What happens when you flip the inputs of an op amp?

[GrahamN-UK]

I could have gone wrong in the calculations, but I don't see it and no one has pointed out where,

Actually, previous posters have pointed out where you went wrong but perhaps they were too subtle for you and their message has been lost in the formulas. So, here's the fundamental point you have missed: what you labelled as the "Voltage Rule of Ideal Op-Amps", Va = Vb, in the first diagram in post #1 IS ONLY VALID FOR NEGATIVE FEEDBACK AMPLIFIERS.

The feedback in a negative feedback amplifier acts to drive Va (in your circuit) to be equal to Vb. In a positive feedback amplifier the feedback acts to drive Va away from Vb. Hence Va is not equal to Vb in a positive feedback amplifier. The Voltage Rule is not valid for a positive feedback amplifier and you cannot use it to analyse a positive feedback amplifier.

The problem with your analysis in the second part of post #1 and in post #13 is that you have still included your Voltage Rule; the assumption that Va = Vb, in your analysis . That is why your analysis gives you the results you know to be wrong.

To analyse the positive feedback rule you'll need to use the formula gneill gave you in post #2: Vout = A(V+ - V-).

Note that putting the wrong formulas into a simulation will produce the wrong outputs, even if it does look very neat.

EE textbooks (like the two I've looked at) derive the equations for the standard inverting input case but always ignore the other way around.

The standard equations are only valid for negative feedback circuits. There is no corresponding equation for a positive feedback amplifier because no corresponding positive feedback amplifier circuit exists. (You should however find an equation for a non-inverting negative feedback amplifier.)

Other posters have pointed out that an op-amp circuit with positive feedback gives you a comparator circuit with hysteresis, also known as a Schmitt trigger circuit. You may find an analysis of this circuit in your textbooks, or online.

What I was kind of hoping for was for someone to describe a more accurate model of an op amp that does account for all that was mentioned here.

You don't need a more accurate op amp model for this analysis. The introductory model is fine as long as you realize why your 'voltage rule of ideal op-amps' applies only to negative feedback circuits.

If you go back and re-read the thread, bearing in mind that your Voltage Rule is only valid for negative feedback amplifiers, you should find that the early posts do actually say this.

 

[LvW]

The feedback in a negative feedback amplifier acts to drive Va (in your circuit) to be equal to Vb. In a positive feedback amplifier the feedback acts to drive Va away from Vb. Hence Va is not equal to Vb in a positive feedback amplifier. The Voltage Rule is not valid for a positive feedback amplifier and you cannot use it to analyse a positive feedback amplifier.

Graham - with all respect, I think you missed the point (and with the last sentence you are wrong).
Let me explain:
* The - simplified and reworded - contents of your second sentence (as quoted above) is: The circuit is unstable because it is unstable. You cannot start your considerations with the sentence ",,,to drive Va away..." and dispense with further calculatons.. That is already the RESULT of a stability check!
* Contrary to your claim, the classical rules for a circuit analysis (KVL,KCL) are indeed valid for the shown circuit - also in case of positive feedback (see my explanations in post #19, 29 and 30).

Example calculation:
Vin-Vp=I*R1; Vp-Vout=I*R2; Vp=Vout/Aol .

It is a simple task to solve this set of equations for Vout. Assuming Aol=1E5 and R1=1k, R2=10k we arrive at Vout=Vin*(-10.001).
Any simulation (DC analysis) will arrive at the same result - if you have a simple or an extended opamp model! This is because in a DC analysis all time-depending (delaying) parts are neglected! And these parts - always present in real circuits - are responsible for instability!

Of course, in reality and knowing that we have an instable circuit, this result cannot be achieved in practice - but this calculation is correct and without any contradiction. It cannot be used to prove that the circuit is unstable. More than that, for R1/(R1+R2)<1E-5 the circuit will be stable!

(Thanks to the Editor for opening this thread again - it is really an interesting problem because it touches the question: Before starting a circuit analysis using KCL and KVL - is it necessary to know something about the circuit behaviour in advance? And: Is the simulator right or wrong ? Who is wrong: The simulation engine or the user? )

 

[GrahamN-UK]

I wrote, in post #31:
"The Voltage Rule is not valid for a positive feedback amplifier and you cannot use it to analyse a positive feedback amplifier."

Graham - with all respect, I think you missed the point (and with the last sentence you are wrong).
Let me explain:
* The - simplified and reworded - contents of your second sentence (as quoted above) is: The circuit is unstable because it is unstable. You cannot start your considerations with the sentence ",,,to drive Va away..." and dispense with further calculatons.. That is already the RESULT of a stability check!
* Contrary to your claim, the classical rules for a circuit analysis (KVL,KCL) are indeed valid for the shown circuit - also in case of positive feedback (see my explanations in post #19, 29 and 30).

I have not mentioned Kirchhoff's circuit laws, let alone claimed they are invalid. My mention of "The Voltage Rule" does not refer to Kirchhoff's voltage law but to what the OP labelled as the "Voltage Rule of Ideal Op-Amps", Va = Vb, in the first diagram in post #1. I stated this in full in the paragraph before the one you quoted. I thought it would be clear, one paragraph later, what I was referring to. Apparently not. Please look back at post #1 and see what the OP used the phrase to describe. It wasn't KVL.

Whether I've missed the point depends on what the OP says in any followup. My hypothesis is that the OP, who is a student, has learned the rule that (using his terms) Va = Vb but has not learnt, for whatever reason, the necessary condition that this only applies in a (correctly functioning) negative feedback amplifier. As he apparently doesn't know this, the gentle and polite reminders offered by various people earlier in the thread haven't been effective. Jim Hardy and myself have offered hopefully clearer posts recently. At the time of writing the OP is yet to see these. Now the moderator has re-opened the thread perhaps we will find out if they answer the OP's question.

This is because in a DC analysis all time-depending (delaying) parts are neglected! And these parts - always present in real circuits - are responsible for instability!

The OP's question is about static analysis of an op-amp circuit. Until he has resolved that I don't want to veer off into dynamic analysis.

(Thanks to the Editor for opening this thread again - it is really an interesting problem because it touches the question: Before starting a circuit analysis using KCL and KVL - is it necessary to know something about the circuit behaviour in advance? And: Is the simulator right or wrong ? Who is wrong: The simulation engine or the user? )

I hope people have some idea of correct circuit operation before they blindly throw equations into a simulator. How else will they know if the results are credible? It would be like throwing numbers into a calculator without any idea of what result you expected. But then I learned my EE before simulators were around (at least for undergraduates) and learned long arithmetic before calculators were around. Perhaps modern students approach things differently?

In this case the OP does know the results of his analysis are wrong and hence he has asked his question to find out where he has gone wrong - a commenable approach.

 

[Abdullah Almosalami]

Actually, previous posters have pointed out where you went wrong but perhaps they were too subtle for you and their message has been lost in the formulas. So, here's the fundamental point you have missed: what you labelled as the "Voltage Rule of Ideal Op-Amps", Va = Vb, in the first diagram in post #1 IS ONLY VALID FOR NEGATIVE FEEDBACK AMPLIFIERS.

I have certainly more than understood that the ideal model only works for negative feedback. What I have repeatedly said is that I want to know if there is a model, perhaps more complex, that does model a general case of either positive or negative feedback. And the idea of enforcing null, as Jim and others have said, conceptually makes sense. But again, what I would like to know is what is an actual model that can put these concepts to actual quantities.

To analyse the positive feedback rule you'll need to use the formula gneill gave you in post #2: Vout = A(V+ - V-).

That is what I did in post #13, which still revealed the same results.

You don't need a more accurate op amp model for this analysis. The introductory model is fine as long as you realize why your 'voltage rule of ideal op-amps' applies only to negative feedback circuits.

If you go back and re-read the thread, bearing in mind that your Voltage Rule is only valid for negative feedback amplifiers, you should find that the early posts do actually say this.

And here is why I must say no. I am not satisfied with just being told that this is how it works. I do want to know the details. Perhaps I may not understand them now but given enough time, I certainly will.

 

[jim hardy]

.

The standard equations are only valid for negative feedback circuits. There is no corresponding equation for a positive feedback amplifier because no corresponding positive feedback amplifier circuit exists. (You should however find an equation for a non-inverting negative feedback amplifier.)

Quite so.

Example calculation:
Vin-Vp=I*R1; Vp-Vout=I*R2; Vp=Vout/Aol .

It is a simple task to solve this set of equations for Vout. Assuming Aol=1E5 and R1=1k, R2=10k we arrive at Vout=Vin*(-10.001).

Quite so as well.
But don't let a beginner think the circuit will perform a linear multiplication .

I too solved for Vout,, took limit as gain approaches infinity...
with infinite gain it's $Vin X (1 - \frac {Rfb + Rin}{Rin} )$
so using your realistic R values and my infinite gain give $Vout = Vin X(1 - \frac {11k}{1k}) = Vin X(- 10)$ , close 'nuff to your actual -10.001

set Vin = 0 and you get = $0 X (1 - \frac {11k}{1k}) = 0$

but don't let OP think the amplifier will be stable there.

The first accidental millivolt (of either polarity )that appears on Vout will be coupled, attenuated elevenfold, to the +input
which will drive Vout in the same direction , and a lot further because of gain
then get coupled back again to + input
causing runaway in that direction..
That's what positive feedback does.
The output will drive to the amplifier's capability and remain there.

So the circuit is not stable as algebra implies.
Nor is it astable as intuition implies
It's bistable, will sit happily at one limit or the other until you apply enough Vin to overwhelm the feedback and make it flip to the other state.

As has been mentioned already, it's the well known Schmitt Trigger and Rin / Rfb sets the hysteresis.

old jim

 

[Abdullah Almosalami]

View attachment 237612

Va = 0

and Vb = Vc + (Vout - Vc) X Ri/(Ri + Rf)

Vout = Avol X (Vb - 0)
Vout = Avol X (Vc + (Vout - Vc) X Ri/(Ri + Rf))

Gain = Vout/Vc = Avol X (1 + (Vout/Vc -1) X Ri/(Ri + Rf))
looks recursive to me. Check my algebra ?
old jim

My point was not that my assumptions were correct, but that the math itself was not wrong, as in there were no algebraic errors.

On your equations, how come you can use voltage division like that? If we can't assume that no current is flowing into the positive input of the op amp, then Ri and Rf are not necessarily in series, and so the voltage division formula doesn't work, right? And so, your equations here are based on the ideal rules.

But even if it were true, then continuing your math, this is what I get:

I'm not sure what the solution to A over here means, and how there can be 2 different amplifications, so perhaps this is a paradox because of some wrong assumptions, given I didn't make algebraic errors.

 

[Abdullah Almosalami]

No - the problem is not the need for a "more accurate model". The opamp models provided by the manufacturers are very complex and accurate enough.
The reason for the observed (surprising) results are the properties of the various analyses.

That is a good point when considering simulations, and I've indeed missed that circuit analysis works under certain assumptions. You're very right. Thank you for bringing that up.

I suppose then would it be far too complex to try and to use an analysis that does also account for the unstable aspect of a circuit before steady-state?

 

[jim hardy]

On your equations, how come you can use voltage division like that? If we can't assume that no current is flowing into the positive input of the op amp, then Ri and Rf are not necessarily in series, and so the voltage division formula doesn't work, right? And so, your equations here are based on the ideal rules.

yes, i used ideal rules...

I'm not sure what the solution to A over here means, and how there can be 2 different amplifications, so perhaps this is a paradox because of some wrong assumptions, given I didn't make algebraic errors.

Your algebra is so good it intimidates me, to be honest. I've not worked through it yet - you only posted it a moment ago.

I had prior knowledge that this circuit is bistable so had hoped my algebra would arrive at a quadratic so there'd be two solutions
but i didn't find one
but my algebra is notoriously strained, I'm math challenged and make dumb mistakes.

The circuit does not perform a linear multiplication
---see my post immediately previous--

i'll have to look at your algebra this evening, expecting company momentarily.

wow i had my algebra worked out ready to post last night but deleted it thinking it was superfluous...

meantime, see what you think of post # 35 where i tried to explain how to work the circuit in one's head.
That is a necessary thought step so we use our math correctly.
Previous posters have tried to explain that the circuit is not stable, ie the math it performs is not a linear multiplication.
So the algebraic expression Vout = VinX(-10) describes not its transfer function but the cusp on which it teeters.

 

[LvW]

I wrote, in post #31:
"The Voltage Rule is not valid for a positive feedback amplifier and you cannot use it to analyse a positive feedback amplifier."
I have not mentioned Kirchhoff's circuit laws, let alone claimed they are invalid. My mention of "The Voltage Rule" does not refer to Kirchhoff's voltage law but to what the OP labelled as the "Voltage Rule of Ideal Op-Amps", Va = Vb,...
......
I hope people have some idea of correct circuit operation before they blindly throw equations into a simulator. How else will they know if the results are credible?

Graham - I do not want to "fight" wth you and perhaps there was a misunderstanding between us.
(1) At first - just a clarification: Any "voltage rule" (whatever it may be) is a result of Kirchhoffs laws. So - as you certainly know - the approach Va=Vb (V+=V-) is nothing else than the result of applying the voltage divider formulas and the superposition rule to the circuit under discussion.
(2) Now my point: Let`s assume that a student has learned that an idealized opamp provides an output signal even if the voltage difference at the input disappears (and - yes - mathematics allow that the product of zero x infinite may give a finite value). More than that, somebody has told him that this rule (Va=Vb) is valid only for negative feedback. However, the student does not simply believe such a claim - instead, he tries to to convince himself if this is true - and WHY !
(3) Therefore, he starts a calculation for resistive positive feedback (as shown in the task description) - and arrives at the result gain= - R2/R1. And now? His question: "Have I made an error?"
I think, in this case, it is not sufficient just to tell him that he has used a formula that is not valid...he will immediately ask: Why not? Is the result wrong? My answer : No - it is not wrong, but unrealistic because of several simplifications (implicitely contained in the calculation approach).
(4) All I wanted to say is that such a static calculation cannot reveal instabilities of the circuit - and I have explained WHY (missing power switch-on, no noise, no supply voltage variations).

For my opinion, it is not enough simply to state that Va=Vb would be not valid for positive feedback (as you did) without giving a corresponding explanation!

However, to be exact: It is valid...and we do not make any mathematical error if we apply such a rule. And - as you know - all the simulation programs do the same!
However, we must know that the result was found under idealized environmental conditions only!
And we know from our experience that the result is unrealistic and cannot be used in practice (because we live in a real world).
And finally, we come to the conclusion, that we should not make use of this rule (Va=Vb) for positive feedback.
(By the way: We should not forget that there are circuits with positive and negative feedback at the same time! In this case, it is important to know which type of feedback is dominating).

Now - do you see the difference in our approach for answering the question?
You are starting with the statement that Va=Vb would be not valid for positive feedback...and I try to explain the scenario and end up with the same conclusion.

Final comment (regarding your last quoted sentence): I completely agree with you. And that was the reason I have mentioned the fact that simulation programs cannot reveal such stability problems in case of Q-point or static DC analyses. Each user must know what he is doing, what the simulator can only do in the various analyses - and that it is absolutely necessary to evaluate the simulation results and check if they are plausible.

(Sorry for the long answer)
Regards
LvW

 

[Abdullah Almosalami]

meantime, see what you think of post # 35 where i tried to explain how to work the circuit in one's head.
That is a necessary thought step so we use our math correctly.
Previous posters have tried to explain that the circuit is not stable, ie the math it performs is not a linear multiplication.
So the algebraic expression Vout = VinX(-10) describes not its transfer function but the cusp on which it teeters.

I am just now getting that. Was working my way down, and replying as I went along, so if I had missed an important point in that post, my bad!

.

The first accidental millivolt (of either polarity )that appears on Vout will be coupled, attenuated elevenfold, to the +input
which will drive Vout in the same direction , and a lot further because of gain
then get coupled back again to + input
causing runaway in that direction..
That's what positive feedback does.
The output will drive to the amplifier's capability and remain there.

So the circuit is not stable as algebra implies.
Nor is it astable as intuition implies
It's bistable, will sit happily at one limit or the other until you apply enough Vin to overwhelm the feedback and make it flip to the other state.

As has been mentioned already, it's the well known Schmitt Trigger and Rin / Rfb sets the hysteresis.

old jim

Ok just read through it. And yes. Actually, even though I kind of got that positive feedback in general in any system would make things unstable, until you mentioned it here and put it into words, I didn't fully realize it, so thanks for that Jim. And I see! The bistable points A=0, and 1 are the stable points because a gain that is not either value would cause the output to go exponential! I suppose however that gains between 0 and 1 would cause the output to exponentially decay to 0, which is still stable (but not useful?), and it is the gains that are greater than 1 that would cause the unstable exponential growth. And actually, going back through the posts, I just now understood what this post was saying:

Yes - of course. A slight positive feedback will result in a stable amplifier (with reduced bandwidth) - as long as the loop gain is <+1.
However, it was my intention not to complicate the answer to the given problem and to be realistic - because the amount of positive feedback for the circuit under discussion is required to be smaller than 1/Aol (Aol=DC open-loop gain) for a stable bias point.

And indeed I haven't yet come across the Schmitt Trigger circuit, and I'm sure I'll have a much better perspective once I get around to studying that.

 

[jim hardy]

I suppose then would it be far too complex to try and to use an analysis that does also account for the unstable aspect of a circuit before steady-state?

it'd be a boolean expression
something like Vin + Vout/11 > 0

have you taken your logic course yet?

Think how you'd write a line of BASIC code to do the comparison.

 

[jim hardy]

Was working my way down, and replying as I went along, so if I had missed an important point in that post, my bad!

i figured that, from reading your posts.
No "Bad" to ascribe to anyone.
Keep up the good work.

and thanks for the kind words.

"We think only through the medium of words. --Languages are true analytical methods. --Algebra, which is adapted to its purpose in every species of expression, in the most simple, most exact, and best manner possible, is at the same time a language and an analytical method. --The art of reasoning is nothing more than a language well arranged."

i have to resolve my words against my algebra , and it's a struggle. I envy folks like you and LvW who are so fluent in the latter.

old jim

 

[Tom.G]

View attachment 233091
Turns out that this is not a good assumption in this case (that Vout is positive).

If we assume some gain $A$ for the op-amp, then since the (-) input is grounded then the (+) input must be at $V_{out}/A$.

Write KCL for the (+) input node, see what you find. What happens when $A$ gets very large?

Here is a copy of an earlier post #9 by @gneill. Look at the op-amp symbol and the formula in it that defines the output voltage source as "A(V⁺ - V^-)".
Does that match what you have implemented in your algebra? If not, why not?

 

[LvW]

Question: For the purpose of finding the gain formula for the attached circuit - is it allowed to set Vp=Vn [or (Vp-Vn)=Vout/Aol] for the most left amplifiier ?

 

[NascentOxygen]

To analyse the positive feedback rule you'll need to use the formula gneill gave you in post #2: Vout = A(V+ - V-).

The relation V_out = A(v⁺ - v^-) is incomplete, it's a simplification. There are strong conditionals attached to it restricting it to only those instances where V_out is not pinned near either rail.

A more complete representation would be V_out = A(v⁺ - v^-) iff (V_{o max}> V_out > V_{o min})

So before applying that simplified rule, you must first establish that V_out is going to lie within the allowable range, and if it doesn't then that rule doesn't apply.

-edited-

 

[jim hardy]

Question: For the purpose of finding the gain formula for the shown circuit -is it allowed to set Vp=Vn for the most left amplifiier?

My old adage:
"Ir is the duty of the operational amplifier circuit designer to wrap his amplifier with a circuit that allows the amplifier to drive its inputs equal,
else his operational amplifier circuit cannot perform the desired operation."

So yes, that is what you should do.

And i just learned a new word from @NascentOxygen

Thanks !

 

[GrahamN-UK]

The relation V_out = A(v⁺ - v^-) is incomplete, it's a simplification. There are strong conditionals attached to it restricting it to only those instances where V_out is not pinned near either rail.

A more complete representation would be V_out = A(v⁺ - v^-) iff (V_{o max}< V_out > V_{o min})

So before applying that simplified rule, you must first establish that V_out is going to lie within the allowable range, and if it doesn't then that rule doesn't apply.

Yes, of course, though I think one of your comparison signs needs turning round:
(V_{o max} > V_out > V_{o min})

This applies to both positive and negative feedback circuits, though the output will be clipped less often in a negative feedback circuit (because operation will be in the linear region as long as the circuit isn't overloaded.)

 

[LvW]

So before applying that simplified rule, you must first establish that V_out is going to lie within the allowable range, and if it doesn't then that rule doesn't apply.

I think, such a "rule" is self-evident as it applies to all amplifiers - not only opamps.

 

[NascentOxygen]

I think, such a "rule" is self-evident as it applies to all amplifiers - not only opamps.

Evidently not so self-evident, as OP is applying it where it is not applicable—in his positive f/b circuit.

 

[NascentOxygen]

Yes, of course, though I think one of your comparison signs needs turning round:
(V_{o max} > V_out > V_{o min})

Indeed, it does need fixing. Thanks.

 

[Nik_2213]

In my experience, accidentally switching a bread-boarded op-amp's inputs generally led to the output slewing towards one or other power rail, then staying there until my embarrassed intervention.

( Not surprisingly, I was very fond of 741 type, which generally survived such abuse. More modern designs may be less forgiving... )

That said, a little, very carefully applied positive feedback may do wonders in appropriate cases. Oscillators, Schmitt Triggers etc etc. Their constraint was often the slew speed, as rail-to-rail working for an over-driven op-amp may be surprisingly slow.

( Analogy was it got to climb out of the hedge and look around... )

Um, yes, I know the 'trad' 741 didn't go rail-to-rail. But it generally went far and fast enough to out-run my 'driver' stuff...

 

[jim hardy]

This applies to both positive and negative feedback circuits, though the output will be clipped less often in a negative feedback circuit (because operation will be in the linear region as long as the circuit isn't overloaded.)

And then there's the little matter of "Latchup"
some op-amps when overdriven severely will reverse their operation, effectively swapping the signs on the input pins
requiring a power off to return to normal.Ahhh the endless tricks Mother Nature has up her sleeve to torture us !.
I guess the price of complexity is its pitfalls.

 

[LvW]

Evidently not so self-evident, as OP is applying it where it is not applicable—in his positive f/b circuit.

That is the question which was discussed here...but you agree that (without knowing much about positive f/b) Kirchoffs laws can be fulfilled without internal contradiction? That was the core of the OPs question.

 

[Nik_2213]

OT:

I second that.

In addition to Sophie's Regenerative Radio Receiver

there's the magnetic amplifier
wherein positive feedback around a low gain element (resembling a transformer) gives it high gain
and that high gain inductive element can then be wrapped with a circuit that let's it enforce a null, ie "operate" .
It was the basis of early submarine reactor instrument systems and an early commercial nuke plant Yankee Rowe...

http://exvacuo.free.fr/div/Sciences/Dossiers/EM/Magnétisme/George B Trinkaus - Magnetic amplifiers.pdf

===

Thank you, thank you for this. I remember seeing old, old pics and circuits of impressively large, electric-driven machinery controlled by a safe, low-voltage rheostat. A simple 'Auxiliary' winding on power transformer / auto-transformer / 'choke' neatly 'throttled' the core's transfer capacity, elegantly regulating the equipment. No scary 'tap changes' required...

IIRC, I got into a rather nasty exchange on AN Other forum before convincing my detractors that such 'Steam Punk' was viable, never mind the 'industry standard' for some decades...

Feral kin to this are the VLF / ELF 'ground currents' induced by solar storms which may overload power-grid transformers. IIRC, wary core re-design for more 'near-DC tolerance', improved cooling and better monitoring (*) have significantly reduced the potential for widespread, oft-cascading mayhem...
;-)

*) As I understand it, remotely measuring such low frequencies contaminating multi-kilovolt power grids is non-trivial...

 

[GrahamN-UK]

I have certainly more than understood that the ideal model only works for negative feedback.

Ah, we need to break down which bits of the ideal model we are talking about.

Wikipedia lists a number of attributes of an ideal op-amp model at:

https://en.wikipedia.org/wiki/Operational_amplifier#Ideal_op-amps

First, it has a bulleted list, of things that apply irrespective of the feedback type - positive, negative or none at all.

Then it summarises the operation in two golden rules:

I In a closed loop the output attempts to do whatever is necessary to make the voltage difference between the inputs zero.

II The inputs draw no current.

Then the next paragraph tells us "The first rule only applies in the usual case where the op-amp is used in a closed-loop design (negative feedback, where there is a signal path of some sort feeding back from the output to the inverting input)."It's this last part about the output attempting to do whatever is necessary to make the voltage difference between the inputs zero or, in terms of your post #1 Va = Vb, that I'm trying to establish your understanding of.

We are agreed that for negative feedback the ideal model says Va = Vb.
What do you consider happens for positive feedback? Is it:

1. The rule is no longer ideal, but Va is still approximately = Vb
2. The rule is completely broken and Va doesn't have to be anywhere near Vb.

I'm afraid I can't read the subscripts in your post #13 but the second part of post #1 suggests you are using case 1 above. Unfortunately this is false; case 2 is what is actually happening and your use of the Va = Vb rule for positive feedback in your analysis is why you are getting the erroneous result.

Jim Hardy in post #35 has provided a narrative description of what happens to the voltages in a positive feedback circuit if the output voltage is disturbed. Can you repeat that exercise for yourself and agree with his description? Can you extend that description yourself to a negative feedback amplifier? If the output voltage has settled at the expected value for a given input voltage what will happen if the output voltage becomes a little more positive? When that feeds back which way will Va shift? Remembering that Va is at the inverting input to the op amp, which way will that shift the output? Which way will the feedback move Va - towards Vb (0 volts) or away from it. Try the exercise again with a negative dip in the output voltage. Does the circuit move back to the expected value or does the feedback move the output voltage further away from the expected value?

If you don't like a thought experiment try a real one. Build the circuit on a breadboard or in a simulator. Put in some input voltages. Measure the output voltage and the voltage on the non-grounded op-amp input for positive and negative feedback. What happens? Does Va = Vb for both types of feedback? What happens to the non-grounded input in positive feedback when the output clips and you keep increasing the magnitude of the input voltage?

What I have repeatedly said is that I want to know if there is a model, perhaps more complex, that does model a general case of either positive or negative feedback. And the idea of enforcing null, as Jim and others have said, conceptually makes sense. But again, what I would like to know is what is an actual model that can put these concepts to actual quantities.

There are two modes of operation for an op-amp:

1 Negative feedback, stable, moderate gain controlled by the input and feedback resistors,
2 Positive feedback, operation as a Schmitt trigger circuit, with hysteresis (change in threshold voltage) controlled by the circuit resistors.

You seem to be looking for an option 3:

3 Positive feedback, stable gain controlled by the input and feedback resistors,

I'm afraid option 3 doesn't exist. There isn't some elusive better op-amp model we are keeping from you that will make this work.

You seem to have a problem understanding feedback. Don't worry; your time as a student is for learning these things. A couple of people in this thread have also mentioned difficulties understanding it. You've already gone down the route of trying ever-increasing amounts of algebra; I'm not sure that any more complex algebraic models are going to help. Try the thought experiments outlined by Jim Hardy and myself to see if you can puzzle out feedback operation. Back them up with actual or simulated results.

And here is why I must say no. I am not satisfied with just being told that this is how it works. I do want to know the details. Perhaps I may not understand them now but given enough time, I certainly will.

There should be more complex material with op-amps later in your course, so cracking basic feedback operation now will help you when you encounter that.

(I won't be able to make any more long contributions to this thread.)

 

[LvW]

There are two modes of operation for an op-amp:

1 Negative feedback, stable, moderate gain controlled by the input and feedback resistors,
2 Positive feedback, operation as a Schmitt trigger circuit, with hysteresis (change in threshold voltage) controlled by the circuit resistors.

You seem to be looking for an option 3:
3 Positive feedback, stable gain controlled by the input and feedback resistors,
I'm afraid option 3 doesn't exist. There isn't some elusive better op-amp model we are keeping from you that will make this work.

(A) Some comments:
1. Stable - only as long as DC stability (of the operating point) is concerned ; on the other hand - negative feedback will always reduce the dynamic stability margin.
3. Option 3 does exist - positive feedback still allows stable amplification if the loop gain magnitude is <1 (feedback factor k< 1/Aol).

(B) General remark:
To me, the most important and most interesting question to be answered is:
Before we start to analyze a circuit, is it necessary to know beforehand whether it will work stable or unstable?
With other words: If we apply the classical methods of circuit analysis, will the result contain any indication that the circuit cannot work as expected because it is unstable (no DC operational point within the linear range or dynamic instability/oscillation) ?

To me, this is an important question because there are many cases in which we cannot predict by visual inspection if a (complex) sysytem is stable or not.
And therefore the question arises (and this thread is a good example) :
Is it really "wrong" to start always with established methods (even if we are not sure about the stability properties)?

My answer: No - we are making no error using classical methods (in our case: (Va-Vb)=1/Aol or even Va=Vb) because a diligent evaluation of the result will show that "something is wrong". Hence, it is not required to know in advance if the circuit will be unstable.
In the present case:
(a) DC analysis (and hand calculations): Negative gain in spite of applying the input voltage to the non-inv. input (a kind of contradiction).
(b) AC analysis: Rising phase response (not as expected assuming a real opamp model with lowpass properties)
(c) Tran analysis: No stable DC operational point within the linear transfer range.

 

[GrahamN-UK]

I wrote:
[OP] seems to be looking for an option 3:
3 Positive feedback, stable gain controlled by the input and feedback resistors,
I'm afraid option 3 doesn't exist. There isn't some elusive better op-amp model we are keeping from you that will make this work.

(A) Some comments:
...
3. Option 3 does exist - positive feedback still allows stable amplification if the loop gain magnitude is <1 (feedback factor k< 1/Aol).

Theoretically so, yes. But a typical open-loop gain for an op-amp at DC/low frequency is around 10^5
e.g. https://en.wikipedia.org/wiki/Open-loop_gain

So it only takes a tiny amount of positive feedback before the loop gain rises above 1 and the Barkhausen stability criterion has been broken. So this isn't a practical situation for an op amp, which is what we are discussing in this thread.

For a single active device amplifier (e.g. a single transistor or valve/tube stage) with a much lower open-loop gain, positive feedback is more practical. Hence its use in regenerative stages in old radios mentioned by sophiecentaur and Jim Hardy in posts #20 & #22.

However, if you look back at old radio journals, you'll find even single device regenerative amplifiers had a reputation for being unstable and easily bursting into oscillation. Hence they disappeared from commercial use once superhet designs became practical. (You can still find hobbyist designs for them.)

 

[LvW]

However, if you look back at old radio journals, you'll find even single device regenerative amplifiers had a reputation for being unstable and easily bursting into oscillation. Hence they disappeared from commercial use once superhet designs became practical. (You can still find hobbyist designs for them.)

There is no need for me to "look back at old radio journals" ...I have listened to such radio devices (no superhet, but "Geradeaus-Empfänger" in German) in the early fifties of the last century. They were called "Goebbels-Schnauze" (Goebbels-snout) after the Nazi minister for public propaganda. There was a tiny knob for introducing a slight positive feedback which was used to narrow the bandwidth for better frequency selection (AM, of course).

 

[GrahamN-UK]

There was a tiny knob for introducing a slight positive feedback which was used to narrow the bandwidth for better frequency selection (AM, of course).

In English, such circuits tend to be known as Q-multipliers. Too much Q and your amplifier becomes an oscillator, of course.

https://en.wikipedia.org/wiki/Q_multiplier

 

[eq1]

I'm not sure what the solution to A over here means, and how there can be 2 different amplifications, so perhaps this is a paradox because of some wrong assumptions, given I didn't make algebraic errors.

There is an algebra error. The fourth step from the bottom uses the substitution Vo=A(Vc) but the correct substitution is Vo=A(Vb), defined in your step 3. (Assuming I am reading your handwriting correctly that this. The subscripts are a bit small on my screen.)

This circuit can be perfectly stable.

The correct way to analyze the operating point is KCL at B. (b-c)/Ri = (o-b)/Rf substitute b=o/A and solve yielding o=A*c*Rf/(Rf+Ri-A*Ri) (thanks wolfram alpha!) As lim A->inf you get Vo=-c*Rf/Ri which is the correct answer.

Spot check in spice if you like. A=10, c=1, Ri=Rf=1K and alpha says Vo=-5/4V

% cat amp.sp
*
E1 o 0 b 0 10
V1 c 0 1
R1 b c 1K
R2 b o 1K
.op
.end

produces

* "simulator" "HSPICE"
* "version" "M-2017.03-2 linux64"
* "format" "HSP"
* "rundate" "14:46:43 01/24/2019 "
* "netlist" "amp.sp "
* "runtitle" "*"
* time = 0.
* temperature = 25.0000
*** BEGIN: Saved Operating Point ***
.option
+ gmindc= 1.0000p
.nodeset
+ b =-125.0000m
+ c = 1.0000
+ o = -1.2500
*** END: Saved Operating Point ***

So spice and wolfram agree.

But you will probably not see this in lab unless you're very careful for many reasons.

For example, these equations use an ideal amplifier which can produce any output and has infinite bandwidth and never turns on or off. You're amplifier likely has A >> 10. Which means Vo will likely want to be outside the power rails which means the physical opamp can't produce that voltage. Unless you make the input very small there will be no stable Vo but it can be done. (Google how to measure an amplifier's offset voltage, which could be positive or negative, for an example)

Most importantly the amplifier has to transition from unpowered to powered which is an AC event, but this is a DC analysis so it doesn't describe that situation. In AC this circuit is likely unstable depending on the specifics of the amplifier's construction. Turn on can be done but it is very tricky to do without getting stuck at one of the power rails.