---
title: "CC amplifier ac equivalent, Rin and Rout"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:cc-ac-equivalent-rin-rout"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/cc-ac-equivalent-rin-rout
site: BCT Engineering
---

# CC amplifier ac equivalent, Rin and Rout

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

- Asked in: 2068 Chaitra, 3+3+2 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-ac-equivalent-rin-rout
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

Draw ac equivalent circuit of common collector amplifier. Find its input and output resistances.

## Short answer

Rin = R1||R2||(β+1)(re + RE||RL); Rout = RE || [re + R's/(β+1)] where R's = Rs||R1||R2. The CC stage thus provides high input and very low output impedance.

## Answer

#### AC equivalent circuit of the common collector amplifier
The collector is tied to VCC, which is ac ground. The input is applied to the base and the output is taken from the emitter. Using the T-model:

```text
 vs--Rs--+----B
         |     \ re   ie = (β+1) ib
       R1||R2   E-----+---- vo
         |            |
        GND        RE||RL
                      |
                     GND
```
R1 and R2 (both connected to ground for ac) are in parallel at the base; R'E = RE || RL.

#### Input resistance
```text
vb = ie (re + R'E) = (β+1) ib (re + R'E)
Rib = (β+1)(re + R'E)
Rin = R1 || R2 || (β+1)(re + RE||RL)
```

#### Output resistance
Set vs = 0 and remove RL; apply a test voltage vx at the emitter. The resistance in the base circuit is R's = Rs || R1 || R2. The base current now flows out of the base, through R's to ground:
```text
ib = -vx/(R's + rπ)   → current drawn from vx by the transistor: -ie = (β+1) vx/(R's + rπ)
Rout = RE || (rπ + R's)/(β+1)  =  RE || [ re + R's/(β+1) ]
```
Hence Rout ≈ re + R's/(β+1), a small value.

#### Example
β = 100, IE = 2 mA (re = 12.5 Ω), RE = 2 kΩ, RL = 2 kΩ, R's = 2 kΩ: Rin = R1||R2||(101 × 1012.5) ≈ R1||R2||102 kΩ; Rout ≈ 2 k || (12.5 + 19.8) = 31.8 Ω.

## Check yourself

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

1. Why is the circuit called a common collector amplifier in the ac equivalent circuit?
   A. The collector is tied to VCC, which is ac ground, so it is common to input and output
   B. The collector is the input terminal
   C. The collector is connected to the load RL
   D. The collector carries the output signal
   Answer: A. VCC is a constant dc supply, so it is ac ground. The input is at the base and the output at the emitter, leaving the collector common. (Key point: AC ground at collector)

2. Looking into the base with emitter resistance R'E = RE||RL, what is the resistance seen by the base (Rib)?
   A. re + R'E
   B. (β+1)(re + R'E)
   C. β R'E
   D. rπ only
   Answer: B. vb = ie(re + R'E) with ie = (β+1)ib, so Rib = (β+1)(re + R'E). The emitter resistance is reflected up by (β+1). (Key point: Input resistance at base)

3. What is the full input resistance Rin of the amplifier?
   A. (β+1)(re + RE||RL) only
   B. R1 + R2 + (β+1)(re + R'E)
   C. R1 || R2 || (β+1)(re + RE||RL)
   D. RE || RL
   Answer: C. The bias resistors R1 and R2 are both at ac ground at the base, so they are in parallel with Rib. (Key point: Total Rin)

4. What is the output resistance of the emitter follower?
   A. RE || [re + R's/(β+1)], about re + R's/(β+1)
   B. RE + re
   C. β re
   D. RC
   Answer: A. The base-circuit resistance R's is divided by (β+1) when seen from the emitter, which makes Rout small. (Key point: Output resistance)

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