Differential gain on BJT differential pair: not symmetric?
Both gain values can be equal only if two conditions are met:

(1) The circuit is 100% symmetric (fulfilled)

(2) Both signal input voltages (between B and E) are equal (not fulfilled).

For a good understanding of the circuits function it is essential to realize that  - for each of the transistors - the voltage Vbe matters (and not only the voltage at the base node).


For Q3: Vbe=(Vc+Vd)-Ve =Vc-Ve +Vd
For Q4: Vbe=Vc-(Vdx+Ve)=Vc-Ve -Vdx.


(Ve is the DC voltage - equal for both transistors - caused by the common DC current source)

What is Vdx? In contrast to Q3, the transistor Q4 receives an input signal (caused by Vd) from the emitter of Q3 (acting as a emitter follower). This voltage Vdx is slighly smaller than Vd (emitter follower gain &lt;1). The voltage at the base of Q4 is the DC voltage Vc only!

Interpretation (for a correct understanding): Both transistors (Q3 and Q4) can be seen as a series connection of a common-collector and a common-base stage.

Result: Thus, the positive input swing for Q3 (Vd) is not identical to the negative input swing for Q4 (Vdx).