Core losses and noise

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)

Vladimir Vassilevsky DSP and Mixed Signal Designs

formatting link

Reply to
Vladimir Vassilevsky
Loading thread data ...

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.

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com
Reply to
Tim Wescott

Duh... Good argument!

I thought that too, before somebody confused me with domain-spin-orbital-phonon gibberish.

Vladimir Vassilevsky DSP and Mixed Signal Designs

formatting link

Reply to
Vladimir Vassilevsky

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.

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com
Reply to
Tim Wescott

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).

--sp

Reply to
Spehro Pefhany

If there's signal across the leads, you can get extra noise from the magnetic domains snapping. Sort of like shot noise.

--

John Larkin         Highland Technology, Inc 

jlarkin at highlandtechnology dot com 
http://www.highlandtechnology.com 

Precision electronic instrumentation 
Picosecond-resolution Digital Delay and Pulse generators 
Custom laser drivers and controllers 
Photonics and fiberoptic TTL data links 
VME thermocouple, LVDT, synchro   acquisition and simulation
Reply to
John Larkin

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

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 USA 
+1 845 480 2058 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

There's apparently flicker noise in some cores as well.

Best regards, Spehro Pefhany

--
"it's the network..."                          "The Journey is the reward" 
speff@interlog.com             Info for manufacturers: http://www.trexon.com 
Embedded software/hardware/analog  Info for designers:  http://www.speff.com
Reply to
Spehro Pefhany

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

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 USA 
+1 845 480 2058 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

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.

George H.

Reply to
George Herold

I rather vaguely remember core losses are from domain wall movement and other domain "demolition"; polarization direction non-lossy and recoverable.

Reply to
Robert Baer

You named only Snap and Pop; how about Crackle (in your Rice Crispies)?

Reply to
Robert Baer

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

--
Deep Friar: a very philosophical monk. 
Website: http://seventransistorlabs.com
Reply to
Tim Williams

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.