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inductance of transformer windings

Discussion in 'Electronic Design' started by Jamie Morken, Oct 16, 2007.

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  1. Jamie Morken

    Jamie Morken Guest


    I have a transformer primary of 17 turns and 4mH inductance.

    There are 4 coils in series (3 taps / 5 leads)

    I am trying to figure out the inductance of the 4 coils.

    coil1 3turns 706uH

    coil2 2turns 471uH

    coil3 4turns 941uH

    coil4 8turns 1.88mH

    that adds up to 4mH but I read before that when you
    double the turns the inductance quadruples, so I am
    not sure if the above calculations are correct.

  2. Inductance of magnetically coupled turns do not add in
    proportion to the turns count. If they are perfectly
    coupled, they add up in proportion to the square of the
    turns count. For small coils, the coupling is certainly not
    perfect (all the flux created by any turn does not surround
    all the other turns). But the squared approximation is
    still a first guess. So lets see what the inductance per
    turn squared is for this coil and then calculate the total
    inductance if all the turns are put in series aiding (all
    creating flux in the same direction with the same current
    passing through all turns).

    Coil 2 471 uH / (2 turns squared) = 118 uH/T^2
    Coil 1 706 uH / (3 turns squared) = 78 uH/T^2
    Coil 3 941 uH / (4 turns squared) = 58 uH/T^2
    Coil 4 1880 uH / (8 turns squared) = 29 uH/T^2

    As the number of turns goes up, the inductance per turn
    squared is going down, a sure sign of poor coupling.

    If we take the inductance factor of the 8 turn coil and use
    it as an upper limit for all 17 turns in series, the
    estimate would be 29 uH * (17 turns squared) = 8.4 mH.

    You say the inductance of the whole 17 turns is 4 mHy. Is
    this a measured value, with all turns in series? It is not
    impossible, but it would indicate that there is essentially
    no magnetic coupling between the individual sections, and
    that would not make it a very effective transformer.
  3. Well, I screwed that up.

    I was taking the 4 coil inductances as measured values, but
    think I see now, that the only known inductance is the total
    of the 17 turns, and you are trying to calculate the section

    So the full coupled assumption would be that inductance
    factor per turns squared would be:
    4000 uH / (17 turns squared) = 13.8 uH / T^2

    So, by the turns squared assumption:
    Coil 2 ((2*T)^2)*13.8 uH / T^2 = 55 uH
    Coil 1 ((3*T)^2)*13.8 uH / T^2 = 124 uH
    Coil 3 ((4*T)^2)*13.8 uH / T^2 = 221 uH
    Coil 4 ((8*T)^2)*13.8 uH / T^2 = 883 uH

    Since the actual flux coupling will be less than perfect,
    these are minimum possible values. At zero coupling your
    figures are what the section inductances would converge to.

    The actual value will be somewhere in between.
  4. Jamie Morken

    Jamie Morken Guest

    Thanks John, I got those same alternative numbers too
    using this turns squared formula I made

    coil_turns(inductance) =
    (1 / (17 / coil_turns(turns))^2) * 4mH

    I guess that the magnetic coupling of the transformer is the important
    thing to know now! What would you estimate it would be for a modern
    planar transformer design?

  5. Jamie Morken

    Jamie Morken Guest

    I put 0.98 in ltspice as a first guess, but I guess there is a way
    to measure the coupling if you have the physical transformer..

  6. Jamie Morken

    Jamie Morken Guest

    The secondary has 2 turns + 5 turns + 5 turns + 2 turns, (14 turns) can
    I calculate its total inductance by:
    secondary turns/primary turns * primary inductance
    = 14/17*4mH = 3.294mH

    And then use the same method above to get the inductance of each winding
    I guess?

    Also does 4mH seem high for only a 17turn primary? This is the number
    they told me but it seems quite high.

  7. neon


    Oct 21, 2006
    Gee he measured inductance at what frequency nobody knows I don't know either interesting
  8. Paul Mathews

    Paul Mathews Guest

    You don't say how you are measuring inductance. You will get very
    different results for an inductor with a core, depending on things
    like measurement frequency and excitation level.
    Paul Mathews
  9. Depending on what the transformer will be used for, it might
    be very important. I would not estimate it at all, but
    measure it.
  10. Jamie Morken wrote:
    The coupling factor in LTspice is a representation of the
    voltage coupling efficiency between inductors. That means
    that if you apply 1 volt per turn to one winding and you get
    out only .98 volts per turn from another winding, then the
    coupling factor between that pair of windings is .98.

    In your case, there will be a different coupling factor
    between each pair of sections, though the values may be very
    close to each other.
  11. I would estimate its total inductance by:
    ((14*T)*13.8 uH/T^2=2.7 mH
    If the turns are all tightly coupled (assuming a coupling
    factor of 1) then this turns squared formula will work for
    any number of turns. Why do you need to calculate the
    inductance of various windings?
    It implies a very high permeability core with no air gap.
    That also means that it will have a high tolerance, since it
    depends so strongly on the material properties and the
    accuracy of the ground contact surfaces and their proper
    assembly. But it also implies a coupling factor very close
    to 1 and good accuracy of the turns squared model of the
    inductance of any number of turns.
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