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Design beta independent CE amplifier (20 V, 2 mA)

Short answer

Firm divider bias: VE = 2 V, VCE = 10 V, I2 = 10 IB. RE = 2V/2.02mA ≈ 990 Ω, RC = 4 kΩ. VB = 2.7 V gives R2 = 13.5 kΩ, R1 ≈ 78.6 kΩ. With RE bypassed, Zin ≈ 1.12 kΩ, Av ≈ –320.

Design β independent type DC biased Common Emitter amplifier, and find its input impedance and voltage gain. Given parameters: VCC = 20 V DC, IC = 2mA and β=100. Use firm biasing method.

Firm biasing means a stiff voltage divider (I2 ≈ 10 IB). For the gain calculation, the emitter resistor is assumed to be bypassed.

Design guidelines used

  1. Emitter voltage VE ≈ VCC/10 (large enough for thermal and bias stability, small enough to leave room for the output swing).
  2. Collector-emitter voltage VCE ≈ VCC/2, so the output can swing equally up and down (Q-point at the centre of the load line).
  3. Divider current I2 = 10 IB, much greater than IB, so the base voltage is fixed and the bias is independent of β.
  4. VB = VE + 0.7 V.

Design

IC  = 2 mA        IB = IC/β = 2 mA/100 = 20 µA
IE  = IC + IB = 2.02 mA
VE  = VCC/10 = 2 V     RE = VE/IE = 2/2.02 mA = 990 Ω
VCE = VCC/2  = 10 V
RC  = (VCC - VE - VCE)/IC = (20 - 2 - 10)/2 mA = 4 kΩ
VB  = VE + 0.7 = 2.7 V
I2  = 10 IB = 200 µA
R2  = VB/I2 = 2.7/200 µA = 13.5 kΩ
R1  = (VCC - VB)/(I2 + IB) = 17.3/220 µA = 78.6 kΩ

Calculated values: R1 ≈ 78.6 kΩ, R2 ≈ 13.5 kΩ, RE ≈ 990 Ω, RC ≈ 4 kΩ (in practice, use the nearest standard resistor values). The emitter resistor RE is bypassed by a large capacitor CE; C1 and C2 are the coupling capacitors.

Check and small-signal results (RE bypassed)

re  = VT/IE = 25 mV/2.02 mA = 12.4 Ω
rπ  = β re = 1238 Ω         gm = IC/VT = 80 mA/V
R1 || R2 = 11.5 kΩ
Rin = R1 || R2 || rπ = 1.12 kΩ
Av  = -gm RC = -80 mA/V × 4 kΩ = -320

Ro ≈ RC = 4 kΩ.

Answer: input impedance Zin ≈ 1.12 kΩ and voltage gain Av ≈ -320 (no load, RE bypassed). If RE were not bypassed, Av = -RC/(re + RE) ≈ -4 and Zin would be higher.

✓ Answer checked by Claude Opus 5.5 (high effort)

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