Output resistance of unbypassed-emitter CE amplifier

The output resistance is found by setting the input source to zero (vs = 0), removing the load and applying a test voltage vx at the output (collector); Rout = vx/ix.

Small-signal circuit (hybrid-π with ro)

 test vx --> C ----+----------+
 ix              ro         gm vπ
                   |          |
 B--rπ--+----------E----------+
 |      |          |
 RB'    vπ         RE
 |                 |
GND               GND

RB' = Rs || R1 || R2 is the resistance from the base to ground (vs = 0). The emitter node sees RE in parallel with the base path (rπ + RB'):

R'E = RE || (rπ + RB')

Derivation

The collector current ix returns to ground through the emitter, so

ve = ix R'E

The base voltage is ve × RB'/(rπ + RB'), so vπ = vb - ve = -a ve, where a = rπ/(rπ + RB') ≤ 1. KCL at the collector:

ix = gm vπ + (vx - ve)/ro = -gm a ve + (vx - ve)/ro
   = -gm a ix R'E + (vx - ix R'E)/ro

ix [1 + gm a R'E + R'E/ro] = vx/ro

Hence the resistance looking into the collector is

Rout(collector) = vx/ix = ro (1 + gm a R'E) + R'E
                ≈ ro (1 + gm R'E)        (for RB' ≪ rπ, so a ≈ 1)

The unbypassed RE (series feedback) raises the collector resistance far above ro.

Output resistance of the amplifier

The collector resistance is in parallel with RC:

Rout = RC || [ro (1 + gm a R'E) + R'E]  ≈ RC

because the bracket is very large (for example ro = 50 kΩ, gm R'E = 10 gives 550 kΩ). If ro is ignored, Rout = RC.

Result: Rout ≈ RC, the same as for the bypassed amplifier; RE does not lower the output resistance.