---
title: "CC amplifier Rin, Rout and why it is used"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:cc-input-output-resistance-why-used"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/cc-input-output-resistance-why-used
site: BCT Engineering
---

# CC amplifier Rin, Rout and why it is used

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

- Asked in: 2073 Shrawan, 2+2+2+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-input-output-resistance-why-used
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

Determine the input resistance and output resistance of CC BJT amplifier circuit. Why common collector configuration is used in amplifier circuit design.

## Short answer

Rin = R1||R2||(β+1)(re + RE||RL); Rout = RE || [re + R's/(β+1)] ≈ re + R's/(β+1). The CC configuration is used as a buffer due to its high input resistance, low output resistance, voltage gain near unity, and high current gain.

## Answer

#### Input resistance
Common collector (emitter follower) with divider bias, load RL, R'E = RE || RL, T-model:
```text
Rib = vb/ib = (β+1)(re + R'E)
Rin = R1 || R2 || (β+1)(re + RE || RL)
```
It is high (tens to hundreds of kΩ) since the emitter resistance is multiplied by (β+1).

#### Output resistance
With the source off, the base sees R's = Rs || R1 || R2; reflected to the emitter divided by (β+1):
```text
Rout = RE || [ re + R's/(β+1) ] ≈ re + R's/(β+1)
```
It is low (tens of ohms). Example: IE = 1 mA, R's = 5 kΩ, β = 100: Rout ≈ 25 + 50 = 75 Ω.

#### Why common collector is used
- **Buffer / impedance matching:** it has high Rin and low Rout, so it takes almost no current from a high-impedance source and drives a low-impedance load without loss of signal.
- **Voltage gain ≈ 1** (R'E/(re + R'E)), with no phase inversion.
- **High current gain** (β+1), which is useful for power output stages.
- Less distortion and a stable gain, because of 100 % series negative feedback (the emitter resistance), and a wide bandwidth.
- Used as the last stage of multistage amplifiers, in voltage regulators and in Darlington pairs.

## Check yourself

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

1. Why is Rin of the CC high?
   A. Because RC is large
   B. Because rpi is large only
   C. The emitter resistance is multiplied by (beta+1) at the base
   D. Because the collector is grounded
   Answer: C. Rib = (beta+1)(re + RE||RL). (Key point: CC input resistance)

2. Why is Rout low?
   A. It is multiplied by beta
   B. RE is small
   C. ro is zero
   D. The source-side resistance is divided by (beta+1) and added to re
   Answer: D. Rout ≈ re + R's/(beta+1), tens of ohms. (Key point: CC output resistance)

3. For IE = 1 mA, R's = 5 kΩ, beta = 100, Rout ≈ ?
   A. About 5 kΩ
   B. About 25 kΩ
   C. About 75 Ω
   D. About 7.5 Ω
   Answer: C. 25 + 5000/100 = 75 Ω. (Key point: Example Rout)

4. Why is CC used in amplifier design?
   A. For low current gain
   B. For phase inversion
   C. For high voltage gain
   D. As a buffer with gain near 1, high Rin, low Rout and high current gain
   Answer: D. It matches high-impedance sources to low-impedance loads. (Key point: Why used)

## Related in Electronics Devices and Circuit

- [CC amplifier ac equivalent, Rin and Rout](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Acc-ac-equivalent-rin-rout) — Electronics Devices and Circuit, chapter 2
- [AC equivalent: Rin and Rout of divider-bias amplifier](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ace-ac-equivalent-rin-rout-100k-20k) — Electronics Devices and Circuit, chapter 2
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- [Common collector amplifier Ri, Ro and gain](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Acc-amplifier-derivation) — Electronics Devices and Circuit, chapter 2, asked 4×
- [CE bypassed amplifier Rin, Rout, Av, Ai](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ace-bypassed-rin-rout-av-ai) — Electronics Devices and Circuit, chapter 2

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