Question

The latching current of thyristors in the single-phase full converter is $I_L=500mA$ and the delay time is $t_d=1.5\mu s$. The converter is supplied from a 120 V, 60 Hz supply and has a load of L = 10 mH and $R=10\Omega$. The converter is operated with a delay angle of $\alpha ={30}^{\circ }$. The minimum value of gate pulse width, $t_G$ is _________

• A. 58 93 $\mu s$
• B. $61.93\mu s$
• C. $57.43\mu s$
• D. $60.43\mu s$

Answer 1

The correct option is D $60.43\mu s$

Given $I_L=500mA=0.5A$

$t_d=1.5\mu s$

$\alpha ={30}^{\circ }$

$L=10mH$

and R = $10\Omega$

Instantaneous value of the input voltage $V_g\left(t\right)=V_{m}sin\omega t$

Where, $V_m=\sqrt{2}\times 120=169.7V$

$V_1=V_{m}sin\omega t$

$=169.7\times sin\frac{\pi }{6}=84.85V$

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

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

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

$t_1\times \left(\frac{di}{dt}\right)=I_L$

or $t_1\times 8485=0.5$

$t_1=58.93\mu s$

Therefore, the minimum width of the gate pulse is

$t_G=t_1+t_d=58.93+1.5$

$=60.43\mu s$