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Question about inductors

  1. May 17, 2015 #1
    If I am making my own inductor following the equation:
     
  2. jcsd
  3. May 17, 2015 #2
    Sorry didn't finish, following the equation L=μN2A/l where L is inductance, μ is the magnetic permeability, N is the number of turns and l is the length, and I have a piece of aluminum as my core with a μ=1.256665x10-6, if the aluminum core is hollow, meaning a conduit pipe, when calculating the inductance would μ still be just the value for aluminum or would it be a combination of aluminum and air since the pipe is hollow. Please answer quickly and thank you for your help in advance!
     
  4. May 17, 2015 #3
    Aren't they the same thing?

    According to the chart on Wikipedia they have the same value out to 5 decimal points. I doubt a handmade inductor will be accurate that far out. I think you will be lucky to be within 5%.

    You should be able to treat an aluminum core as an air core for small currents. (I don't know if it saturates like iron, but I doubt it.) You will be adding core losses though.
     
  5. May 17, 2015 #4
    Alrighty thanks a lot! That should be a lot of help.
     
  6. May 18, 2015 #5
    For AC, the aluminium will have a current induced into it which have its own magnetic field opposing the main one. This will reduce the inductance slightly.
     
  7. May 18, 2015 #6
    How much is slightly?
     
  8. May 18, 2015 #7
    Not sure how to calculate it. Why don't you use a plastic or cardboard former? These materials all have a relative permeability of 1.
     
    Last edited by a moderator: May 18, 2015
  9. May 18, 2015 #8
    Me either. o_O

    It depends on the geometry of the aluminum tube/bar. Nominally the tube will cut through the magnetic flux just like a loop of wire. It will develop a counter flux just like a loop of wire. So it should subtract an amount about equal to a one loop inductor from your multi-loop inductor (but of the length of the tube). Of course that assumes the geometry supports the tube as a full loop cutting through all the flux, which it won't. Still in a typical application it should get the lion's share, so I'd figure it for one loop.

    Are you aware of induction furnaces? It is remotely possible you might be unknowingly building one. They are actually finicky, so it's unlikely, but core losses might be a problem at radio frequencies (~100,000 Hz.).

    If you are concerned about the actual value, you can measure it when you are done. This is not trivial. The easiest way I know is to build a tank circuit with a known capacitance and find the resonant frequency. Then calculate back.

    Almost all non-conductive materials are the same permeability as air. Wouldn't that be easier?
     
  10. May 18, 2015 #9
    True. I'll try that. I'm applying for the Google science fair and need it for tomorrow so this should be good. And my frequency needed is 40kHz so would that still be a problem?
     
  11. May 18, 2015 #10
    Yes, 40kHz will buy you some hefty core losses. How much of a problem that is will depend on your application. If it plugs into the wall and doesn't get hot you should be fine. If it is human powered... Sweat a little more.

    If it's just a science project, don't worry. The worst that can happen is the entire thing burns and you win the pyrotechnic award. :nb)

    No one will care if it uses 1/2% more power than it should.
     
  12. May 19, 2015 #11
    L=μN2A/l is not the correct equation to use, or at least not as you might think it is.

    "l" is the effective magnetic path length. It includes the return path outside the cylinder. Also the permeability over the return path is the permeability of air--which is about the same as the vaccum permeability.

    There's an additional problem with the formula. With a low permeability core, not all the flux passing through one loop will pass through the other N-1 turns, so the N^2 term is too large.

    I would look for an online inductance calculator, or table of various inductor geometries.

    Additionally, i don't think you want to use an aluminum mandrel, or any conductive mandrel. It's effectively a short circuit, transformer secondary winding of one turn. This reduces the Q.
     
    Last edited: May 19, 2015
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