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CE amplifier input resistance and overall gain

Short answer

DC analysis gives IE=2.49mA, re=10Ω. Input resistance is Rin = RB || (β+1)(re+RE) = 680kΩ || 92.1kΩ ≈ 81kΩ. Overall voltage gain (including source) is Avs = -2.19 ≈ -2.2.

Determine input resistance and overall voltage gain of given Common Emitter BJT Amplifier. β=110. [Figure: CE amplifier with +18 V supply; 680 kΩ from +18 V to base; 3 kΩ collector resistor; 0.82 kΩ emitter resistor; source Vs with series Rs = 0.6 kΩ and coupling capacitor C1 at the input; coupling capacitor C2 and load RL = 4.7 kΩ at the output]

The amplifier has fixed (base-resistor) bias with an unbypassed emitter resistor. Find the dc bias first (for re), then the ac parameters.

Given

VCC = 18 V, RB = 680 kΩ, RC = 3 kΩ, RE = 0.82 kΩ, β = 110, Rs = 0.6 kΩ, RL = 4.7 kΩ, VBE = 0.7 V.

DC bias

IB = (VCC - VBE)/(RB + (β+1) RE) = 17.3/(680 k + 111 × 0.82 k)
   = 17.3/771.0 k = 22.4 µA
IC = β IB = 2.47 mA ,    IE = 2.49 mA
re = VT/IE = 25 mV/2.49 mA = 10.0 Ω

Input resistance

Rib = (β+1)(re + RE) = 111 × (10.0 + 820) = 92.1 kΩ
Rin = RB || Rib = 680 k || 92.1 k = 81.1 kΩ

Voltage gain

Av (from base to output) = -(RC || RL)/(re + RE)
RC || RL = 3k × 4.7k/7.7k = 1.831 kΩ
Av = -1831/830 = -2.21

Overall gain, including the source resistance:

Avs = Av × Rin/(Rin + Rs) = -2.21 × 81.1/(81.1 + 0.6) = -2.19

Answer: Rin ≈ 81 kΩ; overall voltage gain ≈ -2.2 (the gain is small because RE is unbypassed).

✓ Answer checked by Claude Opus 5.5 (high effort)

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