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How do I make/describe an element with negative resistance of minus 1 Ohm?

It's I–V characteristic should then be

\$I = -U\$

Is this possible?


  • I need exactly what I wrote. No compromises. The negative resistance should not be used differentially.
  • I need to build such the device. It should not only work grounded, but at any potential.
  • Of course it should be kind of "active".
  • If a power source is required for such a device -- then that is ok. Power sources are allowed :)
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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Voltage Spike
    Dec 25 '20 at 19:27
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You can make such a device using active components. If one end of the resistor is grounded, this simple circuit would work, at least in theory. The op-amp would have to be able to deliver sufficient current and have sufficient voltage swing for whatever voltage you wanted to apply.

schematic

simulate this circuit – Schematic created using CircuitLab

For example, if you apply 5V, the op-amp would have to swing to +10V while sourcing 5A.

An actual negative resistor would be a power source (P = E^2/R) so it's impossible to have such a device that does not use or contain a source of energy.

There are a few devices and circuits that exhibit negative differential resistance- over some region of operation the current drops with increasing voltage (while the total current is always positive). For example, Esaki diodes, discharge tubes, switching power supplies. So in a small signal analysis, resistance can be negative.


Edit:

You've added that you require it to be ungrounded. In that case, merely power it by a battery or an isolated DC-DC converter and use the circuit as shown (with a suitable power op-amp). There are still many "compromises" - as in limitations and non-idealities as there are with any real circuit or element, of course.

Here is a simulation with a 1K resistor for R1 which simulates a -1K resistor.

schematic

simulate this circuit

enter image description here

As you can see the current through the simulated -1K resistor is the mirror of the current through the real 1K resistor R4. I've offset the other side of the resistor by 5V from ground in each case and the sawtooth goes from 0 to 10V so the voltage across each element goes from -5V to +5V. Not shown are the power supplies.

You won't easily be able to simulate the isolated version because op-amp models typically have ground nodes within the model.

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    \$\begingroup\$ @Circuitfantasist in that case write a better answer. \$\endgroup\$
    – mkeith
    Dec 24 '20 at 22:38
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    \$\begingroup\$ @mkeith No, no, don't dare Circuit fantasist to write an answer! \$\endgroup\$ Dec 24 '20 at 23:10
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    \$\begingroup\$ @PeteW it is not enough that the slope of the V/I curve be negative. It must also be linear, or somewhat linear. A resistor means V / I = k, where k is constant. A constant power load means that V * I = k, where k is constant. Not the same thing. \$\endgroup\$
    – mkeith
    Dec 24 '20 at 23:10
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    \$\begingroup\$ @Circuitfantasist It behaves like a -1 ohm resistor with one end grounded. Just as requested. For 0V input (short the input to ground) current is 0. For 1V input, current into point X is -1A. Etc. \$\endgroup\$ Dec 24 '20 at 23:25
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    \$\begingroup\$ @PeteW A switching power supply with a constant load on a regulated output is (ideally) a constant power device. But it's only differential resistance. If you look at the input current at the minimum input voltage at which it will operate, the current will flow into the (+) terminal of the supply. As you increase the voltage the current will decrease from that (high) value to lower values. But it will never get to zero, let alone negative. \$\endgroup\$ Dec 24 '20 at 23:28
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A resistor is a two terminal circuit element that enforces the following relationship between voltage and current:

V / I = R

Where V is the voltage measured across the two terminals, and I is the current flowing through the element. To be a resistor, R must be constant over a range of voltage and currents.

Logically, then, a resistor with resistance of -1 Ohms would be a two terminal element which enforces the following rule:

V / I = -1

I think this explains what a negative resistance is or would be at a theoretical and/or mathematical level. Approximating a negative resistance with electronics is possible (see Spehro's answer). But that is more of a negative resistance emulator than a true negative resistance.

A true negative resistor would deliver power to any voltage source connected to it. The higher the voltage (in magnitude), the more the power. So this is not possible for a passive circuit element. Spehro showed how it can be done with active elements.

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  • \$\begingroup\$ No, this does not explain anything useful to one who wants to make a negative resistor. To make sure of this, ask the OP to make a negative resistor according to your recipe. It is a recipe for mathematicians.They will even write, for their convenience, a/b = c and a/b = -c ...and will be happy... \$\endgroup\$ Dec 24 '20 at 23:32
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    \$\begingroup\$ @Circuitfantasist write your own answer. Nothing is stopping you. Also, please stop speaking for the OP. That is presumptuous. \$\endgroup\$
    – mkeith
    Dec 24 '20 at 23:33
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    \$\begingroup\$ Passivness of an element is not required. Negative resistor should change electromotive force inside it according to applied voltage. \$\endgroup\$
    – Dims
    Dec 24 '20 at 23:45
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    \$\begingroup\$ @Circuitfantasist you are a teacher sitting in on another professor's class and criticizing the curriculum and method of presentation in front of all the students (and the world). Please go teach your own class. That is what I am asking you. \$\endgroup\$
    – mkeith
    Dec 24 '20 at 23:54
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    \$\begingroup\$ I agree with Fantasist. The question posted by OP already showed they understand that a negative resistor would have the relation \$V=-I\$. \$\endgroup\$
    – The Photon
    Dec 25 '20 at 0:22
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Utterly impractical, but just to indulge the puzzle in an abstract sense:

Two howland pumps chould function as a floating "negative resistor" within a bounded range of voltages/currents/frequencies? (for a no-batteries concept)

(Not completely sure I got the +/- right on the Howland pumps but hopefully you get the idea)

PS -- Merry XMAS

enter image description here

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    \$\begingroup\$ Here you have a floating negative resistor electronics.stackexchange.com/questions/460605/… \$\endgroup\$
    – G36
    Dec 26 '20 at 16:30
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    \$\begingroup\$ Nice, though I have a suspicion that without some compensation these things would very much prefer to oscillate. \$\endgroup\$ Dec 26 '20 at 19:16
  • \$\begingroup\$ @G36, It is a great challenge to reveal the idea behind this mess of resistors and op-amps in your answer... It reminds me of an interesting RG discussion, five years ago, when I managed to do it with Antoniou's GIC. I am tempted to do it with this circuit solution… but I am afraid that I will "waste" the rest of Christmas holidays in thinking :) \$\endgroup\$ Dec 27 '20 at 16:14
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Same idea to the answer with the howland's, but with less stuff -- or a bilateral version of the regular NIC /// update /// realized this is the same circuit of G36 in comment above.

enter image description here

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