---
title: "Diode reverse recovery numerical (2.4 µs)"
subject: "Power Electronics"
chapter: 1
key: "semester_7/power_electronics:diode-reverse-recovery-2-4us"
source: https://www.bctengineer.com/q/semester_7/power_electronics/diode-reverse-recovery-2-4us
site: BCT Engineering
---

# Diode reverse recovery numerical (2.4 µs)

Power Electronics · Chapter 1: Characteristics and specification of power electronic devices

- Asked in: 2071 Shrawan, 4 marks
- Video: [Reverse recovery current |  Numerical Example | Gate Question #Power#Diode#Reverse#Recovery#Chara](https://www.youtube.com/watch?v=Qsa4EuUMw30) — Simplified EEE Studies
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/diode-reverse-recovery-2-4us
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-1

## Question

A power diode has a reverse recovery time of 2.4 µs. If di/dt is 30 A/µs, find (i) stored charge (ii) peak inverse current.

## Short answer

Assuming abrupt recovery, stored charge QRR = ½ × di/dt × trr² = ½ × 30 A/µs × (2.4 µs)² = 86.4 µC. Peak inverse current IRR = trr × di/dt = 2.4 µs × 30 A/µs = 72 A.

## Answer

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Assuming abrupt recovery, which expression gives the stored charge Q_RR?
   A. di/dt x t_rr
   B. (1/2)(di/dt) t_rr^2
   C. (1/2)(di/dt)^2 t_rr
   D. sqrt(2 di/dt) t_rr
   Answer: B. The reverse current is a triangle of height I_RR = t_rr di/dt and base t_rr, so Q_RR = (1/2) I_RR t_rr = (1/2)(di/dt) t_rr^2. (Key point: Stored charge formula)

2. For t_rr = 2.4 us and di/dt = 30 A/us, what is Q_RR?
   A. 72 uC
   B. 36 uC
   C. 86.4 uC
   D. 172.8 uC
   Answer: C. Q_RR = 0.5 x 30e6 x (2.4e-6)^2 = 86.4 uC. (Key point: Value of stored charge)

3. What is the peak inverse current I_RR for this diode?
   A. 12.5 A
   B. 36 A
   C. 144 A
   D. 72 A
   Answer: D. I_RR = t_rr x di/dt = 2.4 x 30 = 72 A, which also matches sqrt(2 Q_RR di/dt). (Key point: Peak inverse current)

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