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transformer resistances

Discussion in 'Electronic Basics' started by davidt, Feb 25, 2008.

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  1. davidt

    davidt Guest

    I have an transformer from a old power supply.
    It has 2 red wires and 2 yellow wires.
    I think the red wires are the high voltage (mains) in and the yellow
    the low voltage (12v ?) out,
    To check this I touched a 6V battery across the yellow wires and got a
    high voltage across the red wires (enought to spark across a 3mm gap)
    But when I measure the resistance across the red wires I get 31.1ohms
    and across the yellow 10.4ohms.
    This is not what I expected.
    I thought the high voltage winding would be thicker wire and less
    turns than the low voltage winding and so would have a lower
    Could anyone explain where I am going wrong?
  2. amdx

    amdx Guest

    You have it backwards. The high voltage side will have more turns
    of smaller wire. (it carries less current) The low voltage side will have
    less turns if thicker wire. (it carries more current)
  3. davidt

    davidt Guest

    Yes that makes sense now when think about the current.
    Thanks for your help.
  4. Only smaller wire because the current requirements are smaller. There is no
    restriction on the wire size for a transformer except that it be able to
    handle the current. (so it definitely could have been larger)
  5. John Fields

    John Fields Guest

    Sure, but since what a transformer does is transfer _power_ from one
    winding to another, the product of current and voltage have to be
    the same for both windings.

    So, if the voltage in one winding is higher than in another, then
    the current in the high-voltage winding _must_ be less than in the
    low-voltage winding and it would be pointless to use wire with a
    diameter more than that required to carry the current.

    More precisely, for a lossless transformer:

    Np Ep Is
    ---- = ---- = ----
    Ns Es Ip

    Where: Np is the number of turns on the primary,
    Ns is the number of turns on the secondary,
    Ep is the voltage impressed across the primary,
    Es is the voltage induced across the secondary,
    Ip is the current forced through the primary, and
    Is is the current into the load
  6. Jamie

    Jamie Guest

    The resistant you're measuring has nothing to do with the operation
    of the transformer other than lowering its efficiency and Q. It's simply
    a by product of the gauge wire used along with the length of wire used
    for each winding.

    The number of turns ratio between the Red windings and yellow windings
    tells the whole story along with the gap spacing between them..

    At this point, maybe you should reference some data from the net on
    transformer theories.
  7. John Fields

    John Fields Guest

  8. Jamie

    Jamie Guest

    my point was short and simple. difference in current mode xformers and
    potential types along with turn ratio's. also the construct of the
    cores. With all this put together leads to PF (power factors), current
    ratio's, form factor, wet/dry, efficiency etc..

    Keeping it simple since the poster obviously needs to research some
    transformer basics.
  9. Phil Allison

    Phil Allison Guest


    ** If it is from a 240 volt mains power supply, then the red wires are for
    240 volts AC and the yellow ones are for 140 volts AC. The unit is rated
    for about 75 VA.

    If from a 120 volts supply, the yellow wires are for 70 volts AC and the
    unit is rated for about 25 VA.

    ........ Phil

  10. And why not Np/Ns = D_Rs/D_Rp where D_Rx is the diameter? or even Np/Ns =

    The fact of the matter is that it doesn't matter as long as the as the wire
    can carry the current(which essentially means heat dissipation).

    You can use any wire size in any circumstance and its not *pointless*. It
    depends on the application. True that you pretty never need a larger
    diameter on the secondary than the primary in a step up... but potentially
    there is still the possibility in certain applications. (such as if the
    secondary happens to be in a much much hotter environment)

    The idea transformer equations say nothing about the diameter's of the wire
    used and for good reason. (not to mention that most wire sizes are

    So going off the restance or the diameter of the wire is by no means correct
    and technically has nothing to do with a abstract ideal transformer. (I'm
    not saying you can't use that relationship to find the optimal diameter for
    a given number of turns but it is not necessary)
  11. John Fields

    John Fields Guest

    Because the turns ratio has nothing to do with the resistance ratio.
    Well, no. What _really_ matters is the resistance of the winding,
    since that'll determine the voltage dropped across it, and that
    voltage won't be able to be used by the load. While the I²R losses
    increase the winding's resistance, it's not usually enough to
    matter much. For a clue, take a look at the temperature coefficient
    of resistance of copper and translate that to the change in
    resistance of a winding due to the change in temp. In any case, the
    temp rise because of I²R losses should be taken care of during the
    design of the transformer and the proper wire size chosen to give
    the desired regulation at that maximum temperature.
    You _cannot_ use any wire size in any application and expect the
    application to work. For example, it would be pointless to wind a
    secondary expected to supply 12 volts at 10 amperes with a single
    strand of #40 AWG.
    Precisely. And, for the transformer the OP was talking about,
    (which he was using as a stepup transformer to get the results he
    did) it would have been pointless to have wound the high-voltage
    winding with larger diameter wire than the low-voltage winding.
    Like where??? You're grasping at straws.
  12. No! Why you think he is winding anything? They are already wound and it was
    wound for a specific application. In general you are correct in that for a
    step up the wire diameter will be smaller because there is less current. The
    point I'm trying to make is that it is not a necessary or sufficient
    requirement for step up transformers. maybe in 99.999% of transfomers it
    will hold but even then there is no "formula" to calculate the diameter of
    the secondary from the primary or vice versa.

    I could easily make a transformer that would violate your "principle" and it
    would work fine! Sure it might be cost or space inefficient but the
    tranformer itself doesn't care.

    Now you say "Well if its too small then it will over heat"!! This has
    nothing to do with basic ideal transformer theory which says nothing about
    wire's resistances or anything. Hell, if I wanted I could then just use
    super conducting material and use infinitely thin wire on both sides!!

    I think thats what you are doing. You are trying to make the OP think that
    the diameter matters and gives a good approximation to the turns ratio and
    it doesn't. Go take a transformer and measure the diameters then take the
    ratio and I bet you won't get a good approximation to the turns ratio! I

    Since I know you are chicken I did it for you. I have a tranformer on my
    desk. Its a step down from 120:12(approximately) so its about 10:1.

    The diameter of the primary is 100mil, the diameter of the secondary is 40
    mil.... hmmm, is this 10:1 or 1:10? not even close!

    Of course now you will claim that the person who made the transformer is
    nutz and has no clue about transformer design.
    WTF? I asked why that isn't shown in the books... I'm ASKING you why that
    isn't true because you are essentially claiming it is. Thats why I have
    question marks there.

    Forget it... I can see this is a lost cause.
    Damn, I really thought I put you on my ignore list... along with the other
    morons. Well, you won't escape this time.
  13. John Fields

    John Fields Guest

  14. amdx

    amdx Guest

    So your transformer has #10 wire on the primary and #18 wire on the
    secondary? I think your secondary will have about 60% or less of the
    resistance of the primary.
    Please post the resistances of the primary and secondary windings, that way
    I can see how I misled the questioner.
    What is the use for this transformer? It's not like anything I'm used to
    Thanks, Mike
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