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a busy cat

In determining the polarity of such a circuit I used the following reasoning:

if Vin goes up ID1 goes up and ID2 goes down, ID4 goes down and VG3,VG4 go up because of PMOS, then Vout goes down, if Vout goes down then Vin will go up. So I ended up with positive feedback. Am I making a mistake? Sometimes when I try to understand a current source's relationship with its nodes polarities I get confused. Could you explain this point too?

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  • \$\begingroup\$ Why do you say that if vout goes down, vin will go up? The feedback I see is between M2 and M3 and it is negative... \$\endgroup\$ – Vladimir Cravero Nov 11 '17 at 22:50
  • \$\begingroup\$ If vout goes down, Id2 goes down since it's NMOS, and then Id1 goes up , then Vin will go up again since it's NMOS \$\endgroup\$ – user8925869 Nov 11 '17 at 23:09
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You are correct, the feedback is positive. To check whether a feedback is negative or positive just imagine breaking the loop at a certain point and applying some disturbance on the node. If the returned disturbance is of the same polarity as the applied one then it is positive otherwise it is a negative feedback.
So for the circuit you showed if, for instance, we break the loop at \$V_{out}\$ and imagine the gate potential of M2 going up by a small amount. This would increase its gate source potential and \$I_{dM2}\$ increases. Since more current now flows into ground, the drain potential of M2 will go down. This would reduce the gate potential of M3 causing \$I_{dM3}\$ to go up. Since there is more current through the resistor \$V_{out}\$ goes up (same way as the applied disturbance). So we see that the feedback tries to amplify any disturbance in the circuit so it is positive feedback.
An easier way to do the same thing is to notice that while traversing the loop we move from gate of M2 to its drain, kind of like a common source, which is an inverting stage. From drain of M2 we move to drain of M3 which is again an inverting stage. Since we go through even number of inversions, the output disturbance will be in phase with the applied disturbance and feedback will be positive.

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