There is inductor with magnetic core. Losses in the core can be shown as resistor in parallel to inductor. Would it be correct assumption that Johnson noise of resistor is equvalent to the noise of the core? (small signal situation)
If the core losses are from conductance, or nearly any other mechanism that operates all the way down to zero current and voltage, then the losses in that apparent resistance would pretty much have to equal Johnson noise losses by simple thermodynamic considerations.
If the noise were greater or smaller then you could build (at least theoretically) a perpetual-motion machine powered by the difference in output from a resistor and an inductor wound on a core.
That proportion of the losses which are from conductance would _be_ genuine Johnson noise losses, coupled from the resistance of the core to the winding. Any losses that were from other mechanisms would arguably not be Johnson noise, but they would still have to add up to the same amount in the end or you're back to successfully building perpetual motion machines.
Note that the noise will be proportional to the core temperature, of course.
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Tim Wescott
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Well, the domain-spin-orbital-phonon gibberish is a pretty good reason to be careful about calling it "Johnson" noise, or at least "Johnson- Nyquist" noise. If you take "Johnson noise" to be a consequence of electron movement, and not just any black-body radiation as seen through a resistive load, then domain-spin-orbital-phonon noise isn't "Johnson- Nyquist" noise, per se.
If you get into a hair-splitting contest then you could argue that if it's not electrons jittering around then it isn't Johnson-Nyquist noise, but it's still Johnson noise, because Johnson's measurements would have been the same, even if Nyquist's explanation was different.
This is probably best done with a few drinks under your belt, though, and in the absence of any pretty girls that you may wish to impress with your un-nerdiness.
Note, too, my original caveat: if the core losses do not extend down to zero current or voltage, then they may not follow Johnson noise. I would expect that my thermodynamic argument only works for losses that are present on an unexcited or infinitesimally excited core: thermal noise due to losses that come from a nonlinear B-H curve is beyond my powers to predict.
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Tim Wescott
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Not sure I understand the question. Is your question whether there are other noise sources than Johnson noise? (The loss will contribute Johnson noise, by the fluctuation dissipation theorem).
If so, I've run into Barkhausen noise with a core that is larger than the Johnson noise of the core (and coil).
If there's signal across the leads, you can get extra noise from the magnetic domains snapping. Sort of like shot noise.
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Linear losses will look like a resistor, and by the fluctuation-dissipation theorem, anything that can dissipate power will generate thermal noise. (If that weren't so, you could make energy flow spontaneously from cold to hot.)
The core will also exhibit Barkhausen noise due to the motion of magnetic domains. This used to be a huge effect in magnetic recording, before heads got so small that they're a single domain.
Cheers
Phil Hobbs
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There's apparently flicker noise in some cores as well.
Best regards, Spehro Pefhany
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Interesting. I'm not too surprised, but I'd expect it to be a pretty small effect. Domain wall relaxation, or something like that?
Cheers
Phil Hobbs
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Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
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Good, That fits my understanding.. Dissipation -> noise. So magnetic damping (a conductor moving through a magnetic field, with a velocity dependent term) should look like Johnson noise. (Sorry I'm thinking of this torsional oscillator we make that has a big copper rotor and adjustable permanent magnets for damping.)
We also can rub a string against the rotor.. frictional damping.. that gives a linear decay rather than exponential (as for magnetic damping). That must have a different character of 'noise' (fluctuation), When the rotor stops from frictional damping it's not at equilibrium, it 'parks' where ever the friction stops it. ... so I'm guessing there's then no noise, unless something 'bumps' it.
Impedance mismatch situation -- when it's moving, it looks like an equivalent loss of so-and-so (the mechanical resistance (i.e., ratio of force to velocity) depending on speed), until it stops, in which case it looks like a short circuit of such-and-such stiffness.
There are obvious mechanical sources of excess noise: uneven surfaces exhibit a microscopic cogging effect (or stick-slip, if stiffness is low enough and static friction high enough). A perfectly smooth surface might not exhibit excess noise, although you get into other interesting phenomena when you start talking ideal surfaces (Van der Waals forces, or straight-up molecular forces for that matter). And coatings (e.g., a layer of lube) exhibit various rheological and microfluidic phenomena that make it just as difficult to analyze.
But don't worry, there's always the uncertainty of quantum mechanical processes underlaying everything...
Tim
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