Multi-feedback Opamp Stability - Page 1
Forum section: electronics / beginners

author: veryevil
author_rank: Unknown
timestamp: November 29, 2017, 11:39:11 am
content: Hi, I'm trying to simulate this MFB opamp and look for the gain and phase margins to check for stability.How would I set it up in LTSpice?Thanks

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: November 29, 2017, 04:27:41 pm
content: Gain margin and Phase margin can be found from the bode plots generated by ltspice and by using the cursors.Phase margins can be found when the gain is at 0dB and Gain margins can be found when the phase is at -180 degrees.The phase margins I found by inspection were at 2.61Hz with a phase of -91.9 degrees, and 24.37kHz with a phase of -347.1 degrees.Therefore, the phase margins are Pm = 180 + P = 88.1 degrees and -167.1 degrees.The gain margin is 504.63Hz when the phase is at -180 degrees which correspond to a gain of 29.7dB.Therefore, the gain margin is Gm = 0 - G dB = -29.7dB.Thus it can be concluded that the system is not stable as the gain and phase margin are both negative (Gm is greater than 1).I would highly suggest performing a mathematical analysis of the circuit to produce a transfer function and thus other methods can be used such as root locus, Nyquist plots or routh-hurwitz tables.Anyone please correct me if my conclusion is wrong. I'd normally find stability through analytical methods.EDIT: My conclusion was wrong, oops

author: veryevil
author_rank: Unknown
timestamp: November 30, 2017, 08:01:12 am
content: Hi, Thanks for the analysis. I got the circuit and values fromhttp://sim.okawa-denshi.jp/en/OPtazyuLowkeisan.htmand got the results below:

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: November 30, 2017, 09:43:27 am
content: Awesome, if you take the transfer function in the Laplace domain, Matlab has a bode plotting utility that can be used to find the margins.Matlab confirms that the system is unstable.Note: If a closed loop system is unstable, then the open loop system is guaranteed to be unstable. However, an unstable open loop system may have a stable closed loop response.It is also of note that the topology is known as sallen-key:https://en.wikipedia.org/wiki/Sallen%E2%80%93Key_topologySome online web tools will just produce a filter to user specifications without regards for stability, which seems to be the case here. How would you make the filter stable? Well the easiest option would be to try a different combination of component values (perhaps try a lower gain). Failing that, or if it is a more complicated system or component values are unable to be changed, you can insert a compensation network to correct for the phase change.

author: danadak
author_rank: Super Contributor
timestamp: November 30, 2017, 12:15:00 pm
content: For the math experts here.Bode plot easy way to eval stability, but if I do a LaPlace analysisof Z at any node (does not matter what node I pick, excluding sourceand gnd nodes) in circuit and find a negative real impedance result is thata necessary and sufficient condition to predict instability ?Regards, Dana.

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: November 30, 2017, 12:38:38 pm
content: Quote from: danadak on November 30, 2017, 12:15:00 pmI'm gonna have a little crack at it through intuition and say that it would depend on the node and the direction of the impedance. Wouldn't it matter where you're "looking" into? (i.e. feedback path).If your impedance shifts by 180 degrees, that would equate to a negative magnitude. In a negative feedback configuration, the resultant would be a positive feedback.As for it being a sufficient condition, It hasn't been taught traditionally.I can give a mathematical approach a shot tomorrow when I have sufficient sleep.

author: Alex Nikitin
author_rank: Super Contributor
timestamp: November 30, 2017, 12:58:23 pm
content: The circuit from the first post is actually stable with a good step response. So, where is the error in your calculations?CheersAlex

author: veryevil
author_rank: Unknown
timestamp: November 30, 2017, 01:10:18 pm
content: Quote from: Alex Nikitin on November 30, 2017, 12:58:23 pmDon't tease me! How do you get to that result?

author: Alex Nikitin
author_rank: Super Contributor
timestamp: November 30, 2017, 01:30:07 pm
content: Quote from: veryevil on November 30, 2017, 01:10:18 pmThere are many ways. To build the circuit. Or to run a transient response analysis on it in LT Spice. Or to use the online tool you've linked earlier. Or to read the Bode plot correctly* .CheersAlex*- it may be useful to remember that the overall configuration is inverting

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: November 30, 2017, 01:34:19 pm
content: Quote from: Alex Nikitin on November 30, 2017, 01:30:07 pmIm quite curious, what would the correct way to read the bode plot be? (finding gain and phase margins)Would you not be able to create a bode plot from the transfer function and then find the margins from that? (granted that it is magnitude)EDIT: After inspecting the transfer function through the use of the final value theorem, the transfer function converger to a finite value. So I assume that is why stability can be assured.

author: Alex Nikitin
author_rank: Super Contributor
timestamp: November 30, 2017, 01:54:08 pm
content: Quote from: Nitrousoxide on November 30, 2017, 01:34:19 pmPlease read the footnote I've added to my previous post while you were replying to it.CheersAlex

author: Nitrousoxide
author_rank: Regular Contributor
timestamp: November 30, 2017, 02:08:34 pm
content: Quote from: Alex Nikitin on November 30, 2017, 01:54:08 pmThe negative inverting configuration would only change the phase, not magnitude? Is that not accounted for in the transfer function that is given by the website? If I multiply the TF by -1, the gain margin tools state that the system is stable (am I meant to multiply by -1, I thought that was taken care of?).Regardless, another way to determine stability is if no poles lie on the right-hand side of the s-plane, which they don't (and hence it is stable).I must have misinterpreted the Gm and Pm results. Honestly I prefer to use pole plots and other analytical methods

author: MrAl
author_rank: Super Contributor
timestamp: November 30, 2017, 05:57:11 pm
content: Hi,You can do an analysis with an ideal op amp to gain some insight.It will take a little work but that's life.The response will only be 2nd order without the input cap and the cap for the non inverting input which are not needed anyway for a decent analysis. 2nd order systems are not difficult to work with mathematically.

author: danadak
author_rank: Super Contributor
timestamp: November 30, 2017, 09:38:05 pm
content: Maybe this is a starting point -http://rfic.eecs.berkeley.edu/~niknejad/ee242/pdf/ee242_lect6_twoports.pdfRegards, Dana.

author: MrAl
author_rank: Super Contributor
timestamp: December 01, 2017, 12:58:23 am
content: Here is a time domain analysis for the part of the circuit just after the first capacitor.It looks stable with a step input.

author: veryevil
author_rank: Unknown
timestamp: December 01, 2017, 07:49:05 am
content: I'm going to build this and the second stage I didn't mention today or monday and will see how well it works.

author: MrAl
author_rank: Super Contributor
timestamp: December 01, 2017, 06:03:31 pm
content: Hello again,Here is the step response with the first cap also kept in the circuit. The second plot is that plot and it shoots down first then gradually rises up again. Note that this is the ideal response, and if the supply voltage does not allow -30v then it will clip at some negative voltage (like -10v) and then gradually rise up slowly.In any case it looks stable and it looks mostly like just a slightly modified filter.