Active load doubles differential amplifier gain

In an active-load differential amplifier the collector resistors RC are replaced by a PNP current mirror (Q3, Q4). We compare the single-ended output gain in both cases.

           +VCC          +VCC
            |              |
           Q3 (diode)----- Q4      PNP mirror (active load)
            |              |
            +-- i1    i4 --+---- Vo
            |              |
  V1 --B   Q1             Q2   B-- V2
            |              |
            +------+-------+
                   |
                 I (tail current source)
                   |
                 -VEE

Passive load (RC)

With a differential input vd = v1 − v2, the emitters form an ac ground and each transistor gets ±vd/2:

ic1 = +gm vd/2 ,  ic2 = -gm vd/2
Single-ended output at collector of Q2:
vo = -ic2·RC = gm·RC·vd/2
Ad(passive) = gm RC / 2

Only the signal current of Q2 reaches the output; the signal current of Q1 is wasted in its own RC.

Active load (current mirror)

1. Q1 current increases by gm·vd/2. This current flows through diode-connected Q3.
2. The mirror copies it, so i4 = ic1 = +gm vd/2 flows from Q4 into the output node.
3. Q2 current decreases by gm vd/2, so Q2 draws −gm vd/2 from the output node.
4. Net current pushed into the output node:

io = i4 - ic2 = gm vd/2 - (-gm vd/2) = gm·vd
Output node resistance = ro2 ‖ ro4
vo = gm·vd·(ro2 ‖ ro4)
Ad(active) = gm (ro2 ‖ ro4)

Comparison

If the load resistance is the same in both cases (ro2‖ro4 = RC), then

Ad(active) / Ad(passive) = gm·R / (gm·R/2) = 2

So the active load doubles the gain because the mirror transfers the signal current of Q1 to the output and both halves contribute (full differential-to-single-ended conversion). In practice ro2‖ro4 (about 50 kΩ to 100s of kΩ) is also much larger than any practical RC, so the actual gain is many times higher, e.g. gm = 40 mS and ro2‖ro4 = 50 kΩ gives Ad = 2000. The active load also takes little dc voltage and little chip area.