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Short note: Differential amplifier with active load

Short answer

Using a PNP current mirror as active load replaces collector resistors, giving gain Ad = gm(ro2 ‖ ro4), twice that of passive single‑ended. It converts differential to single‑ended output, saves area, provides high CMRR, but demands a high‑input‑impedance next stage.

Write a short note on differential amplifier with active load.

A differential amplifier with active load uses a current mirror (usually PNP, Q3–Q4) in place of the two collector resistors of the differential pair Q1–Q2.

      +VCC ---- Q3(diode) ==== Q4 ---- +VCC
                  |             |
                  |             +----- vo
       v1 ---B   Q1            Q2   B--- v2
                  +------+------+
                         I (tail source)
                       -VEE

Working: for a differential input vd, ic1 = +gm vd/2 and ic2 = −gm vd/2. Q3 carries ic1 and the mirror copies it into Q4, so the net current into the output node is ic1 − ic2 = gm·vd.

Ad = gm (ro2 ‖ ro4)   (single-ended output)
Compare passive load: Ad = gm·RC/2

Advantages:

  • Gain is twice that of a passive load of equal resistance, and in practice much higher since ro2‖ro4 is large (gain of 1000 or more in one stage).
  • Differential to single-ended conversion without losing half the signal.
  • Small dc voltage drop and small chip area.
  • Good CMRR, since a common-mode change gives equal currents that the mirror cancels at the output.

Limitations: the high output resistance must be followed by a high-input-resistance stage (e.g. Darlington or emitter follower), and the gain depends on ro, which varies with devices. It is used as the input stage of the 741 and most op-amps.

✓ Answer checked by Claude Opus 5.5 (high effort)

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