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https://designers-guide.org/forum/YaBB.pl Design >> Analog Design >> is this circuit have overal -ve feedback? https://designers-guide.org/forum/YaBB.pl?num=1313568156 Message started by raja.cedt on Aug 17th, 2011, 1:02am |
Title: is this circuit have overal -ve feedback? Post by raja.cedt on Aug 17th, 2011, 1:02am hello all, Can any one please tell me is this circuit have -ve feedback around the opamp? i feel gain from the M1,M2 stack will be more than M3,M4 stack, so in this case +ve feedback is more dominant than -ve FB. Thanks. |
Title: Re: is this circuit have overal -ve feedback? Post by Ken Kundert on Aug 17th, 2011, 3:35pm I am having a hard time understanding your question. Can you re-ask the question, this time being careful with your English. Also, please define what you mean by +ve feedback and -ve feedback. -Ken |
Title: Re: is this circuit have overal -ve feedback? Post by wave on Aug 17th, 2011, 5:31pm This particular circuit looks odd, but I've seen similar discussions on Bias circuits. (either a delta VBE bipolar loop, or delta Vgs MOS loop). Often a amplifier is used to control the PMOS current mirrors. (as drawn, I'm not sure the Amp is controlling the circuit much on the Nmos cascodes). Basically M3 is lower impedance than the M2 side. This will influence the gain, and the "polarity" of the op-amp hook up. It's not the most direct circuit to analyze, so I'm assuming/guessing that the question --- is it Positive or Neg. Feedback? Raj - make sense? 8-) ~Wave |
Title: Re: is this circuit have overal -ve feedback? Post by Alexandar on Aug 18th, 2011, 12:22am Well, what you could do: connect a voltage source to the cascode, and sweep it. You observe what happens with the voltage at the top node on the left and on the right. Having these graphs, you can draw the conclusion. |
Title: Re: is this circuit have overal -ve feedback? Post by raja.cedt on Aug 18th, 2011, 12:27am hello guys, ya you are correct for this kind of circuit it's hard to say which one is +ve and which one is -ve. But so as to make operating point stable we ned to ensure if any distrubance occur any where finally loop should restore previous operating point. In that sence i applied some signal at the o/p but it's keep on diverging rather converge. other wise...for get about all the discussion above and just tell me wether opamps polarity is correct or not. Thanks. |
Title: Re: is this circuit have overal -ve feedback? Post by buddypoor on Aug 18th, 2011, 12:49am raja.cedt wrote on Aug 18th, 2011, 12:27am:
Can be answered (with regard to a stable quiescent point) only if we know what will be connected to both opamp inputs - in addition to the shown parts (to the left and to the right, since you have drawn a line). |
Title: Re: is this circuit have overal -ve feedback? Post by raja.cedt on Aug 18th, 2011, 12:59am hello buddy poor, nothing is there, shown is the total circuit. |
Title: Re: is this circuit have overal -ve feedback? Post by buddypoor on Aug 18th, 2011, 6:41am raja.cedt wrote on Aug 18th, 2011, 12:59am:
In this case, can you explain why - in your view - "gain from the M1,M2 stack will be more than M3,M4 stack"? |
Title: Re: is this circuit have overal -ve feedback? Post by raja.cedt on Aug 18th, 2011, 6:53am hello buddypoor, please find the attached fig. So correct me if am wrong any where. Thanks. |
Title: Re: is this circuit have overal -ve feedback? Post by buddypoor on Aug 18th, 2011, 9:50am One comment: opening the loop at the opamp output means that both gates are still wired together. So you have no choice to inject the test signal into the gates of M1 or M4. One question: What is your reference node for the gain calculation? For the loop gain the feedback signal at the opamp inputs are relevant. |
Title: Re: is this circuit have overal -ve feedback? Post by nrk1 on Aug 19th, 2011, 7:16am raja.cedt, There is no (significant) feedback around the opamp. Break the loop at the opamp output, inject a test signal to the gates of M1,4, and see what returns at the opamp output. Assuming transistors with gds=0 and current sources with a finite Rout, the test signal at the gates of M1,4 creates a current of 2/gm*Vtest in M4,3 which is also mirrored to M2. M1 acts as a cascode. So the difference voltage across the opamp inputs is zero. A non-zero gds modifies this calculation only slightly. Bottomline, don't use this circuit. |
Title: Re: is this circuit have overal -ve feedback? Post by raja.cedt on Aug 24th, 2011, 11:00am hello nrk1, thanks for your reply. Ya due to test source you will get gm/2*vtest current in both legs, so due to this voltage at the drain of m4 will be roughly (2r0)*(gm/2), but drain of the m1 will like m/2(gm*ro)*ro. so you will have more swing at drain m1. is any thing wrong in my explanation. i saw this ckt in this paper "Transient Charge Feedforward Driver for High-Speed Current-Mode Data Driving in Active-Matrix OLED Displays". Thanks. |
Title: Re: is this circuit have overal -ve feedback? Post by Rakesh on Aug 24th, 2011, 11:31am Hi, Assume that M4, M1 gds to be zero and M2 and M3 has finite gds. Break the loop at gate of M4 and M1. the current produced in transistors will be (gm/2)Vtest+(gds/2)Vtest on the left side and (gm/2)Vtest + gm/(1+gm/gds)Vtest on the right side. Both sides transistors M1 and M4 acts like a cascode transistor with degeneracy 1/gm and 1/gds. Now (gm/2)Vtest is common to both sides. Since we have at the top ideal current sources the extra current should flow into the out put impedances at the drains of M4 and M1. The voltage developed at the +ve terminal of opamp will be (1/gds) (gm/2 +gds/2)Vtest. The voltage developed on the negative terminal will be (gm/2 + gm/(1+gm/gds))(gm/gds^2)Vtest The difference at the input of the opamp is approximately gm/2(1/gds -gm/gds^2)Vtest + (0.5-gm/gds)Vtest Assuming gm/gds >>>1 we have difference as -gm/2(gm/gds^2)Vtest-(gm/gds)Vtest if the gain of the opamp id A then at the output of opamp we have a loop gain whcih is negative and high due to negative feedback as -gm/2(gm/gds^2)A-(gm/gds)A So i think the signs on the opamp is correct Correct me if i am wrong Thanks |
Title: Re: is this circuit have overal -ve feedback? Post by nrk1 on Aug 24th, 2011, 9:27pm From your earlier post "... from M1,2 stack, the gain is ~ (gmro)^2 ..." is wrong. The gain from gate of M1 is ~ 0 If you are referring to Fig. 3 of the paper you mentioned, there is a diode connected transistor from the right branch to ground and a current source from the left. Clearly the gain to the opamp's + sign is more. (buddypoor had earlier asked what else is connected to the circuit). In that case there is no ambiguity. |
Title: Re: is this circuit have overal -ve feedback? Post by raja.cedt on Sep 1st, 2011, 3:36am hello nrk1, thanks for your note. ya i forgot this. But from erlier post gain from gate of m1 is zero but one thing is m2 gate also moves due to m4,m3, correct me if am wrong. Thanks. |
Title: Re: is this circuit have overal -ve feedback? Post by Ricky Chen on Sep 28th, 2011, 11:07pm raja.cedt wrote on Sep 1st, 2011, 3:36am:
Ya you may be right. 1st look left side arm :- due to del-V increment at the gate it will cause del-I1 current sink at the +ve Node of the OPAMP. So del-V1 reduction in + Node of the OPMAP. 2nd look right side arm:- due to del-v increment at the gate it will cause let say del-I2. But this will be far far smaller than del-I1 as it is degenerated by a current source M3. However due to the current mirror action of the left side ( you can apply supperposition for better understanding) del-I3 current will flow FROM the -ve Node. Now question is whether de-I3+del-I2 > or < del-I1. If > then it is a POSITIVE feed-back. If < then it is a NEGATIVE feedback. If the component generated due to mirroring action del-I3 is quite smaller than del-I1 then you will END up getting NEGATIVE feedback. Ricky- |
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