How is the dc bias set up?

Thevenin: VBB = 5 V, RB = 5 kΩ; IB = (VBB-0.7)/(RB+(beta+1)RE)

Which expression gives Rin?

R1 || R2 || (beta+1)(re + RE||RL) ≈ 4.96 kΩ

RE || [re + (RG||RB)/(beta+1)] ≈ 27.6 Ω

Why is overall gain 0.885 while the base-to-output gain is 0.992?

The source resistance RG forms a divider with Rin: Rin/(Rin+RG)

2078 Bhadra · 8 marks

Emitter follower input/output resistance and gain

Short answer

DC analysis yields IE ≈ 0.994 mA, re ≈ 25.1 Ω. AC analysis gives Rin ≈ 4.96 kΩ, Rout ≈ 27.6 Ω, and overall voltage gain ≈ 0.885.

The bipolar junction transistor parameters for the circuit in figure below are β = 200 and VA = ∞. Determine the input resistance, output resistance and overall voltage gain of the circuit. [Figure: common collector (emitter follower) circuit with Vcc = +10 V; collector connected to Vcc; base divider R1 = 10 kΩ (from Vcc) and R2 = 10 kΩ (to ground); source Vg = 1 V with RG = 600 Ω and a coupling capacitor to the base; emitter resistor RE = 4.3 kΩ to ground; output taken from the emitter through a coupling capacitor to load RL = 10 kΩ]

Find the dc bias (for re), then the ac parameters of the emitter follower.

Given

VCC = 10 V, R1 = R2 = 10 kΩ, RG (source) = 600 Ω, RE = 4.3 kΩ, RL = 10 kΩ, β = 200.

DC bias

Thevenin equivalent of the divider:

VBB = 10 × 10k/20k = 5 V ,    RB = R1 || R2 = 5 kΩ
IB = (VBB - 0.7)/(RB + (β+1) RE) = 4.3/(5 k + 201 × 4.3 k)
   = 4.3/869.3 k = 4.95 µA
IE = (β+1) IB = 0.994 mA
re = 25 mV/0.994 mA = 25.1 Ω

Input resistance

R'E = RE || RL = 4.3k × 10k/14.3k = 3.007 kΩ
Rib = (β+1)(re + R'E) = 201 × (25.1 + 3007) = 609 kΩ
Rin = R1 || R2 || Rib = 5 k || 609 k = 4.96 kΩ

Voltage gain

Av = R'E/(re + R'E) = 3007/3032 = 0.992   (vo/vb)
Overall: vo/Vg = Av × Rin/(Rin + RG) = 0.992 × 4.96/(4.96 + 0.6) = 0.885

Output resistance (looking back at the emitter, RL excluded)

R's = RG || RB = 0.6 k || 5 k = 536 Ω
Rout = RE || [re + R's/(β+1)] = 4.3 k || (25.1 + 2.67) = 27.6 Ω

Answer: Rin ≈ 4.96 kΩ, Rout ≈ 27.6 Ω, overall voltage gain ≈ 0.885 (0.992 from the base to the output).

✓ Answer checked by Claude Opus 5.5 (high effort)

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