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Thread: Using VFD on single phase source

  1. #21
    Join Date
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    Quote Originally Posted by mike_kilroy View Post
    So Gar you are good with 1.5 factor?

    Try this: 10a x100v 3ph produces 10x100x1.73=1730watts. Agree?

    How many amps 1ph for same 1730 watts?

    Still want to stick with 1.5?
    It's 1730 VA, not Watts. But let that pass.
    You cannot apply linear circuit analysis to non-linear circuits.
    Si hoc legere scis nimium eruditionis habes.

  2. #22
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    170515-2332 EDT

    mike_kilroy:

    I said nothing about input power. I approximated the diode average current, and since diode voltage is approximately constant I can approximate total diode power dissipation that in turn approximates diode junction temperature rise.

    If you want average input power in this application, then you need the instantaneous product of the input voltage and current, integrate this over an adequate time, and divide by that time.

    The current waveform is not a sine wave, but a smooth rounded pulse with a slower rising slope than falling slope for a sample circuit I measured. Inductance is causing the rounding.

    .

  3. #23
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    Sep 2008
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    Quote Originally Posted by gar View Post
    170515-2332 EDT

    mike_kilroy:

    I said nothing about input power. I approximated the diode average current, and since diode voltage is approximately constant I can approximate total diode power dissipation that in turn approximates diode junction temperature rise.

    If you want average input power in this application, then you need the instantaneous product of the input voltage and current, integrate this over an adequate time, and divide by that time.

    The current waveform is not a sine wave, but a smooth rounded pulse with a slower rising slope than falling slope for a sample circuit I measured. Inductance is causing the rounding.
    Something like a single phase version of this?

    Si hoc legere scis nimium eruditionis habes.

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