Hello Cyril - I understand this contribution as a kind of "generalization" of your previous statements. Nevertheless - for my opinion, still without sufficient justification, I am sorry.

I have given the following example already earlier: Between wo grounded DC sources of opposite polarity we have two equal resistors. Thus, in the middle we will have a zero potential (not only "virtual" but real ground). According to your logic this is an indication that there must be a negative resistance which "neutralizes" one of the real resistors, right?.

Hence, we are again at the question (already earlier discussed): Are we allowed to say that a DC voltage source is equivalent to a negative resistance?Just because the current through the source has the opposite direction if compared with a "normal" load resistor?

That is the basic question - all other examples (with or without feedback) can be reduced to this question, I think.    

And still my answer is NO. 

Hi Lutz,

I am glad to see that you have not lost interest in the underlying circuit ideas... and we can continue to unveil the mysteries around them... Let me illustrate your example above by an attractive picture from the familiar Circuit Idea story. It represents the voltage distribution along the resistive film of a potentiometer (an analog version of your discrete example of two equal resistors). The funny "negative feedback" game illustrates how the virtual ground is obtained and maintained:

https://en.wikibooks.org/wiki/Circuit_Idea/Walking_along_the_Resistive_Film#A_negative_feedback_game

Let's share the roles and begin to play:) I suggest you to control the left voltage source VIN1... and I - the right VIN2...

If you set +1V from the left, and I set -1V from the right, the situation is as you said above. My voltage VIN2 (-1V) neutralizes the voltage drop across r2 (1V), and there is a virtual ground at the slider. If you look at this point, you will see zero voltage and zero resistance... just a piece of wire... So, you can think of me (VIN2) as of a negative resistor -r2 that has neutralized the positive resistance r2...

But this is not a true negative resistor since, if you begin changing VIN1, I will not change VIN2... and the magic of this "static" virtual ground and negative resistance will disappear... You will see that VIN2 is not a genuine negative resistor but a humble constant voltage source... So, we can think of this arrangement as of a kind of static negative resistance (proposed in the Wikipedia work)...

Now I decide (for some reasons) to be a real negative "resistor"... So I become a dynamic, not a static voltage source - I begin changing VIN2 in the opposite direction so that to keep VIN2 = Vr2... and as a result, zero voltage and zero virtual resistance in the middle point...  So, we can think of this arrangement as of a kind of a real negative resistance considered in the question about the inverting arrangement:

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

So, I would answer to your conclusion...

"Hence, we are again at the question (already earlier discussed): Are we allowed to say that a DC voltage source is equivalent to a negative resistance?Just because the current through the source has the opposite direction if compared with a "normal" load resistor? That is the basic question - all other examples (with or without feedback) can be reduced to this question, I think."

... as follows:

The ordinary (constant, steady, static...) DC voltage source is not a complete (real, true, genuine...) negative resistor... it is just like a negative resistor... To be truly negative resistor, it should be a dynamic source... of course with the addition "the current through the source has the opposite direction if compared with a "normal" load resistor" (the necessary but not sufficient condition for the existence of the true negative resistance)...

The negative feedback is just a possible way of doing such dynamic sources... and this is the idea of this question...

Cyril - we move around a circle. Even if I follow you and your "game". Also this example (game) can be reduced to the basic question: May we consider a voltage source as a negative resistance? When you speak about "zero voltage and zero resistance" at the point of virtual ground, this is nothing else than an alternative wording of Kirchhoff´s voltage law (sum of all voltages within a loop equals zero). And, thus, we arrive automatically at the basic question as mentioned above.

Cyril - I forgot to ask you:

In the given picture, thre is a negative sign in the RED equation, but not in the GREEN equation. What is the reason? Why different colors?

Just another insight about the negative feedback approach for creation of negative impedance...

The active element (amplifier) of the negative feedback system creates a negative (mirror, inverted) copy of the positive element. So it is not steady negative resistance... it follows the positive resistance when changing...

Thus the negative feedback system acts as an "impedance inverter". It is interesting that if the original impedance is negative, it will be converted into positive (a speculation)...

It is interesting to see that Tom Hayes talks about inverse dog or dog-1 in his cheerful story:)

http://www.circuit-fantasia.com/collections/tools/strange-things.html

... and we talk here about a negative dog:) Same or different things are they?

Lutz, let's keep this topic - "Also this example (game) can be reduced to the basic question: May we consider a voltage source as a negative resistance?",  in the related question:

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

Regarding "When you speak about "zero voltage and zero resistance" at the point of virtual ground, this is nothing else than an alternative wording of Kirchhoff´s voltage law (sum of all voltages within a loop equals zero)"...

Yes, exactly... Kirchhoff´s, Ohm's and all basic laws are always valid... but we build more general concepts on them... negative impedance is one of them...

I can not agree with your "And, thus, we arrive automatically at the basic question as mentioned above" since there is a subtle difference between the two arrangements - here the voltage source is dynamic while there - constant...

"In the given picture, thre is a negative sign in the RED equation, but not in the GREEN equation. What is the reason?" - I have just omitted the sign in the GREEN equation...

"Why different colors?" I use the color as an additional attribute of electrical quantities to make my circuit stories colorful. I mark currents with green color (by association with water)... voltages - with red color (by association with pressure)... and resistances - with black (graphite) color. In addition, I draw currents as closed loops to show that any current returns there, where it started:)

It is interesting to see that Tom Hayes talks about inverse dog or dog-1 in his cheerful story.

I think, it is a well-known fact that the transfer function of the non-inverting circuit is nothing else than the inverse of the feedback factor. (...really interesting ?).

I have just omitted the sign in the GREEN equation...

Nevertheless, its surprising because in our discussion signs are mostly important.

"I think, it is a well-known fact that the transfer function of the non-inverting circuit is nothing else than the inverse of the feedback factor. (...really interesting ?)."

Yes Lutz, for me (and not only for me) it was... and it is really interesting...

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

and here:

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

"Nevertheless, its surprising because in our discussion signs are mostly important."

Lutz, maybe it is since, when creating this picture (around  2008), I hadn't the notion that the op-amp output was actually a negative "resistor"... so the current sign was not so important:) BTW, I do not know why, but I do not like to talk about currents in terms of signs... I prefer to say the current enters/exits instead the current is positive/negative... 

I began thinking of the op-amp output as of a negative "resistor" just in the beginning of this year... so this is a fresh insight...

I do not like to talk about currents in terms of signs... I prefer to say the current enters/exits instead the current is positive/negative...

Ohh - I am afraid, this will lead to further confusion. Each current enters a part or a voltage source at one node and leaves it at another node.  How will you describe the direction? In all our calculations we use the commonly agreed positive/negative definitions. Why do you want to deviate from this? Does this really help?

OK Lutz, we will not argue about signs... there are more important things to discuss...

.".......there are more important things to discuss..."

OK - for example the following:

"The ordinary (constant, steady, static...) DC voltage source is not a complete (real, true, genuine...) negative resistor... it is just like a negative resistor...

...is not a complete ...it is just like...? I must admit, I don´t understand.

To be truly negative resistor, it should be a dynamic source... of course with the addition "the current through the source has the opposite direction if compared with a "normal" load resistor" .

What happens when we keep the input voltage for the opamp under discussion constant ?

I  think this is false as an amplifier could even not admit any impedance if it is non linear !

Lutz, I just like to say that if the current through the source has the opposite direction, the load is a current-inversion NIC (INIC). I suggest here to stick to the voltage-inversion NIC (VNIC) since it is simpler and more intuitive. This means, in our "game", to keep the voltage VIN2 with opposite polarity regarding VIN1.

You made me re-think seriously on the other question - if we can consider the ordinary voltage source as a negative resistor:

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

Emmanuel, would you explain in detail your view?

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...  

Then, if we begin changing the resistance (at constant voltage), the current will change hyperbolically... and the ratio V/-I (the source negative resistance -R) will follow the resistance R. So, it seems we can consider the constant voltage source loaded by a varying resistor again as an acting alone negative resistor with "resistance" |-R| = 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,  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...

Now let's make a connection with the op-amp inverting arrangement and our "game"...

The first voltage source is the op-amp output (VIN2 in the game arrangement)... the second source is the input voltage source VIN (VIN1 in the game)... and the resistor R is the network of the both resistors (R1 and R2 in the op-amp arrangement, or r1 and r2 in the game)...

Let's now consider the operation of the game arrangement (it is much more visible and comprehensible to us, humans, than the op-amp arrangement; in addition, it is more flexible... see why below).

The Actor 1 (probably Lutz, because I do not see others willing to take on this thankless job:) changes the input voltage VIN1... but the Actor 2 (probably I) does not react at all (maybe because I аm dozed:)... and its voltage VIN2 stays constant... (the op-amp output is "frozen" like a S&H circuit in the state HOLD). So VIN2 does not show ahy resistance - neither positive nor negative... it has zero resistance... it does not prevent the current... it is just a piece of wire (with respect to the current variations).

Let's examine what are the resistances in the circuit... to present it by their sum (a "resistor type" of KVL:) It consists of three "resistors" in series - negative (VIN1), zero (VIN2) and positive (r1 + r2)... and this sum should be zero. It seems that differentially the negative "resistance" VIN1 is equal to the total positive resistance r1 + r2.

So the conclusion is that in this situation (VIN2 or VOA = const) the voltage source VIN2 or the op-amp output does not show negative resistance; it has zero differential resistance.

But at some point I wake up:) and start to do the thankless job - to change VIN2 so that to keep the zero voltage of the common point (the virtual ground) between the two resistors. As a result, VIN2 = Vr2 and acts as a true negative resistor with negative resistance -r2 that adds to the negative resistance of the input voltage source.

So, now we have a sum of two negative "resistances" (VIN1 and VIN2) equal to the sum of the two positive resistances (r1 + r2)... and the total sum of the resistances in the loop is zero  (KRL:)

KRL ? Kirchhoff´s Resistance Law? (Or CRL?)

Yes Lutz... is there any sense in it or just to keep it as a joke? If there is, it would be a reason to use the negative resistance concept for something usuful...

KRL: "The sum of the negative resistances in any closed loop is equivalent to the sum of the positive resistances in the loop" :)

KRL = KVL/I

By the way, this evening I can't edit my comments. Do you have the same problem? Or it is a problem of my computer?

Here are the two "resistor" versions of KCL and KVL:)

KGL (another form of KCL): "The algebraic sum of the voltage-controlled negative conductances and positive conductances at a node is zero" - KGL = KCL/V

KRL (another form of KVL): "The sum of the current-controlled negative resistances in any closed loop is equivalent to the sum of the positive resistances in the loop" - KRL = KVL/I

---and in case there are only positive and NO negative resistances in the loop? Does this new rule apply?

KVL without voltages? A circuit without voltage sources? IMO it is quite boring and useless:)

The "trick" with negative resistances is that they really are not resistances... they are sources... and we can replace the conventional sources with them...

It seems I have problems with Javascript on my computer, and this is the reason why I cannot edit my comments. I have tried three browsers withоut any positive result. If someone can help me, I would be thankful...

But I have the same problem with another computer (when entering Edit the browser cycles continuously)... It seems this is a problem of the RG software...

But it is interesting that Delete works fine... I just tried it...

Yes - EDIT doesn´t work at the moment.

Prof.Cyril,

After a sufficiently long lull period devoted to Electronics ,I am reverting back to my Basic Electronics/Electrical.

I really wonder at your ingenious way of interpreting the action of the Circuits. I try to elaborate it as follows.

Referring to pre commencement of the Play, I too fully agree with your statements regarding inverting or non inverting mode of Opamp. But I try to explain it in my own different way.

The Feedback Resistance can also be treated as a Transfer Resistance between Input and Output Terminals. As regards to – ve feed back voltage ,it can be considered as + ve feed back Voltage across a – ve Resistance which is in parallel with the Original Transfer Resistance and the Equivalent Resistance between input and output terminals become zero( R in parallel with( -R), giving the property of High Input Impedance to the Opamp. If the gain A = 1 (by Directly connecting the input and output terminals together) we get the Voltage Follower Configuration. Hence I Agree with the statement,” In op-amp inverting circuits, the op-amp output acts as a pure negative impedance element that neutralizes the positive impedance of the element connected between the output and the inverting input. As a result, the combination of both the (positive and negative) elements has zero total impedance (a virtual ground):”.

In op-amp non-inverting circuits, the op-amp output acts as a voltage source with voltage equal to the input voltage and an internal negative impedance that neutralizes the positive impedance of the element connected between the output and the non-inverting input. As a result, the combination of both the (positive and negative) elements has zero total impedance, and the whole input voltage appears across the other positive element connected between the non-inverting input and ground…. Agree with this statement due to Virtual Ground (in Voltage Follower)

The Statement.”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”... Agreed

Coming to the Play : Assumptions made- The first voltage source is the op-amp output (VIN2 in the game arrangement)... the second source is the input voltage source VIN (VIN1 in the game)... and the resistor R is the network of the both resistors (R1 and R2 in the op-amp arrangement, or r1 and r2 in the game)... Vin1 changes Vin2 does not change .

Change takes place in the input voltage VIN1. and voltage VIN2 stays constant VIN2 stays constant... (the op-amp output is "frozen" like a S&H circuit in the state HOLD). So VIN2 does not show any resistance - neither positive nor negative... it has zero resistance... it does not prevent the current... it is just a piece of wire (with respect to the current variations)…..Agreed

Now we can treat this configuration as a Series Circuit with Voltage Source of (Difference of the two voltages), (sum of two – ve Resistances of Op-amp) and ( sum of two Differential Resistors)

Then to change VIN2 so as to keep the zero voltage of the common point (the virtual ground) between the two resistors, we have , VIN2 = Vr2 and it can be treated to be across a true negative resistor with negative resistance( -r2) that adds to the negative resistance of the input voltage source.

So, now we have a sum of two negative "resistances" (VIN1 and VIN2) equal to the sum of the two positive Differential resistances (r1 + r2)... and the total sum of the resistances in the loop is zero (KRL:) since VIN2 = Vr2 .

This condition may be stated in terms of KVL we may also reinterpret is as KRL(or CRL-Cyryl’s Resistance Law): "The sum of the negative resistances in any closed loop is equivalent to the sum of the positive resistances in the loop" :) for Current Controlled Devices like Op-amp.

KRL = KVL/I .Agreed

KGL (or CCL-Cyryl’s Conductance Law) follows from KCL on Similar Lines for Voltage Cotrolled Devices.;

KGL (another form of KCL): "The algebraic sum of the voltage-controlled negative conductances and positive conductances at a node is zero" - KGL = KCL/V

KRL (another form of KVL): "The sum of the current-controlled negative resistances in any closed loop is equivalent to the sum of the positive resistances in the loop" - KRL = KVL/I

Vah re Vah ! Wonderful Statements.

I think I am Right and I am a Cyril Follower like Voltage Follower as I think I have followed the Statements.

P.S.

Dear Prof. Pisupati,

Welcome back to the kingdom of free basic circuit ideas... and thank you for the support! I need both positive and negative responses (feedbacks) to continue thinking about circuits... only the lack of responses (no feedback... silence... muteness...) prevents discussions... kills the enthusiasm and the birth of new ideas...

I especially want to thank you for for your sharing "I am reverting back to my Basic Electronics/Electrical", because the popular belief today is the basic ideas are simple and their modern implementations - complex... and therefore it is not worth thinking about them... But you know the sad truth - that most difficult is to comprehend the "simplest" (basic) truths because they are not just simple but genius simple... So most people do not bother to do this and take them for granted... but it is more a religion than a science...

I am especially impressed by your initial thoughts: "The Feedback Resistance can also be treated as a Transfer Resistance between Input and Output Terminals. As regards to – ve feed back voltage ,it can be considered as + ve feed back Voltage across a – ve Resistance which is in parallel with the Original Transfer Resistance and the Equivalent Resistance between input and output terminals become zero( R in parallel with( -R), giving the property of High Input Impedance to the Opamp. If the gain A = 1 (by Directly connecting the input and output terminals together) we get the Voltage Follower Configuration."... but I need time to realize them... Maybe I will do it next few days since I go to country tomorrow... It is a good idea to illustrate them with some conceptual hand-drawn picture on the whiteboard...

Regarding the "new" Kirchhoff's laws:) I just want to say that prof. Lutz von Wangenheim has a great merit of their birth... although the paradox is that his negative responses played a positive role:) Although joking with "resistor Kirchhoff's laws", I still secretly hope they are correct... and useful in some way...

Thank you Prof. Pisupati for the emotional final salutation! I believe that it will attract more new followers of these underlying, immortal and eternal circuit ideas!

Best regards, Cyril

My request is when we talk about negative resistance to specify what kind it is - current-controlled (voltage source) or voltage-controlled (current source). I recall that in inverting circuits it is of the first type (current-controlled voltage source connected in series to the negative feedback resistor and the input voltage source in such a way to help the input source)...

I would like to add another insight to both "resistor laws":

At least one of the negative resistors in the loop has to have static negative resistance (to be an ordinary voltage source).

Another insight (constraint ensuring a stability.):

The sum of the true (differential) negative resistances/conductances should not exceed the sum of the positive resistances/conductances.

....although the paradox is that his negative responses played a positive role.

Yes - I have made the same experience: When I try to convince somebody I am forced to be very exact and self-critical. As we all know: Teaching something is the best method to learn something.

Quote: "As regards to – ve feed back voltage ,it can be considered as + ve feed back Voltage across a – ve Resistance which is in parallel with the Original Transfer Resistance and the Equivalent Resistance between input and output terminals become zero( R in parallel with( -R), giving the property of High Input Impedance to the Opamp."

I must admit that I am not able to understand the above statement. A negative resistance in parallel "with the original transfer resistance"? More than that, the opamp´s high input impedance is a property of the opamp alone (without any feedback), or did I misunderstand something?

Coming to the two qualifying statements by Prof.Cyril I shall add a third Qualifying statement as in the case of Ohm's Law,"Temperature remaining constant".

P.S.

Too bad - I was hoping for an answer/explanation.

Dear Pisupati,

With your explanation above, you made me begin thinking about the other kind of dynamic resistance - voltage-controlled, since the voltage follower is based on it. Like Lutz, I also have difficulties with the understanding your idea... but it sounds interesting...

This is the dual Miller arrangement where two voltage sources with equal voltages (the input voltage and the op-amp output) are contrary connected via a resistor (the input differential resistance of the op-amp)...

https://en.wikipedia.org/wiki/Miller_theorem#Applications_based_on_subtracting_V2_from_V1#infinite

... and the result is a virtually increased (theoretically, up to infinity) resistance between the op-amp non-inverting input and ground...

In my experience, we use P=V*I =constant to make a judgement.

dp=I*dV+V*dI=0, impedance=V/I=-dV/dI. Namely any circuit that has a constant input P, the input impedance is negative.

Aha... I (Circuit-fantasist ) learned about this phenomenon 8 years ago in my first conversations with Wikipedians about the nature of negative resistance:

https://en.wikipedia.org/wiki/Talk:Negative_resistance/Archive_2#Another_fresh_viewpoint_at_negative_resistance

"A regulated power supply appears as a negative resistance at it's input... You can look at it both intuitively and analytically. An ideal regulator is 100% efficient, so all input power is avaiable at the output, i.e. power is conserved. The only difference is the input/output voltage. If the output voltage is constant and connected to a fixed resistive load, the output power is constant. The input power must also be constant since power is conserved. If the input voltage is increased, the input current drops to maintain constant power. This is negative differential resistance: current goes down when voltage goes up (and vice-versa). It's not hard to relax the 100% efficiency constraint and show that non-ideal regulated supplies also exhibit negative resistance.  Madhu 11:36 pm, 12 July 2006, Wednesday (8 years, 28 days ago)"

"As far as I can see, you mean a "clever" voltage regulator (for example pulse regulator), which does not regulate the output voltage by dissipating the surplus power on an analog regulating element (transistor). Instead, it "observes" the output voltage by a feedback and changes the pulse width thus regulating the current needed. Another more primitive but clear example: someone regulates AC voltage by observing it on the transformer's output (that is, using feedback) and changing the transformer ratio (for example, switching over the winds or moving a slider). Maybe, (hydro) power stations in energetics use this phenomenon when regulate their electrical outputs; maybe, they behave as negative resistors at their hydraulic inputs? Well, let's finalize these thoughts in a conclusion: The input part of the circuit consuming a constant power behaves as a negative resistor. -Circuit-fantasist 9:04 pm, 20 July 2006, Thursday (8 years, 21 days ago)"

But IMO this is an example of a dufferential negative impedance...

Thank you for the participation, Cyril

Dear Pisupati and Lutz,

I am leaving for a few days in the country but I will be with you and our common problems...

Pisupati, although I would really like to see negative resistance in the ubiquitous op-amp follower, I still fail... but attempts continue:) But really you made me seriously begin thinking about this dual (voltage-controlled) dynamic resistance arrangement...

IMO the op-amp output of an op-amp follower is not a negative resistor since it is connected in series but contrary to the input voltage source. So, it is sooner another positive but virtual resistance connected in series to the real postivie resistance of the op-amp differential input.

• When the op-amp output voltage is less than the input voltage (imagine the humble emitter follower), the total resistance will be moderately increased and the current decreased.
• If this voltage is exactly equal to the input voltage (an ideal follower), the total resistance is increased up to infinite and the current is zero.
• And finally, when the op-amp output voltage exceeds the input voltage (INIC), the current reverses its direction and the op-amp (and accordingly, the total) resistance becomes negative.

While the first two cases (increased and infinite resistance) can be implemented by negative feedback, the third (negative resistance) requires an additional positive feedback to go beyond the infinity:) and to reverse the resistance...

Best regards, Cyril

According to Miller Theorem

Zin = ( Z / (1-K)). This becomes infinity, only if K = 1 as in the case of Voltage Follower.

This is what I have stated in the beginning, explaining in terms of R I I (-R) giving the property of High Input Impedance to the Opamp. If the gain A = 1 (by Directly connecting the the input and output terminals together) we get the Voltage Follower Configuration. REq of R I I (-R) =( R * (-R)) / (R + (-R)) is infinityThis corroborates, with your statement ,"and the result is a virtually increased (theoretically, up to infinity) resistance between the op-amp non-inverting input and ground in the case of dynamic resistance - voltage-controlled, since the voltage follower is based on it.

I should have been specific about the input terminal, non inverting,( Since the inverting terminal is directly shorted to output terminal in the case of voltage follower., I have not Specified))

This confusion would not have arisen if the circuit is given, I am not able to give the circuit in the reply to qn. in R.G. Then I have to give a separate File attachment, which I have done in one or two cases earlier where the circuit had to be given specifically. I took the Voltage Follower Circuit

P.S.

The written above is an example of a virtual resistor that is actually not a resistor but a source... it is a contrary connected dynamic source that creates an opposing voltage... and thus impedes the current...

I repeat my suggestion to ask an extremely interesting question about all these kinds of virtual resistors (and not only resistors) in electronics. Here is a possible formulation of such a future question:

How do we create virtual electrical elements in electronics?

Dear Pisuapati,

I also try to present the equivalent infinite impedance as a result of mixing two opposite resistances - positiwe and negative. It is clear that the resistor (the op-amp input resistance here) is the positive impedance... but what is the negative resistance? The op-amp output? I tried to see it there thinking in the order below

We have two voltage sources with the same (e.g., positive) polarity regarding the ground. Their positive terminals are connected (like a bridge) with a positive resistor. The current is:

I = (VIN - VOA)/R = VIN/R - VOA/R = IIN - IOUT = 0

So, as though there is a sum of two equal but opposite currents and the result is zero current, leading to infinite resistance of the whole network composed by the two elements in series (R and the op-amp output). As though the op-amp output acts as a negative resistor producing insted consuming a current... and it should neutralize the positive resistance... But the problem is that both they act in the same way - they oppose the input voltage since the voltage (drops) across them have the same polarity when travelling the loop (+  -...  + -)... and so on... and so forth.................

The problem of this assertion (R - R = 0) is how to present the equivalent infinite resistance as a result of subtraction between two opposite resistances...

Prof.Cyril,

What you have stated is correct,It is a Virtual resistor.There is no Static Resistancr Connected,

In the case of Amplifier action we have the static D.C. resistance decided by the biasing D,C, Voltage and Current. And the Dynamic Resistance is due to the Signal Excursion about the Operating Point,Which is called A.C. Resistance.

In the Inverted Voltage Divider Circuit given R1  and  R2 are Static Resistances.

As indicated by you the Virtual Resistor has to be studied in detail.

P.S.

The view I have taken is that the Opposing Voltage appearing across a static voltage drop,represented as a drop across a static Resistance due to a current(D,C. or A.c) is similar to a self or Mutually induced emf in an inductance.In Electrical Circuits we represent it by means of a equal and opposite voltage source .This opposing voltage Source can be represented by means of a Voltage Drop across a Static Resistance

in wicu opposing current is flowing.

E = -V = (-I) * R = I * (-R)

That is how I brought in the concept of Negative Resistance ie. (-R).

If it is A.C. Voltage drop this (-R) also will be changing in Magnitude .( - ve A.C..Resistance,we my call.

This negative resistance is different from the negative resistance obtained in the negative resistance region of devices kike Tunnel Diode.

P.S.

Exactly Pisuapati... the opposing voltage acts like the emf in an inductor... and this idea is used to create a virtual inductor (gyrator). As you can see from the second picture, the voltage source VS opposes the current IL in the same manner as the real inductor:

https://en.wikipedia.org/wiki/Talk:Gyrator#Demystifying_gyrator_circuits

https://en.wikipedia.org/wiki/Talk:Gyrator#Latest_additions

Maybe you can add a comment in the RG question below?

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

Cyril

This Gyrator Circuit of creating a Virtual Inductor using Capacitors which are Economical and Space conserving is of Immense Use.WE have discussed about Gyrators earlier.

P.S.

Yes, it is true... but it was interesting for me to guess how such virtual inductors were made...

Cyril,  have a nice time in the country - and, if you have time, reconsider the "virtual inductor" as you have shown above. I don´t think that the active circuit behaves like the passive model.

Regards

Lutz 

Dear Lutz,

It would be well to know your doubts when thinking... Really, I have not been thinking on this circuit since those "glorious" Wikipedia times... but I am eager to begin thinking again...

Aha... Edit function already works...

Well - here is my result (active circuit):

Zin=[RL+s(RL*R*C)]/(1+s*RL*C)

I did not expect such a quantitative answer to my qualitative reasoning. What this expression still want to tell us? Where (what) is the problem in the conceptual "bridge" picture?

Any controlled element implies negative impedance. Let an externally controlled resistor R suffer a variation dR. Then the variation dV of the voltage drop on R will always be of opposite sign to the variation of current dI in the circuit. This negative "impedance" may serve as an indicator of an external control [1].

________________

[1] Serge Luryi, "Development of RF equivalent circuit models from physics-based device models", in Future of Microelectronics: The Road Ahead, ed. by S. Luryi, J. M. Xu, and A. Zaslavsky, Wiley Interscience, New York (1999) pp. 463-466.

To S. Luryi: May I ak you for a simple example? Let´s take a voltage controlled FET resistor. For a constant current and when I increase the resistance (dR positive) I expect an increase in voltage (dV positive). Is this a counter example - or not? 

To. L. v. Wangenheim: Constant current is a pathological example. The devil is in the detail of the constant current source. Replace it with a physical element, viz. a battery with an internal resistance - however large - and the counter example disappears.

http://www.ee.sunysb.edu/~serge/164.pdf

To S. Luryi: Unfortunately I have chosen the "constant current example" - and, of course, I agree with you  that this is somewhat "pathological". But this is not the main point. I have no problems to repeat my example with a voltage source driving a current through a voltage-controlled FET-resistor. Am I wrong, or will the current increase  as a result of a voltage increase? Where is a negative resistance?

To L. v. Wangenheim: The negative resistance in this example is the "variational impedance" Rvar = (delta Vch / delta I) in response to a variation delta VG of the gate voltage. The quantity Vch  is the voltage drop from source to drain of the transistor and delta Vch its variation. The quantity delta I is the variation in the circuit current. If you compute Rvar, you will find it negative for any constant value of the internal battery resistance r. 

Best regards, Serge

An equivalent example is a battery source V of internal resistance r driving current I through the controllable resistor R -- that is controlled by its temperature. Lighting a match underneath the resistor and measuring the variations in the current and in the voltage drop on the resistor you will always find them to be of opposite sign.

To S. Luryi: I am very sorry but I cannot follow your explanation. To me, a FET in its ohmic region is comparable to a classic ohmic resistor. And as such it is used in many circuits. Where is the difference to an ohmic resistor? There is a voltage drop between D and S and there is a current Id. And when the current Id increases the voltage VDS also increases because the current-voltage curve has a positive slope, has it not?  Am I wrong ?

To L. v. Wangenheim: I cannot say it more clearly than I did. Just do the calculation. When considering the variation, delta (IR), of the voltage drop on the resistor (in response to a variation delta R), please do not forget both terms.

There is no difference between the example of a FET controlled by gate and a classical ohmic resistor controlled by a candle.

Mr. Luryi, please can you answer two questions:

1.) Is it correct that the V-I characteristic of an ohmic resistor has a positive slope?

2.) If so - how can a positive delta(V) cause a negative delta (I) ?

1) Correct

2) I have already explained that. Resistor at different temperatures is not the same resistor. 

Dear Serge Luryi and  Lutz Wangenheim,

Thank you for the interesting discussion that I found here after my return.  As I understand, you have strayed from the problems of the true negative resistance (impedance) and have begun to discuss the problems of the differential negative resistance (NDR). Before I join in it, I want to remind that there is a question asked specially about NDR... and it needs your attention. So, I have copied your discussion to Page 3 of this more related question, and added my comment there:

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/3

Quote: "2) I have already explained that. Resistor at different temperatures is not the same resistor."

Right, and exactly this is the reason - I think - we are not allowed to "combine" both resistors (with different values) into one single element with specific properties. The resistance of one part (static or differential) is defined using the voltage-to-current characteristic without changing its value during the measurement. 

To L. v. Wangenheim:  We are allowed to define quantities, so long as the definition is clear and free of ambiguities. My definition of the variational impedance associated with a controlled resistor satisfies this requirement and is useful. It can be unambiguously measured. Needless to say, it is not the same thing as the ordinary (or differential, for that matter) resistance of the same element. We have no further disagreement.

To Cyril Mechkov: The point I was making and the variational impedance concept I introduced in the above discussion have nothing to do with NDR. The latter is simply the slope of the V(I) curve, which can sometimes be negative. The NDR commonly occurs in semiconductors and is often useful, e.g. Gunn diodes. There is no diode that has "absolute" negative resistance V/I < 0 (its existence is forbidden by thermodynamics, at least near equilibrium).

To S. Luryi and Cyril M.:

Sorry, but I cannot continue the discussion because of a journey. Will be back Sept. 1st. Adios, Lutz vW.

Nice holiday, Lutz!

Cyril

Serge Luryi,

I continue to think with pleasure on your idea because just such extremely basic concepts are my favorite topic of thinking. However, let's first clarify what exactly your claim is...

If I understood correctly, you claim that besides true negative resistors (over dynamic sources) and differential negative resistors (over dynamic positive resistors), there is another type of negative resistors - "variational resistors"?

If so, there is something confusing in these elements - that they are externally-controlled (2-port) devices... while the first two are self-controlled (1-port) devices...

The resistance is a function that defines the relation between the independent (input) variable (voltage or current) and dependent (output) variable (current or voltage) - I = V/R or V = I*R...

In your arrangement, some input quantity (voltage, current, movement, light...) defines the resistance... and then the resistance defines two dependent (output) variables (current and voltage) - IOUT = V/(R1 + RIN), VOUT = V*RIN/(R1 + RIN)... where RIN is the lower resistor R2 of the voltage divider... and then you talk about the ratio of these two output quantities...

So, I would like to know from where I can see this "variational negative resistance" - from the side of the controlling input source or from the side of the  voltage source supplying the voltage divider...Because, I said it in the question about the negative differential resistance (NDR)...

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

... if we look from the side of the input source, we will see the steady positive resistance of the resistor's input port; if we look from the side of the supply source, we will see the steady positive resistance of the resistor's output port set by the controlling source...

About the signs... I explained this phenomenon in the question about the negative differential resistance:

"Your arrangement is the basic electric circuit - a real voltage source with voltage V and internal resistance Ri drives a variable (somehow-controlled) resistive load RL. We can think of it also as of a voltage divider with fixed upper resistance R1 and variable (controllable) lower resistance R2.

In this situation, if you increase the resistance RL (R2), of course the current through the load will decrease and the voltage across the load will increase... and they will change in opposite directions... but this is not an indication of any negative resistance... this is the well-known 19-century Ohm's "positive" resistance...

We can talk about negative resistance when really the voltage and current change in opposite directions... but in the case when changing one of them as an input quantity, e.g. the (input) voltage... Then, looking from the side of the input voltage source, we will see a negative resistance...

For this to happen, the load should be a dynamic resistor... and moreover to change its resistance vigorously enough..."

to Cyril Mechkov:

Let me repeat myself. I defined a quantity, the variational impedance, that has the dimensionality of resistance, but is NOT resistance of any physical element. It's purpose is to illuminate the nature (and detect the presence) of control. Consider a battery source V of internal resistance r driving current I through the controllable resistor R -- that is controlled by its temperature. Give a variation of temperature in the resistor and measure (or calculate) the resultant variations in the current and in the voltage drop on the resistor. Simple calculation shows you will always find these variations to be of opposite sign. Their ratio -- the "variational impedance" -- is therefore negative and that can be used as a test of the existence of control.

Sorry for bringing this digression to your discussion. I had thought to illuminate the subject by bring forth an amusing and somewhat subtle fact. Instead I confused everybody!

Can I illustrate your explanations by a hand-made picture? I've drawn it on the whiteboard on another occasion but it is also suitable for this case...

Here an op-amp output is the voltage source V, its output resistance ROUT is the internal resistance r and the variable load RL is the controllable resistor R (i.g., controlled by its temperature).

When we decrease RL, its green IV curve rotates counterclockwise and the operating (intersection) point moves up along the blue IV curve (the so-called load line) of the real voltage source with voltage VOA (V) and internal resistance ROUT (r). Note that the voltage drop across the load decreases while the current through it increases (i.e., their variations have opposite signs)... and the load line is inclined to the left (it has a negative slope)... but this does not mean that there is a negative resistance here...

So, my conclusion is that the fact that the variations of the voltage across and current through an externally-controlled resistor have opposite signs does not mean negative resistance. 

IMO the negative resistance phenomenon is a special case of the more general phenomenon named amplification. In the classic analog electronics, we implement it in a quite silly wasteful way:) - by controlling the electrical energy with active elements (variable resistors). They can be externally-controlled (as in conventional 2-port amplifiers) or self-controlled (as in exotic 1-port amplifiers). In the first case, there is no negative resistance; we can observe it only in the second case... An example is the tunnel-diode amplifier...

Maybe, for your goals it is enough to say that the voltage and current change in different directions when the resistance varies... without mentioning negative resistance...