# Phase Controlled Rectifiers

 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   Power Electronics
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   Power Electronics
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
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)

 A 12.45 B 25.32 C 14.25 D 17.64
GATE EE 2022   Power Electronics
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,
$I_{s1}=\frac{4I_o}{\pi} \cos \left ( \frac{\alpha }{2} \right )=\frac{4 \times 15}{\pi} \cos \left ( \frac{45^{\circ} }{2} \right )=17.64A$
 Question 4
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.
 A 0.44 B 0.22 C 0.66 D 0.88
GATE EE 2022   Power Electronics
Question 4 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}
 Question 5
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).
 A 200 B 1600 C 600 D 800
GATE EE 2022   Power Electronics
Question 5 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
$P_{o}=\frac{V_{or}^2}{R} =\frac{200^2}{50}=800W$

There are 5 questions to complete.