---
title: "Step-down chopper variable frequency, 5 ms ON time"
subject: "Power Electronics"
chapter: 4
key: "semester_7/power_electronics:step-down-chopper-variable-frequency-5ms"
source: https://www.bctengineer.com/q/semester_7/power_electronics/step-down-chopper-variable-frequency-5ms
site: BCT Engineering
---

# Step-down chopper variable frequency, 5 ms ON time

Power Electronics · Chapter 4: DC chopper

- Asked in: 2071 Shrawan, 8 marks
- Video: [Chopper Control Techniques | Power Electronics | Lecture 66](https://www.youtube.com/watch?v=gZfUpLEk2rY) — Electronics Maddy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/step-down-chopper-variable-frequency-5ms
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-4

## Question

Explain the operation of a step down chopper in variable frequency mode. If it is operated with on time constant at 5msec. Calculate the OFF time for a duty cycle of 40% and average value of the output voltage.

## Short answer

With constant Ton = 5 ms and k = 0.4, period T = Ton/k = 12.5 ms, Toff = T − Ton = 7.5 ms. Average output voltage Vo = k · Vs = 0.4·Vs; if Vs = 200 V, Vo = 80 V.

## Answer

In a step-down chopper the average output is $V_o = kV_s$ with $k = T_{on}/T$. **Variable-frequency (frequency modulation) control** keeps either $T_{on}$ or $T_{off}$ constant and changes the period $T$ to change $k$.

#### Operation in variable-frequency mode
```text
  +---CH---+--------+
  |        |        |
 Vs      FD^      Load  vo
  |        |        |
  +--------+--------+

 constant Ton, larger T -> lower k
 vo |__      __      __
    |  |____|  |____|  |____
 constant Ton, smaller T -> higher k
 vo |__  __  __  __  __
    |  ||  ||  ||  ||  |
```
- The switch is turned ON for a fixed time $T_{on}$; the next turn-on is delayed (or advanced) to set $T$.
- Increasing $T$ (lower frequency) lowers $k$ and $V_o$; decreasing $T$ raises them.
- **Drawbacks:** the chopping frequency varies over a wide range, so input/output filter design is difficult; low frequency at small $k$ increases current ripple and may make load current discontinuous; possible interference with signalling and telephone lines. Hence constant-frequency (PWM) control is usually preferred.

#### Numerical
**Given:** $T_{on} = 5$ ms (constant), $k = 0.4$.

$$
\begin{aligned}
T &= \frac{T_{on}}{k} = \frac{5}{0.4} = 12.5\ \text{ms} \quad (f = 80\ \text{Hz}) \\
T_{off} &= T - T_{on} = 12.5 - 5 = 7.5\ \text{ms}
\end{aligned}
$$

Average output voltage:

$$
V_o = kV_s = 0.4\,V_s
$$

The supply voltage is not stated; taking $V_s = 200$ V (the value used in the same textbook circuit):

$$
V_o = 0.4\times200 = 80\ \text{V}
$$

**Answer:** $T_{off} = 7.5$ ms; $V_o = 0.4V_s$ ($= 80$ V for $V_s = 200$ V).

## Check yourself

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

1. In variable-frequency mode with constant Ton, increasing T does what?
   A. Lowers k and the output voltage
   B. Raises k
   C. Leaves k unchanged
   D. Raises the frequency
   Answer: A. k = Ton/T, so larger T lowers k. (Key point: Effect of period)

2. For Ton = 5 ms and k = 0.4, Toff is:
   A. 5 ms
   B. 7.5 ms
   C. 12.5 ms
   D. 2 ms
   Answer: B. T = 5/0.4 = 12.5 ms; Toff = 7.5 ms. (Key point: OFF time)

3. What chopping frequency results?
   A. 200 Hz
   B. 125 Hz
   C. 80 Hz
   D. 12.5 Hz
   Answer: C. f = 1/12.5 ms = 80 Hz. (Key point: Frequency)

4. Average output voltage with Vs = 200 V is:
   A. 200 V
   B. 40 V
   C. 500 V
   D. 80 V (generally 0.4 Vs)
   Answer: D. Vo = kVs = 0.4 x 200 = 80 V. (Key point: Average voltage)

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