---
title: "Class B push-pull power and efficiency, 20 V peak"
subject: "Electronics Devices and Circuit"
chapter: 4
key: "semester_3/electronics_devices_and_circuit:class-b-push-pull-numeric-20v-16ohm"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/class-b-push-pull-numeric-20v-16ohm
site: BCT Engineering
---

# Class B push-pull power and efficiency, 20 V peak

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

- Asked in: 2081 Bhadra, 2+2+2 marks
- Video: [Class B Push-Pull Stage: Power Calculations](https://www.youtube.com/watch?v=YltVJ3KzaR4) — Mateo Aboy
- 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-push-pull-numeric-20v-16ohm
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-4

## Question

A Class B push pull amplifier is providing 20 V peak signal (Sine Wave) to a 16 Ω load and power supply is VCC = 30 V. Determine the input power, output power and circuit efficiency.

## Short answer

Output power Po = Vp²/(2RL) = 20²/(2×16) = 12.5 W. DC input power Pi = 2VCC Vp/(π RL) = 2×30×20/(π×16) ≈ 23.87 W. Efficiency η = (π/4)(Vp/VCC) = 52.4%.

## Answer

**Given:** Class B push-pull, peak output voltage Vp = 20 V, load RL = 16 Ω, VCC = 30 V. 

#### 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 = 20/16 = 1.250 A
- **Output power** Po = 20² / (2 × 16) = **12.50 W**
- **Input power** Pi = 2 × 30 × 20 / (π × 16) = **23.87 W**
- **Efficiency** η = 12.50 / 23.87 = **52.4 %**

Check with η = (π/4)(Vp/VCC) = 0.7854 × 20/30 = 52.4 %.

The power lost as heat in the two transistors is Pi − Po = 11.37 W.

**Answer: Pi = 23.87 W, Po = 12.50 W, η = 52.4 %.**

## 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 and RL = 16 Ω?
   A. 25 W
   B. 12.5 W
   C. 20 W
   D. 6.25 W
   Answer: B. Po = Vp^2/(2RL) = 400/32 = 12.5 W. (Key point: Output power)

2. Which expression gives the input power, and what is its value (VCC = 30 V)?
   A. 2 VCC Vp/(π RL) = 23.87 W
   B. VCC Vp/RL = 37.5 W
   C. VCC Vp/(π RL) = 11.9 W
   D. Vp^2/RL = 25 W
   Answer: A. Pi = VCC x 2Im/π with Im = 1.25 A gives 2 x 30 x 20/(π x 16) = 23.87 W. (Key point: Input power)

3. What is the efficiency, and what quick check confirms it?
   A. 78.5%, because Vp is large
   B. 52.4%, from (π/4)(Vp/VCC)
   C. 66.7%, from Vp/VCC
   D. 41.9%, from Po/VCC
   Answer: B. 12.5/23.87 = 52.4%, matching 0.7854 x 20/30. (Key point: Efficiency)

## Close variants set in other exams

- [Class B power and efficiency, 14 V peak, 16 Ω](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Aclass-b-numeric-14v-16ohm) — Electronics Devices and Circuit, chapter 4, asked 2×

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