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Amt of current to saturate ferrite rod

Discussion in 'Electronic Design' started by Paul Nelson, Dec 1, 2011.

  1. Paul Nelson

    Paul Nelson Guest

    Is it technically possible to saturate a garden variety ferrite rod
    (10mm dia. x 120mm long) with a single layer winding?

    This assumes applied DC.

    Paul Nelson
     
  2. mike

    mike Guest

    Yes, to the degree that you can "saturate" any material.
    Isn't saturation an asymptotic property that can't be fully achieved
    this side of infinity?
    So, depends on your definition of saturation.
     
  3. Tim Williams

    Tim Williams Guest

    At least for moderate length rods (roughly 6 mm dia. x 30 mm long, often
    seen in computer SMPSs), the change in inductivity is about half, so
    you'll hardly notice it. The change is just as sharp as any ferrite
    inductor though.

    At an aspect ratio like that, you'll probably see more gradual saturation,
    as the center becomes saturated first and the whole thing progressively
    becomes an air-cored inductor.

    Depending on what you're doing, rods often aren't wound all the way to the
    end, because inductivity drops -- SWAG, fringing sucks down maybe half the
    field out past 80% from the center. You occasionally also see multilayer
    windings, which are great if the wire is thin, but only up to a winding
    height about equal to the diameter. Think of the fields of a bar magnet,
    they loop around pretty well from the ends, but any space within those
    arcs is prime real estate for coil winding purposes.

    Tim
     
  4. Sjouke Burry

    Sjouke Burry Guest

    (Paul Nelson) wrote in @news.tpg.com.au:
    Yes.
     
  5. Joerg

    Joerg Guest


    But I didn't know that even those can be hit with AMT :)
     
  6. admformeto

    admformeto Guest

    If you just want static saturation then use neodymium magnet.
    The electromagnetic saturation will consume some power to the point that the
    winding will start decapitating some heat so to avoid that multi layering
    with thicker wire must be used.

    Mathew Orman

    http://www.faster-than-light.us/
     
  7. John S

    John S Guest

    As others have pointed out, yes.

    Also, as others have pointed out, how will you define saturation?

    Addressing your subject line:

    1 amp. (assumes 100 oersteds) (955 turns of wire over the 120mm)
    or
    10 amps. (assumes 100 oersteds) (95 turns of wire over the 120mm)
    or
    etc.

    Of course, the 100 oersteds is probably overkill for what you're asking,
    but who can tell at this point?
     
  8. Ralph Barone

    Ralph Barone Guest

    And if you need more current, use an edgewound ribbon conductor. I agree,
    the question is vague.
     
  9. legg

    legg Guest

    You might apply the formula:

    B = 4 x pi x 10^-7 x N x I / le

    B = saturation flux density in Teslas ( ferrite ~ 0.3)
    N = number of turns
    I = DC current in Amps
    Le = magnatic path length in Meters
    - ( length of air path between rod ends)

    As the ferrite saturates, le doubles, before B can continue to climb.

    In practical situations, you'd be more concerned with temperature rise
    in the copper. Because of the long gap, inductance and Q values are
    pretty low, until you hit RF frequencies, but are better than
    air-cored structures of comparable size.

    RL

    again server issues. every post is a frigging test....
     
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