# Power Electronics

 Question 1
The single phase rectifier consisting of three thyristors $T_{1}, T_{2}, T_{3}$ and a diode $D_{1}$ feed power to a $10 \mathrm{~A}$ constant current load. $T_{1}$ and $T_{3}$ are fired at $\alpha=60^{\circ}$ and $T_{2}$ is fired at $\alpha=240^{\circ}$. The reference for $\alpha$ is the positive zero crossing of $\mathrm{V}_{\text {in }}$. The average voltage $\mathrm{V}_{o}$ across the load in volts is _____ (Round off to 2 decimal places).

 A 25.35 B 36.47 C 39.79 D 48.64
GATE EE 2023      Phase Controlled Rectifiers
Question 1 Explanation:

$\therefore$ Average output voltage,
$\begin{gathered} \mathrm{V}_{o}=\frac{1}{2 \pi}\left[\int_{\alpha}^{\pi} \mathrm{V}_{\mathrm{m}} \sin \omega t d \omega t+\int_{\pi+\alpha}^{2 \pi+\alpha}-\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t} d \omega t\right] \\ =\frac{V_{m}}{2 \pi}[1+3 \cos \alpha] \end{gathered}$

Put the values,
\begin{aligned} V_{o} & =\frac{100}{2 \pi}\left[1+3 \cos 60^{\circ}\right] \\ & =39.79 \mathrm{~V} \end{aligned}
 Question 2
The circuit shown in the figure has reached steady state with thyristor '$T$' in OFF condition. Assume that the latching and holding currents of the thyristor are zero.
The thyristor is turned ON at $t=0$ sec. The duration in microseconds for which the thyristor would conduct, before it turns off, is ___ (Round off to 2 decimal places).

 A 7.33 B 5.36 C 4.25 D 8.23
GATE EE 2023      Phase Controlled Rectifiers
Question 2 Explanation:
Case (i) $t\lt 0$ sec :

Case (ii) $\mathrm{t} \gt 0$ sec:
Thyristor is ON
Redraw the circuit :

$\mathrm{I}_{4 \Omega}=\frac{100}{4}=25 \mathrm{~A}$
and
\begin{aligned} \mathrm{i}_{\mathrm{C}} & =\mathrm{I}_{\mathrm{P}} \sin \omega_{0} \mathrm{t} \\ & =\mathrm{V}_{\mathrm{S}} \sqrt{\frac{C}{L}} \sin \omega_{0} \mathrm{t} \\ & =100 \sqrt{\frac{1}{4}} \sin \omega_{0} \mathrm{t}=50 \sin \omega_{0} \mathrm{t} \end{aligned}

$\therefore$ Current through thyristor :
$\mathrm{I}_{\mathrm{T}}=25+50 \sin \omega_{0} \mathrm{t}$

When $\mathrm{I}_{\mathrm{T}}=0 \Rightarrow$ Thyristor is turn off.

$0=25+50 \sin \omega_{0} t$
$\Rightarrow \quad \omega_{0} \mathrm{t}=210^{\circ} \times \frac{\pi}{180^{\circ}}$
$\Rightarrow \quad \mathrm{t}=\sqrt{\mathrm{LC}} \times \frac{7 \pi}{6}$
$=\sqrt{10^{-6} \times 4 \times 10^{-6}} \times \frac{7 \pi}{6}$
$=7.33 \mu \mathrm{sec}$

 Question 3
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 Semiconductor Devices and Commutation Techniques
Question 3 Explanation:
I-V characteristic of given switch :

this characteristic shows by options (A) and (D) switch.
 Question 4
The chopper circuit shown in figure (i) feeds power to a 5 A DC constant current source. The switching frequency of the chopper is $100 \mathrm{kHz}$. All the components can be assumed to be ideal. The gate signals of switches $S_{1}$ and $S_{2}$ are shown in figure (ii). Average voltage across the $5 \mathrm{~A}$ current source is

 A $10 \mathrm{~V}$ B $6 \mathrm{~V}$ C $12 \mathrm{~V}$ D $20 \mathrm{~V}$
GATE EE 2023      Choppers
Question 4 Explanation:
Output Voltage Waveform :

$\therefore$ Average voltage, $\mathrm{V}_{0}=\frac{1}{10}[20 \times 3]=6 \mathrm{~V}$
 Question 5
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 5 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}

There are 5 questions to complete.