In all the op-amp inverting circuits (transimpedance amplifier, inverting integrator and differentiator, diode log and antilog converters, etc.), the op-amp compensates the voltage drop across the passive element (e.g., a resistor as in the attachment below) connected between the op-amp output and inverting input by adding an equivalent "mirror" voltage in series:

https://en.wikibooks.org/wiki/Circuit_Idea/Voltage_Compensation

As a result, the same current flows through the two elements - the resistor and the op-amp output, and the same voltage appears across them; so, they process the same energy and they have the same impedance. But while the first of them is a passive element that consumes energy (voltage) from the input voltage source, the second is an active element that adds the same energy (voltage) to the input voltage source. Then, if the first element has a "positive" resistance, the second element will show a negative resistance!

https://en.wikibooks.org/wiki/Circuit_Idea/Voltage_Compensation#Negative_impedance_viewpoint_at_voltage_compensation

So, we can conclude that in all the op-amp inverting circuits, the combination of the op-amp and the power supply actually acts as an element with negative impedance that neutralizes the positive impedance of the element connected between the output and the inverting input. As a result, the whole combination of the "positive" element (the resistor R in the picture) and negative element (the supplied op-amp) behaves just as... a piece of wire with zero impedance!

From this negative impedance viewpoint, in the circuit of a capacitive integrator the op-amp is a "negative capacitor" producing the voltage Vc, in a diode log converter - a "negative diode" producing the voltage Vf, etc.

It is interesting to compare this negative "resistor" with the true negative resistance circuit (voltage-inversion negative impedance converter - VNIC)...

https://en.wikibooks.org/wiki/Circuit_Idea/Revealing_the_Mystery_of_Negative_Impedance

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