Question

The latching current of thyristors in the single phase full converter is $I_L=300mA$ and the delay time is $t_d=1.5\mu sec.$ The converter is supplied from a 120 V, 60 Hz supply and has a load at L = 10 mH and R = 10 $\Omega$. The converter is operated with delay angle of $\alpha ={30}^o$. The minimum value of gate pulse width, $t_G$ is ___$\mu$sec.

• A. 36.86

Answer 1

The correct option is A 36.86
Given,
Latching current, $I_L=300mA$
$\alpha ={30}^o$
$L=10mH$
$R=10\Omega$
$t_d=1.5\mu sec$

The instantaneous value of the input voltage
$V_s\left(t\right)=V_{m}sin\omega t$
$V_m=120\sqrt{2}=169.705V$
$V_1=169.705sin\frac{\pi }{6}=84.85V$

The rate of rise of anode current $\frac{di}{dt}$ at triggering instant will be approximately

$\frac{di}{dt}=\frac{V_1}{L}=\frac{84.85}{10\times {10}^{-3}}=8485A/s$

If $\frac{di}{dt}$ is assumed constant for a short time, for gate triggering the time $t_1$ required for the anode current to rise to level of latching current.

$t_1\left(\frac{di}{dt}\right)=I_L$
$t_1\times 8485=0.3$
$t_1=35.356\mu sec$

Therefore, the minimum width of the gate pulse
$t_G=t_d+t_1=35.356+1.5$
$=36.856=36.86\mu sec$