--- title: "Transformer coupled class B: 78.5% and maximum loss" subject: "Electronics Devices and Circuit" chapter: 4 key: "semester_3/electronics_devices_and_circuit:transformer-class-b-785-max-loss" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/transformer-class-b-785-max-loss site: BCT Engineering --- # Transformer coupled class B: 78.5% and maximum loss Electronics Devices and Circuit · Chapter 4: Output Stages and Power Amplifiers - Asked in: 2072 Chaitra, 3+3+3+3 marks - Video: [Class B Power Amplifier Push Pull Transformer Amplifier | Electronic Devices and Circuits 2 in EXTC](https://www.youtube.com/watch?v=O-AAFXXEqbE) — 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/transformer-class-b-785-max-loss - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-4 ## Question Draw the circuit diagram of transformer coupled class B push pull amplifier and its corresponding characteristic graph. And from graph prove that maximum efficiency is equal to 78.5%. Also find the condition when it has maximum loss. ## Short answer The graph's ac load line from (VCC,0) to (0,Im) shows peak swing equals VCC, so Po = VCC Im/2, Pi = 2VCC Im/π, giving ηmax = π/4 = 78.5%. Maximum power loss in transistors occurs at Vm = 2VCC/π, where efficiency is 50%. ## Answer #### Circuit diagram ```text T1 (input) T2 (output) +--- B1 Q1 (NPN) C1 ---+ | E1 | o----+ || +--------+ | +--------+ || +----+ ) || ( | ( || ) | Vi ) || (-- CT -- GND GND VCC -- CT --( || ) [RL] ) || ( | ( || ) | o----+ || +--------+ | +--------+ || +----+ | E2 | +--- B2 Q2 (NPN) C2 ---+ ``` The centre-tapped input transformer drives Q1 and Q2 in opposite phase. The centre tap of the output primary goes to VCC, and the two collectors go to its ends. Q1 conducts on one half-cycle and Q2 on the other; the secondary delivers the full waveform to RL. #### Characteristic graph (composite characteristic) ```text iC1 ^ | Im * Q1 conducting | \ | \ | \ ac load line, | \ slope -1/R'L 0 +---------*---------+-------> vCE1 | \ | \ Q2 conducting | \ (Q2's axes drawn inverted) | \ -Im + * 0 VCC 2VCC (Q point) ``` The dc load line is nearly vertical through VCE = VCC (the dc resistance of the primary is negligible). The Q point is at (VCC, 0) because class B is biased at cut-off (ICQ = 0). The ac load line of Q1 has slope −1/R'L and runs from the Q point up to the maximum current Im at VCE = 0 (full swing Vm = VCC). Q2's characteristic is identical and is drawn inverted, so the same straight line continues to (2VCC, −Im). #### Proof of 78.5 % from the graph From the graph the swing is limited by VCEmin = 0, so the peak voltage swing is Vm = VCC and the peak current is Im = VCC/R'L. - Po = Vm Im/2 = VCC Im/2 - Each transistor supplies a half-sine pulse of peak Im. The average current per transistor is Im/π, so for the two together Idc = 2Im/π. - Pi = VCC · Idc = 2 VCC Im/π η = Po/Pi = (VCC Im/2) / (2 VCC Im/π) = π/4 = 0.785 **ηmax = 78.5 %** #### Condition for maximum loss Power lost in the two transistors: P2Q = Pi − Po = 2 VCC Vm/(π R'L) − Vm²/(2R'L). For the maximum, dP2Q/dVm = 0: 2 VCC/(π R'L) − Vm/R'L = 0 → **Vm = 2 VCC/π ≈ 0.636 VCC** At this condition, P2Q(max) = 2 VCC²/(π² R'L), so each transistor dissipates VCC²/(π² R'L). The loss is maximum at about 63.6 % of the full output swing, not at full output or at no signal. At this point η = (π/4)(2/π) = 50 %. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Where is the Q point on the characteristic graph? A. At (VCC, 0), since class B is biased at cut-off B. At (VCC, Im) C. At (0, 0) D. At the middle of the ac load line Answer: A. Class B has ICQ = 0 and the vertical dc load line passes through VCE = VCC. (Key point: Q point of class B) 2. From the graph, the full swing limited by VCEmin = 0 gives which result? A. Vm = 2VCC, η = 100% B. Vm = VCC and Im = VCC/R'L, giving η = π/4 = 78.5% C. Vm = VCC/2, η = 39% D. Vm = 0.636 VCC, η = 50% Answer: B. Po = VCC Im/2 and Pi = 2VCC Im/π, so η = π/4. (Key point: Proof from graph) 3. At what output swing is the loss in the transistors maximum? A. Vm = VCC B. Vm = 0 C. Vm = VCC/2 D. Vm = 2VCC/π, about 0.636 VCC Answer: D. Setting dP2Q/dVm = 0 gives Vm = 2VCC/π. (Key point: Condition of maximum loss) 4. At the maximum-loss condition, what are the efficiency and total dissipation? A. 78.5%, 2VCC^2/(π^2 R'L) B. 50%, 2VCC^2/(π^2 R'L) C. 50%, VCC^2/(2R'L) D. 25%, VCC^2/R'L Answer: B. η = (π/4)(2/π) = 50% and P2Q(max) = 2VCC^2/(π^2 R'L); each transistor dissipates half. 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