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.

enter image description here

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:

enter image description here

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?

  • \$\begingroup\$ You're forgetting that he's computing the differential gain, thus, node P is actually grounded for this small-signal analysis. \$\endgroup\$
    – Designalog
    Commented Dec 29, 2022 at 18:47
  • \$\begingroup\$ Differential gain for an asymmetric circuit though, so you can't apply the half-circuit concept directly and assume node P is grounded. However, I think when he shorts Iout to GND, then perhaps you can, since it is symmetric then. \$\endgroup\$ Commented Dec 29, 2022 at 18:49
  • \$\begingroup\$ Even if you do assume node P is AC ground, then Iout would be -gm2Vin/2, his answer is +gm1Vin/2. \$\endgroup\$ Commented Dec 29, 2022 at 18:50

1 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).

  • \$\begingroup\$ Excellent answer. One more thing, would it be correct to assume node P to be AC ground? \$\endgroup\$ Commented Dec 29, 2022 at 20:42
  • \$\begingroup\$ No, not necessarily. It depends on the input signal whether there's a signal on P or not. It's only equal to AC ground when the differential input signal is symmetric around ground. If the input signal is not symmetrical, then there will be a signal on P. \$\endgroup\$ Commented Dec 29, 2022 at 21:04
  • \$\begingroup\$ I believe that is the case here, differential input signals are being applied. \$\endgroup\$ Commented Dec 30, 2022 at 15:00
  • \$\begingroup\$ It's not specified whether the differential signal is symmetric or not. The author assumed it, though. Nevertheless, it's possible to solve this without such an assumption, so it's best to just do that. \$\endgroup\$ Commented Dec 30, 2022 at 20:28

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