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more noob questions - transformers

Discussion in 'Electronic Basics' started by [email protected], Mar 29, 2007.

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  1. Guest

    Ok ... try not to laugh when i ask... tho i only have a few questions
    on designing a home made transformer.. And of course i dont want to
    burn my house down.. so lets start from there.

    I would like to make my own hv transformer and thought i would start
    by pulling a apart a regular 12 volt ac / dc 12 volt 1.5 amp
    transformer and removing the rectification.
    come into and go directly to the core however i would think there is
    some kind of resistor before or hiden around the primary windings.
    First questions is this dont make alot a sense to me as it is
    esentially a short curcuit. However if it does actually work this
    way then the windings are wrapped so that itself is a good resistor
    to precent the short circuit. however that the main reason why I am
    writing all this.
    So my question is if there is a resistor what should it be. ( or how
    can i calculate what it shoudl be) . Now i know there is a billion
    formulas for making good transformers so that why i want to start
    from a premade dc to and remove the rectification so i can run it
    into another transformer to step it back up again. this shoudl
    limit what i shoudl have to calculate for resistors if needed.

    . this is again why i like to know why it works that way from the
    mains as i will have to pass the secondary winding thru the primary
    of the home wound step up transformer which again would be a short
    circuit .. or at least what looks to be. i will put fuses ( and i
    would expect resistors ) between the 2 transformers to prevent too
    much current on my hv winding .
    I plan on striping another laminated core . or at least the secondary
    windings on a second core to wind my prefered step up

    now if i use a 110 ->12vdc transformer and its a 1.5 amp . I will
    choose my second core to be rated around 1 amp and my hv amperage
    should end up in small ma rating for light usage without burn out.

    I reviewed many diferent transformer designs and all they ever give
    is is a direct input from the mains to the windings or there just
    way to complicated as there beyond the scope of anythign basic.

    im looking for such a basic design it just keeps the same frequency
    so that why i thougth stripping the rectificatin should do from a dc
    supply. I can figure out the insulation needed for the layers on my
    core to prevent it from shorting out to the core or itself due to the
    hv so lets not worry about that either.

    If anyone could please let me know if this should really be more
    complicated if it really is that simple.
    Im looking simplest as possible design that will not blow up when I
    plug it into the mains.

    I appreciate any imput reguarding this. I have some HV experience with
    whimhurst and those types of machines however not much when it comes
    to current limiting. Im certianly not going to make something without
    making sure i understand the safty parts of it first.


  2. tempus fugit

    tempus fugit Guest

    There is no resistor in a transformer. There are 2 coils. The one that the
    mains comes in contact with is called the primary. The secondary does not
    make any contact at all with the primary, and therefore not the mains
    either. They are 2 separate coils wrapped around a piece of metal (iron I
    think). Since the mains voltage is AC, and therefore moving rapidly from one
    direction to the other, it induces a current into the secondary. The voltage
    you get out of the secondary depends on the ratio of the # of turns in the
    primary to the # of turns in the secondary. So if you have 10x as many turns
    in the primary than the secondary, there will be 12v (assuming 120v at the
    primary) on the secondary. As far as choosing the type of wire or number of
    wraps, that's a little out of my league, so I'll leave that to teh more
    experienced in the group.

    As always, be very careful when working with mains voltages......
  3. Guest

    Thanks for your reply. Thats the way it looked to me without a
    resistor and i just wanted to make sure as primary going into the
    mains still is not clear to me why it dont short out.
    If thats the case does transformers come some power all the time?
    they must. And if what im doiung is essentially chaining a few
    together . i would think they even comsune more power without a load?
    or is it the other way. because there is no load it prevents
    current on secondary as the equal and opposite fields react as a
    balanced resistor until the current is used? Is that right? if this
    is the case the ammount of efficency loss in transformer i would think
    would be measureable to the constant power drain of a transformer with
    no load?
    Like i said im a noob to this so I just want to make sure i
    understand this the best i can befire i start with this.
    and again thanks for your help.

  4. John Fields

    John Fields Guest

    First off, what you have is _not_ a transformer, it's an AC to DC
    converter or, simply, a "power supply"

    The transformer is the device in the power supply with two leads
    connected to the mains terminals and the other leads connected
    either to four rectifiers (two leads) or to two rectifiers and to
    the output common terminal (three leads)
    No resistor is needed because the inductive reactance of the
    primary winding (with an unloaded secondary) will limit the amount
    of current in the primary to what is needed to magnetize the core
    and supply the various winding and core losses.
    The simplest possible way to get what you want would be for you to
    unwind the primary and count the number of turns as you unwind it.
    Record that number so you won't lose it.

    The output voltage of the transformer can be described by:

    Ep Ns
    Es = -------

    Where: Es is the voltage induced across the secondary,
    Ns is the number of turns on the secondary,
    Ep is the voltage impressed across the primary, and
    Np is the number of turns on the primary.

    so, rearranging to solve for the number of turns on the primary and
    assuming you found 100 turns on the secondary:

    Ep Ns 120V * 100t 12000 Vt
    Np = ------- = ------------- = ---------- = 1000 turns
    Es 12V 12V

    Now, rearranging to solve for the number of turns required on the
    secondary for any secondary voltage desired:

    Es Np
    Ns = -------

    Assuming that you want 1000 volts out of the secondary:

    Es Np 1000V * 1000t 1E6 Vt
    Ns = ------- = --------------- = -------- = 8333 turns
    Ep 120V 1.2E2V

    So now you have the problem that you've got to get about 83 times
    more turns on the secondary than you took off.

    Maybe it'll work, maybe not. Let's try to find out.

    A transformer is a device that transfers _power_ from the primary to
    the secondary, so if your transformer is rated to supply 1.5 amps at
    12V, then that's:

    P = IE

    where: P is power in watts,
    I is current in amperes, and
    E is voltage in volts,


    P = IE = 12V * 1.5A = 18 watts.

    Rearranging to determine how much current your new secondary can

    I = ---

    and for a 1000V secondary, that's:

    P 18W
    I = --- = ------- = 0.018 ampere
    E 1000V

    OK. Now we have to determine what wire size to use and whether it'll
    fit in the space the old secondary used to fit into. Assuming about
    750 circular mils per ampere, your old wire was about #20, with a
    diameter of about 0.032". At the same current density, your new
    wire would have an area of:

    0.018A * 750CM
    A2 = Ik = ---------------- = 13.5 circular mils,

    Which fits in somewhere between #38AWG with an area of 15.72
    circular mils and #39 with an area of 12.47 circular mils.

    However, the enamel (and you'll need double or triple thickness
    because of the high voltage requirement)covering the wire will add
    to the area drastically, (remember that area goes with the _square_
    of the radius) making me think you won't be able to do it.

    Also, I've assumed a perfect transformer with no core or copper
    losses, and that makes it even more likely that you won't be able to
    build a high voltage supply just by rewinding the secondary,
    especially if (as indicated below) you're expecting to reach
    voltages approaching those generated by Wimshurst machines
  5. Hello!
    Th reason the coil or inductor does not short out is beacause the voltage i
    AC. Inductors are like capacitors, they store energy. However real world
    inductors do have some small resistance allso. In the inductor the energy
    is converted to a magnetic field. So when you turn of the power, the
    magnetic field collapses and is converted back to current. But very
    importently the current is now going in the reverse direction! With a
    capacitor the current will not change direction. It`s more like charging a
    So when you have two inductor close to each other the magnetic field in
    one(primary) will be converted a current in then other(secondary). But if
    the secondary inductor has no load, no power will be used there.
    But like i wrote before some of the power will be lost in the inductors
    because of the internal resitance in the wires and thus generate heat. And
    heat is the problem. When you buy a transformer it is rated in VA[Voltage x
    Ampere]. You might think why not use Watt instead? That V*I too. True, but
    only for resistance. For reactive elements like inductor and capacitors
    there is a phaseshift between voltage and current. The "resitance" you see
    in such a element is called impedance Z and is dependent on the frequency of
    the AC signal.

    So when you connect a reactive element in a circuit you must watch out
    because the the current can become quite high. A typical reactive device is
    a electrical motor or transformer. You often have to compensate this by
    inserting a capacitor so the phaseshift is minimal. In big factories that
    use alot of electrical engines this is requirement or the power company will
    cut the power.

    Hope this helped a litle.
    If you want to know more about reactive elements and such you should take a
    look at laplace transform of electrical circuits. Its much easier than using
    diffrental equations :)

    Good luck

  6. Bob Myers

    Bob Myers Guest

    Sorry, no. The current does NOT change direction - in fact, that is one
    of the key things to learn about inductors. (The voltage ACROSS the
    inductor is another story, however.)

    Two simple rules:

    - Capacitors oppose changes in voltage (i.e., a capacitor will "try" to
    maintain the voltage across it at the same level despite changes in the
    current through it), while...

    - Inductors oppose changes in current.

    If you try to interrupt the current flowing through a conductor, energy
    will come back out of the magnetic field (where it has been stored)
    in such a way as to try to keep the current going at the same level.
    This will result in a (possibly very large) voltage spike across the
    inductor, if that's what it required to try to keep the current at the
    same level (as would be the case if you suddenly switched a large
    resistance into a series connection with the inductor, or simply tried
    to open the circuit with a switch - which is why switches that are
    switching large inductive loads very often "arc over").

    This is expressed by the equation:

    V = L(di/dt)

    If you aren't familiar with calculus, all this is saying is that the voltage
    across an inductor is proportional to the time rate of the change of the
    current. Note that if the change is negative, so is the resulting voltage.
    However, an ideal transformer with a purely resistive load does NOT present
    reactive load at the primary, which may be a source of some confusion here.
    apparent reactance seen at the primary (and the resulting phase shift)
    would come
    from the characteristics of the load itself and from imperfect coupling
    the primary and secondary windings (i.e., not all of the field from the
    winding "links" to the secondary, and so that portion of the field accounts
    some inductance apparent from the primary side).
    But again, a transformer per se is not NEARLY as reactive as you
    might think it would be from just looking at the primary winding. Neither
    is a motor. (Can you see why? Think about where the energy in the field
    is going...).

    Bob M.
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