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Cross-over distortion and class B push-pull efficiency

Short answer

Crossover distortion is removed by class AB biasing: a small forward bias via diodes holds both transistors at the edge of conduction, yielding smooth handover. The general efficiency of a class B push-pull stage is η = (π/4)(Vm/VCC), reaching a maximum of 78.5% when Vm = VCC.

How can we solve the cross-over distortion in Class B amplifier? Calculate the general efficiency of class B push-pull Amplifier.

Solving crossover distortion

Crossover distortion appears near the zero crossing of a class B push-pull output, because neither transistor conducts while the input is below about 0.6 to 0.7 V. It is removed by:

  1. Class AB biasing. Give each transistor a small forward bias so that both conduct slightly at zero signal. Conduction angle becomes a little more than 180°, so the hand-over between the two devices is smooth.
  2. Diode or VBE-multiplier bias. Connect two forward-biased diodes between the two bases. Their drop (about 1.3 V) holds each base-emitter junction at the edge of conduction, and tracks temperature changes.
  3. Negative feedback. Put the output stage inside the feedback loop of an op-amp. The high loop gain shrinks the dead zone.
        +VCC
          |
       [R]
          |-----B  Q1 (NPN)
         D1
         D2
          |-----B  Q2 (PNP)
       [R]
          |
        -VCC

General efficiency of class B push-pull

Let the load current be sinusoidal with peak Im, so the peak output voltage is Vm = Im RL.

  • Output power: Po = Im² RL/2 = Vm²/(2RL)
  • Each supply delivers a half-sine current, with average Im/π. For both supplies together: Idc = 2Im/π, so Pi = VCC · 2Im/π = 2 VCC Vm/(π RL)

η = Po/Pi = (Vm²/2RL) / (2 VCC Vm/(π RL))

η = (π/4)(Vm/VCC) × 100 %

The maximum, at Vm = VCC, is π/4 = 78.5 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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