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

# Short note: Differential amplifier with active load

Advanced Electronics · Chapter 1: Operational Amplifier Circuits

- Asked in: 2074 Asoj, 3 marks
- Video: [BJT Diff Amp: Active Load](https://www.youtube.com/watch?v=fJpTHzLuB3k) — Mateo Aboy
- 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-diff-amp-active-load
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-1

## Question

Write a short note on differential amplifier with active load.

## Short answer

Using a PNP current mirror as active load replaces collector resistors, giving gain Ad = gm(ro2 ‖ ro4), twice that of passive single‑ended. It converts differential to single‑ended output, saves area, provides high CMRR, but demands a high‑input‑impedance next stage.

## Answer

A **differential amplifier with active load** uses a current mirror (usually PNP, Q3–Q4) in place of the two collector resistors of the differential pair Q1–Q2.

```text
      +VCC ---- Q3(diode) ==== Q4 ---- +VCC
                  |             |
                  |             +----- vo
       v1 ---B   Q1            Q2   B--- v2
                  +------+------+
                         I (tail source)
                       -VEE
```

**Working:** for a differential input vd, ic1 = +gm vd/2 and ic2 = −gm vd/2. Q3 carries ic1 and the mirror copies it into Q4, so the net current into the output node is ic1 − ic2 = gm·vd.

```text
Ad = gm (ro2 ‖ ro4)   (single-ended output)
Compare passive load: Ad = gm·RC/2
```

**Advantages:**
- Gain is twice that of a passive load of equal resistance, and in practice much higher since ro2‖ro4 is large (gain of 1000 or more in one stage).
- Differential to single-ended conversion without losing half the signal.
- Small dc voltage drop and small chip area.
- Good CMRR, since a common-mode change gives equal currents that the mirror cancels at the output.

**Limitations:** the high output resistance must be followed by a high-input-resistance stage (e.g. Darlington or emitter follower), and the gain depends on ro, which varies with devices. It is used as the input stage of the 741 and most op-amps.

## Check yourself

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

1. What replaces the collector resistors in a differential amp with active load?
   A. A PNP current mirror (Q3, Q4)
   B. A Zener diode
   C. An inductor
   D. A capacitor
   Answer: A. A mirror load serves as the active load. (Key point: Structure)

2. What is the gain with active load (single-ended)?
   A. RC/RE
   B. gm RC/2
   C. gm/ro
   D. gm (ro2 ‖ ro4)
   Answer: D. Net current gm vd flows into ro2 ‖ ro4. (Key point: Gain)

3. What advantage does the mirror load give?
   A. Differential to single-ended conversion without losing half the signal
   B. Lower gain
   C. Larger dc drop
   D. Larger chip area
   Answer: A. The mirror adds Q1's signal to Q2's at the output. (Key point: Benefit)

## Related in Advanced Electronics

- [Active load doubles differential amplifier gain](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aactive-load-gain-twice-passive) — Advanced Electronics, chapter 1, asked 2×
- [Mirror error and active-load differential gain](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Amirror-error-active-load-gain) — Advanced Electronics, chapter 1
- [AC analysis of differential amplifier](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Adifferential-amplifier-ac-analysis) — Advanced Electronics, chapter 1
- [Active load preference and Widlar in IC biasing](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aactive-load-widlar-ic-biasing) — Advanced Electronics, chapter 1

## Related in other subjects

- [Differential amplifier using BJT](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Adiff-amp-bjt-circuit) — Basic Electronics Engineering, chapter 4, asked 2×
- [BJT differential amplifier collector waveforms](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Adiff-amp-bjt-waveform) — Basic Electronics Engineering, chapter 4
- [Why class A amplifier is cooler with load](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Aclass-a-cooler-with-load) — Electronics Devices and Circuit, chapter 4
- [Piezoelectric load cell and charge amplifier](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_and_measurement%3Apiezoelectric-load-cell-charge-numerical) — Instrumentation and Measurement, chapter 7