plese give me a hand of two stage opamp?

[carlson]

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 ?

 

[wholx]

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.

 

[carlson]

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

 

[avinash]

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.

 

[okawa]

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.