SCR di/dt, dv/dt and current ratings numerical

Data: $v_s = \sqrt{2}\times 230 \sin 314t$, so $V_m = 325.27$ V, $\omega = 314$ rad/s; $L = 15$ mH, $R = 2\ \Omega$; snubber $R_s = 10\ \Omega$, $C_s = 0.15\ \mu$F.

(i) Maximum di/dt and dv/dt

The worst case is when the voltage applied to the circuit is at its peak $V_m$, with zero initial current and an uncharged snubber capacitor.

di/dt at turn-on: when the SCR is fired, at $t = 0^+$ the current is zero, so the whole voltage appears across $L$:

$$\left(\frac{di}{dt}\right)_{max} = \frac{V_m}{L} = \frac{325.27}{15 \times 10^{-3}} = 21685\ \text{A/s} = 0.0217\ \text{A}/\mu\text{s}$$

dv/dt while OFF: the SCR is OFF and the snubber is in series with $L$ and $R$:

$$V_m = L\frac{di}{dt} + (R + R_s)\,i + v_C$$

The voltage across the SCR is $v_T = R_s i + v_C$. At $t = 0^+$, $i = 0$ and $v_C = 0$, so

$$\begin{aligned} \frac{dv_T}{dt} &= R_s\frac{di}{dt} + \frac{i}{C_s} = R_s\frac{V_m}{L} \\ &= 10 \times \frac{325.27}{0.015} = 216846\ \text{V/s} \approx 0.217\ \text{V}/\mu\text{s} \end{aligned}$$

(The snubber discharge current through the SCR at turn-on is limited by $R_s$ to $V_m/R_s = 32.5$ A.)

(ii) Average and RMS current of the SCR at $\alpha = 90^\circ$

The circuit is a half-wave controlled rectifier with RL load.

$$\begin{aligned} X_L &= \omega L = 314 \times 0.015 = 4.71\ \Omega \\ Z &= \sqrt{2^2 + 4.71^2} = 5.117\ \Omega \\ \phi &= \tan^{-1}\frac{4.71}{2} = 66.99^\circ \end{aligned}$$

Load (= SCR) current for $\alpha \le \omega t \le \beta$:

$$i = \frac{V_m}{Z}\left[\sin(\omega t - \phi) - \sin(\alpha - \phi)\,e^{-(\omega t - \alpha)/\tan\phi}\right]$$

Setting $i(\beta) = 0$ and solving numerically gives the extinction angle $\beta = 239.58^\circ$.

Average current (from the average output voltage):

$$\begin{aligned} I_{avg} &= \frac{V_m}{2\pi R}(\cos\alpha - \cos\beta) \\ &= \frac{325.27}{2\pi \times 2}(\cos 90^\circ - \cos 239.58^\circ) = 13.11\ \text{A} \end{aligned}$$

RMS current, by integrating $i^2$ from $\alpha$ to $\beta$:

$$I_{rms} = \sqrt{\frac{1}{2\pi}\int_{\alpha}^{\beta} i^2\, d(\omega t)} = 22.47\ \text{A}$$

Answer: $(di/dt)_{max} = 21685$ A/s, $(dv/dt)_{max} \approx 2.17 \times 10^5$ V/s; SCR average current $= 13.11$ A and RMS current $= 22.47$ A at $\alpha = 90^\circ$ (the SCR should be rated above these values with a safety margin).