plese give me a hand of two stage opamp?

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carlson

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I am doing the problems of razavi's book. In the problem 9.18,it says that Vgs3=Vgs5 in order that M5 carries the expected curren when vin=0. why it need to Vgs3=Vgs5 ? what current is "the expected current"? whether it must set Vgs3=Vgs5 in other similar two stage opamps ?
 

hi carlson,
in case Vin = 0, we can find Vx = Vy, thus Vgs3 = Vgs5.
the current expected flowing through M5 is 1mA in this question, which determines the W/L of M5.
 

but I still don't know why 1mA is the expected current and what benefit it have?
 

dear carlson
Vgs3 is made equal to Vgs5 so as to reduce the systematic offset.as already told by wholx that if M3 and M4 are matched and they constitute a current mirror.at Vin=0 I3=I4 Vgs3=Vgs4 hence Vx=Vy indicates Vgs3=Vgs5.
 

sure.
the current density in M5 the same as those in M3 and M4 will reduce the systematic offset, and the same L will reduce the random offset.
 

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