How can I obtain amplifier circuit loop gain? - Page 1
Forum section: electronics / projects

author: Lucky-Luka
author_rank: Unknown
timestamp: January 16, 2020, 08:06:24 pm
content: In the last few days I've tried to understand how to get the loop gain (Aol*ß) of my amplifier circuit. I've received several suggestion. I would like to hear your opinion on this matter. Are these 2 methods equivalent? Are they correct?The results look the same.Cheers

author: Benta
author_rank: Super Contributor
timestamp: January 16, 2020, 09:02:29 pm
content: "Loop gain" makes little sense. You can measure open-loop gain (x - x open) or closed-loop gain (x - x connected). Please clarify.

author: Lucky-Luka
author_rank: Unknown
timestamp: January 16, 2020, 09:09:37 pm
content: I have already said what it is...

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 16, 2020, 09:24:32 pm
content: I typically use the inductor method. As an alternative do the test in two steps. First close the loop and get the node voltages at the input, then open the loop and drive those voltages into the inputs...yielding...hopefully...the same operating point.But to your question, I would say your two methods are mostly equivalent for ac analysis.

author: magic
author_rank: Super Contributor
timestamp: January 16, 2020, 10:34:00 pm
content: You have one of the sources backwards and phase is 180° off.The gigafarad capacitor always seemed intuitively right to me. The latter is weird, and who would want to edit the plots to divide by V(x) every time one switches to AC analysis. I don't like it.Quote from: Benta on January 16, 2020, 09:02:29 pmThere is even a wikipedia article on thatIt makes sense. For example, unity loop gain frequency rather than unity gain frequency is the one you would care about in an amplifier permanently wired for high closed loop gain and decompensated correspondingly. Expect to see this jargon in anything related to audio amplifiers.

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 16, 2020, 11:26:55 pm
content: Quote from: magic on January 16, 2020, 10:34:00 pmIndeed...it does make sense.When designing the compensation for an amplifier, you plot loop gain vs phase to find phase margin. Most of the time the goal is unity-gain stability and for that case open-loop gain and loop gain are equivalent.

author: Cerebus
author_rank: Super Contributor
timestamp: January 17, 2020, 12:04:46 am
content: I see you're using LTSpice.Got to wherever your copy of LTSpice has stored its 'examples' directory and open "LoopGain.asc" - it contains an example of using "Middlebrook's method" in LTSpice. There is also a "LoopGain2.asc" in the same directory showing Tian et al's method of doing essentially the same thing using a slightly different method. I personally found the latter easier to use.

author: H713
author_rank: Regular Contributor
timestamp: January 17, 2020, 07:31:56 am
content: I know this is off-topic, but if you actually plan to build that you will need something considerably beefier than a 2N4401/2N4403 to drive those 2N3055s to any reasonable power level. The current gain of a 2N3055 at high currents is something like 5. Those little TO-92 package drivers aren't going to like that.

author: Lucky-Luka
author_rank: Unknown
timestamp: January 17, 2020, 10:42:04 am
content: what's the purpose of the big inductor?

author: magic
author_rank: Super Contributor
timestamp: January 17, 2020, 11:32:48 am
content: Try withoutThe whole L/C circuit passes output DC level back to the input transistor to establish normal operating conditions, but blocks AC, at least above some nano-hertz frequencies

author: Lucky-Luka
author_rank: Unknown
timestamp: January 17, 2020, 12:17:06 pm
content: I don't understand why you need to block AC

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 17, 2020, 02:54:05 pm
content: Quote from: Lucky-Luka on January 17, 2020, 12:17:06 pmBecause you want to OPEN the loop for frequencies of interest (everything above the nano Hz range, e.g., as magic said).

author: mikerj
author_rank: Super Contributor
timestamp: January 17, 2020, 03:09:05 pm
content: Quote from: Lucky-Luka on January 17, 2020, 12:17:06 pmThe whole idea is to remove the feedback at AC to measure open-loop gain/phase, but retain DC feedback to set the correct operating conditions. If you don't block AC then you are providing negative feedback and therefore won't be measuring open-loop gain.

author: Lucky-Luka
author_rank: Unknown
timestamp: January 17, 2020, 04:56:06 pm
content: 1)Before stating that my request is fundamentally wrong I think you should read my question more carefully Lesolee. Haven't you read the anwers I received?2)The video was to meant to point out what the loop gain is, since another user didn't know it.Thanks everybody for the answers.p.s. LOL. Lesolee's post misteriously disappeared after my reply.

author: bson
author_rank: Unknown
timestamp: January 19, 2020, 12:02:33 am
content: Vf/Vin, where Vf is the feedback voltage.

author: Cerebus
author_rank: Super Contributor
timestamp: January 19, 2020, 01:47:52 am
content: Quote from: bson on January 19, 2020, 12:02:33 amWhat question do you think that is an answer to?

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 19, 2020, 03:49:21 am
content: Quote from: Cerebus on January 19, 2020, 01:47:52 amYou mean.: "To what question do you think that is the answer?"Anyhooo, it is actually the answer to the title of the original post.

author: Lucky-Luka
author_rank: Unknown
timestamp: January 19, 2020, 09:01:36 am
content: Apart from the fact that I've already received and answer, how can I obtain Vf?I don't think that such an answer adds something to the thread.Maybe you can prove me wrong.

author: magic
author_rank: Super Contributor
timestamp: January 19, 2020, 09:54:50 am
content: You can put an "AC 1" source at the normal input of completely unmodified amplifier, run AC analysis and plotV(output)/(V(in+)-V(in-)), where in+,in- are of course the two inputs to the differential pair. Philosophically speaking, that's the open loop gain.It will give about the same plot, but I suspect that it may be slightly more optimistic at high frequencies and that's why people use other methods to asses stability of the circuit. Or maybe I'm wrong and it will be exactly the same

author: Lucky-Luka
author_rank: Unknown
timestamp: January 19, 2020, 10:03:11 am
content: It doesn't look good.

author: Cerebus
author_rank: Super Contributor
timestamp: January 19, 2020, 11:57:51 am
content: Quote from: Wimberleytech on January 19, 2020, 03:49:21 amA "grammar nazi" remark, on a forum where half the people can't spell and one third of the posts don't even even vaguely meet the rules of grammar? You're going to be busy...Moreover, I don't think it's the answer to anything. There's no Vfdefined anywhere here, and if we take the naïve assumption that Vfis the feedback voltage at the negative input then it's obviously wrong. It'd help if the poster clarified themselves.

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 19, 2020, 03:44:41 pm
content: Quote from: Cerebus on January 19, 2020, 11:57:51 amLighten up Cerebus...just having a little fun.Frankly, I thought you were a bit pejorative in your response, so I was just playing along.I am an old Texan...you should be better than I.Anyway, back to the technical stuff. Why do you say Vf/Vin does not represent the loop gain?

author: Cerebus
author_rank: Super Contributor
timestamp: January 19, 2020, 04:47:27 pm
content: Quote from: Wimberleytech on January 19, 2020, 03:44:41 pmWhat is Vf? No, seriously; it's just been introduced out of thin air.If Vfis, as a I said a naïve assumption would be, the voltage at the negative input, then Vf/Vincan't be the loop gain.Let's take the classic feedback equationVout= VinA / (1 + Aβ)Where Voutand Vinare obvious, A is the open loop gain and β is the feedback fraction.The loop gain (by definition) is Aβ, if Vfis as assumed above then Vf= βVout. So (substituting)Vf/ Vin= βVout/ Vin~= 1 and is not = Aβ.The bloody SMF editor stole the keystrokes that should have produced a β and instead decided that I wanted to save a 1/4 written draft message - bastard!

author: Lucky-Luka
author_rank: Unknown
timestamp: January 19, 2020, 05:02:30 pm
content: @CerebusThat's what I was thinking.That's why I think that post doesn't add value to the thread.I have asked a precise question and I have presented 2 solutions.I can accept other solutions but it would be nice to detail them a little bit...Cheersp.s. English is not my first language so be patient

author: Lesolee
author_rank: Regular Contributor
timestamp: January 19, 2020, 05:09:10 pm
content: The questions were:(1) Are they equivalent?Yes, from an theoretical point of view.(2) Are they correct?Yes, with the same caveat.In some sense the ground generator version is less "accurate" since it removes the load of the feedback network, but 22K compared to the 10R load is pretty negligible.In some sense the floating generator version is more complicated since the plot is a ratio rather than a direct reading, but again that is evidently not an issue.

author: Wimberleytech
author_rank: Super Contributor
timestamp: January 19, 2020, 05:10:33 pm
content: Quote from: Cerebus on January 19, 2020, 04:47:27 pmI agree with your analysis given your assumptions. But I interpreted it as the output of Aβ.This forum is due for a LaTex or some other formula entry capability (including greek characters), I think.

author: Cerebus
author_rank: Super Contributor
timestamp: January 19, 2020, 05:25:01 pm
content: Quote from: Wimberleytech on January 19, 2020, 03:44:41 pmNo it was just a straight question. It was, at first inspection, not the answer to "How can I obtain amplifier circuit loop gain?" (this is complicated by the fact that several commenters appear not to know what "loop gain" is) so I scanned back, couldn't see a question (implied or direct) that it appeared to fit as an answer and so asked what question he thought he was answering - I actually thought I'd missed something. Pejorative would have been handed out directly, I'm not known for my subtlety.

author: Cerebus
author_rank: Super Contributor
timestamp: January 19, 2020, 05:30:04 pm
content: Quote from: Wimberleytech on January 19, 2020, 05:10:33 pmThe forum does actually support \$\LaTeX\$ via Mathjax but that is complicated (for me) by three thingsRemembering how to invoke LaTeX on hereRemembering, from my infrequent usage of TeX/LaTeX how to express what I want in the same, andMathJax have deprecated their CDN (that this forum links to) and eventually it will break thus rendering LaTeX indecipherable by us mere morals, or rather by not rendering LaTeXA search of the forum yielded this, which you can probably use as a starting point:Quote from: emece67 on September 05, 2019, 09:32:45 amSo, if I've got this right, I said:\[V_{out} = {V_{in} A \over {1 + A\beta}}\]And I forgot:4. The output comes out too bloody small by default.

author: bson
author_rank: Unknown
timestamp: January 19, 2020, 07:32:44 pm
content: Quote from: Lucky-Luka on January 19, 2020, 09:01:36 amThe feedback voltage Vf is the voltage at the base of Q9. Vin the voltage at the base of Q10. The easiest way to measure them in LTspice is to add labels for those nodes; press F5 and type "Vf" in the box, then press ENTER. This gives you a net label to place: place it on top of the wire that leads to the base of Q9. Repeat for Q10 but call it "Vin" instead. Then run the simulation so you have a plot window. In the plot window, press the A key, and add a trace for "V(Vf)/V(Vin)". This will plot the loop gain.I mostly just assumed you know how to use LTspice already, and are familiar with basic amplifier analysis (since you asked for an analytical quantity).

author: Lucky-Luka
author_rank: Unknown
timestamp: January 19, 2020, 08:00:11 pm
content: The issue is not related to LTSpice... The fact is that... your method doesn't give the loop gain...At least... from my calculations... Vf=Vout*beta; Vout=[A/(1+A*beta)]*Vin; Vf/Vin is just about 1... Did I miss something?