I am examining op amps and I found this pdf very helpful: http://www.aicdesign.org/SCNOTES/2010notes/Lect2UP230_(100327).pdf

In this tutorial, the op amp circuit topology is given as this: op amp topology

and gains of stages of the op amp are given like this:

gain equations

First time I see an equation like this. I thought the (DC)gain of the first stage is: \$gm_{2} * r_{ds2} // r_{ds4}\$

it seems really wrong to me. Can anyone explain why the gain is represented that way in this tutorial?

  • \$\begingroup\$ can you specify the page no. of this tutorial in above pdf \$\endgroup\$
    – Virange
    Nov 30 '15 at 9:31
  • \$\begingroup\$ on page number 4. \$\endgroup\$
    – Alper91
    Nov 30 '15 at 9:33
  • \$\begingroup\$ @Alperözel hope my explanation helps \$\endgroup\$ Nov 30 '15 at 15:26

gds = 1/rds => conductance add up when you have parallel resistances.

G = 1/R

R1 || R2 = 1 / (G1 +G2)

Thus, rds2 || rds4 = 1/ [(gds2 + gds4) ]

So, GAIN for the first stage;

Av = gm2 * (rds2 || rs4) = gm2 / (gds2 + gds4)

here since m1 and m2 are differential pairs ,its ok to assume that their trans conductances are almost same (gm1 =gm2).


Regarding the expression in terms of current ,

The drain resistance (rd) is due to channel length modulation and is given by :

rds = 1/ (lambda * Id) or gds = ( lambda * Id).

So replacing gds, we get;

Av= gm2 / [lambda*(Id2 + Id4) ]


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