- 2074 Asoj · 3 marks
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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