---
title: "Step-down chopper with switch drop: efficiency"
subject: "Power Electronics"
chapter: 4
key: "semester_7/power_electronics:step-down-chopper-efficiency-1v"
source: https://www.bctengineer.com/q/semester_7/power_electronics/step-down-chopper-efficiency-1v
site: BCT Engineering
---

# Step-down chopper with switch drop: efficiency

Power Electronics · Chapter 4: DC chopper

- Asked in: 2076 Chaitra, 8 marks
- Video: [Mathematical problem chopper](https://www.youtube.com/watch?v=LS-xyb8jiPs) — DEPT of EE, GMIT
- 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-efficiency-1v
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-4

## Question

A step down dc chopper has a 20 ohm resistor load and input voltage 220V. When the converter switch is ON, its voltage drop in chopper switch is 1V and the chopper frequency is 2KHz. If the duty cycle is 0.8, calculate: (i) Average value of the output voltage (ii) Efficiency of chopper circuit

## Short answer

With ON-state drop V_ch = 1 V, the effective pulse height is V_s - V_ch = 219 V. Average output voltage V_a = k(V_s - V_ch) = 0.8 × 219 = 175.2 V. Efficiency η = (V_s - V_ch)/V_s = 219/220 = 99.55%.

## Answer

When the chopper switch has an ON-state drop $V_{ch}$, the load gets $(V_s - V_{ch})$ during $T_{on}$, and the switch dissipates power. Efficiency is the ratio of load power to input power.

**Given:** $R = 20\ \Omega$, $V_s = 220$ V, $V_{ch} = 1$ V, $f = 2$ kHz, $k = 0.8$.

#### (i) Average output voltage
$$
\begin{aligned}
V_a &= k\,(V_s - V_{ch}) = 0.8\times(220-1) = 0.8\times219 \\
&= 175.2\ \text{V}
\end{aligned}
$$

(The frequency 2 kHz only fixes $T = 0.5$ ms and $T_{on} = 0.4$ ms; it does not change average values for a resistive load.)

#### (ii) Efficiency
Output power (load voltage is $219$ V for a fraction $k$ of the time):

$$
\begin{aligned}
P_o &= \frac{1}{T}\int_0^{kT}\frac{(V_s - V_{ch})^2}{R}dt = k\,\frac{(V_s-V_{ch})^2}{R} \\
&= 0.8\times\frac{219^2}{20} = 1918.44\ \text{W}
\end{aligned}
$$

Input power (source supplies current $(V_s - V_{ch})/R$ during $kT$):

$$
\begin{aligned}
P_i &= \frac{1}{T}\int_0^{kT}V_s\,\frac{V_s - V_{ch}}{R}dt = k\,\frac{V_s(V_s-V_{ch})}{R} \\
&= 0.8\times\frac{220\times219}{20} = 1927.2\ \text{W}
\end{aligned}
$$

$$
\begin{aligned}
\eta &= \frac{P_o}{P_i} = \frac{V_s - V_{ch}}{V_s} = \frac{1918.44}{1927.2} = \frac{219}{220} \\
&= 0.99545 = 99.55\%
\end{aligned}
$$

Switch loss $= 1927.2 - 1918.44 = 8.76$ W.

**Answer:** $V_a = 175.2$ V; efficiency $\eta = 99.55\%$.

## 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. 175.2 V
   B. 176 V
   C. 220 V
   D. 219 V
   Answer: A. Va = k(Vs - Vch) = 0.8 x 219 = 175.2 V. (Key point: Average voltage with drop)

2. Does the 2 kHz chopping frequency affect the average values for this resistive load?
   A. Yes, higher frequency lowers the average voltage
   B. No, it only fixes T = 0.5 ms and Ton = 0.4 ms
   C. Yes, it changes the efficiency
   D. Yes, it sets the switch drop
   Answer: B. Averages for a resistive load depend on k, not f. (Key point: Effect of frequency)

3. Which expression gives the efficiency?
   A. Vch/Vs
   B. k(Vs - Vch)/Vs
   C. (Vs - Vch)/Vs
   D. Vs/(Vs - Vch)
   Answer: C. Po/Pi = (Vs-Vch)^2/(Vs(Vs-Vch)) = (Vs-Vch)/Vs. (Key point: Efficiency formula)

4. What is the chopper efficiency?
   A. 95.5%
   B. 99.00%
   C. 80%
   D. 99.55%
   Answer: D. 219/220 = 0.99545; switch loss is about 8.76 W. (Key point: Efficiency value)

## Related in Power Electronics

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