Hi Raj,
Quote:1. Since gm1=2I/(2xVod) and I = Vod/Rs, gm1 = 1/Rs. gm2 = 2I/Vod = 2/Rs. So loop gain is 2/Rs*Rs = 2.. I didn't understand the bold section, how did you calculate. I guess here your mixing large signal and small signal, I normally don't use Vov in the loop gain however I could be wrong. Many people have been agreed for 0(no 0dB) loop gain.
So you agree that gm1 = I/V
od2 and gm2 = 2I/V
od2? Then for the loop gain we break the loop at the gate of M2 and go around the loop. It looks like thats what you did, but your impedance looking out from drain of M4 is (1/gm-Rs). I don't see how you arrived at this result. If I just look at M1 and Rs, when I apply a test voltage at gate of M1, the current = gmVtest, so impedance = 1/gm, not (1/gm-Rs), unless you are making some assumption about the output impedance of M1.
One mistake I see that I made though, since I = Vod/Rs, and I = 4K(Vod)
2, then Vod = 1/4KRs. Therefore I = 1/4KRs
2. So the current is indeed fixed.
EDIT:
OK I see my mistake, I forgot to take the voltage drop across Rs (same mistake ywguo made). So yes it becomes 2/Rs*(1/gm-Rs) = 0.
Quote:I guess week +ve feedback has, because if you check another traditional Gm=1/R circuit which is very robust from PVT point of view (forget about 2nd order effects) with less than 1 +ve fb loopgain.
That's a really good question. So for a circuit with weak +ve feedback, what causes it to reach a stable operating point? Well, if you look at a VCO, if the +ve feedback is less than 1, then the output will not build up. So your original statement isn't always true. Perhaps it is a large signal effect (try calculating the large signal loop gain?). Or perhaps it is simply the result of the model parameters. For example suppose I have a series current source feeding a series resistor and diode, and the diode is ON.
---(->)----[diode]----[res]---GND
I
So the voltage will be I.R and the diode will have little effect. Even if the diode changes due to PVT, the voltage will still be I.R. That doesn't mean that there is some feedback mechanism working. Just that the circuit is insensitive to the diode.
I dunno, perhaps I am on the wrong track. what do you think?
regards,
Aaron
EDIT:
So I simulated this using ideal MOS devices and the loop gain is definitely 0...