Dear very respected colleague Cyril,

welcome,

I am happy to see you newly on the RG. Really i missed you. Yes your style in explaining the operation of the analog electronic circuit is like one who wrights a scenario of story. You are a circuit scenarist if i am allowed to create this expression. I wrote this in many occasions while sharing in answering your questions.

Returning to your question, when the colleagues are delayed in answering your question you answered it yourself.

The negative resistance can be formally defined as R-= - V/I so it is a resistance which source the current on its voltage source in the opposite to positive resistance which sink the current from the voltage source. So it is charging the source in the opposite of a positive resistance which discharges the source.

This is the definition.

When we come to implementation we can implement the negative resistance by a single device biased in the negative resistance range or we can create the negative resistance by the negative impedance networks.

In case of using tunnel diode or a thyristor or a unijunction transistor as a negative resistance device, then in this case one has to speak about an equivalent negative differential resistance.

As the negative resistance region is no passing through the zero point of the I-V curve. Here the total resistance of the device is always positive, while only the differential resistance which is the slope of the curve may be positive or negative.

As you showed us the device either shows and N of an S I-V curve with the negative resistance regions are not passing through the point of origin(0,0).

In case of N-curve the device is voltage controlled, while for the S-curve the curve is device is current controlled based on the nature of the device.

Here only the slope of the I=V curve which defines the sign of the resistance. One is then speaks of the differential resistances.

When constructing negative resistances using synthesis circuit one can make the I-V curve pass through the zero point in the negative resistance range and then the slope and the intercept will be both equal such that dV/dI= V/I in case of the a linear I-V curve. In case of nonlinear I-V curve we the total and incremental will be negative but the have different values.

Best wishes

I cannot understand what makes me, all my life, constantly try to comprehend the essence of circuit phenomena...

Here again I thought deeply about the way in which the two types of negative resistance (differential and absolute) solve the same basic task for which they were created - to compensate equivalent positive resistances.

DNR is an element (tunnel diode, Gunn diode...) while ANR is a circuit (VNIC, INIC). Both they add energy to the load with the help of a an additional voltage source (power supply)... but they control it in a different way.

DNR controls the supplying voltage source like an electronic "rheostat" that changes its static resistance in series to the load. ANR controls the supplying voltage source like an "electronic potentiometer" that includes a part of the supplying voltage in series to the load...

Dear very respected colleague Cyril,

welcome,

I am happy to see you newly on the RG. Really i missed you. Yes your style in explaining the operation of the analog electronic circuit is like one who wrights a scenario of story. You are a circuit scenarist if i am allowed to create this expression. I wrote this in many occasions while sharing in answering your questions.

Returning to your question, when the colleagues are delayed in answering your question you answered it yourself.

The negative resistance can be formally defined as R-= - V/I so it is a resistance which source the current on its voltage source in the opposite to positive resistance which sink the current from the voltage source. So it is charging the source in the opposite of a positive resistance which discharges the source.

This is the definition.

When we come to implementation we can implement the negative resistance by a single device biased in the negative resistance range or we can create the negative resistance by the negative impedance networks.

In case of using tunnel diode or a thyristor or a unijunction transistor as a negative resistance device, then in this case one has to speak about an equivalent negative differential resistance.

As the negative resistance region is no passing through the zero point of the I-V curve. Here the total resistance of the device is always positive, while only the differential resistance which is the slope of the curve may be positive or negative.

As you showed us the device either shows and N of an S I-V curve with the negative resistance regions are not passing through the point of origin(0,0).

In case of N-curve the device is voltage controlled, while for the S-curve the curve is device is current controlled based on the nature of the device.

Here only the slope of the I=V curve which defines the sign of the resistance. One is then speaks of the differential resistances.

When constructing negative resistances using synthesis circuit one can make the I-V curve pass through the zero point in the negative resistance range and then the slope and the intercept will be both equal such that dV/dI= V/I in case of the a linear I-V curve. In case of nonlinear I-V curve we the total and incremental will be negative but the have different values.

Best wishes

Dear Abdelhalim,

Let me first say a few words about the human nature of our collaboration here...

You belong to this small group of people who see not only the specific technical issue being discussed but also the person behind it. If most of the contributors to this social network were like you, it would be closer to Facebook, Twitter and other human networks... instead of resembling a computer network …

I am glad to see that you appreciate not only circuit explanations but also the way they are presented. My favorite way to reveal circuit ideas is to build and explore circuits step by step. I name these stories "building scenarios".

Now about the mysterious phenomenon of negative resistance… Your reflections above made me think once again (since the early 90s:) what exactly happens in a circuit of serial connected S-shaped negative resistance and linear positive resistance… how the negative resistance compensates equal part of the positive resistance. I think the mechanism of differential negative resistance (NDR) is more unusual and difficult to explain... then let's start with it...

In this arrangement, two resistors with equal but dual (positive and negative) resistances are driven by a constant current source. The formal explanation is that resistances are mutually destroyed and the resultant resistance of this network is zero. But how does this work? Here is my explanation:

Initially the current source passes constant current through the two resistors and creates constant voltage drops across them so the total voltage drop is constant as well.

If the current increases, the voltage drop across the positive resistor increases since its resistance is constant. However, the voltage drop across the negative resistor decreases since, as I described in a related question, its ohmic resistance does not stay constant but decreases… and even more vigorously than the current increases. Figuratively speaking, it happens that the current is trying to increase the voltage but resistance - to reduce it... and the resistance prevails in this "struggle"...

Now the most important conclusion:

Since the two resistances are equal, the voltage decrease of the negative resistor is equal to the voltage increase of the positive resistor; so the total voltage drop across the whole network does not change when the current increases (like a Zener diode). This means that the network has zero differential resistance… the negative differential resistance has compensated the positive resistance.

But the main question remains - "How does this dynamic but still positive resistor inject energy into circuit?" Still positive resistances consume rather than produce energy...

The idea is quite absurd and paradoxical... but it is widely used in our life and particularly in all kinds of amplifying stages. Here is its essence.

To give an impression that we add energy, we initially introduce significant losses so the resultant energy decreases. Then we begin reducing the losses… the resultant energy increases… and we say that we "produce" energy:) Here are some popular (and funny) examples:

• To look good, we first present ourselves as bad and then gradually reduce our badness:)
• To make ourselves happy, we cause ourselves problems and then we start to solve them. For example, when I was a kid, I had read a book in which the protagonist kicked his toes with a hammer and then was glad that the pain was gone:)
• To please someone very much, we first tell him/her the bad news and then refute it.
• In a transistor amplifying stage, we initially half-open the transistor thus introducing 50% losses. Then we gradually open it thus decreasing the initial losses up to zero (if you do not understand what I am talking about, it is the biasing). We say the transistor "amplifies" that means it somehow adds energy to the circuit...

According to this idea, the negative differential resistor introduces initial losses by its positive ohmic resistance (like the transistor above)... and a voltage drop appears across it (the same as the one across the positive resistor). Then, when the current increases, the negative resistor decreases its ohmic resistance, the initial losses decrease and the total power increases.

So, the idea to "produce" energy is to introduce initial losses and then to decrease them.

Cyril, the only truly active element is always the power supply. Others only distribute its energy - optimally in accordance with the input signal :-) Josef

... "distribute its energy" between them and loads... From this viewpoint, active elements are also loads for the power supply... variable loads...

By the way, can an active element act both as a regulating element and a load?

POINT OF VIEW OF Every POWER SUPPLY: it what eats current is appliance (load) - so all is load :-)

Dear Cyril,

I would like to welcome the respected colleague Josef and second him.

When you synthesize the negative resistance you build an electronic circuit containing amplifiers. They must be DC biased to act as a negative resistance. So, by biasing them they consume DC electric power. The useful output power is that only produced by the negative resistance. This machine is an energy conversing machine which always work with effeciency less than one.

This also valid for the negative differential resistance. One has to bias the device in the negative resistance region which means one has to pay DC power to get a useful power generated by the negative differential resistance.

Also here the conversion efficiency is less than one.

On the other side it is the waveform of the generated energy which caused us to pay this cost to convert from DC to an energy with a specified waveform. The new waveform in spite of expensive it is necessary for the functional performance for example amplifying ac signals or generating oscillations.

Josef, I think you also belong to "the small group of people" above:)

Cyril, it's such a funny problem with that energy.

Its "prakosmic" origin is unknown.

And everything following is just a transport or some form of transformation.

Then the law of conservation of energy applies - except for the damn start :-)))

More about the nature of negative differential resistance...

Besides the question, "How does the negative differential resistance compensate equivalent positive resistance?" another interesting question is, "How does the negative differential resistance 'amplify'?" For example, we would like to know how a tunnel diode amplifier "amplifies". Here is my intuitive explanation without using any formal clichés that hide the truth:

Simply speaking, any voltage amplifying stage is a voltage divider with voltage-controlled transfer ratio that is supplied by the power supply. It is implemented by replacing the one of resistors (usually the grounded) in the ordinary voltage divider by a voltage-controlled resistor (in the more sophisticated complementary stage, even both resistors are replaced). The input voltage controls its resistance from zero to infinity... the divider ratio changes from zero to one… and the voltage across it changes from zero to the supply voltage.

In the classic transistor amplifying stage, the (grounded) resistor is replaced by a transistor. In our case, it is replaced by some kind of negative differential resistor, e.g. a tunnel diode… so the specific question here is, "How does a tunnel diode 'amplify'?"

The general idea is the same. The only difference is that while in the case of the classic transistor amplifier the two voltage sources (input and supply) are separated, in the case of the tunnel diode amplifier they are connected in series and can be considered as one composed voltage source that supplies the voltage divider. Thus the four elements - the input voltage source, power supply, negative differential resistor (tunnel diode) and positive resistor (load) are connected in series (in a loop)... and the same current flows through them. Indeed, some problem of this configuration is that only two of these elements can be grounded... but this is a detail when revealing the main idea:)

As above, when the input voltage varies, the total voltage and common current change; so the negative resistor vigorously changes its resistance... the divider ratio vigorously changes from minimum to maximum… and the voltage across it (and the load) changes within the magnitude of the supply voltage. Note the voltage across the negative resistor changes in an opposite direction to the input voltage in the case of S-shaped NDR and in the same direction in the case of N-shaped NDR; i.e, this is an inverting amplifier in the first case and non-inverting amplifier in the second case.

We can illustrate the circuit operation by a kind of a mechanical analogy if we think of this dynamic voltage divider as of two joined springs (the upper is ordinary but the lower has "differential negative resistance"). In the beginning, we pull up enough the top of this composed spring to bias the middle point. Then, if we slightly pull up the top of this composed spring, the lower spring (suppose acting as S-shaped NDR) will vigorously pull down the middle point... and v.v., if we slightly push down the top of the composed spring, the lower spring will vigorously push up the middle point ("inverting amplifier"). Thus we slightly wiggle the top but the middle point vigorously moves. A very interesting mechanism...

Someone would ask, "Could we remove the power supply?" We can… but then the output voltage would change only within the input voltage… and there will be no amplification. The trick here is that the current is the same but the output voltage is many times higher than the input one (up to the supply voltage)… so there is a power "amplification".

So the tunnel diode amplifier is a dynamic voltage divider with transfer ratio controlled by its supplying voltage.

So far, we have considered two separate circuit techniques based on the NDR phenomenon - compensation and amplification. The question is whether there is any connection between them. I think there is such a connection and we can explain the amplification of the tunnel diode circuit above through the compensation…

Here is a simple explanation:

The negative resistance compensates a part of the positive load resistance and the total resistance decreases. So, at the same changes of the input voltage, the changes of current through and voltage across the load are bigger… hence the amplification.

The general conclusion is that the S-shaped negative differential resistor connected in series to a positive load increases the current variations through it. Note the current magnitude decreases but the current variations increase...

By the way, the same is valid for the ordinary transistor amplifying stage where including a transistor in series to the load (Rc) actually decreases the current but increases its variations...

Besides in terms of currents, we can express the general conclusion above in terms of voltages:

The S-shaped negative differential resistor connected in series to a positive load decreases the instant (static) voltage across it but increases its variations...

... or, of course, in terms of resistances:

The S-shaped negative differential resistor connected in series to a positive load increases the total static resistance of this network (since its static resistance is also positive) but decreases the total differential resistance (since its differential resistance is negative).

It is interesting to explain why, when considering the phenomenon of negative differential resistance, we can use interchangeably the voltage and resistance. Here is my explanation…

In this humble network of two resistors in series, the current is the same… the voltage drops across resistors are proportional to their resistances… so they represent these resistances… Thus we can think of these resistances as voltages (drops)... and we can operate with them (add) instead with resistances.

The voltage drop across the positive resistor (load) is an "ordinary" voltage drop proportional to the varying current and constant resistance. However, the voltage drop across the negative resistor is a bit unusual since it is proportional to both varying variables (current and resistance)... but we think it depends only on the current… and this dependence is linear again...

Both voltage drops have the same positive polarity; only the second voltage decreases when the current increases. Thus it decreases the total voltage drop across the whole network... and, accordingly, the total resistance.

So, in the case of negative differential resistance, "negative" means an opposite direction, not an opposite polarity.

I think that after all these comments, the phenomenon of negative differential resistance has cleared up (at least for me:)

We can clarify it further by comparing it to the phenomenon of true (absolute) negative resistance. That is what I suggest we do...

Hi Cyril

Question: Is there something like a "true absolute negative resistance"?

I mean: A resistor is a two-pole element with a positive resistance value. Is there a two-pole circuit with a negative resistance between both (floating) nodes ?

Or shouldn`t we better speak about an active circuit with an input resistance that is (a) grounded and (b) has a negative voltage-to-current ratio?

(A voltagfe-controlled current source which drives a current back into the controlling voltage).

Yes, Lutz… I understand what you mean - the bizarre "current inversion negative impedance converter" (aka INIC)...

But will you mind if we completely reveal the mystery around its NDR "brother" (e.g. the tunnel diode) before?

OK, let's consider the simplest arrangement - a perfect voltage source loaded with an N-shaped NDR (tunnel diode). The source voltage varies within the negative resistance region.

To explain the operation of this circuit, first imagine the voltage source is loaded by an ordinary resistor with constant resistance. So, if the voltage increases, the current will proportionally increase according to Ohm's law… and the source will experience considerable loading.

Then imagine the voltage source is loaded by a dynamic resistor that increases its resistance when the voltage increases so that the ratio V/R stays constant. This "dynamic load" will consume constant current… and the source will "feel" unloaded (this "feeling" is only in regard to the change).

Finally imagine the voltage source is loaded by a dynamic resistor that increases more significantly its resistance when the voltage increases so that the ratio V/R decreases. This "over-dynamic load" will consume decreasing current… and the source will "feel" a kind of "super-unloaded"... Figuratively speaking, the load will act as an "anti-load".

What can such an "anti-load" be used for? It can compensate an equivalent load. Let's consider an example.

Imagine a voltage divider (potentiometer) is loaded by a resistive load with "positive" resistance RL that decreases the output voltage. To eliminate it, we can connect in parallel a negative differential resistor (tunnel diode) with the same but "negative" resistance -RL. The result will be infinite (differential) resistance. How did this magic?

Here is my explanation (I am thinking for the first time about this arrangement; so excuse me if I am wrong somewhere):

The divider's output is loaded by two resistive loads in parallel - positive and negative (think of this combination as a kind of composed load). The output voltage is in the middle of the negative resistance region.

If the output voltage increases, the current through the composed load (RL||-RL) should increase according to Ohm's law. But the negative load vigorously increases its ohmic resistance and the current through it decreases. It turns out at what magnitude the current increases, with such a magnitude the current through -RL decreases... and the whole load current remains unchanged... The voltage divider will "feel" unloaded as in the second case above.

So, the combination of two equivalent but opposite resistances in parallel is equivalent to a single current-stable dynamic resistor. But what is the benefit of this more complex network?

The problem is that the positive load may already exist and we cannot replace it with something else. Then we complement it with the opposite (negative) "thing" and in this way, we destroy it.

An interesting question in StackExchange related to our discussion caught my attention:

What is the physical meaning of negative resistance?

(https://electronics.stackexchange.com/questions/435418/what-is-the-physical-meaning-of-negative-resistance)

Here is what a newbie has written under the question:

"I am a bit confused about the physical meaning of negative resistance. Mathematically, a component which has negative resistance shows a decreasing voltage across its terminal when the current inside it grows, and vice versa. But how is this physically possible?"

So people still need "physical" explanations...

A week after my last comment, I will compare the dual ways in which the two types of differential negative resistance (S- and N-shaped) neutralize equivalent positive resistances.

Both they initially introduce losses into circuits: the S-shaped negative resistor introduces voltage losses (subracts a voltage drop); the N-shaped resistor introduces current losses (subtracts, diverts current). They do that by changing their resistance in opposite directions: the S-shaped negative resistor increases while the N-shaped resistor decreases its resistance.

After that, they decrease the initial losses to compensate the positive resistance. For this purpose, the S-shaped negative resistor decreases while the N-shaped resistor increases its resistance. As a result: in the first case, the voltage drop across the S-shaped negative resistor decreases... so the total voltage drop across the whole network does not change... and the current increases as though there is no positive resistance in series; in the second case, the current through the N-shaped negative resistors decreases... so the total current through the whole network does not change... and the voltage increases as though there is no positive resistance in parallel.

Now let's turn our attention to the N-shaped negative resistance and complete the dialogue that began 10 days ago... but continued, for some reason not clear to me, as a monologue...

In short, I suggest to compare the ways in which the two types of N-shaped negative resistance (differential and absolute) neutralize equivalent positive resistances...

The N-shaped differential negative resistor connected in parallel to a positive resistor compensates this positive resistance by increasing its resistance when the voltage increases. As a result, the diverted current decreases and the total current stays constant.

The N-shaped absolute negative resistance (INIC) does the same... only by injecting its own additional current... not by diverting the existing current. And, as Lutz noted above, it "pushes" its current back into the input voltage source. But why? What is the point of this?

To answer this pertinent question, let's consider, like in the case of N-shaped differential negative resistor, the simplest INIC arrangement - a perfect voltage source loaded with an N-shaped true negative "resistor" (INIC). Assume the source voltage varies within the negative resistance region.

As above, to explain the operation of this bizarre circuit, first imagine the voltage source is loaded by a grounded resistor R (its lower end is connected to ground). Figuratively speaking, the right end of the resistor is "lower" than its left end and therefore the current is coming out of the source. So, if the voltage increases, the current will proportionally increase according to Ohm's law… and the source will experience considerable loading.

Then let's disconnect the lower end of the resistor from ground and insert there a varying voltage source (like in the attached picture). Make its voltage equal but opposite to the input voltage. The right end of the resistor is at the same "height" as its left end (as though the resistor is stretched between the two sources); so there is no current flowing through the resistor... no current is consumed from the input source... and it "feels" unloaded. This is a "bootstrapping".

Finally, make the opposing voltage twice the input voltage (as in the attached conceptual picture). Now the right end of the resistor is "higher" than its left end. The result is surprising - the load current is the same as in the first case but with an opposite direction! Instead to leave it enters the positive terminal of the input voltage source. The source "feels" a kind of "super-unloaded"... the load acts as an "anti-load".

Is there any sense in this configuration? I do not see any sense especially in the case of an ideal voltage source… Then what can such a negative "anti-load" be used for? It can compensate an equivalent positive load connected to the input source. Let's consider an example.

Remember the voltage divider (potentiometer) above. It was loaded by a resistive load with "positive" resistance RL that decreased the output voltage. To eliminate it, now we can connect in parallel a true N-shaped negative resistor (INIC) with the same but "negative" resistance -RL. The result will be infinite resistance. What is this magic? Here is my explanation:

The divider output is loaded by two resistive loads in parallel - positive and negative (think of this combination as a kind of composed load). The output voltage is in the middle of the negative resistance region.

If the output voltage VL increases, the positive load should consume proportional current VL/RL from the divider output. But the doubled voltage source passes the same current (2VL - VL)/R = VL/R through the load. The network of a resistor (R = RL) and doubled voltage source (V = 2VL) acts as a negative "resistor". It provides the whole load current so the voltage divider has the "feeling" that it is not loaded… it works at perfect load conditions (open circuit).

This was a conceptual INIC. It can be made by a non-inverting amplifier with а fixed gain of 2...

... or, most often, by an op-amp non-inverting amplifier.

Before concluding this interesting "discussion", I want to make a little emotional digression…

With these explanations at the lowest possible level, I wanted not only to truly explain these circuit phenomena veiled in mysticism... but also to show the strength of this approach to the formalism that currently prevails. Because, one thing is to "explain" the incomprehensible electronic circuit with ready-made clichés such as "differential resistance", "positive feedback", "small-signal values", "stability", etc. ... and another thing is to reduce it to an elementary circuit of two electrical elements...

I think that figuring out basic circuit ideas in such a notional "philosophical" way - through much imagination, searching for life associations and analogies, thought experiments with simpler equivalent circuits… and other heuristic means, is no less important than designing specific chips and putting them into practice. The mere fact that few people can think in this productive way - the way inventors think, means they at least deserve respect... even just because they do it selflessly... and share with everyone...

At least they do not deserve silence...

Hi Cyril......I am not sure if the following fits well with the discussion:

I think, each working oscillator circuit (even if designed as a loop consisting of an amplifier and a passive fedback network) can be seen also as a parallel combination between a positive and a negative resistance - however, at one single frequency only. In some cases, such a view can simplify the analysis of transistor-based oscillator circuits.

Hi, Lutz!

As far as I can remember, this is the configuration in the case of an LC oscillator where an N-shaped negative "resistor" (INIC) connected in parallel to an LC tank compensates its positive resistance (the resistive losses in the inductor).

A really good example… thank you!

The support of just one from 15+ million RG members was enough to stimulate my imagination… and I started thinking again about what exactly an INIC does when annihilating (making infinitely high) equivalent positive resistance...

Cyril - not only in case of an LC oscillator....this applies to ALL oscillator circuits (including RC versions). However, sometimes such a view makes not much sense and does not help, but it is interesting to know that such an interpretation is always possible.

In my opinion, such an interpretation means to think in terms of resistance instead of voltage or currents.

The advantage of this approach is manifested in cases where, in certain electronic circuits with two terminals, the current is proportional to voltage or the voltage is proportional to current. This allows us, instead of considering them as something vastly different from passive resistors, to represent them as dual to them.

As a result of the interaction between such "normal" (positive) and "dual" (negative) resistors, zero or infinite resistance is often obtained, which simplifies the analysis.

This powerful approach can also be applied to other electrical elements - nonlinear resistors, capacitors, inductors, etc.

My feeling that I was in an uplifting creative inspiration that would lead to another insight turned out to be true...

Riding a bike for more than an hour at University Park, I was able to realize something that had been bothering me for years - what exactly the INIC has been doing in both famous circuits with negative resistance named "Howland current source" and "Load canceller" (compensated voltage divider above)... what is common and what is different between them...

Coming back to the subject of harmonic oscillators - the fact that all oscillator circuits can also be interpreted as a parallel combination of a passive impedance and an (active) negative impedance makes the distinction between two-pole and four-pole oscillators obsolete. Or - am I wrong?

This seems to be because the negative "resistor" is actually a 1-port amplifier and its input and output are located on this single port.

But I think we have to differentiate between "circuit with negative resistance" and "circuit that acts as a circuit with negative resistance". I will give an example that I often use but usually others do not understand me:

In an op-amp inverting amplifier, the op-amp output is connected in series to the resistor R2. The op-amp output produces voltage equal to the voltage drop across R2 and, in this way, it compensates the voltage drop. The result is the virtual ground at the inverting input.

Seen from the side, it looks like the op-amp is acting like an S-shaped negative resistor (VNIC) with resistance -R2 that neutralizes the positive resistance R2... and the result is zero resistance at the inverting input. But this is not entirely true because the true negative resistor (even if it is a circuit) has only two terminals... and in this case it would produce voltage across itself based on the current flowing through it... i.e., it would act as a "2-terminal current-to-voltage converter".

The op-amp here produces the same voltage but in a little different way - it "observes" the virtual ground and changes its output voltage so that to keep it always zero... i. e., it does it by means of negative feedback. For this purpose, it needs a third wire - the negative op-amp input, to "sense" the result of the voltage subtraction.

That is why, I say that in the op-amp inverting amplifier (transimpedance amplifier) the op-amp output "acts as a negative resistor"... not "it is a negative resistor":

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#share

So I think that more frequently in oscillator circuits we observe circuits acting as negative resistors (having more than two terminals) than genuine negative resistance circuits (that would have only two terminals).

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

By the way, I decided to look for a more up-to-date material on the Howland current source... and here is what I found (written just a month ago):

https://circuitdigest.com/electronic-circuits/howland-current-pump-circuit