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I was doing some measurements for a coursework. Op-amp, LM741 in an x10 inversor configuration.

enter image description here

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

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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:

enter image description here

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:

enter image description here

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.

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  • \$\begingroup\$ +1 @Rojo This is how you'd model it. probably about 3pF of capacitance, but the model may not be accurate enough to see the effect. \$\endgroup\$ – placeholder Oct 12 '12 at 16:17
  • \$\begingroup\$ @ThePhoton +1 /ME surprised... I wanted to suggest (almost) the same, edited my answer, and somehow it didn't make it here... I didn't draw though. \$\endgroup\$ – ppeterka Oct 12 '12 at 20:01
  • \$\begingroup\$ +1, I haven't had time or brains to really think this through but it looks promising. I'll probably be accepting this, thanks a lot \$\endgroup\$ – Rojo Oct 14 '12 at 22:20
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Reading your description one can tell you're in a country that has 50 Hz mains power.

You don't say how you're putting the circuit together, breadboarding or otherwise. But there is no reason to have such high resistor values. The Johnson noise will also be too high and the main issue is that your wiring is picking up stay fields and the high resistance is preventing the op-amp from being able to dominate the aggressor signal.

Further improvements you can do:

  • place a resistor equivalent to the parallel resistance 1M||10M from the non-inverting input to ground.
  • shield the inverting input to reduce the chance of pickup.
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  • \$\begingroup\$ +1 R1 could be 1k and R2 10k for example, possibly multiply both by 10 if the source needs less loading. Hard to see any requirement for any more. \$\endgroup\$ – user207421 Oct 12 '12 at 7:58
  • \$\begingroup\$ Thanks +1. Added an edit with more clarification. Your first suggestion seems ingenious, and the second one suggests the resistors are to blame for the pickup, right? If so, how would you model it? I mean, what's the path of that extra current, and does it depend on the resistance values? \$\endgroup\$ – Rojo Oct 12 '12 at 10:31
  • \$\begingroup\$ The current is not dependant on the resistor values, but the voltage induced by the current is (Ohm's law). The wires in a circuit act always as antennae too. This is not shown on PSPice, as this is the effect of your environment on your circuit. \$\endgroup\$ – ppeterka Oct 12 '12 at 13:14
  • \$\begingroup\$ @ppeterka thanks. I'm still thinking about how to model it however \$\endgroup\$ – Rojo Oct 12 '12 at 13:25
  • \$\begingroup\$ So it is an extra current that travels through both resistors whose value depends on the environment, the wiring, orientation of the circuit, etc, but not on the resistance values... \$\endgroup\$ – Rojo Oct 12 '12 at 13:31
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As others have noted, the circuit is picking up the mains frequency of 50Hz.

You explained that you are doing this for your education, so I would recommend something that has educational value: try moving around the circuit, and measuring the amplitude of the 50Hz noise. You'll find that not only the place, but even the orientation of the circuit can play a role in how the noise is received and amplified. You'll probably also find other frequencies appearing near other electronic devices...

Or even, for educational purposes you can touch the input to various objects (not mains, or other hayardous voltages of course...), like your hand.

Also: no, it does NOT work 'fine' with 1k-10k resistors - just the noise is a lot smaller! Turn up the sensitivity on the scope, and you have a chance to see it.

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  • \$\begingroup\$ Thanks, +1 I'll be playing with that! Yeah, I know the noise is still there but small with 1K-10K, I just wanted to point out that the effect probably can't be modeled as a "voltage noise" because depending on the resistor values the voltage noise amplitude changed a lot \$\endgroup\$ – Rojo Oct 12 '12 at 13:32
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50Hz noise will be from the mains. Since you are learning you will probably be powering your circuit from a lab voltage source. If you look to about 10kHz you should see some more distortion, provided your noise floor is low enough, this lower peak at higher frequency is the signature of the power switching devices in your power source.

The noise is in your circuit is lower when you use lower value resistors because your circuit is dominated by thermal noise generated within the resistors. In most cases higher resistor values mean higher noise power.

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