--- title: "Transformer coupled class A efficiency" subject: "Electronics Devices and Circuit" chapter: 4 key: "semester_3/electronics_devices_and_circuit:class-a-transformer-coupled" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/class-a-transformer-coupled site: BCT Engineering --- # Transformer coupled class A efficiency Electronics Devices and Circuit · Chapter 4: Output Stages and Power Amplifiers - Asked in: 2082 Shrawan, 2+4 marks; 2075 Ashwin, 5 marks; 2069 Chaitra, 6 marks (asked 3 times) - Video: [Transformer Coupled Class A Power Amplifier - High Power Amplifiers - Applied Electronics](https://www.youtube.com/watch?v=YrO4FXKg5Hw) — Ekeeda - 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-transformer-coupled - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-4 ## Question Draw the circuit diagram of transformer coupled class A power amplifier and derive the expression for its general efficiency. ## Short answer Transformer-coupled class A amplifier uses a load-matched transformer; dc drop is negligible so VCEQ ≈ VCC. General efficiency is η = 50% × [(VCEmax − VCEmin)/(VCEmax + VCEmin)]², with a maximum of 50% when VCEmax = 2VCC. ## Answer In a transformer-coupled class A amplifier the load RL is connected to the collector through a transformer. The transformer matches the load to the transistor: the load seen by the collector is R'L = (N1/N2)² RL. The dc resistance of the primary is almost zero, so the dc drop is small and VCEQ ≈ VCC. #### Circuit diagram ```text +VCC | +-----------+ | | T (N1:N2) [R1] )||(------+ | N1 )||( N2 | | )||( [RL] | | (------+ | C Vi o-||-+----------B Q (NPN) | E [R2] | | +------+ | | | GND [RE] [CE] | | GND GND ``` (R1, R2: base divider; RE with CE: stabilisation; primary N1 between VCC and the collector; load RL on the secondary N2.) #### Derivation of efficiency The ac load line has slope −1/R'L through the Q point (VCEQ = VCC, ICQ). The collector voltage swings between VCEmax and VCEmin. VCC = (VCEmax + VCEmin)/2 (Q point in the middle of the swing) **Output power (ac):** Po = (VCEmax − VCEmin)² / (8 R'L) **Input power (dc):** Pi = VCC ICQ. Since the dc resistance of the primary is zero, the ICQ for the maximum swing is ICQ = VCC/R'L. Pi = VCC²/R'L = (VCEmax + VCEmin)²/(4 R'L) **General efficiency:** η = Po/Pi = [(VCEmax − VCEmin)² / (8R'L)] × [4R'L/(VCEmax + VCEmin)²] **η = 50 % × [(VCEmax − VCEmin)/(VCEmax + VCEmin)]²** For the largest swing, VCEmin = 0 and VCEmax = 2VCC, giving η = 50 %. So the maximum efficiency of the transformer-coupled class A amplifier is **50 %**, double the series-fed value (25 %). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What load does the collector see through the transformer? A. R'L = (N1/N2)^2 RL B. R'L = (N2/N1)^2 RL C. R'L = (N1/N2) RL D. R'L = RL Answer: A. The transformer reflects the secondary load to the primary by the square of the turns ratio N1/N2. (Key point: Load matching) 2. Why is VCEQ approximately VCC in this amplifier? A. Because RE is very large B. Because the primary has almost zero dc resistance, so the dc drop is small C. Because the capacitor CE blocks dc D. Because the load is open Answer: B. With negligible dc resistance in the primary, the dc voltage drop is small and the whole VCC appears across the transistor. (Key point: Q point) 3. Which expression gives the ac output power in the derivation? A. (VCEmax + VCEmin)^2/(8R'L) B. VCC^2/R'L C. (VCEmax - VCEmin)^2/(8R'L) D. (VCEmax - VCEmin)^2/(2R'L) Answer: C. Po = (VCEmax - VCEmin)^2/(8R'L), using the peak-to-peak swing of the collector voltage. (Key point: Output power) 4. What is the general efficiency expression and its maximum? A. 25% x [(VCEmax - VCEmin)/(VCEmax + VCEmin)], max 25% B. 100% x [(VCEmax - VCEmin)/(VCEmax + VCEmin)]^2, max 100% C. 78.5% x [(VCEmax - VCEmin)/(VCEmax + VCEmin)], max 78.5% D. 50% x [(VCEmax - VCEmin)/(VCEmax + VCEmin)]^2, max 50% Answer: D. Dividing Po by Pi = (VCEmax + VCEmin)^2/(4R'L) gives 50% x the squared ratio, and with VCEmin = 0 the ratio is 1, so 50%. (Key point: General efficiency) ## Close variants set in other exams - [Class B power amplifier efficiency](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Aclass-b-efficiency-derivation) — Electronics Devices and Circuit, chapter 4, asked 8× ## Related in Electronics Devices and Circuit - [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× - [Transformer coupled class B, max efficiency, crossover](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Atransformer-class-b-efficiency-crossover) — Electronics Devices and Circuit, chapter 4 - [Transformer coupled class B: 78.5% and maximum loss](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Atransformer-class-b-785-max-loss) — Electronics Devices and Circuit, chapter 4 - [Transformer coupled push-pull efficiency, 20 V supply](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Aclass-b-transformer-numeric-20v) — Electronics Devices and Circuit, chapter 4 - [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× ## Related in other subjects - [Isolation amplifier applications; optically coupled circuit](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aisolation-applications-optical) — Advanced Electronics, chapter 4 - [Ideal transformer and coupled circuit model](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aideal-transformer-coupled-circuit-model) — Electrical Machines I, chapter 2 - [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