---
title: "Design divider-bias CE amplifier with gain -90"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:ce-design-gain-minus-90"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-design-gain-minus-90
site: BCT Engineering
---

# Design divider-bias CE amplifier with gain -90

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

- Asked in: 2075 Ashwin, 8 marks
- Video: [Quick Guide: Designing A BJT Common Emitter Amplifier](https://www.youtube.com/watch?v=mRla_j3Hf-A) — trevortjes / The Repair Café
- 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-design-gain-minus-90
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

Design voltage divider biased common emitter BJT amplifier to get voltage gain of -90. Assume β=100 and Vcc=+12V.

## Short answer

With IC = 1 mA, VE = 1.2 V, VCE = 6 V: RC = 4.8 kΩ, RE1 ≈ 28 Ω (unbypassed), RE2 ≈ 1.16 kΩ (bypassed). Bias resistors: R1 ≈ 92 kΩ, R2 ≈ 19 kΩ. Gain is -90, stable due to RE1.

## Answer

A bypassed CE stage has Av = -RC/re. To obtain a *specific* gain of -90 and a stable value, a small unbypassed emitter resistor RE1 is kept in the ac path, with RE2 bypassed: Av = -RC/(re + RE1).

Given: VCC = 12 V, β = 100, Av = -90.

#### Guidelines
VE = VCC/10 = 1.2 V; VCE = VCC/2 = 6 V; I2 = 10 IB; VB = VE + 0.7 V. Choose IC = 1 mA (a convenient value).

#### Design
```text
IC = 1 mA ,   IB = 10 µA ,   IE = 1.01 mA
RE (total) = VE/IE = 1.2/1.01 mA = 1.19 kΩ
RC = (VCC - VE - VCE)/IC = (12 - 1.2 - 6)/1 mA = 4.8 kΩ
re = 25 mV/1.01 mA = 24.75 Ω

Gain: |Av| = RC/(re + RE1) = 90
   re + RE1 = 4800/90 = 53.3 Ω   →   RE1 = 53.3 - 24.75 = 28.6 Ω (use 27 Ω to 30 Ω)
RE2 = RE - RE1 = 1.19 k - 0.029 k = 1.16 kΩ (use 1.2 kΩ), bypassed by CE

VB = 1.2 + 0.7 = 1.9 V ,   I2 = 10 IB = 100 µA
R2 = 1.9/100 µA = 19 kΩ
R1 = (12 - 1.9)/110 µA = 91.8 kΩ
```
Final: R1 ≈ 92 kΩ (91 kΩ), R2 = 19 kΩ (18 kΩ), RC = 4.8 kΩ, RE1 ≈ 28 Ω, RE2 ≈ 1.16 kΩ with bypass capacitor CE.

#### Check
```text
Av = -4800/(24.75 + 28.6) = -90.0
Rin = R1 || R2 || (β+1)(re + RE1) = 15.7 k || 5.39 k = 4.0 kΩ
```
(Here the gain is for no load. A load RL reduces it to -(RC||RL)/(re + RE1); RE1 also makes the gain stable against changes in β and temperature.)

## Check yourself

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

1. Why is RE split into RE1 and RE2?
   A. To eliminate RC
   B. To make the circuit non-inverting
   C. RE1 unbypassed sets a definite stable gain; RE2 bypassed keeps dc stability without cutting gain
   D. To increase beta
   Answer: C. Av = -RC/(re+RE1) so the gain is set by resistors, while RE total gives dc bias stability. (Key point: Why split RE)

2. What is the required re + RE1?
   A. RC/90 = 53.3 Ω
   B. RC/9 = 533 Ω
   C. RC × 90 = 432 kΩ
   D. RC = 4.8 kΩ
   Answer: A. |Av| = RC/(re+RE1) = 90 gives re+RE1 = 4800/90 = 53.3 Ω. (Key point: Gain equation)

3. What is RE1, given re = 24.75 Ω?
   A. About 28.6 Ω
   B. About 53.3 Ω
   C. About 24.75 Ω
   D. About 1.19 kΩ
   Answer: A. RE1 = 53.3 - 24.75 ≈ 28.6 Ω. (Key point: RE1 value)

4. With VE = 1.2 V, VB = 1.9 V and I2 = 10 IB = 100 µA, R2 is:
   A. 190 kΩ
   B. 1.9 kΩ
   C. 19 kΩ
   D. 12 kΩ
   Answer: C. R2 = 1.9 V / 100 µA = 19 kΩ. (Key point: Divider design)

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