I was doing some measurements for a coursework.
Op-amp, LM741 in an x10 inversor configuration.
When the resistances in use become high enough, 1MOhm and 10MOhm, the output is correct but it has a 50Hz noise on top, of amplitude that would suggest an extra 80nA through the resistors.
That effect isn't seen while simulating on PSpice, but it is expected behaviour since other people measured it too.
Now, the datasheet gives 80nA as the bias current, so I, who still don't know much of how an op-amp works internally, suspected it was to blame.
However, I've been told that the bias current is always DC so it can't have anything to do with that, and that it has to do with something acting like an antenna, but no clear explanation was offered to me.
Is that true? I have no clue what's going on and would appreciate orientation
EDIT
The objective is just educational, it's to understand the behaviour. With 1k-10k resistors it works perfectly. 50Hz is the mains frequency so the noise is surely somehow due to that. But I'd like to know more why that noise happens, how to model it in a way that it explains the fact that it's there with 1MOhm resistors but not with 1K. Also, if it's related to some parameter in the LM741 datasheet that explains the amplitude of the noise
Best Answer
I won't try to answer the whole question, but I'll try to explain why you're getting pick-up, and why the resistor values matter.
The pick-up, is, like other said, coming from your mains voltage. It might be coming from a small ripple leaking through your power supply regulators to your circuit rails, or it might just be coupled in from nearby power lines. In any case, the fact that changing the resistor values affects it strongly suggests this is capacative pick-up:
The main idea is that electric field from someplace with a 50 Hz signal on it is terminating on the copper connected to the inverting input of your op-amp. This has the same effect as if there's a (very tiny) capacitor between that source and your op-amp input.
So why do the resistor values matter?
I'll assume that the driving circuit supplying the desired signal at the "in" port of your circuit has a low impedance. And we know the output of the op-amp has a low impedance, and the inputs have very high impedance. So the AC equivalent circuit for the interference source reaching the op-amp input looks like this:
Here, the op-amp inverting input is connected to the mid-point where the capacitor connects to the two resistors (just like in the real circuit). We can immediately see this is a voltage divider. For a fixed frequency, the capacitor will have a fixed impedance: 1 / (2 * pi * f * C). So as you increase the resistor values, you'll directly increase the signal seen by the op-amp, up until the parallel resistance is much more than the capacitor's impedance and you are near 100% coupling.
This divider effect explains why we often hear about shielding or isolating the "high-impedance nodes" in op-amp circuit designs.