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ac/dc power supply efficiency

Discussion in 'Electronic Design' started by [email protected], Aug 5, 2013.

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  1. Guest

    Hello to all,
    making measurements on an ac/dc power supply (230Vac -> 24Vdc) I noticed that the cos phi is 0.56.
    The active power is 65W and the reactive power is about the same value.
    Now, I'm sure my load for power supply is 59W at 24Vdc (resistive load)
    How would you compute the efficiency of the power supply, i.e. power of the load / active power or power of the load / (active+reactive power)?
    Obviously there is a big difference. The current absorbed by the network is the sum of the two components related to active and reactive power, then I would suggest the latter calculation . Thank you. Greetings. Andrea
     
  2. Phil Allison

    Phil Allison Guest

    ** No it is not !!!

    The ratio of watts to "VA" may well be 0.56 though - aka the "power
    factor".


    ** There is simply no "reactive power" involved.

    ** Easy enough to compute.

    ** Where do folk get these whacky ideas ??????

    Efficiency has NOTHING to do with any of this nonsense.

    ** No such conclusion is "obvious" at all.

    Efficiency = power input / power output.

    The DC power supply you allude to could easily be close to 100% efficient.

    Many are.

    Think again.




    ..... Phil
     
  3. Guest

    Phil I'm not agree with you: Efficiency = power input / power output.
    If so efficiency > 100%

    Efficiency is power output/power input.
    My question is if I have to compute power input like the sum of active and reactive power

    Andrea
     
  4. Guest

    no, it isn't. it is power output/power input
    otherwise efficiency would be more than 100% because input power is always greater than output power

    My question is if I have to consider input power as the sum of active plus reactive power
    Andrea
     
  5. Phil Allison

    Phil Allison Guest


    ** Slip of the tongue................


    ** That is YOUR delusion.

    Cos "reactive power " is imaginary and totally non existent in the example
    you are considering.



    .... Phil
     
  6. Phil Allison

    Phil Allison Guest


    ** Now you sound like a fucking, nut case troll.

    Piss off.
     
  7. John S

    John S Guest

    Using your numbers, your efficiency is 59/65 or about 90.8 %.
     
  8. Phil Allison

    Phil Allison Guest

    "John S"
    ** Correct.

    Andrea is another fucking TROLL


    .... Phil
     
  9. amdx

    amdx Guest


    We interrupt this SED thread to apologize for the rantings of Phil.
    He lives to show he is right without teaching the OP why.

    Nut Case Acknowledgement Committee
     
  10. Phil Allison

    Phil Allison Guest

    "amdx = fuckhead & troll "

    ** There is no " 'we" - just one, anonymous, smug, egomaniac ****.


    ** One cannot teach a pig to sing - nor a trolling fool.

    And YOU are both.

    You shit for brains porker.



    ..... Phil
     
  11. amdx

    amdx Guest

    You must recall my previous request that you be creative in you
    responses. Most of this was new, very good on you.
    Thanks, Mikek

    porker? lol
     
  12. whit3rd

    whit3rd Guest

    What is the computation FOR? If you're sizing a fuse or circuit breaker, it
    might be useful to know the input in VA terms, as well as in real-power terms.
     
  13. Guest

    I have to deal with a certain number of loads feeded by that kind of power supply, so the amount of reactive power affects the choice of circuit breakers and power consumption from the wall.

    I'm seriously thinking to replace the power supplies with others that have a better cos fi (0,96 vs. 0,56) to deal with less current from the wall (and also less harmonics)
     
  14. John S

    John S Guest

    Forget "reactive power". Measure the input current with a true RMS meter
    and size your breakers and wiring based on that.
     
  15. Guest


    I've done this, I obtain 0,72A @ 230Vac as input current.
    I'll use this data to size breakers and wiring.
    Using a different power supply, with a good active pfc, the current drops to 0,37A @ 230Vac for the same system - the second power supply is obviouslymore expensive but I'm wondering if reactive power will be figured out as a cost by the customer that know current consumption
     
  16. Guest

    Fuses are actual rated with the I²t parameter, so you really want to
    know the power supply peak current.

    In a simple capacitor input power supply, current is drawn only just
    before the positive or negative voltage peak. The half cycle is 180
    degrees long, but the actual conduction angle might be as short as 18
    degrees i.e 1 ms at 50 Hz.

    However, the total charge for this half cycle must be transferred
    during the conduction angle, i.e in this case 10 times the average
    current.

    Assuming a resistive load drawing 10 Arms (2.3 [email protected] V), the fuse
    heating would be 1 A²s during each half cycle. However, if the same
    charge needs to be transferred during 18 degrees (1 ms), the peak
    current would be 100 A and the fuse heating 10 A²s, thus blowing the
    fuse long before 2.3 kW power.

    Putting such peaky loads on all three phases would put the _sum_ of
    all three phases into the neutral wire, since the current peaks do not
    overlap. The neutral wire heating would be up to 9 times the heating
    of an individual phase conductor.

    Only with overlapping currents will some (or all) phase currents
    cancel out in the neutral conductor.

    The conduction angle depends on the source impedance of the feeding
    electric network and the size of the capacitor in the power supply. In
    practice, conduction angles would be slightly longer than 1/10 of the
    half cycle time.

    You really need to measure the peak current or conduction angle using
    a current transformer and oscilloscope.
    In the EU, power supplies greater than 75 W needs to have a good power
    factor, requiring some PFC circuitry. It appears that your supply has
    been designed to be just below that limit and hence does not need a
    PFC.
    At least in EU, buy one that is rated above 75 W and the power factor
    is OK.
     
  17. John S

    John S Guest

    I think you have an instrumentation problem. Read Phil Allison's early
    response to you for understanding.

    If your input power is 65W and your input VA is 230*.72, then your power
    factor is 65/166 or about .4 which does not agree with your initial post
    of .56 (Cosine whatever).

    Additionally, using the PFC, the output power is 65W and your input
    power is .37*230 (85W) so the efficiency is now 76% (maybe).

    One or more of your instruments may not be capable of measuring what you
    need. Use caution when making measurements. Check your instruments'
    specifications to see if they can handle the waveform factors. Do you
    have an oscilloscope?

    Cheers,
    John S
     
  18. Phil Allison

    Phil Allison Guest


    ** FORGET the term "reactive power " - it is meaningless for your
    situation where there is NO reactive current.
    ** You are obsessed with serious miscomprehensions.

    If you have 65W load with a *power factor* of 0.56 that load is 65/0.56 =
    116VA.

    So the RMS current draw is just under 1 amp at 120VAC.

    Inrush surge current will be the thing that gets you with such multiple
    loads and you do not even know what it is.



    .... Phil
     
  19. Phil Allison

    Phil Allison Guest

    ** Only in relation to short surges.
    ** Which happens only at switch on !!!!

    ** Absolute CRAP !!!!!!!!!!!

    Fuses and breakers are rated with the RMS value of the current and that is
    all you need to know.

    ** More absolute CRAP !!

    Piss off FOOL !!!!!!


    ... Phil
     
  20. Jasen Betts

    Jasen Betts Guest

    no. becauae they don't overlap, only three times the phase wire heating,
     
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