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Efficiency of Transformer

Discussion in 'Electronic Basics' started by [email protected], Feb 28, 2005.

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  1. Guest

    I am currently attempting a project on the efficiency of a transformer.
    As I am using low voltage and current inputs I am recieving low
    efficiency level eg. below 15% I am wondering if this is normal and
    possible solutions to the problem. Any good website's on transformers
    and their efficiency.

    Cheers
     
  2. Andrew Holme

    Andrew Holme Guest

    What frequency? What power level? Please describe your test circuit.
    How are you measuring efficiency?
     
  3. What frequency and power level are we talking about? Very different
    problems at 60 Hz than at 10 MHz.
     
  4. Jamie

    Jamie Guest

    thats a wind open area.
    the first problem to be concerned about is the
    DC resistance in the primary and secondary windings.
    the primary winding dc resistance can be scaled to the
    secondary windings in the same manner as the voltage
    is scaled.
    for example, if you had lets say 30 ohms of DC resistance
    on the primary and the ratio was 10:1, for a rough guess
    you could assume that you will have an effective resistor
    drop in the secondary load circuit of 3 ohms.
    then talking into account the Dc resistance in the
    secondary winding compounds the problem.
    then you have the saturation problem of the transformer
    that also plays a roll in the efficiency along with the way
    its would on each other, capacitance built up, core type etc.

    if you need a nice PDF to look at here you go.
    http://www.midcom-inc.com/pdf/TN69.pdf

    and if your bored
    http://eccentrix.com/members/jfcole/water_transformer.htm
     
  5. Guest

    I think charles is using a 12 volt ac signal, with a current of 10mA i
    dont know the frequency he is using

    roy
     
  6. roy davidson

    roy davidson Guest

    I think charles is using a 12 volt ac signal, with a current of 10mA i
    dont know the frequency he is using


    roy
     
  7. Guest

    Hi I am using a 12v supply and a frequecy of 5KHz

    I am using formula current*voltage*cosθ/2 for power at input

    Using P=IV at output

    Cheers
     
  8. Jamie

    Jamie Guest

    i don't remember dividing the angle by 2?

    P= I*V*cos(PhaseAngle);

    or am i missing something here?
     
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