Power Electronics

Question 1
The steady state current flowing through the inductor of a DC-DC buck boost converter is given in the figure below. If the peak-to-peak ripple in the output voltage of the converter is 1 V, then the value of the output capacitor, in \mu F , is _______________. (round off to nearest integer)

A
124
B
148
C
165
D
182
GATE EE 2022      Choppers
Question 1 Explanation: 
We have for buck boost converter,
\begin{aligned} \Delta V_c&= \frac{I_o}{fC}\\ where, \; \alpha &=\frac{T_{ON}}{T}\frac{20}{20+30}=0.4 \\ I_{L(Avg)}&=\frac{16+12}{2}=14A \end{aligned}
For buck-boost converter,
\begin{aligned} \frac{I_{o(Avg)}}{I_{L(Avg)}}&=\frac{1}{1-\alpha }\\ \Rightarrow I_L&=I_o(1-\alpha )\\ &=14(1-0.4)=8.4A\\ Given: \; \Delta V_c&=1V\\ Therefore, \; 1&=\frac{0.4 \times 8.4 \times 50 \times 1^{-6}}{C}\\ C&=168\mu F \end{aligned}
Question 2
A 3-phase grid-connected voltage source converter with DC link voltage of 1000 V is switched using sinusoidal Pulse Width Modulation (PWM) technique. If the grid phase current is 10 A and the 3-phase complex power supplied by the converter is given by (-4000 - j3000) VA, then the modulation index used in sinusoidal PWM is ___________. (round off to two decimal places)
A
0.23
B
0.47
C
0.64
D
0.87
GATE EE 2022      Inverters
Question 2 Explanation: 
Apparent power, S = (-4000 - j3000) VA
or S=5000\angle -143.13^{\circ}VA
S=\sqrt{3}V_LI_L
V_L=\frac{5000}{\sqrt{3} \times 10}=288.675V
We have,
Peak, V_{01}=\sqrt{3}M_A\frac{V_s}{2}
Put the values
\sqrt{2}\times 288.675=\sqrt{3}M_A \times \frac{1000}{2}
\Rightarrow \; M_A=0.471
Question 3
For the ideal AC-DC rectifier circuit shown in the figure below, the load current magnitude is I_{dc} = 15 A and is ripple free. The thyristors are fired with a delay angle of 45^{\circ}. The amplitude of the fundamental component of the source current, in amperes, is __________. (round off to two decimal places)

A
12.45
B
25.32
C
14.25
D
17.64
GATE EE 2022      Phase Controlled Rectifiers
Question 3 Explanation: 
Given rectifier circuit is a 1-\phi semiconverter Waveform of source current:

\therefore \; i_{sn}=\sum_{n=1,3,5...}^{\infty }\frac{4I_o}{n\pi} \cos \left ( \frac{n\alpha }{2} \right ) \sin (n\omega t+\phi _n)
Now, fundamental component,
I_{s1}=\frac{4I_o}{\pi} \cos \left ( \frac{\alpha }{2} \right )=\frac{4 \times 15}{\pi} \cos \left ( \frac{45^{\circ} }{2} \right )=17.64A
Question 4
Consider an ideal full-bridge single-phase DC-AC inverter with a DC bus voltage magnitude of 1000 V. The inverter output voltage v(t) shown below, is obtained when diagonal switches of the inverter are switched with 50 % duty cycle. The inverter feeds a load with a sinusoidal current given by, i(t)=10 \sin (\omega t-\frac{\pi}{3})A, where \omega =\frac{2\pi}{T}. The active power, in watts, delivered to the load is _________. (round off to nearest integer)

A
2154
B
3254
C
3181
D
4578
GATE EE 2022      Inverters
Question 4 Explanation: 
For 1-\phi invertor, RMS value of fundamental component
V_{01}=\frac{2\sqrt{2}V_s}{\pi}
Now, Power output
=V_{01}I_{or} \cos \frac{\pi}{3}=\frac{2\sqrt{2}}{\pi} \times 1000 \times \frac{10}{\sqrt{2}} \times \frac{1}{2}=3183.098W
Question 5
The voltage at the input of an AC-DC rectifier is given by v(t)=230\sqrt{2}\sin \omega t where \omega = 2 \pi \times 50 rad/s. The input current drawn by the rectifier is given by
i(t)=10\sin \left (\omega t -\frac{\pi}{3} \right )+4\sin \left (3\omega t -\frac{\pi}{6} \right )+3\sin \left (5\omega t -\frac{\pi}{3} \right )
The input power factor, (rounded off to two decimal places), is, _________ lag.
A
0.44
B
0.22
C
0.66
D
0.88
GATE EE 2022      Phase Controlled Rectifiers
Question 5 Explanation: 
We have,
\begin{aligned} PF&=g.FDF \\ g&=\frac{I_{{s1}}}{I_{sr}} \\ &= \frac{10/\sqrt{2}}{\sqrt{\left ( \frac{10}{\sqrt{2}} \right )^2+\left ( \frac{4}{\sqrt{2}} \right )^2+\left ( \frac{3}{\sqrt{2}} \right )^2}}\\ &=0.894 \\ FDF&= \cos \frac{\pi}{3}=0.5 \\ PF&=0.894 \times 0.5=0.447 \end{aligned}
Question 6
A single-phase full-bridge diode rectifier feeds a resistive load of 50 \Omega from a 200 V, 50 Hz single phase AC supply. If the diodes are ideal, then the active power, in watts, drawn by the load is _____________. (round off to nearest integer).
A
200
B
1600
C
600
D
800
GATE EE 2022      Phase Controlled Rectifiers
Question 6 Explanation: 
For 1-\phi full bridge diode rectifier:
V_{or}=\frac{V_m}{\sqrt{2}} =\frac{\sqrt{2} \times 200}{\sqrt{2}}=200V
Active power drawn by the load
P_{o}=\frac{V_{or}^2}{R} =\frac{200^2}{50}=800W
Question 7
A charger supplies 100 W at 20 V for charging the battery of a laptop. The power devices, used in the converter inside the charger, operate at a switching frequency of 200 kHz. Which power device is best suited for this purpose?
A
IGBT
B
Thyristor
C
MOSFET
D
BJT
GATE EE 2022      Power Semiconductor Devices and Commutation Techniques
Question 7 Explanation: 
Ratings of different power devices:
BJT : 1200 V, 800 A, (10 - 20)kHz
SCR : 10000 V, 3000 A
IGBT : 1200 V, 500 A, 50 kHz
MOSFET : 500 V, 140 A, 1 MHz
Question 8
A single-phase full-bridge inverter fed by a \text{325 V DC} produces a symmetric quasi-square waveform across 'ab' as shown. To achieve a modulation index of 0.8, the angle \theta expressed in degrees should be _______________. (Round off to 2 decimal places.)
(Modulation index is defined as the ratio of the peak of the fundamental component of V_{ab} to the applied \text{DC} value.)

A
51.1
B
28.06
C
58.3
D
69.24
GATE EE 2021      Inverters
Question 8 Explanation: 

\begin{aligned} \widehat{V_{01}} &=m_{a} V_{S}=0.8 \times 325=260 \mathrm{~V} \\ \widehat{V_{01}} &=\frac{4 V_{S}}{\pi} \sin d=260 \\ \frac{4(325)}{\pi} \sin d &=260 \\ \sin d &=\frac{260 \times \pi}{4 \times 325}=0.628 \\ d &=38.9 \\ \therefore \qquad \qquad \theta &=\frac{\pi}{2}-d=90^{\circ}-38.9=51.1 \end{aligned}
Question 9
Consider the buck-boost converter shown. Switch Q is operating at \text{25 kHz} and 0.75 duty-cycle. Assume diode and switch to be ideal. Under steady-state condition, the average current flowing through the indicator is ____________A.

A
15
B
18
C
30
D
24
GATE EE 2021      Choppers
Question 9 Explanation: 


\alpha=0.75, \quad f=25 \mathrm{kHz}
Assume continous conduction:
\begin{aligned} V_{0}&=\frac{\alpha V_{s}}{1-\alpha}=\frac{0.75 \times 20}{1-0.75} \\ V_{0}&=60 \mathrm{~V}\\ I_{0} &=\frac{V_{0}}{R}=\frac{60}{10}=6 \mathrm{~A} \\ I_{L} &=\frac{I_{0}}{1-\alpha}=\frac{6}{1-0.75}=24 \mathrm{~A} \\ \Delta I_{L} &=\frac{\alpha V_{s}}{f_{L}}=\frac{0.75 \times 60}{25 \times 10^{3} \times\left(1 \times 10^{-3}\right)}=1.8 \mathrm{~A} \\ I_{L \min } &=I_{L}-\frac{\Delta I_{L}}{2}=24-\frac{1.8}{2}=24-0.9 \\ \left(I_{L \min }=23.1 \mathrm{~A}\right) &>0 \end{aligned}
\therefore Continous conduction assumption is correct.
I_{L}=24 \mathrm{~A}
Question 10
Consider the boost converter shown. Switch Q is operating at \text{25 kHz} with a duty cycle of 0.6. Assume the diode and switch to be ideal. Under steady-state condition, the average resistance R_{\text{in}} as seen by the source is __________ \Omega.
(Round off to 2 decimal places).

A
1.25
B
1.6
C
2.2
D
3.45
GATE EE 2021      Choppers
Question 10 Explanation: 

Checking for continuous conduction mode
\begin{aligned} \Delta I_{L} &=\frac{\alpha V_{S}}{f L}=\frac{0.6 \times 15}{25 \times 10^{3} \times 1 \times 10^{-3}}=0.36 \mathrm{~A} \\ \frac{\Delta I_{L}}{2} &=0.18 \mathrm{~A} \\ I_{L, \min } &=I_{L}-\frac{\Delta I_{L}}{2}=I_{S}-\frac{\Delta I_{L}}{2} \\ &=(9.375-0.18)=9.195>0 \end{aligned}
As it is continuous conduction
\begin{aligned} V_{0}&=\frac{V_{S}}{1-\alpha}=\frac{15}{1-0.6}=37.5 \mathrm{~V} \\ I_{0}&=\frac{V_{0}}{R}=\frac{37.5}{10}=3.75 \mathrm{~V} \\ \frac{V_{0}}{V_{S}}&=\frac{I_{S}}{I_{0}}=\frac{1}{1-\alpha} \\ I_{S}&=\frac{I_{0}}{1-\alpha}=\frac{3.75}{1-0.6}=9.375 \mathrm{~A} \\ R_{\text {in }}&=\frac{V_{S}}{I_{S}}=\frac{15}{9.375}=1.6 \Omega \end{aligned}


There are 10 questions to complete.