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calculating number of turns on a transformer

Discussion in 'Electronic Design' started by Jamie Morken, Oct 24, 2007.

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  1. Jamie Morken

    Jamie Morken Guest


    I have a flyback transformer design in ltspice with a 1mH primary coil,
    I would like to see how many turns this would require on this transformer:

    core size: E375

    falco part#: 1831-331-002

    cross section area Ac(cm^2) =0.840

    magnetic path length (cm) = 6.94

    WaAc(cm^4) = 0.856

    core volume(cm^3) = 5.830

    this info is from:

    Is there a formula to calculate the required number of primary
    turns to get about 1mH inductance from this data? Thanks for
    any help.

  2. Jamie Morken

    Jamie Morken Guest

    I know the flyback configuration will decrease the inductance
    too, I guess this makes it harder..
  3. Tam/WB2TT

    Tam/WB2TT Guest

    They can boil all this information down into one constant, sometimes called
    Al, given in uH/100 turns. You might want to see if they publesh that for
    your core.

  4. Eeyore

    Eeyore Guest

    I suggest you get a copy of Epcos's Ferrite Magnetic Designer program.

  5. John Larkin

    John Larkin Guest

    Call them and ask what's the Al value. It's amazing that many
    magnetics suppliers don't furnish this. They could also furnish a
    number of other handy electrical and thermal values that are a
    nuisance to calculate, like gauss per ampere-turn and temperature rise
    per watt.

  6. Jamie Morken

    Jamie Morken Guest

    Thanks I downloaded it, looks like a neat program!

  7. Jamie Morken

    Jamie Morken Guest

    I guess also they might have Al values for different core gaps, or can
    this be guesstimated too? :) I don't know much about using core gaps
    for flybacks, is it even necessary to do if you have a core big enough
    that it won't saturate without a gap?

  8. legg

    legg Guest

    You would not normally need AL values for a ferrite core used in a
    flyback, as this is determined by the gap required to store the energy

    L = uo x N^2 x Ae / Lg

    L = inductance in henries

    uo = permeability of free space = 4 x pi x 10E-7

    N = turns count

    Ae = cross-sectional area at the gap in meters^2

    Lg - length of the gap im meters.


    You have to determine N based on peak core flux change, limited by
    core loss or core saturation at the frequency and duty cycle you are
    preparing or able to use. To do this you need more core material
    information concerning it's power loss characteristics, whether the
    material is ferrite or powdered material.

    As powdered material has a distributed gap, it will not normally be
    merchandised without reference to it's permeability or the part's AL

    For ferrite parts, if the frequency is lowish or the part is very
    small, the flux density will likely be determined by the saturation

    In this case,

    Nmin > V x t / ( Bsat x Ae )

    Nmin = minimum turns

    V = applied voltage in volts

    t = period of applied voltage in seconds

    Bsat = saturation flux density in Teslas

    Ae = minimum cross-sectional area of the ferrite material in meters^2

    ( Bsat of ferrite ~ 0.33T @ room temperature )

    You seem already to have determined that 1mH of primary inductance is
    desirable, by some method or other. Note that depending on the
    operating frequency, a certain peak current will be expected in this
    primary inductance in order to deliver the required output power.

    This is determined by the rough formula

    P = L x Ip^2 x f / 2

    P = delivered power and all losses in Watts

    L = primary inductance in Henries

    Ip = peak primary current in Amps

    f = pulse repetition rate in Hertz

    Please do some reading. The old Unitrode/ Texas Instrument app notes
    cover flyback converters and flyback transformers pretty clearly.

  9. Genome

    Genome Guest


    So... In the limit as Lg tends to zero inductorance tends to infinity?

    What is life like when you get dialed in as low quality?

  10. legg

    legg Guest

    When the ratio between the lengths of the gap path and the ferrite
    path lenths approaches 10E-7, then the permeability of the ferrite
    comes into effect.

    A gapless ferrite core isn't much use in storing energy, which is what
    is intended in a flyback circuit. Whether a 20mW circuit can be said
    to have a topology is another thing. erp.

    I thought I might get a respnse from Master Jamie, but I guess he's
    just fooling around again.

    Why didn't you offer a link to your own blither. I recall it being
    modestly illuminatin'.

  11. Jamie Morken

    Jamie Morken Guest

    I did some testing with an ETD29 ferrite core:

    Ferroxcube ETD29 core and bobbins bought from Farnell

    3C90 ferrite material

    AL = 2350 +-25% (with no airgap)
    Ue = 1850 (with no airgap)

    (wound with 22 gauge wire, 25 turns from one end of bobbin to the other end)

    25 turns no airgap gives 1.8mH

    25 turns with 1 layer kapton tape airgap gives 0.446mH (0.002" airgap)

    25 turns with 2 layer kapton tape airgap gives 0.260mH (0.004" airgap)

    25 turns with 3 layer kapton tape airgap gives 0.192mH (0.006" airgap)

    now how many turns do we need with these kapton tape airgaps to get 1mH

    fullcoil_turns =

    1layer of kapton tape:
    fullcoil_turns = sqrt((1mH*(25^2))/0.446mH)
    = 37 turns for 1mH with 1 layer kapton tape airgap

    2layers of kapton tape:
    fullcoil_turns = sqrt((1mH*(25^2))/0.257mH)
    = 49 turns for 1mH with 2 layers kapton tape airgap

    3layers of kapton tape:
    fullcoil_turns = sqrt((1mH*(25^2))/0.192mH)
    = 57 turns for 1mH with 3 layers kapton tape airgap

    So how big should I make the airgap? :)

  12. legg

    legg Guest

    What is your

    a - input voltage (at drop-out)
    b - operating frequency
    c - maximum duty cycle

    Core saturation limiting is best handled by a current limiting control

    For core-loss limited applications, determine the permissible core
    loss; assuming 1degC surface rise in every square centimeter for every
    milliwatt dissipated. (+/-20% )

    All of the milliwatts are generated by the total core volume, giving a
    core loss density. (mW/cm^3 = Kw/m^3)

    This loss density will correspond to a peak flux density at a specific
    operating frequency in the core material's published characteristics.

    Is 1mH capable of storing your power requirement ?

    The peak primary current must be achievable for the same operating
    conditions as the peak flux calculation.

    V = L x di / dt

    V = minimum input voltage

    di = minimum required current peak

    dt = maximum conduction period (at frequency and duty cycle limit)

    L = maximum primary inductance.

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