Question 1 |
A semiconductor switch needs to block voltage V of only one polarity (V \gt 0) during OFF state as shown in figure (i) and carry current in both directions during ON state as shown in figure (ii). Which of the following switch combination(s) will realize the same?
A | |
B | |
C | |
D |
Question 1 Explanation:
I-V characteristic of given switch :
this characteristic shows by options (A) and (D) switch.
this characteristic shows by options (A) and (D) switch.
Question 2 |
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 2 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 3 |
A resistor and a capacitor are connected in series to a 10 V dc supply through a switch.
The switch is closed at t=0, and the capacitor voltage is found to cross 0 V at t=0.4 \tau,
where \tau is the circuit time constant. The absolute value of percentage change required
in the initial capacitor voltage if the zero crossing has to happen at t=0.2 \tau is _______
(rounded off to 2 decimal places).
24.24 | |
78.83 | |
12.45 | |
54.99 |
Question 3 Explanation:
If initial charge polarities on the capacitor is opposite to the supply voltage then only
the capacitor voltage crosses the zero line.
\begin{aligned} V_{c}(t) \; \Rightarrow \;& \text{Final value} \\ &+ (\text{Initial value - Final value}) e^{-t/\tau }\\ 0&=10+(-V_{0}-10)e^{-0.4} \\ 10&=(V_{0}+10)e^{-0.4} \\ V_{0}&=4.918 V \\ \text{Now, } t&=0.2\tau \\ 0&=10+(-{V_{0}}'-10)e^{-0.2} \\ {V_{0}}'&=2.214 \\ \%\text{change in voltage} &= \frac{4.918-2.214}{4.918}\times 100 \%\\ &=54.99\% \end{aligned}
\begin{aligned} V_{c}(t) \; \Rightarrow \;& \text{Final value} \\ &+ (\text{Initial value - Final value}) e^{-t/\tau }\\ 0&=10+(-V_{0}-10)e^{-0.4} \\ 10&=(V_{0}+10)e^{-0.4} \\ V_{0}&=4.918 V \\ \text{Now, } t&=0.2\tau \\ 0&=10+(-{V_{0}}'-10)e^{-0.2} \\ {V_{0}}'&=2.214 \\ \%\text{change in voltage} &= \frac{4.918-2.214}{4.918}\times 100 \%\\ &=54.99\% \end{aligned}
Question 4 |
A non-ideal diode is biased with a voltage of -0.03 V, and a diode current of I_1 is
measured. The thermal voltage is 26 mV and the ideality factor for the diode is 15/13.
The voltage, in V, at which the measured current increases to 1.5I_1 is closest to:
-0.02 | |
-0.09 | |
-1.5 | |
-4.5 |
Question 4 Explanation:
\begin{aligned}
I_{1}&=i_{0}\left [ e^{\frac{0.03}{15/13\times 26mV}}-1 \right ] \\ V_{D}&=\text{ -ve '1' can not be neglected}\\&\text{in diode current equation} \\ I_{1}&=I_{0}[e^{-30mV/30mV}-1] \\ &=I_{0}[e^{-1}-1] \\ &=-0.64\: I_{0} \\ 1.5I_{1}&=I_{0}[e^{V_{D}/30mV}-1] \\ -1.5\times 0.64 I_{0}&=I_{0}[e^{V_{D}/30mV}-1]\\-0.96&=e^{V_{D2}/30mV}-1\\1-0.96&=e^{V_{D2}/30mV}\\0.04&=e^{V_{D2}/30mV}\\ 30 mV \ln (0.04)&=V_{D}\\ V_{D}&=-0.09 V
\end{aligned}
Question 5 |
A single-phase inverter is fed from a 100 V dc source and is controlled using a quasisquare wave modulation scheme to produce an output waveform, v(t). as shown. The
angle \sigma is adjusted to entirely eliminate the 3^{rd} harmonic component from the output
voltage. Under this condition, for v(t), the magnitude of the 5^{th} harmonic component as
a percentage of the magnitude of the fundamental component is _______(rounded off
to 2 decimal places).
15 | |
10 | |
20 | |
25 |
Question 5 Explanation:
Using result,
\begin{aligned} V_{n}&=\frac{4V_{s}}{n\pi }\cos n\sigma\ \text{For, } V_{3}=0\\ \cos 3\sigma &=0 \\ 3\sigma &=\frac{\pi }{2} \\ \sigma &=\frac{\pi }{6}\\ \text{Now, } \frac{V_{5}}{V_{1}}&=\frac{\cos 5\sigma }{5\cos\sigma }\\&=\frac{\cos 5\pi /6}{5\cos \pi /6}=-\frac{1}{5} \\ \% \left | \frac{V_{5}}{V_{1}} \right |&=\frac{1}{5}\times 100=20\% \end{aligned}
\begin{aligned} V_{n}&=\frac{4V_{s}}{n\pi }\cos n\sigma\ \text{For, } V_{3}=0\\ \cos 3\sigma &=0 \\ 3\sigma &=\frac{\pi }{2} \\ \sigma &=\frac{\pi }{6}\\ \text{Now, } \frac{V_{5}}{V_{1}}&=\frac{\cos 5\sigma }{5\cos\sigma }\\&=\frac{\cos 5\pi /6}{5\cos \pi /6}=-\frac{1}{5} \\ \% \left | \frac{V_{5}}{V_{1}} \right |&=\frac{1}{5}\times 100=20\% \end{aligned}
There are 5 questions to complete.