---
title: "Step-down chopper: fundamental of load current"
subject: "Power Electronics"
chapter: 4
key: "semester_7/power_electronics:step-down-chopper-fundamental-load-current"
source: https://www.bctengineer.com/q/semester_7/power_electronics/step-down-chopper-fundamental-load-current
site: BCT Engineering
---

# Step-down chopper: fundamental of load current

Power Electronics · Chapter 4: DC chopper

- Asked in: 2075 Asoj, 8 marks
- Video: [Principle of Step Down Chopper | Power Electronics | Lecture 67](https://www.youtube.com/watch?v=YRtExI37dt0) — 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-fundamental-load-current
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-4

## Question

Determine the average value of output voltage and the fundamental component of the load current of step down chopper having input voltage of 220V DC, load resistance 20 Ω and duty cycle of 45%. Given: chopping frequency of chopper is 1 kHz.

## Short answer

Average output voltage V_a = kV_s = 0.45 × 220 = 99 V. Fundamental load current peak from Fourier series: I_1m = (2V_s/π) sin(πk) / R = (440/π) sin(81°) / 20 ≈ 6.917 A (rms 4.891 A).

## Answer

The output of a step-down chopper with resistive load is a rectangular pulse train (height $V_s$, width $kT$). Its Fourier series gives the dc (average) value and the harmonics; the fundamental is at the chopping frequency.

**Given:** $V_s = 220$ V, $R = 20\ \Omega$, $k = 0.45$, $f = 1$ kHz.

#### Average output voltage
$$
V_a = kV_s = 0.45\times220 = 99\ \text{V}, \qquad I_a = \frac{99}{20} = 4.95\ \text{A}
$$

#### Fourier series of the output voltage
With the pulse from $t = 0$ to $kT$:

$$
v_0(t) = kV_s + \sum_{n=1}^{\infty}\frac{V_s}{n\pi}\Big[\sin(2n\pi k)\cos(n\omega t) + \big(1-\cos(2n\pi k)\big)\sin(n\omega t)\Big]
$$

Peak of the $n$th harmonic:

$$
V_{nm} = \frac{V_s}{n\pi}\sqrt{\sin^2(2n\pi k) + \big(1-\cos 2n\pi k\big)^2} = \frac{2V_s}{n\pi}\sin(n\pi k)
$$

#### Fundamental component ($n = 1$)
$$
\begin{aligned}
V_{1m} &= \frac{2V_s}{\pi}\sin(\pi k) = \frac{2\times220}{\pi}\sin(0.45\times180^\circ) \\
&= 140.06\times\sin 81^\circ = 140.06\times0.98769 = 138.33\ \text{V} \\
V_{1,rms} &= \frac{138.33}{\sqrt2} = 97.82\ \text{V}
\end{aligned}
$$

Fundamental load current (resistive load):

$$
\begin{aligned}
I_{1m} &= \frac{V_{1m}}{R} = \frac{138.33}{20} = 6.917\ \text{A} \\
I_{1,rms} &= \frac{6.917}{\sqrt2} = 4.891\ \text{A}
\end{aligned}
$$

Phase: $\tan\phi = \dfrac{\sin 2\pi k}{1-\cos 2\pi k} = \tan(90^\circ - 81^\circ)$, so $\phi = 9^\circ$:

$$
i_1(t) = 6.917\sin(2000\pi t + 9^\circ)\ \text{A}
$$

**Answer:** $V_a = 99$ V; fundamental load current $= 6.917$ A peak ($4.891$ A rms) at 1 kHz.

## Check yourself

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

1. What is the average output voltage?
   A. 99 V
   B. 220 V
   C. 45 V
   D. 4.95 V
   Answer: A. Va = k Vs = 0.45 x 220 = 99 V (average current 4.95 A). (Key point: Average voltage)

2. At what frequency is the fundamental component of the output?
   A. 50 Hz
   B. The chopping frequency, 1 kHz
   C. The 0 Hz dc value
   D. 2 kHz
   Answer: B. The fundamental of the pulse train is at the chopping frequency. (Key point: Fundamental frequency)

3. What is the peak of the fundamental voltage?
   A. k Vs
   B. (Vs/pi) sin(2 pi k)
   C. (2Vs/pi) sin(pi k)
   D. Vs sin(pi k)
   Answer: C. From V_nm = (2Vs/n pi) sin(n pi k) at n = 1. (Key point: Fundamental amplitude formula)

4. What is the peak fundamental load current?
   A. 4.95 A
   B. 11 A
   C. 138.33 A
   D. 6.917 A
   Answer: D. V1m = 138.33 V; I1m = 138.33/20 = 6.917 A (resistive load). (Key point: Fundamental current)

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