--- title: "Cross-over distortion and class B push-pull efficiency" subject: "Electronics Devices and Circuit" chapter: 4 key: "semester_3/electronics_devices_and_circuit:crossover-distortion-class-b-efficiency" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/crossover-distortion-class-b-efficiency site: BCT Engineering --- # Cross-over distortion and class B push-pull efficiency Electronics Devices and Circuit · Chapter 4: Output Stages and Power Amplifiers - Asked in: 2081 Baisakh, 2+4 marks - Video: [Cross Over Distortion - Elimination of Cross Over Distortion in Class B Amplifier](https://www.youtube.com/watch?v=NwkooG8bwHw) — Padmasri Naban - 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/crossover-distortion-class-b-efficiency - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-4 ## Question How can we solve the cross-over distortion in Class B amplifier? Calculate the general efficiency of class B push-pull Amplifier. ## 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. ## Answer #### 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. ```text +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 %**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does class AB biasing remove crossover distortion? A. It gives each transistor a small forward bias so both conduct slightly at zero signal B. It raises the supply voltage C. It removes one transistor D. It lowers the load resistance Answer: A. A small forward bias makes the conduction angle slightly over 180 degrees, so the handover is smooth. (Key point: Class AB biasing) 2. Why are two forward-biased diodes used between the two bases? A. They amplify the signal B. Their drop (about 1.3 V) holds each junction at the edge of conduction and tracks temperature C. They limit the supply current to zero D. They convert the output to dc Answer: B. The diodes set the bias voltage and, being thermally coupled, follow temperature changes. (Key point: Diode bias) 3. What is the general efficiency of a class B push-pull amplifier? A. (π/4)(Vm/VCC) x 100%, max 78.5% B. (π/2)(Vm/VCC) x 100%, max 157% C. (Vm/VCC)^2 x 25% D. 50% x (Vm/VCC) Answer: A. Po = Vm^2/(2RL) divided by Pi = 2VCC Vm/(π RL) gives (π/4)(Vm/VCC), maximum 78.5% at Vm = VCC. 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