Reverse recovery: show IRR = √(2·QRR·di/dt)

Reverse recovery: when a conducting diode is switched to reverse bias, its current does not stop at zero. The minority carriers stored in the junction must first be removed, so the current falls through zero, flows in the reverse direction for a short time and then decays to zero.

  iD
   ^ IF
   |-------.
   |        \  slope = di/dt
   |         \
 --+----------\--------------------> t
   |           \      .--------
   |            \   .'
   |   - IRR ....\.'   tb
   |          ta  |<->|
   |         |<-->|
   |         |<---trr--->|

• $t_a$: time from current zero to the peak reverse current $I_{RR}$, while charge stored in the depletion region is removed.
• $t_b$: time for the reverse current to decay from $I_{RR}$ to about $0.25 I_{RR}$, while charge in the bulk is removed.
• Reverse recovery time $t_{rr} = t_a + t_b$; softness factor $S = t_b / t_a$.
• Peak reverse current: $I_{RR} = t_a \dfrac{di}{dt}$.
• Reverse recovery (stored) charge $Q_{RR}$ is the area under the reverse current, approximately a triangle:

$$Q_{RR} = \tfrac{1}{2} I_{RR}\, t_{rr}$$

Derivation of $I_{RR} = \sqrt{2Q_{RR}\,di/dt}$: for an abrupt-recovery diode $t_b$ is very small, so $t_{rr} \approx t_a$. Then

$$\begin{aligned} I_{RR} &= t_a \frac{di}{dt} \approx t_{rr}\frac{di}{dt} \\ Q_{RR} &= \tfrac{1}{2} I_{RR} t_{rr} = \tfrac{1}{2}\frac{di}{dt}\, t_{rr}^2 \\ \Rightarrow t_{rr} &= \sqrt{\frac{2 Q_{RR}}{di/dt}} \\ I_{RR} &= \frac{di}{dt}\sqrt{\frac{2 Q_{RR}}{di/dt}} = \sqrt{2 Q_{RR}\frac{di}{dt}} \end{aligned}$$

Also $t_{rr} = \sqrt{2Q_{RR}/(di/dt)}$, so $Q_{RR}$ and $I_{RR}$ both rise with the rate of fall of forward current. Example: $Q_{RR} = 50\ \mu$C and $di/dt = 20$ A/µs give $I_{RR} = \sqrt{2 \times 50 \times 10^{-6} \times 20 \times 10^{6}} = 44.7$ A.