Johnson–Nyquist Noise increases with resistance - so does it become infinite for an open circuit? Specifically why is the input to an oscilloscope that is not connected to anything not completely saturated by Johnson–Nyquist Noise (assuming the resistance through air from oscilloscope input to GND is near infinite).

Johnson–Nyquist Noise Equation from Wikipedia:

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  • \$\begingroup\$ What is the bandwidth \$\Delta f\$of the oscilloscope input, assuming infinite source resistance and (say) 30pF capacitance to ground? \$\endgroup\$ Dec 20, 2019 at 2:07
  • 2
    \$\begingroup\$ Johnson-Nyquist noise is usually related to a concept of thermal vibrations of charge carriers in conductors. If you want to study similar (though more as quantum fluctuations in this case) effects through air, you could read Current Noise in Tunnel Junctions. It's really interesting and they do need to deal with it. \$\endgroup\$
    – jonk
    Dec 20, 2019 at 2:08
  • \$\begingroup\$ Back to your topic. What do you think is the sensitivity of your oscilloscope to thermal charge motion in air? Or, put another way, suppose you divided these supposed voltage fluctuations by the resistance to get the referred-to-current noise and then apply that to the probe resistance and capacitance? \$\endgroup\$
    – jonk
    Dec 20, 2019 at 2:09
  • \$\begingroup\$ By the way, this whole area is extremely interesting. I think Boltzmann's original paper is kind of hard to understand, when compared to Shannon's papers that use his concepts to develop information theory. (Shannon is who made me understand Boltzmann for the first time.) The equipartition theorem in thermodynamics places, on average, half the energy in potential energy and half of it in kinetic. So there is a direct relationship between Johnson-Nyquist noise and kT/C noise, because of this. I really recommend you spend some time here. It's really good stuff. \$\endgroup\$
    – jonk
    Dec 20, 2019 at 2:21
  • \$\begingroup\$ Some papers to read are from an MIT class: Mathematical Theory of Claude Shannon and Information Theory. Please go though them. But I also recommend The Mathematical Theory of Communication, by Shannon and Weaver. Plus Shannon's original papers from around 1948. +1. It will be interesting to see an answer speaking broadly and not just answering an Oscilloscope question. \$\endgroup\$
    – jonk
    Dec 20, 2019 at 2:32

2 Answers 2


The big factor that you didn't consider is that "free space" has capacitance and therefore, an open circuit of (say) 1,000 GΩ shunted with (say) 1 pF (for example) has a bandwidth of: -

$$\dfrac{1}{2\pi RC}$$

Numerically this is 0.159 Hz and the equivalent noise bandwidth is that multiplied by \$\pi/2\$. This restricts noise voltage very significantly.

Johnson–Nyquist Noise increases with resistance - so does it become infinite for an open circuit?

No it doesn't because of the effect of capacitance.


It is important to note that even though the noise voltage increases with resistance, the noise power does not. Since power is equal to Vn²/R, the total power is independent of R and only depends on T and bandwidth. Further analysis shows that there is another factor that goes to zero as the bandwidth increases. This prevents the power from becoming infinite. You can only get infinite voltage with an infinite resistance. However, if you try to measure this voltage with any kind of voltmeter which has a finite input resistance, you will not get an infinite voltage. Most oscilloscopes, for example, have an input resistance of 1 megohm so this will limit the observable noise voltage.


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