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3phase rectifier

  • 03-02-2015 6:44pm
    #1
    kildarejohn
    Registered Users, Registered Users 2 Posts: 328 ✭✭


    Considering a small alternative energy project using a 3 phase alternator with bridge rectifier for battery charging. Rectifiers will result in each phase winding conducting for only 1/3rd of the cycle. Anyone know a formula for calculating the RMS current in the alternator for a particular DC output current?


Comments

  • #2
    2011
    Moderators, Home & Garden Moderators, Technology & Internet Moderators, Regional East Moderators Posts: 12,641 Mod ✭✭✭✭2011


    kildarejohn wrote: »
    Considering a small alternative energy project using a 3 phase alternator with bridge rectifier for battery charging. Rectifiers will result in each phase winding conducting for only 1/3rd of the cycle. Anyone know a formula for calculating the RMS current in the alternator for a particular DC output current?

    If the AC current to is a sine wave then the RMS value = the peak value x 0.707

    The RMS current is considered to be the equivalent of the DC current.


  • #3
    Bruthal
    Closed Accounts Posts: 13,422 ✭✭✭✭Bruthal


    Id say for 3 phase full wave rectifying, it will be higher than the 0.707


  • #4
    2011
    Moderators, Home & Garden Moderators, Technology & Internet Moderators, Regional East Moderators Posts: 12,641 Mod ✭✭✭✭2011


    Bruthal wrote: »
    Id say for 3 phase full wave rectifying, it will be higher than the 0.707

    My reply was in terms of a single phase, I should have stated that.

    You could calculate it this way:
    Power in=power out + losses

    Power in = root3 x line voltage x line current x power factor

    Estimate the losses, subtract it from power in ....


  • #5
    Bruthal
    Closed Accounts Posts: 13,422 ✭✭✭✭Bruthal


    2011 wrote: »
    My reply was in terms of a single phase, I should have stated that.

    You could calculate it this way:
    Power in=power out + losses

    Power in = root3 x line voltage x line current x power factor

    Estimate the losses, subtract it from power in ....

    That will mean the dc voltage will be higher than the ac line voltage, which is the case. Probably something like Vac x root 2.

    But the OP says that each phase winding will only be conducting for 1/3 of the cycle due to the rectifier.

    I would have thought it will conduct the entire cycle except when below the diode forward voltage, which is below 1v usually.


  • #6
    2011
    Moderators, Home & Garden Moderators, Technology & Internet Moderators, Regional East Moderators Posts: 12,641 Mod ✭✭✭✭2011


    Bruthal wrote: »
    But the OP says that each phase winding will only be conducting for 1/3 of the cycle due to the rectifier.

    The reduction in output would have to be estimated.
    I would have thought it will conduct the entire cycle except when below the diode forward voltage, which is below 1v usually.

    Me too, I guess that this is an issue that applies to this rectifier.


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  • #7
    Bruthal
    Closed Accounts Posts: 13,422 ✭✭✭✭Bruthal


    2011 wrote: »
    Me too, I guess that this is an issue that applies to this rectifier.

    Seems unusual if its a bridge for full wave all the same.


  • #8
    kildarejohn
    Registered Users, Registered Users 2 Posts: 328 ✭✭kildarejohn


    I found the following in the IET online forum-
    "for half wave single phase the DC current = 0.318 X Peak AC current
    Full wave single phase the DC current = 0.635 X Peak AC current
    Full wave three phase the DC current = 0.955 X peak AC current.

    therefore
    22A /0.707 x 0.955 = 29.7A"

    I haven't worked out the theory behind it, but intuitively this seems about right to me.


  • #9
    Bruthal
    Closed Accounts Posts: 13,422 ✭✭✭✭Bruthal


    kildarejohn wrote: »
    I found the following in the IET online forum-
    "for half wave single phase the DC current = 0.318 X Peak AC current
    Full wave single phase the DC current = 0.635 X Peak AC current
    Full wave three phase the DC current = 0.955 X peak AC current.

    therefore
    22A /0.707 x 0.955 = 29.7A"

    I haven't worked out the theory behind it, but intuitively this seems about right to me.
    It probably is right by the look of it. 22 is RMS amps. Divide by 0.707 for peak amps. Multiply by 3/pi (0.955) for DC side amps

    It also equates to 22 x 1.35. Not too far off the ac side x root 2.


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