---
title: "CE amplifier input, output resistance and gain (RE1, RE2)"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:ce-partial-bypass-numeric"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-partial-bypass-numeric
site: BCT Engineering
---

# CE amplifier input, output resistance and gain (RE1, RE2)

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

- Asked in: 2074 Chaitra, 8 marks
- Video: [re Model (Solved Problem 2)](https://www.youtube.com/watch?v=NWif9O4MTE4) — Neso Academy
- 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-partial-bypass-numeric
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

Determine the input resistance, output resistance and overall voltage gain of the circuit given below.
[Figure: CE amplifier with Vcc = +10 V; R1 = 47 kΩ from Vcc to base; R2 = 10 kΩ from base to ground; RC = 4.7 kΩ; emitter resistors RE1 = 470 Ω in series with RE2 = 470 Ω, with RE2 bypassed by C3 = 100 μF; β = 150; source Vs with Rs = 600 Ω and coupling capacitor C1 = 10 μF at the input; coupling capacitor C2 = 10 μF and load RL = 47 kΩ at the output]

## Short answer

DC analysis gives IC ≈ 1.05 mA, re ≈ 23.6 Ω. With unbypassed RE1 = 470 Ω, Rin ≈ 7.4 kΩ, Rout ≈ 4.7 kΩ, overall voltage gain Avs ≈ -8.0.

## Answer

Here RE1 = 470 Ω is unbypassed and RE2 = 470 Ω is bypassed by C3, so only RE1 appears in the ac path; both are in the dc path.

#### Given
VCC = 10 V, R1 = 47 kΩ, R2 = 10 kΩ, RC = 4.7 kΩ, RE1 = RE2 = 470 Ω, β = 150, Rs = 600 Ω, RL = 47 kΩ.

#### DC bias
```text
VBB = VCC R2/(R1 + R2) = 10 × 10/57 = 1.754 V
RB  = R1 || R2 = 8.246 kΩ
IB  = (VBB - 0.7)/(RB + (β+1)(RE1 + RE2)) = 1.054/(8.246 k + 151 × 940)
    = 1.054/150.2 k = 7.02 µA
IC  = β IB = 1.053 mA ,   IE = 1.06 mA
re  = 25 mV/1.06 mA = 23.6 Ω
```

#### Input resistance
```text
Rib = (β+1)(re + RE1) = 151 × (23.6 + 470) = 74.5 kΩ
Rin = R1 || R2 || Rib = 8.246 k || 74.5 k = 7.42 kΩ
```

#### Output resistance
```text
Rout = RC = 4.7 kΩ    (ro neglected)
```

#### Voltage gain
```text
RC || RL = 4.7 k || 47 k = 4.27 kΩ
Av = -(RC||RL)/(re + RE1) = -4270/493.6 = -8.65

Overall: Avs = Av · Rin/(Rin + Rs) = -8.65 × 7.42/(7.42 + 0.6) = -8.0
```
**Answer: Rin ≈ 7.4 kΩ, Rout ≈ 4.7 kΩ, overall gain Avs ≈ -8.0** (-8.65 from the base to the output).

## Check yourself

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

1. Which emitter resistor appears in the ac path?
   A. Only RE2
   B. Only RE1; RE2 is bypassed by C3
   C. Both RE1 and RE2
   D. Neither
   Answer: B. The bypass capacitor shorts RE2 at signal frequency, leaving RE1. (Key point: What appears in ac)

2. Which resistance reflected into the base loop is used for IB?
   A. RC
   B. beta RE2
   C. (beta+1)(RE1+RE2)
   D. (beta+1) RE1 only
   Answer: C. Both emitter resistors carry dc current, so both are included. (Key point: DC bias)

3. How is Rin found?
   A. R1 || R2 || rpi
   B. R1 + R2
   C. (beta+1) RE2
   D. R1 || R2 || (beta+1)(re + RE1) ≈ 7.42 kΩ
   Answer: D. Rib = 151 × 493.6 ≈ 74.5 kΩ in parallel with 8.246 kΩ. (Key point: Rin)

4. How is overall gain calculated?
   A. -gm RC
   B. +1
   C. Av = -(RC||RL)/(re+RE1) = -8.65, times Rin/(Rin+Rs) = -8.0
   D. -RC/(RE1+RE2)
   Answer: C. The source-resistance divider reduces -8.65 to about -8.0. (Key point: Gain)

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