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Hysteretic Core Loss

  1. Feb 25, 2010 #1
    I've been attempting to spec a toroidal inductor--what core size, what material and how many turns? I happen to be looking at some made of the material Kool Mu.

    The material if ferrite, so I presume most of the core loss is hysteretic BH loss, as ferrite doesn't conduct well.

    Anyway, all I am given is the core loss as a function of frequency and Bmax. They presume that the DC current is zero and the current is entirely AC.

    But my design requires the core be capable of a large DC current that varies slowly and sinusoidally from -20A to 20A and a small AC current of about 2 amps peak to peak at a much higher frequency. (The low frequency is 60 Hz, and the high frequency is about 100 Khz.)

    It's the high frequency component that contributes directly to the the lion's share to core heating.

    Is there a way to infer the core loss when there is a large DC current component when you are given the AC data alone?
     
  2. jcsd
  3. Feb 25, 2010 #2

    berkeman

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    It seems unusual to use a toroid when you have "DC" current components. It also seem sunusual to use a ferrite material for low frequencies like that. Why did you make those choices?

    The coil will have to be able to handle the full currents (20A) without saturating. Whether it's a DC current or a slowly-varying AC current.
     
  4. Feb 25, 2010 #3

    uart

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    Hi berkeman. I think the "Kool Mu" cores he's referring to are actually powered iron rather than ferrite, which means they have higher working flux density plus a distributed air-gap.

    Still it's an interesting question about the effect of DC offset on hysteresis loss. I don't know the answer to that one.
     
  5. Feb 25, 2010 #4
    Core loss usually includes both hysteretic (BH curve) losses and eddy current losses. In the OP application, the objective of the toroid design would be to limit the magnetic field to about Bmax Tesla for I =22 amps of magnetizing current.

    Using Bmax = μμ0H/L = μμ0NI/L

    where μ = relative permeability, μ0 = 4 pi x 10-7 Henrys per meter, and L = (inner) circumference of toroid, we get for I=22 amps:

    μ = 36,000·Bmax·L/N

    So for Bmax=0.4 Tesla, L= 0.3 meters and N = 20, μ ≈220.

    Several vendors sell gapped or powdered iron or "kool=mu" toroids with the above characteristics. The toroid core material, dimensions, and N need to be determined before the core μ can be specified.

    See for example:

    http://www.arnoldmagnetics.com/mtc/pdf/SoftMag.pdf#search=%22permeabilities%22 [Broken]

    Bob S
     
    Last edited by a moderator: May 4, 2017
  6. Feb 25, 2010 #5
    I think uart's right about the core material. The cores are made of powdered something-or-other so the initial permeability, depending on formulation, ranging from about 25 to 125.

    The application requires filtering the 60Hz component from a pulse width modulated 100Hz square wave using an LC filter.

    The area within the BH hysterysis curve represents energy loss. Multiplied by the frequency taken around the loop, this is power loss.

    (Is BH is in units of energy?)

    A DC offset current gives an offset H0. The hysterysis loop results from the smaller AC current component circulating around B0,H0.

    Mu is not constant but varies as a function of H.

    The manufacturer supplies parametric curves for power loss vs. Bmax and frequency. Given this information, is it possible to infer power loss at some other operating point B0,H0?
     
  7. Feb 26, 2010 #6
    Here's the web page from Mag Inc. that I've been using for data.

    http://www.mag-inc.com/products/powder_cores/kool_mu" [Broken]

    I'm beginning to think that hysteretic loss may only a function of B_peak_to_peak and frequency since excursion of H doesn't seem to be quoted by various manufacturers. Either that, or publishing a lot of power loss curves for various DC offsets is not fashionable.

    Looking into the units involved, BH has units of energy per unit volume in SI units which makes sense.
     
    Last edited by a moderator: May 4, 2017
  8. Feb 27, 2010 #7
    The equation developed by Steinmetz for hysteretic power loss (for soft iron or transformer steel) is

    Physt = const x f x B1.8 watts per cubic meter or per kilogram.
    B has units volt-seconds per m2, and H has units amp-turns per meter, so BH has units volt-amp-seconds per meter3, or joules per cubic meter.

    Bob S
     
  9. Feb 28, 2010 #8
    The parametric curve quoted by Mag Inc. for their Kool Mu compostion is PL mW/cm^2 = B2.00 Gauss F1.60 Hertz.

    Mag Inc. left out the leading constant for whatever reason, but it's reasonably consistant with Steinmetz's equation for = these other magnetic materials tested.

    However, the experimental constaints are absent in both cases: are the excursion of H and B centered about B=0 and H=0 in the quoted parametrically fitted curves?, and does it make substancial difference if they are not?


    (Somewhere in my random internet search for an answer I came across an explanation of the mechanism for core heating. It was explained as magnetic domains jumping crystaline boundries. This makes some sense, [ignoring other interesting properties of magnetic domain size restrictions] as it would induce sudden latice distortions that would manifest as heat.

    I think that the independence upon the H field makes sense if the mechanism driving domain orinentation is a function of the B field independent of mu.)
     
    Last edited: Feb 28, 2010
  10. Feb 28, 2010 #9

    dlgoff

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    I found this which might be of interest:

    attachment.php?attachmentid=23998&stc=1&d=1267372386.jpg

    http://homepages.eee.strath.ac.uk/~bwwilliams/Book/Chapter%2017old.pdf" [Broken]
     

    Attached Files:

    Last edited by a moderator: May 4, 2017
  11. Feb 28, 2010 #10
    The low frequency is 60 Hz, and the high frequency is about 100 Khz.)

    The application requires filtering the 60Hz component from a pulse width modulated 100Hz square wave using an LC filter.

    Is it 100 Khz or 100 Hz?

    The conventional LC filter is a low pass filter.
    This would preserve the 60 Hz component and get rid of the 100 Khz frequency.

    A CL filter might work. If a CL filter is used, there would be very little 60 Hz current in the inductor.
     
  12. Mar 1, 2010 #11
    100 KHz, and that's the idea. Although, in this case it's usually called an LC filter. The input nodes drive across an inductor and a capacitor in series, and the output is taken across the capacitor.
     
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