---
title: "CE amplifier Rin, Rout, Av, Ai (470k, 2.2k, 0.56k)"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:ce-emitter-resistor-470k-numeric"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-emitter-resistor-470k-numeric
site: BCT Engineering
---

# CE amplifier Rin, Rout, Av, Ai (470k, 2.2k, 0.56k)

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

- Asked in: 2068 Chaitra, 2+2+2+2 marks
- Video: [BJT Amplifier Solved Problem | Quiz # 290](https://www.youtube.com/watch?v=yeh-AFtR5MQ) — ALL ABOUT ELECTRONICS - Quiz
- 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-emitter-resistor-470k-numeric
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

For the figure shown below with β = 120 find (a) input impedance (b) output impedance (c) voltage gain (d) current gain. Use small signal model.
[Figure: BJT Q1 amplifier with 20 V supply; 470 kΩ from supply to base; 2.2 kΩ collector resistor with output Vo through coupling capacitor C2; 0.56 kΩ emitter resistor to ground; source Vs through coupling capacitor C1 at the base]

## Short answer

DC analysis yields re = 5.76 Ω. Zin = 470 kΩ || (β+1)(re+0.56kΩ) ≈ 59.8 kΩ, Zout = 2.2 kΩ, Av = -RC/(re+RE) ≈ -3.9, Ai = β·RB/(RB+Rib) ≈ 105.

## Answer

The amplifier uses fixed (base-resistor) bias with an unbypassed emitter resistor and no load (the output is taken from the collector).

#### Given
VCC = 20 V, RB = 470 kΩ, RC = 2.2 kΩ, RE = 0.56 kΩ, β = 120, VBE = 0.7 V.

#### DC bias
```text
IB = (VCC - 0.7)/(RB + (β+1) RE) = 19.3/(470 k + 121 × 0.56 k)
   = 19.3/537.8 k = 35.9 µA
IC = 4.31 mA ,   IE = 4.34 mA
re = 25 mV/4.34 mA = 5.76 Ω ,   rπ = β re = 691 Ω
```

#### Small-signal model
Using the T-model with re and the unbypassed RE in the emitter, the resistance seen at the base is

```text
Rib = (β+1)(re + RE) = 121 × (5.76 + 560) = 68.5 kΩ
```

#### (a) Input impedance
```text
Zin = RB || Rib = 470 k || 68.5 k = 59.8 kΩ
```

#### (b) Output impedance
```text
Zout = RC = 2.2 kΩ    (ro neglected)
```

#### (c) Voltage gain
```text
Av = -RC/(re + RE) = -2200/565.8 = -3.89
```

#### (d) Current gain
Take the output current as the collector current and the input current as the current drawn from the source. Then ib = ii × RB/(RB + Rib) and ic = β ib:
```text
Ai = β RB/(RB + Rib) = 120 × 470/(470 + 68.5) = 105
```
(Check: |Av| Zin/RC = 3.89 × 59.8 k/2.2 k ≈ 105.7; the small difference arises because Av takes α ≈ 1.)

**Answer: Zin ≈ 59.8 kΩ, Zout ≈ 2.2 kΩ, Av ≈ -3.9, Ai ≈ 105.**

## Check yourself

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

1. How is IB found for fixed bias with RE?
   A. VCC/RE
   B. (VCC-0.7)/(RB + (beta+1)RE) ≈ 35.9 µA
   C. (VCC-0.7)/RB
   D. (VCC-0.7)/(RB+RC)
   Answer: B. RE is reflected by (beta+1) in the base loop. (Key point: Base current)

2. What is Zin?
   A. RB ≈ 470 kΩ
   B. RB || (beta+1)(re+RE) ≈ 59.8 kΩ
   C. rpi ≈ 691 Ω
   D. RC ≈ 2.2 kΩ
   Answer: B. Rib = 121 × 565.8 = 68.5 kΩ in parallel with 470 kΩ. (Key point: Zin)

3. Av =?
   A. -RC/(re+RE) ≈ -3.9
   B. -gm RC ≈ -380
   C. -beta
   D. +1
   Answer: A. 2200/565.8 = 3.89, inverting. (Key point: Voltage gain)

4. How is Ai obtained?
   A. RB/RC
   B. Av × RC/Zin
   C. beta RB/(RB+Rib) ≈ 105
   D. beta = 120 exactly
   Answer: C. Only the fraction RB/(RB+Rib) of source current enters the base, then ic = beta ib. (Key point: Current gain)

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