---
title: "Emitter follower input/output resistance and gain"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:emitter-follower-rin-rout-gain-numeric"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/emitter-follower-rin-rout-gain-numeric
site: BCT Engineering
---

# Emitter follower input/output resistance and gain

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

- Asked in: 2078 Bhadra, 8 marks
- Video: [Emitter-Follower Configuration (Part 1)](https://www.youtube.com/watch?v=tTPRbtJJV0o) — 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/emitter-follower-rin-rout-gain-numeric
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

The bipolar junction transistor parameters for the circuit in figure below are β = 200 and VA = ∞. Determine the input resistance, output resistance and overall voltage gain of the circuit.
[Figure: common collector (emitter follower) circuit with Vcc = +10 V; collector connected to Vcc; base divider R1 = 10 kΩ (from Vcc) and R2 = 10 kΩ (to ground); source Vg = 1 V with RG = 600 Ω and a coupling capacitor to the base; emitter resistor RE = 4.3 kΩ to ground; output taken from the emitter through a coupling capacitor to load RL = 10 kΩ]

## Short answer

DC analysis yields IE ≈ 0.994 mA, re ≈ 25.1 Ω. AC analysis gives Rin ≈ 4.96 kΩ, Rout ≈ 27.6 Ω, and overall voltage gain ≈ 0.885.

## Answer

Find the dc bias (for re), then the ac parameters of the emitter follower.

#### Given
VCC = 10 V, R1 = R2 = 10 kΩ, RG (source) = 600 Ω, RE = 4.3 kΩ, RL = 10 kΩ, β = 200.

#### DC bias
Thevenin equivalent of the divider:
```text
VBB = 10 × 10k/20k = 5 V ,    RB = R1 || R2 = 5 kΩ
IB = (VBB - 0.7)/(RB + (β+1) RE) = 4.3/(5 k + 201 × 4.3 k)
   = 4.3/869.3 k = 4.95 µA
IE = (β+1) IB = 0.994 mA
re = 25 mV/0.994 mA = 25.1 Ω
```

#### Input resistance
```text
R'E = RE || RL = 4.3k × 10k/14.3k = 3.007 kΩ
Rib = (β+1)(re + R'E) = 201 × (25.1 + 3007) = 609 kΩ
Rin = R1 || R2 || Rib = 5 k || 609 k = 4.96 kΩ
```

#### Voltage gain
```text
Av = R'E/(re + R'E) = 3007/3032 = 0.992   (vo/vb)
Overall: vo/Vg = Av × Rin/(Rin + RG) = 0.992 × 4.96/(4.96 + 0.6) = 0.885
```

#### Output resistance (looking back at the emitter, RL excluded)
```text
R's = RG || RB = 0.6 k || 5 k = 536 Ω
Rout = RE || [re + R's/(β+1)] = 4.3 k || (25.1 + 2.67) = 27.6 Ω
```

**Answer: Rin ≈ 4.96 kΩ, Rout ≈ 27.6 Ω, overall voltage gain ≈ 0.885** (0.992 from the base to the output).

## Check yourself

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

1. How is the dc bias set up?
   A. IB = 0.7/RE
   B. IB = VBB/RL
   C. Thevenin: VBB = 5 V, RB = 5 kΩ; IB = (VBB-0.7)/(RB+(beta+1)RE)
   D. IB = VCC/R1
   Answer: C. The divider is replaced by VBB and RB, and RE is reflected by (beta+1). (Key point: Thevenin base bias)

2. Which expression gives Rin?
   A. R1 || R2 || (beta+1)(re + RE||RL) ≈ 4.96 kΩ
   B. re + RE
   C. R1 + R2 + RE
   D. (beta+1) re only
   Answer: A. The ac emitter load is RE||RL ≈ 3 kΩ; reflected it is 609 kΩ, in parallel with 5 kΩ. (Key point: Input resistance)

3. How is Rout found?
   A. RG ≈ 600 Ω
   B. RB/(beta+1) only = 25 Ω
   C. RE || [re + (RG||RB)/(beta+1)] ≈ 27.6 Ω
   D. RE + RL ≈ 14.3 kΩ
   Answer: C. The source side resistance seen at the base is divided by (beta+1) and added to re. (Key point: Output resistance)

4. Why is overall gain 0.885 while the base-to-output gain is 0.992?
   A. Because of RL only
   B. The source resistance RG forms a divider with Rin: Rin/(Rin+RG)
   C. Because beta is 200
   D. Because of ro
   Answer: B. Overall gain = 0.992 × 4.96/(4.96+0.6) ≈ 0.885. (Key point: Overall gain)

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