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)
124 | |
148 | |
165 | |
182 |
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}
\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)
0.23 | |
0.47 | |
0.64 | |
0.87 |
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
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)
12.45 | |
25.32 | |
14.25 | |
17.64 |
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
\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)
2154 | |
3254 | |
3181 | |
4578 |
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
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.
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.
0.44 | |
0.22 | |
0.66 | |
0.88 |
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}
\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).
200 | |
1600 | |
600 | |
800 |
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
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?
IGBT | |
Thyristor | |
MOSFET | |
BJT |
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
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.)
(Modulation index is defined as the ratio of the peak of the fundamental component of V_{ab} to the applied \text{DC} value.)
51.1 | |
28.06 | |
58.3 | |
69.24 |
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.
15 | |
18 | |
30 | |
24 |
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).
(Round off to 2 decimal places).
1.25 | |
1.6 | |
2.2 | |
3.45 |
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.