Question

# What is "negative differential resistance"? How is it implemented? How does it operate? What is its relationship with the true negative resistance?

When we see somewhere written "negative resistance", we are never sure what exactly it is, just because it has two varieties - "true" and "differential"... and the former is not a resistance:) We have already started a discussion about the true (absolute) negative resistance
and its op-amp implementation (NIC)
Now it remains only to clarify what the negative differential resistance is. To help the discussion, here are my speculations about the amazing phenomenon.
If we increase the voltage across an ordinary ohmic resistor, the current flowing through it increases proportionally according to Ohm's law (Iout = Vin/R). Also, if we increase the current through the resistor, the voltage across it increases as well (Vout = Iin.R). Thus the voltage and the current change in the same directions. But there are mysterious two-terminal electronic components (negative differential resistors, shortly NDR), having just the opposite behavior in the middle part of their IV curves - the voltage across and the current through them change in opposite directions. Some of them - neon lamps, thyristors, have an S-shaped IV curve while other - tunnel, Gunn and lambda diode, have an N-shaped IV curve (see the attached picture).
Although the negative differential resistance seems to be a mystic phenomenon, it is actually based on an extremely simple, clear and intuitive trick - the powerful "dynamizing" idea that is brought here to the utmost degree. IMO a differential negative resistor is nothing more than a kind of "self-varying", dynamic resistor that changes extremely its instant (ohmic, chordal) resistance depending on the current passing through the resistor or on the voltage applied across it... the differential negative resistor is actually an "over-dynamic" resistor.
Negative differential resistors are dynamic but still positive resistors. They have different kinds of resistance in the three parts of their IV curves (located in the 1st or the 3th quadrant) - "positive" in the end parts and negative in the middle part. The three parts form the whole IV curve that, depending on the NDR behavior in the middle part, can be S- or N-shaped. When the input quantity (no matter current or voltage) increases, the S-shaped NDR decreases while the N-shaped NDR increases its instant resistance.
Each of the two NDR can be driven both by voltage and current. To operate in a linear mode, an S-shaped NDR has to be driven by current while an N-shaped NDR - by voltage (the attached picture); v.v., to operate in a bi-stable mode, an S-shaped NDR has to be driven by voltage while an N-shaped NDR - by current. So, in contrast to the widespread but misleading viewpoint, there are no particular current controlled (CCNR) and voltage controlled (VCNR) negative resistors - there are only S-shaped and N-shaped NDR, and each of them can be controlled both by a current and voltage.
In the middle negative resistance region, negative differential resistors behave as two-terminal active elements (such as transistors). They cannot be used independently; they need an additional power supply to be connected. Thus the combination of the negative differential resistor and the power supply can be considered as another kind of a true negative resistor (the other kind of a true negative resistor is a combination of a constant "positive" resistor and a varying voltage source - this is the case in a NIC). From another viewpoint, this combination can be thought as of an electrical source with negative internal resistance. It is a usual practice to think of a negative differential resistor as of a true negative resistor implicitly assuming the existence of a power supply...
As I have told in the question about the true negative impedance, I wasted a few years of my life to struggle with orthodox wikipedians inhabiting the electronics Wikipedia with the idea to tell the truth about the negative (differential) resistance in the respective Wikipedia article... but I could not... All my edits and imagine finally - the whole article about the negative differential resistance, were removed! Here is a short history of this "epic battle" (I was writing under the user names Circuit-fantasist and Circuit dreamer):
and some my insights shared with wikipedians
But I did not give up and after a year, I told the truth about the negative differential resistance in Wikibooks:

13th Aug, 2019
Cyril Mechkov
Technical University of Sofia
1 Recommendation

1st Sep, 2013
Cyril Mechkov
Technical University of Sofia
Maybe it would be interesting to see how a tunnel diode can amplify?
1 Recommendation
2nd Sep, 2013
Tolga Soyata
George Mason University
hey Cyril, never mind my "just LC can do negative impedance" discussion ... TUNNEL DIODEs can do it too ! I remember this from literally 25 years ago back when I was in undergraduate. Since tunnel diodes can achieve negative resistance, they can be used as oscillators. Check this out:
But, since you have to BIAS them in their operating region where they exhibit negative differential resistance, this biasing circuitry costs energy. So, OHHH NOOO . Another proof, there ain't no free electrons :) You gotta pay for them :)
2nd Sep, 2013
Tolga Soyata
George Mason University
What I mean is : the negative resistance is only within a certain region. To get the diode into that region, you have to have spent energy and brought them there. So, negative resistance is not really giving you free electrons. It is just GIVING BACK the electrons in a controlled way. They are , in the end, your electrons, that you put there !
1 Recommendation
2nd Sep, 2013
Cyril Mechkov
Technical University of Sofia
Hi Tolga! It is wonderful that there are yong people like you with a sense of humor to enliven the "sterile" RG scientific discussions! Circuitry can be fun especially if we make associations with similar situations in our daily life. So let's not forget that we (still) are human beings, not (already) computers and smile as much as possible in our discussions:):):)
About the energy costs... IMO the main losses are into the very negative differential resistor (the tunnel diode in the picture above) since it is just a kind of a resistor dissipating energy in the surrounding area. We may think of the tunnel diode as of a "two-terminal transistor". From the energetic viewpoint, both the devices are passive.
I completely secong your "...the negative resistance is only within a certain region. To get the diode into that region, you have to have spent energy and brought them there..." I hope that you will be pleasantly surprised if you take a look at my reasoning on this issue:
Regards, Cyril
3rd Sep, 2013
Lauri Kettunen
University of Jyväskylä
To take sides on this kind of questions, the first step is to define what is meant by resistance. But, in short, think of Poyntings's theorem and the integral of J.E over a volume. If the integral of J.E is greater than zero, this is the amount of power transferred into heat/ohmic losses in the volume. The other way around, if the integral of J.E is negative, then the volume is a generator.
Next, think of circuit theory. The integral of J.E over a resistor yields the power going into ohmic losses. This power equals to the product UI, and the resistance is the ratio U/I. In case of a resistor, resistance is positive. Next, think of a battery in a DC-circuit. In this case the integral of J.E over the battery is negative and the product UI is the power taken from the batter, i.e., from the generator. Now, if you want to look at the ratio U/I, you'll find that it is negative in this case, and you may interpret this such that the battery is a generator in your system.
2 Recommendations
3rd Sep, 2013
Cyril Mechkov
Technical University of Sofia
OK... but we consider here only the negative differential resistance that is only a resistance... and the element is a resistor... there is no power source... If we include the power source, the combination of the two elements will constitute a true (absolute) negative resistor...
IMO we can assemble a true negative resistor by connecting in series two elements as follows:
* constant ohmic resistor + variable voltage source (NIC)
* variable resistor + constant voltage source (tunnel-diode amplifier)
3rd Sep, 2013
Tolga Soyata
George Mason University
Lauri, I like the Poynting approach. It is one of the many great theorems that says "you can't generate energy out of nowhere :) The net energy is zero, whether it is magnetic energy, electric energy, or whatever else !
The NEGATIVE RESISTANCE is effectively GENERATING ENERGY. You apply 10 Volts, and DRAW -2 mA. So, you are consuming -20 mW, i.e., GENERATING 20 milli Watts of power. But, it is happening on an ISOLATED PIECE OF THE CIRCUIT, at the expense of creating losses on THE ENTIRE CIRCUIT, or, even ON ANOTHER PIECE OF THE CIRCUIT.
Think about this example: I can have an electric cooler in my car. If I just analyze what happens INSIDE the COOLER. Amazing ! It is GENERATING ENERGY since it is cooling inside the cooler. Well, if I look at the entire car, the cooler did this at the expense of either taking energy out of the battery, or consuming gas.
Expanding on Lauri's Poynting Theorem explanation, and explaining in a little simpler way, the mechanism that creates NEGATIVE RESISTANCE is actually LOSING ENERGY ! Why ? Because, the OPAMPs etc ... that allow you to create the effect of negative resistance in an isolated part of the circuit are consuming more energy than what you generate from the effects of negative resistance ... And, while that energy used by OPAMPS and resistors in the circuit are getting lost as heat (as Poynting says :), you are temporarily gaining energy in another part of the circuit (<= what you gained).
Did I explain this right Lauri ?
1 Recommendation
3rd Sep, 2013
Lauri Kettunen
University of Jyväskylä
Let's make a thought experiment: At first you charge a big capacitor and put it into a black box. The only thing what I can see are two connectors on the surface of the black box. Next, I'll connect a resistor to the connectors and by measuring the differential voltage and differential current to find out that the ratio delta U/delta I is negative over the black box and positive over the resistor. Consequently, for a moment the black box appears as a power source in my eyes.
This boils down to Thevenin-Norton theorem and how a system appears when lookin at the voltages and currents on the connectors to the system. Taking this view the idea of negative differential resistance has to do with energy balance.
1 Recommendation
3rd Sep, 2013
Tolga Soyata
George Mason University
Yes ... And, continuing that discussion, if you "observe" the power sources and power sinks that way (black boxes) and add them up, you get a big ZERO :) right ?
3rd Sep, 2013
Lauri Kettunen
University of Jyväskylä
Tolga, yes, I share the same view with you; energy conservation is the corner stone of all physics.
1 Recommendation
3rd Sep, 2013
Cyril Mechkov
Technical University of Sofia
Obviously, the true negative resistance is more interesting to discuss:) Yet I want to remind you again that there is a special issue for him
and its op-amp implementation (NIC)
Tolga, you can find a lot of answers to your questions if you figure it out the circuit of an op-amp NIC. These stories can help you:
http://www.circuit-
but let's discuss them where they belong.
Well, I will repeat here the simple truth about true negative resistors (NICs) - they are just sources... self-variable sources... functional converters: VNIC is a 1-port current-to-voltage functional converter (voltage source - see the attached picture); INIC is a 1-port voltage-to-current functional converter (current source). The function is set by a sample "positive impedance" element (resistor, capacitor, diode...) inside the NIC.
Tolga, about your "...the mechanism that creates NEGATIVE RESISTANCE is actually LOSING ENERGY..." - true negative resistors do not obligatory dissipate energy. Like ordinary steady electrical sources, they convert some non-electrical into electrical energy... and it is possible to do that (almost) without losses (e.g., by applying a PWM). It is possible but difficult for implementation (I have not yet an idea how to do it; maybe you have?:)
Will you say the battery dissipates energy... has losses...? But if you think of the consumed non-electrical energy as of losses, then I agree with you...
1 Recommendation
9th Oct, 2013
Gana Nath Dash
Sambalpur University
Hello All, It is tempting to associate myself in the discussion on "Negative differential resistance". It is true that a device can show negative resistance in certain region of its current-voltage characteristic - like that of a Tunnel diode. It can also be available as a dynamic effect by manipulating a phase difference of pi/2 to 3pi/2 between the AC voltage and current - as in IMPATT, MITATT, TUNNETT and the QWITT diodes. A third way of getting negative resistance is through what is called "inter-valley transfer of electrons" in semiconductors like GaAs and InP which forms the basis of the Gunn diode. Negative resistance can in this way be obtained in different ways all of which are meant to harness power. This does not mean that it will generate power on its own. it simply does not dissipate power and help us convert power to our need.
Another point I would like to add is that Negative resistance is a dynamic effect - which is available for a given frequency range of the AC or it can be the differential resistance which can be negative. The static resistance of a conductor can never be negative - this would mean producing power out of nothing and it is prevented by the second law of thermodynamics.
Thank you all for sharing your views
G.N. Dash
1 Recommendation
9th Oct, 2013
Cyril Mechkov
Technical University of Sofia
Hi, Gananath! Thank you for the contribution. Your thoughts are interesting and valuable. I would only note that the negative differential resistor still dissipates power.
Also, we can make the static resistance (true) negative if we add in series a voltage that is proportional to and higher than the voltage drop across the resistor (the idea of the negative impedance converter with voltage inversion VNIC):
Regards, Cyril
1 Recommendation
11th Oct, 2013
Gana Nath Dash
Sambalpur University
Hi Cyril! Thank you for providing the link. But I observed the conclusion from that is no different from what I asserted. I copy them below
________________________________________________________________
"True negative versus differential negative resistors
True negative resistors are electronic circuits while negative differential resistors can be elements (components) as well as circuits.
Both the negative resistors are dynamic electronic elements (circuits).
True negative resistors are dynamic electrical sources while negative differential resistors are just dynamic resistors that cannot be used independently; they may be used in combination with electrical sources to build true negative resistors."
________________________________________________________________
My point is that DC resistor of an isolated element cannot be negative - that is because it will amount to generating power out of nothing.
we can try several methods to make the resistance negative, to accomplish some specific task or for academic purpose. but we should consider the power of the whole circuit - not the negative element only. If the total power is negative (that is power is produced) then probably the whole electrical engineering has to be rewritten. In this context I fully support Tolga.
But anyway I appreciate the different circuit methods attempted for negative resistance in the wiki book that you suggested. They are interesting pieces of innovative attempts.
Thanks
Gananath
2 Recommendations
11th Oct, 2013
Tolga Soyata
George Mason University
Thanks Gananath,
Negative differential resistor sounds MAGICAL, but, in the end, you are making the OPAMP immitate the behavior of "when the voltage goes up, the current goes down" ... for which you are expanding energy ... In the end, if you count the energy you are spending, and add this current to the current that SUPPOSEDLY WENT DOWN , in the end, the total current WENT UP. So, it is not that NEGATIVE after all :)
1 Recommendation
11th Oct, 2013
Cyril Mechkov
Technical University of Sofia
Thank you Gananath and Tolga for your interesting thoughts!
Gananath, we can think of the negative differential resistor as of a "two-terminal transistor" (i.e., an active element that controlls the power by vigorously changing its instant resistance). It does it in the same manner - by dissipating the redundant power (consumed by the power supply) to the surroundings. Thus the combination of the NDR + power source can be considered as a two-terminal (one-port) amplifier. After this analogy we can apply all the knowledge from the field of amplifiers to explain the NDR arrangement.
Theoretically, a true (absolute) negative resistor should not dissipate power since it is a true source (sources do not dissipate energy, right?) But the practical circuit have an internal "positive" resistance to "sense" the current and produce a voltage proportional to this current... so they dissipate power...
Tolga, IMO there is nothing magical in a negative differential resistor; it is just a two-terminal element - an EXTREMELY DYNAMIC RESISTOR... nothing else...:)
Similarly, there is nothing magical in a true negative "resistor" ; it is just a two-terminal op-amp circuit - an EXTREMELY DYNAMIC SOURCE... nothing else...:)
Usually, negative differential resistors are elements (e.g., tunnel and Gunn diodes, neon lamps, etc.) and rarely - circuits made by transistors. I have not yet seen an op-amp circuit implementation of a negative differential resistance. The reason is obvious - the op-amp output is a voltage "source" while the transistor output is resistive...
21st Jun, 2014
Simone Orcioni
Università Politecnica delle Marche
Dear Cyril,
I have read your interpretation of negative differential resistor and I'll try to synthetize and formalize it in the case of a N-shaped (VCNR).
"Let us to have a nonlinear function i = f(v) that for some v exhibits g=di/dv < 0.
I can obtain the same behaviour by thinking that exists a positive dynamic resistor, I'll call it a Voltage Controlled Resistor i = R(v) v.
To make this true it is sufficient to take R(v) = f(v)/v.
The new i = R(v) v has obviously di/dv = g < 0. "
Clearly I can make the same for the interval when g>0.
The main drawback of this interpretation is that of each of your "linear functions" i = R(v) v, it exists only one point (v, i).
All these points meet the real characteristic i = f(v).
So why to limit to dynamic resistors?
In a similar manner I can say that in reality the negative differential resistance is caused by dynamic parabolics functions i = a(v) v^2.
sim
21st Jun, 2014
Simone Orcioni
Università Politecnica delle Marche
Dear Cyril,
"An S-shaped negative differential resistor (compare with S-shaped NR).
(these elements are frequently named "current controlled negative resistors" - CCNR, although they can be controlled by voltage as well)"
I suppose that the same holds for the N-shaped or VCNR, ie that it is possible to control it in current.
In the past we designed and built a circuit with the behaviour of a VCNR.
You can see it in the attached file in Fig. 9 and the characteristic in Fig. 10.
(The characteristic is S-shaped since we put the current in abscissa because we were interested to the hystesis in current).
To obtain Fig. 9 we needed to built an additional circuit to control it in voltage e measure the current. If it was possible to obtain the characteristic by controlling the circuit in current we would have saved a lot of time.
So it is really possible to control a VCNR in current and a CCNR in voltage?
Maybe it depends on the "interpretation" of the verb "to control" ?
sim
21st Jun, 2014
Cyril Mechkov
Technical University of Sofia
Simone, thanks... but I will answer shortly since I am on holiday until the end of the next week... I simply mean the bistable mode of NDR where it acts as a latch:
21st Jun, 2014
Simone Orcioni
Università Politecnica delle Marche
Dear Cyril,
enjoy your holiday and see you soon.
sim
16th Aug, 2014
Cyril Mechkov
Technical University of Sofia
I have copied this related discussion about NDR from Pages 6 and 7 of the question below:
---------------------------------------------------------------------------------
Serge Luryi: "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..."
Lutz Wangenheim: "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?"
Serge Luryi: "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."
Lutz Wangenheim: "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?"
Serge Luryi: "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... 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."
Lutz Wangenheim: "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?"
Serge Luryi: "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."
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) ?"
Serge Luryi: "1) Correct; 2) I have already explained that. Resistor at different temperatures is not the same resistor..."
16th Aug, 2014
Cyril Mechkov
Technical University of Sofia
Dear Serge Luryi,
I am really fascinated by your way of thinking... maybe because I think in a similar imaginative way... but I, as well as Lutz Wangenheim, can not share your basic concept : "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..." Here are my objections.
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... I have thoroughly considered this "trick" (in your manner of thinking) in the Wikibooks story below...
... and generalized the trick here...
... but I will consider it again with pleasure...
16th Aug, 2014
Cyril Mechkov
Technical University of Sofia
Well, let's do it as an imaginary "voltage-divider game" - you will change the input voltage VIN; I will change the resistance R2 (or v.v., if you prefer:) A will use a similar setup from the Wikibooks story where the voltage source is perfect (without Rin):
First imagine that when you increase VIN, I do not react... and R2 stays steady... Both the current through (I = V/(R1 + R2) and the output voltage across R2 (VOUT = V*R2/(R1 + R2) proportionally increase (change in the same directions)... what is an indication of a positive resistance...
Then imagine that when you increase VIN, I begin simultaneously increasing R2... I can do it moderately... and, as a result, the voltage increases but the current increases more lazily...
... "exactly"... and the voltage increases... but the current does not change...
... or enormously... and the situation becomes most interesting - the current decreases but the voltage VOUT increases even more significantly than VIN... and this is a the so desired negative resistance...
16th Aug, 2014
Cyril Mechkov
Technical University of Sofia
So, my final conclusion is that to see a negative differential resistance, really the positive resistance should be controlled... but not by another port. It has to be controlled by the same port... it has to be "self-controlled"...
In my opinion, we should not talk about negative resistance when the voltage and current change in opposite directions... but because we change the positive resistance as an input quantity... In this case, looking from the side of the input controlling voltage source, we will not see any negative resistance... we will see only the positive resistance of the input load port...
1 Recommendation
16th Aug, 2014
Erik Lindberg
Technical University of Denmark (emeritus)
Dear All
In basic circuit theory a resistor is a two-terminal element for which the element voltage (branch voltage) is a function of the element current (branch current) i.e. V = R(I). If the resistor is linear we have V = R*I  or R = V/I which is the law of Ohm.
For a general nonlinear resistor we may introduce the instant static value equal to the ratio V/I and the instant dynamic value equal to the instant slope of the curve (the tangent). These values may be negative.
The resistor may be interpreted as a current controlled voltage source and we may talk about input or transfer components if we generalize to four terminal components.
The conductor is the dual of the resistor i.e. I = G(V).
16th Aug, 2014
Cyril Mechkov
Technical University of Sofia
Hi Erik,
Thank you for the comment. I would like only to remember that the static negative resistance can be negative only if the element contains an internal source... or if it is a source...
BTW what does "general nonlinear resistor" mean?
16th Aug, 2014
Erik Lindberg
Technical University of Denmark (emeritus)
Hi Cyril,
concerning:
"BTW what does "general nonlinear resistor" mean ?"
The voltage V is an arbitrary function R of the current I i.e. we have an arbitrary curve in the IV-plane. This may include sources and problems with multiple solutions.
2 Recommendations
31st Dec, 2016
Simon Fraser University
Negative resistance should cause cooling in the circuit
11th Aug, 2019
Monjul Saikia
North Eastern Regional Institute of Science and Technology
13th Aug, 2019
Cyril Mechkov
Technical University of Sofia