Gₘ of differential stage CMOS

Question

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 A_v = -G_m·R_out where G_m is the short circuit transconductance with the output shorted to GND. He derives that I_out to that shorted AC ground point is g_m1·V_in/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?

Answer 1

The last circuit you derived (with the two series-connected current sources) is correct. Just ignore the fact that it looks weird and do the math. No simplifications needed, and you don't have to assume that node P is ground, either (which isn't necessarily true).

Using Kirchhoff's Voltage Law on the original circuit, we obtain:

Eq. 1: Vin = Vgs1 - Vgs2

The current through the two series-connected VCCS is the same, so we obtain:

Eq. 2: gm1 Vgs1 = - gm2 Vgs2

Since the MOSFETs are identical (equal gm), we can simplify this to:

Eq. 3: Vgs1 = -Vgs2

Plugging this back into Equation 1, we get:

Eq. 4: Vin = -2 Vgs2

In other words:

Eq. 5: Vgs2 = -Vin/2

So, as a result, the drain current of M2 is:

Eq. 6: Iout = -gm2 Vgs2 = -gm2 Vin/2 = -gm1 Vin/2

The reason why the negative sign is missing in the book is either because the author forgot it, or because they only calculated the absolute value of Av, ignoring the sign (as can be seen at the start of the paragraph you marked yellow).