This sim has the ideal equivalent on the left of the voltage source and a practical circuit on its right. Change the value of the voltage source in the middle and see that the currents through the top resistors equal to -1 times the current they would have if the resistors were just connected to GND.
Both sides represent a negative 1kOhm resistor set by the resistors on the top of each side (impedance is simply the negative value of the top resistor’s value). The circuit on the right has limitations on its effective range (0V to +6V here) because of the real opamp’s output being restricted to its power rails (just +12V to GND here).
The idea of a negative resistance has *some* applications but I just thought it was neat to fiddle with. It works by applying a higher voltage on the opposite end of a resistor than on the input to create a reverse current flow than what would be expected for a regular resistor connected to GND; ie. a 3V source across a 1kOhm resistor should source 3mA through it to GND while it would in theory sink 3mA coming from a -1kOhm resistor. This is achieved by applying double the source voltage on the other end of the resistor to force the same magnitude of current through the resistor in reverse since eg. (3V-6V)/1kOhm=-3mA. By using a non-inverting opamp with a gain of 2 on the other end of a resistor we can simulate the negative resistor equivalent of that resistor value. This illusion breaks down if we go above or below 1/2 the opamp’s power rails however since it would no longer be able to drive enough current to match the equivalent ideal negative resistor.
