--- title: "Common collector amplifier Ri, Ro and gain" subject: "Electronics Devices and Circuit" chapter: 2 key: "semester_3/electronics_devices_and_circuit:cc-amplifier-derivation" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/cc-amplifier-derivation site: BCT Engineering --- # Common collector amplifier Ri, Ro and gain Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor - Asked in: 2081 Bhadra, 8 marks; 2075 Ashwin, 8 marks; 2072 Kartik, 6 marks; 2070 Chaitra, 8 marks (asked 4 times) - Video: [Small Signal Analysis of CC Amplifier - Emitter Follower - Common Collector](https://www.youtube.com/watch?v=-VhOVbJgPio) — 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/cc-amplifier-derivation - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2 ## Question Derive the expression for input resistance, output resistance, voltage gain of a voltage divider bias common collector amplifier. ## Short answer Input resistance: Rin = R1∥R2∥(β+1)(re+RE∥RL), high. Voltage gain: Av = (RE∥RL)/(re+RE∥RL) ≈ 1. Output resistance: Rout = RE∥[re + (Rs∥R1∥R2)/(β+1)], low. The stage is a buffer. ## Answer #### Circuit and ac equivalent Common collector (emitter follower) with voltage divider bias: input at the base through C1, output from the emitter through C2 to RL, collector at ac ground (VCC). ```text Vs--Rs--||--+--------B--\ R1||R2 |> (ib) re model E--+--||--RL RE --- ``` At ac, the supply is ground: R1, R2 go from the base to ground; RE and RL (from the emitter) are in parallel. Let R'E = RE || RL and RB = R1 || R2. Use the T-model: emitter current ie = (β+1) ib, emitter resistance re = VT/IE. #### Input resistance Looking into the base: vb = ie (re + R'E) = (β+1) ib (re + R'E), so ```text Rib = vb/ib = (β+1)(re + R'E) ≈ β (re + R'E) Rin = RB || Rib = R1 || R2 || (β+1)(re + RE||RL) ``` This is large because the small emitter resistance is multiplied by (β+1). #### Voltage gain ```text vo = ie R'E , vi = ie (re + R'E) Av = vo/vi = R'E / (re + R'E) ≈ 1 (since R'E ≫ re) ``` With a source of resistance Rs: Avs = Av · Rin/(Rin + Rs). The gain is positive (no phase shift), slightly less than 1. #### Output resistance Set vs = 0 and look back from the emitter (RL removed). The base sees Rs || R1 || R2 = R's, so ```text Rout = RE || [ re + R's/(β+1) ] ≈ re + R's/(β+1) ``` It is small: tens of ohms for typical values. #### Current gain ```text Ai = io/ii = Av · Rin/RL ≈ (β+1) · RB/(RB + Rib) · RE/(RE + RL) ``` #### Summary | Parameter | Result | |---|---| | Rin | R1 ‖ R2 ‖ (β+1)(re + R'E) (high) | | Rout | RE ‖ [re + R's/(β+1)] (low) | | Av | R'E/(re + R'E) ≈ 1 | | Phase | 0° (in phase) | Hence the stage is a buffer: high input and low output resistance with unity voltage gain. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is Rin of the CC amplifier? A. R1||R2||(beta+1)(re + RE||RL) B. RE||RL C. R1 + R2 + re D. re Answer: A. Looking into the base the emitter resistance is multiplied by (beta+1), in parallel with the bias network. (Key point: Input resistance) 2. What is Av? A. beta B. -R'E/re C. R'E/(re + R'E), approx 1 D. -gm RC Answer: C. vo = ie R'E and vi = ie(re + R'E). (Key point: Voltage gain) 3. What is Rout? A. RC B. R1||R2 C. RE||[re + R's/(beta+1)], low D. (beta+1)RE Answer: C. The source-side resistance is divided by (beta+1), making Rout small. (Key point: Output resistance) 4. What is the phase of the output and the role of the stage? A. In phase; high Rout B. 90 degrees; oscillator C. In phase; buffer with high Rin, low Rout D. 180 degrees; amplifier Answer: C. It is a non-inverting buffer. 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