Thyristor protection and diode reverse recovery numerical

A thyristor is protected against high $dv/dt$ by an RC snubber connected across it and against high $di/dt$ by a small inductor in series with it.

dv/dt protection

• In forward blocking, junction $J_2$ acts like a capacitor; a fast rising voltage drives a current $C_j\,dv/dt$ that can falsely trigger the SCR.
• An $R_s$-$C_s$ snubber across the SCR makes the voltage rise slowly while $C_s$ charges ($dv/dt \approx R_s V_s/L$). $R_s$ limits the capacitor discharge current when the SCR turns ON.

di/dt protection

• At turn-on, conduction starts near the gate and spreads slowly; a very fast current rise creates hot spots.
• A series inductor $L_s$ limits $di/dt = V_s/L_s$. A strong, fast gate pulse also helps by turning ON a larger area at once.

         Ls (di/dt)
   o----/\/\/\/----+-----------+
                   |           |
                 A |           Rs
                  SCR          |
                 K |           Cs
                   |           |
   o---------------+-----------+
       (RC snubber across SCR)

Numerical: reverse recovery of a power diode

Given: $t_{rr} = 3.9\ \mu$s, $di/dt = 50$ A/µs, softness factor $S = t_b/t_a = 0.3$.

$$\begin{aligned} t_{rr} &= t_a + t_b = t_a(1 + S) \\ t_a &= \frac{3.9}{1.3} = 3.0\ \mu\text{s}, \qquad t_b = 0.3 \times 3 = 0.9\ \mu\text{s} \\ I_{RR} &= t_a\frac{di}{dt} = 3.0\ \mu\text{s} \times 50\ \text{A}/\mu\text{s} = 150\ \text{A} \\ Q_{RR} &= \tfrac{1}{2} I_{RR}\, t_{rr} = \tfrac{1}{2} \times 150 \times 3.9 \times 10^{-6} = 292.5\ \mu\text{C} \end{aligned}$$

Answer: $I_{RR} = 150$ A and $Q_{RR} = 292.5\ \mu$C.