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I am confused as to when a resistor is considered to be in the signal path and when its not when it comes to opamp noise calculations. For instance, take the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

a very similar circuit is published in Douglas Self's book, he mentions that the only resistor in the signal path (on the non-inverting input) is the 100ohm resistor R3, so R1 and R2 do not contribute to the noise. It was my understanding that a resistor can be modeled like an ideal or noise-less resistor in series with a noise generator, so what I would think is that for instance, if I replace R1 with a noise generator given by \$\sqrt{4KTBR} \$ with R1 in series, then that noise generator should be amplified by the noise gain of the opamp. Why is R1 and R2 not in the signal path then?

The author also mentions the following circuit which is a simple inverting amplifier with a resistor on the non-inverting input to compensate for bias currents.

schematic

simulate this circuit

In this case the author mentions that resitor R3 causes noise, so I dont get it, in both circuits there is a resistor connected to the non-inverting input, yet in the first circuit it does not produce noise but it does produce noise in the second circuit, so how do I know when a resistor produces noise (in the signal path) and when not? it doesnt seem very intuitive.

Edit: I simulated the first circuit and ran a noise analysis, what I found is that if R3 is small valued, then varying the value of R1 or R2 does not affect the noise output and the noise depends only on R3 (plus the feedback resistors and opamp noise, etc.. Im just focusing on the non-inverting input), however if R3 is not small, then the value of R1 or R2 does affect the noise output, however, I believe this is because of the voltage divider effect is attenuating the noise of the first resistor R3, not because R1 or R2 are contributing to the total noise output, so yeah, to add to the confusion, it would seem that only R3 is in the signal path of the non-inverting input and R1 and R2 contribute no thermal noise to the output, I dont understand this. Doing a simpler simulation, just connecting a resistor to the non-inverting input in parallel with a voltage source shows that the EIN of the resistor is 0, effectively contributing no noise to the output.

I also simulated the second circuit and in fact R3 (of the 2nd circuit) does affect the noise output if I vary its value. So my observations are: shunt resistors in the non-inverting input when using as non inverting amplifier do not contribute to noise, whilst a resistor in the non-inverting input when using as an inverting amplifier does contribute to noise.

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  • \$\begingroup\$ Well, either he made a mistake or you are not reading it correctly. Your intuition is right. \$\endgroup\$ – Edgar Brown Dec 5 '18 at 3:41
  • \$\begingroup\$ Hi Edgar, please look at my edited question. Thanks! \$\endgroup\$ – S.s. Dec 5 '18 at 4:20
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All resistors contribute to noise. Looking at the circuit more carefully I notice two things.

  1. The noise sources of R1 and R2 encounter a very large divider providing about 60dB of attenuation to their noise.
  2. Those noise sources encounter a very strong low-pass filter further dropping their noise spectrum.

But that has nothing to do with “being in the signal path”, if R3 was larger and the capacitors smaller, then their noise contribution would dominate, while keeping the same topology and frequency response.

“Being in the signal path” is at best some sort of mnemonic device that leads you to the right design choices (I.e., make R3 small), but it is not a circuit analysis tool.

When a resistor is “in the signal path” it really means that both it and the signal are equally affected, and there is nothing you can do in the design to reduce that noise that will not also affect the signal. So you should make that noise contribution as small as possible, or try to avoid it altogether.

I have designed low-noise circuits in which some of the major noise contributors are the current source transistors biasing the transistor at the tail of differential pairs (on the opposite corner of the IC). There really can’t be something further away from the “signal path” than that.

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  • \$\begingroup\$ I have simulated a similar circuit with only a resistor in the non-inverting input and a voltage source in parallel, the results are the same, no contribution to total noise, so even with no low pass filter or voltage divider the results are the same. \$\endgroup\$ – S.s. Dec 5 '18 at 13:14
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    \$\begingroup\$ @S.s. You don’t need a low-pass filter when you explicitly put a divider with a 0ohm resistor. What remains there to low-pass? Use a current source instead and see what happens. \$\endgroup\$ – Edgar Brown Dec 5 '18 at 13:16
  • \$\begingroup\$ Yes, I think it just clicked, so you mean that without R3 theres only the voltage source in parallel with a resistor effectively shorting R1 to ground? \$\endgroup\$ – S.s. Dec 5 '18 at 13:52
  • \$\begingroup\$ @S.s. Yup Exactly. \$\endgroup\$ – Edgar Brown Dec 5 '18 at 14:00
  • \$\begingroup\$ I've done phase-locked-loops where the phasenoise was dominated by the current-references to the charge-pumps. All in all, twas a low-noise design, with several primary contributors already pushed way down; examining the MATLAB model of the contributors to evaluate the newly dominant noise (I did not need to reach that clean), the charge-pump biasing would have been next. Funny that all the cures I'd implemented basically were: BURN MORE CURRENT or increase the node capacitance. \$\endgroup\$ – analogsystemsrf Dec 5 '18 at 18:07
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In the 1st circuit, since C1 is high enough, we can assume it as a short:

schematic

simulate this circuit – Schematic created using CircuitLab

You can model a resistor with either;

  • a noise-less resistor with a noise generator voltage source in series,
  • or a noise-less resistor with a noise generator current source in parallel.

So, if we replace R2 and R1 with the 2nd model, we'll see that total noise generator current source, n1 + n2, will see a heavy low-pass filter (C2 = 100nF & R2||R1 = 69k, yielding fC = 23Hz):

schematic

simulate this circuit

Thus the noise will most likely be filtered (Please remember that a parallel RC filter works only if driven by current source).

In the second circuit, if you replace the resistor with noise model then you'll see that there's no filtering. Probably that's why the noise totally reflects to the output.

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  • \$\begingroup\$ Please note that I have simulated the circuit without the capacitors, for instance only R1 and R3, and the results are the same, no noise, so there must be something else at work rather than a filter... \$\endgroup\$ – S.s. Dec 5 '18 at 13:11

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