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I have modeled a non-ideal op amp as an ideal op amp with a noise current I_n flowing between its inputs:

enter image description here

In this circuit, I have also modeled a non-ideal voltage source as an ideal voltage source plus a non-zero source impedance R1. Finally, R2,R3 are the only real resistors in this circuit, and control the feedback of the op amp.

As an exercise in op amp noise, I will compute the theoretical rms value of noise of this circuit, and compare it to experimental values, so I HAVE to get this calculation right (please assume that the above circuit is correct, I have omitted some details in order to ask this question.).

Now here is the question: what is the rms value of the noise in V_out due to the presence of I_n in this circuit? I argue that, since I_n will flow towards lower voltage, it will run into R_2, and draw current from R1, causing an error of I_n * (R2 + R1) * gain.

However, according p.4 of this document (section header is "CURRENT NOISE"): http://www.ti.com/lit/an/sboa060/sboa060.pdf, the noise should be I_n * R3. Perhaps, along with your answer, could you demonstrate why I am wrong and this document is right?

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Assume V= 0 for now. Since the op-amp is ideal (other than noise), the inputs are always at the same potential (due to the the output voltage changing and through presumed negative feedback). If you have a current I_n then, in order for the inputs to be at the same potential, a current must be flowing through R3 equal to I_n. The resistors R1 + R2 across the inputs do not have any effect since the voltage across them is always zero (if R1 and R2 were both shorts you'd have a 0/0 situation).

However, the way you have drawn this circuit (floating V) no current can possibly flow through R3 because you don't have a ground anywhere. So no matter how high the output voltage gets, no current flows. With an ideal op-amp and ideal parts this creates another paradox.

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To begin with, you have not produced the same circuit as the one you are having problems with. If you will refer to figure 3 of that document, you will see that the circuit is

schematic

simulate this circuit – Schematic created using CircuitLab

Using the noise model if the link, and keeping in mind that this is a noise current analysis (noise voltage is assumed zero), the circuit becomes

schematic

simulate this circuit

and the output noise is equal to Inoise* R1$.

Assuming that your original circuit should have been grounded at the + input, and assuming V is zero, then (for an ideal op amp) the presence of R1 and R2 is irrelevant, since the feedback through R3 will be adjusted to keep the voltage across the inputs zero. As a result, the feedback current will equal Inoise, and the output voltage will be Inoise * R3.

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