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How can i figure the current loss of a power supply?

  1. Aug 8, 2009 #1
    OK, so if I'm inputting 120V ac @ 1.5A equals 180W
    and my output is 19V dc @ 3.16A equals 60.04W
    simple subtraction says 116W is left. that can't be right can it?
    am i doing this wrong or am i actually losing 116W of power?
    Last edited: Aug 8, 2009
  2. jcsd
  3. Aug 8, 2009 #2


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  4. Aug 8, 2009 #3
    no, I'm trying to figure the efficiency of the circuit.
    according to the service sticker ( laptop power cube ) the input is 120v 1.6A
    and the output is 19v 3.16A,
    120 x 1.6 is 180W rms
    and 19v x 3.16A is 60.04W rms right?
    so there's about 120W of power being wasted unless i'm not doing this right lol
  5. Aug 8, 2009 #4


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    Those are ratings, not actual usage values. If you want to know the actual circuit efficiency at various loads, you have to measure them.
  6. Aug 8, 2009 #5

    Watt= E X I X cosine of the phase angle (For sine waves)
  7. Aug 8, 2009 #6
    Is the maximum input rating 120V @ 1.6A = 192VA, and the maximum output rating 19V dc @ 3.16A = 60VA? Apparently this is a very reactive (and maybe a transformerless) circuit. VA or volt-amps is not a measure of power if the power factor is low.
    I recently measured both the VA and watts of a 1/4 horsepower induction motor without any load. The input was 415VA and 95 watts. So the power factor in this case is 0.23.
  8. Aug 9, 2009 #7


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    You are measuring 60 watts DC output, so that is one thing you know for sure.

    You can say the input can't be any less than 60 watts.
    You can guess that maybe the power supply is 60% efficient (just an example) and then the input would be about 100 watts. In that case, the current would be (100 watts / 120 volts) or 0.833 Amps.
    If it was 80 % efficient, the input power would be 75 watts and the current at 120 volts would be 0.625 Amps.

    The sticker current has little meaning except to give you an idea of the current under heavily loaded output conditions. It won't draw those currents unless you put a much larger load on the output (or it is very inefficient).

    So, unless you measure the input (carefully!) and find a problem, you don't have a problem. You haven't mentioned overheating or smoke effects, so everything is probably working OK.


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