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CE amplifier input, output resistance and gain (RE1, RE2)

Short answer

DC analysis gives IC ≈ 1.05 mA, re ≈ 23.6 Ω. With unbypassed RE1 = 470 Ω, Rin ≈ 7.4 kΩ, Rout ≈ 4.7 kΩ, overall voltage gain Avs ≈ -8.0.

Determine the input resistance, output resistance and overall voltage gain of the circuit given below. [Figure: CE amplifier with Vcc = +10 V; R1 = 47 kΩ from Vcc to base; R2 = 10 kΩ from base to ground; RC = 4.7 kΩ; emitter resistors RE1 = 470 Ω in series with RE2 = 470 Ω, with RE2 bypassed by C3 = 100 μF; β = 150; source Vs with Rs = 600 Ω and coupling capacitor C1 = 10 μF at the input; coupling capacitor C2 = 10 μF and load RL = 47 kΩ at the output]

Here RE1 = 470 Ω is unbypassed and RE2 = 470 Ω is bypassed by C3, so only RE1 appears in the ac path; both are in the dc path.

Given

VCC = 10 V, R1 = 47 kΩ, R2 = 10 kΩ, RC = 4.7 kΩ, RE1 = RE2 = 470 Ω, β = 150, Rs = 600 Ω, RL = 47 kΩ.

DC bias

VBB = VCC R2/(R1 + R2) = 10 × 10/57 = 1.754 V
RB  = R1 || R2 = 8.246 kΩ
IB  = (VBB - 0.7)/(RB + (β+1)(RE1 + RE2)) = 1.054/(8.246 k + 151 × 940)
    = 1.054/150.2 k = 7.02 µA
IC  = β IB = 1.053 mA ,   IE = 1.06 mA
re  = 25 mV/1.06 mA = 23.6 Ω

Input resistance

Rib = (β+1)(re + RE1) = 151 × (23.6 + 470) = 74.5 kΩ
Rin = R1 || R2 || Rib = 8.246 k || 74.5 k = 7.42 kΩ

Output resistance

Rout = RC = 4.7 kΩ    (ro neglected)

Voltage gain

RC || RL = 4.7 k || 47 k = 4.27 kΩ
Av = -(RC||RL)/(re + RE1) = -4270/493.6 = -8.65

Overall: Avs = Av · Rin/(Rin + Rs) = -8.65 × 7.42/(7.42 + 0.6) = -8.0

Answer: Rin ≈ 7.4 kΩ, Rout ≈ 4.7 kΩ, overall gain Avs ≈ -8.0 (-8.65 from the base to the output).

✓ Answer checked by Claude Opus 5.5 (high effort)

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