---
title: "Derive Ai, Rin and Rout of emitter follower"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:cc-ai-rin-rout-derivation"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/cc-ai-rin-rout-derivation
site: BCT Engineering
---

# Derive Ai, Rin and Rout of emitter follower

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

- Asked in: 2071 Chaitra, 6 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-ai-rin-rout-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

Derive expressions for Ai, Rin and Rout.
[Figure: common collector (emitter follower) amplifier with +Vcc; base divider R1 (from Vcc) and R2 (to ground); emitter resistor RE = RL; input Vin at the base and output Vo at the emitter]

## Short answer

Rin = R1||R2||(β+1)(re+RE). Current gain Ai = (β+1) RB/(RB+Rib) where RB = R1||R2 and Rib = (β+1)(re+RE). Output resistance Rout = RE || [re + (Rs||RB)/(β+1)].

## Answer

The emitter follower (common collector) has a voltage divider R1, R2 at the base; the output vo is taken across the emitter resistor RE (= the load RL). Replace the transistor by the T-model (re in the emitter, ie = (β+1) ib); the collector and VCC are at ac ground. Let RB = R1 || R2.

```text
 ii --->    ib
 vin--+--+----B
      |  |     \ re
     RB  |      E----+--- vo = ie RE
      |           RE
     GND          GND
```

#### Input resistance
```text
vb = ie (re + RE) = (β+1) ib (re + RE)
Rib = vb/ib = (β+1)(re + RE)
Rin = RB || Rib = R1 || R2 || (β+1)(re + RE)
```

#### Current gain
The signal current ii divides between RB and the base:
```text
ib = ii × RB/(RB + Rib)
io = ie = (β+1) ib   (current in RE = load)

Ai = io/ii = (β+1) RB/(RB + Rib)
```
If RB ≫ Rib, Ai ≈ (β+1). For example β = 100, RB = 10 kΩ, Rib = 100 kΩ: Ai = 101 × 10/110 = 9.2.

(In terms of the voltage gain: Ai = Av Rin/RE.)

#### Output resistance
Source off (vin = 0); the base resistance is R's = Rs || RB. A test voltage vx at the emitter drives ie = vx/(re + R's/(β+1)):
```text
Rout = RE || [ re + R's/(β+1) ]  ≈ re + R's/(β+1)
```

The voltage gain is Av = RE/(RE + re) ≈ 1.

| Parameter | Expression |
|---|---|
| Rin | R1‖R2‖(β+1)(re+RE) |
| Ai | (β+1) RB/(RB + Rib) |
| Rout | RE‖[re + R's/(β+1)] |

## Check yourself

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

1. What is the input resistance of the emitter follower with RE as load?
   A. R1 || R2 || (β+1)(re + RE)
   B. R1 || R2 || re
   C. (β+1) RE only
   D. R1 + R2
   Answer: A. vb = ie(re+RE) so Rib = (β+1)(re+RE), in parallel with RB = R1||R2. (Key point: Rin)

2. What is Ai in terms of RB and Rib?
   A. Ai = (β+1) RB/(RB + Rib)
   B. Ai = β
   C. Ai = RB/Rib
   D. Ai = (β+1)(RB + Rib)
   Answer: A. ib = ii RB/(RB+Rib) by current division, and io = ie = (β+1) ib. (Key point: Current gain)

3. When does Ai approach (β+1)?
   A. When RB << Rib
   B. When RB >> Rib
   C. When RE = 0
   D. When β is small
   Answer: B. If RB is much greater than Rib, nearly all the input current enters the base. (Key point: Large RB case)

4. What is Rout of the emitter follower?
   A. RE || [re + R's/(β+1)]
   B. RE + re
   C. β re
   D. R's (β+1)
   Answer: A. The source-side resistance R's is divided by (β+1) at the emitter, giving a low output resistance. (Key point: Rout)

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