Question

A voltage commutated chopper operating at 1 kHz is used to control the speed of dc motor as shown in figure. The load current is assumed to be constant at 10A.
The minimum time in $\mu$sec for which the SCR M should be ON is

Answer 1

Initially, main thyristor(m)and auxiliary thyristor (A) are off and capacitor is assumed charged to voltage V with upper plate positive. When 'm' is turned on at t=0, source voltage V is applied across load and load current $I_o$ begins to flow which is assumed to remain constsnt.
With 'm' ON at t=0, another oscillatory circuit consisting of C,m,L and D is formed where the capacitor current is given by
$i_c=V\sqrt{\frac{C}{L}}sin{\omega }_o=I_{p}sin{\omega }_{o}t$



when, ${\omega }_{0}t=\pi ,i_c=0$.
Between $0<t<\frac{\pi }{{\omega }_0},i_{T1}=I_0+I_{p}sin{\omega }_{0}t$capacitor voltage charges from +$V_{s}to-V_s$ sinusoidaily and the lower plate becomes positive.
At${\omega }_{0}t=\pi ,i_c=0_i,i_{T1}=I_0$ and V${}_c=-V_s$.
At $t_t$, thyristor A is turned on and capacitor voltage V applies a reverse voltage across thyristor in and SCR 'm' is turned off. The load current is now carried by C and SCR A.
So, minimum time for which SCR 'm' should be ON.
$T_{min}=\frac{\pi }{{\omega }_o}=\pi \sqrt{LC}$
$=\pi \sqrt{2\times {10}^{-10}\times 1\times {10}^{-6}}=140\mu s$