Hello, this problem is from (Razavi's Design of Analog CMOS Integrated Circuit, p.79, problem 3.31)
Before asking, I'd like to tell you i've tried and couldn't find answer yet.
1) tried 2 solution : KCL, Superposition
2) find solution on internet : it include only problem ~3.28
addition : using 1), I wanted to simplify answer, so I set some conditions. yes,it didn't work.
both of mine included 3 numerator, but answer has only 2.
Its ok not solving problem, just want to know how to solve it. - already spent almost 40m over...
"all DC sources are replaced by it internal impedances. It means voltage sources are shorted to ground, while current sources are open"
In your case VB has to be shorted to ground and VDD is the only AC sources used in calculations. This mean gm1 is removed, while gm2 is driven by VDD.
Do you see the answer now?
What do you mean?
Small signal circuit is shown textbook previously, this problem just has changed +Vin location.
I used it, so i guess my circuit is correct
"all DC sources are replaced by it internal impedances. It means voltage sources are shorted to ground, while current sources are open"
In your case VB has to be shorted to ground and VDD is the only AC sources used in calculations. This mean gm1 is removed, while gm2 is driven by VDD.
Do you see the answer now?
"all DC sources are replaced by it internal impedances. It means voltage sources are shorted to ground, while current sources are open"
In your case VB has to be shorted to ground and VDD is the only AC sources used in calculations. This mean gm1 is removed, while gm2 is driven by VDD.
Do you see the answer now?