---
title: "Series inductor for di/dt limit"
subject: "Power Electronics"
chapter: 1
key: "semester_7/power_electronics:series-inductor-didt-limit"
source: https://www.bctengineer.com/q/semester_7/power_electronics/series-inductor-didt-limit
site: BCT Engineering
---

# Series inductor for di/dt limit

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

- Asked in: 2070 Asar, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/series-inductor-didt-limit
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-1

## Question

Fig.1c shows the single phase half-wave controlled rectifier with di/dt protection provided by the series inductor. The load is a purely inductive having an inductance of 100 mH. Calculate the value of Ls so that di/dt is limited to 500A/sec. [Figure: Vs = 220 V source, series inductor Ls, thyristor and the load in series]

## Short answer

At firing, di/dt = Vm/(Ls+L). With Vm = 311 V and limit 500 A/s, Ls+L >= 0.622 H; load L = 0.1 H, so Ls >= 0.522 H.

## Answer

In the circuit the series inductor $L_s$ and the load inductance $L$ are in series with the thyristor. At the instant of firing the current is zero, so the whole supply voltage appears across the total inductance and
$$
\frac{di}{dt} = \frac{v_s}{L_s + L}
$$
The worst case is firing at the peak of the supply. Assumption: 220 V is the rms value.

$$
\begin{aligned}
V_m &= \sqrt{2} \times 220 = 311.13\ \text{V} \\
\frac{V_m}{L_s + L} &\le 500\ \text{A/s} \\
L_s + L &\ge \frac{311.13}{500} = 0.6223\ \text{H} \\
L_s &\ge 0.6223 - 0.100 = 0.5223\ \text{H}
\end{aligned}
$$

**Answer:** $L_s \approx 0.522$ H (522 mH).

## Check yourself

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

1. At the instant of firing, with zero initial current, what limits di/dt?
   A. Only the series inductor Ls
   B. Only the load resistance
   C. The total inductance Ls + L, so di/dt = vs/(Ls + L)
   D. The thyristor holding current
   Answer: C. The whole supply voltage appears across the total series inductance at the firing instant. (Key point: Initial di/dt)

2. Which voltage is used for the worst-case di/dt?
   A. The rms value 220 V
   B. The average value of the supply
   C. Half of the rms value
   D. The peak value Vm = sqrt(2) x 220 = 311.13 V
   Answer: D. Firing at the supply peak gives the largest initial di/dt, so Vm is used. (Key point: Worst-case firing instant)

3. What is the required Ls for di/dt <= 500 A/s with a 100 mH load?
   A. 0.522 H
   B. 0.622 H
   C. 0.100 H
   D. 0.311 H
   Answer: A. Ls + L >= 311.13/500 = 0.6223 H, so Ls >= 0.6223 - 0.100 = 0.522 H. (Key point: Result for Ls)

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