Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Diode bridge question

  1. Jun 29, 2009 #1
    Hello,

    I made a circuit to convert AC to DC. It looks like this.

    http://openbookproject.net/electricCircuits/Semi/03267.png [Broken]

    My output DC voltage is around 5 volts. Could someone please explain to me how I could figure out the current running through each of the phases at a given time given a certain DC current (For example lets say 1 A DC)?

    I'm tempted to say that each phase would have 0.33 A_rms going through it (assuming 1 A DC), but I'm not sure if that is right.

    An equation or any help would be most helpful. Thanks

    -edit, I've also though that, given the following picture, maybe only 1 phase at any given time has current going through it, and it it is equal to the DC current. This seems like a better solution, but I'm not sure.

    03269.png
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 29, 2009 #2

    MATLABdude

    User Avatar
    Science Advisor

    Yeah, your output is not going to be 5VDC, unless that 3-phase AC source is actually a transformer or 3 (plugged into your 3-phase mains) with a 5.6 Vpk.
     
  4. Jun 29, 2009 #3
    Oh sorry, our power is coming from an AC motor being run as a generator, not from the wall. During tests we were getting 5 VDC from our circuit. Basically, I know the thermal current that the motor windings can handle, which is why I want to figure out how much current is going through each phase, so I can determine how much load I can put across the generator without frying it.
     
  5. Jun 30, 2009 #4

    MATLABdude

    User Avatar
    Science Advisor

    Ah, well that being the case, you can use Ohm's law, as you suggest and find the current going through your load, which will be the same as the current supplied by each of the windings at their peak output. Remember that in a 3-phase supply, only one leg / phase primarily supplies the current at any given moment. That being said, in your example, each phase would be supplying approximately 1A at its peak. Remember that Vpk = sqrt(2) * Vrms.

    If you *don't* have a simple resistive load, I'd suggest hooking the device up to your bench power supply (set at 5VDC) and seeing how much current it draws as a maximum and/or on average (which number is more important depends on how often the maxima occur, and how high they get).

    EDIT: There's probably a formula you can find in any circuit analysis textbook relating I_pk supplied to the load to the I_pk supplied by any given phase, but using V_pk supplied should give you a pretty close number.
     
    Last edited: Jun 30, 2009
  6. Jun 30, 2009 #5
    Ok, so more or less the current going through any 1 phase at a given time is either 0 or the current being pulled from DC. At any given time, 2 phases are supplying no current while the other is supplying all the current. Is that correct for the most part?
     
  7. Jun 30, 2009 #6

    MATLABdude

    User Avatar
    Science Advisor

    At peak, the other coils are sinking current (as your circuit is drawn, current has to flow in loops!) But since it's a sinusoid, each phase oscillates between supplying most of the current, and sinking most of the current, and in a continuous fashion (i.e. not supplying all the current half the time, and then sinking all the current the other half).
     
  8. Jun 30, 2009 #7
    great thanks! Ok, so basically by looking at the 2nd graph I provided, it seems that when one phase is supplying all the current, the other two are sinking about half of that current. So for 1/3 of the time, any phase is sourcing all the current, and for the other 2/3 of the time, it is sinking half the current. If you average that over an entire cycle, a single phase see 2/3 of the DC amperage going through it I think.
     
    Last edited: Jun 30, 2009
  9. Jun 30, 2009 #8

    MATLABdude

    User Avatar
    Science Advisor

    Something like that. You could just use all the DC current as a worst-case maximum, and/or to give yourself some margin. (You just need to put a less than, or greater than sign in front of whatever answer you get in the end, as appropriate)
     
  10. Jul 2, 2009 #9
    This is a great thing you can do on spice :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Diode bridge question
  1. Diode Question (Replies: 16)

Loading...