When thyristor $T_{1}$ is conducting, capacitor C is charged to d.c. supply voltage $E_{\mathrm{dc}}$ through the resistor $R_{2}$ . Now, when $T_{2}$ is triggered, a voltage twice the d.c. supply voltage $E_{\mathrm{dc}}$ is applied to the $R_{1} C$ series circuit so that current through the circuit is,

$$i=\frac{2 E_{\mathrm{dc}}}{R_{1}} e^{-t / R_{1} C}-----(1)$$

Therefore, the voltage across the thyristor $T_{1}$ is

$$E_{T_{1}}=E_{\mathrm{dc}}-i R_{1}=E_{\mathrm{dc}}-\frac{2 E_{\mathrm{dc}}}{R_{1}} e^{-t / R_{1} C} \cdot R_{1}=E_{\mathrm{dc}}\left(1-2 e^{-t / R_{1} C}\right)$$

For making thyristor $T_{1}$ OFF, the capacitor voltage must be equal to the voltage $E_{t_{1}}$ .

$$\therefore \quad E_{c}=E_{\mathrm{dc}}\left(1-2 e^{-t / R_{1} C}\right)-----(2)$$

Let $t=t_{\text { off }}$ when $E_{c}=0$

$\therefore \quad$ Equation $(2)$ becomes

$$0=E_{\mathrm{dc}}\left(1-2 e^{-\mathrm{t} _\mathrm{{off}} / R_{1} C}\right) \ OR \ 0=1-2 e^{-\mathrm{t} _\mathrm{{off}} / R_{1} C}-----(3)$$

$$\therefore \quad t_{\text { off }}=0.6931 R_{1} C-----(4)$$

OR

$$C=1.44 \frac{t_{\mathrm{off}}}{R_{1}}$$

So, from Eq.(4), $R_{1}$ and C must be such that the turn-off time of SCR $T_{1},$ $t_{\text { off }}$ is satisfied.

The maximum allowable $\frac{dV}{\mathrm{d} t}$ rating for SCR $T_{1}$ may be obtained from the SCR $T_{1}$ data sheet.

The maximum $\frac{\mathrm{d} V}{\mathrm{d} t}$ across $T_{1}$ using the commutating components is given by,

$$\frac{\mathrm{d} V}{\mathrm{d} t}_{(\max )}\gt\frac{2 E_{\mathrm{dc}}}{R_{1} C}$$