in capacitor
Q53.
The input voltage to a buck- boost converter is 10 V. The switch is operating with a
duty ratio of 0.3 and the switching frequency is 25 kHz. The filter inductance is 150
µH and filter capacitance is 220 µF. The average load current is 1.2 A. Determine,
(a) The average output voltage (b) The peak to peak ripple in output voltage (c)
The peak to peak ripple in inductor current (d) The peak and average current of the
switch
*Q54.
A switched mode power converter is shown in Fig. The switches S are ON during
DTs and the switches S´ are ON during (1-D)Ts
S
Ig
+
Vg
L
S′
V0+∆V0
IL+∆IL
S′
I0
S
C
R
Fig for Q54
(a) Evaluate the steady state performance of the circuit. Assume the switches,
inductors and capacitors are to be ideal
(b) Indicate how the voltage conversion ratio will be modified if the inductor has a
resistance of RL
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 23 of 120
Version Code: PEQBTNC06
*Q55.
Consider the circuit given in Fig. Carry out the steady state analysis for the same
and evaluate the following
Fig for Q55
(a) Output voltage (b) Average input current (c) Output power (d) Efficiency (e)
Power dissipation in the MOSFET and the diode
*Q56.
Figure P13 shows a boost converter cascaded by a buck converter. The switches S
−
and S are ON during DTS and (1-D)TS respectively.
Fig for Q56
(a) Evaluate the steady state currents in L1 and L2 in terms of I0 and D.
(b) Evaluate the steady state voltages across C1 and C2 in terms of Vg and D
(c) Evaluate the current ripples in L1 and L2
(d) Evaluate the voltage ripple in C1 and C2
Q57.
A DC-DC converter circuit is shown in Fig. It consists of on active switch (S1) and
three passive switches D1, D2 and D3. It has four energy storage elements - two
inductors (L1, L2) and two capacitors (C1, C2). Consider that the currents through
the inductor and voltage across the capacitors are all continuous. The switch S1 is on
during Ton and off during Toff . The duty ratio of S1 may be designated as D. The
switch drops may be taken to be zero during conduction.
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 24 of 120
Version Code: PEQBTNC06
Fig for Q57
(a) Indicate the duty ratios of the three diodes D1, D2 and D3.
(b) Evaluate the steady-state inductor currents (I1, I2) and the steady state
capacitor voltages (VC1, VC2).
(c) Evaluate the voltage conversion ratio Vo/Vg.
(d) Sketch the steady-state waveforms of (I1, I2; VC1, and VC2).
(e) Evaluate the ripple currents ∆I1 and ∆I2 in terms of Vg, D, L1, L2 and R.
(f) Evaluate the ripple voltages ∆VC1 and ∆VC2 in terms of Vg, D, L1, L2, C1, C2, and
R.
(g) Calculate L1, L2, C1 and C2 by considering the circuit data as Vg = 100 V, D = 0.6,
R = 12 Ω and Ts = 20 µs.
Assume ripple in capacitor voltage is 1% of its average value and ripple in inductor
current is 10 % of its average value
Q58.
In the buck converter shown the diode D has a lead inductance of 0.2µH and a
reverse recovery change of 10 µC at iD =10A.
Fig for Q58
Find peak current through active switch.
Q59.
The following Figure shows a PI controller and its asymptotic magnitude bode plot.
Select R1, R2, and C. make any suitable assumptions if necessary
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 25 of 120
Version Code: PEQBTNC06
Fig for Q59
*Q60.
A fly back converter operating at a duty ratio of 0.3 is shown in the following Fig.
The transistor ON state drop is 1 V. The diode ON state drop is 0.7 V. The
resistance of the inductor windings is 0.5 and 0.25 for the primary and secondary
respectively.
Fig for Q60
Evaluate the voltage conversion ratio and efficiency of the converter
Q61
In a flyback converter, the required output voltage is 100 V for a nominal input
voltage of 12 V. If the switch is operating at D = 0.5
(a) Find the turns ratio of flyback transformer. Assume voltage drop across switch
is 0.8 V and diode is 0.8 V
(b) Find minimum and maximum values of D, if input voltage varies from 10 to 14
V, by maintaining V0 be constant. Assume the switching frequency of 2 kHz
(c) Find the value of Ls on secondary winding so that secondary current is just
continuous at the minimum value of D calculated in part (b). Consider load
resistance of 100 Ω
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 26 of 120
Version Code: PEQBTNC06
Q62.
A fly-back converter is to be designed to operate in just-continuous conduction mode
when the input dc is at its minimum expected voltage of 200 volt and when the load
draws maximum power. The load voltage is regulated at 16 volts. What should be
the primary to secondary turns ratio (N1/N2) of the transformer if the switch duty
ratio is limited to 80 %. Neglect ON-state voltage drop across switch and diodes
Q63.
The average output voltage flyback converter is 24 V at a resistive load of 0.8 Ω. The
duty cycle ratio is 0.5 and switching frequency is 1 kHz. The ON state voltage drops
of BJT and Diode are VT = 1.2 V and VD = 0.7 V. The turns ratio of transformer is
=>
= 0.25. Find the efficiency of the converter
=
?
Q64.
Find maximum voltage stress of the switch in the primary winding and diode in the
tertiary winding if the forward converter-transformer has 10 primary turns and 15
tertiary turns and the maximum input dc voltage is 300 V
Q65.
If the turns ratio of the primary and tertiary windings of the forward transformer
are in the ratio of 1:2, what is the maximum duty ratio at which the converter can
be operated? Corresponding to this duty ratio, what should be the minimum ratio of
secondary to primary turns if the input dc supply is 400 V and the required output
voltage is 15 V. Neglect switch and diode conduction voltage drops.
Q66.
Fig for Q66
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 27 of 120
Version Code: PEQBTNC06
A forward converter is operating at the boundary of continuous and discontinuous
conduction. The switch is operating at 100 kHz. Assume µ = ∞ for the core so that
energy recovery winding is ignored. A load of 10 A at 20 V is being supplied.
∆AB
(a) Find the inductance value
(b) Find peak to peak ripple in output voltage as % of average output voltage (
AB
)
*Q67.
A forward converter operating at a duty ratio of 0.3 is shown in the following Fig.
The transistor while ON drops a voltage of 1.0 V, and the diode while ON drops a
voltage of 0.7 V.
Fig for Q67
Evaluate the output voltage and efficiency of the converter.
*Q68.
A forward converter operating at a duty ratio of 0.4 is shown in the following Fig.
Assume the components to be ideal.
Fig for Q68
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
Page 28 of 120
Sketch the following waveforms under steady state.
(a) Inductor current. (b) Secondary current. (c) Primary current. (d) Output
voltage.
Q69.
A single phase full bridge VSI is fed from 230 V dc. In the output voltage waveform,
only fundamental component of voltage is considered.
(a) Determine the RMS current ratings of switches and diode of the bridge for the
following types of loads:
(i) R = 2 Ω (ii) ωL = 2 Ω
(b) Find also the repetitive peak voltage that may appear across switches in part (a)
Q70.
A single phase full bridge VSI delivers power to RLC load with R = 3 Ω and XL = 12
Ω. The bridge operates with periodicity of 0.2 ms. Calculate the value of C so that
load commutation is achieved for the SCRs. Turn off time for thyristors is 12 µs and
consider factor of safety 2. Assume that load current contains only fundamental
component.
Q71.
A single phase full bridge VSI delivers power at 50 Hz to RLC load with R = 5 Ω, L
= 0.3 H and C = 50 µF. The dc input voltage is 220 V.
Evaluate,
(a) Expression for load current up to 5th harmonic
(b) Power delivered to load and fundamental power
(c) The RMS and peak currents of each switch
(d) Conduction time of switches and diodes by considering only fundamental
components
Q72.
A single phase full bridge VSI delivers power to a load of R = 12 Ω and L = 0.04 H
from a 400 V DC source. If the inverter operates at a frequency of 50 Hz, determine
the power delivered to the load for
(a) Square wave operation
(b) Quasi square wave operation with an on period of 0.6 of a cycle
(c) Two symmetrically spaced pulses per half cycle with an on period of 0.6 of a
cycle
Q73.
A single phase current source inverter (CSI) with ideal switches has the following
data:
Source current = 30 A, frequency = 500 Hz, and pure capacitive load = 20 µF
For this inverter, Evaluate:
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 29 of 120
Version Code: PEQBTNC06
(a) The circuit turn off time
(b) The maximum value of reverse voltage that appears across switches
Q74.
A single phase capacitor commutated CSI connected to the load R has the following
data:
R = 40 Ω, C = 50 µF, Source current = 40 A, frequency = 500 Hz. Evaluate,
(a) Express the load current as function of time and its value at t = 0 and t = T/2
(b) The circuit turn off time
Q75.
A 3-phase 1200 mode inverter feeds a star connected load of R = 5 Ω. DC source
voltage is 230 V and output frequency is 50 Hz.
(a) Express the line to line output voltage, line to neutral output voltage and line
current in fourier series up to 11th harmonic components.
(b) RMS values of line to line and line to neutral voltages
(c) RMS values of line to line and line to neutral voltages at fundamental frequency
(d) THD for line current
(e) Load power and average value of source current
(f) Average and RMS value of switch currents
Q76.
SCR T in the figure below is initially OFF and is triggered with a single pulse of
EE
EE
width 10 s. It is given that C = D F G μH and K = D F G μF. Assume latching and
holding currents are zero and initial conditions L and C are zero.
(a) Evaluate the conduction time of SCR T
(b) Voltage across device and capacitor after SCR is turned OFF
Fig for Q76
Q77.
A circuit employing current commutation as shown below has C = 20 µF and L = 3
µH. Initially capacitor is charged towards the source voltage (=230 V dc).
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 30 of 120
Version Code: PEQBTNC06
Fig for Q77
Determine the conduction time for auxiliary SCR (TA) and circuit turn off time for
main SCR (TM) in case constant load current is (a) 300 A and (b) 60 A
Q78.
In the circuit shown in the Figure below, has commutating elements L = 20 µH and
C = 40 µF are connected in series with the load resistance of R = 1 Ω.
Fig for Q78
Check whether self commutation or load commutation, would occur or not. Find also
conduction time of SCR
Q79.
For the circuit shown in Fig, (dv/dt) rating of thysristor T is 400 V/µs. and its
junction capacitance is 25 pF. Switch S is closed at t = 0.
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 31 of 120
Version Code: PEQBTNC06
Fig for Q79
(a) Calculate the value of Cs so that thyristor T is not turned on due to dv/dt
(b) In case maximum current through thyristor in above circuit is limited to 40 A,
determine the value of Rs
Q80.
For illustrating complementary commutation, the following circuit is employed
where Vdc = 200 V and R1 = 10 Ω.
Fig for Q80
(a) Find the value of capacitor so that T1 is commutated in 50 µs.
(b) It is required that SCR T2 is turned off naturally when current through it falls
below holding current of 4 mA. Find the value of R2.
Q81.
In the complementary scheme of commutation, Source voltage is 200 V dc, R1 = 10 Ω
and R2 = 100 Ω. Evaluate,
(a) Peak value of current through SCRs T1 and T2
(b) Capacitance value C if each SCR has turn off time of 40 µs. Take a factor of
safety 2. Justify if you make any assumptions
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 32 of 120
Version Code: PEQBTNC06
*Q82.
(a) A separately excited dc motor is represented in block diagram form as show in
Fig. Fill all the blocks with usual and your convenient notation
TWL
Vt(s)
?
Te(s)
Ia(s)
?
?
Q(s)
-Ea(s)
?
Fig for Q82-a
(b) From the block diagram given in part (a), find G1 (s) and G2 (s) as per the
definitions given below
#O "
#O "
M " = N
Q
V%W M " = N
Q
P "
XYZ "
RST UE
A[ UE
(c) Now, express G1 (s) in terms of machine mechanical and electrical time
constants which are defined as below:
]^ _
C^
\O =
V%W \b =
`R `a
]^
Where, KT and KE are torque and electrical constant
(d) The dc motor under consideration is having the following data.
T rated = 10 N-m ; Nrated = 3700 rpm KT = 0.5 N-m/A ; KE = 53V/1000 rpm;
Ra = 0.37 Ω, 6e = 4.05ms, 6m = 11.7 ms
If this motor is controlled from a power electronic converter, Evaluate the terminal
voltage required (in steady state) if motor is required to deliver a torque of 5 Nm at
a speed of 1500 rpm
(e)
If G1(s) in the given statement is expressed as M " =
/cd
efD
gh
kg
Ge g
ij
ij
Then find the values of D and ωn by using data given in part (d). Now, plot
the asymptotic magnitude and phase plot of G1 (s) by means of bode plot. Then find
out Phase margin and Gain margin with approximate hand calculations or using
MATLAB
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 33 of 120
Version Code: PEQBTNC06
(f) A PI controller is used in the speed control loop (shown in Fig) to obtain the
transfer function of the following form:
*
Q (s)
*
Σ
Kp
ω (s)
Fw(s)
−
1
2o
"
1 + " D# G + #p
p
Where D = 0.5 and ωn = 300 rad/s
lY " =
#"
=
# ∗ "
ω(s)
1
s
Q(s)
Fig for Q82-f
(g) Draw the bode plot of closed loop transfer function lq " = q∗ f if the gain KP =
qf
60 is used for the proportional position regulator.
(h) What is the bandwidth of the above closed loop systems (Hint: This is indication
of speed of the response)
Q83.
A 200 V, 1450 rpm, 100 A separately excited dc motor has an armature resistance of
0.04 Ω. The machine is driven by a 3-ϕ half controlled rectifier operating from a 3-ϕ
220 V, 50 Hz supply. The motor operates at rated speed and rated load torque.
Assuming continuous conduction to evaluate (a) Firing angle of the converter (b)
RMS value of fundamental input current (c) Fundamental power factor (d) THD in
source current
Q84.
A 3-ϕ full converter is feeding a 100 HP, 400 V, 1500 rpm separately excited dc
motor having armature resistance of 0.1 Ω and filter choke which is connected in
series with armature to maintain constant current of 175 A. The bridge is connected
to a 3-ϕ, 400 V, 50 Hz supply. The source has an inductance of 0.5 mH and back emf
constant of the machine is 0.25 V/rpm. Evaluate (a) Firing angle (b) Overlap angle
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
Page 34 of 120
Q85.
A 200 V, 875 rpm, 150 A separately excited dc motor has an armature resistance of
0.06 Ω. It is fed from a single phase fully controlled rectifier with an ac source
voltage of 220 V, 50 Hz. Assuming continuous conduction calculate,
(a) Firing angle for rated motor torque and at 750 rpm
(b) Firing angle for rated motor torque and at (-500) rpm
(c) Motor speed for α = 1600 and rated torque
Q86.
A 220 V, 1500 rpm, 50 A separately excited dc motor with armature resistance of
0.5 Ω is fed from a 3-ϕ full converter. Available ac source has a line voltage of 440 V,
50 Hz. A Y-∆ connected transformer is used to feed the full converter so that motor
rated terminal voltage equals the rated voltage when converter firing angle is zero.
(a) Calculate the turns ratio of transformer (between phase windings of primary
and secondary)
(b) Determine the firing angle of the converter when (i) motor is running at 1200
rpm and rated torque (ii) motor is running at -800 rpm and twice the rated torque
Q87.
A 230V, 960 rpm and 200 A separately excited dc motor with armature resistance of
0.02 Ω. The motor is fed from basic chopper circuit which provides both motoring
and baking operation. The source has a voltage of 230 V. Assuming continuous
conduction
(a) Calculate the duty ratio of the chopper for motoring operation at rated torque
and 350 rpm
(b) Calculate the duty ratio of the chopper for braking operation at rated torque and
350 rpm
(c) If maximum duty ratio of chopper is limited to 0.95 and maximum permissible
motor current is twice the rated. Calculate maximum permissible motor speed
obtainable without field weakening and power fed to the source
(d)If motor field is also controlled in part (c), calculate filed current as a fraction of
its rated value for a speed of 1200 rpm
Q88.
A four pole, 10 HP, 460 V Induction motor is supplying its rated power to a
centrifugal pump type of load at a 60 Hz frequency. Its rated speed is 1746 rpm
(a) Calculate its speed, slip frequency, and slip when it is supplied by a 230 V, 30 Hz
source
(b) If the starting torque is required to be 150 % of the rated torque for a constant
air gap flux, find the starting frequency that need to apply to the motor
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 35 of 120
Version Code: PEQBTNC06
Objective Questions
Practice Test - 1
Q1.
Which of the following power semi conductor device is popularly using in wind and solar
power converters?
(A) SCR
(B) GTO
(C) IGBT
(D) BJT
Q2.
Which of the following MOSFET is most suitable as power electronic switch?
(A)
N – channel depletion MOSFET
(B)
P – channel depletion MOSFET
(C)
N – channel enhancement MOSFET
(D)
P – channel enhancement MOSFET
Q3.
Consider the following statements and choose the correct option.
Statements about ideal switches:
1.
In OFF state, current flowing through ideal switch is zero.
2.
In ON state, voltage across ideal switch is zero.
3.
The ideal switch need finite energy to switch ON/OFF or OFF/ON.
4.
The switch can be turned ON and OFF instantaneously
(A) All statements are true
(B) Only 1, 2 and 4 true
(C) Only 1, 2 and 3 are true
(D) None
04.
Match List – I (Transfer characteristics) and List – II (Devices) and select the correct
option.
List – I
List – II
ID
(1)
(2)
ID
(P)
(Q)
D
G
G
VGs
ID
(3)
(4)
S
S
VGs
D
(R)
ID
(S)
G
G
S
VGs
(A)
(C)
P
1
4
Q
3
1
R
2
3
S
VGs
S
4
2
(B)
(D)
P
2
1
Q
1
2
D
R
4
3
S
3
4
Q5.
A BJT with a device drop of 1.2 V is carrying a current which is shown below.
i
10A
http://electrical-mentor.blogspot.in/
0
Copyright © Reserved 2014
(µs)
5
15 20
30
Page 36 of 120
Version Code: PEQBTNC06
(A) 5 W
(B) 5.5 W
(C) 6 W
(D) 8 W
Q6.
Which of the following device (s) will be considered as bipolar and unidirectional switch.
1. SCR
2. Symmetrical GTO
3. Asymmetrical GTO
4. BJT in series with diode
(A) Only 1, 2, 3
(B) 1, 3 and 4 (C) 1, 2, 3, 4
(D) 1, 2 and 4
Q7.
During forward conduction, a thyristor has static V – I characteristic as shown by a straight
line in given figure. Find the average power loss in the thyristor in case thyristor is
carrying a constant current of 80 A for one half cycle.
Ia
(A) 70.4 W
(B) 11.6 W
(C) 40 W
(D) 50 W
100A
Va
0.8V 2.0V
Q8.
The switching waveform for a power transistor are shown in below figure
In case, Ics=80A, Vcc=220V, ton=1.5 µs and toff=4 µs then power loss in the transistor during
turn ON is
ic
(A) 8.8 W
(B) 23.46 W
(C) 32.6 W
(D) None
ICs
t
υCE
VCC
t
ton
tof
T
Q9.
For a power diode, the reverse recovery time is 3.9 µs and the rate of diode current decay is
50 A/µs. For a softness factor of 0.3, Find the storage charge.
(A) 350 µC
(B) 292.5 µC
(C) 150 µC
(D) 200 µC
Q10.
Turn on time of an SCR in series with RL circuit can be reduced by
(A) Increasing R
(B) Decreasing R
(C) Increasing L
(D) Decreasing L
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 37 of 120
Version Code: PEQBTNC06
Q11.
For dynamic equalizing circuit used for series connected SCRs, the selection of C is based
on
(A) Reverse recovery characteristics
(B) Turn – on characteristics
(C) Turn – off characteristics
(D) Rise time characteristics
The circuit symbol for GTO is
Q12.
A
A
A
A
G
G
G
G
K
(Q)
(P)
K
(R)
G
(A) P, Q and S are correct
(C) Only Q & S are correct
K
(S)
(B) Only P & Q are correct
(D) All are symbols are GTO
Q13.
Input power factor of any rectifier circuit can be defined as below:
(A) IPF =
Actual power input to the rectifier
Apparent power input to the rectifier
(B) IPF = (distortion factor) × (displacement factor)
(C) IPF =
DPF
1 + (THD )
2
(D) All the above
Q14.
A single phase semi converter is operated from 230 V, 50 Hz AC supply and operated with a
firing angle of
π
. The load on the converter is highly inductive with a resistance of 15.53 Ω
3
and load current is ripple free. RMS value of the fundamental source current will be.
(A) 10A
(B) 9A
(C) 5A
(D) 7.8A
Q15.
A single phase fully controlled rectifier is supplying power to a purely resistive load of 100
Ω and the bridge is triggered with α= 90°. The source voltage is 200 V, 50 Hz. The power
delivered to the load is given by
(A) 1 kW
(B) 2 kW
(C) 3 kW
(D) 4 kW
Q16.
The DC equivalent circuit of a single phase full converter is shown below. The net average
output voltage is available across terminals X and Y. Find the source inductance by
assuming input frequency of 50 Hz.
x
I0
http://electrical-mentor.blogspot.in/
2Vm
cosα
π
= 220V
V0=200V
yCopyright © Reserved 2014
Page 38 of 120
Version Code: PEQBTNC06
(A) 1 H
(B) 0.2 H
(C) 0.01 H
(D) none
Q17.
Match List – I (1-φ bridge configuration) with list – II (Average output voltage) and choose
the correct option. (Assume I0 is constant)
List – I (1-φ
φ bridge)
(P) with 4 SCRs
List – II [Average output voltage]
2Vm
π
2Vm
2. V0 =
cosα
π
V
3. V0 = m [3 + cos α ]
2π
V
4. V0 = m [1 + cos α ]
1. V0 =
(Q) with 2 SCRs + 2 diodes
(R) with 1 SCR + 3 Diodes
(S) with 4 diodes
π
(A)
(C)
P
2
4
Q
4
2
R
1
3
S
3
1
(B)
(D)
P
2
4
Q
4
2
R
3
1
S
1
3
Q18.
Which of the following PE converter circuits can be used to operate the DC machine in
regenerative braking mode? Select your choices from the given options.
1. 1-φ full converter
2. 1-φ semi converter
3. 3-φ full converter
4. 3-φ semi converter
5. Type E chopper
6. 1-φ full bridge VSI
Options:
(A) 1, 3, 5 and 6 only
(B) 1, 2, 3, 4 only
(C) 1, 3 and 5 only
(D) All can be used
Q19.
In controlled rectifiers the nature of load current i.e. whether load current is continuous or
discontinuous
(A)
does not depend on type of load and firing angle delay
(B)
depends on type of load and firing angle delay
(C)
depends only on the type of load
(D)
depends only on the firing angle delay
Q20.
If α=90o in the following circuit, the RMS value of the output voltage is
(A) 230 V
(B) 230 2 V
http://electrical-mentor.blogspot.in/
(C) 115 V
(D) 115 2 V
Copyright © Reserved 2014
Version Code: PEQBTNC06
Page 39 of 120
T1
230V ∼
D3
D1
R
T2
+
υ0
−
50Hz
D4
D2
Q21.
A 230 V, 50 Hz one-pulse SCR controlled converter is triggered at a firing angle of 40o and
load current extinguishes at an angle of 210o. Find the circuit turn off time (in second) if
load R=5 Ω and L=2 mH
(A) 120
(B) (1/120)
(C) 60
(D) (1/60)
Q22.
A four quadrant operation of the DC machine requires
(A)
Two full converters is series
(B)
Two full converters connected back to back
(C)
Two full converters connected in parallel
(D)
Two semi converters connected back to back.
Q23.
The output ripple frequency of 3-φ semi converter depends on
(A) Input frequency
(B) firing angle
(C) source inductance
(D) both A&B
Q24.
A 3-φ full converter feeds power to a resistive load of 10 Ω. For a firing angle of 45o, the load
takes 1 kW. Find the magnitude of the line Voltage
(A) 100 2 V
(B) 100 3 V
(C) 200 V
(D) 200 3 V
Q25.
A 3-φ half wave controlled rectifier is delivering power to highly inductive load with
continuous load current. Which of the following equation can be used to find average output
voltage when firing angle is 450?
(A) V0 =
3 3Vmp
[1 + cos α]
2π
3Vmp 
π 

(C) V0 =
1 + cos α + 

2π 
6 

(B) V0 =
3 3Vmp
cos α
2π
3 3Vmp 
π 

(D) V0 =
1 + cos α + 

2π 
6 

Q26.
A 3-φ full converter is supplying power to an inductive load and current of 31.42 A. The
RMS value of fundamental input current will be
(A) 24.5 A
(B) 31.42A
(C) 25.65A
(D) None
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 40 of 120
Version Code: PEQBTNC06
Q27.
In 12 pulse rectifier, the lowest harmonic in the input current will be
(A) 3rd
(B) 5th
(C) 7th
(D) 11th
Q28.
Which of following rectifier can only be used as both 3-pulse and 6-pulse rectifier?
(A) 1-φ full converter
(B) 3-φ full converter
(C) 3-φ half wave converter
(D) 3-φ half controlled converter
Q29.
In step up chopper (boost converter), the average output voltage when duty cycle ratio=1.
(A)
infinity
(B)
zero
(C)
equal to source voltage
(D)
less than infinity but greater than source voltage
Q30.
In which of the following DC-DC converter, the nature of power conversion will be of
current to voltage type
(A) Buck converter
(B) Boost converter
(C) Buck-Boost converter
(D) fly back converter (Isolated type)
Q31.
A chopper circuit shown in the figure is operating at a switching frequency of 25 kHz with a
duty cycle ratio of 0.25 The peak to peak ripple in the inductor current will be
sw
+
12V
(A) 1 A
(B) 1.8 A
120µH
V0
−
(C) 0.5 A
(D)0.8 A
Q32.
The fundamental difference between transformer and inductor used in power conversion
circuits is
(A)
both have same purpose
(B)
Inductor is used to smoothen energy flow where as transformer will provide
electrical isolation and voltage levels matching.
(C)
Inductor will store energy whereas transformer will not store energy
(D)
both B & C are correct
Q33.
A type – A chopper is operating at 2 kHz from a 100 V dc source has a load time constant of
5 ms and load resistance of 10 Ω. Find the max value of inductor current for a mean load
voltage of 50V.
(A) 5.25A
(B) 5.025A
(C) 5.125A
(D) 4.875A
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 41 of 120
Version Code: PEQBTNC06
Q34.
A diode is connected in series with LC circuit as shown in fig. Assume the capacitor is
initially charged to a voltage of -50 V.
sw
D
L=0.2mH
C=10µF
230V
The voltage across capacitor at the time diode turns off is
(A) 230 V
(B) -280 V
(C) -220 V
(D) 510 V
Q35.
Find the conduction time of SCR in the following circuit. Assume L and C are initially
relaxed
100Ω 5mH
1µF
300V
(A) 0.314 ms
(B) 0.314 µs
(C) 3.14 ms
(D) none
Q36.
A complementary commutation scheme is shown in fig.
R1
R2
T1
T2
Vdc
If Vdc = 200V, R1 = 10Ω and R2 = 100Ω determine peak value of current through T2.
(A) 24A
(B) 42A
(C) 21A
(D) 12A
Q37.
In the circuit shown in figure, if Vdc = 200 V, C = 4 µF and L = 16 µH and R = 20 Ω. The
peak value of current through T1 and D can respectively be
T1
Vdc
D
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 42 of 120
Version Code: PEQBTNC06
(A) 110 A, 100 A
(B) 110 A, 10 A
(C) 10 A, 110 A
(D) 100 A, 110 A
Q38.
In 3-φ full bridge VSI, which of the following voltage will have 3rd harmonic in 180°
operation
(A) line voltage
(B) phase voltage
(C) pole voltage
(D) none
Q39.
A single phase full bridge VSI is feeding power to a resistive load of 5 Ω. The fundamental
output voltage is found to be 200 V (rms). Find the rms value of switch and diode currents
at fundamental frequency
(A) Isw,1 = 20 2A and ID,1 = 10 2A
(B) Isw,1 = 20 2A and ID,1 = 0A
(C) Isw,1 = 10 2A and ID,1 = 0A
(D) Isw,1 = 10 2A and ID,1 = 5 2A
Q40.
The source voltage of a 3 – φ full bridge VSI is 200 V. The rms value of phase voltage in
120° operation will be
(A) 40.82V
(B) 20.41V
(C) 81.64V
(D) 141.42V
Q41.
A single phase full bridge VSI is operating in 180° square operation. The phase angle
between the pole voltages is 45°. The RMS value of the output voltage between two poles is
A B
100V
(A) 100 V
(B) 100 ×
π
V
4
(C) 200 V
(D) 50 V
Q42.
A single phase full bridge VSI has a source voltage of 200 V. The load consists of RLC in
series where R = 1Ω, ωL = 7Ω and
1
= 6Ω . Identify the fundamental component of load
ωC
current from the following.
(A) 180 sin (ωt + 45°)
(B) 180 sin(ωt - 45°)
(C) 127.3 sin(ωt-45°)
(D) 127.3 sin(ωt+45°)
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 43 of 120
Version Code: PEQBTNC06
Q43.
The operating points of three power electronic switches on VI plane is shown below
Consider the following statements regarding the switches P, Q, R
1. P is most suitable for VSI
2. P is most suitable for CSI
3. Q is the most suitable for VSI
4. Q is most suitable for CSI
5. P, Q and R can be used in either VSI or CSI
Now, select the correct option from the following
(A) only 2 & 3 are correct
(B) only 1 & 4 are correct
(C) 2, 3 and 5 are correct
(D) All are correct
Q44.
In single pulse modulation used in PWM inverters, Vdc is the input dc voltage. For
eliminating third harmonic, the magnitudes of rms value of fundamental component of
output voltage and pulse width are respectively
4Vdc
,60 o
π
2 2
(C)
Vdc , 60 o
π
(A)
(B)
(D)
6
π
Vdc ,120 o
4
Vdc ,120 o
π
Q45.
A PWM inverter is capable of producing the following type of output voltage
(A)
variable in magnitude and frequency
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 44 of 120
Version Code: PEQBTNC06
(B)
(C)
(D)
variable voltage, fixed frequency
Fixed voltage, variable frequency
Fixed voltage, fixed frequency
(2)
Q46.
Consider the following circuits
Load
(1)
Load
SCR
∼
∼
(3)
TRIAC
(4)
Load
Load
∼
MOSFET/
∼
BJT/
IGBT
From the above circuits which one can be considered as AC voltage regulator
(A) only 1 & 3 (B) only 1, 2 and 3
(C) only 1
(D) All circuits
Q47.
An induction motor is required to run at a very low speed around 25 to 40 rpm from 50 Hz
source. Which of the following circuit is most suitable for this application
(A) step up cycloconverter
(B) inverter
(C) step down cycloconverter
(D) All the above
Q48.
A load consisting of R = 10 Ω and ωL = 10 Ω is being fed from 230 V, 50 Hz source through a
1-φ AC voltage controller. For a firing angle delay of 45°, the rms value of load current will
be
(A) 23A
(B)
23
2
A
(C) >
23
2
A
(D) <
23
2
A
Q49.
A dc motor is driven from a 3-φ full converter draws a dc line current of 10A with negligible
ripple. The rms value of thyristor current will be
(A) 10 A
(B) 7.07 A
(C) 5.77 A
(D) 17.32 A
Q50.
Which of the following chopper circuit can be used to drive the dc motor in regenerative
braking is
1. Type – A
2. Type – B 3. Type – C 4. Type – D
5. Type - E
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 45 of 120
Version Code: PEQBTNC06
(A) only 2
(B) only 2, 3 and 4
(C) all except 1
(D) all circuits
Q51.
A separately excited dc motor fed through a single phase semi converter runs at a speed of
1200 rpm. when ac supply voltage is 220 V, 50 Hz and the motor counter emf is 90 V. The
firing angle is 90° and armature resistance is 1 Ω. Find the average armature current
(A) 7A
(B) 8A
(C) 9A
(D) 10A
Q52.
In the speed control of dc motors in servo applications, where response time is very critical,
which of the following PE converter circuit is most preferable
(A) DC-DC converter (switched mode)
(B) phase controlled rectifiers
(C) Dual converters
(D) Any one is suitable
Q53.
A separately excited dc motor is driven from a 3-φ full converter. The armature current is
ripple free. Find 3rd and 5th harmonic components of line currents as a % of the fundamental
component respectively
(A) 0% and 20%
(B) 0% and -20%
(C) 20% and 0%
(D) 0% and -40%
Directions: the following items consist of two statements; one labeled as Assertion (A) and
the other as Reason (R). You are to examine these two statements carefully and select the
correct answers to these items using the codes given below.
(A)
Both A and R are true and R is the correct explanation of A
(B)
Both A and R are true but R is not the
correct explanation of A.
(C)
A is true but R is false
(D)
A is false but R is true
Q54.
Assertion (A): In SCR, latching current corresponds to turn ON process.
Reason (R): In SCR, holding current corresponds to turn OFF process.
Q55.
Assertion (A): Semi converters are not suitable for braking applications.
Reason (R): V0 is always +ve for all values ‘α’ in semi converters.
Q56.
Assertion (A): In rectifiers, input power factor can be improved by using freewheeling
diode across load.
Reason (R): With freewheeling, some power can fed back to the source.
Q57.
Assertion (A): The output ripple frequency in 3-φ rectifiers is less than 1-φ rectifier.
Reason (R): Output voltage of 3-φ rectifier will have more number of pulses than 1-φ
rectifier.
Q58.
Assertion (A): MOSFET is most suitable switch in DC – DC converters than SCR.
Reason (R): Ripple content is less when tsw is high in DC-DC converters.
Q59.
Assertion (A): MOSFET is most preferable switch CSI.
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
Page 46 of 120
Reason (R): Switches with anti parallel diodes should not be used in CSI.
Q60.
Assertion (A): 1 - φ triac can be used in fan regulators.
Reason (R): Voltage control method is effective in low power 1-φ induction motors.
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 47 of 120
Version Code: PEQBTNC06
Objective Questions
Practice Test - 2
Q1.
Consider the following statements
(1) The ON state voltage drop of GTO is higher than that of SCR.
(2) GTO can be used for more switching frequency than that of SCR.
(3) SCR can handle large currents than GTO.
(4) GTO can turn ON and OFF from same gate.
Choose the appropriate options.
(a) Only 1 is true
(c) 1,2 and 4 are true
(b) 1,2 and 3 are true
(d) all statements are true
Q2.
In the following single-phase diode Rectifier circuit, the average and RMS current
rating of the diode will be respectively
(a) IDav = 10.35 A and IDr = 12.68 A
(c) IDav = 7.32 A and IDr = 10.35 A
(b) IDav = 14.64 A and IDr = 10.35 A
(d) IDav = 10.35 A and IDr = 7.32 A
Q3.
The output voltage of a 3-phase voltage source inverter contains 5th and 7th
harmonics. Assume the output is balanced.
If Va = V1m sin (ωt) +V5m sin (5ωt) +V7msin(ωt) then Vb can be expressed as
2π 

(a) Vb = V1msin  ωt −
 +V5msin(5ωt)+V7msin(7ωt)
3 

2π 
2π 
2π 



(b) Vb = V1m sin  ωt −
 +V5msin  5ωt +
 + V7 m sin  7ωt −

3 
3 
3 



2π 
2π 
2π 



(c) Vb = V1m sin  ωt −
 + V5 m sin  5ωt −
 + V7 m sin  7ωt +

3 
3 
3 



(d) None of these
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 48 of 120
Version Code: PEQBTNC06
Q4.
In a single phase semi converter, THD in source current is found to be 31%. Then
firing angle could be
(a) 30°
(b) 40°
(c) 45°
(d) 60°
Q5.
A three phase half wave phase controlled rectifier is operated from a 3-phase star
connected 400 V, 50 Hz supply and the load resistance is R = 10 Ω If it is required to
obtain an average output voltage of 50% of maximum possible output voltage then
the converter circuit need to operate at α =
(a) 135°
(b) 67.7°
(c) 60°
(d) 30°
06.
A boost converter is shown in the figure. The DC source voltage is 100 V and load
resistance is 10 Ω. Assume that the inductor has an internal resistance of 0.5 Ω.
The range of duty cycle which can give the stable operation of the converter circuit
is
(a) 0< D< 0.78
(b) 0.78<D< 1
(c) 0.8<D<1
(d) 0<D<1
Q7.
A single phase full bridge VSI has a source voltage of 200 V DC. The load consists
1
of RLC in series where R = 1Ω, ωL = 6Ω and
= 7Ω . Identify the fundamental
ωC
component of the load current from the following.
(a) 180 sin (ωt+45°)
(b) 180 sin(ωt−45°)
(c) 127.3 sin (ωt−45°)
(d) 127.3 sin (ωt+45°)
Q8.
A separately excited DC Motor is driven from a 3 phase full converter. The
armature current is ripple free. Find 3rd and 5th harmonic components of line
currents as a percentage of the fundamental component respectively
(a) 0% and 20%
(b) 0% and -20%
(c) 20% and 0%
(d) 0% and −40%
Q9.
A buck converter circuit is shown in the figure
Vdc is input voltage and v0 is the output voltage. Consider iL and v0 as state
variables. If R = 1Ω, L = 1H and C = 0.1 F then the nature of the response will be
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 49 of 120
Version Code: PEQBTNC06
(a) Under damped
(c) Over damped
(b) Critically damped
(d) Undamped
Common Data for Questions 10 & 11
A single phase semi converter is operated from 230 V, 50 Hz AC supply and with a
firing angle of (π/4). The load on the converter is highly inductive with a resistance
of 17.765 ohm and load current is ripple free
Q10.
The RMS value of the freewheeling current is
(a) 10 A
(b) 3.33A
(c) 5 A
Q11.
RMS value of fundamental source current will be
(a) 10 A
(b) 9 A
(c) 5 A
(d) None
(d) 8.315 A
Q12.
A single phase full bridge inverter has RLC load of R = 4Ω, L = 35 mH and C = 155
µF. The DC input voltage is 230 V and the fundamental output frequency is 50 Hz.
The conduction time of the diodes is (consider only fundamental components)
(a) 6.264 ms
(b) 2.5 ms
(c) 3.736 ms
(d) None
Q13.
In the following rectifier circuit, the load current will be maximum at ωt =
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 50 of 120
Version Code: PEQBTNC06
(a) 90°
(b) 135°
(c) 45°
(d) Load current does not have max
Q14.
Regarding Buck-Boost converter, which of the statement is true
1. It will operate as Buck converter when 0<D<0.5
2. It will operate as Boost converter when 0.5<D<1.0 and stable throughout this
range
3. It will operate as Boost converter when 0.5<D<Dmax and unstable if D>Dmax.
Select the options as below:
(a) only 1 & 3 are true
(b) all are true
(c) only 1&2 are true
(d)
None of these
Q15.
In the given circuit a thyristor converter is feeding a resistor R, The power
consumed by R in the absence of SCR is P. In the presence of SCR, the power
consumed by R at α = 90° would be reduced by a factor of
(a) 1/2
(b) 1/4
(c) 1/6
(d) 1/12
Q16.
A 3-φ full converter feeds power to a resistive load of 10 Ω. For a firing angle of 30°
the load takes 5 kW. Find the magnitude of per phase input supply voltage
(a) 230 V
(b) 188 V
(c) 108 V
(d) 335.5V
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 51 of 120
Version Code: PEQBTNC06
Q17.
A 1−φ fully controlled rectifier is supplying power to a resistive load of 35 Ω as
shown in figure. If the bridge is triggered at 90°, the average current drawn by the
load is given by
(a) 0 A
(b)
2A
(c) 2 2A
(d)
1
2
A
Q18.
A 3−φ Fully controlled rectifier is operated from a 3−φ star connected 400 V, 50 Hz
AC supply and the load resistance is R = 10 Ω. A large inductance is connected in
series with the load to maintain ripple free load current. If it is required to obtain
an average output voltage of 86.66 % of the maximum possible output voltage. Find
the firing angle
(a) 30°
(b) 60°
(c) 45°
(d) 90°
Q19.
A Power converter is shown in the figure has two power switching devices namely X
and Y. The source voltage is 50 V. The inductor current is steady 5 A without any
ripple.
On the V-I plane, identify the correct operating points of switches from the given
options.
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
Page 52 of 120
(d) None
Q20.
A single phase AC voltage Regulator has a load resistance of 10 Ω. Input voltage is
200 V, 50 Hz. If SCRs are triggered at π/2 with symmetrical triggering scheme, find
the RMS value of SCR current.
(a) 0A
(b) 20 A
(c) 4.5 A
(d) 10 A
Q21.
The following chopper circuit is operating at a switching frequency of 1 kHz with a
duty cycle ratio of 50%. Assume a voltage drop of 2 V across the switch when it is
ON. Find the converter circuit efficiency.
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
Page 53 of 120
(a) 95%
(b) 98%
(c) 100%
(d) 99%
Q22.
A buck converter is shown in the figure. The switch is operating at 25 kHz. From an
experiment, it is observed that ∆V0 = 20 mV and ∆IL = 0.8 A.
For an average output voltage of 5V, find the value of capacitance
(b) 145. 8 µF
(c) 150 µF
(d) None
(a) 200 µF
Q23.
A single phase full bridge VSI is feeding power to a resistive load of 2 Ω. The
fundamental output voltage is found to be 230 V (rms). Find the RMS value of
switch and diode currents at fundamental frequency
(a) Isw1 = 81.33 A
ID1 = 0A
(b) Isw1 = 54.818 A
ID1 = 0A
ID1 = 17.328 A
(c) Isw1 = 54.818 A
(d) None of these
Q24.
The reverse recovery time of a diode is 3 µs and the rate of fall of the diode current
is 3 A/µs. The peak value of reverse current of the diode is given by
(a) 135 A
(b) 90 A
(c) 100 A
(d) 8100 A
Q25.
In a resonant commutation circuit, supply voltage is 200 V. Load current is 10 A
and the device turn off time is 30 µs. The ratio of peak resonant current to load
current is 1.5. Find the value of L&C
(a) L = 2.66 mH & C = 1.5 µF
(b) L = 1.5 mH & C = 0.266 µF
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 54 of 120
120
Version Code: PEQBTNC06
(c) L = 0.266 mH & C = 1.5 µF
(d) None of these
Common Data for Questions 26 &
27
Two 3-phase bridge rectifiers are connected as shown below.
Q26.
The total number of pulses in the output voltage at the terminals A & B is
(a) 6
(b) 12
(c) 3
(d) 24
Q27.
The lowest harmonic in the input current is
(a) 3
(b) 5
(c) 11
(d) 13
Q28.
A 220 V, 60 A DC series motor having a combined resistance of armature and field
of 0.15 Ω is controlled in regenerative braking mode through a DC-DC converter.
The DC source voltage is 220 V. Motor constant is 0.05 V-s/rad. The average motor
armature current is rated and ripple free. Find the speed during regenerative
braking
(a) 32.33 rad/s
(b) 47 rad/s
(c) 74 rad/s
(d) None
Q29.
A 3-phase full bridge VSI delivers power to a resistive load from a 450 V DC source.
For a star connected balance load of 10 Ω/ph. Find the RMS value of the load
current under 180° conduction mode
(a) 18.708 A
(b) 13.23 A
(c) 18.371 A
(d) 21.213 A
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
Page 55 of 120
Q30.
A thyristor is used in an application carrying half sinusoidal current of period of 1
msec and a peak of 100 A as shown in fig. The thyristor may modeled during
conduction to have a constant voltage drop of 1.1 V and a dynamic resistance of 8
mΩ. Evaluate the average conduction loss in the device
(a) 35 W
(b) 0.4 W
(c) 55 W
(d) 20 W
Q31.
The wave form in the following figure shows the periodic current thorugh a powerswitching device in a switching converter application.
A BJT with a device drop of 1.2 V and MOSFET with an ON state resistance of 150
mΩ are considered for this application. The conduction loss in BJT and MOSFET
are respectively
(a) 6W & 3.75 W (b) 6 W & 6.67 W (c) 3.75w & 6 W
(d) 6.67 W&6 W
Q32.
In the following circuit, MOSFET Q is switched at 100 kHz with a duty ratio of 0.5.
MOSFET is having an ON state resistance of 1 Ω when it is ON. Find average
conduction losses in MOSFET
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 56 of 120
Version Code: PEQBTNC06
(a) 20.41 W
(b) 41.67 W
(c) 12.5 W
(d) 4.1667 W
Q33.
A 3-phase full converter supplying power to inductive load with ripple free current
is shown in fig. All positive group devices are represented with P1P2,P3 and all
negative group devices are represented with N1, N2, N3 as shown below.
By assuming υYB = Vml sinωt and α = 60° the following load voltage is obtained.
Which of the following statement is true as per the given output voltage waveform
(a) x1 = v RB , x 2 = vYR & x3 = vYB
(b) x 1 = v RY , x 2 = v YB & x 3 = v BY
(c) x1 = vRB , x2 = vYR & x3 = vBY
(d) x 1 = v RY . x 2 = v YB & x 3 = v BR
Q34.
A forward converter is operating at boundary of continuous & discontinuous
conduction. The switch is operating at 100 kHz. Assume µr of core is ∞ so that
energy recovery in winding can be neglected. A load of 10 A at 20V is being supplied
as shown is fig. Find peak to peak ripple in output voltage
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
Page 57 of 120
(a) 0.25 V
(b) 2.5V
(c) 0.025 V
(d) 1.25 V
Q35.
Which of the following converter circuit representation is equivalent to Buck-Boost
converter?
Q36.
A power electronic switch is realized by SCR and diode as shown below. Identify the
static operating points on V-I plane
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
Page 58 of 120
Q37.
Which of the following switch can be used as four quadrant switch [used in matrix
converter]
Q38.
The center-tap full-wave single-phase rectifier circuit uses 2 diodes as shown in the
given figure. The rms voltage across each diode is
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 59 of 120
Version Code: PEQBTNC06
(a) 790.7 V
(b) 395.3 V
(c) 280 V
(d) 201.3 V
Q39.
In the following resonant pulse communication scheme, find the voltage across main
SCR when it gets turned off.
(a) 460 V
(b) 177 V
(c) 174.37 V
(d) 230 V
Q40.
An RL load is connected to DC voltage source of 220 V through a diode as shown
below. A free wheeling diode is connected a cross the load to recover the trapped
energy. Assume that switch is closed for 100 µs and then opened. Find the final
energy stored in the inductor by assuming negligible load resistance
(a) 1 J
(b) 0.5 J
http://electrical-mentor.blogspot.in/
(c) 1.5 J
(d) 1.1 J
Copyright © Reserved 2014
Page 60 of 120
Version Code: PEQBTNC06
Objective Questions
Practice Test – 3
Q1.
A single phase full converter charges a battery which offers a constant value of E = 12V. A
resistor R is inserted to limit the battery charging current. The supply voltage to the bridge
is 40 V, 50 Hz, Ac. Consider a voltage drop of 1 V conducting SCRs. Assume that pair of
SCRs fired continuously. Find the value of R in case battery charging current is 10 A.
(a) 2 Ω
(b) 2.4 Ω
(c) 4 Ω
(d) 8 Ω
Q2.
In the circuit shown in Fig. switch S is open and a current of 20 A is flowing through FD, R
& L. If switch S is closed at t = 0, the expression for current flowing through the switch is
given by
sw
+
R
10Ω
L
10mH
FD
220V
20A
−
Fig.
(a) i( t ) = 22 + 2e −1000 t
(b) i( t ) = 2 − 22e −1000 t
(c) i( t ) = 22 − 2e −1000 t
(d) i( t ) = 0
Q3.
The switching waveform for a power electronic switch is shown in fig. Find the energy loss
during turn ON and turn OFF transistors
isw
80A
t
vsw
220V
t
tON
tOFF
1.5µs
4µs
(a) E ON = 4.4 mJ : E OFF = 13.2 mJ
(b) E ON = 4.4 mJ : E OFF = 11.73 mJ
(c) E ON = 13.2 mJ : E OFF = 4.4 mJ
(d) E ON = 11.73 mJ : E OFF = 4.4 mJ
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 61 of 120
Version Code: PEQBTNC06
Q4.
A power diode is modeled in ON state as shown in figure. Consider it is ideal in blocking
and switching durations
A
K
⇒
A
K
Rd
Vd
The above diode is capable of dissipating 75W in ON state. For square wave operation, it is
rated for peak current of 100A and 135 A at duty ratios of 0.5 and 0.33 respectively. Find
the diode parameters Rd and Vd
(a) R d = 5mΩ : Vd = 0.98V
(b) R d = 0.98Ω : Vd = 5mV
(c) R d = 3mΩ : Vd = 0.5V
(d) Not a valid model.
Hint: Diode current waveforms are shown below
i
i
I=100A
I
I=135A
I
D=0.5
D=0.33
Q5.
A composite switch used in a power converter is shown in figure. The periodic current
through the switch is also shown. Evaluate the total power loss in the composite switch.
i
12A
t
5
10
(µs)
20
25
Rds=0.1Ω
B
A
VON=0.8V
(a) 4.8 W
(b) 3.6 W
(c) 8.4 W
(d) None
Q6.
A battery is to be charged from single phase un controlled bridge rectifier. On full
discharge, the battery voltage is 10.2 V and on full charge it is 12.7 V. The battery internal
resistance is 0.1 Ω. Find the input voltage to the rectifier so that the battery charging
current under full charging condition is 10%
(a) 14.415 V
(b) 41.41 V
(c) 24.41 V
(d) 48 V
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 62 of 120
Version Code: PEQBTNC06
Q7.
A 3 - φ full converter bridge is supplying power to a purely resistive load of 10 Ω. The input
to the bridge is 400 V, 50 Hz, 3 - φ AC. If the bridge is triggered at
α = 900. Find power
delivered to the load.
(a) 0 W
(b) 1384 W
(c) 1117.6 W
(d) 524 W
Q8.
A single phase diode bridge is delivering power to purely resistive load of 10 Ω. The input to
the bridge is 230 V, 50 Hz, 1 - φ AC. Find the rms value of second harmonic component of
load current.
(a) 0A
(b) 8.76 A
(c) 9.76 A
(d) 10.76 A
Q9.
A single phase full converter feeds power to RL load with R = 6Ω, L = 6mH. The AC source
voltage is 230 V, 50 Hz. In case one of the 4 SCRs gets open circuited due to a fault, find the
average load current on the assumption of load current is continuous by taking α = 600
(a) 12.94 A
(b) 8.62 A
(c) 17.25 A
(d) None
Q10.
Consider the following circuits
(P)
(R)
(Q)
MOSFET/
∼
∼
BJT/
∼
From the above circuits, which one will operate as AC voltage regulator
(a) only P
(b) only P & Q
(c) P, Q and R (d) None
Common Data for Questions 11 & 12
An AC voltage regulator and diode bridge rectifier are connected together as shown below.
T1
D1
D3
i0
+
A
R
230V
∼
50Hz
B
T2
D4
V0
−
D2
Q11.
Find power delivered to load in case α = 600 and R = 10 Ω
(a) 4251.8 W
(b) 2125.8 W
(c) 1062.75 W
http://electrical-mentor.blogspot.in/
(d) 8502 W
Copyright © Reserved 2014
Page 63 of 120
Version Code: PEQBTNC06
Q12.
In case, if D3 in above circuit is open circuited then find power delivered to load.
(a) 4251.8 W
(b) 2125.8 W
(c) 1062.75 W (d) 8502 W
Q13.
The input voltage given to a converter and current drawn by converter are expresses as
r = 300 "$%100s + 100 "$%300s
s
s
s
)r = 10 "$% D100s − G + 5 "$% D300s + G + 2 "$% D500s − G
3
4
6
Find input power factor of the converter
(a) 0.44 lag
(b) 0.6 lag
(c) 0.707 lag
(d) 0.522 lag
Q14.
A separately excited DC motor is driven from a 3-phase full converter. The armature
current is ripple free. Find 3rd and 5th harmonic components of line currents as a % of the
fundamental component respectively.
(a) 0% and 20%
(b) 20% and 0%
(c) 0% and −20%
(d) 0% and 40%
Q15.
A pure inductance of
2
Henry is connected to a single phase full bridge and bridge is
π
operated with firing angle of 1200. Input supply to the bridge is 200V, 1φ, 50 Hz AC. Find
the peak current flowing through load.
(a) 1A
(b) 0.5 A
(c) 2A
(d) unstable system
Q16.
In the flowing DC – DC converter circuit, if the switch is operated at 20 KHz with duty
ratio of 0.5. Find the energy transferred from V1 to V2
V1
(a) 940 J
(b) 94 J
(c) 0.47 J
(d) 0.047 J
+100V
L=100µH
V2
sw
D
+300V
Q17.
A 3 - φ full converter supplying power to an inductive load with ripple free load current is
sown in the figure.
P1
P2
P3
I0
+
V0
−
R
Y
B
N1
http://electrical-mentor.blogspot.in/
N2
N3
Copyright © Reserved 2014
Page 64 of 120
Version Code: PEQBTNC06
By assuming VYB = Vm sin ωt and with α = 600. The following output voltage is obtained.
V0
x2
x1
60
x3
ωt
0
0
360
180
0
Which of the following statements are true as per the given output voltage waveform.
(b) x1 = VRB , x2 = VYR , x3 = VBR
(a) x1 = V RY , x 2 = VYB , x3 = V RB
(d) x1 = V RB , x 2 = VYR , x3 = V BY
(c) x1 = VRY , x 2 = VYB , x3 = VBR
Q18.
In single phase semi converter, THD in source current is found to be 31%. Then firing angle
could be
(a) 300
(b) 400
(c) 450
(d) 600
Q19.
In the following DC – DC converter circuit the switch is operating at frequency 10 kHz.
When the switch is at position 1, the inductor stores energy
for a period of 50 µs and release energy is 20 µs when the switch is moved to position 2.
Find ratio of V1 to V2.
1
V1
(a)
2
5
(b)
5
2
2
V2
L
(c)
7
2
(d)
2
7
Q20.
A 3 - φ voltage source inverter has a star connected load of R = 5Ω, and L = 23 mH. The
inverter fundamental frequency is 60Hz. And DC input voltage is 220 V consider that
inverter output voltage is free form 11th and higher harmonics. Find the rms value of line
current in load
(a) 9.91 A
(b) 13.02A
(c) 7.75 A
(d) None
Q21.
Consider the following DC – DC converter circuit.
100mH
220V
D
sw
+
10µF
1KΩ
V0
−
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 65 of 120
Version Code: PEQBTNC06
Switch is operating at a duty cycle ratio of 0.5. Find the settling time for the response to

reach within 2% tolerance band H int : t s =

(a) 50 ms
4
ξω n

s

(b) 79 ms
(c) 100 ms
(d) 10 ms
Statement for Linked Answer Questions 22 & 23 (Numerical Type)
A DC Motor transfer function defined as follows:
G 1 (s) =
ω(s )
Vt (s ) T
=
L =0
1
K E s .Tm .Te + sTm + 1
[
2
]
Where
Tm = mechanical time constant =
Te = electrical time constant =
RaJ
KTE
La
Ra
KT & KE = are torque and emf constants respectively
Machine data is as follows:
T rated = 10 N-m ; nrated = 3700 rpm
KT = 0.5 N-m/A ; KE = 53V/1000 rpm
Ra = 0.37 Ω, Te = 4.05ms, Tm = 11.7 ms
Q22.
This motor is controlling from a power electronic converter. Determine the terminal voltage
in steady state if the motor is required to deliver a torque of 5 N-m at a speed of 1500 rpm
…………
Q23.
If G1(s) in the given statement is expressed as
G 1 (s ) =
1/ K E
2sD s 2
1+
+
ω n ω 2n
Then D and ωn will be respectively
Q24.
A single phase full bridge converter is used to regulate DC output voltage. The rms value of
AC input is 230V, 50Hz. And load current I0 = 10A. In below figure Y-axis represents active
and reactive powers with and without free wheeling diode and X-axis represents firing
angle.
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
2500
Version Code: PEQBTNC06
2000
Page 66 of 120
1500
1000
500
0
0
-500
30
45
60
75
90
105 120 135 150 165 180
-1000
-1500
-2000
-2500
W
Match the following
List – I
1. W
2. X
3. Y
4. Z
Codes:
1
(a)
A
(b)
D
(c)
B
(d)
A
X
Y
Z
List - II
A.
Active power with freewheeling diode
B.
Reactive power without freewheeling diode
C.
Active power without freewheeling diode
D.
Reactive power with freewheeling diode
2
B
C
C
B
3
C
B
A
D
4
D
A
D
C
Q25.
A single phase full bridge VSI has an RLC load with R = 10 Ω, L = 31.5 mH and C =112
µF.The inverter fundamental output frequency is 50 Hz and DC input voltage is 220V. This
VSI is controlling with sine triangular PWM technique with 5 pulses per half cycle and
harmonics of order higher than 9 and above are negligible effect. Find power delivered to
the load (consider harmonic power if present).
(a) 1639.55 W
(b) 885.45 W
(c) 1770.9 W
(d) 3279.1 W
Q26.
A commutation circuit for a thyristor is shown in fig. Determine the available turn off time
of the circuit if Vdc = 100 V, R = 10 Ω and C = 10 µF voltage across capacitor before SCR T2
is fired, is Vdc with polarity as shown.
T1
C
−
+
VCO
R
Vdc
T2
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 67 of 120
Version Code: PEQBTNC06
(a) 69.3 µs
(b) 69.3 ms
(c) 0.693 µs
(d) 0 s
Common Data for Questions 27 & 28
A 220 V, 1450 rpm, 100 A separately excited DC motor has am armature resistance of 0.04
Ω. the motor is driven by a 3 - φ half controlled converter operating from a 3 - φ, 220 V, 50
Hz AC supply. The motor operates at rated speed and rated load torque. Assume
continuous conduction.
Q27.
RMS value of fundamental component of input current
(a) 70 A
(b) 63.87A
(c) 26.67 A
Q28.
Total harmonic distortion in the input current is
(a) 60%
(b) 31%
(c) 48.43%
(d) None
(d) 65.5%
Statement for Linked Answer Questions 29 & 30
A single phase full bridge VSI is operating in 1800 square operation mode with source
voltage 100 V – 0 – 100 V. The phase angle between the pole voltages is 450.
Q29.
RMS value of the output voltage between two poles at fundamental frequency
(a) 34.44 V
(b) 50 V
(c) 68.9 V
(d) 127.28 V
Q30.
RMS value of switch current at fundamental frequency for a resistive load of 10 Ω
(a) 4.87 A
(b) 6.89 A
(c) 9.74 A
(d) 2.445 A
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 68 of 120
Version Code: PEQBTNC06
Objective Questions
Practice Test - 4
Q1.
In which of the following devices, conductivity modulation phenomenon is absent
(A) Power diode
(B) SCR
(C) Power MOSFET
(D) Power BJT
Q2.
A power MOSFET rated for 25 A, carries a periodic current of 20 A as shown in
figure. The ON-state resistance of the MOSFET is 0.1 Ω. What is the average ONstate loss in the MOSFET? i
20 A
π
(A) 15 W
2π
(B) 20 W
3π
4π
5π
ωt
(C) 25 W
(D) 10 W
Q3.
In the following circuit, the ammeter reads 12.68 A when it is of MI type. What
would be the ammeter reading if it is of MC type
D
A
230 V,
∼
0.1 H
50 Hz
(A) 12.68 A
(B) 10 A
(C) 10.35 A
(D) Ammeter will damage
Q4.
The average input and output voltages of chopper circuit are 100 V and 50 V
respectively. The inductor current wave form is as shown in figure.
iL
IL max
100 50
L L
Ton
http://electrical-mentor.blogspot.in/
t
T
Copyright © Reserved 2014
Page 69 of 120
Version Code: PEQBTNC06
What could be the chopper circuit?
(A) step down chopper
(C) step up/down chopper
(B) step down/up chopper
(D) step up chopper
Q5.
Which of the following converter circuit operation will be unstable for large duty
cycle ratios?
(A) Buck converter
(B) Boost converter
(C) Buck-Boost converter
(D) Both (B) & (C)
Q6.
A diode in anti parallel with the controlled switch like IGBT is used in VSI to
(A) prevent reversal of dc link current
(B) control a non-unity power factor load at the output.
(C) protect the circuit against accidental reversal of dc bus polarity.
(D) None of the above
Q7.
A single phase full bridge VSI is fed from DC source such that fundamental
component of output voltage is 200 V. Find average value of source current when
load is R = 10 Ω. Assume all harmonics absent.
(A) 10A
(B) 20 A
(C) 40 A
(D) None
Q8.
A 3-phase induction motor operates under constant volt/hertz control. At 50 Hz
supply, the motor current and its pf are 30 A and 0.3 lag at the time of starting.
These values at the time of starting at 25 Hz supply would be
(A) Motor current > 30 A and pf >0.3 (B) Motor current < 30 A and pf < 0.3
(C) Motor current < 30 A and pf > 0.3 (D) Motor current > 30 A and pf < 0.3
Q9.
A single phase ACVR feeds a pure resistive load. Each SCR conducts for angle of γ
when operating at firing angle α > 90o. If the load is replaced by a pure inductance
keeping the firing angle same as before. The conduction angle of each SCR would be
γ
γ
(A) 2 γ
(B) γ
(C)
(D)
2
3
Q10.
The fundamental input power factor of single phase full converter is
1
when it is
2
operating with a firing angle of 40o. Assume the load current is constant and ripple
free. What is the overlap angle in degree ________
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 70 of 120
Version Code: PEQBTNC06
Q11.
The fourier series for voltage and current of a power electronic circuit in pu are
given below:
υ(t) = 1.2 sin(ωt)+0.33sin(3ωt) +0.2sin(5ωt)
i(t) = 0.6 sin(ωt+30o)+0.1sin(5ωt+45o)+0.1sin(7ωt+60o)
The average power is given by (in p.u)
(A) 0.312
(B) 0.007
(C) 0.32
(D) Voltage equation is wrong.
Q12.
Two six pulse converters, used for a bipolar HVDC transmission system (shown in
figure) are rated at 1000 MW, ±200 kV
+
+
υ01
–
+
υ0
υ02
–
–
The RMS current rating of each thyristor will be
(A) 2500 A
(B) 1443.4 A
(C) 2041.2 A
(D) 0 A
Q13.
In the following chopper circuit, The IGBT Q is switched at 10 kHz. The circuit is
operated in steady state at the boundary of continuous and discontinuous inductor
current. If IGBT has a constant voltage drop of 0.5 V when it is ON, Find the
conduction loss in IGBT.
Q
100 V
(A) 853 W
(B) 16 W
iL
100 µH
(C) 32 W
400 V
(D) None
Q14.
In a boost converter, the duty ratio is adjust to regulate the output voltage V0 at 48
V. the input voltage varies in a wide range from 12 to 36 V. the maximum output
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 71 of 120
Version Code: PEQBTNC06
power is 120 W. 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.
(B) 12 µH
(C) 20 µH
(D) < 9 µH
(A) 9 µH
Q15.
In the circuit shown in fig, capacitor C is initially charged to Vdc = 200 V with
polarity as indicated. Find the circuit turn off time for main SCR (T1) after it is
voltage commutated by SCR (TA) load current is constant at 40 A and C = 10 µF.
T1
–
C
+
200 V
(A) 20 µs
(B) 50 µs
Load
TA
(C) 25 µs
(D) Insufficient data
Q16.
A separately excited dc motor is fed from 220 V dc source through a step down
chopper operating at 400 Hz. The load torque is 30 N-m at a speed of 1000 rpm. The
motor has ra = 0, La = 20 mH. Machine constant is 1.5 V-s/rad. Calculate minimum
and maximum values of armature current.
(A) Ia,max = 22.8 A and Ia min = 17.2 A
(B) Ia,max = 22.8 A and Ia min = 18.48 A
(C) Ia,max = 17.2 A and Ia min = 11.57A
(D) None
Q17.
A 220 V, 1450 rpm, 100 A separately excited dc motor has an armature resistance of
0.05 Ω.the machine is driven by a 3–φ half controlled converter operating from a 3–
φ, 220 V, 50 Hz supply. The motor operates at rated speed and rated load torque.
Assume continuous conduction. % THD in source current at AC mains is given by
(A) 70.5 %
(B) 68.2 %
(C) 31 %
(D) None
Q18.
A single phase full bridge VSI delivers power to a series RLC load with R = 2 Ω and
ωL = 10 Ω. The periodic time T = 0.1 ms. What value of C should be used to has load
commutation consider tq = 10µ s and factor of safety 1.5.
(A) 12.48 µF
(B) 1.248 µF
(C0 12.48 mF
(D) None
Q19.
A 3-ph square wave inverter feeds a balanced 3-phase inductance type of load. The
worst case load phase current (peak magnitude) is expected to be 100 A and the
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 72 of 120
Version Code: PEQBTNC06
worst case dc input voltage is expected to be 600 V. The diodes of inverter will be
subjected the following peak voltage and current stresses.
(A) 600 V, 100 A
(B) 600 V, 70.7 A
(C) 424 V, 70.7 A
(D) 424 V, 100 A
Q20.
A PWM inverter is operated from a dc link voltage of 600 V. the maximum rms line
to line voltage (fundamental component) will be less than or equal to
(A) 600 V
(B) 490 V
(C) 467 V
(D) 405 V
Q21.
A single phase full bridge VSI, fed from 230 V dc is connected to load of R = 10 Ω
and L = 0.03 H. fundamental frequency of inverter output is 50 Hz. Inverter is
operating with a quasi-square wave output with an on-period of 0.5 of a cycle. Find
the rms value of load current at fundamental frequency.
(A) 15.07 A
(B) 10.65 A
(C) 9.15 A
(D) None
Q22.
A single phase force commutated CSI is shown in the figure.
T3
T1 50 µ F
A
40 A
B
T4 40 Ω
T2
Output frequency of the inverter is 500 Hz. Find circuit turn off time
(B) 750.25 µs
(C) 200 µs
(A) 438.14 µs
(D) None
Q23.
dV
rating of SCR is 350 V/µs and its junction
dt
capacitance, is 20 pF. Switch S is closed at t = 0. Calculate to value of Cs so that
dV
SCR T is not turn ON due to
dt
20 Ω
For the circuit shown in figure,
S
D
250 V
Rs
Cj
T
Cs
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 73 of 120
Version Code: PEQBTNC06
(A) 0.0357 µF
(B) 0.025 µF
(C) 2.5 µF
(D) 3.5 µF
Q24.
A 3-ph fully controlled rectifier is supplying power to a purely resistive load of 10 Ω.
Input to the bridge is 230 V, 50 Hz. The power delivered to the load in watts when α
= 90o is _______ (in W)
Q25.
A single phase CSI (with ideal switches) has the following data:
I = 30 A, f = 500 Hz and load of pure capacitance = 20 µF. Find the peak value of
reverse voltage that appears across thyristors in volts ______
Q26.
In the following circuit. The RMS value of load current in amps by assuming α= 90o
is ______ (in amp)
D1
T1
D3
A
200 V ∼
Io
 10 

Ω
 2
T2
B
50 Hz
D4
D5
Common Data for Questions 27 & 28
A three phase fully controlled converter operates from a 3-phase 230 V, 50 Hz
supply through a Y- ∆ transformer to supply a 220 V, 600 rpm, 500 A separately
excited DC Motor. The motor has an armature resistance of 0.02 Ω.
Q27.
What should be the transformer turns ratio (from primary to secondary) such that
the converter produces rated motor terminal voltage at 0o firing angle.
(A) 1:1.2267
(B) 1:1.5
(C) 1:1.789
(D) 1:0.8152
28.
The above converter is now used to brake the motor regeneratively in the reverse
direction. If thyristors are to be provided with a minimum turn off time of 100 µs,
what is the maximum reverse speed at which rated braking torque can be
produced?
(A) 600 rpm
(B) 657 rpm
(C) 700 rpm
(D) 625 rpm
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 74 of 120
Version Code: PEQBTNC06
Statement for Linked Answer Questions 29 & 30
The chopper below controls a dc machine with an armature inductance La = 0.2 mH.
The armature resistance can be neglected. The armature current is 5 A. fs = 30 kHz
and D = 0.8.
id
ia
La
i0
+
Vdc +–
υ0 =V0
+ ea
–
–
Q29.
The average output voltage V0, equals to 200 V. Calculate the ripple in armature
current.
(A) 8.332 A
(B) 2.5 A
(C) 6.667 A
(D) 1.6675 A
Q30.
The load on the dc machine is now reduced and Ia,max = 2 A. The current is now
discontinuous. What is the back emf voltage Ea,?
(A) 250 V
(B) 200 V
(C) 175 V
(D) 235 V
Q31.
A single phase asymmetrical semi-converter feeds an RL load with R = 10 Ω and
large L so that load current is current stiff. The source voltage to the bridge is 200
V, 50 Hz. For a firing angle of 30° , the RMS value of diode current will be _______
A.
Q32.
Consider a current stiff load with free wheeling diode across it. Which of the
following rectifier circuit will have 3 pulses in the output voltage?
(B) 3-φ full converter with α = 90°
(A) 1-φ full converter with α = 90°
(C) 3-φ semi converter with α = 90°
(D) Both (B) & (C)
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 75 of 120
Version Code: PEQBTNC06
Q33.
In the 3-φ inverter circuit shown, the load is balanced and gating scheme is 120°
conduction mode. All the switches are ideal. If the dc source voltage is 200 V, the
power consumed by 3-φ load is
S1
S3
S5
S4
S6
S2
15 Ω
15 Ω
Vdc = 200V
15 Ω
3-ph Balanced load
A) 5.33 kW
(B) 3 kW
(C) 4 kW
(D) 1.33 kW
Q34.
The source voltage to the AC voltage regulator shown in the figure is 220 V, 50 Hz.
If SCRs are triggered at 120° symmetrically in every half cycle, then RMS value of
the fundamental component of source current (is) will be
T1
is
220 V
∼
i0
j 1.2284 Ω
T2
50 Hz
(A) 57 A
(B) 70 A
(C) 40 A
(D) 0 A
Q35.
In a Buck-Boost converter operating at 20 kHz, L = 0.05 mH. The output capacitor
C is sufficiently large and Vdc = 15 V. The output is to be regulated at 10 V and the
converter is supplying a load of 10 W. Calculate the duty ratio D
(A) 0.4
(B) 0.2
(C) 0.3
(D) None
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 76 of 120
Version Code: PEQBTNC06
Interview questions
ELECTRICAL ENGINEERING
Note:
The following interview questions have been collected from so many students from
my previous batches and consolidated as below.
Try to get answers from yourself by referring standard text books, have a group
discussion among your friends. Do not directly go to teacher without some home work
as you may not get the same questions. It is recommended the reader to get answers
on your own by applying fundamentals so that you can expose deeper into the subject
and you can increase your confidence levels. Cracking these exams is very simple if
you have confidence with subject knowledge
IIT Kanpur
Power & Control
• Questions on control circuits & stability issues.
• Basics of VSI (Voltage source inverter) in details, its wave form for various
types of loads, its analysis for different modes of conduction.
• Devices (power electronic), their characteristics eg. Difference between a
signal and power devices, Bundle conductors, corona losses, surge
impedance
• Induction motor T-S characteristics, Synchronous motors and generators,
applications of dc motors.
• They gave different references & inputs to comparator and ask about the
output.
• They asked circuit diagram of buck, buck-boost and various wave forms.
• They gave some circuit in which there was freewheeling diode and asked
about its operation in detail
• Equivalent diagram of induction motor and significance of different
elements in equivalent diagram.
• On stability analysis (power system)…. They gave a circuit in which a
single line to ground fault had occurred. They asked to draw the swing
curve, indicate accelerating and decelerating areas, asked how rotor angle
is increased physically, what is the physical meaning of accelerating and
decelerating areas.
• Explain Ferranti effect. They asked in very details about Ferranti effect.
• Why hydro power generators rotate at slower speed than thermal (steam)
generators?
• Questions on load-flow studies & electric drives, from the basics but tricky.
• One of the Profs drew a circuit of dc-dc converter and asked to identify that.
Be very clear about buck, buck-boost, boost converters, inverters.
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
•
•
•
Page 77 of 120
In control system they asked very basic questions of bode plot, nyquist,
polar plot and asked to draw them with a given transfer function (not very
tough).
They’ll first ask your area of interest. Depending on that they’ll ask
questions on that field and related fields. Nevertheless, prepare for the
whole syllabus.
They’ll ask you to answer each question by drawing circuit, writing
equations...etc……on the black board.
IIT Delhi
Power Electronics, Electrical Machines & Drives
• Questions on Induction motor principles & BE/B.Tech project
• Draw four quadrant chopper circuit without SCR?
• Why more than six pulse is used in HVDC converters?
• Torque-slip characteristics of Induction Motor
• Questions about switching devices for PE circuits
• How to select switch based on application?
• V & inverted V curve of SM.
• Inverters and Devices in much detail, they used to ask the typical rating of
the devices.
• First they asked about B.Tech project in very much details,
• They asked about linear system, linear equations……then asked whether
Swing equation is linear or not and asked each and every question out of it.
• Asked other very basic questions from transmission line, distribution
system etc.
• If you hold a good GATE score (/rank) and fairly good SGPA, they will
primarily focus on your B.Tech project. If it is not the case, they will ask
everything you have read in B.Tech (that was the case with one of my
friend).
Control & Automation
• What are advanced controllers?? (they intended to ask about fuzzy and
ANN)
• Comment on stability of a given Nyquist plot.
• Parameters in root locus
• Something about bode plot
• Basic control system & Lead controller
• Define stability, and stable system
• What is impulse response?
• Linear and nonlinear systems, causal and anti-causal systems
• Different stability criteria
• Transfer function definition and why initial conditions are zero while
defining transfer function
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 78 of 120
Version Code: PEQBTNC06
•
Derive the transfer function for DC motor and draw its block diagram.
While answering this question, they will mostly observe your approach
IIT Bombay
Microelectronics (RA)
• Working of NMOS.
• Some questions on OP-AMP.
• Questions mostly on B Tech project
Power Electronics & Power Systems (RA)
• Some Questions in Mathematics.
• Draw the full bridge rectifier diagram.
• Explain commutation process.
• Questions on basic circuit theory.
• Questions based on operation and speed control of induction motor,
transformer OC and SC test.
Control & Computing (RA)
• Questions were not from the B.Tech syllabus.
• Majority were on Computer & its peripherals. ( for a Network Admin in
electrical dept)
IIT Madras
Communication Systems (M.S. RA)
• Basic questions on DFT, DTFT.
• Purpose of transforms like DFT, DTFT etc.
Power Electronics
• They will start with basic concepts of Inductor and capacitor.
• Most of the questions will be on DC-DC converters and their various
quadrants of operation.
• Explain the basic difference between 4-Q chopper and single phase VSI
• If we give 3-ph AC supply to IM and TF. Rotation magnetic field will not
produce in TF but in IM, it will produce. Explain with phasor diagrams
•
IIT Kharagpur
Power Electronics
• Draw torque-slip characteristics of IM and locate the normal stable operating
point. Now if the two of the supply terminals suddenly reversed, then where
it will shift after t=0+
• Draw the buck converter circuit diagram and derive output voltage equation
from energy conversion principles. If resistor is removed and capacitor is
ideal, then what is the output voltage
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 79 of 120
Version Code: PEQBTNC06
•
•
•
Draw separately exciter DC motor circuit. Now, connect the motor in such a
way that it should rotate at half the rated speed at no load. Do not touch field
winding as it is constant
What is the time constant of 1/(s+1) and give any physical example for such
system. Obtain the step response and what is the physical meaning of the
response plot
Draw series RLC circuit and by taking I as reference, draw phasor diagram
for VR, VL, VC, Vtotal
IIT Roorkee
Written exam
Imp point: No negative marking.
• Electrical Drives: Questions on constant power and constant torque drives
...e.g they changed the armature voltage and were asking about the
dynamics of the machine.
• Instrumentation: Questions on Increasing the range of
Ammeter/Voltmeter.
• Questions on Laplace transform.
• Control Systems: Bode plot , Stability...(easy questions, just refer
Nagarath Gopal and Ogatta books of Control Systems).
• Application of KCL in a circuit containing inductor and/or capacitor.
• Switching circuits (capacitor and Inductor transient and steady state
concepts).
• Result was a combination of GATE score & the exam conducted.
• Direct counseling thereafter.
IISc Bangalore
Electrical Communication Engineering [M.S.]
• Basic questions on mathematics, signals and system, communication and
probability theory
• Only basic fundamentals and no questions from advanced technology. ( Very
strict on fundamentals)
• Panel often asks to explain things in a physical sense rather than putting
some equations and solving them.
For example:
Physically how will you interpret the process of convolution?
• Syllabus of the interview will be clearly mentioned before the interview.
• There will be some 5 topics mentioned and out of those one has to select 3
topics and questions will be asked from these topics only.
Electrical Engineering
• Questions on B Tech project & electronics.
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 80 of 120
Version Code: PEQBTNC06
Interview questions from IIT Madras (Dec-2013)
(Note: The following questions are collected from Madhuri Depuru, while she attending for
MS interview in IIT Madras during the winter session for 2013-14 academic year)
1. If you are given a pure sinusoidal wave and a perfect DC is required from it. What is
the circuit you will employ?
2. Draw the most simple rectifier circuit you know
3. Derive Load current equation for RL load in rectifier circuits using both Laplace
transform and differential equations
4. What is the most advanced PE equipment employed in motor drives as per your
knowledge?
5. Can a full wave bridge rectifier be used as an inverter? Can you use it without
changing the source?
6. How can you justify pulsating DC output you are getting out of rectifier can serve
the purpose of rectification?
7. What is the physical form inverter we are using in homes?
8. What happens if an inductor of small value if connected to the source given in VSI?
9. Even though DC motors have efficient speed control capabilities, we are using
Induction motors for most of the applications. Why?
10. Why supply frequency is only 50 or 60 HZ?
11. Why not diode bridge circuit cannot be used as an inverter?
12. Draw the waveforms for RL load in VSI circuits?
13. Draw the waveform for RC load? Will you use RC load parallel combination or series
combination? Explain in detail
14. What happens if 10A rms, current source is given to LC parallel circuit where initial
stored energies in L and C are zero?
15. What are the practical applications of KCL, KVL and ohms laws?
16. Is there any particular situation in circuit, ohm’s law cannot be applicable?
17. What is the need of cyclo converters?
18. Explain everything you know about three phase cyclo converter by drawing as much
as waveforms you can draw. (Note: Professor in the panel asked her to draw
waveforms on board…She realized the pain I am having daily in my class….I am
kidding… I am enjoying daily in work)
19. If we are happy with half wave rectifier circuits, why full wave rectifier circuits are
invented?
20. What are the typical ratings of SCR?
21. Do you know about Matrix converter?
22. Tell me the practical applications of PE you have seen? (My 3 hours introduction
class is the answer for this question)
23. What is DC transformer?
24. Will PE equipment affect the stability of any machine employed in electric drives?
25. What is the resistance between primary and secondary windings of the transformer?
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
Page 81 of 120
26. Why only semiconductor devices in PE circuits will operate as switches?
27. Do you know anything about secondary breakdown of BJT?
28. What happened if mechanical switches are used in place of SCRs in converters?
29. Finally why do you want to join in power electronics why not in power systems in
M.S/M.Tech?
Written test questions:
1. Explain SCR characteristics as bipolar and unidirectional switch?
2. A single phase full converter input is sinusoidal voltage. And load current is 4 A of
constant. Find ripple factor in output voltage, fundamental rms source current,
fundamental power factor, THD. Derive all the equations used
3. Explain with derivations, how to find active power, pf when voltage and current are
having harmonics
4. A constant DC voltage source is connected to and RL draw load through a diode.
Draw current and voltage across all components. Derive all the equations
One important question: Finally, one of the professors in panel asked her “Who is your PE
faculty and where did you learn power electronics”
As the candidate answered all the questions, interview panel appreciated her
understanding in the subject
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
Page 82 of 120
One more student (Anvesh Reddy) interview experience during M.Tech
admission in IIT KGP and BARC in 2013
I am giving here directly in his hand writing to motivate you
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
http://electrical-mentor.blogspot.in/
Page 83 of 120
Copyright © Reserved 2014
Page 84 of 120
Version Code: PEQBTNC06
Some more standard interview questions in Electrical
Engineering:
1. What is the basic difference between Transformer and Inductor? What are the
applications of these two in power electronic converter circuits?
2. What do you understand by step down and step up transformers?
3. Draw the flux waveform in a transformer when its primary is excited by voltage
waveform of (a) Sinusoidal (b) Square (c) Trapezoidal
4. How can you measure the losses in a transformer? How can you minimize these
losses?
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 85 of 120
Version Code: PEQBTNC06
5. Is it possible to operate a single phase 110 V, 50 Hz transformer satisfactorily at
200 Hz? If yes, explain how?
6. How are the transformer losses affected if pf of a given load is varied?
7. How is voltage regulation of transformer affected by a change in its operating
frequency?
8. Two mutually coupled coils act as an ideal transformer. Find the nature of its
reluctance.
9. What happens in a transformer if its core is made from high permeability
ferromagnetic material?
10. What are the basic difference between hot rolled and CRGO laminations used in
a transformer?
11. What are the three basic principles for the electromechanical energy conversion?
12. For an electromechanical energy conversion process, reaction of coupling
magnetic field on electrical or mechanical system is essential. Explain in simple
words.
13. How is the reluctance torque produced in a rotating machine?
14. Salient pole alternators are more suitable for low speed operation. Explain?
15. Non salient pole alternators are more suitable for high speed operation.
Explain?
16. Can a dc generator be converted into alternator? Justify your answer with
reasons
17. What is the meaning of “electrical angle”? How is it different from mechanical
angle?
18. What are the methods of reducing space harmonics in a machine?
19. How is a sinusoidal voltage developed in a synchronous machine?
20. Why should the output voltage from a poly phase alternator be sinusoidal?
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 86 of 120
Version Code: PEQBTNC06
21. What is the magnetization curve in reference to a dc shunt generator? Explain
the procedure to get it
22. What do you mean by GNA and MNA in a dc machine?
23. What is the motor best suitable for traction applications? Any other alternative
you can suggest
24. Draw the block diagram for a DC motor and then explain speed control of dc
motor
25. Explain armature reaction in (a) DC machine (b) Synchronous machine
26. What is meant by alternator on infinite bus bars?
27. Does a change in the excitation of a synchronous motor affect its speed and
power factor?
28. What is meant by synchronizing power?
29. What are different methods to find out voltage regulation of an alternator?
Which one is the best one?
30. What is the purpose of damper winding used in salient pole type synchronous
machine?
31. Name three important characteristics of a three phase synchronous motor not
found in induction motor?
32. What is synchronous condenser? Do you know any equipment which will do the
same job without rotating parts?
33. It is desirable that the incoming machine should be little fast at the time of
synchronization. Explain with phasor diagrams?
34. With 3 ph balanced sinusoidal voltages, a rotating magnetic field will produce in
3-phase induction motor but not in 3 phase transformer. Explain mathematically
with neat diagrams
35. Name the two types of three phase induction motors. What are the differences in
construction between the two?
36. How can the direction of the following machines can reverse
(a) dc motor
(b) Induction motor
(c) Synchronous motor
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 87 of 120
Version Code: PEQBTNC06
37. How does the change in supply voltage and frequency affect the performance of
three phase induction motor?
38. Explain why the starting currents are very high in reference to
(a) dc motor
(b) Induction motor
39. What is induction generator? Name any one popular application in present days
40. Where do we prefer the use of lap windings and wave windings?
41. What is meant by balanced system and phase sequence in three phase systems?
42. Explain two watt meter method in the measurement of power in a three phase
system. What is the nature of pf when, (a) one wattmeter reads zero (b) two
wattmeter readings are equal but opposite sign
43. What is the basic definition of power system stability? Explain various terms
you are familiar in stability?
44. What is per unit system? What is the use of it?
45. Explain the synchronous machine behavior when three phase short circuit is
created at the terminals
46. What is swing equation and inertia constant?
47. Explain the following terms:
(a) Synchronous reactance
(b) Transient reactance
(c) Sub transient reactance
48. What is meant by capability curve in synchronous machine?
49. What is meant by surge impedance loading in a transmission line?
50. What is operating frequency in India? How many regional grids in India. Do you
know any country in the world is operating with two different frequencies?
51. Explain mathematically, the transfer of power (active and reactive) between two
active sources shown below:
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 88 of 120
Version Code: PEQBTNC06
52. What is load flow problem and explain various buses used in load flow study
53. What is the simple algorithm used in load flow study
54. Explain the following concepts with neat sketches
(a) Transient stability
(b) Small signal stability
(c) Voltage stability
55. What will happen if excitation system in a synchronous generator fails? How
can you detect and protect the systems from excitation failures
56. What will happen if prime mover to a synchronous generator fails when it is
connected to infinite bus?
57. When HVDC transmission system is preferable. Do you know any HVDC links
in India? (Note: to get the answer, visit www.powergridindia.com)
58. Explain various components of HVDC transmission system
59. What do you meant by FACTs? Name any FACTs controllers that you are
familiar with
60. Explain the function of SVC and draw simple schematic
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
Page 89 of 120
Answers for Conventional Questions
L = 0.1 H
C = 0.05 F
(a) 10 mH (b) 20 dB and 10 Ω
(a) 100 krad/s, 10 krad/s and 1000 krad/s (b) 50 krad/s
(a) 5 A and 8.165 A (b) 5 W
25 W
(a) 23.75 A and 1.2 A (b) 28.5 W
55 W
(a) 5 A (b) 6.67 A (c) 6 W with BJT and 6.67 W with MOSFET
(a)Vt = 0.98 V; Rd = 0.005 Ω (b) θJC = 1.25 0C/W; θCA = 1 0C/W
(a) Vt = 0.771 V and Rd = (1/70) Ω (b) 10.577 W
(a) 72 W (b) 3 W (c) 0.4167%
(a) 4.5 A and 6.93 A (b) 4.8 W, 3.6 W and total loss is 8.4 W
2.8 W
(a) Eon= 133.33 µJ; Eoff = 266.67 µJ (b) Eon= 400 µJ; Eoff = 800 µJ (c) 40 W
and 120 W
Q16. Eon = 60 mJ and Eoff = 60 mJ
Q17. 79.69 0C
Q18. (a) 2A and 8 A (b) 3.2 W and 12.8 W (c) 118.4 0C
Q19. (b) 107.8 W (c) 21.6 0C
Q20. C is ideal upto a frequency of 18.4 kHz
Q21. (a) 12.79 A (b) 58.1 A
Q1.
Q2.
Q3.
Q4.
Q5.
Q6.
Q7.
Q8.
Q9.
Q10.
Q11.
Q12.
Q13.
Q14.
Q15.
Q22.
Q23.
Q24.
Q25.
Q26.
Q27.
***
L1 = 325.25 µH, L2 = 827.9 µH, L12 = L21 = 177.41 µH
***
***
77.6 µF
(a) 1.28 mH (b) 5.14 mH
Q28. (a) 10 V (b) Iload = 25 A
Q29. ****
Q30. 20A, 31.416 A, 628.32 V and 13.96 kVA
Q31. (a) L =70√2 mH; V1 = 0 V and V2 = 0 V (b) V1 = 0 V; V2 = 220 V and A = 12.25
A
Q32.
Q33.
Q34.
Q35.
Q36.
Q37.
Q38.
Q39.
(a) V1 = √2 V ; V2 =200√2 V (b) V2 =200√3 V
(a) 2.5467 Ω (b) 49.42 V (c) 4.167
3.1416 hr
***
(a) 650 V (b) PIV rating will increase (c) 325 V (d) 41.55 X 103A2s
***
10A and 1835.6 W
1A
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 90 of 120
Version Code: PEQBTNC06
Q40. (a) 7243 W (b) 4 kW
Q41. (a) 67.70 (b) 7.765 A and 10.477 A
Q42. (a) 232 V and 133.15 V (b) 10A & 17.32 A; 565.68 V (c) 19 W (d) 232 V, 154 V;
0 A, 0 A and 7.5 A, 15 A
Q43. 50.110
Q44. 4 kW, 2.31 kVAR and 0.827 lag
Q45. 19.95 A
Q46. (a) 12 V (b) 44.370 and 0.3344 ms (c) 129.60
Q47. (a) 33.33 A & 57.735 A (b) 622.25 V (c) 81.65 A
Q48.
Q49.
Q50.
Q51.
Q52.
Q53.
Q54.
Q55.
Q56.
Q57.
Q58.
Q59.
Q60.
Q61.
Q62.
Q63.
Q64.
Q65.
Q66.
Q67.
Q68.
Q69.
Q70.
Q71.
Q72.
Q73.
Q74.
Q75.
***
(a) (i) 341.16 V (ii) 157.1 V (iii) 157.1 V (iv) 209.44 V (v) 104.72 V (vi)104.72 V
1443.4 A and 104.7 kV
(a) 50% (b) 228.57 µH (c) 350 µF
(a) 50% (b) 0.833 A (c) 1.416 A (d) 21.93 mV
(a) 4.29 V (b) 65.45 mV (c) 0.8 A (d) 2.11 A and 0.514 A
***
***
***
(g) L1 = 2.67 mH, L2 = 960 µH, C1 = 24 µF and C2 = 2 µF
24.14 A
vw
vg
=4;
***
vw = 25
(a) N2/N1 = 9 (b) 0.46 < D < 0.55 (c) 7.35 mH
50:1
96%
Vsw = 500 V and Vd = 750 V
1/3 and 1/9
(a) 6 µH (b) 1.25 %
***
***
(a) {i} 73.211 A and 0 A; {ii} 51.776 A & 51.776 A (b) {i} 230 V {ii} 230 V
2.148 µF
(b) 204.54 W & 204.12 W (c) 9.044 A & 4.522 A (d) 5.513 ms & 4.487 ms
(a) 5285.56 W (b) 3400.96 W (c) 2706.34 W
(a) 500 µs (b) 750 V
(a) 40x1 − 1.245yz{|EE} ~, -9.8 A and 9.8 A; (b) 438.14 µs
(b) 162.635 V and 93.897 V (c) 155.304 V and 89.665 V (d) 28.15 % (e)
5154.86 W and 22.412 A (f) 7.667 A and 13.28 A
Q76. (a) 100 µs (b) VT = -15 V and VC = 30 V
Q77. (a) 24.335 µs and tc = 13.23 µs (b) 24.335 µs and tc = 76.273 µs
Q78. Yes and 125.664 µs
Q79. (a) 0.025 µF (b) 10 Ω
Q80. (a) 7.2135 µF (b) 50 kΩ
Q81. (a) IT1,max = 24 A and IT2,max = 42 A (b) C = 11.542 µF
Q82. ***
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
Q83.
Q84.
Q85.
Q86.
Q87.
Q88.
Page 91 of 120
(a) 69.70 (b) 63.98 A (c) 0.821 lag (d) 70.3 %
(a) 39.20 (b) 8.30
(a) 29.30 (b) 1200 (c) -893.2 rpm
(a) 1.559: 1 (b) (i) 34.650 (ii) 104.20
(a) 0.376 (b) 0.34 (c) 962 rpm and 87.4 kW (c) 0.8
(a) 900 rpm, 0.45 Hz, 1.5 % (b) 2.7 Hz
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 92 of 120
Version Code: PEQBTNC06
Answers for Objective Questions
Test -1
1-C
2-C
3-B
4-B
5-C
6-D
7-A
8-A
9-B
10 - D
11 - A
12 - C
13 - D
14 -D
15 - B
16 - B
17 -B
18 -C
19 -B
20 -D
21 -B
22 -B
23 -D
24 -A
25 -B
26 -A
27 -D
28 -D
29 -B
30 -B
31 -A
32 -D
33 -C
34 -D
35 -A
36 -B
37 -A
38 -C
39 -B
40- C
41 - D
42 - B
43 - A
44 -B
45 -A
46 - D
47 - C
48 - B
49 - C
50 - C
51 - C
52 - A
53 - B
54 - B
55 - A
56 - C
57 - D
58 - A
59 - D
60 -A
Test -2
1-C
2-A
3-B
4-D
5-B
6-A
7-A
8-B
9-C
10 - C
11 - D
12 - C
13 - C
14 -A
15 - B
16 - C
17 -C
18 -A
19 -B
20 -D
21 -D
22 -A
23 -A
24 -B
25 -C
26 -B
27 -C
28 -B
29 -D
30 -C
31 -B
32 -D
33 -D
34 -A
35 -C
36 -B
37 -D
38 -B
39 -C
40- D
Test -3
1-B
2-C
3-B
4-A
5-C
6-A
7-B
8-C
9-B
10 - C
11 - A
12 - B
13 - D
14 -C
15 - A
16 - D
17 -D
18 -D
19 -B
20 -A
21 -B
24 - B
25 - B
26 -A
27 -B
28 -D
29 -C
30 -A
22
83.2 V
23
D = 0.85 and 145 rad/s
7–B
17 – B
27 - A
8–C
18 – B
28 – B
9–A
19 – A
29 – C
41922
20 - C
30 - D
1–C
11 – C
21 – B
31:12.8
2–B
12 – B
22 – A
32 - C
3–C
13 – B
23 – A
33 - C
4–B
14 – D
Test-4
5–D
6–B
15 – B 16 – A
24 : 457.6
25 - 750
34 - B
35 - C
http://electrical-mentor.blogspot.in/
26 – 20
Copyright © Reserved 2014
Page 93 of 120
Version Code: PEQBTNC06
How to design a closed control system for
power electronic converters
In this document, the control requirements of the dc-dc converter are stated and are
related to the frequency domain performance indices such as the loop gain, crossover frequency, dc loop-gain, and phase margin of the loop gain and so on. The
closed loop controller design is briefly outlined and then demonstrated through the
example of a buck and boost converter.
Control requirements of closed loop system:
The control specification of the converter will be in two parts.
•
•
Steady state accuracy
Settling time and allowed transient overshoot in the event of disturbances or
command changes.
The steady state error is related to the dc loop gain [G(0)H(0)] of the closed loop
system. The steady state error is approximately 1/ G(0)H(0).
For example, a dc loop gain of 100 will result in a steady state error of about 1%.
The settling time and transient overshoot are related to the 0 dB crossover
frequency of the loop gain and the phase margin (PM). If ωc is 0 dB crossover
frequency of the loop gain then the settling time (for a stable system) will be about
to 3/ωc to 4/ωc
The approximate transient overshoot is related to the phase margin (PM) of the loop
gain according to the following table.
PM (deg)
Overshoot (%)
300
37%
350
30%
400
25%
450
16%
500
9%
550
5%
600
1%
For acceptable transient overshoot, the phase margin may be taken as 450. The first
design step in closed loop controller design is to convert the control specification to
the following:
1. Desired DC loop gain (to meet the steady state error)
2. Desired ωc(to meet the settling time)
3. Desired phase margin (to meet the transient overshoot)
Compensator structure is shown in the following figure
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 94 of 120
Version Code: PEQBTNC06
The first stage H1(s) of the compensator achieves the desired bandwidth ωc and the
desired phase margin and the second stage is designed to meet the desired steady
state error.
The following steps will make the design procedure to be simple:
The important rule that is used here is that, if the loop gain crosses 0 dB (unity
gain) with a single slope (-20dB/decade), then the closed loop system will be stable.
The reason is that the phase gain of a function crossing 0 dB with a single slope at a
frequency of ωc is approximately the same as the function K=1/ωc(s) and is equal to 900. This argument is valid only when the loop gain is a minimum phase function.
The actual phase angle will depend on the poles and zeroes nearest to the crossover
frequency. With the above simple rule in mind, the compensator function H1(s) is
selected to be simple lead-lag compensator.
 " = `
"
1 + D# G
€
"
1 + # ƒ
‚
The purpose of is to make the slope of crossover section of the loop gain to -20
dB/decade near the desired crossover frequency, and to improve the phase margin
•
•
•
•
If G(s) is a first order system in the vicinity of ωc, then H1 may be just K1
If G(s) is a second order system in the vicinity of ωc, then select ωz1 and ωp1
such that ωz1 < ωc < ωp1
If G(s) is a second order system with pole pair ω0, then ωz1 may be taken as
ω0 and ωp1 is usually several times higher than ωz1 (about ωp1 = 10 to 100
times ωz1)
Now K1 may be selected to meet the requirements of ωc and PM
The structure of second system can be considered as follows:
M" =
http://electrical-mentor.blogspot.in/
`
"
"
1 + „# + #E
E
Copyright © Reserved 2014
Page 95 of 120
Version Code: PEQBTNC06
The next part of the compensator H2(s) is needed to meet the steady state error
specification. If G(0)H1(0) is already compatible with the steady state error, then
H2(s) is 1. However, if G(0)H1(0) is not compatible with the desired steady state
error, H2(s) is different from unity. The conditions on H2(s) are
•
•
•
G(0)H1(0)H2(0) = dc loop gain
H2(s) must not affect the gain & phase margin already designed. Or in the
other words, phase and magnitude gain of H2(s) in the vicinity of ωc must be
00 and 0dB respectively.
A PI controller H2 (s) of the following form will satisfies the above
requirements.
"
1+
#€
 " =
"
#€
Then the overall compensator is
"
#
€
€
† ‡
ˆ
" = `
"
"
1 + # ƒ
#€
‚
"
1 + D# G
1+
And this H(s) can be realized using either operational amplifier circuits or
DSP controller or FPGA controller
The control transfer functions for three basic DC-DC converters are giving
here for quick reference:
Buck:
‰E
" = ‹
WŠ
Boost:
1
C
Œ1 + " ] + " CK
‰E
‹
" =
1 − o
WŠ
Ž1 + "
Buck-Boost:
‰E
‹
" =
1 − o
WŠ
Ž1 + "
1−"
C
]1 − o
C
CK
+ "

1 − o
]1 − o
1 − "o
C
]1 − o
C
CK
+ "

1 − o
]1 − o
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 96 of 120
Version Code: PEQBTNC06
Compensator design example 1
(For buck converter)
A buck converter transfer function is given as follows:
M" =
1 +
40
"
2"
+
ƒ
100s 100s
Q: Design a compensator with acceptable transient and steady state response
parameters as follows:
a) Phase Margin and Gain Margin of loop gain should be greater than 450 and 20 dB
respectively
b) Steady state error should not be more than 5%
c) Settling time should not be more than 1 s
d) Peak overshoot should not be more 5 %
Justify all the design requirements with suitable response plots
Note: Derivation of transfer function G(s) is out of scope of this document. You can
refer any standard text book in PE
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 97 of 120
Version Code: PEQBTNC06
Compensator Design procedure with bode plot approach:
Step1.
Let us draw the asymptotic gain bode plot of G(s)
Fig 1
Observations from Fig 1:
M" =
40
"
ƒ
100s
0dB cross over frequency of open loop plant transfer function is around 2000 rad/s
and phase margin is around 18.20. The consequence of this is peak overshoot in the
response will be very high. That can be observed in the following step response.
http://electrical-mentor.blogspot.in/
2"
1 + 100s +
Copyright © Reserved 2014
Page 98 of 120
Version Code: PEQBTNC06
Step response of G/(1+G) is shown in Fig 2:
Fig 2
From the step response of Fig 2, we can observe that transient response
specifications are not satisfied
Hence compensator is required
As G(s) is a second order system with a complex pole pair ω0 then ωz1 in
compensator may be taken as ω0 and ωp1 is usually as several times ωz1 (10 to 50
time)
Standard form of H(s) is as follows:
" =
http://electrical-mentor.blogspot.in/
"
1+#
€
"
1+#
‚
Copyright © Reserved 2014
Page 99 of 120
Version Code: PEQBTNC06
Step2.
Let us draw the asymptotic gain bode plot of compensator H(s)
"
100s
" =
"
1+
40000s
1+
Fig 3
H(s) is basically a lead-lag compensator and it can improve the PM to this plant
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 100 of 120
Version Code: PEQBTNC06
Step3.
Let us draw the asymptotic gain bode plot of G(s)H(s)
Note: Here G(s)H(s) is loop gain
Fig 4
Observations from loop gain bode plot (Fig 4):
1. Closed loop system is stable as loop gain is having PM = 90.50, it is the indication
of peak overshoot of CLTF in time response is less than 1%
2. 0 dB cross over frequency is 5 rad/s it means settling will be satisfied which can
justified in step response of CLTF as shown below
3. Loop gain is having no issues with GM
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 101 of 120
Version Code: PEQBTNC06
Step4:
Verification of response in time domain:
Step response of output voltage V(t) with unity feedback is shown inn Fig 5
Fig 5
As settling time and steady state errors specifications are satisfied, there is no need
to use PI regulator in the low frequency region
Cross checking other possibilities:
Suppose in H(s), if we take pole at 10000π instead of 40000π. Then also stability
condition will satisfy but settling time will not satisfy. Let us see these two
observations from bode plot of loop and step response with the follow modifications
 " =
And
k
ewBB
k
e‘BBBB
 " =
k
ewBB
k
ewBBBB
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 102 of 120
Version Code: PEQBTNC06
Fig 6
Step response of V(t) with compensator with two pole locations in the compensator:
Fig 7
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
Page 103 of 120
From step response shown in Fig 7, we can observe that compensator with poles at
40000π and 10000π are satisfying the stability conditions. But, compensator with
pole at 40000π (blue color) is giving settling time of 0.74 s whereas compensator
with pole 10000π (green color) is giving settling time of 2.99 s.
The same point can be observed from bode plot of loop gain shown in Fig 6; 0dB
cross over frequency with H1 is more than H2 that means bandwidth is better with
H1
Therefore " =
k
wBB
k
e
‘BBBB
e
will give satisfied response in closed loop operation
If we select suitable circuit components to realize this transfer function or proper
controller hardware will satisfy response in closed loop operation
Hence compensator design is complete in all aspects.
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 104 of 120
Version Code: PEQBTNC06
Compensator design example 2
(For boost converter)
A boost converter has the following transfer function
"
1 − 4000s
M" = 40
2"
" 1+
+D
G
80s
80s
Q: Design a compensator with acceptable transient and steady state response
parameters as follows:
a) Phase Margin and Gain Margin of loop gain should be greater than 450 and 20 dB
respectively
b) Steady state error should not be more than 5%
c) Settling time should not be more than 1 s
d) Peak overshoot should not be more 5 %
Justify all the design requirements with suitable response plots
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 105 of 120
Version Code: PEQBTNC06
Compensator Design procedure with bode plot approach:
Step1.
Let us draw the asymptotic gain bode plot of G(s)
Fig 8
Note: From the transfer function G(s) of boost converter it can be observed that it
has one RHP zero. Due to this compensator design is little difficult. It cannot give
stable operation in closed loop for the entire range
From the bode plot of G(s) shown in Fig 8, PM and GM are not satisfied values
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 106 of 120
Version Code: PEQBTNC06
Step response of G/(1+G) [Unity FB closed loop without compensator] is shown in
Fig 9
Fig 9
Observations:
From Fig9, it can be observed that even though steady state response is satisfying
due to feedback, peak overshoot of 74% is not acceptable. This is because PM of loop
gain without controller is 140 which very less (look into Fig 8).
Hence compensator is required
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 107 of 120
Version Code: PEQBTNC06
Step 2: Bode plot of compensator H(s):
Let us consider a compensator has a transfer function of
"
D1 + 80sG
" =
"
D1 + 4000sG
Fig 10
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 108 of 120
Version Code: PEQBTNC06
Step 3:
Bode diagram of G(s)H(s) is shown in Fig 11:
Fig 11
This loop gain is not having satisfied PM and GM. Therefore, if we shift the pole to
somewhat high value, we can achieve this
Let us consider compensator transfer function as below:
"
D1 + 80s G
" = `
"
D1 + 40000s G
Where K1 = 0.1
And bode plot of new loop gain is as shown in Fig 12:
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 109 of 120
Version Code: PEQBTNC06
Fig 12
Now PM and GM are satisfied. Therefore, BW and peak over shoot conditions will
be satisfied.
But as low frequency gain is around 12 dB, it will not satisfy the steady state error.
Which can understand from the time response of v(t) shown in Fig 13
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 110 of 120
Version Code: PEQBTNC06
Fig 13
From the curve shown in Fig 13, we can say that SS error is more than 20%. To
reduce this steady state error, let us introduce PI regulator of the following transfer
function with pole at very low frequency say 500 rad/sec
"
1+
500
"
500
And bode plot of this PI regulator is shown in Fig 14
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 111 of 120
Version Code: PEQBTNC06
Fig 14
Now the final compensator is as follows:
"
"
D1 + 80s G
D1 +
G
500 ”ℎy–y ` = 0.1
" = `
“
"
"
D1 + 40000s G
D
G
500
Circuit realization of this compensator is as shown in Fig 15:
Fig 15
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 112 of 120
Version Code: PEQBTNC06
Step4: Bode plot of loop gain and step response of output
The final bode plot of loop gain [G(s) H(s)] is as shown in Fig 16:
Fig 16
Observations:
1. Loop gain is having reasonable GM of 22 dB.
2. Loop gain is having reasonable PM of 730 which will give satisfied peak overshoot
3. 0 dB cross over frequency is around 1000 rad/s means good BW and hence
response will have small settling time
3. Low frequency gain of more than 42 dB will give acceptable steady state error
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 113 of 120
Version Code: PEQBTNC06
We can observe all the above conclusions in the step response of the output voltage
as shown in Fig 17:
Fig 17
Note: Buck Boost converter will also have similar transfer function that boost
converter (with RHP zero) and hence design procedure is same as boost converter.
Hence not repeating
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 114 of 120
Version Code: PEQBTNC06
Useful units for Electrical Engineering
Expression in
terms of SI
base units
Quantity
Time
Electrical current
Electrical capacitance
Electrical charge
Name of
Unit/Description
second
ampere
farad
coulomb
Symbol
s
A
F
C
Electrical conductance
Electrical inductance
Electrical potential
Electrical resistance
Force
Frequency
Magnetic flux
siemens
henry
volt
ohm
newton
hertz
weber
S
H
V
Ω
N
Hz
Wb
Magnetic flux density
Power or radiant flux
Pressure
Work, energy, heat
tesla
watt
pascal
joule
T
W
Pa
J
Wb/m2
J/s
Electrical charge density coulomb per cubic meter
C/m3
m-3 s A
Electric field strength
volt per meter
V/m
m kg s-3 A-1
Electric flux density
coulomb per square meter
C/m2
m-2 s A
Energy density
joule per cubic meter
J/m3
m-1 kg s-2
Moment of inertia
kilogram meter squared
kg.m2
kg.m2
Torque
Newton meter
N .m
m2 kg s-2
Permeability
henry per meter
H/m
m kg s-2 A-2
Permitivity
farad per meter
F/m
m-3 kg-1 s4 A2
kW
10-3 m2 kg s-3
Power
kilowatt
Permeability of free space —E = 4s “ 10˜ H/m
C/V
A/V
W/A
V/A
Vs
Nm
Permittivity of free space ™E = 8.854 “ 10 F/m
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Version Code: PEQBTNC06
Page 115 of 120
Common mistakes in units display – Observe carefully
and follow the standard practice*
Wrong
V = 1000 Vmax
100volts
10 Amps
10 KWs
1 Ohm or 1 Farad or 1 Henry
1 mµF
1milli-gram
1 ampere/meter
10 % (m/m) or 10 % (by weight)
20 mL H2O/kg (or) 20 mL of water/kg
35 × 48 cm
1 MHz – 10 MHz or 1 to 10 MHz
20 ºC – 30 ºC or 20 to 30 ºC
123 ± 2 g
70 ± 5 %
240 V ± 10 % (one cannot add
240 V and 10 %)
kilogram/m3, kg/cubic meter,
kilogram/cubic meter, kg per m3, or
kilogram per meter3
m = five kilograms or m = five kg
the current was 15 amperes
a 25-kg sphere
an angle of 2 º3 '4 "
Sin x, Cos x, Tan x
It is 75 cm. long
l = 10 m 23 cm 4 mm
the resistance is 100 Ω/square
µr = 1.2 µ
Correct
Vmax = 1000 V
100 volt or 100 V
10 A
10 kW
(a) 1 ohm or 1 farad or 1 henry
(b) 1 Ω or 1 F or 1 H
1 nF
1 milligram or 1 mg
1 ampere per meter or 1 A/m
a mass fraction of 10 %
the water content is 20 mL/kg
35 cm × 48 cm
1MHz to 10 MHz or (1 to 10) MHz
20 ºC to 30 ºC or (20 to 30) ºC
123 g ± 2 g or (123 ± 2) g
70 % ± 5 % or (70 ± 5) %
240 × (1 ± 10 %) V
kg/m3, kg · m−3, or kilogram per cubic
meter
m = 5 kg
the current was 15 A
a 25 kg sphere
an angle of 2º3'4"
sin x, cos x, tan x
Its length is 75 cm or It is 75 cm long
l = 10.234 m
the resistance per square is 100 Ω
µr = 1.2 × 10-6
*Note: The above table is prepared by referring Guide for the use of International System of
Units (SI) released by NIST (National Institute of Standards and Technology, Department of
Commerce USA) as special publication in 2008
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 116 of 120
Version Code: PEQBTNC06
Useful Mathematical Formulae
1. Trigonometric functions:
sinš ± œ = sin š cos œ ± cos š sin œ
cosš ± œ = cos š cos œ ∓ sin š sin œ
sin 2š = 2 sin š cos š
cos 2š = 1 − 2 sin š = 2 cos š − 1
sin š + sin œ = 2 sin
sin š − sin œ = 2 cos
š+œ
š−œ
cos
2
2
š+œ
š−œ
sin
2
2
cos š + cos œ = 2 cos
š−œ
š+œ
cos
2
2
cos š − cos œ = −2 sin
š−œ
š−œ
sin
2
2
sin š cos œ =
1
x sinš + œ + sinš − œ~
2
cos š cos œ =
1
xcosš + œ + cosš − œ~
2
cos š sin œ =
sin š sin œ =
1
xsinš + œ − sinš − œ~
2
1
xcosš − œ − cosš + œ~
2
2. Integration functions:
žy
^Ÿ
y ^Ÿ
sin z Wz = V +
ž y ^Ÿ cos z Wz =
ž sin % z Wz =
ž cos % z Wz =
y ^Ÿ
V +
V sin z − cos z
V cos z + sin z
z sin 2%z
−
2
4%
z sin 2%z
+
2
4%
ž sin ¡z . sin %z Wz =
sin¡ − %z sin¡ + %z
−
¢!– ¡ ≠ %
2¡ − %
2¡ + %
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 117 of 120
Version Code: PEQBTNC06
ž cos ¡z . cos %z Wz =
sin¡ − %z
sin¡ + %z
+
¢!– ¡ ≠ %
2¡ − %
2¡ + %
ž sin ¡z . cos %z Wz =
cos¡ − %z cos¡ + %z
−
2¡ − %
2¡ + %
sin %z
ž sin ¡z . cos %z Wz =
2%
ž z cos %z Wz =
z sin %z cos %z
+
%
%
ž z sin %z Wz = −
z cos %z sin %z
+
%
%
Integration by parts: ¤ ¥ W = ¥ − ¤ W¥
3. Leibnitz’s linear 1st order differential equation:
¦
Ÿ
+ §¨ = „ Where P, Q are functions of x
). l = y ¤ © Ÿ , then solution is ¨ ). l = ¤ „ ). l Wz + K
4. Fourier series:
A function f (t) is said to be periodic of time if f (t+T) = f(t), then f (t) can be
expanded in Fourier series as under:
¢ = VE + ∑«
pUxVp cos %# +
VE =
R
¤E ¢ W =
R
R
¬
p
sin %#~, where
¤E ¢ # W#
¬
¬
2
1
Vp =
ž ¢ cos %# W =
ž ¢ # cos %# W#
X
s
p
E
R
E
¬
2
1
=
ž ¢ sin %# W =
ž ¢ # sin %# W#
X
s
And
E
E
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 118 of 120
Version Code: PEQBTNC06
Use of symmetry in Fourier series
Symmetry
Even
Condition required
f(-t) = f(t)
Odd
f(-t) = - f(t)
Half – wave
f(-t) = - f[t+(T/2)]
Even Quarter
wave
Even and Half wave
Odd Quarter
wave
Odd and Half wave
¬
2
= 0 and Vp = ž ¢# cos%# W#
s
an and bn
p
Vp = 0 and
Vp =
p
p
E
¬
2
= ž ¢# sin%# W#
s
E
= 0 for even ‘n’
¬
2
Vp = ž ¢# cos%# W#
s
E
p
= 0 for all ′n′
p
²4
= s ž ¢# si %%#W# for odd ′n′N
± E
°0
for even ′n′
¬⁄
²4
Vp = s ž ¢# cos%#W# for odd ′n′N
± E
°0
for even ′n′
Vp = 0 ¢!– V´´ ′%′
¬⁄
5. Laplace transforms:
ℒx¶~ = 1
ℒx1~ =
ℒx~ =
1
"
1
"
ℒx p ~ =
%!
" pe
ℒ xy ±^P ~ =
ℒ xy ^P ~ =
1
"∓V
1
" + V
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 119 of 120
Version Code: PEQBTNC06
ℒxsin #~ =
ℒxcos #~ =
#
" + #
"
" + #
ℒxy ^P sin #~ =
ℒxy ^P cos #~ =
P
ℒ Ž1 − y R  =
#
" + V + # "+V
" + V + # 1
"" + X
1
y ^P − y ¹P
º=
ℒ¸
" + V" + −V
ℒ ŒP »¢¼ = ". ½¾ − ¢0
ℒ x¤ ¢ W~ =
½¾
f
+
¿ Àw E
f
Where ¢ 0 = N¤ ¢ W |PUE and ½¾ = ℒx¢~
Note: The last two equations can be proved by writing the basic definition of Laplace
transform and then apply integration by parts
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014
Page 120 of 120
Version Code: PEQBTNC06
You reached end of the Question Bank
Thank You
http://electrical-mentor.blogspot.in/
Copyright © Reserved 2014