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question about winding a coil to correct inductance

Discussion in 'Electronic Design' started by Ben, Nov 3, 2003.

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  1. Ben

    Ben Guest

    Hi there,

    further to my previous question I have now wound a coil to act as an
    antenna for my RFID application. I don't really understand the physics
    behind this, hence I don't really understand the following result, but
    I'm really hoping someone can explain it to me!

    Now according to the instructions of the circuit board which drives my
    antenna, the antenna needs to be resonate at 125kHz, which apparently
    will happen if the antenna has an inductance of 1.62mH. So, to find out
    how many turns the coil needs, according to these instructions, I should
    use the formula:

    N=sqrt( L*(3d+9l+10e)/(0.08*d^2) )

    Where N is number of coils, L is desired inductance, d is coil diameter,
    l is coil width, and e is coil thickness.

    I want a coil of width 10cm, diameter 1.9cm, width around 0.3cm. This
    results in me needing 837 turns....

    So, 1000 turns later (just to be on the safe side) I end up with a coil
    which gives me a frequency of more like 75kHz when I plug it into the
    board. I put in too many turns because you can easily remove turns but
    you can't put them back in once you've cut the wire. But, from what I
    thought I understood, more turns would equal more inductance would equal
    a greater frequency? And in any case, 20% more turns wouldn't make a 50%
    smaller frequency would it? So, can any tell me what I have done wrong
    or which bit I don't understand?

    Thanks very much,

    Ben Kenward
    Department of Zoology
    Oxford University
  2. Ian Stirling

    Ian Stirling Guest

    Look at the resonant frequency equation.
    More turns = more inductance = lower reactance = lower freq.
  3. John Fields

    John Fields Guest

    Increasing the number of turns increases the inductance, and since

    f = ------------
    2pi sqrt(LC)

    you can see that if L increases and everything else stays the same, f
    will also increase.

    Also, if L doubles for a given C, f won't double; it'll only increase by
    the difference in the _square roots_ of the different LC products.
  4. John Fields

    John Fields Guest

  5. Ian Stirling

    Ian Stirling Guest

    I'd like to say that you'r an idiot, and should not correct completely
    correct statements.
    Unfortunately, you are of course completely right.
    I have absolutely no idea what I was thinking.
    Plus, for flat coils, the simple formula is at best approximate, and
    using it as a first guess then measuring is probably the easiest route.
  6. Joe Legris

    Joe Legris Guest

    Congratulations! At least your oscillator oscillates. However, 20% more
    turns should give 41% more inductance resulting in a frequency 20%
    lower, not higher.

    This is because the resonant frequency of a coil and capacitor is
    1/2pi*SQRT(L*C) but the inductance is proportional to the square of the
    number of turns on the coil. Therefore the frequency is inversely
    proportional to the number of turns.

    There is apparent confusion about the coil "width" and "thickness" in
    that you call them both "width". Which is which?

    What kind of capacitor are you using? Its value may off by 20% or more
    and may vary significantly with temperature.

    Joe Legris
    Check out my Ebay listings:
    Brand new electronic components in lots of 10+
  7. Where'd you learn how to do math? Try that statement out with any simple
    integer in the bottom half of a fraction, and see how well it works...
  8. John Fields

    John Fields Guest

    Twasn't a problem with the math, it was a problem with the English.

    "f will also increase" should have read "f will decrease".

  9. Decrease!
  10. John Fields

    John Fields Guest

  11. default

    default Guest

    Sounds roughly correct. Inductance increases with the square of the
    turns, more inductance equals lower frequency: what you have . . .

    There's somewhat more going on than the formulas tell you. The
    location of nearby metal, the wire diameter, length:diameter ratio,
    and insulation all play a role in the actual inductance - most
    inductance calculations are simplified or only work well within very
    specific limits.

    There's also the fact that you are concerned with the frequency, which
    in turn, will be affected by other circuit parameters - like voltage
    to the oscillator, capacitors across the coil, internal capacitance of
    the driving circuitry, self resonance of the coil, temperature affects
    on the components etc..

    I take it this antenna also serves to determine the frequency?

    You could have wound the correct coil and lowered the frequency with a
    cap across the coil or some ferrite material for a core or a strip of
    aluminum foil near/on the coil.
  12. I can't seem to get anywhere near 837 turns for 1.62mH.
    Are the defs of d/l/e correct, and what are the units?
    F-resonant is approximately proportional to the reciprocal
    of the turns-count. So if 1000 turns resonated at 75KHz
    then it might be closer to 600 turns for 125KHz.
  13. Ben

    Ben Guest

    Thanks a lot for all your responses, folks... It seems I got it the
    wrong way round - after removing turns, I was eventually able to get
    into the frequency range, and get my antenna to read an RFID tag! It
    seems this is quite a trial and error thing - the eventual optimum
    number of turns for maximum read range was a fair way off the number
    given by the maths.


    Ben Kenward
    Department of Zoology
    Oxford University
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