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Derive Ai, Rin and Rout of emitter follower

Short answer

Rin = R1||R2||(β+1)(re+RE). Current gain Ai = (β+1) RB/(RB+Rib) where RB = R1||R2 and Rib = (β+1)(re+RE). Output resistance Rout = RE || [re + (Rs||RB)/(β+1)].

Derive expressions for Ai, Rin and Rout. [Figure: common collector (emitter follower) amplifier with +Vcc; base divider R1 (from Vcc) and R2 (to ground); emitter resistor RE = RL; input Vin at the base and output Vo at the emitter]

The emitter follower (common collector) has a voltage divider R1, R2 at the base; the output vo is taken across the emitter resistor RE (= the load RL). Replace the transistor by the T-model (re in the emitter, ie = (β+1) ib); the collector and VCC are at ac ground. Let RB = R1 || R2.

 ii --->    ib
 vin--+--+----B
      |  |     \ re
     RB  |      E----+--- vo = ie RE
      |           RE
     GND          GND

Input resistance

vb = ie (re + RE) = (β+1) ib (re + RE)
Rib = vb/ib = (β+1)(re + RE)
Rin = RB || Rib = R1 || R2 || (β+1)(re + RE)

Current gain

The signal current ii divides between RB and the base:

ib = ii × RB/(RB + Rib)
io = ie = (β+1) ib   (current in RE = load)

Ai = io/ii = (β+1) RB/(RB + Rib)

If RB ≫ Rib, Ai ≈ (β+1). For example β = 100, RB = 10 kΩ, Rib = 100 kΩ: Ai = 101 × 10/110 = 9.2.

(In terms of the voltage gain: Ai = Av Rin/RE.)

Output resistance

Source off (vin = 0); the base resistance is R's = Rs || RB. A test voltage vx at the emitter drives ie = vx/(re + R's/(β+1)):

Rout = RE || [ re + R's/(β+1) ]  ≈ re + R's/(β+1)

The voltage gain is Av = RE/(RE + re) ≈ 1.

ParameterExpression
RinR1‖R2‖(β+1)(re+RE)
Ai(β+1) RB/(RB + Rib)
RoutRE‖[re + R's/(β+1)]

✓ Answer checked by Claude Opus 5.5 (high effort)

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