Question

R1 RL R2 Vs Given: Vs = 110 VRMS at 60 Hz R1 = 2 ko R2 is 100 ka IGT = 10 mA If the Conduction angle is desired to be 150°, what should R2 be set to? R = 13.49 kohm R = 5.78 kohm R = 5.71 kohm R = 13.56 kohm

Answer 1

Ans)

Given conduction angle  $\beta =150^{o}$

firing angle is given as

- 180º – B= 180º – 150º = 30°

g= 30°

The input voltage at this angle is

$V_{s,\alpha }=\sqrt{2}V_{s,rms}sin(\alpha )=\sqrt{2}*110*sin(30^{o})$

$V_{s,\alpha }=77.782$

Now equation for gate current is given as

$I_{GT}=\frac{V_{s,\alpha }}{R_{1}+R_{2}}=\frac{77.782}{2k+R_{2}}=10\: \: mA$ given

$2k+R_{2}=\frac{77.782}{10m}$

$R_{2}=7.78\: k-2k=5.78\: k\Omega$

$\mathbf{R_{2}=5.78\: k\Omega }$ so answer is R= 5.78 k ohms

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