help me to understand class AB input diff pair

[Jenifer_gao]

class ab site:edaboard.com

Hi All:

I am listening to the Berkeley lecture of EECS140 about the stability. There is a circuit about Class AB Input Stage Cross Coupled Differential Pair, which is attached. I don't quite understand how this circuit works. Because on the schematic, node A and node C has the same voltage Vi+, and node B and node D has the same voltage of Vi-. But I think from node A to node C, it has a voltage drop about 2Vdsat+2Vt, and the same thing from node B to D. And I don't quite understand the function of the level shift here.
Can anybody explain it? Thanks.

J

 

[okguy]

I think that Vi+ is not physically connected to C (resp. Vi- to D).
That's only the ac part that is equal due to the 2.Vt shift.
BTW, did you try to simulate it?

 

[Jenifer_gao]

No, I don't simulate it yet. Yes I should try

 

[sunking]

Pay attention to the symbol: Vi. Its means complex variable, phasor or rms value of the signal

 

[Btrend]

those are really "small signals" , so the notations are vi and not Vi

 

[nathan]

The level shift is needed to keep the core devices (i.e., M6 and M3 or M7 and M2) properly biased. Infact, for a given bias current you need to guarantee a DVgs+Vth each for strong inversion condition.

Note that M1 and M4 act as a source follower here (almost ideal) such that, from a small signal standpoint, you have same vi+/- at the input of the core devices.

Then each device from the core amp modulates a current equal to gm*vi+/- ... now you can do your math.

nathan

 

[rajath]

Yeah. The function of M1/M4 is to achieve a high input impedance, and that of M5/M8 is to shift the Dc level of the input signal so that M6/M7 are properly biased. What you mentioned regarding the voltage drop from point A to point C is right, but that is the DC voltage drop. The small signal voltage at both points is the same as the source follows the small signal variation of the gate and vice-versa.