Differential stage

I've doubts about this circuit. The 2 transistors are matched in saturation, and the book I follow neglects channel modulation. The author considers as input signals VGS1 = VGS0 + Vin/2 and VGS2 = VGS0 - Vin/2 where VGS0 is the common mode and Vin is the differential signal. He considers as output of the circuit the difference between the two currents, that is I1 - I2

Now, the author writes:


If I make some computation, I get a different expression of I1+I2, which contains this quantity: [(Vin)^2]/2

If I neglect this term, I get the same expression of the book, so the first question is: Why does the author neglect this term? Then the author makes a simplified small signal circuit, which is the following:

small signal circuit

He applies a common mode signal and he gets that the common mode current is:

iCM = (gm x vCM)/(1 + 2 X gm x r1)

Then he says that iCM is approximately equal to vin/(2 x r1) That's the second question: why does he do this approximation? You can find the book reference here: enter image description here enter image description here enter image description here enter image description here

Thank you! Stefano

  • \$\begingroup\$ Differential amplifiers are usually used in cases where Vin is very small (for example, in an op amp, it's ideally zero--the virtual ground). The square of a very small quantity is very very small, negligible in comparison to the rest of the system. I don't know if that's the full reason the author left it out (I only skimmed the text, I admit), but it seems likely. This is a technique used very often in engineering. \$\endgroup\$
    – Hearth
    May 8, 2018 at 17:54
  • \$\begingroup\$ Thank you for your answer. Yes, I suspected that this was the reason, but I was not sure. Thus, do we study small signal circuits of the different op-amp stages because the negative feedback (which is usually used in op-amps) sets the input differential voltage approximately to zero? Thank you \$\endgroup\$
    – Stefanino
    May 8, 2018 at 18:39

1 Answer 1


It's probably ok to assume that \$2*g_m*R_1 \gg 1\$, right? Try using that to simplify the expression for common-mode current.

  • \$\begingroup\$ Hi, thanks for your answer. Yes, the approximation you refer to is ok, I don't understand why vin = vcm , where vin is the differential input and vcm is the common mode voltage. Thank you \$\endgroup\$
    – Stefanino
    May 8, 2018 at 18:35

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