how to work "single ended differential pair"

[020170]

**broken link removed**

In my knowledge, single ended differential pair may work as follow.

first, input of the M1 result in output wave at M1's source ,
that is, M1 is CD amp.

second, M1's output wave come into M2's source, that is, CG amp.

third, in my knowledge, M2 is served as a two-type amp, that is, CG amp and CS amp.

at last, Vout wave is completly differential, between Vgs1 and Vgs2.

my thought look like simple and intuitive, but there are some weird problem.

1. M1 and M2's source terminal is completly AC ground, virtual ground
but how does M1's input come into M2's source?

2. If I connected Capacitance LOAD at Vout, M1 and M2's source voltage is following M2's input. but why??

thanks

 

[angelote]

This circuit is an OTA. You have to think in currents.

id3=id4 becauso M4 has the same plarization than M3.

id1=id3, so id4=id1

iout=id4-id2

id1≈g·vi+
id2≈g·vi-

vi = (vi+) - (vi-)

Then, iout=g·vi

vout depends on load[/quote]