---
title: "CE bypassed amplifier Rin, Rout, Av, Ai"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:ce-bypassed-rin-rout-av-ai"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-bypassed-rin-rout-av-ai
site: BCT Engineering
---

# CE bypassed amplifier Rin, Rout, Av, Ai

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

- Asked in: 2078 Kartik, 8 marks
- Video: [Small Signal Analysis of CE Amplifier using Hybrid pi model](https://www.youtube.com/watch?v=SYvf2DVSkh8) — 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/ce-bypassed-rin-rout-av-ai
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

Derive the expression for Rin, Rout, Av and Ai in CE capacitor bypassed amplifier.

## Short answer

For a bypassed CE amplifier, Rin = R1||R2||rπ, Rout ≈ RC, Av = -gm(RC||RL) ≈ -(RC||RL)/re, and Ai = Av·Rin/RL. The bypass capacitor removes RE from the ac path, maximizing gain.

## Answer

#### Small-signal model
For the CE amplifier with voltage divider bias and RE bypassed by CE (so the emitter is at ac ground), the hybrid-π model has: rπ between B and E, the source gm vπ with ro between C and E, the bias resistors RB = R1 || R2 at the input, and RC || RL at the output.

```text
 vs--Rs--+--B--rπ--E(gnd)      C--+--+
         |       vπ  gm vπ(ro)----+  RC||RL
       R1||R2                         |
         GND                         GND
```
gm = IC/VT, rπ = β/gm.

#### Input resistance
```text
Rin = R1 || R2 || rπ
```

#### Output resistance
Set vs = 0: then vπ = 0 and the dependent source is zero, so
```text
Rout = RC || ro  ≈ RC
```

#### Voltage gain
```text
vo = -gm vπ (RC || RL || ro) ,   vπ = vi (neglecting Rs)

Av = -gm (RC || RL)  ≈ -(RC || RL)/re
```
With a signal source: Avs = Av · Rin/(Rin + Rs).

#### Current gain
```text
Ai = io/ii = (vo/RL)/(vi/Rin) = Av · Rin/RL

Ai = -gm (RC||RL) Rin/RL
```
When RB ≫ rπ, Rin ≈ rπ and gm rπ = β, so Ai ≈ -β RC/(RC + RL); this approaches -β when RL ≪ RC.

| Parameter | Expression |
|---|---|
| Rin | R1‖R2‖rπ (moderate) |
| Rout | RC (moderate to high) |
| Av | -gm(RC‖RL) (high, inverting) |
| Ai | Av·Rin/RL (high) |

## Check yourself

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

1. What is Rin?
   A. R1 || R2 || rpi
   B. (beta+1) RE
   C. R1 + R2
   D. RC || RL
   Answer: A. With RE bypassed, the input sees the bias resistors in parallel with rpi. (Key point: Input resistance)

2. Why is Rout ≈ RC?
   A. Because RL is infinite
   B. Setting vs = 0 gives vpi = 0, so the dependent source is dead and only RC||ro remains
   C. Because re is large
   D. Because gm = 0 always
   Answer: B. With the source off the controlled current is zero. (Key point: Output resistance)

3. What is Av?
   A. -beta
   B. -RC/RE
   C. -gm (RC||RL), inverting
   D. +gm RC
   Answer: C. vo = -gm vpi (RC||RL) with vpi = vi. (Key point: Voltage gain)

4. What is Ai?
   A. Av · Rin/RL
   B. beta always
   C. Rin/RC
   D. Av · RL/Rin
   Answer: A. Ai = io/ii = (vo/RL)/(vi/Rin) = Av Rin/RL. (Key point: Current gain)

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