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CE amplifier Rin, Rout, Av, Ai (470k, 2.2k, 0.56k)

Short answer

DC analysis yields re = 5.76 Ω. Zin = 470 kΩ || (β+1)(re+0.56kΩ) ≈ 59.8 kΩ, Zout = 2.2 kΩ, Av = -RC/(re+RE) ≈ -3.9, Ai = β·RB/(RB+Rib) ≈ 105.

For the figure shown below with β = 120 find (a) input impedance (b) output impedance (c) voltage gain (d) current gain. Use small signal model. [Figure: BJT Q1 amplifier with 20 V supply; 470 kΩ from supply to base; 2.2 kΩ collector resistor with output Vo through coupling capacitor C2; 0.56 kΩ emitter resistor to ground; source Vs through coupling capacitor C1 at the base]

The amplifier uses fixed (base-resistor) bias with an unbypassed emitter resistor and no load (the output is taken from the collector).

Given

VCC = 20 V, RB = 470 kΩ, RC = 2.2 kΩ, RE = 0.56 kΩ, β = 120, VBE = 0.7 V.

DC bias

IB = (VCC - 0.7)/(RB + (β+1) RE) = 19.3/(470 k + 121 × 0.56 k)
   = 19.3/537.8 k = 35.9 µA
IC = 4.31 mA ,   IE = 4.34 mA
re = 25 mV/4.34 mA = 5.76 Ω ,   rπ = β re = 691 Ω

Small-signal model

Using the T-model with re and the unbypassed RE in the emitter, the resistance seen at the base is

Rib = (β+1)(re + RE) = 121 × (5.76 + 560) = 68.5 kΩ

(a) Input impedance

Zin = RB || Rib = 470 k || 68.5 k = 59.8 kΩ

(b) Output impedance

Zout = RC = 2.2 kΩ    (ro neglected)

(c) Voltage gain

Av = -RC/(re + RE) = -2200/565.8 = -3.89

(d) Current gain

Take the output current as the collector current and the input current as the current drawn from the source. Then ib = ii × RB/(RB + Rib) and ic = β ib:

Ai = β RB/(RB + Rib) = 120 × 470/(470 + 68.5) = 105

(Check: |Av| Zin/RC = 3.89 × 59.8 k/2.2 k ≈ 105.7; the small difference arises because Av takes α ≈ 1.)

Answer: Zin ≈ 59.8 kΩ, Zout ≈ 2.2 kΩ, Av ≈ -3.9, Ai ≈ 105.

✓ Answer checked by Claude Opus 5.5 (high effort)

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