I have to withdraw from this discussion, as it's clear that others seem 

to want to indulge in a strange sort of "modeling" that goes beyond the

simple facts. Indeed, I find it distressing that at this very late date in the

development of op amps (my company was founded on them, almost

50 years ago)  there is still anything resembling a mystery left to discuss.

I have a feeling that academics love to invent their own brand of magic,

in a desperate attempt to appear "original".

Please, gentlemen, set aside your magic wands, throw wide open your

classroom windows, breathe in the beautiful simplicity of your subject

matter, and thus "inspired" feel the power to move on to REALLY hard

problems. 

Barrie

The interpretation assumes ideal opamp and i think may not be suitable for circuit analysis.?

Baiju, thank you for the response. You give me the hope that there is still (young) people in RG who are interested in great circuit phenomena...

Regarding your first comment, I will repeat it again... and again... and again... When we talk about (pure) ideas, we always have in mind ideal components. Only when we begin to realize our ideas, we should begin to think about the realities and limitations of this world... Otherwise, we will never come up with something new...

So, in the case of an ideal op-amp, the negative impedance of its output (this sounds a bit strange:) would be exactly equal to the "positive" impedance connected between the output and the inverting input (the virtual ground will be exactly zero volts). In the case of a real op-amp (e.g., with a gain of 100000), it will differ from it with 1/100000 and the virtual ground will be about 100 µV. If you want to compensate even this negligible error, you probably will guess how to make the op-amp to increase its output voltage with this value... and thus you will arrive at the exotic VNIC... Can you guess how? If you want, we can discuss it here...

BTW, this week, my students and I conducted exactly this experiment in the laboratory. The two voltmeters in the attached picture show the voltage drop across the "positive" resistor and the opposite voltage across the negative "resistor". The middle voltmeter in the movie below shows the result of the resistance neutralization (the virtual ground) while the current continously changes (above 4.5 volts the op-amp is saturated):

http://www.circuit-fantasia.com/ResearchGate/Investigating_the_operation_of_VNIC.AVI

About whether this view is useful for circuit analysis ... I can not answer now ... I still enjoy this my insight and do not want to spoil my mood with practical aspects of its application:)

I totally agree with you and certainly impressed by the originality of your thoughts. I was only augmenting it.

Dear Baiju,

Yesterday, I discussed this unconventional negative impedance viewpoint at op-amp inverting circuits with my students. Тhus we turned the ordinary lecture in an exciting experience (for them and me:) I calmed those few curious boys (future computer specialists and your colleagues) that besides them and their teacher:), there is at least another inquisitive young man (from more than 100,000 specialists and scientists in the field of circuitry in RG) who shares this view...

I conducted this exciting lecture in a small nice room with my favorite 3-section blackboard where I exposed my ideas in a fully colorful way... Svetoslav (one of these boys) made the photo below by his phone...

Dear Mechkov,

Your thoughts have forced me to have a revisit with Franco.

The principle of virtual ground forces an opposite current/voltage to nullify the input voltage reflected at the inverting terminal. it is this effect of the negative feedback, resulting in the cancellation that appears as a negative impedance?

Dear Cyril,

you know that in circuit theory I'm less unconventional or rather quite conservative.

I saw your wiki link, that let's say, does not have the virtue of the synthesis.

Anyway following the reasoning in your above posts (same voltage, opposite current = negative resistance), also the VG in the below circuit expresses this unconventional negative resistance.

I know my reasoning seems dull so I will persevere and will demonstrate a thing that everyone knows, that this resistance is zero.

The circuit in your blue circle has three terminal so I'm a little confused which port should have negative resistance.

So if you intend, as I imagine, the output terminal and ground you know too well

that is zero.

Since you love schematics I have made one too, for the ideal OPAMP in its linear region.

As you can see in the schematic i1=i2; i1=0 (VR1=0); i2=0; vt=0; Rin = vt/it = 0;

I think it would be very hard for you regarding impedance to describe an op amp with a single component. I won't even go as far as talking sign I will just say module, input should be huge and output is small or as Simone Orcioni has explained even ideally cero, while mathematical abstractions sometimes do have their time and place. In this case I don't this oversimplification attempt will get you anything.

To view them as negative components is interesting I guess, but in the end it is just a mathematical abstraction that happens in the particular case you are handling. You cannot take that "negative" component and connect it to another circuit for example(do correct me if I am wrong) I guess it is a good way to show students(like me) what would a negative impedance imply somehow.

Dear Juan Jesus,

I would take advantage you are a student to show (as you rightly say) when mathematical abstractions does "have their time and place" and when a real negative impedance can be "connected to another circuit".

In the following, when the OPAMP is in linear region:

i1 = vt/R1 = i2 ; => vo = vt(1+ R2/R1) ;

(vt - vo)/R = it ; vt/R - vt/R - vt R2/(R*R1) = it ;

Ri = vt/it = - R*R1/R2 ;

other regions are easier to analyze.

Now that you have found a nonlienar bipole with current controlled characteristic with a region with negative resistance you can use it to make an astable multivibrator.

Dear colleagues,

Sorry for the delay... I had some problems and was not able to reply to your comments but I was thinking continuously on them... I shall now answer them consecutively...

Baiju,

I spend much of my life:) to reveal the mystery of virtual ground and to convince people that there is nothing supernatural in this circuit phenomenon. Here are some of my attempts:

https://en.wikibooks.org/wiki/Talk:Circuit_Idea/Virtual_Ground (an old Wikipedia discussion)

http://circuit-fantasia.com/circuit_stories/inventing_circuits/virtual_ground/virtual_ground.htm

Simply speaking, the virvtual ground is a point with an "artificial" zero voltage while the real ground is a point with a "natural" zero voltage. For example, we can create this "magic" point somewhere on a resistive film (e.g., inside a potentiometer or even a bare resistor) or between two resistors by making two voltage sources oppose each other:

https://en.wikibooks.org/wiki/Circuit_Idea/Walking_along_the_Resistive_Film#Creating_a_virtual_ground (a fancy story that I created together with my students in 2008)

Figuratively speaking, when the one (left) source "pulls up this point, the other other (riight) "pulls down" it so that it stays "immovable" (with zero voltage). Actually, the virtual ground voltage is a weighted sum of the two opposite voltage (i.e., a difference). Thus the right (following) voltage Vin2 is always equal and opposite to the voltage drop Vr2 across the right resistor r2... so it can be produced by a negative resistor with "resistance" -r2... or, by an op-amp with a "parallel-parallel" negative feedback... the result would be almost the same...

Dear Simone,

About your first comment... IMO in this world there is a room for both conventional and unconventional thinkers and they should not oppose each other but rather complement... However, the problem of the conventional "thinking" is that most often it does not involve thinking and understanding but sooner knowing and reproducting ready-made circuit solutions. Thus practically, "conventional thinking" people know circuits but do not understand them (maybe the most typical example are orthodox Wikipedians blindly conveying the knowledge).

Understanding is a deeply personal... and therefore unconventional process... To understand a circuit means to see the fundamental circuit ideas behind it, to make a connection between the particular circuit and other well-known solutions... even to see non-electrical manifestations of this ideas in the routine...

https://www.researchgate.net/post/What_does_to_understand_a_circuit_mean#share

... and finally to generalize these ideas into universal principles as Prof. Lutz Wangenheim suggests in the question about CFA...

https://www.researchgate.net/post/What_is_the_truth_about_the_exotic_current_feedback_amplifier_Is_it_something_new_or_just_a_well_known_old_Is_it_really_a_current_feedback_device

... to use later exactly for the purposes of the synthesis mentioned by you...

Simone, now about your thought, "Anyway following the reasoning in your above posts (same voltage, opposite current = negative resistance)..."

First let me note that "same voltage, opposite current = negative resistance" applies to the dual negative resistance circuit INIC (current-inversion negative impedance converter). Here "same current, opposite voltage = negative resistance" is valid, and we consider VNIC (voltage-inversion negative impedance converter).

Now about "also the VG in the below circuit expresses this unconventional negative resistance". IMO this issue deserves a separate question and I intend to ask it... but I will begin disproving this misconseption still here...

As far as I understand, this "negative resistance viewpoint at voltage sources" is the following. A voltage source is connected to a load (a resistor)... so the voltage V (VG in your figure) across them and the current I through them are the same... and therefore the ratio V/I (the resistance) for each element is the same... Thus the resistor has a resistance R = V/I and the voltage source has a "negative resistance" -R = V/-I... and the sum of the two resistances (voltages, according to KVL) is zero... See also the work of this Wikipedia guy (it is assembled entirely by else's thoughts extracted from reputable sources):

https://en.wikipedia.org/wiki/User:Chetvorno/work6#Negative_static_or_.22absolute.22_resistance

This concept sounds temptingly simple but...

In your arrangement, there is only one "main" voltage source and one resistor (the load)... and this is the possibly simplest electric circuit still from 19-th century - a source driving a load...

But "negative resistance" is a "supplemental" concept... It implies another (supplemental, "helping") voltage source (BH in the attached picture)... and this is not an ordinary constant but "dynamic" voltage source whose voltage is proportional to the current flowing through it (a 2-terminal current-to voltage converter)... and so it will act as a negative resistor with resistance -Ri. This negative resistance compensates another positive resistance Ri (e.g., the source internal resistance or the line resistance) thus giving as a result zero total resistance between the main source VIN and the load RL... and this 4-component circuit is reduced to your initial 2-component circuit... The sense of this "trick" is that the unwanted resistance Ri (the voltage drop across it) is neutralized by an equivalent voltage.

If this supplemental voltage source would an ordinary constant voltage source, it would still compensate the voltage drop across Ri... but only for one value of the current; maybe because of that they name this kind of "negative resistance" with the name "static negative resistance". Really, it can compensate also the relatively steady voltage drop across a constant-voltage nonlinear resistor (diode, LED, Zener diode, etc)... but this is just another special case...

Note that, in contrast with your "negative resistor", this voltage source will not independently produce voltage if there is no input voltage VIN; it starts acting after the main (input) voltage source begins increasing its voltage from zero.

IMO the word "resistor"/"negative resistor" has the meaning of something that resists/"helps" the current flowing through it... so it implies some initial current produced by another (main, input) voltage source... Therefore, this main source is simply a source, not a negative resistor... and maybe this viewpoint is just a misconseption as many others in the field of negative impedance phenomena... or maybe not? Maybe we can see somewhere examples of "selfexciting" negative resistors? Maybe Prof. Lutz Wangenheim would help us to see them in his favorite area...

Now let me dispel your confusion - "The circuit in your blue circle has three terminal so I'm a little confused which port should have negative resistance".

Yes Simone, this "op-amp negative resistor" is not a real negative resistor; it only behaves as (mimics) the genuine 2-terminal negative resistor (VNIC):

https://en.wikibooks.org/wiki/Circuit_Idea/Revealing_the_Mystery_of_Negative_Impedance#Mimicking_the_true_S-shaped_negative_resistance

The output of the op-amp (including the power supply) is the port having negative resistance; it is connected in series to the "positive" resistance that should be annihilated... and the current flows through this port. The inverting input of the op-amp serves only as a high impedance "sense" giving a chance to the op-amp to adjust its output voltage equal to the voltage drop across the positive resistance. So, if you want to convince someone that the op-amp is a negative resistor in this arrangement, you should carefully camouflage this third wire:)

As for your equivalent circuit of inverter implemented with an ideal op-amp and Juan's "input should be huge and output is small or as Simone Orcioni has explained even ideally cero"... sorry, but I can not understand what you mean and what the problem is... I do not see any problem with this arrangement...

Now about the second part of Juan's insertion:

"To view them as negative components is interesting I guess, but in the end it is just a mathematical abstraction that happens in the particular case you are handling. You cannot take that "negative" component and connect it to another circuit for example(do correct me if I am wrong) I guess it is a good way to show students(like me) what would a negative impedance imply somehow."

Exactly, Juan... you are right! And your way of thinking is remarkable! I needed you last Thursday when we, my students and I, were converting an inverting amplifier into a VNIC... first on the whiteboard...

... and then on the laboratory setup (MICROLAB system)...

So Juan, I cannot detach this "op-amp negative resistor" from the "positive" resistor connected between the op-amp output and its inverting input... they are sticked together like Siamese twins:) I can give you only the result of the neutralization - just a "piece of wire" (aha, maybe you and Simone wanted to tell me just that above)...

But fortunately, in many cases you will not want a "pure" negative resistance separated from the positive resistance; you probably will want exactly this pair of two opposite and mutually neutralized resistances (impedances)...

If you still want a separate negative "resistor", you should make the op-amp overcompensate the positive resistance... i.e., to convert the ordinary op-amp inverting circuit into the mysterious VNIC... Then a voltage (the inverted VRi) will appear across the op-amp inputs... and the virtual ground will "move" somewhere towards the input voltage source (between two adjacent resistors, inside a resistor or even inside the input source if it is imperfect)... as we observed it last week in the laboratory...

And finally, I want to show to Yuan, as he is a curious student, the power of the intuitive approach to reveal not only that something is done somehow... but how and why it is done exactly in this way. In the Wikibooks stories below, I have shown, step-by-step, what and how the op-amp does to create the mystic negative resistance in the middle part of the S-shaped IV curve of a VNIC:

https://en.wikibooks.org/wiki/Circuit_Idea/Revealing_the_Mystery_of_Negative_Impedance#An_N-shaped_negative_resistor_.28compare_with_N-shaped_NDR.29

https://en.wikibooks.org/wiki/Circuit_Idea/Linear_Mode_of_Voltage_Inversion_NIC

https://en.wikibooks.org/wiki/Circuit_Idea/Negative_Impedance_Converter

(Simone, to justify the name "S-shaped", X should represent V and Y - I; also, the inputs of the op-amp in the astable multivibrator circuit should be swapped)

Dear Cyril,

I allow myself to criticize but I don't want to acritically "oppose".

I have seen your response but it is sunday night, it's late, so I'll answer next days.

Anyway, about your last post, I know S-shaped bipole but I prefer to use current controlled (to bo loaded with a capacito) or voltage controlled (to be loaded with an inductor) bipole, so that their static characteristic is always a function.

Quote Prof. Cyril M.: "Maybe Prof. Lutz Wangenheim would help us to see them in his favorite area..."

I do not know if I can contribute much that is new about "negative resistance".

To me, this is not a mystical circuit at all ("mystical VNIC"?) but simply a voltage controlled current source - with a current that flows through the controlling source. Applying the definition of resistance to this arrangement, one comes to the concept of negative resistance (in contrast to a conventional positive resistor which can be seen as a voltage controlled current sink). That's all.

I think I should add something:

In one of the foregoing contributions it was said that the "concept of neg. resistance" can also be regarded as a "mathematical construct" only. For my opinion, this sounds reasonible because we are not FORCED to use this concept for describing the observed effects.

I rather think that the occurence of a negative resistance effect is always connected with POSITIVE feedback. Hence, we always can analyze such circuits applying the feedback theory.

This becomes obvious when analyzing oscillator circuits. We know that we can discriminate between two basic classes: Two-pole and four-pole oscillators. However, it is always possible to convert one class into the other one and vice versa (redrawing and/or new interpretation).

Sometimes such a conversion is not easy and, hence, does not help too much to understand the function of the circuit. And that is the reason, both concepts do exist in parallel.

@Lutz von Wangenheim: "simply a voltage controlled current source - with a current that flows through the controlling source"

for duality I think that also a "current controlled voltage source connected to the terminal of controlling current" can work.

@ Simone O: Do you have any circuitry in your mind?

@ Cyril: Either I am too old or to stupid, but I cannot follow your assertion that the opamp in your examples (see your first post) acts as a negative resistor. However, I must admit that I didn´t go through all the referenced links.

Therefore: Cyril, please can you explain this assertion with plain simple words? To me, in your example circuits the opamp is a negative voltage source driving a current which is in accordance with our classical definition for current directions (current sink).

More than that, combining a classical (passive) resistor with an (actively realized) negative resistance of the same value results either in a total resistance of zero (series) or infinite (parallel). I cannot see that one of these two cases applies to your example circuits.

My colleagues (friends), I read your writing above and I can not believe that, in RG is still possible to happen such a benevolent, open and honest discussion, in which no one is afraid to speak his mind even the risk being wrong ... that's what I have dreamed of and sought for so many years...

Oh, Lutz ... I must first apologize that I was not able to thank you for the wonderful suggestion in the other issue regarding CFA. I actually found a very powerful explanation of the basic idea there... but I decided to write something here ... and the ideas started popping up one after another... and I always could not get back to CFA... But I have finally clarified myself the negative resistance phenomenon, and what would be interesting for you, in the oscillatror circuits (Wien bridge oscillator)...

Let me explain really "with plain simple words" anything said by me, but you know that it takes time and a place ... Let's clarify once and for all the circumstances surrounding this so simple, but shrouded in mystery phenomenon... I just want to ask you not to be offended by the fact that I often put links to materials created by me and also attach images. This saves me the need to draw much on the whiteboard (which is generally better but there is no time to do it), and also allows readers to get better acquainted with my speculations...

@Lutz: this, for examples?

P.S. in the figure at bottom right, the abscissa should be -Ic.

But I don't find the original to correct it.

https://www.researchgate.net/file.PostFileLoader.html?id=5392e3dcd11b8b165b8b458f&key=3deec5392e3dc9fe18

Well Lutz, what is the main idea in this class of circuits? This is always the most important thing - to see the main idea.... and only then its possible implementations...

In my opinion, the main idea is to compensate the undesired voltage drop across a "positive element" (consumer) with an equivalent voltage produced by a "negative element" (source) connected in series... so, we can name it shortly "voltage compensation". As a result of this action, the overall voltage across this network of two elements will be zero... and if the one side of the network is grounded, the other will be so called "virtual ground".

To realize this idea, we should create somehow a "mirror copy" of the undesired voltage drop by including an additional varying voltage source. Broadly speaking, we can do it in two possible ways - with and without negative feedback.

"With negative feedback" simply means to adjust (even manually) the additional voltage so that the overall voltage across this network to be zero (to keep up the virtual ground). It seems it is the perfect "voltage compensation" since it does not depend on the value of the positive resistance (impedance)... and also, because of the huge op-amp open-loop gain, the virtual ground voltage is almost zero (but still it couldn't be exactly zero). This widely used "voltage compensation" techniques requires a third wire to sense the result of the comparison (virtual ground)...

"Without negative feedback" can be simply implemented by "copying" the undesired voltage drop by a voltage follower (with a differential input or ungrounded). As above, the compensating voltage does not depend on the value of the positive resistance (impedance) but it depends on the follower's gain...

Then "without negative feedback" voltage compensation can be implemented (in some special cases) simply by inverting the input voltage (not the undesired voltage drop)... This is a quite "mechanical" technique that requres another wire to "sense" the input voltage...

And finally, "without negative feedback" voltage compensation can be implemented in the most exotic way by using the current as a kind of a "transmission" that connects the two voltages. That is why, there is no need of a third wire and the compensating negative resistor is a 2-terminal element. Note that in this arrangement there is no negative feedback and to obtain a perfect voltage compensation, the two resistances should be equal...

Regardless of the type of compensation, the result is the same - the removal of the undesired voltage. But in the first case (inverting circuit) the op-amp achieves this result directly by adding an equal voltage while in the second case (negative resistor) it achieves the same result indirectly, by adding an equal negative resistance. Figuratively speaking, in the first case, it is "sure" that it has achieved its goal, while in the second case it only "implies" it... and if its negative resistance is not exactly equal to the positive, it is wrong...

Hello Cyril,

I must admit that I still have trouble in understanding. Maybe I'm too practical-oriented (and/or simple -minded). Therefore, a very simple example is perhaps helpful:

Inverting operational amplifier circuit with a closed-loop gain Acl = -2.

(Input: Vi=+1V and output Vo=-2V).

My questions (using your terms):

1.) Which voltage is "undesired" and WHY?

2.) How is this voltage "compensated" (meaning of the term compensation: Elimination of undesired effects).

3.) Which and where is the "overall voltage across this network" which turns out to be zero?

(Answering these questions will help me to understand your terms "undesired", "compensation", "overall voltage", "positive element", "negative element").

My interpretation/description of this circuit is as follows (and I kindly ask you to tell me if something is wrong):

An equivalent circuitry (based on an ideal opamp) is simply a series connection of two resistors (R, 2R) which is fed by two voltage sources (1V at the left node and -2V at the right node). As th result, the current is 3V/3R and the potential between both resistors is zero (virtual ground).

Finally, my original question was why /how - in your sight - the inverting opamp acts as a "negative resistor".

An excellent question, dear Lutz... but it requires an excellent answer as well:) You have the unique ability to ask very clear and precise questions!

I began preparing my answers to all your questions above... but until I finish, please have a look at the attached picture. It shows how my students and I converted a transimpedance amplifier into a voltage-inversion negative impedance converter (VNIC) a week ago. I drew it today on the whiteboard (after the final lab on microcomputers) and snapped it especially for all you contributing in this exciting discussion...

First, the voltage divider R4-R5 was unconnected so R3 and OA formed a transimpedance amplifer that was keeping up a virtual ground between R2 and R3 (point 3). Thus the positive resistance R3 was neutralized by the negative output "resistance" of OA. After we connected the divider R4-R5, the transimpedance amplifier became negative "resistor" (VINIC). Now the positive resistance R2 was neutralized by the negative resistance of the VNIC, and the virtual ground moved from point 3 to point 2. As a result, only the positive resistance R1 remained in the circuit... all to the right became zero (the "mix" of R2, R3, R4, R5 and OA behaved just as a "piece of wire")...

Now about your last "naive" questions (it would be well if our comments were numbered, dated or somehow labeled to cite them... but who cares about our desires here?)

In my opinion, this is a matter of interpretation ... Depending on how our brain is initially arranged by Mother Nature (formal, creative, imaginative, philosophical...) each of us will see what he/she can and want to see in this circuit...

So we can see here simply four elements in series - two resistors and two sources (or one composite) ... or a voltage divider driven by the composite source... or a resistive summer that sums the veighted voltages of the two sources ... or an inverting amplifier, in which the op-amp has produced such output voltage V2 so that the voltage at the middle point between the two resistors is zero (virtual ground)... or a shorted current source built by V1 and R1... or cascaded voltage-to-current and current-to-voltage converters... or something else...

My mind is arranged in such an odd way:) that I see, as in a "movie", how the same problem is solved in all these inverting circuits - how to make (in a clever way) the input resistor R1 act as a perfect voltage-to-current converter. I have spent (and continue spending) a lot of effort, time and nerves to make people see what I see. Here are only some of my stories:) where this problem is initially posed, and then the evolution of the same powerful idea is revealed in a form of "building scenarios":

https://en.wikibooks.org/wiki/Circuit_Idea/Voltage_Compensation#The_general_structure

https://en.wikipedia.org/wiki/Miller_theorem#Applications_based_on_adding_V2_to_V1

http://www.circuit-fantasia.com/circuit_stories/inventing_circuits/active_v-to-i_converter/active_v-to-i_converter.htm

https://en.wikibooks.org/wiki/Circuit_Idea/Passive_Voltage-to-Current_Converter#Imperfections

https://en.wikibooks.org/wiki/Circuit_Idea/Op-amp_Inverting_Voltage-to-Current_Converter#Problem:_The_real_load_affects_the_current

http://www.circuit-fantasia.com/circuit_stories/building_circuits/ammeter/op-amp_ammeter/op-amp_ammeter.htm

https://en.wikibooks.org/wiki/Circuit_Idea/Passive_Current-to-Voltage_Converter#Imperfections

fantasia.com/circuit_stories/inventing_circuits/transimpedance_amplifier/transimpedance_amplifier.htm

http://www.circuit-fantasia.com/circuit_stories/inventing_circuits/active_i-to-v_converter/active_i-to-v_converter.htm

http://www.circuit-

https://en.wikibooks.org/wiki/Circuit_Idea/Op-amp_Inverting_Current-to-Voltage_Converter#Problem:_The_internal_resistor_affects_the_current

.........

Hi Cyril - thank you for responding to my questions.

But I have the impression that an answer to my "naive" questions - numbered with 1...3 - is not as easy as I had hoped so..

Therefore, your references to several wiki contributions?

But I hope you understand that some of your assertions (as cited, for example, in my questions) require additional explanations.

New example:

Quote: "Thus the positive resistance R3 what neutralized by the negative output "resistance"of OA."

Operational amplifier with "negative output resistance" - without any explanation?

Have I missed this up to now? I am somewhat confused.

Well Lutz, here are my "naive" answers to your questions:) They are based on a heuristic procedure where we solve the problems and build step-by-step the inverting circuit (the convential approch is to try to find the answer in the final circuit solution).

1. In the inverting configuration (e.g., an inverting amplifier), the input voltage source and the Element 1(the first resistor R1) constitute a simple current source that, to work properly, needs to be shorted. But we want to consume and convert this current into voltage; that is why we connect the Element 2 (the second resistor R2). Its resistance introduces this "undesired" (for the input source) voltage drop VR2.

https://en.wikibooks.org/wiki/Circuit_Idea/Voltage_Compensation#A_problem_appears

2. To solve the problem, we have to remove this voltage (to zero the resistance R2). But we (the load) need it... it is "desired" for us... so we cannot simply zero (short) it. We must do it in a rather paradoxical (at first glance) way - this voltage (resistance) must both exist (seen from the side of the input source) and do not exist (seen from the load side).

We can solve this contradiction by "virtually zeroing" R2. For this purpose, we add an additional voltage equal to the voltage drop VR2 thus compensating (removing, destroing, neutralizing, annihilaing...) it... and then use the compensating voltage VR2 as an inverted output.

3. The mentioned network consists of two elements in series - the resistor R2 and the voltage "source" VOUT. The first element R2 subtracts ("eats", consumes...) a voltage (drop) VR2; the second element VOUT adds the same voltage VR2. So, the "overall voltage" across the network and the overall resistance are zero... i.e., it behaves as a "virtual piece of wire":)

...

The op-amp output produces voltage VOUT = VR while the resistor R2 (R in the attached figure) consumes voltage VR... and the same current flows through them... So the op-amp output can be considered as a negative "resistor" with resistance -R2.

As we discussed above, this negative resistance cannot be used alone; it is firmly coupled with the positive resistance R2 and neutralizes it. We can see and use only the result of this neutralization (zero voltage and resistance, or the virtual ground)...

Dear Cyril,

I have read the following link:

https://en.wikipedia.org/wiki/User:Chetvorno/work6#Negative_static_or_.22absolute.22_resistance

and now I agree that, with the definition of static or chordal resistance R=v/i in opposition of the differential resistance r = dv/di, a Voltage generator loaded with a passive resistor shows a negative chordal resistance.

I don't really understant the utility of this, except for showing that in this case the generator dissipate a negative power (that could also be shown in a plain manner by calculating the power from its definition, anyway ...)

But now all the pages and schematic on the output negative resistance of the inverting amplifier with OPAmp, IMHO, can be summarized with the following syllogism:

A voltage source loaded with a passive resistor exhibits chordal resistance;

the output of the inverting amplifier with OPAmp behaves as a voltage source

then the output of the inverting amplifier exhibits negative chordal resistance.

I don't see the usefulness but I agree.

All my circuits and related considerations about negative resistance of course concerned the differential resistance r = dv/di.

kind regards

sim

P.S. the graphic in https://www.researchgate.net/file.PostFileLoader.html?id=5393831ed685cc757e8b465e&key=ef3175393831e73932

in your answer to a my point in the inverting amplifier, should regard the inverting amplifier object of the discussion but is quite singolar. The OPAMP is always in saturation.

Simone, thanks for your thoughts... I continue thinking about the question if the ordinary voltage source possesses negative resistance... and if a negative resistor can act alone... and I have the feeling that I almost realized it...

My last speculation about the usefulness of the odd negative resistance approach versus the more conventional negative feedback approach is that it is simpler - the circuit can be analyzed by the simple addition and subtraction instead multiplication and division... like logarithmic functions...

About the attached circuit - it is an ordinary transimpedance amplifier but powered by floating (ungrouded) power supply (because the load is grounded in this example). The op-amp adjusts its output voltage (between its output and the local ground) equal and opposite to the voltage drop across the resistor Ri so that their difference is always zero. The op-amp will saturate if the voltage drop across Ri becomes bigger than the maximum output voltage.

Here is an example of a "static negative resistor" that I used a year ago in the microcomputer lab. The enigmatic black "tablet" at the top of the prototyping board is a 1.5V alkaline battery element (to get it, I opened a 12V A27 battery for a car remote control). It is connected in series to the LED used as a load of a 3-bit hand-made DAC controlled by Port B of HC11 microcontroller. I camouflaged the tablet with a heat shrink tape and present it as a new electronic component:) ... and then asked them what is inside the tablet... and finally replaced it with an op-amp to obtain a transimpedance amplifier...

Hi Cyril, thank you for additional explanations.

And the good news is: OK - now I got it.

Correct me if I am wrong: The voltage you call „overall voltage“ is simply the sum of two voltages connected in a closed loop, right? And, of course, according to Kirchhoff´s voltage law (KVL) this sum is always zero.

Let´s take the most simple example consisting of a voltage source driving a current through a connected resistor. Of course, the loop voltage is zero.

But my question is:

Does it make sense to interpret this scenario as a connection of a positive and a negative resistance of the same value? More than that, are voltage and current directions of the voltage source in accordance with the DEFINIONS of a negative resistance? If this would be true, we could treat each voltage source in each circuit as a negative resistance, couldn´t we? I have some doubts if this would make sense.

*Quote Cyril:

"My last speculation about the usefulness of the odd negative resistance approach versus the more conventional negative feedback approach is that it is simpler - the circuit can be analyzed by the simple addition and subtraction instead multiplication and division".

My comment: Is it really simpler? What do your students think about it? We should not forget, that the purpose of the circuit under discussion is amplification. And amplification means: Multiplication.

As another argument in favour to the feedback approach: All the properties of the amplifier circuit (gain, bandwidth, stability,..) are determined by the feedback network (loop gain). Hence, we NEED the feedback approach for designing and analyzing such an amplifier.

Regards

Lutz vW

@Cyril about https://www.researchgate.net/file.PostFileLoader.html?id=5398af40d3df3e6d7f8b4569&key=f43a2e2a-7765-4e52-a645-5427f0920fb2.

Thank you for answer.

In my last reply, in a hurry I have connected RL and V+ to Vo.

Yes, it does. And it is often used in this manner. For example a high quality factor simulated inductor for narrow band resonant circuit filters. Look for the Gyrator at wikipedia...

"Yes, it does" ???

What does this mean? What are you referring to?

Simone, today it was interesting in the lab since a few of my students brought elements and a prototyping board to experiment with the well-known circuit of a generator using Schmitt trigger... actually it was your VNIC circuit above...

At the beginning, their circuit was not working because they were supplied it with a single source without an artificial midpoint. After we split the source (by a voltage divider), it began to act according to our expectations. Then we analyzed its operation in terms of negative impedance phenomenon...

Dear Lutz, I continue thinking about these extremely simple concepts but still I can not give fully convincing answers to these simple questions ... I need to think more on all this...

About the connection between negative resistance and negative feedback approach I would remind about a similar concept in transistor amplifying stages - presenting the bipolar transistor as a "transconductance amplifier"... "Transconductance" instead "gain" may sound strange but it is useful concept there...

To me, it does not sound "strange" at all. Remember the OTA concept (operational transconductance amplifier) working as a nearly ideal VCCS. To me, the BJT (as a stand-alone unit) is nothing else than a non-ideal VCCS (input and output resistances finite instead of infinite).

I would like to make a provocative addition:

To see a BJT as a VCCS will sound strange only for somebody who still believes that the BJT is an element which is controlled by the base current rather than the base-emitter voltage..

Lutz, the OTA concept can be the next extremely interesting issue... But first we have to finish the demystification of the CFA...

You know I like "provocations" and frequently use them both in our discussions and mental circuit experiments... So, as a (positive) reaction to your "provocation", I will add another simple argument in favor of the "voltage-controlled BJT" concept:

I think we can make the base current as we want small increasing the beta of the transistor right up to a superbeta... or applying an emitter degeneration... but (the range of) the base-emitter voltage stays always the same. Therefore, the current plays a secondary role towards the voltage...

As I said above, the "static negative resistance" deserves a separate question:

https://www.researchgate.net/post/Are_electrical_sources_elements_with_static_negative_impedance_If_so_is_there_any_benefit_from_this_concept

Lutz, I am referring to the main question.

Cyril, the bipolar transistor is very interesting when connected in reverse polarity (interchanged collector-emitter). It exhibit a negative impedance. You can change a slope by resistor connected between base and collector. Or by capacitor connected instead that resistor You can make a simple oscillator. Old transistors work better in this way probably due to different technological processes used in the past.

But maybe, this is an example of a differential negative resistor ("negistor")?

Hi Samuel, thank you very much for your interesting contribution. I must admit, I have never heard about these BJT applications. Did you already try to show these effects via SPICE simulation? Can you show us a corresponding circuit?

Dear Cyril and Lutz,

Please take a look here

Erik, it looks very interesting... Only, I do not unserstand why the gain A is (should be) time-variant...

About the mysterious "negistor"...

http://www.keelynet.com/zpe/negistor.htmhttp://www.keelynet.com/zpe/negistor.htm

... that is still a differential negative resistor...

Dear Cyril,

concerning: " I do not understand why the gain A is (should be) time-variant..."

When you analyze your circuits with op.amps.you simply observe it. !!

Do you believe that an op.amp. has a constant time invariant gain ???

Dear Cyril,

concerning:

http://www.keelynet.com/zpe/negistor.htmhttp://www.keelynet.com/zpe/negistor.htm

site not found, error 404

Dear Erik, when talking about basic concepts, we should assume an ideal op-amp...

Here is an unbroken link:

http://www.keelynet.com/zpe/negistor.htm

Dear Cyril,

Thanks a lot for the negistor link.

Concerning basic concepts the ideal op amp is useless in connection with oscillators. Because of virtual short circuit of the input terminals the concepts of positive and negative feed-back disappear !!! The signals we observe are the step response from the power supply input. An oscillator do not have a time-invariant DC bias point as we normally assume for our "linear" circuits (amplifiers, filters, etc.). Please prove that the ideal op-amp is an inverting amplifier as we normally just assume ;-)

Gentlemen (Cyril, Simone, Samuel, Erik,...),

I am afraid, the discussion get´s slowly somewhat "out of control". The main subject (title) concerns the question if systems with negative feedback "behave" as negative impedances - and if such an approach can be helpful.

I think, we all know how negative impedances are defined and how such units can be realized (NIC, GIC).

And we know relevant applications (Samuel Piovarci did mention simulated inductors and gyrators; and Erik has explained again the classical NIC realization ).

In the middle part of the discussion, we (including myself) even were discussing BJT properties. And - in this context - we came up with an uncommon application of the BJT which can be used as a negative resistor (new for me and interesting).

But, for my opinion, all these contributions did not at all touch the open questions related to the main subjects: Is it possible and does it make sense (for a better understanding) to interpret the classical inverting opamp amplifier as a "negative impedance element"?

I would only note that there is a special question about the negative differential resistance:

https://www.researchgate.net/post/What_is_negative_differential_resistance_How_is_it_implemented_How_does_it_operate_What_is_its_relationship_with_the_true_negative_resistance

Dear Cyril I just added an aswer to that question.

sim

Dear Lutz,

Thank you for your comments.

I agree that the discussion get´s somewhat "out of control".

Cite: All these contributions did not at all touch the open questions related to the main subjects: Is it possible and does it make sense (for a better understanding) to interpret the

classical inverting opamp amplifier as a "negative impedance element"?

A negative impedance element is a two terminal element (input port). The ideal classical inverting opamp is very useful in case of linear circuits. It is a four terminal element so the

answer is that it is not possible and it does not make sense to interpret this element as a negative impedance element.

The paradox is that positive and negative feed-back disappear because of virtual short circuit of the input terminals.

We deal with "input" elements i.e. two terminals which may be interpreted as controlled sources (VCVS, VCCS, CCVS, CCCS). The control is with signal (Voltage or Current) or time-derivative of signal.

We deal with "transfer" elements i.e. four terminals (two ports) which also may be interpreted as controlled sources. The negative resistor realized by means of an opamp and three resistors is a three terminal element because of the ground node in connection with the opamp power source.

The "input" and the "transfer" elements may be negative impedance elements.

Cite: "The paradox is that positive and negative feed-back disappear because of virtual short circuit of the input terminals."

Hi Erik - as you can imagine (you know me since several years) I agree to your main conclusions, of course. However, I do not completely understand the above cited sentence. I suppose, it is related to the classical NIC configuration (with opamp).

For me, it is vice versa: We have those feedback path´s - and as a result we observe that there is very small voltage across the inputs (called "virtual short", although it is not a short).

"A negative impedance element is a two terminal element (input port). The ideal classical inverting opamp is very useful in case of linear circuits. It is a four terminal element so the answer is that it is not possible and it does not make sense to interpret this element as a negative impedance element."

I completely agree with the above sentence: a resistor (either negative or positive) is a two-terminal element, whereas the opamp is a four-terminal device. Here, what happens in the output port (Vout=-Vr) is related to what happens to the input control terminals (+ and -) and is not a property ("negative resistance") of the opamp output in itself. As a matter of fact, the equivalent output resistance of the opamp, that is evaluated switching off al independent sources, is zero and is not negative.

On the other hand, it can be said that the "opamp and feedback resistor" block, in the ideal case, shows a zero equivalent resistance between the inverting opamp terminal (-) and the reference voltage. This is due to the effect of feedback that keeps v+-v- very close to zero by setting the output voltage that is required for that purpose.

Dear and respected all, who are involved in the discussion,

Here i would just like to share my own understanding regarding the original discussion that " whether a inverting op-amp in -ve' feedback can be considered as -ve' impedance synthesizer or not "

Well my judgement says no , it can't be considered so because the negative impedance is basically that device which when given a voltage at its input, pushes out current from its input instead of drawing it.

But here the system which we are talking about does not do so. It draws current when a reference voltage is applied at its input because of the virtual ground at the -ve' input terminal of the op-amp. Thus both the applied voltage and the produced input current appear to be in same phase and hence we can't consider it as -ve' impedance.

I do not know why our world is designed in such a way that even the nicest things, including my rest, eventually end... but I hope that our interesting discussions here, in RG, will never end:) Although my schedule was tightly filled with all sorts of health and entertaining procedures, I never stopped thinking (even in the mineral water:) on these so interesting questions...  Oh, a nice surprise from the RG team -  a picture viewer... thank you:) 

Watching the reactions of the participants here, I once again convinced how powerful dominant ideas are... how people obey them without question... and how people defend fiercely these doctrines.. if only someone tries to revise them... Conventional viewpoints are as religious dogmas... and new ideas are like a heresy that should be severely punished ... So I am grateful that I was born now and not in the Middle Ages because I would be sent to the stake dozens of times:)

Of course, it was just a humor... but now let me answer convincingly to your comments... and so to defend my "op-amp negative impedance" concept ...

Dear Lutz,

Really, our discussion(s) deviate(s) from time to time in other directions, but you well know that there is nothing wrong with that. At such moments, I usually formulate a new question and so I try to redirect that part of the discussion there. For example, the questions below are closely related to the current question...

https://www.researchgate.net/post/Are_electrical_sources_elements_with_static_negative_impedance_If_so_is_there_any_benefit_from_this_concept

https://www.researchgate.net/post/What_is_the_basic_idea_behind_the_negative_impedance_converter_How_is_it_implemented_How_does_it_operate_What_does_the_op-amp_do_in_this_circuit

https://www.researchgate.net/post/What_is_negative_impedance_Does_it_exist_If_so_how_can_elements_with_negative_impedance_be_implemented_Are_they_passive_or_active

https://www.researchgate.net/post/How_do_we_convert_imperfect_passive_circuits_into_op-amp_inverting_circuits_with_negative_feedback_What_does_the_op-amp_really_do_in_these_circuits

https://www.researchgate.net/post/Is_there_any_connection_between_the_humble_resistor_and_the_transimpedance_amplifier_What_does_the_op-amp_really_do_in_this_electronic_circuit

... and we can move some of our comments there...

Dear Erik,

I very much respect your lateral thinking... and I think we are very much alike in this tendency to present things in a novel way... but this most often brings us more trouble than positive emotions:) About your assertion below...

"A negative impedance element is a two terminal element (input port). The ideal classical inverting opamp is very useful in case of linear circuits. It is a four terminal element so the answer is that it is not possible and it does not make sense to interpret this element as a negative impedance element."

... I would clarify that here, saying "op-amp", I actually mean "the output part (port) of the op-amp". So, if there was enough space for a longer question, I would say:

"Does the output part of the op-amp in all the inverting circuits with negative feedback behave as a negative impedance element (negative "resistor", "capacitor", etc)?"

Note that in op-amp inverting circuits, only the op-amp output (port) affects the input source (through the negative feedback network)... the op-amp input does not affect the input source... In op-amp non-inverting circuits neither the output nor the input affects the input source (since the feedback network is not connected between the output and the input)...

Erik, now about your next thought, "The paradox is that positive and negative feed-back disappear because of virtual short circuit of the input terminals." This subject is more relevant to the question about NIC but it is interesting to begin discussing it here.

We can think of the 4-resistor NIC circuit (including the source input resistance) as of a bridge circuit supplied by the op-amp output voltage... and driving the op-amp differential input... Or we can think of it as of two voltage dividers connected to the inverting and non-inverting inputs. Really, if this bridge was perfectly balanced (R1/(R1+R2) = R3/(R3+R4), the difference between the negative and positive feedback network will be zero (perfect virtual ground). But, as Lutz has noted above, "there is very small voltage across the inputs"... and the polarity of this voltage is such that there is a small negative feedback...

I suppose this is (was) a very popular measurement arrangement; for example, I have considered it in the Wikibooks story about the Deborah Chung "apparent negative resistance" sensation:

https://en.wikibooks.org/wiki/Circuit_Idea/Deborah_Chung%27s_%22Apparent_Negative_Resistance%22#Presenting_the_carbon_fiber_network_as_a_bridge_circuit

In a NIC, the bridge circuit converts the single-ended op-amp output voltage into a differential op-amp input voltage. Depending on the proportion between the two divider ratios (>, < or =), this voltage-to-voltage converter can have inverting (negative feedback), noninverting (positive feedback) or zero (no feedback) transfer ratio. in the latter case (a balanced bridge), the four resistors do not play any role and the whole circuit is equivalent to the bare (open loop) op-amp...

And finally Erik, I cannot agree with your assertion that a NIC is a three-terminal element:

"The negative resistor realized by means of an opamp and three resistors is a three terminal element because of the ground node in connection with the opamp power source."

An op-amp is a 4-terminal (2-port) device but a NIC made by the same op-amp is a 2-terminal (1-port) circuit (look for example at the attached circuit diagram of a VNIC, where the input current source is connected across the single circuit port). Of course, the op-amp output (port) can be used as a second (output) port... and it is frequently used as in the case of the Deboo integrator... but this is not obligatory... The negative impedance appears at the input port (between the inverting input and the ground for the attached VNIC circuit)...

Dear Cyril,

I agree with your third answer to Erik post, but I did not completly understand the previous.

So I think that the challenge of Erik:

"The paradox is that positive and negative feed-back disappear because of virtual short circuit of the input terminals."

deserves further attention.

I agree with Erik about the interchangeability of the terminals of the OPAMP in the behaviour of NIC; but this is valid only when the OPAMP works in "linear region" (vd=0).

When the OPAMP saturates, the behaviour is likeably different.

Please see the attached file.

Hi Simone,

Thank you for your interesting insertion that interrupted my monologue and turned it into a dialogue:) This matter is so interesting to me that I am ready to discuss it ad infinitum:) BTW, if you can remember, I started this initiative (unveiling the mystery around NIC) in the beginning of this year by two stories written in parallel in ResearchGate and Wikibooks:

https://www.researchgate.net/post/What_is_the_basic_idea_behind_the_negative_impedance_converter_How_is_it_implemented_How_does_it_operate_What_does_the_op-amp_do_in_this_circuit

https://en.wikibooks.org/wiki/Circuit_Idea/Negative_Impedance_Converter

Yes, the Erik's assertions are original and really deserve further attention. I will only illustrate them by two conceptual "bridge" pictures...

... and their op-amp implementations... and will continue answering the remaining questions raised during my vacation...

With these pictures I also comment the NIC circuit diagrams in your pdf file above (the input source in the picture A should be a current type, and in the picture B - voltage type)...

Dear Crovetti, thank you for the insertion... it allows me to develop further my idea...

First about your thought, "I completely agree with the above sentence: a resistor (either negative or positive) is a two-terminal element, whereas the opamp is a four-terminal device."

Really, an op-amp is a 4-terminal device (component) having an input port and output port. And we can enrich it with additional components (e.g., feedback networks) thus obtaining another 4-terminal device (circuit). But, in some cases, we can use only two terminals (one port) without noticing the other pair of terminals. Some examples can be various op-amp constant voltage and current sources... or a transimpedance amplifier "without output" (i.e., its input is used as a virtual ground output). In the latter case we can even do not notice the genuine ground and so to consider this circuit as 1-terminal.

Now about your main assertion:

"Here, what happens in the output port (Vout=-Vr) is related to what happens to the input control terminals (+ and -) and is not a property ("negative resistance") of the opamp output in itself. As a matter of fact, the equivalent output resistance of the opamp, that is evaluated switching off al independent sources, is zero and is not negative."

Crovetti, the op-amp output resistance would be zero, if the op-amp output voltage stayed constant. But it does not stay constant... it changes in the opposite direction proportionally to the current flowing through it... and this is exactly a virtual negative "resistor" created by inverting the voltage... And to convince you and other "non-believers" in this fact, I just drew on the small whiteboard at home how the IV curve with a negative slope of this "op-amp negative resistor" is obtained (I feed last hope that the negative slope of this curve will convince you that the element having such an IV curve is an element with negative resistance:)

Actually, this is the typical Miller arrangement where the input voltage source is connected through a resistor to another varying voltage source (the op-amp output here):

https://en.wikipedia.org/wiki/Miller_theorem#Implementation

For concreteness, imagine a transimpedance amplifier (an op-amp with a resistor R connected between its output and inverting input) driven by an input voltage source Vin with internal resistance Ri. On the graphical representation above (superimposed IV curves), the green lines (slanted to the left) represent the moving IV curve of a real (input) voltage source with voltage Vin and total internal resistance Ri + R (I have combined the two resistances in series into one). The vertical red lines represent the moving IV curve of the op-amp output "voltage source" Voa. The blue line shows the trajectory of the intersection point (of the green and red lines)... and this is the S-shaped IV curve of the so controversial "op-amp negative resistor"...

In the beginning, assume that the input voltage is negative enough so the op-amp is saturated to the positive rail (Voa = +Vsat). If we begin increasing the input voltage Vin, its green IV curve begins moving (translating) to the right. Тhe op-amp remains saturated, and its red IV curve stays immovable in the rightmost position. The intersection (operating) point moves from point 1 to 2 thus picturing the bottom vertical part of the S-shaped IV curve. In this region, the op-amp output resistance is really zero as Crovetti said above.

The magic of the negative resistance appears after the point 2. The input voltage continues increasing and its red IV curve continues moving to the right. The op-amp enters the linear mode and begins acting - it begins vigorously moving its vertical red IV curve in the opposite direction (to the left). As a result, the operating point changes its trajectory to the left and up thus picturing the blue line with a negative slope (the IV curve of a negative resistor). Thus, in this faerie region (2-6), the input voltage and current increase but the op-amp output voltage decreases. The polarity of the op-amp output voltage is opposite to the voltage drop across the resistor but it is in the same direction with the input voltage (travelling the loop)... what is the behavior of a current-controlled negative resistor (VNIC)..

At point 6, the op-amp saturates again but now to the negative rail (Voa = -Vsat)... the magic of the negative resistance ceases... and the op-amp output resistance becomes zero again... The intersection (operating) point moves from point 6 to 7 thus picturing the top vertical part of the S-shaped IV curve...

Thus, during the holidays, I managed to get to the final insight that the operational amplifier in op-amp inverting configurations represents the simplest negative impedance converter with voltage inversion (VNIC). More precisely, it is a kind of a "negative impedance mirror" (NIM:) since it creates a voltage Voa that is a "mirror copy" of the voltage drop VR = I.R... what is nothing else than a negative resistor with "resistance" -R... In this arrangement, the current flows through two elements in series: a voltage drop VR = I.R is lost in the first element... so, it is a positive resistor; then a voltage VOA = VR = I.R is added by the second element... so, it is a negative resistor...

It is interesting to see that in the Miller arrangement, we always have a sum of two ingredients - positive and negative (we can think of them either in terms of resistances or voltages). If the positive one dominates, we have virtually decreased effective resistance... when they are equal, we have zero effective resistance... if the negative resistance dominates, we have negative effective resistance...

Dear Shuvadeep,

In your comment above you talk about the dual "current inversion negative impedance converter" (INIC) while here we discuss the "voltage inversion negative impedance converter" (VNIC). Regards, Cyril.

It seems this question is closely related to the question about the static negative impedance...

https://www.researchgate.net/post/Are_electrical_sources_elements_with_static_negative_impedance_If_so_is_there_any_benefit_from_this_concept?

... since in both the arrangements there are two elements in series with equal voltages - a source and a resistor... so their difference gives zero voltage - a ground. But here this ground is a virtual ground...

... while there it is a real ground...  

Hi Cyril - I am back from a 10-days trip to the eastern part of Turkey (very close to mount Ararat) and I like to share my opinion:

Quote Cyril: "In this arrangement, the current flows through two elements in series: a voltage drop VR = I.R is lost in the first element... so, it is a positive resistor; then a voltage VOA = VR = I.R is added by the second element... so, it is a negative resistor..."

I must admit that I cannot follow this explanation (....so, it is a negative resistor). Where is the justification?

To me, the left node of the feedback resistor sees zero voltage and the right node sees  -1V (in case of +1V at the amplifier input). Hence, the  current goes from left to the right - in full accordance with our definition of a positive resistor. 

So - where is a NEGATIVE resistor? 

(Comment: I appreciate the recent introduced editing capabilities).

Welcome Lutz! I see your break in the exotic places of Turkey is good for your creative enthusiasm:) I still prefer to go to the romantic ancient Sozopol to which I feel never-ending nostalgia...

Now about your question, "Where is the negative resistor?" The answer is simple - the op-amp output (including the ground as a second terminal) is the negative resistor.

As you correctly said, "the left node of the feedback resistor sees zero voltage and the right node sees  -1V (in case of +1V at the amplifier input). Hence, the  current goes from left to the right - in full accordance with our definition of a positive resistor."

Yes, the current goes through the feedback resistor creating a positive potential at its input terminal; so it is a  positive resistor. Then the current goes through (enters) the op-amp output creating the same but negative potential at its input terminal; so it is a  negative "resistor". The sum of the two equal resistances - positive and negative, gives zero total resistance...  zero voltage across the network of the both elements... and zero voltage at the left node of the feedback resistor (a virtual ground)...

Cyril - thus, you are saying that between the opamp´s output node and ground a negative resistance can be identified, correct?

However, for my opinion, this is in contrast to your own statement that a constant voltage source (CVS) cannot be interpreted as a negative resistance.

See https://www.researchgate.net/post/Are_electrical_sources_elements_with_static_negative_impedance_If_so_is_there_any_benefit_from_this_concept?_tpcectx=qa_overview_following

Exactly Lutz, the "resistance" (actually, the voltage) between the op-amp output node and ground is negative.

The op-amp output is not a constant voltage source when the input current (voltage) varies... so the op-amp output voltage is not constant; it varies proportionally to the current when the input voltage varies (the Miller effect). So there is no contradiction with the other question where the voltage stays constant.

An example... Imagine you vary the input voltage (as above) - the op-amp output voltage varies thus representing negative resistance. Then fix the input voltage at some value, connect a load to the op-amp output and vary its resistance - the output current will vary but the output voltage will not vary... the op-amp output will act as a constant voltage source with zero resistance...

Another example (from above) is when the op-amp output voltage reaches the supply rails it stops changing and the op-amp output begins acting as a constant voltage source with zero resistance. These moments form the vertical sections 1-2 and 6-7 of the op-amp's output S-shaped IV curve (in the conventional NIC they are inclined to the right because of the resistor R connected in series to the op-amp output)...

How nice is that there are at least two "circuit philosophers" in this crazy world where all know... but only few understand...

Cyril - I am sorry, but I do not understand your assertions. And - for my opinion - there is no real justification of these assertions.

Quote 1:.. the "resistance" (actually, the voltage) between the op-amp output node and ground is negative.

Quote 2: Imagine you vary the input voltage (as above) - the op-amp output voltage varies thus representing negative resistance.

I am aware of the fact that the output voltage changes as a result of an input voltage change. That is normal amplifier behaviour. But where is a negatve-resistive effect?

Dear Lutz,

Thank you for your support to this so interesting discussion. We have no choice - once we started this topic, we should definitely get to a final conclusion. I really count on your opinion as a distinguished professional in this field. I am a little upset because three hours ago I fell into the epicenter of an unprecedented hail in Sofia (ice chunks were the size of an egg) and my car was hit... but I'll try to collect my thoughts and answer...

Yes Lutz, in every amplifier "the output voltage changes as a result of an input voltage change"... but here, in op-amp inverting circuits (with shunt-shunt negative feedback), there is an additional factor - the output voltage "source" is connected in series with the input source (in the same direction) and with the feedback resistor. So, when the input voltage varies, the output voltage varies as well and affects ("helps") the very input voltage... it destroys the "positive" resistance R in the feedback... so it acts as an equivalent negative resistance R...

We can think of the transimpedance part of the inverting circuit as of a network of two elements in series - a "positive" resistor R and an op-amp output. The current through them is the same; so, we can think of both elements in terms of voltages or in terms of resistances:

So, this network is a pair of two "opposite" elements (positive and negative) but with equivalent quantities (currents, voltages and resistances). These two paired elements are inseparable as Siamese twins:) because the second is a result of (follows) the first... and destroys it... and we see the result of this neutralization (the virtual ground) at the inverting input...

The op-amp output voltage is a mirror (negative) copy of the voltage drop across the positive resistor R; so it is a negative resistor -R. What else be an "element" (circuit) creating a voltage V = I.R compared with an element  consuming the same voltage if not a negative "resistor"? Of course, it can be named "current-to-voltage converter" or "transimpedance amplifier"... or something else.... but it is a more special 2-terminal element since it is connected in the same input loop...

Cyril - I still cannot follow your explanation.

Let´s make a simple equivalent circuit:

1.) Assuming an inverting opamp based amplifier with two equal resistors leading to a gain of "-1". The input voltage is Vin=+1V.

2.) Now, the opamp ouput and the feedback resistor R behave like a simple series connection of a resistor R and a voltage source of Vout=-1 V.  This series combination is grounded at both sides (virtual ground on the left side).

3.) Of course, Kirchhoff´s voltage law is fulfilled because we have a current through R which is Vout/R (directed from left to right.). A very basic circuit.

4.) Question 1: Where is a negative resistance?

5.) Question 2: "the resistance ROA = -VOA/I of the op-amp output is negative"

My view: When a resistor R is connected to a voltage source V this resistor solely determines the current I. Thus, you cannot interpret the ratio V/I as an internal resistance (as a property) of the source. 

Just an interesting fact... In this inverting arrangement, the resistor R2 (R in your comment) does not determine the current I; it is determined by the input part (VIN, R1) of the circuit - I = VIN/R1. The resistor R determines the op-amp output voltage VOA = -I.R2. So, if we vary R2, VOA will vary as well... and the ratio VOA/R2 = I will stay constant. In terms of resistance, R2 - ROA = 0... so ROA = R2... and the transimpedance part is a kind of a "resistance mirror"...

Good morning, Cyril!

Yes - that´s true. Thus, the opamp output can be modelled as a current-controlled current source (controlled by the input current). Interestingly, this concept immediately leads to another representation of an inverting opamp amplifier based on a "pathological" element which can be found in the literature as "nullor" (combination of nullator and norator). 

Current-controlled current source? Maybe when R2 is the load?

Yes - because the input current (Iin through R1 from left to right) is identical to the output current (Iout through R2 also from left to right) and the output voltage is developped across R2.

Hence, the output part of the whole circuit can be modelled as a controlled-current source - controlled by Iin - with the value (Iout=-Iin) and the load R2.

Thus: Vout=(Iout)*R2= (-Iin)*R2=(-Vin/R1)*R2=-(R2/R1)*Vin.

Hmm... an odd CCCS... since the output current is actually the input current... this is a kind of a current follower...

Dear Cyril and Lutz,

A resistor is a current controlled voltage source  CCVS.

We measure the current in an element between node 1 and node 2.

We measure the voltage between node 3 and node 4 :    V(3,4) = R(I(1,2))

i.e. the voltage V is a function  R  of   I  :   V=R(I)

If we (a) short circuit nodes 1 and 3  and  (b) short circuit nodes 2 and 3

we have a two-terminal element R = V/I which may be positive or negative

according to the instant values of V and I

Have a nice and warm summer

All the best to all of you

ERIK

Dear Erik,

a CCVS is a two port. At the port 1 you have a short circuit, at the port 2 a voltage generator v2 = R*i1.

If you want to model a resistor as a CCVS you must measure the current flowing in a short placed in series with the voltage generator.

So port1 and port2 must be placed is series not in parallel.

Dually you can model it as a VVCS, with the port1 (an open) , that controls the current generator,  placed in parallel with the current generator itself (port2).

sim

Dear Simone,

concerning:

" a CCVS is a two port. At the port 1 you have a short circuit, at the port 2 a voltage generator v2 = R*i1."

At port 1 you have an element and NOT a short circuit. You may of course add

a short circuit to measure the current.

concerning:

"So port1 and port2 must be placed is series not in parallel."

Why  ?   When you couple the two ports in parallel the transfer resistance

becomes an input resistance !