cmos differential mode equvalent circuit

[akbarza]

hi
in the below figure,what's the differential mode equivalent circuit?
explanation that as you can see the m5 is in saturation mode, but m4 can be in saturation or triode mode.
in common mode, what is equal circuit?
thanks

 

[frankrose]

In diff. mode a simple voltage conrolled current source is the small signal equvivalent model with resistive and capacitive load:
Gm=Gm2,3 RL=1/(Gds2+Gds4) CL~Cds4+Cdg4+Cdb4+Cdb2+Cds2+Cdg2
If M4 is in triode region, the RL=1/(Gds2+Gm4)~1/Gm4, and model is the same.

Common-mode model from Razavi (RSS=1/(Gds1)):

 

[akbarza]

In diff. mode a simple voltage conrolled current source is the small signal equvivalent model with resistive and capacitive load:
Gm=Gm2,3 RL=1/(Gds2+Gds4) CL~Cds4+Cdg4+Cdb4+Cdb2+Cds2+Cdg2
If M4 is in triode region, the RL=1/(Gds2+Gm4)~1/Gm4, and model is the same.

Common-mode model from Razavi (RSS=1/(Gds1)):

View attachment 158056

i think the drawn circuit with me is different from your.if there is in my pic the connecton line that connects vout+ and vout- in your pic,then your answer was true.

 

[FvM]

Review Razavi, Design and Analysis of Integrated Circuits to understand why the short can be made in common mode equivalent circuit.

 

[frankrose]

Yes, everything is written down and explained in the literature which you are curious about, please read it before write anything. But the point is that when common-mode gain is analised the drains of the diff.pair vary in-phase with quasi same amplitude, so they can be connected together theoretically by a wire ( fig.(a) above ), which simplify the model creation ( fig.(b) above ). And to prevent an other misunderstanding, when I wrote RSS=1/Gds1, I meant Gds1 is the output conductance of M1 from your posted picture, not from I shared.

 

[akbarza]

hi
thanks for answers
i accept your answer about common mode, but my principle question is about differential mode.
anyone can explain about differential equivalent circuit of above pic?