---
title: "CC amplifier small-signal model, Rin and gain"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:cc-small-signal-rin-gain"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/cc-small-signal-rin-gain
site: BCT Engineering
---

# CC amplifier small-signal model, Rin and gain

Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor

- Asked in: 2070 Ashad, 2+3+3 marks
- 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-small-signal-rin-gain
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

Draw the small signal model for Common Collector Amplifier and find its input resistance and voltage gain.

## Short answer

The small-signal model has rπ between base and emitter, with a dependent source gm vπ from collector to emitter. Input resistance Rin = R1||R2||[rπ+(β+1)(RE||RL)]. Voltage gain Av = (β+1)R'E/[rπ+(β+1)R'E] ≈ R'E/(re+R'E) ≈ 1.

## Answer

#### Small-signal (hybrid-π) model of the common collector amplifier
The collector is tied to VCC (ac ground). The base is the input, and the emitter the output (with RE || RL).

```text
 vs--Rs--+----B--rπ--E-----+---- vo
         |     |           |
       R1||R2  +-( ) gm vπ-+   (current source from C(gnd) to E)
         |                RE||RL
        GND                |
                          GND
```
Here vπ = vb - vo, and the controlled source gm vπ feeds the emitter.

#### Input resistance
The emitter current is ie = ib + gm vπ = (β+1) ib. Let R'E = RE || RL. Then
```text
vb = ib rπ + ie R'E = ib [rπ + (β+1) R'E]
Rib = rπ + (β+1) R'E  = (β+1)(re + R'E)       [rπ = (β+1) re]
Rin = R1 || R2 || [rπ + (β+1)(RE || RL)]
```

#### Voltage gain
```text
vo = ie R'E = (β+1) ib R'E ,   vb = ib [rπ + (β+1) R'E]

Av = vo/vb = (β+1) R'E / [rπ + (β+1) R'E]  = R'E/(re + R'E)  ≈ 1
```
Including the source resistance: Avs = Av · Rin/(Rin + Rs).

#### Example
IE = 1 mA (re = 25 Ω), β = 100, R'E = 2 kΩ: Rib = 101 × 2025 = 204.5 kΩ, Av = 2000/2025 = 0.988.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In the hybrid-π model of the common collector, what is vπ?
   A. vb - vo
   B. vo
   C. vb
   D. vs
   Answer: A. vπ is across rπ, which is between base and emitter, so vπ = vb - vo. (Key point: vπ in CC)

2. What is the resistance looking into the base?
   A. rπ + (β+1)R'E
   B. rπ + R'E
   C. β R'E
   D. (β+1) rπ
   Answer: A. vb = ib rπ + ie R'E with ie = (β+1)ib, giving Rib = rπ + (β+1)R'E. (Key point: Rib)

3. What is the voltage gain Av?
   A. -gm R'E
   B. R'E/(re + R'E), slightly below 1
   C. β
   D. (re + R'E)/R'E
   Answer: B. Av = (β+1)R'E/[rπ + (β+1)R'E] = R'E/(re + R'E), about 1 and non-inverting. (Key point: Voltage gain)

4. How is the gain from the source computed?
   A. Avs = Av Rin/(Rin + Rs)
   B. Avs = Av + Rs
   C. Avs = Av Rs/Rin
   D. Avs = Av
   Answer: A. Rs and Rin form a voltage divider at the input. (Key point: Source loading)

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