---
title: "Class B efficiency, 20 V peak into 8 Ω, 25 V supply"
subject: "Electronics Devices and Circuit"
chapter: 4
key: "semester_3/electronics_devices_and_circuit:class-b-numeric-20v-8ohm"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/class-b-numeric-20v-8ohm
site: BCT Engineering
---

# Class B efficiency, 20 V peak into 8 Ω, 25 V supply

Electronics Devices and Circuit · Chapter 4: Output Stages and Power Amplifiers

- Asked in: 2068 Chaitra, 4 marks
- Video: [Class B Power Amplifier - High Power Amplifiers - Applied Electronics](https://www.youtube.com/watch?v=T9HvHYsbC3g) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/class-b-numeric-20v-8ohm
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-4

## Question

A class B audio amplifier is providing 20V peak sine wave signal to 8Ω speaker with power supply of 25V (=VCC). At what efficiency is it operating?

## Short answer

Output power Po = 20²/(2×8) = 25 W; input power Pi = 2×25×20/(π×8) ≈ 39.79 W; efficiency η = (π/4)(20/25) = 62.8%.

## Answer

**Given:** Class B push-pull, peak output voltage Vp = 20 V, load RL = 8 Ω, VCC = 25 V. The amplifier drives an 8 Ω speaker.

#### Formulas
- Peak load current: Im = Vp/RL
- Output power: Po = Vp²/(2RL)
- Input (dc) power: Pi = VCC · Idc = 2 VCC Vp/(π RL), because each supply gives an average current Im/π per half-cycle, and the two together give 2Im/π
- Efficiency: η = Po/Pi × 100 %

#### Calculation
- Im = 20/8 = 2.500 A
- **Output power** Po = 20² / (2 × 8) = **25.00 W**
- **Input power** Pi = 2 × 25 × 20 / (π × 8) = **39.79 W**
- **Efficiency** η = 25.00 / 39.79 = **62.8 %**

Check with η = (π/4)(Vp/VCC) = 0.7854 × 20/25 = 62.8 %.

The power lost as heat in the two transistors is Pi − Po = 14.79 W.

**Answer: Pi = 39.79 W, Po = 25.00 W, η = 62.8 %.**

## Check yourself

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

1. What is the output power for Vp = 20 V into 8 Ω?
   A. 50 W
   B. 12.5 W
   C. 25 W
   D. 2.5 W
   Answer: C. Po = Vp^2/(2RL) = 400/16 = 25 W. (Key point: Output power)

2. What is the input power with VCC = 25 V?
   A. 62.5 W
   B. 50 W
   C. 19.9 W
   D. 39.79 W
   Answer: D. Pi = 2 VCC Vp/(π RL) = 2 x 25 x 20/(π x 8) = 39.79 W. (Key point: Input power)

3. At what efficiency is it operating?
   A. 62.8%
   B. 78.5%
   C. 80%
   D. 50%
   Answer: A. 25/39.79 = 62.8%, matching (π/4)(20/25). (Key point: Efficiency)

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