--- title: "Class B power amplifier efficiency" subject: "Electronics Devices and Circuit" chapter: 4 key: "semester_3/electronics_devices_and_circuit:class-b-efficiency-derivation" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/class-b-efficiency-derivation site: BCT Engineering --- # Class B power amplifier efficiency Electronics Devices and Circuit · Chapter 4: Output Stages and Power Amplifiers - Asked in: 2083 Baisakh, 2+2+2 marks; 2082 Baisakh, 4+2 marks; 2079 Bhadra, 8 marks; 2075 Ashwin, 5 marks; 2073 Shrawan, 4+4 marks; 2068 Chaitra, 1+3+4 marks (asked 6 times) - Video: [Lec-160 Efficiency of Class B Power Amplifier-Derivation | A E | R K Classes | Hindi |](https://www.youtube.com/watch?v=ZvE2_kd_lZ8) — R K Classes - 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/class-b-efficiency-derivation - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-4 ## Question Draw the circuit diagram of class B power amplifier and derive the expression for its general efficiency. ## Short answer In a complementary-symmetry class B amplifier, each transistor conducts half cycle. General efficiency is η = (π/4)(Vm/VCC), reaching maximum 78.5% when Vm = VCC. ## Answer A **class B** power amplifier biases each transistor at cut-off, so each conducts for 180° (half of the cycle). Two complementary transistors work in push-pull: the NPN amplifies the positive half-cycle and the PNP the negative half. #### Circuit diagram (complementary-symmetry class B) ```text +VCC | C / Vi o--+--B Q1 (NPN) | E \ | +-----||----+ | | | | E / [RL] +--B Q2 (PNP) | C \ GND | -VCC ``` #### Derivation of efficiency Let the load current be sinusoidal with peak Im (and peak output voltage Vm = Im RL). **Output (ac) power:** Po = Vm Im/2 = Im² RL/2 = Vm²/(2RL) **Input (dc) power:** Each transistor draws a half-sine current pulse from its supply. The average of a half-sine of peak Im is Im/π. Both supplies together: Idc = 2Im/π, Pi = VCC · Idc = 2 VCC Im/π **Efficiency:** η = Po/Pi = (Im² RL/2) / (2 VCC Im/π) = π Im RL/(4 VCC) = (π/4)(Vm/VCC) So **η = 78.5 % × (Vm/VCC)** (general expression η = (π/4)(Vm/VCC)). #### Maximum efficiency The largest undistorted peak output is Vm = VCC (ignoring VCE(sat)): ηmax = π/4 = 0.785 → **78.5 %** #### Remarks - Power dissipated in the two transistors: P2Q = Pi − Po. It is maximum at Vm = 2VCC/π, not at full swing, where P2Q(max) = 2VCC²/(π² RL). - Class B is much more efficient than class A (25 % or 50 %), but has crossover distortion. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In class B push-pull, each transistor conducts for: A. 180° B. 360° C. 90° D. Less than 90° Answer: A. Each transistor is biased at cut-off and handles one half-cycle. (Key point: Conduction) 2. The average dc current from both supplies is: A. 2Im/π B. Im/2 C. Im/π per supply total only D. Im Answer: A. Each half-sine pulse averages Im/π; together Idc = 2Im/π and Pi = 2VCC Im/π. (Key point: Input power) 3. The general efficiency is: A. (π/4)(Vm/VCC) B. (π/2)(Vm/VCC) C. (π/4)(VCC/Vm) D. Vm/(4 VCC) Answer: A. η = Po/Pi = (Im^2 RL/2)/(2VCC Im/π) = (π/4)(Vm/VCC). (Key point: General efficiency) 4. With Vm = VCC, ηmax is: A. 78.5% B. 50% C. 25% D. 100% Answer: A. π/4 = 0.785. (Key point: Maximum efficiency) ## Close variants set in other exams - [Transformer coupled class A efficiency](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Aclass-a-transformer-coupled) — Electronics Devices and Circuit, chapter 4, asked 8× ## Related in Electronics Devices and Circuit - [Series fed Class A amplifier maximum efficiency](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Aclass-a-series-fed-max-efficiency) — Electronics Devices and Circuit, chapter 4, asked 9× - [Class B push-pull power and efficiency, 20 V peak](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Aclass-b-push-pull-numeric-20v-16ohm) — Electronics Devices and Circuit, chapter 4, asked 2× - [Class B power and efficiency, 14 V peak, 16 Ω](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Aclass-b-numeric-14v-16ohm) — Electronics Devices and Circuit, chapter 4, asked 2× - [Design class B amplifier 20 W into 8 Ω](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Aclass-b-design-20w-8ohm) — Electronics Devices and Circuit, chapter 4 - [Cross-over distortion and class B push-pull efficiency](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Acrossover-distortion-class-b-efficiency) — Electronics Devices and Circuit, chapter 4 - [Transformer coupled class B push-pull, 25π % efficiency](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Atransformer-class-b-max-efficiency) — Electronics Devices and Circuit, chapter 4, asked 6× ## Related in other subjects - [General FM expression; FM using phase modulator](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-general-expression-phase-modulator) — Communication Systems, chapter 3 - [General vs one-class CF](https://www.bctengineer.com/question.md?key=semester_8%2Finformation_system%3Ageneral-vs-one-class-cf) — Information System, chapter 7 - [Transformer losses and efficiency expression](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Atransformer-losses-efficiency-derivation) — Electrical Machine, chapter 2, asked 2× - [Transformer losses and efficiency expression](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atransformer-losses-efficiency-expression) — Electrical Machines, chapter 2