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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?

Somewhere I have read that an example of component with negative resistance is a voltage source. But I do not understand this statement, since a voltage source is a component which at most shows a (positive) internal resistance.

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  • \$\begingroup\$ Maybe if you see a circuit with two resistors in series (voltage divider), having in the middle 2.5V, a component with negative resistance can be said to 'add voltage' instead of removing voltage... but I leave a real answer to the experts here ;-) \$\endgroup\$ – Michel Keijzers Apr 25 at 16:11
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    \$\begingroup\$ Minus R will provide power, not dissipate power. \$\endgroup\$ – analogsystemsrf Apr 25 at 16:41
  • \$\begingroup\$ Meh, there's no such thing as negative resistance. It's an artifice of improperly (IMO) applying Ohm's Law to something non-linear (not resistor-like). If we flip it into conductance, you are saying something has a negative conductance, i.e. its conductance goes below 0 (below total insulator - in other words current flow induces a reverse voltage). Such a device is not a good fit for Ohm's Law. \$\endgroup\$ – Harper Apr 25 at 23:04
  • \$\begingroup\$ There are two types 'S' & 'N'. en.wikibooks.org/wiki/Circuit_Idea/… \$\endgroup\$ – Optionparty Apr 26 at 3:10
  • \$\begingroup\$ You can make your own negative resistor. Go to sparkbangbuzz.com/els/ntype-nr-el.htm \$\endgroup\$ – richard1941 Apr 26 at 5:41
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There are a number of mechanisms that result in a region where locally increasing voltage results in locally decreasing current. For example, an Esaki (tunnel) diode.

enter image description here

A common example would be a switching power supply with a steady load. Assuming the efficiency is more-or-less constant, increasing the input voltage results in less current being drawn. It is always consuming energy though.

A stand-alone component that exhibits negative resistance (rather than negative differential resistance) is not possible without some kind of energy source within the component, otherwise it would violate conservation of energy (\$P = E^2/R\$) and negative P would indicate it is acting as a power source.


If you want to play with a negative resistance effect, one way (assuming you don't mind one end being grounded) is to use a negative impedance converter:

schematic

simulate this circuit – Schematic created using CircuitLab

The above circuit acts like a -10K resistor with one end grounded (within its linear range), and works down to about zero volts. Any power it produces comes from the op-amp supplies.

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    \$\begingroup\$ That is really a fine choice of an example device you picked. \$\endgroup\$ – The Photon Apr 25 at 16:23
  • \$\begingroup\$ @ThePhoton LOL, great minds and all that. \$\endgroup\$ – Spehro Pefhany Apr 25 at 16:45
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    \$\begingroup\$ @J... No, it really is negative differential resistance. You put a stiff voltage across it and keep it from oscillating the current will follow that curve. See, for example, DC Characterization of Tunnel Diodes Under Stable Non-Oscillatory Circuit Conditions by Wang et al. \$\endgroup\$ – Spehro Pefhany Apr 25 at 18:31
  • \$\begingroup\$ This is an example of a "type N" device. There are also "type S" devices. \$\endgroup\$ – richard1941 Apr 26 at 5:46
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In this context, we have to discriminate between (1) pure differential (dynamic) neg. resistances (as shown in the examples of the other answers) and (b) a static negative resistance.

For a differential neg. resistance (rdiff) the current CHANGES are negativ, for a static neg. resistance the CURRENT itself has a negative sign.

My following answer concerns only the static negative resistor:

Such an element does not "consume" a current - driven by a voltage source, but - the other way round - it drives a current (prop. to the voltage) in an opposite direction into the voltage source.

Hence. it is a voltage-controlled current source. For such circuits only active realisations are possible (using transistors or - in most cases - opamps). The most popular circuit is the NIC (Negative-Impedance Converter).

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Anything that drops in voltage with a rise in current has a negative resistance.

Power sources have this property. The passive components with incremental negative resistance include; any gas discharge bulb or arc, Avalanche effect diodes, Tunnel Diodes, SCR's during trigger phase.

https://en.wikipedia.org/wiki/Negative_resistance

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But how is this physically possible?

Some components, like Esaki diodes and glow tubes, have an I-V curve that is entirely in the I and III quadrants, but has a negative slope region over a limited range. In this region, a small-signal model of the device will have negative resistance.

enter image description here

(image source)

In the Esaki diode, this behavior is caused by tunneling current that is possible at low bias but not at higher bias voltage.

It's also possible to make an op-amp circuit with negative input resistance over a limited range. There the I-V curve can even pass through the II and IV quadrants since power can be supplied from the op-amp's power terminals.

Somewhere I have read that an example of component with negative resistance is a voltage source.

Looking at the input side of a regulated switching supply with a fixed load, it will often appear as a negative resistance.

This is because it is a constant power load. If the input voltage drops, the regulator circuit will increase the current drawn in order to continue supplying the load with the desired output voltage.

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Somewhere I have read that an example of component with negative resistance is a voltage source. But I do not understand this statement, since a voltage source is a component which at most shows a (positive) internal resistance.

Perhaps a voltage source is mentioned, because we all know that an ideal voltage source should have zero internal resistance: a good one will have a small positive resistance, to which is added any wire resistance going to the load.

For an electronically regulated supply, it is possible to force output resistance past zero into negative resistance region. This is done by routing some of the load current so that regulating voltage node is adjusted in such a direction that output voltage is forced up. An example of the common LM317 regulator having negative output resistance is shown below - beware, some loads produce wild results:

schematic

simulate this circuit – Schematic created using CircuitLab
Using the built-in circuit simulator, \$ R_{load} \$ was swept from 5 ohms up to 15 ohms:

  • at 5 ohms, voltage drop across Rload is 4.322V

  • at 15 ohms, voltage drop across Rload is 3.993V

The result of that 1-ohm resistor, (and the direction of Rload's current going through it) forces this voltage supply to have negative resistance: at heavier loads, voltage across the load resistor goes up.

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Shown here is a "Typ-A" NIC block. The grounded resistor (impedance) R3 is converted into a negative resistor (impedance) with a conversion factor (-R1/R2). This typ is short-circuit.stable. (An open-circuit stable NIC results for interchanged opamp inputs).

schematic

simulate this circuit – Schematic created using CircuitLab

Comments: The shown NIC is stable as long as the source resistance of the voltage source (not shown in the figure) is smaller than R1. These NIC blocks are use for undamping filters, oscillators and other systems with unwanted positive (parasitic) resistances. Mathematically, they can be treated as "normal" resistors in series and parallel combinations - however, with a negative sign, of course.

A very popular application is the "NIC integrator" (or "Deboo integrator"), where an NIC block is connected to the common node of a simple R-C lowpass. In this case, the NIC can compensate the pos. resistor R - thus resembling a current source which loads the intergating capacitor.

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  • \$\begingroup\$ Why did you answer twice? \$\endgroup\$ – pipe Apr 25 at 17:58
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    \$\begingroup\$ It was by accident.....I have tried to include the figure (later) - and suddenly there were two answers... \$\endgroup\$ – LvW Apr 25 at 18:32
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A perfect negative resistor is impossible, but a device can have negative resistance characteristics over a limited range.

The resistance of a non-linear device varies and at a given voltage the equivalent resistance is equal to the slope of the line. If the slope is negative in a range, that range has negative resistance.

enter image description here

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  • \$\begingroup\$ Mattmann944...I think it is important to add that your example concerns a DIFFERENTIAL (dynamic) negative resistance only!! Each working point on your "neg. Resistance" curve resembles a POSITIVE static resistance. More than that, a "perfect" negative resistor is possible, indeed (however, as perfect as each electronic part can be....). No ohmic resistor is "perfect". \$\endgroup\$ – LvW Apr 25 at 19:25
  • \$\begingroup\$ Yes, your answer is technically more correct than mine. The OP doesn't appear to be a college student, so I tried to keep it simple. I have only seen negative resistance used in the differential sense. Most of the Wikipedia article is devoted to differential. I did say slope, which implies differential. \$\endgroup\$ – Mattman944 Apr 25 at 22:40
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DC-DC Converter inputs are a good example of a negative resistance. As voltage goes down, current increases to provide the same power output. Also a negative resistance can be created by an op amp circuit.

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In a simple way, resistance is the ratio between voltage and current, if you plot the voltage versus the current present in a certain component, the resistance will appear as the slope between these variables. In a physic way, a positive resistance means that if the voltage of a component rises, the current that flows by also rises, otherwise, a negative resistance means that when the voltage of a component rises, the current declines.

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