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The Designer's Guide Community Forum
https://designers-guide.org/forum/YaBB.pl Design >> Mixed-Signal Design >> Opamp Based Bandgap Stability issue https://designers-guide.org/forum/YaBB.pl?num=1259152108 Message started by Leila on Nov 25th, 2009, 4:28am |
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Title: Opamp Based Bandgap Stability issue Post by Leila on Nov 25th, 2009, 4:28am Can somebody explain this to me :-/ Opamp based bandgaps usually have both positive and negative feedback loops. Care must be taken to make sure the negative feedback loop has higher gain than the positive feedback loop. Stability analysis is performed by breaking the loop at a point that is common to both positive and negative feedback loops. I have an stability issue with the bandgap and try to run an stability test but Can not figure it out how? I appreciate your kind advices. |
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Title: Re: Opamp Based Bandgap Stability issue Post by raja.cedt on Nov 25th, 2009, 5:53am hi, what stability issue you has faced? by the way where did you opened? i guess you have been opened at the op amp output? Thanks, Rajasekhar. |
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Title: Re: Opamp Based Bandgap Stability issue Post by loose-electron on Nov 26th, 2009, 5:01pm Two control loops? I dont think so. In terms of a classic control system, the input of the difference amplifier is the voltages on the (two different size) PN junctions. One control loop, one dominant pole in the system. |
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Title: Re: Opamp Based Bandgap Stability issue Post by ywguo on Nov 27th, 2009, 6:44am Hi Jerry, I think there are two loops in a conventional bandgap reference. A few years ago I wondered which node should be tied to the inverted input of opamp. At last I decide to tie the non-inverted input to n1 because it is AC ground, it is emitter of a transistor. The voltage at node n1 is set by the voltage across a PN junction. n1 is desired to be AC ground compared with n2. Now I have an idea that the conventional bandgap reference consists of two loops. One is marked with red line, the other is marked with blue line. The first loop is negative feedback, and the second loop is postive feedback. Obviously the first loop dominates because n1 is a low impedance point, n1 is tied to the emitter of Q1. Is it correct? Thanks Yawei |
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Title: Re: Opamp Based Bandgap Stability issue Post by raja.cedt on Nov 27th, 2009, 11:12pm hi Yawei, ya it's correct, but you cann't say that n1 is sinal ground because i feel we have to say that it should be represented by it's small signal equivalent (dynamic resistance)..but you are correct compared to n1 n2 should be high impedance node. @loose-electron: i did't understand why you are saying that it's single loop..because wht i feel you are feedback to both terminals of the opamp. Thanks, Rajasekhar. |
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Title: Re: Opamp Based Bandgap Stability issue Post by thechopper on Nov 28th, 2009, 9:39am Hi, I also agree with Yawei...generally speaking in multiloops ckts where there is one positive feedback loop one must make sure the -ve loops have higher gain than the positive ones to ensure stability. Having both loops one node in common at the opamp output, the circuit can be broken at such point and thus analyzed as a single loop one (may be this is what Jerry wanted to convey). @Raja: I do not think Yawei meant that n1 is and AC ground, but just a lower impedance node compared to n2, low enough in order to justify connecting it to the positive input to the opamp. Obviously there is some dynamic impedance associated to it but certainly lower than the one seen at n2. Best Tosei |
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Title: Re: Opamp Based Bandgap Stability issue Post by mowiehowie on Nov 11th, 2010, 9:39am when you guys say "breaking the loop" you mean breaking by inserting an "AC CUT" ? cant i put an AC cut in both positive and negative feedbacks, and then insert an AC source at the negative feedback input and measure the frequency response will be M2's drain minus M1's drain. |
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Title: Re: Opamp Based Bandgap Stability issue Post by Mayank on Nov 11th, 2010, 11:46pm Quote:
Quote:
Placing a separate ac cut in the other loop(one in which ac test src has nt been inserted) will not allow that loop's effect to show up in the plot because you broke the second loop, but you are not feeding the test voltage into that loop at exactly the same pt where you are placing it in the first loop. This scheme can provide you the LGs of both loops separately, but combined effect cannot be analysed. Breaking the loop at the common pt. & inserting a test ac src there will give you common injection. Thus, you can test the return voltage at opamp output & be worried about the overall stability of this dual-feedback architecture instead of probing both loops separately. --M |
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Title: Re: Opamp Based Bandgap Stability issue Post by Ken Kundert on Nov 12th, 2010, 10:00am In this situation, rather than consider two single-ended loops, it is better to decompose the signals into their differential and common-mode components. First determine whether the loop is stable for purely differential perturbations, then determine if it is stable with common-mode perturbations (generally an easier criteria to meet because the common-mode gain is so much lower than the differential mode gain). Oh, and you aren't really breaking the loop, right? Breaking the loop is a conceptual concept that only really works for purely ideal components (zero output impedance, infinite input impedance, etc.). You cannot really do it in practice and expect to get the right answer. -Ken |
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Title: Re: Opamp Based Bandgap Stability issue Post by yvkrishna on Nov 23rd, 2010, 1:08am Quote:
You could do that with adding a differential voltage AC input at both the inputs of opamp and analysing the differential output for stability. Regards, yvkrishna |
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Title: Re: Opamp Based Bandgap Stability issue Post by mowiehowie on Nov 30th, 2010, 11:47am thank you all for the answers. i could make my bg circuit stable utilising the analysis i mentioned above, by adjusting the phase and gain margins of the [positive loop gain - minus negative loop gain]. to be sure about stability a good way is to start-up the circuit with several vdd ramps over PVT variations. maybe this can be helpful to someone. |
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Title: Re: Opamp Based Bandgap Stability issue Post by Dan Clement on Jun 12th, 2011, 9:25am Besides just AC analysis, I also like to force some transient disturbances and evaluate the response visually. |
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Title: Re: Opamp Based Bandgap Stability issue Post by RobG on Jun 13th, 2011, 8:05am Dan Clement wrote on Jun 12th, 2011, 9:25am:
You could also farm out that work to the guy up north ;) |
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Title: Re: Opamp Based Bandgap Stability issue Post by loose-electron on Jun 14th, 2011, 1:43pm Some semantics going on here - If you look at the analysis - you get a drive component associated with the common mode FB and then the differential FB. The net result is presented to a singular differencing amplifier. Two paths back? Yes sort of. As for "breaking the loop" - again - an approximation can be done by opeing the loop in a simulation, but there are much better methods that have been pointed out by Ken. |
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Title: Re: Opamp Based Bandgap Stability issue Post by vp1953 on Jun 16th, 2011, 2:19pm In addition to the three loops to consider for stability, the overall negative feedback should be greater than the positive feedback at all frequencies for the circuit to work. |
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Title: Re: Opamp Based Bandgap Stability issue Post by surreyian on Jul 4th, 2012, 9:26am raja.cedt wrote on Nov 27th, 2009, 11:12pm:
hello i dont quite understand why n2 is high impedance node. Can someone help me understand how we decide which input of the op-amp to connect to the bandgap circuit. is it because the impedance seen is (R1+R2)/(R1+R2+R3), compared to the impedance seen at n1 R1/R2. Assuming all have the same impedance. So high impedance = high gain, is that why it is tied to the inverting input. |
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Title: Re: Opamp Based Bandgap Stability issue Post by tzg6sa on Jul 9th, 2012, 3:26am Think at the gains from the amps output! G1=1/(1+gmd*R1) G2=(R3+1/[m*gmd])/(R1/m+R3+1/[m*gmd])=(1+m*gmd*R3)/(1+gmd*R1+m*gmd*R3) where gmd: gm of the diode; m: the ratio of the currents in the two legs. G1<G2 always, since 1/n=1/n*(n+x)/(n+x)=1/(n+x)+(x/n)/(n+x)<(1+x)/(n+x) Therefore G2 has to be connected in a way to form the negative gain loop: inverting terminal of the amp. Simple, right? |
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Title: Re: Opamp Based Bandgap Stability issue Post by Kevin Aylward on Jul 13th, 2013, 9:23am An AC analyses of a band gap is misleading and does not actually explain what is happening. There is a DC voltage/current that is technically always a “negative” loop in the sense that if too high it makes an output less, and too low it makes an output higher. However, it is in a notional positive feedback loop. To correctly understand this, the delta Vbe circuit must be correctly understood, as shown here. http://www.kevinaylward.co.uk/ee/ptat/PTAT.xht . It’s absolutely fundamental to understanding how a really bandgap works. |
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Title: Re: Opamp Based Bandgap Stability issue Post by Shajo on Feb 16th, 2016, 4:25am Hi all, I'm having a tough time deriving equation 3.17 from the circuit in the attachment. Please do point me in the right direction. |
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Title: Re: Opamp Based Bandgap Stability issue Post by blue111 on Dec 12th, 2016, 12:51am hi, Shajo, may I know from which book the screenshot is taken from ? Thanks! |
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