Electronic – A questions about negative resistance

input-impedancenegative-resistanceoperational-amplifier

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

5=I*(150-100)

I = 0.1 A , Is that right ?

Edit

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,

schematic

simulate this circuit – Schematic created using CircuitLab

Best Answer

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.