---
title: "Design common base amplifier (15 V, 1.5 mA)"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:cb-amplifier-design-15v"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/cb-amplifier-design-15v
site: BCT Engineering
---

# Design common base amplifier (15 V, 1.5 mA)

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

- Asked in: 2075 Chaitra, 7 marks
- 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/cb-amplifier-design-15v
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

Design a common base amplifier circuit using β independent method. Given parameters are Vcc=15V, IE=1.5mA, β=100 and input and output impedances are comparatively large. Use appropriate guideline to support your design.

## Short answer

Using VE = VCC/10, VCE = VCC/2, and I2 = 5IB: RE = 1 kΩ, RC ≈ 4 kΩ, R1 ≈ 144 kΩ, R2 ≈ 30 kΩ. The stage provides Av ≈ 242, Rin ≈ 16 Ω, Rout ≈ 4 kΩ.

## Answer

This is a common base amplifier with the base held at a fixed voltage by a divider (β-independent bias) and grounded for ac by a capacitor. A relatively large input and output impedance is obtained with large RE and RC (so a modest bias current) and a divider current at the lower end of the range (I2 = 5 IB).

Given: VCC = 15 V, IE = 1.5 mA, β = 100.

#### Guidelines
1. VE ≈ VCC/10 for bias stability.
2. VCE ≈ VCC/2 for maximum symmetrical swing.
3. Divider current I2 = 5 IB (≫ IB), so VB is fixed and the bias is β independent.
4. VB = VE + 0.7 V; a large capacitor CB from the base to ground makes the base an ac ground.

#### Design
```text
IE = 1.5 mA     IC = α IE = 1.485 mA     IB = IE/(β+1) = 14.9 µA
VE = VCC/10 = 1.5 V      RE = VE/IE = 1 kΩ
VCE = VCC/2 = 7.5 V     RC = (VCC - VE - VCE)/IC = (15 - 1.5 - 7.5)/1.485 mA = 4.04 kΩ
VB = VE + 0.7 = 2.2 V
I2 = 5 IB = 74.3 µA
R2 = VB/I2 = 29.6 kΩ
R1 = (VCC - VB)/(I2 + IB) = 12.8/89.1 µA = 144 kΩ
```
Choose R1 ≈ 144 kΩ, R2 ≈ 29.6 kΩ, RE ≈ 1 kΩ, RC ≈ 4.04 kΩ, and a large bypass capacitor CB; C1 and C2 are coupling capacitors.

#### Small-signal check (ac equivalent: input at the emitter, output at the collector)
```text
re  = VT/IE = 25 mV/1.5 mA = 16.7 Ω
Rin = RE || re = 16.4 Ω      (low, as expected for CB)
Rout ≈ RC = 4.04 kΩ
Av  = RC/re = 4.04 kΩ/16.7 Ω = 242   (no load, non-inverting)
```

**Answer:** R1 ≈ 144 kΩ, R2 ≈ 30 kΩ, RE = 1 kΩ, RC ≈ 4 kΩ; Av ≈ 242, Rin ≈ 16 Ω (the low Rin is inherent in a CB stage), Rout ≈ 4 kΩ.

## Check yourself

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

1. For VCC = 15 V, IE = 1.5 mA, what are VE and RE?
   A. VE = 2.2 V, RE = 1.47 kΩ
   B. VE = 7.5 V, RE = 5 kΩ
   C. VE = 1.5 V, RE = 10 kΩ
   D. VE = 1.5 V, RE = 1 kΩ
   Answer: D. VE = VCC/10 = 1.5 V and RE = 1.5/1.5 mA = 1 kΩ. (Key point: VE and RE)

2. What is RC?
   A. 7.5 kΩ
   B. 1 kΩ
   C. 4.04 kΩ from (15-1.5-7.5)/1.485 mA
   D. 5 kΩ
   Answer: C. RC = 6 V / 1.485 mA = 4.04 kΩ. (Key point: RC)

3. With VB = 2.2 V and I2 = 5 IB = 74.3 µA, R2 and R1 are:
   A. R2 ≈ 29.6 kΩ, R1 ≈ 144 kΩ
   B. R2 ≈ 144 kΩ, R1 ≈ 29.6 kΩ
   C. R2 ≈ 2.96 kΩ, R1 ≈ 14 kΩ
   D. R2 ≈ 15 kΩ, R1 ≈ 15 kΩ
   Answer: A. R2 = 2.2/74.3 µA = 29.6 kΩ; R1 = 12.8/89.1 µA = 144 kΩ. (Key point: Divider)

4. Why is Rin only about 16 Ω?
   A. R1 is too large
   B. Inherent to CB: the input looks into the emitter, Rin ≈ re
   C. RE was chosen too small
   D. beta is low
   Answer: B. The CB input resistance is about re = VT/IE regardless of design. (Key point: Low Rin)

## Close variants set in other exams

- [Design common base amplifier, beta independent](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Acb-amplifier-design-beta-independent) — Electronics Devices and Circuit, chapter 2, asked 2×

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