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The thermal noise power from a resistor is p = 4kTR where: k is the Boltzman constant, T is temperature, R is resistance.

Since an open circuit has infinite resistance, it would produce infinite noise--which, of course, is ridiculous. Where has my thinking gone astray? Does it have to do with the fact that any practical circuit has capacitance filtering the noise and also that to get that noise there must be some degree of impedance matching? That is, a very high resistance produces tons of noise, but is easily filtered by a tiny capacitance and is typically very mismatched impedance-wise.

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    \$\begingroup\$ An open circuit has no matter with which to generate the noise. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 23 '16 at 9:16
  • \$\begingroup\$ Well yes, there are certainly corner cases (for example, the equation also breaks down at very high frequencies), but consider the extreme case of a very high value resistor. \$\endgroup\$ – Digiproc Jan 23 '16 at 9:19
  • \$\begingroup\$ And an open circuit current source produces infinite voltage and destroys the known universe. Be afraid. \$\endgroup\$ – Andy aka Jan 23 '16 at 11:56
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The only way to develop any noise power from this infinite noise voltage source is to draw current from it.

In other words, to connect it in parallel with a finite load resistance R.

And infinite resistance in parallel wth R is just ... R.

So the voltage noise produced by the overall network is just that of the load resistance R.

Which happens any time you leave an amplifier input open-circuit (without an antenna wire).

Measuring (or listening to) the noise output of a high gain low noise amplifier with the input (a) open circuit, (b) terminated by the usual source impedance, (c) short circuited will show decreasing noise levels.

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  • \$\begingroup\$ OK, so I think I was on the right track in my first guess: naming that term "power" in the formula is misleading. Its really voltage squared--so of course it would be proportional to resistance. But the resistance is also the impedance, so the actual power is NOT proportional to resistance. \$\endgroup\$ – Digiproc Jan 23 '16 at 17:33
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I don't think there is a broken corner case or paradox here, either in reality or in theory.

Consider what it means to actually have a very large resistor. Is it that weird to have large numbers of voltage noise in an environment where voltage does little? That is not the same as large numbers of current noise (since voltage noise is less than linear, RMS current noise must decrease in R) nor the same as large numbers of power noise (which should remain constant as R changes - I agree it'd be insane if power noise grew toward infinity with R).

Can you measure any pathological behavior of a big resistor with high voltage noise as the resistance approaches infinity? You can connect one leg of the resistor to ground and another leg to a measuring device, but what is the measuring device, a magical amplifier that somehow has even higher input impedance? From what I understand, voltage amplifiers with unreasonably high input impedance don't exist, even though they are generally desired, because they wouldn't function very well (for example, the thermal noise from the input impedance would be quite high).

If you leave the input of an actual op-amp buffer floating (i.e. at a near-infinite resistance to the nearest conductor), you could expect it to either measure an extremely random voltage, or to measure ground or rail voltage or something like that depending on the internal construction (because the input impedance i.e. impedance between input and ground/rail/whatever is not actually infinite, and in fact less than the impedance of some air in the room). The latter seems to be what happens in reality but I don't think the former would break intuition either.

Also, from a physical perspective, thermal noise doesn't care what the resistance is, it is just randomly adding energy to electrons. That should directly modify power noise independent of R. And as mentioned, power noise being constant in R is equivalent to voltage noise increasing in R.

I think it is easy to become really confused in electronics by thinking of voltage as if it were a fundamental quantity, just because so many circuits store and manipulate information in terms of voltage. Power is far more fundamental and generally does weird things wrt physical intuition less often ("wtf, a passive device can turn 10V into 1000V?!").

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  • \$\begingroup\$ Right. See my comment under the answer I marked as answer. (I wish one could mark two as answer). So I +1'd this one (but I've not enough points for it to show). \$\endgroup\$ – Digiproc Jan 23 '16 at 17:37
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Your equation is incorrect.

The noise voltage is proportional to resistance. The noise power is proportional only to the bandwidth over which you measure it.

And the values we're talking about here are very tiny. Even over a bandwidth of 1 GHz, it's only -84 dBm — about 4 picowatts — at room temperature.

Remember, we're just talking about the random thermal motion of electrons. Any field produced by that motion tends to reverse it.

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  • \$\begingroup\$ +1. I agree its incorrect now (see my comment under the answer I marked as answer). That formula is all over the internet, and that's what confused me! \$\endgroup\$ – Digiproc Jan 23 '16 at 17:40

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