---
title: "Diode reverse recovery numerical (3 µs, 30 A/µs)"
subject: "Power Electronics"
chapter: 1
key: "semester_7/power_electronics:diode-reverse-recovery-3us"
source: https://www.bctengineer.com/q/semester_7/power_electronics/diode-reverse-recovery-3us
site: BCT Engineering
---

# Diode reverse recovery numerical (3 µs, 30 A/µs)

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

- Asked in: 2079 Baishakh, 8 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-3us
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-1

## Question

Explain the reverse recovery characteristics of Diode with necessary waveforms. The reverse recovery time of a diode is 3 µs and the rate of fall of the diode current is 30 A/µs. Determine storage charge and peak reverse current.

## Short answer

When a conducting diode is reverse biased, stored charge causes a reverse current flow for the recovery time trr. Assuming abrupt recovery, stored charge Qrr = 0.5 × (di/dt) × trr² = 0.5 × 30 A/µs × (3 µs)² = 135 µC, and peak reverse current Irr = trr × di/dt = 90 A.

## Answer

**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.

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

#### Numerical

**Given:** $t_{rr} = 3\ \mu$s, $di/dt = 30$ A/µs. No softness factor is given, so the usual assumption of abrupt recovery ($t_b \ll t_a$, so $t_{rr} \approx t_a$) is used.

$$
\begin{aligned}
Q_{RR} &= \tfrac{1}{2}\frac{di}{dt}\, t_{rr}^2 = \tfrac{1}{2} \times 30 \times 10^{6} \times (3 \times 10^{-6})^2 \\
&= 135 \times 10^{-6}\ \text{C} = 135\ \mu\text{C} \\
I_{RR} &= \sqrt{2 Q_{RR}\frac{di}{dt}} = \sqrt{2 \times 135 \times 10^{-6} \times 30 \times 10^{6}} \\
&= \sqrt{8100} = 90\ \text{A}
\end{aligned}
$$

(Check: $I_{RR} = t_{rr}\,di/dt = 3 \times 30 = 90$ A.)

**Answer:** stored charge $Q_{RR} = 135\ \mu$C, peak reverse current $I_{RR} = 90$ A.

## Check yourself

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

1. Why does the diode current flow in the reverse direction for a short time after the diode is switched to reverse bias?
   A. The reverse supply drives a large leakage current
   B. Stored minority carriers in the junction must first be removed
   C. The depletion layer becomes thinner
   D. The diode capacitance charges through the load only
   Answer: B. The current continues until the stored minority carriers are removed, so it goes negative before decaying to zero. (Key point: Why reverse current flows)

2. How is the peak reverse current related to the rate of fall of current?
   A. IRR = ta * di/dt
   B. IRR = di/dt / ta
   C. IRR = tb / ta
   D. IRR = 0.25 * di/dt
   Answer: A. The current falls at slope di/dt for time ta to reach the peak, so IRR = ta*(di/dt). (Key point: Peak reverse current formula)

3. With trr = 3 us, di/dt = 30 A/us and abrupt recovery (trr ~ ta), what is the stored charge QRR?
   A. 90 uC
   B. 270 uC
   C. 135 uC
   D. 45 uC
   Answer: C. QRR = 1/2 * (di/dt) * trr^2 = 0.5 * 30e6 * (3e-6)^2 = 135 uC. (Key point: Stored charge result)

4. For the same data, what is the peak reverse current IRR?
   A. 90 A
   B. 45 A
   C. 135 A
   D. 10 A
   Answer: A. IRR = trr * di/dt = 3 us * 30 A/us = 90 A, which agrees with sqrt(2*QRR*di/dt). (Key point: Peak reverse current result)

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- [Reverse recovery: show IRR = √(2·QRR·di/dt)](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Areverse-recovery-irr-formula) — Power Electronics, chapter 1

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