# Power Semiconductor Devices and Commutation Techniques

 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 A B B C C D D
GATE EE 2023   Power Electronics
Question 1 Explanation:
I-V characteristic of given 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?
 A IGBT B Thyristor C MOSFET D BJT
GATE EE 2022   Power Electronics
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

 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).
 A 24.24 B 78.83 C 12.45 D 54.99
GATE EE 2020   Power Electronics
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}
 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 1.5$I_1$ is closest to:
 A -0.02 B -0.09 C -1.5 D -4.5
GATE EE 2020   Power Electronics
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).
 A 15 B 10 C 20 D 25
GATE EE 2020   Power Electronics
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}

There are 5 questions to complete.