"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 suppose, you are not surprised to read my answer: No - I don´t think so.

But it would be interesting to hear other opinions.

I received an email from RG that this question is modified by an editor. I note that the topics are changed with more specific and less popular. Do you think that the new topics are relevant to this issue? Static electricity...spectroscopy? I don't think so...

Well, I have corrected the topics with more related... Oh, another surprise from RG team - we can format the text... very well... thank you:)

It seems this question is closely related to the question about the "op-amp negative impedance"...

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

... 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 real ground...

... while there it is a virtual ground... 

Dear Lutz,

Regarding your questions, "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 would say:

No, it does not "make sense to interpret this scenario as a connection of a positive and a negative resistance of the same value"...

Yes,  the voltage and current directions of the voltage source are in accordance with the DEFINIONS of a negative resistance... Both the elements are voltage sources... but the first is a constant voltage source while the second is a varying voltage source. Inside both the sources, the current flows from the negative to the positive terminal while in the resistor it flows from the positive to the negative terminal... So there is no problem with the voltage and current signs and directions... there is no difference between the two arrangements in this respect... The difference is between the shapes of the IV curves of the two sources... let's see why in the graphical representation...

1. We have connected two elements (a sorce and a load) to each other; so the voltage VA across and the current IA through them are equal... and they are presented by the operating point A on the common coordinate system V(X), I(Y).

2. This operating point is a result of an intersection between the IV curves of the two elements. The load is an ordinary ohmic resistor; so its IV curve is a straight line inclined to the right...

3. But what is the other IV curve? We don't know... it can be linear, nonlinear, time-dependent... whatever... and we will not know what it is until we begin changing something in this circuit... Well, let's vary the resistance R... its IV curve begins rotating and the trajectory of the intersection (operating) point pictures the IV curve of the other element (the source). If it is a constant voltage source, the IV curve will be a vertical line representing zero (not 'static negative') resistance.

Of course (just a little regression), with the same "quantity varying" technique we can obtain the R IV curve. This is the idea of this graphical presentation - one of the two superimposed IV curves moves and pictures the other curve and v.v. We can name the moving curve "scannig IV curve" and the immovable curve - "scanned IV curve".

4. If the voltage source is not simply a constant voltage source but the more sophisticated self-varying voltage source whose voltage is proportional to the current through it (i.e., a VNIC) the trajectory of the intersection (operating) point will picture the inclined IV curve of a voltage source with negative internal resistance -R (simply, a negative resistor)...

So, the so called "static negative resistance" is an absolutely wrong concept and the Wikipedia material below only confirms what folly is to blindly transfer knowledge without thinking...

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

The truth is that all these elements (CVS, CCS, VNIC, INIC...) are sources whose common feature is that they inject energy into circuits. They can do this nice work having various IV characterstics - vertical, horizontal, inclined to the left or right... etc... To distinguish them, we call them with different names corresponding to the form of their IV curves - constant voltage source, constant current source, real voltage source, real current source, voltage source with negative resistance... etc...

So, the CVS is a constant voltage source having zero internal resistance; it is not a negative resistor having negative internal resistance... they are different devices... If it was a negative resistor, its IV curve would be inclined... not vertical...

The negative resistor is a kind of a "functional" voltage or current source mirroring the IV characteristic of the respective passive element (resistor)...

OK - now I can fully agree  (that a voltage source CVS behaves NOT like a negative resistor).

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Dear Lutz.

I also continue thinking about this concept named "static negative resistance"... My last insight about it is:

Indeed, we could define the ratio V/I for every 2-terminal electrical element including the voltage source, capacitor, inductor... We can do it even when we do not know what the element actually is (considering it as a black box). When we know that this element is a resistor, we name it "resistance"... but is it correct to name it in such a way when the element is not a resistor or a negative resistor (i.e., when its IV curve is an inclined line? I think NO!

If we do it, we should know that this is a chordal resistance. For example, in the case of a constant voltage source, this will be a line inclined to the opposite direction (towards the resistor's line). But if we change the load resistance, the intersection (operating) point will move and picture a vertical line which is the differential IV curve of the voltage source...

But still... "chordal resistance of a charged capacitor" sounds too strange...

The next comment is written by Lutz von Wangenheim (copied from the question below):

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

"Indeed, we could define the ratio V/I for every 2-terminal electrical element including the voltage source, capacitor, inductor... We can do it even when we do not know what the element actually is (considering it as a black box). When we know that this element is a resistor, we name it "resistance"... but is it correct to name it in such a way when the element is not a resistor or a negative resistor (i.e., when its IV curve is an inclined line? I think NO!".

Hi Cyril - yes, agreed. We can measure a resistance R by applying a voltage and measure the current. Can we do the same with a voltage source as a test object to find its property V/I ? No. Nevertheless, there is a current I through the source and a voltage V across the source. But it is not correct to attribute this current to the source (alone) because it is determined by the load. Thus, does it make sense to define a ratio V/I for the voltage source? I doubt.

The next comment is written by Erik Lindberg (copied from the question below):

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Dear Cyril and Lutz,

I still do not understand the concept of a "chordal resistance". Please provide me a definition i.e. an equivalent scheme and the corresponding equations. To me a resistance is a 2-terminal element for which we have an equation describing the voltage of the element as function of the current through the element (CCVS) V=R(I). In case we have a characteristic with more than one voltage value for one current value I prefer to handle the resistor as a conductor (VCCS) I=G(V) so we can avoid the "jumping" problems.

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Let's see what Wikipedia says:

https://en.wikipedia.org/wiki/Electrical_resistance_and_conductance#Static_and_differential_resistance

The next comment is written by Erik Lindberg (copied from the question below):

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Thank you so much Cyril !

Static conductance B (I/V). Also called chordal or instant DC conductance - This corresponds to the usual definition of conductance; the current divided by the voltage.

Dynamic conductance C (dI/dV) correspondig to the slope - the derivative of current with respect to voltage. Also called the instant dynamic conductance.

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Yes Erik... more precisely, they are conductances... But still, is there any sense in such a concept as a "source chordal resistance" (RA in the figure below)?

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

And here is an "animation" - when we increase Ri the "chordal resistance" RA increases in the same manner... Do not take seriously all this... rather think of it as of a some kind of fun:)

But this illustration is true and very useful if we replace VA with a Zener diode to illustrate how it keeps up a constant voltage VA across its terminals. In this case, the blue inclined line represents the chordal (static, ohmic, instant...) resistance RA of the Zener diode, and the vertical red line is its differential IV curve representing its zero differential resistance...

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

But we discuss the case with one source connected to a resistor...

So the question is, "What does the blue inclined line represent - the 'source chordal resistance'... or simply the resistance R?"

The next comment is written by Lutz von Wangenheim (copied from the question below):

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Cyril - according to your own explanations - it should represent both:

The slope 1/R represents

* the ohmic law for the load resistor R, and (at the same time)

* the negative "chordal resistance" (-1/Rc): hence: R=-Rc.

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Yes Lutz... but is there any sense of this representation? And why the 'source chordal resistance' is negative... since it still has the same straight line inclined to the right?

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Here is the graphical representation of the "source static resistance" drawn by the Wikiguy:

https://commons.wikimedia.org/wiki/File:Battery_IV_curve_showing_negative_static_resistance.svg

Lutz and Erik, frankly, do you understand something from it? Since I do not understand anything...

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Really, I remembered, when we superimposed the two IV curves - the black one of a load resistor (RL) and the red one of a voltage source (V) with negative internal resistance (-Ri), both they were inclined to right...

A question: Can we remove the constant voltage source V? If yes, its IV curve (red) should begin from the coordinate origin, right? And the operating point A will stay there as well? And both voltage and current will be zero? And what would be the use of this arrangement (a single negative resistor drives a resistor)?

The next comment is written by Erik Lindberg (copied from the question below):

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Dear Cyril,

concerning:

"Here is the graphical representation of the "source static resistance" drawn by the Wikiguy:

https://commons.wikimedia.org/wiki/File:Battery_IV_curve_showing_negative_static_resistance.svg

Lutz and Erik, frankly, do you understand something from it? Since I do not understand anything..."

If the current in the load R of the voltage source is negative the source is charged and we have a negative static resistor.

The next comment is written by Lutz von Wangenheim (copied from the question below):

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Cyril and Erik - I think the secret behind the negative values is as follows:

* In a load resistor the positive direction of the current is defined (it is only a DEFINITION!) from the positive potential to the negative potential. That means: Current and voltage have the same direction across/through the resistor.

* Now - applying the same definitions to the voltage-to-current ratio of the source we arrive at a negative current and a negative "chordal" resistance.

* However, I think it is questionable if the directions defined exclusively for a LOAD (in German: Verbraucher-Zählpfeilsystem) may be applied also for sources

The next comment is written by Erik Lindberg (copied from the question below):

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

For a 2-terminal element somewhere in an equivalent scheme of an electrical/electronic circuit we define a first node positive and a second node negative. We define the positive current of through element from the positive node to the negative node.

In case of a voltage source the current is negative if the source deliver a positive current to the load.

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

But let us return to our basic questions, "What is a negative resistor?"... and "Is the voltage source a negative resistor?"

Dear Erik, thank you for the explanations in the pdf file. The current and voltages there are completely real... "fictive" is only the "source chordal resistance" RA. I know the picture is not true... I have drawn it only as a "provocation":)

But let's now thoroughly consider the graphical representation in the work of the Wikipedian:

https://commons.wikimedia.org/wiki/File:Battery_IV_curve_showing_negative_static_resistance.svg

This picture represents only the IV curve of the voltage source. To understand the "source static resistance" idea, we need to see the whole picture - the graphical representation of the both elements (the source and the load) in a form of two superimposed IV curves.

So, I would like to ask you (Erik and Lutz): How is this IV curve obtained (what is the laboratory set-up)? Where the second IV cuve (of the load resistor) passes? Please superimpose the two IV curves to see where the operating point is... since it would be interesting to me to see how the load resistor IV curve (having a positive slope) will pass through the origin... and will intersect the source IV curve in its "negative resistance" part...

It seems my attempt to move this discussion to the more related question about "static" negative resistance was unsuccessful:) The "op-amp negative resistance" (the subject of this discussion) is much more real implementation of the true negative resistance idea...

The next comment is written by Lutz von Wangenheim (copied from the question below):

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Cyril - because the above I-V characteristic shows a Voc point (open circuit) as well as a point named with Isc (short circuit) I assume that we have a fixed voltage source and a variable load, OK?

This is in contrast to the "normal" I-V diagram with a fixed load and a rising voltage. In this case, I do not understand how the voltage can be larger than Voc (for rising I).

Hence, I do not understand the meaning of the whole diagram. Perhaps, only the red portion of the curve is to be considered?

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Dear Lutz,

I always read with pleasure your depth reflections... I suppose that you have already guessed what is the secret of the laboratory setup for measuring the whole IV curve of the voltage source with a voltage VOC and internal resistance RS - the load with resistance RL should contain another voltage source VL. When varying this additional voltage (or maybe, the resistance RL as you suggest), the source IV curve will move (rotate) and picture the whole IV curve of the voltage source VOC. Depending on the relation between the magnitudes of the two voltages, the current I will be I > 0 (when VL > VOC), I = 0 (when VL = VOC) and I < 0 (when VL < VOC).

This was my answer to your surprise "how the voltage can be larger than Voc (for rising I)." Only, I do not see negative resistance here. What I see are only two voltage sources (VOC and VL) with their positive internal resistances (RS and RL)...

I suppose the author considers the red section of the IV curve as a negative part since the voltage and the current have opposite signs (positive voltage and negative current). But is this sufficient to have a negative resistance? We should discover...

The next comment is written by Lutz von Wangenheim (copied from the question below):

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Cyril - could it be that we are trying to understand something which simply is false ("garbage")? At least, some additional explanations from the author are missing - and I don´t know if it makes sense to guess how to interprete his drawing. I cannot imagine that a second (hidden!) voltage source was used.

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

I think so... Some (maybe, reputable) authors have proposed that formal concept and people like this WP editor transmit it further without understanding... so, as you said, they miss important details... and the concept is distorted as in the phone game... and becomes even more absurd... I do not know why (maybe it is a kind of some obsession:) I keep looking for a convincing explanation for its nonsense and still can not find it...

I feel there is some absurd here in the following... The voltage source is considered as a negative copy of the resistor... as though the resistor determines the source... the resistor is the "master" and the source is the "slave"... But actually the causality is opposite - the source is the "master" and the resistor is the "slave":)

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

The second voltage (it can be current as well) source is necessary to measure the source IV curve in this part where we should reverse the current through the source (to "push" the current into the source)... and after the moment when the source is shorted and we want to continue exploring the curve... For this purpose, the "load" can be only a varying voltage source (if there is at least small internal resistance Ri) or a current source. Then its IV curve (the blue line) wil be vertical/horizontal. I have thought many times over this graphoanalytical method for visualizing the circuit operation... and widely use it when understanding and explaining circuits...

Here I would like to share one of my insights about this way of graphical presentation: No matter what the slopes of the two superimposed curves are; it is only important they to be opposite... or, in other words, one of them to be inverted...

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Another discrepancy - the Wikipedia picture of the IV curve shows that the "source static negative resistance" is equal to its internal resistance RS while the circuit diagram shows that it is equal to the load resistance RL...

Only see how many kinds of resistances are defined in the text below the picture representing the source IV curve:

"The current-voltage characteristic (IV curve) of an ideal battery, showing negative static resistance (red) in its normal operating range."

"The battery is modeled as a voltage source in series with a resistance Rs representing the battery's internal resistance.

"The battery's static resistance at a given point on its IV curve is equal to the inverse slope of the line from the origin to the point."

"The battery has positive differential resistance, equal to its internal resistance."

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Today, after a 1-hour bicycling in the park accompanied by a hard thinking:), I managed to finally find an answer to the question that torments us so long:

Are the voltage source and the op-amp output the same negative impedance elements such a VNIC (voltage inversion negative impedance converter)?

Here is my answer:

Yes, they are.... but to varying degrees... If I have to sort them in an ascending order of their quality, then we would have this:

voltage source ---> op-amp output ---> VNIC

Let's see why... For simplicity, I will consider the resistance instead the more general impedance... and will use static as a synomim of chordal (ohmic, linear) resistance...

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

1. The voltage source has a negative static and positive differential resistance.

This is the worst example of negative resistance only suitable for people who do not have even an elementary intuitive understanding of electrical circuits - a kind of "pure theorists" (which, in engineering, actually means "poor theorists":)

We can see such a negative resistance if we begin blindly measuring the voltage across and current through the elements in an electrical circuit... and composing the ratio V/I for each element... Thus we will see in some cases that the current enters the element... and the voltage at its input terminal is positive... So, we conclude that this is an element with positive resistance R = V/I...

In other cases, we will see that the current enters the element... but the voltage at its input terminal is negative... So, we conclude that this is an element with negative resistance R = -V/I...

In the possibly most elementary electrical circuit - a voltage source driving a resistor, we will see both the cases: the current enters the resistor... and the voltage at its input terminal is positive; then it enters the voltage source... but the voltage at its input terminal is negative... So, we will (can) conclude that these are two elements with equal but opposite resistances - positive (R = V/I) and negative (R = -V/I)... that netralize each other and the result is zero resistance (the real ground in the second picture below)... We can show this on the graphical representation below by drawing two lines one over the other (since |RS| = |RL|) that pass throgh the coordinate origin and cross somewhere the red pat of the source IV curve. The first line will represent the negative static source resistance RS, and the second - the positive resistance RL of the resistor...

But there is a problem - our circuit of two elements (a source and resistor) is a closed system... and we cannot use it... We cannot use neither the "negative source resistor"... nor the whole circuit of the two elements in series... because we can not be put inside the ground between the two elements... If we still do this (including other source there), we will see that the voltage ("negative resistance RS") of the first source will no longer be equal to the voltage drop (positive resistance RL) of the resistor when the current varies...

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

2. The op-amp (inverting ) output has both static and differential negative resistance... but that cannot be used independently.

This is already a good example of true negative resistance suitable for people having a good intuitive understanding of electrical circuits - a kind of "circuit philosophers" (which actually means "poor practitioners and theorists... but good thinkers":))

As in the case of the voltage source above, when measuring the voltage across and current through the op-amp output, we will see that the current enters the op-amp output (represented by the voltage source in the first picture below)... but the voltage at the "input" output terminal is negative... So, as above, we conclude that this is an "element" with negative resistance... and there are two elements with opposite resistances - positive (R2 = VR2/I) and negative (ROA = -VOA/I) = VR2/I... that netralize each other... and the result is zero resistance (the virtual ground in the second picture below)...

Note that now the lines representing the static and differential negative resistance of the op-amp output match... and we can show this on the graphical representation (the third picture) below by drawing the two lines one over the other...

Again there is a problem but now it is less - we cannot use independently this negative resistance... we can use only the whole circuit of the two elements in series... the result of the neutralization (the zero total resistance, the virtual ground)... Now we can be put inside the virtual ground between the two elements... and we have done it by including the input voltage source and the resistor R1. In contrast to above, we will see that the voltage ("negative resistance ROA") of the op-amp output is always equal to the voltage drop (positive resistance R2) of the resistor R2 when the current varies...

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

3. The VNIC has both static and differential negative resistance... and they can be used independently.

This is the best example of a really true negative resistance excellent for people who are both good circuit practitioners and theorists... perfect professionals... but not obligatory good thinkers":)))

The difference with the arrangements above is that here we measure the voltage across and current through the whole circuit... and it behaves as a true negative resistor... that neutralizes a part of the resistance of our input circuit... and the result is a decreased resistance...

As in the case of the "op-amp negative resistance", the lines representing the static and differential negative resistance of the VNIC match... and we can show this on the graphical representation (the second picture) below by drawing the two lines one over the other...

There is no any problem now - we cannot use independently this negative resistance...In contrast to above, if we have connected a positive resistor (R1) to the input, we will see that the voltage (negative resistance) of the VNIC is always equal to the voltage drop (positive resistance R1) of the resistor R1 when the current varies...

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

My final conclusion is:

True negative resistance elements (such as op-amp inverting output and VNIC) obligatory have (equivalent) negative differential resistance and negative static resistance. Ordinary voltage sources have positive differential resistance and negative static resistance; so they are inferior negative resistance elements.

Differential negative resistance elements (such as neon lamps) have negative differential resistance and positive static resistance.

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Today I saw that Google has set our discussion in second place (of total 2,240,000 results) on the result list when entering the string "static negative resistance"...

https://www.google.com/#q=static+negative+resistance&safe=off

... and in sixth place (of total 90,200 results) when entering "op-amp inverting amplifier negative resistance"...

https://www.google.com/#q=op-amp+inverting+amplifier+negative+resistance&safe=off

... which is still some kind of recognition for our efforts...

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Maybe you find it difficult to comment on the issue because there is an embarrassing duplication of "Rs" in the pictures of the Wikipedia material?

Thus in the text below the first figure, Rs represents the battery internal resistance...

http://commons.wikimedia.org/wiki/File:Battery_IV_curve_showing_negative_static_resistance.svg

"The battery is modeled as a voltage source in series with a resistance Rs representing the battery's internal resistance... The battery's static resistance at a given point on its IV curve is equal to the inverse slope of the line from the origin to the point. The battery has positive differential resistance, equal to its internal resistance."

... while in the next figure below, Rs represents the static resistance of a power source:

http://commons.wikimedia.org/wiki/File:Battery_and_resistor_circuit_2.svg

"Proof that the static resistance of a power source (RS) such as a battery is negative, and is equal to the negative of the static resistance of the attached load (RL)."

To remove the duplication, I suggest to designate the static negative resistance of the source with Rs, and its internal resistance - with Ri. Well, let's now present graphically the three typical cases of "static negative resistance" at VL = 0, VL >0 and VL < 0:

• The black line (with the red middle part) represents the internal resistance Ri of the voltage source.
• The blue line represents the resistance RL of the load. It is biased from the coordinate origin with the value of the load voltage VL.
• The green line represents the static (chordal) resistance RS of the voltage source.

Since the voltage V across and the current I through the two elements connected to each other (the real voltage source and Ihe load) are the same, their IV curves are superimposed on a common coordinate system. The intersection (operating) point represents these common quantities (V and I).

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

1. VL = 0. This is the case considered in the Wikipedia work - a real voltage source with an open-circuit voltage VOC and internal resistance Ri drives a resistor load with resistance RL.

As the load voltage VL is zero (there is no voltage source in the load), the IV (blue) curve of the "positive" ohmic resistor Ri passes through the origin and intersects somewhere the red part of the source IV curve. The IV (green) curve of the static (chordal) resistance of the source has a negative slope and lies on the RL IV curve since the two ratios V/I = |RS| = |RL|) are equal.

The current leaves the positive terminal of the voltage source VOC and enters the positive terminal of the load resistor RL; so the source is source and the load is load:)

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

2. VL > VOC. As I said above, to measure the source IV curve in its top part (above the X axis), we should connect an additional positive voltage source VL in series to the load resistor... thus we can reverse ("push" back) the current through the main voltage source.

As the load voltage VL exceeds the input voltage VOC, the IV (blue) curve of the "positive" ohmic resistor Ri is moved to right and intersects somewhere the top black part of the source IV curve. Now the IV (green) curve of the static resistance RS of the source has a positive slope and does not match the RL IV curve since the two ratios V/I (RS and RL) are not equal... so the voltage source has a positive static resistance here.

Now the current leaves the positive terminal of the load network and enters the positive terminal of the voltage source VOC; so the source is load and the load is source:)

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

3. VL < 0. Similarly, to measure the source IV curve in its bottom part (left of the Y axis), we should connect an additional negative voltage source VL in series to the load resistor... thus we can reverse the voltage across the load.

As the load voltage VL goes down below zero, the IV (blue) curve of the "positive" ohmic resistor Ri is moved to left and intersects somewhere the bottom black part of the source IV curve. The IV (green) curve of the static resistance RS of the source has a positive slope again and does not match the RL IV curve as above since the two ratios V/I (RS and RL) are not equal... so the voltage source has again a positive static resistance.

Interesting... what is source and what load here?

The next comment is written by Lutz von Wangenheim (copied from the question below):

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Cyril - as I have mentioned already before, I have severe doubts if we are allowed to apply the definition of a resistor to the V-I ratio of a voltage source. And - more important - does it make sense to do that?

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Dear Lutz,

I began to expose my view on this matter in Page 17 of this discussion (now we can link to certain pages of our discussions which is definitely a convenience; even greater convenience would be if we could link to certain comments)...

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc/17

... but I continue thinking about it...

The questions are: What does the ratio V/I (for an arbitrary element) express? Can we always name it "resistance"? What does it mean when named "resistance" and the element is not a resistor but something else - a source, capacitor, etc.?

The next comment is written by Lutz von Wangenheim (copied from the question below):

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Dear Cyril - on page 17 you have a simple circuit consisting of a parallel combination of a dc source and an ohmic resistor resulting in Rs=-RL.

If this would be true (if the dc source could be handled as a negative resitance) this parallel combination should have an overall resistance of infinite. Is this the case? Obviously not.

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Infinite or zero resistance? I think zero...

In my opinion, although it sounds quite strange, this possibly simplest Ohm's arrangement (in the first picture below) possesses something similar to current-controlled negative resistance (VNIC). Tо be convinced of this speculation, let's do an experiment...

First, let's break the circuit in the lower part (between the lower ends of the elements), connect an ammeter and measure the current flowing through it. Then let's replace the ammeter with a current source producing the same current as the measured one as it is shown in the second picture below. The total voltage (according to KVL) would be zero (a virtual ground at the bottom end of the resistor)... the total resistance of the network also would be zero...

Maybe the problem is that you consider the two elements as parallel connected while I consider them as series connected. Actually, they are neither parallel nor series connected; they are connected in a simpler (the most elementary) way...

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

But I continue thinking that the authors mentioned in the Wikipedia work consider the simplest Ohm's circuit as two equal but opposite resistors (positive and negative) connected in series (travelling along the loop). As though they have applied a kind of "resistance" KVL", RL - RS = 0... (KRL:) but not KCL ( RL - RS -> infinity)...

Current-controlled negative resistors are implemented by voltage sources in series while voltage-controlled negative resistors - by current sources in parallel. So, if this arrangement possessed some negative resistance, it should be the first type...

BTW what do you think about my explanation of why the input of the transimpedance amplifier behaves as an inductor at very high frequencies? It is based on the generalized Miller theorem:

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

The next comment is written by Lutz von Wangenheim (copied from the question below):

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Cyril - as you know, I have problems with your interpretation.

* If the left symbol is a voltage source (driving the current I) it has NO internal resistance Rs=-RL ,

* If the left symbol is a negative resistance -Rs, where is a voltage source driving the current?

Copied from the question below:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Yes, Lutz... these were the problems of this worst example of negative resistance only suitable for "pure theorists" :)

https://www.researchgate.net/post/Are_electrical_sources_elements_with_static_negative_impedance_If_so_is_there_any_benefit_from_this_concept/4

I would first specify that this interpretation is not mine; it is of this Wikipedian (more precisely, of the cited "reputable" sources in his work). Initially, I have rejected it but now I am inclined to accept some its aspects with some reservations...

We have not discussed this detail - the role of the internal source resistance Ri. IMO there is no connection between the internal resistance Ri and the static negative resistance RS of the source. Both the kinds of voltage sources (an ideal voltage source with Ri = 0 and a real voltage source with some Ri) possess the so called "static negative resistance". The only difference in the graphical representation below would be the slope of the black Ri IV curve - it would be vertical in the case of an ideal voltage source.

BTW this Wikipedian has made an error confusing RS (static negative resistance) with Ri (internal resistance) in his both figures in Wikimedia Commons:

https://commons.wikimedia.org/wiki/File:Battery_IV_curve_showing_negative_static_resistance.svg

https://commons.wikimedia.org/wiki/File:Battery_and_resistor_circuit_2.svg

RS has nothibg to do with Ri. I would correct it but this person will react negatively... if you want, you can do it (everyone can edit Wikimedia:)

So, regarding your first insertion...

"* If the left symbol is a voltage source (driving the current I) it has NO internal resistance Rs=-RL" - there is no problem

... and regarding the second thought...

"* If the left symbol is a negative resistance -Rs, where is a voltage source driving the current?" - the absurd is that the same "voltage source drives the current" and serves as a negative resistor... It's like a vicious circle...

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

--------------------------------------------------------------------------------

Yes, Lutz... these were the problems of this worst example of negative resistance only suitable for "pure theorists" :)

https://www.researchgate.net/post/Are_electrical_sources_elements_with_static_negative_impedance_If_so_is_there_any_benefit_from_this_concept/4

I would first specify that this interpretation is not mine; it is of this Wikipedian (more precisely, of the cited "reputable" sources in his work). Initially, I have rejected it but now I am inclined to accept some its aspects with some reservations...

We have not discussed this detail - the role of the internal source resistance Ri. IMO there is no connection between the internal resistance Ri and the static negative resistance RS of the source. Both the kinds of voltage sources (an ideal voltage source with Ri = 0 and a real voltage source with some Ri) possess the so called "static negative resistance". The only difference in the graphical representation below would be the slope of the black Ri IV curve - it would be vertical in the case of an ideal voltage source.

BTW this Wikipedian has made an error confusing RS (static negative resistance) with Ri (internal resistance) in his both figures in Wikimedia Commons:

https://commons.wikimedia.org/wiki/File:Battery_IV_curve_showing_negative_static_resistance.svg

https://commons.wikimedia.org/wiki/File:Battery_and_resistor_circuit_2.svg

RS has nothibg to do with Ri. I would correct it but this person will react negatively... if you want, you can do it (everyone can edit Wikimedia:)

(I remember again thay there is a mistake in the attached figure below - the blue line represents RL but not Ri; I can't correct it as it does not exist on my whiteboard)

So, regarding your first insertion...

"* If the left symbol is a voltage source (driving the current I) it has NO internal resistance Rs=-RL" - there is no problem

... and regarding the second thought...

"* If the left symbol is a negative resistance -Rs, where is a voltage source driving the current?" - the absurd is that the same "voltage source drives the current" and serves as a negative resistor... It's like a vicious circle...

To me - a resistor is an element (passive, or active in the negative case) that does REACT with a current upon an external voltage source - thereby fulfilling Ohm´s law. As I have mentioned earlier - I think, this is not the case for the schematic above if the dc source is considered as a negative resistor.

... or an element that does REACT with a voltage upon an external current source - thereby fulfilling Ohm´s law in its dual form V = I.R... and this is the case discussed in:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc

....and this is the case.....

Really? Can you justify it?

I have already justified it in a form of four conclusions at Page 19:

https://www.researchgate.net/post/Does_the_op-amp_in_all_the_inverting_circuits_with_negative_feedback_behave_as_a_negative_impedance_element_negative_resistor_capacitor_etc/19

Well, Lutz... let's put the question in a reverse way: I have shown why the op-amp output possesses negative impedance; you now show why it does not possess negative impedance:)

Lutz, I continue to think on your thought that "a resistor is an element (passive, or active in the negative case) that does REACT with a current upon an external voltage source"... and my addition that it can be also an element that does REACT with a voltage upon an external current source...

Yes, you (we) are absolutely right... let's consider the two dual resistor connections...

1. A resistor in series: the positive resistor prevents the current by subtracting a voltage drop from while the negative resistor helps the already existing (external) current by adding a voltage to the exciting voltage source.

2. A resistor in parallel: the positive resistor prevents the voltage by subtracting a current from while the negative resistor helps the already existing (external) voltage by adding a current to the exciting current source.

And so, is the ordinary voltage source a negative resistor? Or more generally, does it possess negative impedance?

Despite all our objections to this claim, we must recognize that still there is a formal but very simple and strong argument in its favor: as the current can be negative (-I)... their product (the power P = -I*V) can be negative... and therefore their ratio (the resistance R = V/-I) could be negative as well... So, as a voltage source has a negative current, it has a negative power and also a negative resistance...

If in the simplest (Ohm's) arrangement - a voltage source drives a resistor, we begin changing the voltage, the current will change proportionally... and their ratio (its negative resistance -R) will stay constant... So, it seems we can consider the varying voltage source as an acting alone negative resistor with "resistance" -R whose absolute value is equal to the value of the load resistance R...

So it seems we can make a conclusion that, in the simplest (Ohm's) arrangement of only two elements - a voltage source and a resistor R, we can consider the ordinary voltage source as a negative resistor with resistance -R...

But when we include another source, the situation becomes quite different...

Now, when we begin changing its voltage, again the current will change proportionally (to the difference between the two voltages)... but the first voltage source will not change its voltage (it is a constant-voltage source)... so its ratio V1/-I (its negative resistance -R) will be zero... So, in this case, the first voltage source has steady zero differential "resistance"... and it is not a negative resistor...

we must recognize that still there is a formal but very simple and strong argument in its favor: as the current can be negative (-I).

My counter argument: The sign of a current (positive or negative) is simply a matter of DEFINITION. And it is commonly agreed that the direction of a current is considered to be positive if it goes OUTSIDE the voltage source (through the load) from the positive potential to the negative potential of the source. This implies, of course, that this current within this source goes from "-" to "+". From this, it follows (because it is the same current in both cases) that this current must NOT be considered as "negative".

Yes Lutz, that is why I do not like to say "negative current"... But here is more important that the two quantities (voltage and current) have opposite signs... so their product and ratio are negative...

Maybe this is better:

"... we must recognize that still there is a formal but very simple and strong argument in its favor: as the voltage can be negative (-V) while the current is positive (I)... their product (the power P = I*-V) can be negative... and therefore their ratio (the resistance R = -V/I) could be negative as well... So, as a voltage source has a negative (with respect to the current) voltage , it has a negative power and also a negative resistance..."

Let´s stick to your simple example (battery and load resistor).  Indpendent how you see the battery (reverse or not, positive or negative) my argument still holds: The current which is considered to be positive goes within the voltage source from "-" to "+". Although you do not like the term "negative current" you are using a minus sign in your calculations.  And this is - as far as I think - not in accordance with our common definitions.

Lutz, all the (op-amp) examples of these questions consider a curent-controlled negative "resistor" where a positive current enters the negative terminal of a voltage source (the op-amp output).

Cyril - may I propose not to jump from one example to another one? I am afraid, otherwise we will not come to any conclusion.

Therefore, I quote again from one of your last contributions:

So it seems we can make a conclusion that, in the simplest (Ohm's) arrangement of only two elements - a voltage source and a resistor R, we can consider the ordinary voltage source as a negative resistor with resistance -R...

My response was: There is only one current flowing through the loop which is defined as positive through the voltage source from "-" to "+" and outside the source (through the resistor) from "+" to"-".  

Two questions:

(a) Is this correct? (b) What is the justification for a negative sign?

Yes, the current is positive... but they say the voltage is negative since the current enters the negative terminal of the element (source) while, in the case of the resistor, the current enters the positive terminal of the element (resistor)...

IMO this is quite artificial, formal and sterile... and is a kind of a vicious circle... since the current is produced by the very source, not by some external source. But in the case of a true negative resistor (including the inverting op-amp output), everything is fine since the current is produced by an external (the input) source... There the current enters the negative terminal of the second source (the negative resistor)...

...but they say the voltage is negative since the current enters the negative terminal of the element (source)

Who says that? Until our discussion about this subject has started  I never have heard  or read about such an assertion.

Dear Lutz,

From my student years I know that the voltage across and the current through a voltage source have opposite signs... so the power is negative, while the voltage drop across and the current through the load (resistor) have the same signs... so the power is positive... Of course this is a convention that allows us to distinguish the source from the load... IMOmore important for young students is to have an intuitive notion about this difference...

Dear Cyril - does this mean we/you have now an answer to the main question :

Are electrical sources elements with "static negative impedance"? If so, is there any benefit from this concept?

I have not found the answer to this question... and I continue thinking...

But Lutz, if really the two "new" Kirchhoff's laws triumph, I suggest you we share the glory:) because in these  discussions (mainly between us) this idea was born. New circuit laws are not born so often from the Ohm's and Kirchhoff's time... and they deserve to be published in a reputable source (e.g., IEEE Transactions on Circuits:) At least we should ask two dual questions in RG to see the reactions of the "luminaries" in electrical engineering:

1. Can we formulate a Kirchhoff´s law for conductances (KGL) based on the Kirchhoff´s law for currents (KCL)?

2. Can we formulate a Kirchhoff´s law for resistances (KRL) based on the Kirchhoff´s law for voltages (KVL)?

Regards, Cyril

If we decide that we will consider the voltage-to-current ratio of a voltage source as a resistor, I think - yes - we can formulate such a KRL. In this case, it will follow automatically from the KVL.

However (of course, now comes an "however") - regarding the second part of the question (...benefit from this concept) I have severe doubts.

For example. we shouldn´t overlook the fact, that in case of two voltage sources connected in series (but with opposite signs) one of the sources (the smaller one) suddenly again will have a positive resistance. Thus, the question if a voltage source "has" a positive or a negative resistance depends on the configuration. And that, certainly will not simplify anything. But the KRL will be satisfied always.

Very interesting observation - negative resistors have their own polarities while positive resistors adopt the polarity forced on them by the current... Congratulations, Lutz!

Obviously, negative resistors are more complex elements than the positive resistors - there is only one type of positive resistor while there are two types of (true, absolute) negative resistors - current- and voltage-controlled... and they can be connected in two possible ways - in the same or in the opposite direction... and their behavior depends on the way of connection... very interesting...

In addition, true negative resistors are too capricious - they can operate in a stable (linear) or in an unstable (bistable) mode... and all this is because these "resistors" are not resistors:)... they are sources... but a special kind of sources... they are active circuits...

Dear Lutz and other colleagues,

I spent a week in a small village in the center of the country, away from the city noise and without an internet... i.e., almost in the conditions in which Kirchoff formulated his famous laws nearly two centuries ago:) Temperatures were unusually high (almost 40 degrees) and during the hot afternoons I hid in a tightly closed room with a big fan and I wrote these lines on my laptop (only this gadget broke the illusion of the Kirchoff's time:)...

Lutz, I decided it is time to inform the scientific community about our idea to modify the Kirchhoff's laws that was born in the process of heated discussions. I suggest you do it in the form of an "imaginary article" that to write together as an open project... and then offer it in a proper magazine (at least, we can place it in our profiles as raw data)...

Such an open project can be a web page (module) in Wikibooks but the problem there is that every casual visitor can edit it. So I preferred to upload it in my Google cloud storage, where we can edit it together... and even it can be available to other RG members... Here is the link to the initial (uncompleted, raw) version that can be edited by anyone who wishes to do so:

https://docs.google.com/document/d/1s-TlRXC3U6cv-MHL5Oc0R6VX2bUBkCxV36Bp-WIKHHI/edit?usp=sharing

I suggest you we continue to comment here, in our discussions, of what we can write in the article. This can serve as a good certificate of our personal contributions as well...

As regards the authorship, I think the idea is mainly the result of discussions between you and me here... but if someone else has a claim to authorship, we can include him/her. I think the order of writing the author names corresponds to their contributions at the moment... but we can change it dynamically...

After these organizational issues only remains for us to see if the idea is still true and if it is worth:) I suggest to do it in a separate question:

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

Regards, Cyril

HimCyril - I am back. But due to some personal circumstances presently I have not enough time   to study your new contribution. Sorry about that.

Regards

Lutz

Welcome Lutz! I hope you rested well. I will rest next week when I hope for a good weather.

Regards,

Cyril

Cyril - thak you. I was on a ship and have visited Gdansk, Riga, Tallin, St. Petersburg, Helsinki, Stockholm. Not really relaxing- but interesting. All the best for your vacation.

Finally, I have asked the extremely interesting question about how to create virtual electrical elements in electronics:

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