---
title: "Short note: Differential amplifiers"
subject: "Advanced Electronics"
chapter: 1
key: "semester_4/advanced_electronics:short-note-differential-amplifier"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/short-note-differential-amplifier
site: BCT Engineering
---

# Short note: Differential amplifiers

Advanced Electronics · Chapter 1: Operational Amplifier Circuits

- Asked in: 2073 Shrawan, 3 marks; 2072 Chaitra, 3 marks; 2068 Chaitra, 3 marks (asked 3 times)
- Video: [Differential Amplifiers, Part 1](https://www.youtube.com/watch?v=_pvlk_SlyDo) — Electronics with Professor Fiore
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/short-note-differential-amplifier
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-1

## Question

Write a short note on differential amplifiers.

## Short answer

A differential amplifier amplifies the difference between two inputs while rejecting common‑mode signals. With a BJT pair, double‑ended gain Ad = gmRC, single‑ended Ad = gmRC/2; common‑mode gain Acm ≈ –RC/(2RE); CMRR ≈ gmRE. Direct coupling, temperature stability, and noise rejection make it the standard op‑amp input stage.

## Answer

A **differential amplifier** is a two-input amplifier that amplifies the *difference* between its two input voltages and rejects any signal common to both. It is the input stage of every op-amp.

```text
          +VCC      +VCC
           |          |
           RC         RC
           |--Vo1  Vo2--|
        C1 |          | C2
  V1 --B Q1            Q2 B-- V2
         E1|          |E2
           +----+-----+
                |
           RE or current source
                |
              -VEE
```

**Output:** Vo = Ad·Vd + Acm·Vcm, where Vd = V1 − V2 (difference signal) and Vcm = (V1 + V2)/2 (common-mode signal).

**Key results (BJT pair, gm = IC/VT):**
- Differential gain, double-ended output: Ad = gm·RC; single-ended: gm·RC/2.
- Common-mode gain (single-ended): Acm ≈ −RC/(2RE), small when RE is large.
- CMRR = |Ad/Acm| ≈ gm·RE (single-ended). A constant-current tail (very large RE) gives very high CMRR.
- Differential input resistance Rid = 2rπ.

**Configurations:** dual-input balanced output, dual-input unbalanced output, single-input balanced output and single-input unbalanced output.

**Features and uses:** direct coupling (no capacitors, so it amplifies down to dc), good temperature stability because drifts in both halves cancel, rejection of noise and hum picked up equally by both inputs, and use as the input stage of op-amps, instrumentation amplifiers and comparators.

## Check yourself

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

1. What does a differential amplifier amplify?
   A. Only the common-mode signal
   B. Only dc signals
   C. The difference between its two inputs and rejects the common signal
   D. The sum of its two inputs
   Answer: C. It amplifies Vd = V1 − V2 and rejects signals common to both inputs. (Key point: Purpose)

2. How can CMRR of a BJT differential amplifier be increased?
   A. Use unequal collector resistors
   B. Use a very large RE, e.g. a constant-current tail
   C. Reduce gm
   D. Use a small tail resistance RE
   Answer: B. CMRR ≈ gm·RE (single-ended), so a very large RE or a current-source tail gives very high CMRR. (Key point: CMRR)

3. For a BJT pair, what is the double-ended differential gain Ad?
   A. gm·RC
   B. gm·RC/2
   C. 2rπ
   D. RC/(2RE)
   Answer: A. Double-ended output gives Ad = gm·RC; single-ended gives half of this. (Key point: Gain)