Raja,

  Why would you think that Gm=1/R? Shouldn't the open
loop gain be something like, gm_M4 * (go_M1 || go_M4)?
The output resistance of M1 and M4 is gds of a transistor
in saturation, not Gm. The output resistance of M1 should
be much larger than Rs so it can be ignored for small
signal analysis. Of course, you can't ignore it for large
signal analysis. So to me at first glance it seems like the
loop gain should be similar to the loop gain for a one pole
op-amp without a cascode. Certainly the open loop gain will
be low, but it should not be less than one, unless the power
supply voltage is so low the circuit does not setup properly.

                                                            Sheldon

@ywguo: Here feedback taken from drain of M1 means there is drop trough Rs, I guess you miss this point.
@Shedlon: I have even included gds of M1 still same numerator, except minor difference in denominator.

Thanks,
Raj.

Sir,
Thanks for your reply. if posb could you please explain about large signal transfer curve, I didn't understand the parabola concept.. One more thing the other circuit has +ve feedback with loop gain less than 1 so it is stable. My basic question is can +ve fb with loop gain less than 1 has de-sensitive properties like -ve fb??

Thanks,
raj.

Hi,

Assume the current through the two branches is the same, but the RHS NMOS is 4x larger. Using simple device equations, since Ids ∝ (Vod)², Vod of RHS NMOS should be half of LHS NMOS for the same current. Assume 1xVod refers to RHS NMOS. So LHS NMOS has 2xVod, and voltage drop across Rs is 1xVod. Since gm1=2I/(2xVod) and I = Vod/Rs, gm1 = 1/Rs. gm2 = 2I/Vod = 2/Rs. So loop gain is 2/Rs*Rs = 2.

BTW, how do you get a loop gain of 0? Is that 0dB = 1?

I don't think weak positive or negative feedback has any desensitization property. So I assume your question is, why does the circuit stabilize to a fixed operating point? You are assuming there must be some feedback mechanism?

Personally, I wonder how well this circuit actually works, since I haven't seen it before. If those current mirrors are ideal, then I don't think this circuit has only 1 operating point. Seems it only requires that Vod,m1 = 2Vod,m2. So as long as the transistors follow square law behavior and the  voltages never saturate, the circuit looks like it could take any current. The feedback mechanism looks like it could be a large signal one, something like in a VCO, where the saturating voltage swing causes the loop gain to drop. In this case, the drain voltage can't rise indefinitely...

Did I make a mistake somewhere? I assumed very simplistic equations...


regards,
Aaron

Hi Raj,


Quote:


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 = 1/4KRs². 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:


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...

Hi Raj.,


I dunno. I'm sure some people would argue that you're stretching the definition of feedback here. As you say, any time the transfer function looks like x/(y+z) you can divide by y and it will become x/y/(1+z). Then you can model it as a negative feedback system. But that doesn't guarantee that any signal is fed back from the output to the input...

Its interesting to think about anyway.

For the diode and resistor, it depends on the size of the resistor and what you mean by sensitive. Insensitive IMO means it is not much affected, but doesn't mean that it is completely not affected. I'm simply pointing out that there are scenarios where a circuit has a fixed operating point and is insensitive to PVT, but there is no -ve feedback.


regards,
Aaron

Sir,
Thanks for your reply. I agree here there is desensitisation just because no loop gain. In case of increase in temp or PVT change current automatically change to adjust gm, I feel this is restoring mechanism or desensitisation. Could you please tell me where I am wrong here..

Thanks,
Raj.