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

[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.

 

[jim hardy]

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.)

here's the bottom five steps. Can you highlight your observation ?

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.

I too solved for Vout using KVL for b, took limit as gain approaches infinity...
with infinite gain it's$Vin X (1 - \frac {Rfb + Rin}{Rin} )$ which resolves to the same expression as yours
and only proves that Kirchoff's Laws agree.

This circuit can be perfectly stable.

i strongly disagree. See post 35.

Sitting on a cusp is not stable.
and it is a disservice to a beginner to infer otherwise.

This circuit does not develop any restoring forces to return it to equilibrium..
It does the opposite, develops a disrupting force that drives it away from equilibrium toward infinity
but it hits the power supply rail first.

It's a disservice to a beginner to claim otherwise.

Vo=-c*Rf/Ri is NOT the mathematical operation that this circuit performs. To perform that one requires negative feedback.

What this circuit does mathematically is a logical computation with only two output states,
True and False,
represented by the maximum and minimum output values of Vout

"output state" = Vin + (Vout-Vin)X(Rin/(Rfb +Rin) > 0

where "output state" is a logical variable and the voltages (and R's) are real ones.
and (Vout-Vin)X(Rin/(Rfb +Rin) is just the voltage across Rin

Ahhhh, nostalgia.
I always modeled my circuits in Basic on my TI-99/4A,
and that Basic would assign to "output state" integer value,
either 0 (no bits set) for FALSE or -1 (all bits set for TRUE. .
That Basic was interesting , you could mix logicals and reals in an equation which was sometimes handy. But that's another story.

Anybody see algebra goofs or typos ? Please advise.

old jim

 

[alan123hk]

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 ?

It is allowed to set (Vp=Vn) or (Vout=A(Vp-Vn)) for performing the circuit analysis.
Vout = A(Vp-Vn) is the basic definition of an ideal operational amplifier.
Vp = Vn is based on the very high open-loop gain, the potential difference between its inputs tend to zero when a feedback network is implemented.

Before we start to analyze a circuit, is it necessary to know beforehand whether it will work stable or unstable?

There is no need to know it as you have already given the answer on #56.
I totally agree with your analysis.
We find the steady-state result for the given circuit by Kirchhoff's laws but no guarantee for stability, therefore, a verification of loop gain < 1 is necessary afterward. If the loop gain > 1, then the circuit is unstable, It cannot be used as a stable linear amplifier with the relationship given by steady-state analysis.

For example, the single stage non-inverting positive feedback amplifier that OP mentioned on #1, the feedback factor is obviously Ri/(Ri+Rf), let's assume the gain of the ideal Op Amp be G, then loop gain = G*Ri/(Ri+Rf), which must be less than 1 for stable operation, namely G*Ri/(Ri+Rf)<1 or G<(Rf/Ri+1). Therefore, when G becomes positive infinity, the amplifier is unstable for any positive values of Rf and Ri.

On the other hand, the single stage inverting negative feedback amplifier has loop gain = -G*Ri/(Ri+Rf) which must be less than 1 for stable operation, namely -G*Ri/(Ri+Rf)<1, or -G<(Rf/Ri+1), so that it must be stable for any positive values of Rf and Ri since the left hand side is either 0 or any negative number.

 

[alan123hk]

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

This is indeed a Schmitt Trigger circuit. The critical design equation is also related to Vout = -(R2/R1)*Vin.
Since OP's main focus seems to be related to the contradiation of equation derivation, I just mean that Vout = -(R2/R1)*Vin given by steady-state analysis doesn't mean we can realize a stable linear amplifier with that relation.

 

[eq1]

here's the bottom five steps. Can you highlight your observation ?

I'm on an iphone so I can't mark up the image (or at least I don't know how to) but in second line from the top A*Vo/Vc is reduced to A*A in the line below. This is not a valid substitution because vo=A(Vb-Va)=A*Vb

This circuit does not develop any restoring forces to return it to equilibrium..
It does the opposite, develops a disrupting force that drives it away from equilibrium toward infinity
but it hits the power supply rail first.

The circuit as drawn is absolutely stable because an ideal opamp has infinite bandwidth and therefore has no phase delay so it can't be unstable.

For a real opamp... is the circuit stable? My answer is: it depends!

What does it depend on? Which opamp obviously (as I'm sure you know, attenuators exist too!), how it is powered, and what the valid input is (say I restrict the input to +-1mV, power from +-10V, and have A=1000, no hitting the rail here), is it internally compensated, is it even voltage mode, etc. etc.

But we also need to agree on the definition of stability. I understand the intuitive appeal of that definition but, with respect to Merriam-Webster, I would interject that it has limited value. For example, is a resistive divider stable? I would argue it is but I am not sure how I use that definition to demonstrate it. I think a better, but still intuitive, definition is "bounded input gives bounded output" [1] but there are other choices. Wikipedia lists 8! [2] Also note, by this definition an oscillator is stable. (Probably by Merriam's too as I think they conceded periodic motion as a type of equilibrium.)

[1] https://en.wikipedia.org/wiki/BIBO_stability
[2] https://en.wikipedia.org/wiki/Stability_theory

 

[alan123hk]

The circuit as drawn is absolutely stable because an ideal opamp has infinite bandwidth and therefore has no phase delay so it can't be unstable

I agree that an ideal opamp has infinite bandwidth and therefore has no phase delay it must be unstable. We refer to the dynamic system when talking about stability. If the system does not have any time dependent characteristic, then it has no stability issues.

Therefore, I just try to approximate the dynamic system when I apply the loop gain to analyze the system which does not include any time-dependent characteristic.

This reminds me of the question raised by OP, we can't describe the behavior of dynamic system in reality without using a dynamic model.

To get closer to reality, I think it would be helpful to add a little delay (td) to the ideal OpAmp, just redefine the equation to Vout(t) = Gain*Vin(t-td), and then we should be able to get the desired results through analysis or simulation. There are of course many other methods, such as adding RC delay circuit at the output and so on.

 

[LvW]

......
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.

When a circuit with feedback is unstable - it is unstable and, hence, cannot be used as a linear amplifier...and it does not matter if the input signal is small or large or if it is DC or AC...
* DC instability (static) means: No fixed (stable) DC operating point within the linear transfer range
* AC instability (dynamic) means: Self-excitement (oscillations) around a stable bias point.

 

[sophiecentaur]

The circuit as drawn is absolutely stable because an ideal opamp has infinite bandwidth and therefore has no phase delay so it can't be unstable.

It's surely not so much a matter of stability as Linearity. The output of any amplifier will eventually be limited by the power supply volts. When there is positive feedback, the input volts need to go far enough 'the other way' to pull the output volts away from the supply rail and onto the other one. It is hard to remove the time dependent aspect of the circuit and the input signal would have to change fast enough (and with prediction) to bring the output volts back to Zero.
As has been mentioned already, it's the problem of balancing a pencil on its point - but worse; it involves getting the pencil upright from a near-horizontal position and then catching and maintaining it in the vertical position.

 

[eq1]

As has been mentioned already, it's the problem of balancing a pencil on its point

Yes. I agree. It's exactly analogous to that. The problem originally presented is the equivalent of asking is there a mathematical solution to balancing an idealized pencil on an idealized surface to which I answer, yes, there is a mathematical solution to that idealized problem. Does that mean I can balance a pencil on my desk right now? No, it does not. But that idealized problem does have a solution.

it involves getting the pencil upright from a near-horizontal position and then catching and maintaining it in the vertical position.

Maybe. Then again. Maybe this pencil is in 0g and it's already upright. We don't know because those details have not been provided so why assume them. And that's my point and why I answer it depends. If one puts a 741 in place of the ideal opamp will it be stable? No definitely not. So maybe don't put a 741 there. What circuits are legal to place there? Unknown. The problem does not specify that.

 

[LvW]

As has been mentioned already, it's the problem of balancing a pencil on its point - but worse; it involves getting the pencil upright from a near-horizontal position and then catching and maintaining it in the vertical position.

There is a similar problem:
There are some circuits with feedback fulfilling Barkhausens oscillation criterion - however, the circuit does NOT oscillate (real opamp model), but goes into saturation. But for an IDEAL model it oscillates (in simulation). Tricky situation - in particular to find an explanantion.

 

[jim hardy]

i repeat

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.

 

[jim hardy]

I just mean that Vout = -(R2/R1)*Vin given by steady-state analysis doesn't mean we can realize a stable linear amplifier with that relation.

Maybe. Then again. Maybe this pencil is in 0g and it's already upright. We don't know because those details have not been provided so why assume them. And that's my point and why I answer it depends. If one puts a 741 in place of the ideal opamp will it be stable? No definitely not. So maybe don't put a 741 there. What circuits are legal to place there? Unknown. The problem does not specify that.

Fair enough.

I think we all know how the circuit operates
and we realize that

That formula is true only at two infinitesimal points
It does not represent describe_{(is a better verb-jh)} the circuit's behavior at any other point.
So it cannot be used to evaluate the circuit for stability, linearity, or prediction of behavior.
In short it does not represent the circuit under discussion.

In other words, you can't use a single linear equation to describe a discontinuous function even if a computer simulation tells you it's okay.
You've got to either bound Vo as @NascentOxygen did in post #45
or use an inequality instead.

And as teachers we should encourage beginners to recognize that.

No more semantics and hair splitting on this one for me.

Over and out.

old jim

and @ still won't autocomplete when i click it, until after about six tries and a preview.

 

[sophiecentaur]

In other words, you can't use a single linear equation to describe a discontinuous function even if a computer simulation tells you it's okay.

Precisely and precisely. A good computer simulation could, however, recognise what's happening and spot the situation. The sort of simulation you usually come across is just like a student whose good at Maths but who never built anything.

 

[jim hardy]

I'm on an iphone so I can't mark up the image (or at least I don't know how to) but in second line from the top A*Vo/Vc is reduced to A*A in the line below. This is not a valid substitution because vo=A(Vb-Va)=A*Vb

thanks, now i see it

shoulda been $A=\frac{Vo}{Vb}$ instead.old jim

ps
@alan123hk Great graphic, it broke down the communication barrier.
Nicely Done !

 

[Abdullah Almosalami]

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-

Well sure. The saturation points are there, but it doesn't change my question.

 

[jim hardy]

but it doesn't change my question.

So my question then is what is the difference?

Are you asking "When you flip the inputs ?"

Meaning that the linear and continuous function inside the red rectangle no longer describes the behavior of your circuit
so any conclusions drawn from it are irrelevant.

You now need to describe the circuit with a nonlinear discontinuous function.

The plots of the two functions just happen to cross at two points as shown .

That's the difference.

 

[LvW]

Well sure. The saturation points are there, but it doesn't change my question.

Abdullah A. - does that mean that you have the feeling that your question was not yet answered? (In spite of more than 70 contributions?)

 

[Abdullah Almosalami]

Abdullah A. - does that mean that you have the feeling that your question was not yet answered? (In spite of more than 70 contributions?)

Oh my bad I was just working my way down the responses and I'm not even all the way through yet. But certainly my question has been answered a while back at this point! And with much gratitude for the amount of knowledge put in here beyond my expectations! I might even revisit this post after I've gone through more coursework and into the details of the internal circuitry of amplifiers and such to look at the responses from a new perspective once more.

 

[Averagesupernova]

I really don't understand why this thread has the momentum that it seems to. Switching the inputs of an op-amp from inverting negative feedback to positive feedback as shown in the very first post of this thread does in no way imply that their analysis should be the same. If one cannot accept this from the beginning then one needs to be prepared for a difficult road ahead. That is not to say that I have not been fooled by some strange op-amp configurations. I certainly have.

 

[jim hardy]

I really don't understand why this thread has the momentum that it seems to.

For me it has been a months long search for that good sentence.
But it took @alan123 's picture in post 63 to unlock my alleged brain.

I think perhaps many of us people who are drawn to science struggle for words ? I know i do.
I struggle for math, too
best i can do for this circuit is

$Vout = 12 X \frac{(\frac{Vout}{2} + {Vin}) } {abs(\frac{Vout}{2} + {Vin}) }$

......

but i digress... what counts is we got there. It took all of us, though.

Good thread !

old jim