# Power Electronics Miscellaneous

 Question 1
A double pulse measurement for an inductively loaded circuit controlled by the IGBT switch is carried out to evaluate the reverse recovery characteristics of the diode. D, represented approximately as a piecewise linear plot of current vs time at diode turn-off. $L_{par}$ is a parasitic inductance due to the wiring of the circuit, and is in series with the diode. The point on the plot (indicate your choice by entering 1. 2, 3 or 4) at which the IGBT experiences the highest current stress is ______. A 1 B 2 C 3 D 4
GATE EE 2020   Power Electronics
Question 1 Explanation: Using KCL, $I_s=I_L-I_D$
$\therefore \; I_s$ is maximum when, $I_D$ is minimum, i.e. at point 3.
Therefore, IGBT experiences highest current stress at point 3.
 Question 2
For the circuit shown in the figure below, assume that diodes $D_1,D_2$ and $D_3$ are ideal. The DC components of voltages $v_1 \; and \; v_2$, respectively are
 A 0 V and 1 V B -0.5 V and 0.5 V C 1 V and 0.5 V D 1 V and 1 V
GATE EE 2017-SET-1   Power Electronics
Question 2 Explanation: \begin{aligned} V_{2\; avg}&=\frac{V_m}{\pi}=\frac{\pi/2}{\pi}=\frac{1}{2}=0.5V \\ V_{1\; avg} &=\frac{1}{2 \pi}[ \int_{0}^{\pi}\frac{\pi}{2}\sin 100 \pi t\cdot d(\omega t)\\ &+\int_{\pi}^{2\pi} \pi \sin 100 \pi t\cdot d(\omega t) ]\\ &= -0.5V \end{aligned}

 Question 3
Consider a HVDC link which uses thyristor based line-commutated converters as shown in the figure. For a power flow of 750 MW from System 1 to System 2, the voltages at the two ends, and the current, are given by: $V_{1}$=500 kV, $V_{2}$=485 kV and $V_{3}$=1.5 kA. If the direction of power flow is to be reversed (that is, from System 2 to System 1) without changing the electrical connections, then which one of the following combinations is feasible? A $V_{1}=-500 kV,V_{2}=-485 kV$ and latex]I=1.5kA[/latex] B $V_{1}=-485 kV,V_{2}=500 kV$ and latex]I=1.5kA[/latex] C $V_{1}=500 kV,V_{2}=-485 kV$ and latex]I=-1.5kA[/latex] D $V_{1}=-500 kV,V_{2}=-485 kV$ and latex]I=-1.5kA[/latex]
GATE EE 2015-SET-1   Power Electronics
Question 3 Explanation:
To maintain the direction of power flow from system 2 to system 1, the voltage $V_1=-485kV$ and voltage $V_2=500kV$ and $I=1.5kA$.
Since, current cannot flow in reverse direction. Option (B) is correct answer.
 Question 4
The SCR in the circuit shown has a latching current of 40 mA. A gate pulse of 50 $\mu$s is applied to the SCR. The maximum value of R in $\Omega$ to ensure successful firing of the SCR is ______. A 4050 B 5560 C 6060 D 8015
GATE EE 2014-SET-2   Power Electronics
Question 4 Explanation:
Let us assume the SCR is conducting, \begin{aligned} &I_{ss}=\frac{100}{500}=0.2A\\ &[\because \text{inductor will be dhort circuited in DC}]\\ &i(t)=I_{ss}(1-e^{-t/\tau })\\ &\tau =\frac{L}{R}=\frac{200 \times 10^{-3}}{500}\\ &\;\;=4 \times 10^{-4}\; sec\\ &\text{Given }t=50 \times 10^{-6}\; sec\\ &\therefore \; i(t)=0.2\left ( 1-e^{\frac{50 \times 10^{-6}}{4 \times 10^{-4}}} \right )=23.5A \end{aligned} $V=I \times R$
$R=\frac{V}{I}=\frac{100}{16.5 \times 10^{-3}} =6060 \Omega$
 Question 5
A single-phase SCR based ac regulator is feeding power to a load consisting of 5 $\Omega$ resistance and 16 mH inductance. The input supply is 230 V, 50 Hz ac. The maximum firing angle at which the voltage across the device becomes zero all throughout and the rms value of current through SCR, under this operating condition, are
 A 30$^{\circ}$ and 46 A B 30$^{\circ}$ and 23 A C 45$^{\circ}$ and 23 A D 45$^{\circ}$ and 32 A
GATE EE 2014-SET-2   Power Electronics
Question 5 Explanation:
The maximum firing angle at which the voltage across the device becomes $'\phi '$ = load angle.
\begin{aligned} \phi &= \tan ^{-1} \left ( \frac{\omega L}{R} \right )\\ &= \tan^{-1}\left ( \frac{2 \pi \times 50 \times 16 \times 10^{-3}}{5} \right ) \\ \phi &=45.15\simeq 45^{\circ} \end{aligned}
Rms value of current through SCR is
\begin{aligned} I_{T_{rms}} &=\sqrt{\left [ \frac{1}{2 \pi}\int_{\alpha }^{\pi+\alpha }\left ( \frac{V_m}{z}\sin (\omega t-\phi ) \right )^2 d(\omega t) \right ]}\\ &=\sqrt{\frac{V_m^2}{2 \pi z^2} \int_{\alpha }^{\pi+\alpha } \left [ \frac{1-\cos 2(\omega t-\phi )}{2} \right ] d(\omega t)}\\ &=\sqrt{\frac{V_m^2}{2 \times 2 \pi z^2} \left [\pi- \left.\begin{matrix} \frac{\sin 2(\omega t-\phi )}{2} \end{matrix}\right|_\alpha ^{(\pi+\alpha )} \right ] }\\ &=\frac{V_m}{2z \sqrt{\pi} } \sqrt{\left [ \pi+\frac{-\sin 2(\pi+\alpha -\alpha )+\sin 2(\alpha -\alpha )}{2} \right ] }\\ &=\frac{V_m}{2z \sqrt{\pi} }\sqrt{\pi}=\frac{V_m}{2z}\\ I_{T_{rms}}&=\frac{230\sqrt{2}}{2 \times \sqrt{5^2(2 \pi \times 50 \times 16 \times 10^{-3})^2}}\\ &=22.93\simeq 23A\\ I_{T_{rms}}&=23A \end{aligned}

There are 5 questions to complete.