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AC equivalent: Rin and Rout of divider-bias amplifier

Short answer

DC bias gives re ≈ 16.2 Ω. With unbypassed RE, the base sees (β+1)(re+RE) ≈ 102.6 kΩ. Rin = R1||R2||102.6 kΩ ≈ 14.3 kΩ; Rout is simply RC = 5 kΩ.

Draw the ac equivalent circuit for given circuit and find its input and output resistances. Assume β = 100 for the BJT. [Figure: BJT amplifier with Vcc = +15 V; base divider R1 = 100 kΩ from Vcc and R2 = 20 kΩ to ground; collector resistor RC = 5 kΩ with output Vo at the collector; emitter resistor RE = 1 kΩ to ground; input vi at the base]

The circuit has a base divider (R1 = 100 kΩ, R2 = 20 kΩ), RC = 5 kΩ, and RE = 1 kΩ in the emitter, which is assumed unbypassed (no bypass capacitor is shown). VCC = 15 V, β = 100.

DC bias (to find re)

VBB = 15 × 20/120 = 2.5 V ,   RB = 100 k || 20 k = 16.67 kΩ
IB = (2.5 - 0.7)/(16.67 k + 101 × 1 k) = 1.8/117.7 k = 15.3 µA
IC = 1.53 mA ,   IE = 1.545 mA
re = 25 mV/1.545 mA = 16.2 Ω

AC equivalent circuit

VCC is replaced by ground: R1 and R2 are in parallel from the base to ground; the transistor T-model (re in the emitter path, with current source α ie to the collector); RE in series from the emitter to ground; RC from the collector to ground; the output is at the collector.

 vi--+----B--------C---+--- vo
     |     \ re   (α ie)|
   R1||R2   E           RC
     |      |           |
    GND     RE         GND
            |
           GND

Input resistance

Rib = (β+1)(re + RE) = 101 × (16.2 + 1000) = 102.6 kΩ
Rin = R1 || R2 || Rib = 16.67 k || 102.6 k = 14.3 kΩ

Output resistance

Looking into the collector, the transistor is a current source (ro = ∞), so

Rout = RC = 5 kΩ

(Voltage gain, if needed: Av = -RC/(re + RE) = -4.9.)

Answer: Rin ≈ 14.3 kΩ, Rout ≈ 5 kΩ.

✓ Answer checked by Claude Opus 5.5 (high effort)

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