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Effect of input CM voltage on Differential Pair

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Newbie level 2
Sep 29, 2005
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I have a question regarding the input common mode voltage on the gain of a simple differential amplifier with a tail current source.

From what I have read so far, the differential Gm of the circuit is a function of (W/L) of the input transistor and the tail current. This equation implies that if I am to increase the common mode input level, the differential gain would stay the same provided that I keep everything in saturation. Is this correct?

If I increase the common mode input level, Vgs will increase, correct? Does that not increase the gm of the input transistor? If the gm of the input transistor pair increases.... doesn't that increase the gain of the circuit? Why does increasing input common mode voltage not have the same effect as increasing (W/L) of the input pair?

Consider the simple condition, that Av=gm*rds, gm=2Id/Vov, ro=1/(lambda*Id), so Av=2/(lambda*Vov). From this, we can see that the gain is determined not by Id, but Vov. Vov changed with W/L.

First, Gm of the circuit is a function of (W/L) of the input transistor and the tail current. Correct, provided the transistor bias conditions are constant. i.e., your overdrive value is constant. you can refer the graph in razavi text book.
Second, is also correct. Because if you increase the common mode input level, there is no need to change in differential gain. differential gain is the gain we get when a small excitation given at input. common mode voltage is the common voltage at two inputs. both are different and independent.
third, your point is correct. but when u increase Vgs, Vds also changes and it compensates the increase in Vgs to mainitain the constant current in the current equation. But increse in (W/L) is independent of any other parameter, and so gm increases.
hope i cleared ur doubt.

Thank you for the replies.

I think I understand what is going on now. I was thinking that increasing the common mode at the input will directly increase Vgs... which is not true because the input pairs act like source followers and Vgs will stay relatively constant.

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