---
title: "Thyristor protection and diode reverse recovery numerical"
subject: "Power Electronics"
chapter: 1
key: "semester_7/power_electronics:dvdt-didt-protection-reverse-recovery"
source: https://www.bctengineer.com/q/semester_7/power_electronics/dvdt-didt-protection-reverse-recovery
site: BCT Engineering
---

# Thyristor protection and diode reverse recovery numerical

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

- Asked in: 2082 Baishakh, 4+4 marks
- Video: [di/dt & dv/dt PROTECTION OF A THYRISTOR](https://www.youtube.com/watch?v=weAPDep1aA4) — KK CONCEPTS OF ELECTRICAL ENGINEERING
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/dvdt-didt-protection-reverse-recovery
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-1

## Question

Describe the dv/dt protection and di/dt protection methods for a Thyristors. For power diodes, the reverse recovery time is 3.9 µs and the rate of diode current decay is 50 A/µs. For a softness factor of 0.3, calculate the peak inverse current (Irr) and storage charge (Qrr).

## Short answer

dv/dt protection uses an RC snubber across the thyristor to slow voltage rise; di/dt protection uses a series inductor to limit current rise at turn-on. Given trr=3.9 µs, di/dt=50 A/µs, S=0.3: ta=3.0 µs, tb=0.9 µs, so IRR=150 A and QRR=292.5 µC.

## Answer

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.

```text
         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.

## Check yourself

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

1. Which pair correctly matches protection method to thyristor stress?
   A. Series inductor for dv/dt, RC snubber for di/dt
   B. RC snubber for dv/dt, series inductor for di/dt
   C. RC snubber for both
   D. Series inductor for both
   Answer: B. An RC snubber across the SCR limits dv/dt, while a series inductor limits di/dt = Vs/Ls. (Key point: Protection methods)

2. With trr = 3.9 us and softness factor S = tb/ta = 0.3, what is ta?
   A. 0.9 us
   B. 3.0 us
   C. 3.9 us
   D. 1.17 us
   Answer: B. trr = ta(1+S), so ta = 3.9/1.3 = 3.0 us and tb = 0.9 us. (Key point: Splitting trr)

3. What is the peak reverse current IRR for ta = 3.0 us and di/dt = 50 A/us?
   A. 15 A
   B. 50 A
   C. 195 A
   D. 150 A
   Answer: D. IRR = ta x di/dt = 3.0 x 50 = 150 A. (Key point: Peak reverse current)

4. Using QRR = 1/2 IRR trr, what is the storage charge?
   A. 292.5 uC
   B. 585 uC
   C. 450 uC
   D. 146 uC
   Answer: A. QRR = 0.5 x 150 A x 3.9 us = 292.5 uC. (Key point: Stored charge)

## Related in Power Electronics

- [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
- [Thyristor di/dt and dv/dt protection](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athyristor-didt-dvdt-protection) — Power Electronics, chapter 1, asked 4×
- [SCR di/dt, dv/dt and current ratings numerical](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ascr-snubber-didt-dvdt-numerical) — Power Electronics, chapter 1
- [Diode reverse recovery numerical (2.4 µs)](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Adiode-reverse-recovery-2-4us) — Power Electronics, chapter 1
- [di/dt, dv/dt protection and gate circuit design](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Aprotection-gate-design-factors) — Power Electronics, chapter 1
- [Why high di/dt and dv/dt harm a thyristor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ahigh-didt-dvdt-protection) — Power Electronics, chapter 1

## Related in other subjects

- [Reverse breakdown effect in diodes](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Areverse-breakdown-diodes) — Electronics Devices and Circuit, chapter 6, asked 2×