0
\$\begingroup\$

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:

Computation

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

\$\endgroup\$
  • \$\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 '18 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 '18 at 18:39
0
\$\begingroup\$

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.

\$\endgroup\$
  • \$\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 '18 at 18:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.