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Transformer Load Question

Discussion in 'Electronic Basics' started by Guy Paddock, Jan 17, 2004.

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  1. Guy Paddock

    Guy Paddock Guest

    I have a 110v to 25v step-down transformer that I'm trying to have drive an
    electromagnetic coil. The original coil this transformer drove was rated for
    4 amps and 25 volts. My digital multimeter tells me that the transformer
    itself puts out around 27-28 volts, actually. Anyways, I am building the new
    coil myself. I have already wound it and measured its impedance. According
    to the original rating, the original electromagnetic coil should have had an
    impedance of 6.25 ohms. My coil, at the moment, has an impedance of about 8
    ohms. My coil will be applied differently than the original coil however; I
    need mine to repell something, not just attract it, so I got a 4 amp, 50PIV
    Full-Wave Bridge Rectifier at RadioShack and connected it between the
    transformer and the coil. Currently, the transformer powers the coil as
    required, with specific magnetic poles, and it is capable of repelling other
    magnets, but it only functions for a short time (5-10 minutes). During this
    time, the transformer gets very hot, and the coil gets a bit warm as well.

    My question is this: shouldn't the fact that my coil has a higher impedance
    mean that there should be LESS of a load on the transformer? If the first
    coil had a 4 amp draw, and had an impedance of 6.25 ohms, then mine, with an
    8 ohm impedance, should only have a draw of 3.125 amps (at rated voltage).
    If this is so, why am I experiencing a heat issue?

    TIA
    --Guy
     
  2. Ian Bell

    Ian Bell Guest

    If this is the unloaded output volts then this would be about right if the
    transformer has the typical 10% regulation at full load - 10% of 28 is just
    under 3 and 28 minus 3 equals 25.


    Ian
     
  3. You should be more specific becasue you use impedance and resistance and it
    is confusing. If the original coil was AC and your homebrew is DC you're
    talking about apples and oranges., still fruit but big difference.
    hank wd5jfr
     
  4. Guy Paddock

    Guy Paddock Guest

    I do not mention resistance. Besides, impedance and resistance produce the
    same effect, do they not? Are they not the same? Resistance slows the
    current flowing through a given point in the circuit, and I was under the
    impression that impedance measures the AMOUNT of resistance.

    More so, does the fact that the circuit is converted from AC to DC impact it
    much? The rectifier simply uses a diode bridge internally, forcing the
    current to flow in one direction, so I don't see how forcing the electrons
    to flow in one direction for each cycle impacts it. Is there much of a
    voltage or amperage loss associated with it?

    --Guy
     
  5. Resistance is an effect that is independent of time, so it works the
    same for DC and any frequency of AC. But coils have an inductive
    impedance component in addition to the resistive component. And the
    effect of inductance is certainly a function of time. Inductance adds
    no impedance for DC, while it adds to the resistance (vectorially at
    right angles) a reactance that is proportional to frequency. For
    example, if your 8 ohm (resistive) coil had an inductance of 100
    millihenries, it would have a total impedance at 60 Hz or square root
    (100^2 + (2*pi*60*.1)^2) = 107 ohms. But if the inductance were 1
    henry, then the total AC impedance would be square root(100^2 +
    (2*pi*60*1)^2)= 390 ohms. Both coils would have only 100 ohms
    impedance for DC. So DC coils generally have more turns of finer wire
    to draw the same current as coils driven by AC. And the higher
    resistance DC coil will produce more heat with that same current, due
    to this higher resistance. Holding back current with inductance does
    not produce heat, but doing so with resistance does.
     
  6. Guy Paddock

    Guy Paddock Guest

    So I would need to remove some turns from my coil in order to create less
    heat?

    --Guy
     
  7. Guy Paddock

    Guy Paddock Guest

    Also, how do I know how many henries my coil is?

    --Guy
     
  8. Guy Paddock

    Guy Paddock Guest

    Perhaps my problem is in the way my coil is wound? Is there a place I could
    get a coil that would meet my specs?

    Or, perhaps there is some other way to keep the transformer from
    overheating?

    --Guy
     
  9. No. You need a coil with higher resistance. The current the coil
    will pass is V/R. The power the coil will produce is V^2/R or I^2*R.
    More turns of finer wire will pass less current. The field strength
    is proportional to the current and to the number of turns.

    So if you use wire with half the cross sectional area, about twice the
    turns will fit in the same space, while the resistance will be about 4
    times higher (a factor of 2 increase for the smaller area, and another
    factor of 2 for twice the number of turns, and so the length). So the
    field strength will be about half (1/4 the current but twice the
    turns). The heat, in both the transformer and the coil will both
    drop. The transformer winding losses will drop by a factor of 16
    (because the windings still have the same resistance, but the current
    will be about 1/4, by I^2*R. But the coil heat will drop to only 1/4,
    because the current is 1/4 but the resistance is times 4.
     
  10. If you can measure its resistance and also its current draw with a
    measured applied voltage, you can calculate the inductance from the
    impedance formula.

    I=V/(sqrt(R^2 + (2*pi*F*L)^2), where F is the AC frequency, and L is
    the inductance in henries.
     
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