# Gm of MOS differential stage

Can someone explain why current Iout in Fig 5.23b is simply gm1Vin/2? What about the current from M2?

I have annotated what I think in the figure.

Your annotations correctly show the small-signal current magnitude of $$\g_mv_{in}/2\$$ on each branch of the differential pair, but I think you have an opposite assumption for sign - the text appears to assume that the input to M2 swings up ($$\v_{in}/2\$$ over AC ground) while the input to M1 swings low by the same magnitude below AC ground.

Then, M4 is a constant current source (seeing as it is the output side of a current mirror biased by a constant current) meaning it contributes essentially zero small-signal current; by KCL this means that the output current must also be $$\g_mv_{in}/2\$$.

You may be confusing this structure with this one, which mirrors the current from one branch into the other:

Schematic and annotations are my own work reproduced from this answer of mine

which does have a system transimpedance of $$\g_m\$$ rather than $$\g_m/2\$$.

• But in small signal the Iss is open circuit. won’t the currents have to come from the Iout ground via the Rout of the FETs? Commented Sep 12, 2022 at 21:33
• @AlfroJang80 No, that same $g_mv_{in}/2$ that's coming up the right branch is exactly the same that's coming down the left branch. KCL is satisfied at the node where the tail current meets the two sources. Commented Sep 12, 2022 at 21:49
• @nanofard Transistor M1 wants to push gmvin/2 and transistor M2 wants to push -gmvin/2, they are acting independently. If r_out of the transistors are neglected, then it makes sense that these have to be equal since KCL has to be satisfied at node P. However if the output resistance of the transistors is included, there is a parallel path across the transistors, so it's not necessary gmvin/2 from one to be equal to -gmvin/2 of the other. Commented Sep 15, 2022 at 18:25
• @AlfroJang80 The output impedance is indeed neglected, and can be neglected in your given circuit: the drain of M2 is at small-signal ground (since it's shorted to measure the current), and the drain of M1 is also at small-signal ground (it's connected to an unchanging rail). Commented Sep 15, 2022 at 18:32
• Excellent point. Now your answer makes complete sense. Thank you Commented Sep 15, 2022 at 18:34