Editing Operational amplifier (section)

==== Differential amplifier ====
A differential voltage ''V''<sub>in</sub> at the op amp inputs (pins 3 and 2, respectively) gives rise to a small differential current in the bases of Q1 and Q2 ''i''<sub>in</sub>&nbsp;≈ ''V''<sub>in</sub>&nbsp;/ (2''h''<sub>ie</sub>''h''<sub>fe</sub>). This differential base current causes a change in the differential collector current in each leg by ''i''<sub>in</sub>''h''<sub>fe</sub>. Introducing the transconductance of Q1, ''g''<sub>''m''</sub>&nbsp;= ''h''<sub>fe</sub>&nbsp;/ ''h''<sub>ie</sub>, the (small-signal) current at the base of Q15 (the input of the voltage gain stage) is ''V''<sub>in</sub>''g''<sub>''m''</sub>&nbsp;/ 2.

This portion of the op amp cleverly changes a differential signal at the op amp inputs to a single-ended signal at the base of Q15, and in a way that avoids wastefully discarding the signal in either leg. To see how, notice that a small negative change in voltage at the inverting input (Q2 base) drives it out of conduction, and this incremental decrease in current passes directly from Q4 collector to its emitter, resulting in a decrease in base drive for Q15. On the other hand, a small positive change in voltage at the non-inverting input (Q1 base) drives this transistor into conduction, reflected in an increase in current at the collector of Q3. This current drives Q7 further into conduction, which turns on current mirror Q5/Q6. Thus, the increase in Q3 emitter current is mirrored in an increase in Q6 collector current; the increased collector currents shunts more from the collector node and results in a decrease in base drive current for Q15. Besides avoiding wasting 3&nbsp;dB of gain here, this technique decreases common-mode gain and feedthrough of power supply noise.