Skip to main content

Design common base amplifier (15 V, 1.5 mA)

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.

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.

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

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)

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electronics Devices and Circuit chapter 2 questions →