AC analysis of differential amplifier

A differential amplifier amplifies the difference vd = v1 − v2 and rejects the common-mode signal vcm = (v1 + v2)/2. Its ac analysis is done by splitting any input into these two parts and using the half-circuit method.

         +VCC           +VCC
          |               |
          RC              RC
          +-- vo1   vo2 --+
          |               |
 v1 ---B Q1              Q2 B--- v2
          |               |
          +-------+-------+
                  | E
                  RE  (tail)
                  |
                -VEE

Differential-mode analysis

For v1 = +vd/2 and v2 = −vd/2, the change in IE1 equals minus the change in IE2, so the current through RE stays constant and the emitter node is ac ground.

Half circuit (CE stage, emitter grounded):
  ic1 = gm·vd/2 ,  ic2 = -gm·vd/2
  vo1 = -gm·RC·vd/2 ,  vo2 = +gm·RC·vd/2
Double-ended: Ad = (vo2 - vo1)/vd = gm·RC
Single-ended: Ad = gm·RC/2
Differential input resistance: Rid = 2rπ

Common-mode analysis

For v1 = v2 = vcm, both emitter currents change together, so the current in RE is doubled. Each half sees an emitter resistance of 2RE.

Half circuit: CE stage with 2RE in the emitter
  Acm = vo1/vcm = -gm·RC / (1 + 2gm·RE) ≈ -RC/(2RE)
Common-mode input resistance ≈ β·(2RE) per side
Double-ended output: vo2 - vo1 = 0 (ideally Acm = 0)

CMRR and total output

Single-ended:
CMRR = |Ad/Acm| = (gm·RC/2)/(RC/2RE) ≈ gm·RE
CMRR(dB) = 20 log10 |Ad/Acm|
vo = Ad·vd + Acm·vcm

For a high CMRR, RE should be very large; this is why the tail resistor is replaced by a constant current source (current mirror), whose output resistance is hundreds of kΩ. Using double-ended output with matched halves also cancels the common-mode output.