Homework issue - Non-ideal op-amp - Page 1
Forum section: electronics / beginners

author: kickit2
author_rank: Contributor
timestamp: April 08, 2017, 07:42:21 pm
content: Hello everyone,Working some homework on non-ideal op-amps and I'm stuck. Basically, something is wrong with my node voltage equations as my system of equations finds no solution.https://drive.google.com/file/d/0B9QMsEp5mU1kc2RYUWIyYnN0S3c/view?usp=sharingis an image of my work thus far - at this point I'm just tryin to get a functional system of equations, ie - node voltage equations that work. I greatly appreciate any helpThanks!

author: IanB
author_rank: Super Contributor
timestamp: April 08, 2017, 07:58:54 pm
content: Just responding here as an amateur, but here are my thoughts:You can write three equations in three unknowns:Eq 1: a current balance around VNEq 2: a current balance around VOEq 3: the output voltage of the amplifier VAThe unknowns are VN, VO and VA.It seems to me that it should be possible to solve those three equations for the three voltages, and after that determine the currents (if necessary).

author: kickit2
author_rank: Contributor
timestamp: April 08, 2017, 08:03:29 pm
content: I have done exactly that, but for Vg, VN, and Vo - problem is I have an error somewhere in one of the equations which makes the system have no solution. The issue I'm having it finding my "face palm" moment to where the equations can produce a solution.

author: IanB
author_rank: Super Contributor
timestamp: April 08, 2017, 08:22:42 pm
content: Surely your face palm moment is to treat Vg as an unknown voltage, when it is shown as a voltage source and must therefore be a known input to the system?If you do what you have done, my guess is that the system will turn out to be singular and have no solution.The voltage you don't know, and should therefore include in the equations, is the output voltage of the amplifier.By the way, when you say "I have done exactly that,but...", it means you have not, in fact, done "exactly that". You have done something else, and therein lies your problem.

author: ebastler
author_rank: Super Contributor
timestamp: April 08, 2017, 08:32:35 pm
content: I am not sure I read your third equation correctly -- is that A * V_delta = Vo, where V_delta = Vp - Vn?If so, where do you get that from? Aren't you postulating the behavior of anidealopamp there?

author: kickit2
author_rank: Contributor
timestamp: April 08, 2017, 08:42:01 pm
content: Thats the definition of a "non-ideal" model. See attachment for how a non-ideal op-amp is ideally modeled (ass odd as that sentance sounds).

author: kickit2
author_rank: Contributor
timestamp: April 08, 2017, 08:45:16 pm
content: If I'm following you, what your saying is to assume that Vg is some known value - perhaps 1 to make things easy. The idea is to create a transfer function that would apply for all values of Vg where H=Vo/Vg. In this case, setting Vg to oneshouldn'thurt anything, as dividing anything by one doesn't change it. Right?

author: ebastler
author_rank: Super Contributor
timestamp: April 08, 2017, 08:49:45 pm
content: Quote from: kickit2 on April 08, 2017, 08:42:01 pmBut your Vo point is not at the output of the "internal" amplifier (the diamond-shaped entity). There is a voltage divider, composed of the internal 5k and the external 500 Ohm, right?EDIT: To paraphrase, the third equation you gave is valid for the voltage directly at the output of the "internal" amplifier, but not for Vo. It would be correct only for Ro = 0.

author: kickit2
author_rank: Contributor
timestamp: April 08, 2017, 08:53:45 pm
content: Itisa voltage divider, but not as simple as a single current path. That's why node Vo is defined and the currents in and out are summed to equal zero. If the system had no feedback (and no 100k strapping back to Vn) then you would be correct. In this case, the current through Rf (100k) must be considered too. But the 500 and 5k are both factored into the "Vo" equation.I should add that I've email my professor about this, and I'll post the error for future reference if it isn't highlighted prior to then.

author: ebastler
author_rank: Super Contributor
timestamp: April 08, 2017, 08:59:33 pm
content: Quote from: kickit2 on April 08, 2017, 08:53:45 pmI understand and trust your second equation (the current balance at point Vo).But I don't trust your third equation, A * (Vp-Vn) = Vo.'nuf said; maybe I'm missing your point.But it's your equations which don't have a solution, so one of those three might be wrong...

author: kickit2
author_rank: Contributor
timestamp: April 08, 2017, 09:06:52 pm
content: Quote from: IanB on April 08, 2017, 07:58:54 pmThere is no Va - The unknowns are Vn,Vo,VgQuote from: IanB on April 08, 2017, 08:22:42 pmthe "but" came in where I was showing that I was solving for the unknowns in question, and that the unknowns stated by the last reply were incorrect.

author: kickit2
author_rank: Contributor
timestamp: April 08, 2017, 09:44:09 pm
content: That's an amazing point. The note set states that Vo=AVd... but now that you point that out, that probably only happens when there is no other load on Vo, whichwouldmean that 0 current is flowing though Ro, so no voltage drop... hummm.... thats gotta be it. I have to move away from that assumption that is given, as there will be current flow, and therefor a voltage drop. Awesome - thats a step in the right direction.So now I need to find a third equation.. Actually, combining with IanB's comment = assuming Vg=1 turns this into a two unknown two equation problem... lets see

author: ebastler
author_rank: Super Contributor
timestamp: April 08, 2017, 09:50:54 pm
content: Quote from: kickit2 on April 08, 2017, 09:44:09 pmYes, that's what I was trying to get across all along. Actually, the note you had attached does not talk about Vo at all. It only talks about the voltage on theother(internal) side of Ro. So the third equation you used is invalid.It seems that Vn and Vo are the only two unknown voltages in your system. So why do you need a third equation anyway?EDIT: This ties back to what IanB said in his second post, by the way. You don't have to solve for Vg, since this is a known input; only two unknowns need to be dealt with.

author: kickit2
author_rank: Contributor
timestamp: April 08, 2017, 10:02:33 pm
content: The noteset that I posted was not the note set I was working off of, although it is equivalent. The attached image is of the one I'm actually using - and it directly states that AVd=Vo - but if you separate yourself from that "fact" and look closer - it doesn't show any feedback, meaning that current though Ro=0. Additional, the idea was to find a transfer function, which works for any Vg, so Vg is an unknown. BUT, if I would have remembered that H=Vo/Vg, I would have seen that setting Vg to one would eliminate it as an unknown, which not changing H. In the end, it was exactly what I though it was - a stupid mistake of me overlooking something. I defiantly appreciate you guys looking this over for me. Sorry I didn't understand your point at first, but for whatever reason, seeing it graphically made it click and question my assumption of the noteset directly applying.

author: ebastler
author_rank: Super Contributor
timestamp: April 08, 2017, 10:08:21 pm
content: Indeed, that other note is a bit misleading. Only works for Ro=0 (which can't be assumed for the non-ideal op-amp), or for the situation without any load on the output (which is not too relevant in practice).Please see my previous post -- I think youdon'tneed a third equation!

author: IanB
author_rank: Super Contributor
timestamp: April 08, 2017, 10:21:08 pm
content: Quote from: kickit2 on April 08, 2017, 09:06:52 pmI think you've got it now, but Va is the voltage on the output of the amplifier, upstream of the 5k output impedance.Therefore, you have the following three equations:Current balance around Vn:(Vg - Vn) / 5 + (Vp - Vn) / 500 + (Vo - Vn) / 100 = 0Current balance around Vo:(Va - Vo) / 5 - (Vo - Vn) / 100 - (Vo - 0) / 0.5 = 0Amplifier voltage Va:Va = A(Vp - Vn)I know Va isn't labeled on the diagram, and the third equation can technically be substituted into the second equation, but introducing Va like this can help to clarify the analysis and help avoid errors.

author: kickit2
author_rank: Contributor
timestamp: April 08, 2017, 10:26:44 pm
content: I do see what you mean now - I wasn't aware that you were referring to the dependent voltage source as Va, and assumed you had made a typo. In any case, I understand where I went wrong now with assumptions - thanks again for all your help.

author: rstofer
author_rank: Super Contributor
timestamp: April 09, 2017, 02:25:33 am
content: I dumped your equations into wxMaxima (hopefully without typos) and there is a solution. It shows Vn as a very small number and Vo as much larger.

author: Ratch
author_rank: Regular Contributor
timestamp: April 09, 2017, 03:22:00 am
content: Quote from: kickit2 on April 08, 2017, 07:42:21 pmTwo equations and two unknowns. Ask if you have any questions.Ratch

author: MrAl
author_rank: Super Contributor
timestamp: April 10, 2017, 01:03:14 pm
content: Quote from: kickit2 on April 08, 2017, 07:42:21 pmHi,Maybe you just need some more practice with nodal analysis and dependent sources.Also, you can write an equation for all the resistors and that will get you started. That's because they form one big voltage divider with a different voltage at each end and taps that are important to the values that you want to solve for.Also, yes the internal gain is Vout=A*(vp-vn) and that is always the case for that kind of dependent source. The output resistance 5k is in series with that, so of course the actual output Vo should be either lower or higher but not the same as the internal Vout.