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Power dissipation calculation uncertainty

Discussion in 'Electronic Equipment' started by Roque, Aug 7, 2003.

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  1. Roque

    Roque Guest

    If I have a power supply similar to a computer power supply where it's
    rated at 300W, what is the proper way to figure out what my actual power
    usage is when I have this system running under its maximum load?

    Here's what I've done so far....

    I have a power supply that provides 24.5 volts and 5.1 volts DC output,
    I find I can run a total current of 9 amps continuous, or 20 amps peak
    before tripping the internal protection.

    When I put a current probe on the 24.5 and 5.1 volt lines,
    using a digital scope with math functions, I get 8 amps RMS on the 24 volt
    line and 1.15 amps RMS on the 5.1 volt line, in my worst case use of the
    machine.

    I'm trying to decide what power dissipation I have...

    With my scope, I saved 30,000 points of voltage and current data against
    the same time line, and I
    imported this V, I data to a spreadsheet....and created a new calculation
    where I multiplied V * I for each data point...for instantanious power per
    sample. I then squared each V * I data value, summed it, divided by 30,000
    samples, and took the square root...hoping to find RMS power.

    With 8 amps at 24.5 volts and 1.15 amps at 5.1 volts, I found about 200W
    power dissipation...

    Does this make sense? I'm not experienced with "power" electronics to be
    confident about it so I wanted to seek advice.

    Thanks.
     
  2. A E

    A E Guest

    WOW, talk about *overkill*! :)
    You don't need to calculate RMS when dealing with DC, you just multiply I by V.
    So (24.5 x 8) + (5.1 x 1.15 ) = 201.865W.
     
  3. Fred Abse

    Fred Abse Guest

    What's a guy who couldn't see that doing with an oscilloscope capable of
    doing what he did with it?
     
  4. although the unit is rated at a total of 300W, don't you think that
    the wattage on each tap may be different? The 24V line may be rated up
    to 300W (12A) but I would be very suprised if the 5V line could
    deliver anything like the 60A that would be required to get 300W @ 5V.

    I would suppose that the unit is designed to run some light
    electronics off the 5V line and some kind of heavier electric
    actuators off the 24V line. With these assumptions I would expect the
    5V line to only be rated at a few Amps, while the 24V line would be
    rated much higher. You would only need to exceed the limit on the
    lowest rated line to trip the protection circuit (say, by drawing 2A
    on the 5V line that may only be rated for 1.5A)

    - Jeff Dutky
     
  5. A E

    A E Guest

    Yeah, so? The OP clearly stated the current on each rail, the numbers didn't trigger any
    alarms for me.
     
  6. A E

    A E Guest

    Bah, have you been in a university lately? The kids there have equipment that
    took a building full of engineers to design, and is used for the most trivial
    tasks. That's life, electronics is a commodity now.
     
  7. Ghost Chip

    Ghost Chip Guest

    If it's reality, then does it have to make sense?

    "The difference between fiction and reality is that fiction has to make
    sense." Tom Clancey.

    The 300W is probably the input power to the supply at the rated line
    voltage. Some goes to heat etc. If the supply is 75% efficient, then 225W
    will get to the loads.
    Zoramy
     
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