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

[Abdullah Almosalami]

So, from my textbook, and what seems to be standard, an inverting amplifier circuit goes something like this:

However, when I switch the terminals of the op amp and follow through with the equations, I get the same V_out.

So my question then is what is the difference? I know there is other circuitry involved with the op amp internally, so I guess that's involved but I haven't gotten there yet.

 

[gneill]

When you flip the inputs you no longer have a negative feedback situation (where a portion of the output is fed back to the negative input of the op amp). The mechanism that forces $v^+$ equal to $v^-$ is no longer present. So indeed, the internal circuitry of the op amp is going to produce something different.

Think of the op amp as a dependent voltage source with some amplification factor A so that $V_o = A(v^+ - v^-)$ and reanalyze the circuit.

 

[berkeman]

So my question then is what is the difference?

Good for you to ask questions like this and try to work out the answer. Keep on doing that, and you will go far.

In addition to the comments by @gneill have a look at this wikipedia page about the circuit you have stumbled upon. Pay particular attention to how to set the hysteresis voltage in such circuits -- that's an important practical part of learning to use comparators.

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

 

[gneill]

Hmm. I think it's not quite a comparator circuit as the output does influence the input. It may be counter-intuitive at first glance, but the circuit will have a fixed gain and surprisingly, it turns out to be the same as for the negative feedback scenario.

My point was that you shouldn't rely on the "voltage rule of ideal op-amps" when setting out to analyze a scenario that doesn't clearly have negative feedback, and would be better served by going back to basics for a first pass at the circuit.

 

[DaveE]

The ideal op-amp assumption that Va = Vb is only valid if the amplifier is operating with negative feed back so that the high gain can adjust Va in the correct direction. If you remove the feedback, or change its polarity, you will create a comparator, not a linear amplifier, and the input voltages can vary greatly. So before you can apply the ideal amplifier approximations, you must verify that the circuit has the correct form of negative feedback.
You can do a more detailed (and accurate) analysis if you remove the Va = Vb assumption, and add in the amplifier gain Av as Vo = Av*(Vb-Va). Then observe what happens if you let Av get really large.
The ideal op-amp approximations (for a voltage amplifier) are:
1) Infinite gain. This implies equal input voltages for negative feedback.
2) Infinite input impedance at both inputs.
3) Zero output impedance.
None of these are always true in practice, but they allow an approximate solution to quickly understand the function of the circuit and a good starting point for an analysis of the non-ideal effects when each/any of the assumptions breaks down.

 

[jim hardy]

"Operational" nickname grew out of the early days when they were used to perform some continuous mathematical operation like add subtract or find logarithm...
Op-amps were expensive vacuum tube gizmos so logic (Boolean math) operations would more likely be done by relays.

But the name "operational" stuck to the amplifier even though that term describes its intended use.
It is the duty of the "operational circuit" designer to wrap his "amplifier" with a circuit that let's it force its inputs equal.
Otherwise he has built something other than an "Operational Amplifier Circuit" - perhaps a comparator or logic gate..

Splitting hairs i know, but it's an important distinction.

I was lucky enough to grow up on this early Philbrick-Nexus manual
http://www.waynekirkwood.com/images/pdf/Applications_Manual_for_Operational_Amplifiers_Part_1.pdf

If it doesn't "enforce a null" it's not "operational".

see also http://www.philbrickarchive.org/

any help ?

old jim

 

[Tom.G]

You seem to have gotten the signs a bit confused.

Cheers,
Tom

 

[Averagesupernova]

If it doesn't "enforce a null" it's not "operational".

A very good way to look at op-amps.

 

[gneill]

You seem to have gotten the signs a bit confused.
View attachment 233079

Cheers,
Tom

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?

 

[LvW]

Some comments:

(1) The shown calculation in the first post (second case) contains no errors - however, one should stumble over the fact that the output goes negative although the non-inv. input is connected to the input signal.
(2) Even a simulation program (DC or AC simulation) will confirm the result.
(3) But we all know that - in reality - the result is wrong.
(4) Where is the error? The circuit with positive feedback would be a stable amplifier only in case the following (idealized) conditions would be met:
* No noise, neither internally nor externally.
* Absolutely stable supply voltages
* No power-switch-on transients.
(5) The mentioned calculation/simulations assume these unrealistic idealized conditions (like a large ball is riding upon another smaller ball).

 

[sophiecentaur]

When I first came upon Feedback, I couldn't imagine following the signal round and round the loop. I asked the "how does it know?" question to myself. The secret is to tale the steady state situation when things have settled down after switch on, and believe in the equations - I don't think you can avoid some maths when dealing with feedback.
You can ease into the dynamic aspects of feedback eventually.

 

[jim hardy]

The secret is to tale the steady state situation when things have settled down after switch on, and believe in the equations - I don't think you can avoid some maths when dealing with feedback.

Seconded ! With Great Vigor !

It is very counterintuitive and was a major stumbling block for me.

That's why i hammer on this phrase :

It is the duty of the "operational circuit" designer to wrap his "amplifier" with a circuit that let's it force its inputs equal.

it reminds me to write the KVL equations for both of the amplifier's input pins and set them equal.

 

[Abdullah Almosalami]

When you flip the inputs you no longer have a negative feedback situation (where a portion of the output is fed back to the negative input of the op amp). The mechanism that forces $v^+$ equal to $v^-$ is no longer present. So indeed, the internal circuitry of the op amp is going to produce something different.

Think of the op amp as a dependent voltage source with some amplification factor A so that $V_o = A(v^+ - v^-)$ and reanalyze the circuit.

I know this question and your reply was a while back now but hopefully you're still around. I finally came around to analyzing the circuit with the input resistor, dependent voltage source, and output resistance of the op amp, and I found that it still acts the same way pretty much, with a tiny detail of difference. So here's what I did:

1) Regular Inverting Op Amp Circuit:

2) Now flip the Op Amp inputs:

You see from both final equations (bottom of each picture), they are almost exactly the same except the "AR2" term in the first case is negative and in the second case positive.

How does this affect things? If I put this up on Desmos, and set some values for R1, R2, Ri, Ro, and A, I get the following:

They are pretty much the same line. And if I tried to see more exact values, I find that there is indeed a very small difference, but it's essentially insignificant:

So it seems like even in the more detailed model, flipping the op amp's inputs does not change the output much at all. I suppose perhaps I'd have to go down to the CMOS level next to analyze once more. What do you all think?

 

[Abdullah Almosalami]

Some comments:

(1) The shown calculation in the first post (second case) contains no errors - however, one should stumble over the fact that the output goes negative although the non-inv. input is connected to the input signal.
(2) Even a simulation program (DC or AC simulation) will confirm the result.
(3) But we all know that - in reality - the result is wrong.
(4) Where is the error? The circuit with positive feedback would be a stable amplifier only in case the following (idealized) conditions would be met:
* No noise, neither internally nor externally.
* Absolutely stable supply voltages
* No power-switch-on transients.
(5) The mentioned calculation/simulations assume these unrealistic idealized conditions (like a large ball is riding upon another smaller ball).

You mention the idealized conditions in which the positive feedback would work. How do you know this? Is this from a much more detailed op amp model?

 

[sophiecentaur]

I found that it still acts the same way pretty much

You do know that there is a huge difference in the behaviour of an Op Amp with positive and with negative feedback so you must have an error if your calculations tell you otherwise. You should question your initial equations (the blue ones) because I don't understand where they come from. I can't read your handwriting easily enough but you have the original. All EE textbooks derive how OP amps behave in circuits so why not just refer to yours?

 

[jim hardy]

However, when I switch the terminals of the op amp and follow through with the equations, I get the same Vout.

That's because of algebra.
$Vout = -Vin \left( \frac{ Rfb}{ Rin} \right)$ is the only relation that makes Vb equal Va ..
That's what @LvW was demonstrating.

But your circuit when connected with positive feedback does not allow the amplifier to cause that relationship. .

It is the duty of the "operational circuit" designer to wrap his "amplifier" with a circuit that let's it force its inputs equal.
Otherwise he has built something other than an "Operational Amplifier Circuit" - perhaps a comparator or logic gate..

When you swapped those inputs you changed from a linear multiplier circuit, $\left( \frac{ Rfb}{ Rin} \right)$,
to a primitive sort of logical comparator circuit that does the Boolean operation ' out = (c greater than a) ' ...

 

[alan123hk]

They are pretty much the same line. And if I tried to see more exact values, I find that there is indeed a very small difference, but it's essentially insignificant:

one should stumble over the fact that the output goes negative although the non-inv. input is connected to the input signal

I tried to describe the situation in a more concise way as follows: -

Great and brilliant, you have shown a special and interesting state of the op amp I have never thought of before, although this is just a mathematical calculation that would not happen in reality.

 

[sophiecentaur]

Great and brilliant, you have shown a special and interesting state of the op amp I have never thought of before, although this is just a mathematical calculation that would not happen in reality.

"Not in reality"? It's the basis of the very familiar Schmidt Trigger circuit. Lookitup.

 

[LvW]

You mention the idealized conditions in which the positive feedback would work. How do you know this? Is this from a much more detailed op amp model?

Where did you read "...would work.." ?
I never have stated that such a circuit (positive feedback) would work as an amplifier.
What I have tried to explain was the following:
Using the known equations for analyzing such a circuit (KVL, KCL) we arrive at a set of equations which result in an amplified signal.
No inner contradiction during calculations. And all the simulation programs give the same result - if you are doing a DC analysis or an AC analysis.

Why? Because in these analysis (and for our hand calculations) we have neglected all the REAL WORLD influences: Noise, power supply fluctuations and - most important - power.switch-on transients. Hence, such a circuit will never work in reality. But we must know that only an analysis in the time domain (TRAN analysis using a real opamp model) will reveal such a kind of instability - not a TRAN analysis with an ideal VCVS model nor a DC analysis or an AC analysis (however, the phase response shows an "unnormal" behaviour).

 

[sophiecentaur]

I never have stated that such a circuit (positive feedback) would work as an amplifier.

But positive feedback has been used in linear amplifiers. The Tuned Radio Frequency receiver of a few years ago used positive feedback to produce a high gain, narrow band amplifier. The precise amount of feedback needed fine adjustment or the gain would be too low - or the device would oscillate. If the feedback parameter is low enough, you can get increased gain and linear behaviour - but not when an OP amp is the basic amplifying device. Take a low gain transistor or valve and you can get useable circuit gain with the right circuit components.

 

[LvW]

But positive feedback has been used in linear amplifiers..

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.

 

[jim hardy]

But positive feedback has been used in linear amplifiers.

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

 

[sophiecentaur]

However, it was my intention not to complicate the answer

The OP really needs to be shown the complexities and realities in order to resolve his apparent paradox. I haven't checked the calculations but there must be some hidden constraints about the actual validity of the answer in most circumstances, I guess.

 

[anorlunda]

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...

Not only that, the magnetic amplifier was the basis of the famous GE sign on the building where I once worked in Schenectady. It has 1399 bulbs. It slowly dims on and off in a 10 second cycle. In 1926, when it was first lit, smoothly modulating nearly 100 kW was a significant accomplishment. There was a second sign on the other side of the plant operating 180 degrees out of phase with the first sign's cycle. The engineers were very proud that the sum of the two power demands was constant through the whole cycle. The on-site power plant's capacity in 1926 may not have been much bigger than 100 kW.

If the building still exists 7 years from now, it will be the 100th anniversary for the sign.

p.s. In that picture, I can see several burned out bulbs. Tsk tsk. That would never have been allowed in my day.

 

[Abdullah Almosalami]

You do know that there is a huge difference in the behaviour of an Op Amp with positive and with negative feedback so you must have an error if your calculations tell you otherwise. You should question your initial equations (the blue ones) because I don't understand where they come from. I can't read your handwriting easily enough but you have the original. All EE textbooks derive how OP amps behave in circuits so why not just refer to yours?

I could have gone wrong in the calculations, but I don't see it and no one has pointed out where, and 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. I do know that it doesn't work in reality, so 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.

 

[Abdullah Almosalami]

Where did you read "...would work.." ?
I never have stated that such a circuit (positive feedback) would work as an amplifier.

(4) Where is the error? The circuit with positive feedback would be a stable amplifier only in case the following (idealized) conditions would be met:
* No noise, neither internally nor externally.
* Absolutely stable supply voltages
* No power-switch-on transients.

I was asking where did you get reasoning about the idealized conditions. As in, based on what more detailed model of the op amp that would now account for transients and noise and such. That's more what I was looking. I understand that switching the outputs doesn't work the same as seen from the equations shown so far, and it seems most of the replies are saying the mathematical model here is just ideal and doesn't work in reality, but to me that's a push to say alright then, well, what is a more accurate mathematical model? Perhaps more complex but something that could be used for analysis, and then maybe from that explore other cases.

 

[jim hardy]

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

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

 

[jim hardy]

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

$\frac{Vout - Vb}{Rf} = \frac{Vb - Vc}{Ri}$

$Vb$ and $Vout$ as in previous post

 

[LvW]

I was asking where did you get reasoning about the idealized conditions. As in, based on what more detailed model of the op amp that would now account for transients and noise and such. That's more what I was looking. I understand that switching the outputs doesn't work the same as seen from the equations shown so far, and it seems most of the replies are saying the mathematical model here is just ideal and doesn't work in reality, but to me that's a push to say alright then, well, what is a more accurate mathematical model? Perhaps more complex but something that could be used for analysis, and then maybe from that explore other cases.

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.

Ask yourself: What happens when a circuit with feedback tends to self-excitement ? The output voltage increases - and this increase is coupled back to the input supporting this increase ...We can observe that this process needs a certain (in many cases a very short) time due to unavoidable delays in real opamp models ...and, therefore, we can demonstrate this effect in a time domain simulation only (TRAN analysis).

In contrast, it is not the task of AC analyses to take into account such timely effects - these analyses must reveal the frequency-dependent properties of a circuit only and, hence, must assume steady-state conditions. For this reason, this analysis must assume that the supply voltages are switched on "long tiome ago". The same applies to the DC analysis , which - by its nature - cannot predict what the circuit perhaps will do in the near future because all delaying elements (like capacitors) do not contribute to the DC behaviour.

In summary - even in case of an unstable circuit, the simulation programm will always find a fixed DC operational point - and a following AC analysis shows a frequency response with gain. Remember, the AC analysis is a small-signal analysis - it is based only on the slope of a theoretical tangent applied in the DC operating point. These results are not surprising because our hand calculation would give exact the same results.
The only difference is that we - hopefully - will not start to perform such calculation by hand because we KNOW in advance that the circuit will be unstable.

 

[LvW]

The OP really needs to be shown the complexities and realities in order to resolve his apparent paradox. I haven't checked the calculations but there must be some hidden constraints about the actual validity of the answer in most circumstances, I guess.

I don`t think it is a "paradox".
If we wouldn`t know about the problems of such positive feedback we would start calculations by hand (KVL,KCL) - and we would get a fine result.
Why? Because in our calculations we have assumed steady-state conditions. Hence, it is not a "paradox" but we simply have made an error.
If one looks at the characteristics and differences of the individual analyses (DC, AC, TRAN), it becomes clear that only an analysis in the time domain can reveal the tendency to self-excitation. Hence, the simulation programs makes no "error" while finding a DC bias point with a corresponding tangent (basis for AC analyses).
The error is on our side because we have wrongly interpreted the results of the analyses.

 

[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.