---
title: "Gate pulse duration in cycloconverter with inductive load"
subject: "Power Electronics"
chapter: 6
key: "semester_7/power_electronics:cyclo-short-gate-pulse-inductive"
source: https://www.bctengineer.com/q/semester_7/power_electronics/cyclo-short-gate-pulse-inductive
site: BCT Engineering
---

# Gate pulse duration in cycloconverter with inductive load

Power Electronics · Chapter 6: AC voltage controller

- Asked in: 2081 Bhadra, 3 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/cyclo-short-gate-pulse-inductive
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-6

## Question

Why short duration of gating pulse is not suitable for the inductive load in AC cyclo-converter?

## Short answer

With inductive load, the outgoing SCR still conducts when the incoming is fired, so the incoming is reverse biased and a short pulse ends before it becomes forward biased. Use a long gate pulse or pulse train to ensure turn-on.

## Answer

With an inductive load, the load current lags the voltage, so the outgoing thyristor is still conducting when the incoming thyristor is fired.

- Suppose the firing angle $\alpha$ is less than the load angle $\phi$. When the incoming SCR (say P2 or T2) is fired, the outgoing SCR is still carrying current until the extinction angle $\beta > \pi + \alpha$.
- The conducting SCR clamps the voltage, so the incoming SCR is **reverse biased** (or has no forward voltage) at the instant of the gate pulse.
- A **short pulse** ends before the outgoing SCR stops conducting, so the incoming SCR never turns on. It then waits for the next pulse, one full cycle later.
- Result: only one SCR (one group) conducts, the output becomes unsymmetrical, a DC component appears in the load and transformer, and frequency control is lost.

**Remedy:** use a **long (wide) gate pulse** lasting up to $\pi$ (or a high-frequency **pulse train**) so the incoming thyristor turns on as soon as the outgoing one stops and it becomes forward biased.

## Check yourself

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

1. With an inductive load, why can a short gate pulse fail to fire the incoming SCR?
   A. The pulse amplitude is too high
   B. The outgoing SCR still conducts, so the incoming SCR is reverse biased or has no forward voltage when pulsed
   C. The load current leads the voltage
   D. The gate has no cathode
   Answer: B. The load current lags the voltage, so the outgoing SCR is still on when the incoming one is fired. (Key point: Reverse-biased incoming SCR)

2. What is a result of the incoming SCR failing to fire?
   A. Output frequency doubles
   B. The output voltage becomes zero in both groups
   C. Only one group conducts, giving an unsymmetrical output with a DC component and loss of frequency control
   D. The supply is short circuited
   Answer: C. The incoming SCR waits a full cycle for the next pulse, so the output becomes unsymmetrical and DC appears. (Key point: Consequence)

3. What is the remedy?
   A. A long gate pulse up to pi or a pulse train
   B. A smaller firing angle
   C. A larger load resistance
   D. A higher supply frequency
   Answer: A. A wide pulse or pulse train keeps the gate signal present until the incoming SCR becomes forward biased. (Key point: Remedy)

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