--- title: "Series fed Class A amplifier maximum efficiency" subject: "Electronics Devices and Circuit" chapter: 4 key: "semester_3/electronics_devices_and_circuit:class-a-series-fed-max-efficiency" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/class-a-series-fed-max-efficiency site: BCT Engineering --- # Series fed Class A amplifier maximum efficiency Electronics Devices and Circuit · Chapter 4: Output Stages and Power Amplifiers - Asked in: 2081 Bhadra, 1+4 marks; 2079 Baisakh, 6 marks; 2079 Bhadra, 3+4 marks; 2076 Chaitra, 3+3 marks; 2076 Ashwin, 3+3 marks; 2074 Chaitra, 6 marks; 2071 Shrawan, 6 marks; 2071 Chaitra, 3+3 marks; 2070 Ashad, 1+2+3 marks (asked 9 times) - Video: [Class-A Power Amplifier | Series fed Class-A Derivation | Power Amplifiers | Class-6](https://www.youtube.com/watch?v=cm8VFZE5Z-g) — VivTronics - 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-a-series-fed-max-efficiency - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-4 ## Question Draw a series fed Class A amplifier and derive its maximum efficiency. ## Short answer For a series-fed class A amplifier, the load is in series with the collector. Maximum efficiency is 25% when the Q point is at VCEQ = VCC/2 and ICQ = VCC/(2RC). ## Answer A series-fed class A amplifier has the load (RC) connected in series with the transistor from the supply, and the base bias is given through a resistor RB. The transistor conducts for the full 360° of the input cycle. #### Circuit diagram ```text +VCC | +-------+-------+ | | [RB] [RC] (load) | | | +-----o Vo | | | C Vi o--||-+--------------B (NPN) E | GND ``` #### Derivation of efficiency **Input power (dc):** Pi = VCC · ICQ (the signal does not change the average current in a class A stage). **Output power (ac)**, for a sinusoidal swing with peak collector current Ic(p) and peak output voltage Vc(p) = Ic(p) RC: Po = Vc(p) Ic(p)/2 = Ic(p)² RC/2 Efficiency η = Po/Pi × 100% **Maximum swing:** For the largest undistorted swing, the Q point is set at the middle of the load line: VCEQ = VCC/2 and ICQ = VCC/(2RC). The peak current then equals the quiescent current: Ic(p) = ICQ = VCC/(2RC), and Vc(p) = VCC/2. Po(max) = (VCC/2)(VCC/2RC)/2 = VCC²/(8RC) Pi = VCC · ICQ = VCC · VCC/(2RC) = VCC²/(2RC) ηmax = Po/Pi = [VCC²/(8RC)] / [VCC²/(2RC)] = 1/4 **Maximum efficiency = 25 %** In general, for a peak-to-peak output swing Vpp: η = 25 (Vpp/VCC)² %, which reaches 25 % when Vpp = VCC. Practical efficiency is lower, because the signal swing never reaches the full supply. The efficiency is low because a constant dc current is drawn, and power is dissipated in RC and the transistor even when there is no signal. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The dc input power of a class A stage is: A. VCC ICQ B. VCC^2/RC only at full swing C. Vm Im/2 D. 0 Answer: A. The signal does not change average current, so Pi = VCC*ICQ. (Key point: Input power) 2. For maximum undistorted swing the Q-point is set at: A. VCEQ = VCC/2, ICQ = VCC/(2RC) B. VCEQ = VCC C. ICQ = 0 D. VCEQ = 0 Answer: A. The midpoint of the load line gives equal swing each way. (Key point: Q-point for max swing) 3. The maximum efficiency of a series-fed class A amplifier is: A. 25% B. 50% C. 78.5% D. 100% Answer: A. Po = VCC^2/(8RC), Pi = VCC^2/(2RC), giving 1/4. 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