operational-amplifier
I'm still very new to electrical engineering and at the moment trying to get my head around the concept of the ideal op amp.
In Fundamentals of Electric Circuits by Alexander and Sadiku the following diagram is given as a "current-to-voltage converter".
simulate this circuit – Schematic created using CircuitLab
The reader is asked to prove $$v=-IR,$$ which I think I did. But what do we need the op amp for? Wouldn't the following diagram be exactly the same?
The node voltage equations are not so difficult. But it's really a tedious work to type the equations. I got the same result with you, maybe we are both wrong :).
The output current from the op-amp (as depicted in the picture in the question) is that current needed to keep the inverting input at ground potential.
So, with 1V at R1 (left hand side), there has to be -1V at the output to make the inverting input zero volts.
This means the current is -1V/100R = -10 mA.
If R2 were (say) double the value (200 ohms), the voltage at the output is -2V but the current remains the same.
In fact the current flowing in R2 is that current flowing thru R1 due to the input voltage. Remember the op-amp forces the inverting node always equal the non-inverting node and, if the non-inverting node is 0V then R1's current is simply input volts/R1. This is what is meant by a virtual earth amplifier.
Best Answer
Your op-amp is a vital clog in the wheel in this circuit. Since the op-amp has infinite resistance, the current I entirely flows to the alternate path through the resistor R. Note that since the + end of the op-amp is at ground and because of the infinite gain of the op-amp, the - end is also held to ground. So the drop across the resistor is V = IR. So effectively, the output of the op-amp will be held at V = IR as the output is connected between the point V which is the other end of the resistor and ground. So, it is the characteristics of the op-amp which are enabling the functionality of a current to voltage converter here.