---
title: "Design common base amplifier (24 V, 1 mA, beta 200)"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:cb-amplifier-design-24v"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/cb-amplifier-design-24v
site: BCT Engineering
---

# Design common base amplifier (24 V, 1 mA, beta 200)

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

- Asked in: 2071 Shrawan, 5+3 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-24v
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

Design Common Base Amplifier using β-independent dc biasing method. Use appropriate guideline to support your design. Given parameters are: VCC = 24VDC, IC = 1 mA and β = 200. Also find its voltage gain by using its ac equivalent circuit.

## Short answer

Use voltage-divider bias with VE ≈ VCC/10 = 2.4 V, VCE ≈ VCC/2 = 12 V, and I2 = 10 IB. Calculated: RE = 2.39 kΩ, RC = 9.6 kΩ, R1 ≈ 380 kΩ, R2 ≈ 62 kΩ. Voltage gain Av = RC/re = 9.6 kΩ / 24.9 Ω ≈ +386.

## Answer

This is a common base amplifier with voltage-divider (β-independent) bias. The base is grounded for ac through CB.

Given: VCC = 24 V, IC = 1 mA, β = 200.

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

#### Design
```text
IE = 1.005 mA     IC = α IE = 1 mA     IB = IE/(β+1) = 5 µA
VE = VCC/10 = 2.4 V      RE = VE/IE = 2.39 kΩ
VCE = VCC/2 = 12 V     RC = (VCC - VE - VCE)/IC = (24 - 2.4 - 12)/1 mA = 9.6 kΩ
VB = 3.1 V
I2 = 10 IB = 50 µA
R2 = VB/I2 = 62 kΩ
R1 = (VCC - VB)/(I2 + IB) = 20.9/55 µA = 380 kΩ
```
Choose R1 ≈ 380 kΩ, R2 ≈ 62 kΩ, RE ≈ 2.39 kΩ, RC ≈ 9.6 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/1.005 mA = 24.9 Ω
Rin = RE || re = 24.6 Ω      (low, as expected for CB)
Rout ≈ RC = 9.6 kΩ
Av  = RC/re = 9.6 kΩ/24.9 Ω = 386   (no load, non-inverting)
```

#### Voltage gain from the ac equivalent circuit
In the T-model, the emitter (input) is driven through the resistance re, and the collector current α ie flows in RC:
```text
vi = -ie re ,    vo = -α ie RC  (output across RC; unloaded)
Av = vo/vi = α RC / re ≈ RC/re = 9.6 kΩ / 24.9 Ω ≈ +386
```
**Answer:** R1 ≈ 380 kΩ, R2 ≈ 62 kΩ, RE ≈ 2.39 kΩ, RC = 9.6 kΩ; Av ≈ +386 (in phase), Rin ≈ 25 Ω.

## Check yourself

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

1. How is the base made an ac ground in the CB design?
   A. Large bypass capacitor CB from base to ground
   B. Connecting the base to VCC directly
   C. Removing R2
   D. Adding RE in series
   Answer: A. CB shorts the base to ground for ac while the divider sets the dc base voltage. (Key point: Base grounded for ac)

2. With IC = 1 mA and β = 200, what is IB?
   A. 5 µA
   B. 10 µA
   C. 1 µA
   D. 50 µA
   Answer: A. IE = 1.005 mA, so IB = IE/(β+1) = 5 µA. (Key point: Base current)

3. What is the input resistance of the CB amplifier?
   A. Very high, about β re
   B. Low, about re (RE || re = 24.6 Ω)
   C. RC
   D. R1 || R2
   Answer: B. The input is at the emitter, so the resistance is RE || re, which is low. (Key point: Input resistance CB)

4. What is the unloaded voltage gain and its phase?
   A. -RC/re, inverting
   B. +RC/re, about +386, non-inverting
   C. -gm RE
   D. +1
   Answer: B. Av = αRC/re ≈ 9.6 kΩ/24.9 Ω ≈ 386, in phase with the input. (Key point: Voltage gain)

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