This is a follow-up question to Gm of MOS differential stage
The author of Design of Analog CMOS Integrated Circuits is trying to compute the gain of Fig 5.23(a) in the following extract.
Since the circuit is not symmetrical, he uses a method of Av = -Gm·Rout where Gm is the short circuit transconductance with the output shorted to GND. He derives that Iout to that shorted AC ground point is gm1·Vin/2.
I am having difficulty understanding that. If I replace the devices M1 and M2 with their VCCS small-signal equivalent assuming zero channel length modulation, I get the following:
There are two current sources in series with each other whose values depend on the common source node voltage. How can I even solve this? Can that common source node voltage be considered AC ground?