Question

The circuit in the figure is a current commutated DC - DC chopper where, $T{h}_M$ is the main SCR and $T{h}_{AUX}$ is the auxiliary SCR. The load current is constant at 10 A. $T{h}_M$ is ON. $T{h}_{AUX}$ is triggered at t = 0. $T_{hM}$ is turned OFF between.

Answer 1

To turn-off the main thyristor, in current commutated chopper, Entire main thyristor's current should transfer to capacitor.

For that, the capacitor polarities should reverse. To get this, we turn-on auxiliary SCR.

To calculate the starting point of turn-off time of SCR,

= Time to get capacitor reversed ($t_1$) + Time to get capacitor current equal to load current $t_q$




$w{t}_1=\pi$

where $w=\frac{1}{\sqrt{LC}}$

$t_1=\pi \sqrt{LC}$

$t_1=\pi \times \sqrt{10\times 25.28}$

$t_1=49.95\mu sec$

For $t_q$

$i_c=i_o$

$v_{dc}\sqrt{\frac{c}{L}}sin\omega t_2=i_o$

$230\times \sqrt{\frac{10}{25.28}}sin\omega t_2=10$

$\omega t_2=0.069radians$

$t_2=1.097=1\mu sec$

= $t_1+t_2$

= 51.04 $\mu sec$ is the starting time of the turn off of the main thyristor