calculating number of turns on a transformer

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

  1. Jamie Morken

    Jamie Morken Guest

    Hi,

    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: http://www.falco.com/download/SMPSSelGuide.pdf

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

    cheers,
    Jamie
    Jamie Morken, Oct 24, 2007
    #1
    1. Advertising

  2. Jamie Morken

    Jamie Morken Guest

    Jamie Morken wrote:
    > Hi,
    >
    > 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: http://www.falco.com/download/SMPSSelGuide.pdf
    >
    > Is there a formula to calculate the required number of primary
    > turns to get about 1mH inductance from this data? Thanks for
    > any help.
    >
    > cheers,
    > Jamie


    I know the flyback configuration will decrease the inductance
    too, I guess this makes it harder..
    Jamie Morken, Oct 24, 2007
    #2
    1. Advertising

  3. Jamie Morken

    Tam/WB2TT Guest

    "Jamie Morken" <> wrote in message
    news:%TPTi.139170$Da.65705@pd7urf1no...
    > Hi,
    >
    > 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: http://www.falco.com/download/SMPSSelGuide.pdf
    >
    > Is there a formula to calculate the required number of primary
    > turns to get about 1mH inductance from this data? Thanks for
    > any help.
    >
    > cheers,
    > Jamie

    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.

    Tam
    Tam/WB2TT, Oct 25, 2007
    #3
  4. Jamie Morken

    Eeyore Guest

    Jamie Morken wrote:

    > Hi,
    >
    > 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: http://www.falco.com/download/SMPSSelGuide.pdf
    >
    > Is there a formula to calculate the required number of primary
    > turns to get about 1mH inductance from this data? Thanks for
    > any help.


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

    Graham
    Eeyore, Oct 25, 2007
    #4
  5. Jamie Morken

    John Larkin Guest

    On Wed, 24 Oct 2007 22:43:07 GMT, Jamie Morken <>
    wrote:

    >Hi,
    >
    >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: http://www.falco.com/download/SMPSSelGuide.pdf
    >
    >Is there a formula to calculate the required number of primary
    >turns to get about 1mH inductance from this data? Thanks for
    >any help.
    >
    >cheers,
    >Jamie


    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.

    John
    John Larkin, Oct 25, 2007
    #5
  6. Jamie Morken

    Jamie Morken Guest

    Eeyore wrote:
    >
    > Jamie Morken wrote:
    >
    >> Hi,
    >>
    >> 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: http://www.falco.com/download/SMPSSelGuide.pdf
    >>
    >> Is there a formula to calculate the required number of primary
    >> turns to get about 1mH inductance from this data? Thanks for
    >> any help.

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


    Thanks I downloaded it, looks like a neat program!

    cheers,
    Jamie

    >
    > Graham
    >
    Jamie Morken, Oct 26, 2007
    #6
  7. Jamie Morken

    Jamie Morken Guest

    John Larkin wrote:
    > On Wed, 24 Oct 2007 22:43:07 GMT, Jamie Morken <>
    > wrote:
    >
    >> Hi,
    >>
    >> 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: http://www.falco.com/download/SMPSSelGuide.pdf
    >>
    >> Is there a formula to calculate the required number of primary
    >> turns to get about 1mH inductance from this data? Thanks for
    >> any help.
    >>
    >> cheers,
    >> Jamie

    >
    > 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.


    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?

    cheers,
    Jamie

    >
    > John
    >
    >
    >
    Jamie Morken, Oct 26, 2007
    #7
  8. Jamie Morken

    legg Guest

    On Wed, 24 Oct 2007 22:43:07 GMT, Jamie Morken <>
    wrote:

    >Hi,
    >
    >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: http://www.falco.com/download/SMPSSelGuide.pdf
    >
    >Is there a formula to calculate the required number of primary
    >turns to get about 1mH inductance from this data? Thanks for
    >any help.


    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
    needed.

    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
    value

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

    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.

    RL
    legg, Oct 27, 2007
    #8
  9. Jamie Morken

    Genome Guest

    On Oct 27, 12:02 am, legg <> wrote:
    > On Wed, 24 Oct 2007 22:43:07 GMT, Jamie Morken <>
    > wrote:
    >
    >
    > 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
    > needed.
    >
    > 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.
    >


    Blork.

    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?

    DNA
    Genome, Oct 27, 2007
    #9
  10. Jamie Morken

    legg Guest

    On Sat, 27 Oct 2007 14:56:47 -0700, Genome <>
    wrote:

    >On Oct 27, 12:02 am, legg <> wrote:
    >> On Wed, 24 Oct 2007 22:43:07 GMT, Jamie Morken <>
    >> wrote:
    >>
    >>
    >> 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
    >> needed.
    >>
    >> 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.
    >>

    >
    >Blork.
    >
    >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?
    >
    >DNA



    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'.

    RL
    legg, Oct 28, 2007
    #10
  11. Jamie Morken

    Jamie Morken Guest

    legg wrote:
    > On Sat, 27 Oct 2007 14:56:47 -0700, Genome <>
    > wrote:
    >
    >> On Oct 27, 12:02 am, legg <> wrote:
    >>> On Wed, 24 Oct 2007 22:43:07 GMT, Jamie Morken <>
    >>> wrote:
    >>>
    >>>
    >>> 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
    >>> needed.
    >>>
    >>> 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.
    >>>

    >> Blork.
    >>
    >> 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?
    >>
    >> DNA

    >
    >
    > 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'.
    >
    > RL


    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
    inductance?


    fullcoil_turns =
    sqrt((fullcoil_inductance*(coil_turns(turns)^2))/coil_turns(inductance))


    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? :)

    cheers,
    Jamie
    Jamie Morken, Oct 29, 2007
    #11
  12. Jamie Morken

    legg Guest

    On Mon, 29 Oct 2007 23:09:17 GMT, Jamie Morken <>
    wrote:

    <snip>
    >
    >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
    >inductance?
    >
    >
    >fullcoil_turns =
    >sqrt((fullcoil_inductance*(coil_turns(turns)^2))/coil_turns(inductance))
    >
    >
    >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? :)
    >


    What is your

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

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


    Core saturation limiting is best handled by a current limiting control
    topology.

    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 ?

    >>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


    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.

    RL
    legg, Oct 30, 2007
    #12
    1. Advertising

Want to reply to this thread or ask your own question?

It takes just 2 minutes to sign up (and it's free!). Just click the sign up button to choose a username and then you can ask your own questions on the forum.
Similar Threads
  1. Watson A.Name - \Watt Sun, the Dark Remover\

    Estimating the Number of Turns of an Inductor

    Watson A.Name - \Watt Sun, the Dark Remover\, Jun 5, 2004, in forum: Electronic Components
    Replies:
    90
    Views:
    2,211
    Watson A.Name - \Watt Sun, the Dark Remover\
    Jun 14, 2004
  2. Al Borowski
    Replies:
    22
    Views:
    1,748
    John Woodgate
    Apr 12, 2004
  3. Watson A.Name - \Watt Sun, the Dark Remover\

    Estimating the Number of Turns of an Inductor

    Watson A.Name - \Watt Sun, the Dark Remover\, Jun 5, 2004, in forum: Electronic Design
    Replies:
    90
    Views:
    1,310
    Watson A.Name - \Watt Sun, the Dark Remover\
    Jun 14, 2004
  4. Brad
    Replies:
    0
    Views:
    642
  5. Optik
    Replies:
    14
    Views:
    3,901
    Optik
    Nov 11, 2011
Loading...

Share This Page