--- title: "AC equivalent of CE amplifier, unbypassed RE" subject: "Electronics Devices and Circuit" chapter: 2 key: "semester_3/electronics_devices_and_circuit:ce-unbypassed-ac-equivalent" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-unbypassed-ac-equivalent site: BCT Engineering --- # AC equivalent of CE amplifier, unbypassed RE Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor - Asked in: 2080 Bhadra, 4 marks; 2072 Chaitra, 8 marks (asked 2 times) - Video: [CE Amplifier with Unbypassed RE - Small Signal Analysis using Hybrid Equivalent Model](https://www.youtube.com/watch?v=XcDXJ1S2Sfk) — 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/ce-unbypassed-ac-equivalent - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2 ## Question Draw the ac equivalent circuit for a CE amplifier (emitter resistor unbypassed) and derive the expressions for voltage gain, input impedance. ## Short answer CE AC equivalent with RE unbypassed: short capacitors, ground VCC, T-model with re, RE in emitter. Zin = R1||R2||[(β+1)(re+RE)]. Av = –(RC||RL)/(re+RE). Zo ≈ RC. ## Answer #### AC equivalent circuit (RE unbypassed) With the coupling capacitors shorted and VCC grounded, the circuit is: source vs, bias resistors RB = R1 || R2 from the base to ground, the transistor (T-model: re in the emitter), RE in the emitter lead to ground, and RC (with RL) from the collector to ground. ```text vs--+----B--\|C----+---- vo | | | RC||RL R1||R2 re | | | GND GND E | RE | GND ``` The T-model: the base current ib = ie/(β+1); the collector current source is α ie ≈ ie. #### Input impedance ```text vb = ie (re + RE) = (β+1) ib (re + RE) Rib = vb/ib = (β+1)(re + RE) (or rπ + (β+1)RE) Zin = R1 || R2 || (β+1)(re + RE) ``` #### Voltage gain Output: vo = -ic (RC || RL) ≈ -ie (RC || RL); input: vb = ie (re + RE). So ```text Av = vo/vb = -(RC || RL) / (re + RE) ``` (Exact: Av = -β(RC||RL)/[rπ + (β+1)RE]). If RE ≫ re, Av ≈ -(RC||RL)/RE, which depends only on resistors, so it is stable but lower than with a bypass capacitor (where Av = -(RC||RL)/re). #### Other results - Output impedance (ro neglected): Zo ≈ RC. - Higher Rin than the bypassed circuit, lower gain, and less distortion because of local negative feedback by RE. | Quantity | Unbypassed RE | |---|---| | Av | -(RC‖RL)/(re + RE) | | Zin | R1‖R2‖(β+1)(re+RE) | | Zo | RC | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is Zin? A. RE B. R1||R2||(beta+1)(re + RE) C. R1||R2||re D. (beta+1)re Answer: B. The unbypassed RE is multiplied by (beta+1). (Key point: Input impedance) 2. What is Av? A. -gm RC B. (RC||RL)/re C. -beta D. -(RC||RL)/(re + RE) Answer: D. vo = -ie(RC||RL) and vb = ie(re + RE). (Key point: Voltage gain) 3. How does unbypassed RE affect gain and Rin? A. Lower gain, higher Rin, less distortion B. Higher gain, higher Rin C. No effect D. Higher gain, lower Rin Answer: A. RE gives local negative feedback. (Key point: Effect of RE) 4. If RE much greater than re, Av is approximately? A. -(RC||RL)/re B. -gm RC C. -(RC||RL)/RE D. 1 Answer: C. Gain then depends only on resistors. 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