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Power Amplifiers Class B

Push pull operation, crossover distortion, efficiency 78.5%.

Darshan N
Updated: 7 April 2026
7 min read

A Class B push-pull amplifier uses two transistors that take turns, each conducting for exactly half the input cycle. The TDA2003 car audio amplifier chip and most smartphone speaker drivers use Class B or AB topology because they can deliver watts of audio power without burning the battery.

Class B Push-Pull Amplifier+VCC (15V)Q1 NPNBD139VinVoutRL=8ΩQ2 PNPBD140-VCC (-15V)Crossoverdistortionnear Vin=0η_max= 78.5%
Figure 1: Class B push-pull configuration showing complementary transistor pair and output node

Core Concept

In Class B, both transistors are biased at cut-off with no input signal. Q1 (NPN, BD139) conducts only when Vin is positive. Q2 (PNP, BD140) conducts only when Vin is negative. The load RL sees a reconstructed full sine wave as the two halves are pushed and pulled alternately.

Because there is no quiescent current, there is almost no power wasted at idle. Maximum theoretical efficiency reaches 78.5%, compared to 25% for Class A. This is why Class B and its variant Class AB dominate audio power amplifiers from the LM386 to the TDA7294.

The cost is crossover distortion. Near zero volts, both transistors are off simultaneously because each needs about 0.6V of forward bias to conduct. This creates a notch in the output waveform around every zero crossing. Class AB solves this by adding a small forward bias (two diodes in the bias network) so the transistors overlap slightly near zero.

Key Equations

DC power consumed (sinusoidal input, peak voltage Vm): PDC = (2 × VCC × Im) / π where Im = Vm / RL is the peak load current.

AC output power: Pac = Vm² / (2 × RL)

Maximum efficiency: η_max = π/4 ≈ 78.5% at Vm = VCC

Power dissipated in each transistor: Pdiss = PDC - Pac Split equally between Q1 and Q2.

Maximum transistor dissipation (occurs at Vm = 2VCC/π): Pdiss_max = VCC² / (π² × RL)

Example
Given:
  VCC = 15V
  RL = 8 Ω (speaker load)
  Input: full-swing sinusoid, Vm = 15V (peak)

Why this formula:
  Class B efficiency is maximum when Vm reaches VCC.
  Both halves combine so PDC uses the average of a half-wave rectified sine.

Formula:
  Im = Vm / RL
  PDC = (2 × VCC × Im) / π
  Pac = Vm² / (2 × RL)
  η = Pac / PDC × 100%

Substitution:
  Im = 15 / 8 = 1.875 A
  PDC = (2 × 15 × 1.875) / π = 56.25 / 3.1416 = 17.90 W
  Pac = (15)² / (2 × 8) = 225 / 16 = 14.06 W

Calculation:
  η = 14.06 / 17.90 × 100% = 78.5%

  Total transistor dissipation = 17.90 - 14.06 = 3.84 W
  Each transistor dissipates 3.84 / 2 = 1.92 W

Final Answer:
  PDC = 17.90 W
  Pac = 14.06 W
  Efficiency = 78.5%
  Each transistor dissipates 1.92 W at full output
Exam Tip: GATE frequently tests Class B efficiency. Maximum efficiency is π/4 = 78.54%, which occurs when output peak voltage equals VCC. Do not confuse this with maximum transistor dissipation, which occurs at Vm = 2VCC/π (about 64% of VCC), not at full swing. Students often write that peak signal causes peak transistor heating, but peak dissipation is at Vm = 2VCC/π, and at Vm = VCC the transistor actually dissipates less.

Key Properties

  • Conduction angle is 180 degrees per transistor. Each transistor conducts for exactly one half cycle.
  • Maximum theoretical efficiency is 78.5% (π/4), much better than Class A's 25%.
  • Zero quiescent current at idle, so no power is wasted when no audio signal is present.
  • Crossover distortion appears near every zero crossing because both transistors are cut off at Vin ≈ 0.
  • Class AB is the practical solution: a small forward bias (0.6–1.2V from two diodes) keeps both transistors barely conducting at idle, eliminating crossover distortion.
  • Maximum transistor dissipation occurs at Vm = 2VCC/π, not at maximum output power.
  • Requires a complementary pair: one NPN (BD139, 2N3055) and one PNP (BD140, MJ2955) with matched characteristics.

Quick Revision

  • Conduction angle: 180° per transistor (NPN for +ve half, PNP for -ve half).
  • Max efficiency: 78.5% at Vm = VCC.
  • PDC = 2VCC×Im/π; Pac = Vm²/(2RL).
  • Zero quiescent current, so idle power dissipation is zero.
  • Crossover distortion is the main weakness, fixed in Class AB by small forward bias.
  • Peak transistor dissipation at Vm = 2VCC/π ≈ 0.637 × VCC.
  • Complementary symmetry pair required: one NPN and one PNP with same current/voltage ratings.
  • Exam trap: writing that maximum transistor dissipation occurs at maximum output swing (Vm = VCC). It actually occurs at Vm = 2VCC/π, which is less than VCC.

Class B Efficiency

Examine push-pull configurations and distortion.

Question 1 of 3

Q1.What is the theoretical maximum efficiency of a Class B push-pull amplifier?