Testing Power supplys

Discussion in 'Hardware' started by unholy, Feb 18, 2006.

  1. unholy

    unholy

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    I have started to change the testing method of testing power supplies to show everyone exactly how good a power supply is.
    The new graph shows how jittery the power supply is when Idle and loaded.
    heres a look at the Sunbeam Nuuo 550 watt 12v1 rail, Idle/Load
    This is measured over time, using a RS232 Multimeter :)

    Please give comments on if its a good idea :)

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    Last edited: Feb 18, 2006
    unholy, Feb 18, 2006
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  2. unholy

    unholy

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    No comments? it must be a good idea then :p
    unholy, Feb 19, 2006
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  3. unholy

    The Modfather

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    No comments because nobody knows what you're saying.
    The Modfather, Feb 20, 2006
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  4. unholy

    Ian Administrator

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    Just seen this now :) Looks like a good way to test it to me, I wonder what a cheapy power supply would do?
    Ian, Feb 20, 2006
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  5. unholy

    bigal

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    A hundredth of a volt jump isn't much to worry about. We would need to define standard thresholds, like a 10th of a volt. Try changing the scale from 12.00 to 12.0 and see how it looks.
    bigal, Feb 21, 2006
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  6. unholy

    unholy

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    Sounds like a plan :)
    unholy, Feb 22, 2006
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  7. unholy

    The Modfather

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    Here's a question. How can I calculate the effeciency rating for a PSU to compare it against the manufacturers claims? I need to know the right way to do this.
    The Modfather, Feb 22, 2006
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  8. unholy

    bigal

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    You somehow need to measure the power loss occuring within the PSU to determine efficiency. Maybe you can use several current meters concurrently - one on the input line, and one on each output line (all voltage rails). It would be tough to perform, as you need to load down all rails at their rated power levels. I would not try to measure this rating without the right equipment.
    bigal, Feb 23, 2006
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  9. unholy

    The Modfather

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    Yeah, I didn't think it'd be easy to do. Drac tried to do that with a PSU we're reviewing two versions of and couldn't match the rating the mfg. gave for it. I just felt something was wrong with how he was doing it and he wasn't 100% sure so he decided not to post his findings, which was a good idea. I thought if there WAS a way I'd try to track it down and do it with the one I'm testing.
    The Modfather, Feb 24, 2006
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  10. unholy

    bigal

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    I bet even with the best equipment we could use, you would still get some losses (of power consumption measurement) on the order of 1-2%. You almost need a sealed chamber that's capable of maintaining a controlled environment around the PSU. If you had that, perhaps you could calcuate power loss due to thermal load increase. In fact, that might be the easiest way of getting an accurate reading. All you need is the right formula...
    bigal, Feb 24, 2006
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  11. unholy

    unholy

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    Efficenty is AC in/DC Out, and theres ur percentage :) because AC in Wattage/Ampage is going to be more then the DC Out, because of the lose when the power supply converts the power.

    To be honest, its something pretty hard to determine.

    the forumla would be

    a = ac watts in
    d1 = dc 12v1 watts
    d2 = dc 12v2 watts
    d3 = dc 5v watts
    d4 = dc 3.3v watts
    l = lose of amps through heat energy and noise and so forth.

    Technically if the Power supply had a efficency of 100% the formula would be

    a = d1 + d2 + d3 + d4

    But because thats not the fact.

    a = d1 + d2 + d3 + d4 + l
    a-l = d1 + d2 + d3 + d4

    -l = d1 + d2 + d3 + d4 - a
    l = d1 - d2 - d3 - d4 + a

    and to get the eff.

    l/a = percent wasted
    100 - (l/a) = Efficency

    *Note IT MUST BE IN WATTS, otherwise problems will occur where u say 10amps at 240v is the same as 10amps at 3.3volt*
    unholy, Feb 25, 2006
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  12. unholy

    bigal

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    You know enough to be dangerous....
    bigal, Feb 25, 2006
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  13. unholy

    The Modfather

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    Ok so I've got a 560w PSU with three rails that's claiming an effeciency of 83%, what's the formula?
    The Modfather, Feb 25, 2006
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  14. unholy

    bigal

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    Better yet, what's the loss? It's all about determining the power loss....
    bigal, Feb 25, 2006
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  15. unholy

    unholy

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    IF u load it with 560watts.

    a = AC Power in watts
    da = DC Power
    l = lose

    l = a - da
    l = a - 560

    because l is a percentage 83% = 560watts

    560 / 0.83 = 674.69879518072289156626506024096 Watts AC input

    l = 674 - 560
    l = 114 watt lose

    You would be using 674.69 watts when running that machine :) at 560 watts obiviously but thats pretty hard to reach :p
    unholy, Feb 26, 2006
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  16. unholy

    bigal

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    But he want's to determine the real efficiency, to verify the manufacturer's claim. You can't just plug in the stated value and tell him he's using 675 Watts when it's fully loaded down. Maybe the efficiency is really only 78%. How would you know?
    bigal, Feb 26, 2006
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  17. unholy

    The Modfather

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    Yep, that'd be the problem. Also how can you load it to it's full capacity and know it's exactly the right load?
    The Modfather, Feb 26, 2006
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  18. unholy

    bigal

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    It would help to determine the internal operating temperature of the PSU under near-full loading conditions. I say this because most PSU specifications are claimed at a balmy 25C, and they drop substantially at a more realistic 40C. When I worked in Reliabililty Engineering at Hughes, we had to go through a "derating" exercise for each component, using guildline forumulas to determine the real "maximum" power dissipation values in a real-world (or out of this world) environment. I believe a few companies post specs that their PSU's will adhere to above and beyond 25C. PC Power and Cooling is one of those companies.

    Without knowing anything about a particular PSU, I would derate the "official" power dissipation claim for each voltage rail about 25% to adjust for the real-world operating temperature that's expected in the case. From that reduced power dissipation, you can estimate the power resistor loading by using Ohm's Law (Power (WATTS) = Voltage (VOLTS) * Current (AMPS)). If you know the power and voltage, you can solve for the desired current. Since current also equals Voltage / Resistance, you can solve for Resistance, the value you need to load down the PSU voltage rail.


    reference: http://www.seatw.com/resistor/resistor-ohms-law.htm
    bigal, Feb 27, 2006
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  19. unholy

    unholy

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    Although i doubt u will be able to find the right resistors you need speciall for like 40 Amps :|

    this is where the idea of light blubs comes into play :)

    You have an array of Light Bulbs each have a certain reistance, all linked so u can gradually turn them on and load the power supply up to full. By the way make sure ur only loading to what the rail supports not the overall output :p

    Have a Anmeter before the Power supply and after on each rail :) Add up the wattages and it should equal the DC power output or after transformation power consumption. and use the following info from above in my previous posts.

    I will be making my test power supply, but i will be doing more then you want to do :p, i will also be Testing Eff over Load, and Power over Load over time, and so forth :p

    I will also be making a Guide on how to make it :)
    unholy, Feb 28, 2006
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  20. unholy

    The Modfather

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    Ok, in other words, there's no really effective way for a reviewer to verify the effeciency of a PSU without going into debt and taking an E.E. course.

    Okay, thanks. :)
    The Modfather, Feb 28, 2006
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