why Pmos current source can increase differential gain?

[diego.fan]

I don't understand the last sentence in red. Differential gain should be Rd*gm I think, but how to get the final conclusion? Thanks

 

[erikl]

Differential gain should be Rd*gm ...

True - but this concerns gm of M1 or M2.

For a given |VGSP - VTHP| , this translates to a factor of five reduction in the transconductance of M3 and M4

...and this translates to nearly a factor of 5 increase of the output impedance of M3 and M4, which essentially determines the gain.
Now Rd = 1/(gm(M3)+gm(M5)) , and

gain ≈ gm(M1)/(gm(M3)+gm(M5))​

M5 & M6 - with the same Vds as M3 & M4 - however can be operated with an (absolute) lower |Vb| than |Vgs=Vds|, hence will have a considerably lower gm, respectively higher Rd.

In practice - depending on the selected aspect ratio between M5/M3 (M6/M4) and the chosen inversion mode - one probably can't expect this full factor of five in gain enhancement.