Diode reverse recovery numerical (2.4 µs)

When a conducting power diode is reverse biased, its stored charge makes it conduct in reverse for the reverse recovery time $t_{rr}$. Assuming abrupt recovery ($t_b \ll t_a$, so $t_{rr} \approx t_a$), the reverse current is a triangle of height $I_{RR}$ and base $t_{rr}$.

Given: $t_{rr} = 2.4\ \mu$s, $di/dt = 30$ A/µs.

(i) Stored charge

$$\begin{aligned} Q_{RR} &= \tfrac{1}{2}\frac{di}{dt}\,t_{rr}^2 = \tfrac{1}{2} \times 30 \times 10^{6} \times (2.4 \times 10^{-6})^2 \\ &= 86.4 \times 10^{-6}\ \text{C} = 86.4\ \mu\text{C} \end{aligned}$$

(ii) Peak inverse current

$$I_{RR} = \sqrt{2Q_{RR}\frac{di}{dt}} = \sqrt{2 \times 86.4 \times 10^{-6} \times 30 \times 10^{6}} = 72\ \text{A}$$

(Check: $I_{RR} = t_{rr}\,di/dt = 2.4 \times 30 = 72$ A.)

Answer: $Q_{RR} = 86.4\ \mu$C and $I_{RR} = 72$ A.