A forced commutated thyristorized step - down chopper is shown in the figure. Neglect the ON - state drop across the power devices. Assume that the capacitor is initially charged to 50 V with the polarity shown in the figure. The load current (I_L) can be assumed to be constant at 10 A. Initially, Th_M is ON and Th_A is OFF. The turn – off time available to Th_M in microseconds, when Th_A is triggered, is _______(rounded off to the nearest integer)

The correct answer is 50.

1. Circuit Parameters and Identification from the Diagram:
Based on the provided circuit diagram and the problem description:
- Input Source Voltage: $V_s=50 V$
- Commutating Capacitor: $C=10 \mathrm{\mu F}$ (as labeled next to the capacitor in the image)
- Constant Load Current: $I_L=10 A$
- Initial Capacitor Voltage: $V_c=50 V$ (with the top plate positive and bottom plate negative, indicated by the "+" and "-" signs on the capacitor in the image)

2. Circuit Operation and Commutation Process:
Initially, the main thyristor ${\mathrm{Th}}_M$ is ON and conducting the load current. The auxiliary thyristor ${\mathrm{Th}}_A$ is OFF.
When ${\mathrm{Th}}_A$ is triggered to turn off the main thyristor:
- The auxiliary thyristor ${\mathrm{Th}}_A$ turns ON, connecting the bottom plate of the capacitor to the positive terminal of the 50 V DC source.
- Since the capacitor is charged to 50 V (top plate positive), the potential at the cathode of ${\mathrm{Th}}_M$ is raised to:
$V_{\mathrm{cathode}}=V_s+V_c=50+50=100 V$
- Since the anode of ${\mathrm{Th}}_M$ is connected to the 50 V supply, the voltage across the main thyristor becomes:
$V_{{\mathrm{Th}}_M}=50-100=-50 V$
This reverse voltage immediately turns ${\mathrm{Th}}_M$ OFF.

3. Calculation of Turn-off Time available to ${\mathrm{Th}}_M$:
After ${\mathrm{Th}}_M$ turns OFF, the constant load current $I_L$ flows through the capacitor and ${\mathrm{Th}}_A$ , discharging the capacitor.
The main thyristor ${\mathrm{Th}}_M$ remains reverse-biased until the capacitor voltage $v_c\left(t\right)$ drops to 0.
The available turn-off time (circuit turn-off time, $t_c$ ) is therefore the time required for the constant current $I_L$ to reduce the capacitor voltage from 50 V to 0 V:
$t_c=\frac{C·V_c}{I_L}$

Substituting the given parameters:
$t_c=\frac{10\times {10}^{-6} F\times 50 V}{10 A}$
$t_c=50\times {10}^{-6} \text{seconds}=50 \mathrm{\mu s}$