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CC amplifier ac equivalent, Rin and Rout

Short answer

Rin = R1||R2||(β+1)(re + RE||RL); Rout = RE || [re + R's/(β+1)] where R's = Rs||R1||R2. The CC stage thus provides high input and very low output impedance.

Draw ac equivalent circuit of common collector amplifier. Find its input and output resistances.

AC equivalent circuit of the common collector amplifier

The collector is tied to VCC, which is ac ground. The input is applied to the base and the output is taken from the emitter. Using the T-model:

 vs--Rs--+----B
         |     \ re   ie = (β+1) ib
       R1||R2   E-----+---- vo
         |            |
        GND        RE||RL
                      |
                     GND

R1 and R2 (both connected to ground for ac) are in parallel at the base; R'E = RE || RL.

Input resistance

vb = ie (re + R'E) = (β+1) ib (re + R'E)
Rib = (β+1)(re + R'E)
Rin = R1 || R2 || (β+1)(re + RE||RL)

Output resistance

Set vs = 0 and remove RL; apply a test voltage vx at the emitter. The resistance in the base circuit is R's = Rs || R1 || R2. The base current now flows out of the base, through R's to ground:

ib = -vx/(R's + rπ)   → current drawn from vx by the transistor: -ie = (β+1) vx/(R's + rπ)
Rout = RE || (rπ + R's)/(β+1)  =  RE || [ re + R's/(β+1) ]

Hence Rout ≈ re + R's/(β+1), a small value.

Example

β = 100, IE = 2 mA (re = 12.5 Ω), RE = 2 kΩ, RL = 2 kΩ, R's = 2 kΩ: Rin = R1||R2||(101 × 1012.5) ≈ R1||R2||102 kΩ; Rout ≈ 2 k || (12.5 + 19.8) = 31.8 Ω.

✓ Answer checked by Claude Opus 5.5 (high effort)

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