In the given rectifier, the delay angle of the thyristor \({T_1}\)_{} measured from the positive going zero crossing of \({V_s}\;\) is \(30^\circ\). If the input voltage \({V_s}\) is \(100\;sin\left( {100\pi t} \right)\;V\), the average voltage across \(R\) (in volt) under steady-state is_______.

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GATE EE 2015 Official Paper: Shift 2

CT 1: Ratio and Proportion

2846

10 Questions
16 Marks
30 Mins

__ Concept__:

A diode is forward biased when its voltage is anywhere on the +_{ve} voltage side of the origin.

During the positive half cycle of an A.C. sinusoidal waveform, the thyristor is forward biased (anode positive).

**Explanation:**

Given information,

supply voltage (V_{s}) = 100 sin(100πt)

firing angle (α) = 30°

__ Case I:__ For 0 < ωt < α

None of the device working

so, output voltage (V_{0}) = 0

__ Case II:__ For α < ωt < π,

T_{1}D_{2} : forward biased

D_{3}D_{4} : Reversed biased

given circuit can be redrawn as,

using KVL,

V_{0} = V_{s}

__ Case III:__ For π < ωt < 2π

D_{3}D_{4} : forward biased

T_{1}D_{2}: Reversed biased

using KVL, V_{0} = -V_{s}

Average output voltage is given as,

\(\rm V_0 = \frac{1}{2 \pi} \left[\displaystyle\int_{\pi/6}^\pi V_m \sin (\omega t) d(\omega t) + \displaystyle\int_{\pi}^{2\pi} -V_m \sin (\omega t) d(\omega t) \right]\)

\(\Rightarrow \rm \rm V_0 = \frac{1}{2 \pi} \left[\displaystyle\int_{\pi/6}^\pi 100 \sin (\omega t) d(\omega t) + \displaystyle\int_{\pi}^{2\pi} -100 \sin (\omega t) d(\omega t) \right]\)

\( \rm \Rightarrow \rm \rm V_0 = \frac{1}{2 \pi} \left[ -100 \left[ \cos \omega t \right]_{\pi/6}^\pi + 100 \left[ \cos \omega t \right]_{\pi}^{2\pi} \right]\)

\(\Rightarrow \rm V_0 = \frac{1}{2 \pi} \left[ -100 \left(\cos \pi - \cos \frac{\pi}{6} \right) + 100 ( \cos 2 \pi - \cos \pi) \right]\)

\(\Rightarrow \rm V_0 = \frac{1}{2 \pi} \left[ -100 \left( -1 - \frac{\sqrt 3}{2} \right) + 100(1+1) \right]\)

∴ V_{0} = 61.53 V