Based on my very basic understanding of the Johnson noise, it's not just a DC phenomena, but should change with frequency in a system, where there is a frequency dependent, real component to the impedance. I think both the skin effect and eddy current losses contribute to the effective resistance of an inductor, so Johnson noise should increase at higher frequencies. Am I being dumb?

  • 2
    $\begingroup$ No, you are not dumb, you got it exactly right. Thermal noise (i.e. the irreducible noise component that is based on the system being in equilibrium with a thermal bath) voltage and current depend on the real part of the impedance of the noise source, i.e. the equivalent noise resistance does depend on skin effect and eddy current losses and is frequency dependent. I hope you are aware that the total noise power is independent of the resistance, though? Higher equivalent noise resistance means increase in noise voltage and decrease in noise current and vice versa. $\endgroup$
    – CuriousOne
    Jun 14, 2015 at 22:03
  • $\begingroup$ @CuriousOne Thank you for your answer! I was thinking purely about the noise voltage, thank you for clarifying its power. I guess radiative losses also add to the noise resistance? $\endgroup$
    – Someone
    Jun 15, 2015 at 7:30
  • 1
    $\begingroup$ Yes, whatever causes losses will change the effective noise resistance and noise power, but you have to keep in mind that one also has to know the temperature, which for radiative losses has one component given by the temperature of space and then there is the effective antenna temperature, so that complicates things further. $\endgroup$
    – CuriousOne
    Jun 15, 2015 at 8:58

1 Answer 1


As stated in the comments, any loss in the system will contribute to the Johnson noise, so you are right about the skin effect and the Eddy current.

I want to add that, interestingly enough, this apply not only to electric circuits, but to other linear dissipative systems. A very interesting paper from 1951, Irreversibility and Generalized Noise, proves it and provides the following examples:

  1. Brownian motion
  2. Dipole radiation
  3. Acoustic radiation in a gas

I am a electronic engineer, and I stumbled upon this fact while designing a circuit interface for a CMUT, a very low-noise capacitive transducer. As it turns out, the model for such a transducer can be used to predict the electric noise coming from it, although electrically the CMUT is only a capacitor. In facts, the real part of the impedance of the device is mostly due to the acoustic media (air, water, oil, and whatnot), because that is where the energy dissipation happens.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.