Everybody knows what a "current to voltage" converter is, a basic application configuration for op amps.
This is a very level basic question. See the picture. What is wrong? Have I missed something?
Just to emphasize the importance of ("hidden" or not) wiring ...
This shows where all the current flows and how the 10uA finds it's way back to your ground connection:
For the op amp output to sink the 10uA of current, it has to send it down the negative supply pin, hence it returns to your source via the ground connection of the negative supply.
If you put in a negative voltage (V4 < 0) so that the current flows the other way, then the output goes positive to drive the current back through the feedback resistor, and this current is sourced from the positive supply (V6) and is returned to your voltage source via it's ground connection.
simulate this circuit – Schematic created using CircuitLab
Figure 1. I1 doesn't go into the inverting input due to its high impedance. It goes through the feedback resistor, R1.
The current from the non-inverting input is just the bias current and has little, if anything, to do with the current on the inverting input.
That's basically how a transimpedance amplifier works; the output keeps the inverting node at the same voltage as the non-inverting node. The output takes that current (or drives that current if you prefer). The input impedance of the OP27 is a gazillion ohms. However, the OP27 does have input bias currents of circa 100 nA and these might pose a significant error if it's important.
I have redraw a little my schematic to show where this current is "gone".
It is a great part of the answer, but it lacks "something".
Remember that we are in "simulation" and that some currents sources are hidden in the model of op amp we use.
So the answer is not "complete". I think the true answer is in the "laboratory" where we can measure really everything (not always simple).
For adjusting "offset", make V4 = 0 in microcap v12 "schematic", click on "Dynamic DC", click on "define Voffset ..." then change through arrows up or down until Vout is nearest 0 V. Then remake V4 = 10 V.
Your current will not emerge from the non-inverting input, because of the high input impedance. For the same reason, no current will enter the inverting input. This is the path that your input current will take:
Q1 and Q2 represent the push-pull output stage of the opamp. Note that current will branch at the output of the opamp, some to be sunk by the opamp itself via Q2, and some will be sunk by whatever load you connect to the output.
The input current emerges at the negative supply pin of the opamp. The current emerging there will also include the opamp's own operating current, as well as the input current being "measured".
If the input current is negative (in the opposite direction) it will follow this path instead:
This time the opamp (and whatever load you have connected) will be sourcing current for the input to sink.
I read these answers full of technical details and wonder how it is possible not to reveal the simple but brilliant idea behind this op-amp circuit consisting only of a resistor and op-amp? I realized it 30 years ago (Fig. 1) and with its help I was able to understand and explain many other op-amp circuits.
Fig. 1. A conceptual picture of an active current-to-voltage converter from my archive (1992). Here is the translated text:
May 14, 1992. "Ideal" current-to-voltage converter (a possible explanation by an opposite voltage). The current-sensing resistor RI creates a "harmful" voltage drop VR (it is necessary but undesired; there is a contradiction). We can destroy it by an "anti voltage" V(E)anti that is subtracted from VR (it is an inverse copy of VR). It can be implemented by an op-amp A that adjusts Vanti so that VR - Vanti = 0 ("active copy" principle).
The idea is obvious: To measure the current I by a voltmeter, we break the circuit, insert a resistor RI and measure the voltage drop across it (VR = I.RI). But this voltage introduces an error since it is subtracted from the input "current-creating" voltage VIN and the current decreases. So we decide to destroy it by adding an equivalent voltage V = VR. For this purpose, we break again the circuit and insert a varying voltage source producing the compensating voltage V. This voltage is added to the input voltage and the error is eliminated - Fig. 2.
Fig. 2. Full conceptual circuit of four elements in a loop (the picture is taken from a similar story about the inverting amplifier). Note the two voltages are summed in a series manner.
So the (properly supplied) op-amp output acts as a small variable "battery" connected in series to the resistor (Fig. 3) that adds the compensating voltage VOUT = I.R in the circuit. And, of course, the input current will flow through this "battery".
Fig. 3. Op-amp implementation of the idea
To close the current path, we have to draw the respective power source - negative if the input voltage is positive (like in Fig. 3) and positive if the input voltage is negative.
It is interesting to see how the op-amp copies the voltage drop VR at its output. According to KVL, we can see in Fig. 2 and Fig. 3 a loop of three voltages - VR, VOUT and VA. The op-amp changes VOUT so that to keep VA zero (negative feedback). As a result, VOUT = VR.
Another clever trick is that we use the compensating voltage as a buffered, grounded and inverted output voltage (the last feature is a "gift" that is not always desired).
To illustrate the circuit operation in a more attractive way, we can "geometrically" draw the circuit diagram - Fig. 4.
Fig. 4. A "geometrical" representation of the op-amp current-to-voltage converter
In this representation, the "positive circuit part" is drawn above the zero voltage level (ground) and the "negative circuit part" is drawn below the ground. The voltages are represented by voltage bars in red and the currents - by current loops in green and blue.
Note something very important - the input current (in green) does not flow through the load. The load current (in blue) is provided only by the negative supply source, i.e., the load does not consume current from the input voltage source. This is a big advantage of the active op-amp circuit compared with the passive one (resistor).
The power of this step-by-step building approach is that it shows the circuit evolution from the humble 1-resistor passive circuit to the more sophisticated op-amp circuit. We see this is not a new circuit; it is an improved old circuit. So the active op-amp current-to-voltage converter consists of a passive current-to-voltage converter and helping op-amp.
If we are curious enough, we can see a similarity between the resistor R and the op-amp output - there is the same voltage I.R across them; so both behave as resistors. But while the resistor subtracts its voltage drop from the input voltage, the op-amp adds its output voltage to it.
So the op-amp output acts as a negative "resistor" with resistance -R that neutralizes the positive resistance R. The whole circuit (resistor and op-amp) behaves as a "piece of wire"... and the input current flows through this "artificial wire" - Fig. 5.
Fig. 5. The transimpedance amplifier presented as a "piece of wire"
As a rule, we know the passive is bad and the active is good. But here the passive "circuit" (resistor) has a very significant advantage over the active - it allows to measure currents of large magnitude.
The problem with the active op-amp current-to-voltage converter is that the current passes through its output stage... and the latter must withstand it. That is why the ammeters inside multimeters are not made with the op-amp circuit, no matter how perfect it is, but with the simple passive circuit (a humble resistor between the inputs).
The input current does not go through the feedback resistor. It is nullified by current coming backwards through the feedback resistor from the output. OK, the two resistors do look like a voltage divider between two different potentials, so in one sense the input current does go to the output.
But . . .
The opamp has lotsa gain, and the action of negative feedback drives the inverting input to behave as a virtual ground. This changes the analysis of the circuit. the two resistors do form a voltage divider, but having a point in the middle of the divider that does not move breaks that current into two related but not in the same way currents. The output does whatever it takes to make the two input voltages equal, so whenever the loop is closed, the two currents are equal. But they are not in the same direction.
In your case the two resistors are of equal value so the conversion factor through the circuit is 1 volt per amp. If R4 were 100 K, the conversion would be 10 V per amp, but the two currents still would be the same value, and the net current at the inverting input still would be 0.
Note, all of this assumes a theoretically perfect opamp with zero input bias current and zero input offset voltage.
