---
title: "Class A series-fed efficiency, 12 V supply"
subject: "Electronics Devices and Circuit"
chapter: 4
key: "semester_3/electronics_devices_and_circuit:class-a-series-fed-efficiency-numeric"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/class-a-series-fed-efficiency-numeric
site: BCT Engineering
---

# Class A series-fed efficiency, 12 V supply

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

- Asked in: 2082 Baisakh, 3+3 marks
- Video: [AEC#32 series fed Class A power amplifier || ECAcademy](https://www.youtube.com/watch?v=PxdhIyooeJM) — EC Academy
- 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-a-series-fed-efficiency-numeric
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-4

## Question

Calculate the efficiency of a class A series-fed amplifier for a supply of 12 V and outputs of: (a) V(P-P) = 12 V and (b) V(P-P) = 6 V.

## Short answer

Efficiency η = (VCE(p-p)²)/(4VCC²) = 25% × (VCEpp/VCC)². (a) For 12 V p-p output, η = 25%. (b) For 6 V p-p, η = 6.25%.

## Answer

**Given:** VCC = 12 V, series-fed class A stage.

**Assumption:** The Q point is set at the centre of the load line: VCEQ = VCC/2 = 6 V, ICQ = VCC/(2RC). The load resistance RC is not given, but it cancels in the efficiency.

#### Formulas
Pi = VCC · ICQ = VCC²/(2RC)

Po = VCE(p-p)² / (8 RC)  (output power for a sine swing of peak-to-peak value VCE(p-p))

η = Po/Pi = VCE(p-p)² / (4 VCC²) = 25 % × (VCE(p-p)/VCC)²

#### (a) VCE(p-p) = 12 V
η = 25 × (12/12)² = **25 %**

(Steps: Po/Pi = 144 / (4 × 144) = 0.25.)

This is the largest possible swing (full supply range 0 V to 12 V), so it is the maximum efficiency of a series-fed class A amplifier.

#### (b) VCE(p-p) = 6 V
η = 25 × (6/12)² = 25 × 0.25 = **6.25 %**

(Steps: Po/Pi = 36 / (4 × 144) = 0.0625.)

#### Comparison
| VCE(p-p) | VCE(p-p)/VCC | η |
|---|---|---|
| 12 V | 1.0 | 25 % |
| 6 V | 0.5 | 6.25 % |

Halving the swing reduces the output power to one quarter, while the input power stays constant, so the efficiency falls to one quarter.

**Answer: (a) 25 %, (b) 6.25 %.**

## Check yourself

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

1. For a series-fed class A stage with Q point at the centre, which formula applies?
   A. η = 25% x (VCE(p-p)/VCC)^2
   B. η = 50% x (VCE(p-p)/VCC)
   C. η = 78.5% x (VCE(p-p)/VCC)
   D. η = 25% x (VCE(p-p)/VCC)
   Answer: A. Po/Pi = VCE(p-p)^2/(4VCC^2), and RC cancels, so η = 25% x (swing/VCC)^2. (Key point: Efficiency formula)

2. What is the efficiency for VCE(p-p) = 12 V with VCC = 12 V?
   A. 12.5%
   B. 50%
   C. 6.25%
   D. 25%
   Answer: D. η = 25 x (12/12)^2 = 25%, the maximum for a series-fed class A amplifier. (Key point: Full swing result)

3. What is the efficiency for VCE(p-p) = 6 V, and why does it fall?
   A. 6.25%, because Po falls to one quarter while Pi stays constant
   B. 12.5%, because Po halves
   C. 6.25%, because Pi falls to one quarter
   D. 25%, because it is independent of swing
   Answer: A. Output power goes as swing squared, so halving the swing quarters Po while Pi = VCC ICQ is unchanged: 25% / 4 = 6.25%. (Key point: Effect of halving swing)

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