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

# Quasi complementary-symmetry class AB amplifier

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

- Asked in: 2078 Bhadra, 3 marks
- Video: [Power Amplifier(Part 7): Class AB/B Amplifier (Diode Biasing & CE Driver)](https://www.youtube.com/watch?v=CUV8E9WH2VQ) — conceptech2018
- 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/quasi-complementary-class-ab
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-4

## Question

Draw the circuit diagram of Quasi complementary-symmetry class AB amplifier using diodes.

## Short answer

Quasi complementary symmetry uses two NPN power transistors: an NPN Darlington for the upper half and a PNP driver with NPN power transistor (Sziklai pair) for the lower half, acting as a high-power PNP. Diodes provide bias for class AB operation, reducing crossover distortion.

## Answer

A quasi-complementary class AB amplifier is used when a matched pair of NPN and PNP power transistors is not available. Both output (power) transistors are NPN. The upper half is an NPN Darlington pair, and the lower half is a PNP driver feeding an NPN power transistor (a feedback or Sziklai pair, which acts like a high-power PNP).

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

- Q1-Q2: NPN Darlington (upper half). Q1 base is at the top of the diode string; Q2 emitter goes through Re to the output.
- Q3-Q4: PNP driver Q3 (emitter to the output) whose collector drives the base of the NPN power transistor Q4 (collector to the output, emitter to −VCC). This pair is the lower half.
- R1, R2 and the diodes form the bias chain between the two driver bases; the input is coupled in through Cc. Exact wiring and the number of diodes vary between textbooks.

#### Working
- The diodes (or a VBE multiplier) develop enough voltage to forward bias the base-emitter junctions in the signal path. There are three junctions between the two driver bases (Q1, Q2 and Q3), so three diodes are used (some texts show two diodes plus a small resistor). This sets a small quiescent current, which removes crossover distortion (class AB).
- Positive half-cycle: Q1-Q2 conduct and source current into RL from +VCC.
- Negative half-cycle: Q3-Q4 conduct and sink current from RL to −VCC. The combination Q3-Q4 acts as a single PNP transistor with current gain of about β3 β4.
- Small emitter resistors Re give thermal stability and some current limiting.

## Check yourself

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

1. Why is a quasi-complementary stage used?
   A. Because a matched NPN-PNP power pair is not available
   B. To raise the efficiency to 100%
   C. To avoid using diodes
   D. Because it needs no supply
   Answer: A. Both output transistors are NPN; the lower half uses a PNP driver plus NPN power device to act like a PNP. (Key point: Why quasi-complementary)

2. What do D1, D2 achieve?
   A. They limit the output voltage
   B. They rectify the signal
   C. They forward bias the base-emitter junctions to set a small quiescent current, removing crossover distortion
   D. They replace Re
   Answer: C. The diode voltage sets a small idle current so both halves conduct around zero, i.e. class AB. (Key point: Role of diodes)

3. What is the effective gain of the lower half (Q3-Q4)?
   A. β3 + β4
   B. β3 β4, acting as a single PNP
   C. β4 only
   D. Unity
   Answer: B. The PNP driver feeding the NPN power transistor acts as a single PNP with gain β3 β4. (Key point: Lower half gain)

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