---
title: "Complementary-symmetry class AB with Darlington pair"
subject: "Electronics Devices and Circuit"
chapter: 4
key: "semester_3/electronics_devices_and_circuit:complementary-symmetry-class-ab"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/complementary-symmetry-class-ab
site: BCT Engineering
---

# Complementary-symmetry class AB with Darlington pair

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

- Asked in: 2076 Chaitra, 3 marks; 2076 Ashwin, 3 marks; 2075 Ashwin, 3 marks; 2069 Chaitra, 2 marks; 2071 Chaitra, 3 marks; 2070 Chaitra, 3 marks (asked 6 times)
- Video: [Power Amplifiers Complementary Symmetry and Class C](https://www.youtube.com/watch?v=Jslr23Vo3lA) — Gopi Krishna
- 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/complementary-symmetry-class-ab
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-4

## Question

Draw the circuit diagram of Complementary-Symmetry class-AB amplifier using Darlington pair transistors.

## Short answer

The Darlington complementary-symmetry class AB uses NPN and PNP Darlington pairs. A diode string provides 4VBE bias to eliminate crossover distortion, and small emitter resistors give thermal stability.

## Answer

A complementary-symmetry class AB amplifier uses an NPN and a PNP power transistor in push-pull, with the bases biased slightly above cut-off to remove crossover distortion. When high current gain is needed, each power transistor is replaced by a **Darlington pair**, so that the driver can be small.

#### Circuit diagram
```text
  +VCC ---+------------------+------------------+
          |                  |                  |
        [R1]                 C                  C
          |                  |                  |
          +--------------B  Q3 (NPN)            |
          |                  E --------------B  Q1 (NPN)
         D1                                     E
         D2                                     |
 Vi o-||--+                                    [Re]
         D3                                     |
         D4                                     +--------o Vo
          |                                     |        |
          |                                    [Re]     [RL]
          |                                     |        |
          |                                     E       GND
          |                  E --------------B  Q2 (PNP)
          +--------------B  Q4 (PNP)            C
          |                  C                  |
        [R2]                 |                  |
          |                  |                  |
  -VCC ---+------------------+------------------+
```

(Q3–Q1 form the upper NPN Darlington and Q4–Q2 the lower PNP Darlington. In each pair the emitter of the first transistor drives the base of the second, and the collectors are joined to the supply.)

#### Working
- The diode string D1–D4 (or a VBE multiplier), carrying current through R1 and R2, provides about 4 × 0.7 V of bias: one diode drop for each of the four base-emitter junctions between the two input bases. This keeps each Darlington just conducting, giving class AB operation without crossover distortion.
- For a positive half-cycle the upper NPN Darlington conducts and sources current to RL; for a negative half-cycle the lower PNP Darlington sinks current from RL.
- The total current gain of each pair is β ≈ β1 β2, so the input source needs to supply only a small base current.
- Small emitter resistors (Re) give thermal stability and short-circuit protection.

## Check yourself

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

1. Why is each power transistor replaced by a Darlington pair?
   A. To increase the supply voltage
   B. To obtain high current gain so the driver can be small
   C. To remove the need for biasing
   D. To make the circuit class C
   Answer: B. A Darlington pair has gain β1 β2, so only a small base current is needed from the source. (Key point: Why Darlington pairs)

2. What do D1 and D2 (or a VBE multiplier) do in the circuit?
   A. Rectify the input signal
   B. Protect against reverse supply
   C. Bias the base-emitter junctions so each Darlington just conducts, removing crossover distortion
   D. Reduce the load current
   Answer: C. They provide the bias drop for all the base-emitter junctions, giving class AB operation with no dead zone. (Key point: Role of diodes)

3. During the negative half-cycle of the input, which part conducts?
   A. The upper NPN Darlington, sourcing current
   B. Both Darlingtons equally
   C. Neither conducts
   D. The lower PNP Darlington, sinking current from the load
   Answer: D. The lower PNP Darlington conducts on the negative half-cycle and sinks current from RL to -VCC. (Key point: Half-cycle operation)

## Related in Electronics Devices and Circuit

- [Quasi complementary-symmetry class AB amplifier](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Aquasi-complementary-class-ab) — Electronics Devices and Circuit, chapter 4
- [Crossover distortion and quasi-complementary class AB](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Acrossover-quasi-complementary-class-ab) — Electronics Devices and Circuit, chapter 4, asked 2×

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