I would like to look at negative resistance from a practical point of view, not just theory. Would you answer my questions,please?

The op-amp, R1, R2, and R3 behave as a negative resistance: Rn = - 100 ohms.

R4 is in series with the negative resistance.

1- If I constructed this circuit on a breadboard, Will the negative resistor work as a generator and injects current through AA batteries and damage them? Or the current just stops?

2- If the supply was a transformer with bridge rectifier, Would the bridge rectifier prevent the current form interning the supply?

3- If the op-amp IC is supplied with 12 volts or any value that is different from input voltage, Will that affect any parts of the circuit ?

4- If I changed the value of R4 to be 150 ohms, The current will be:

V = I*R


I = 0.1 A , Is that right ?


Rn is the equivalent negative resistance, on of its terminals appears at the non inverting input of the op-amp and the other terminal appears at the ground.

What I mean by AA batter or transformer is we remove the 5 volts supply and put an AA battery or a transformer instead.

I think the output terminal is also the non inverting input because the circuit behave just like a resistance.

For more information about negative impedance converter :https://en.wikipedia.org/wiki/Negative_impedance_converter

Thank you,


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ I think you'll want to indicate where in the circuit you would expect \$R_n\$ to appear. Also, if you're interested in how an output might behave, like pumping current through a battery, you should indicate the output node and also indicate how the battery would be connected (+ to node x and - to node y). \$\endgroup\$ – scanny Sep 30 '15 at 21:15
  • \$\begingroup\$ I've made some edits in the question. please, read it and I'm sorry about my english :) \$\endgroup\$ – Michael George Sep 30 '15 at 22:47
  • \$\begingroup\$ As you try and push current into the non-inverting node the opamp pushes back. (through R3) until the opamp hits one of it's power rails. If you drive it from a voltage source as you've shown... I think it hits the power rail no matter what... but I'm not sure. Have you built it? \$\endgroup\$ – George Herold Oct 1 '15 at 0:00
  • \$\begingroup\$ I've not built it, yet. I'm very busy those days doing something else for my collage :) \$\endgroup\$ – Michael George Oct 1 '15 at 12:46

In this case you could just sum R4 and the negative resistance to get the supply current. But if you work out the equations you would get:

\$I_s\$ = \$\frac{\text{R2} \text{ Vg}}{\text{R2} \text{ R4}-\text{R1} \text{ R3}}\$

which is, for the values given is \$I_s\$ = -0.1A , current into the source. This would require a fair amount of voltage out of the OpAmp though.

\$V_o\$ = \$\frac{\text{R3} \text{ Vg} (\text{R1}+\text{R2})}{\text{R1} \text{ R3}-\text{R2} \text{ R4}}\$

For the values given, OpAmp output voltage \$V_o\$ = 20V.

Of course, it all blows up if R4 = Rneg.


Okay, I take it you're identifying the 5V voltage source as the AA batteries you mention. Note that alkaline AA batteries produce 1.5V (nominal) each, so it could be 4.5V or 6V, but not 5V.

Also note that R4 does not appear in the Wikipedia circuit for a negative resistance converter, so you would need to do a network analysis to determine what the formulas would be and if indeed it still behaved like a negative resistance.

If you removed R4, (and perhaps if you didn't), current would indeed flow "backward", into the positive terminal of the voltage source as opposed to out of it. This would require of course that the V+ and V- supply to the op amp was high enough. What would be happening is the \$V_{out}\$ of the op-amp would be a higher voltage, say 10V, and current would flow from the output to the voltage source at the rate:

$$I_{rev} = \frac{V_{out} - 5V}{150\Omega}$$

Whether that current was enough to damage the batteries is a separate question; perhaps "How much reverse current can an alkaline AA battery withstand without damage?"

The op amp supply voltage will not affect the circuit behavior, unless it is insufficient to produce the required \$V_{out}\$. You'll have to do the math on that yourself, based on the particular op amp used and how close it can get to its rails. But if it works at say +/-10V, it would work the same at +/-12V or +/-15V.


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