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Whats the rule for a Secondary of a transformer VS Rectification metho

  1. Sep 21, 2013 #1
    I seem to remember my lecturer talking about Transformers and Rectification methods and there ratings for different modes of operation (25 Years ago).

    I bought a book called Understanding DC Power Supplies By Barry Davies and I cant find any reference to my question in it.

    The question is if you use a Full-wave rectifier using a center tap transformer and 2 diodes, can each of the transformers secondary rating be equal 1/2 the output current needed because each winding is only supping the power for 1/2 the time?
     
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  3. Sep 21, 2013 #2

    vk6kro

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    In a full wave rectified transformer, each half of the transformer secondary winding conducts for only half the time.

    So, for example, if the winding was rated for 5 amps then it could deliver 10 amps if it was only delivering it for half the time.
    The winding can cool down between bursts of conduction.
     
    Last edited: Sep 21, 2013
  4. Sep 21, 2013 #3

    Averagesupernova

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    This is only true if it is a center tap common secondary. A full wave bridge that utilizes the whole secondary at once will be limited to its rating.
     
  5. Sep 21, 2013 #4

    jim hardy

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  6. Sep 28, 2013 #5
    Could I assume if I had a Full Wave Center Tap Bridge and loaded the windings 180 ° out of phase I may be able to draw I D.C = 1.27 x Sec. I A.C ?
     
  7. Sep 28, 2013 #6

    The Electrician

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    Are you really going to have a purely resistive load? Are you not going to filter the rectifier output with a capacitor to obtain nearly pure DC?
     
  8. Sep 28, 2013 #7

    jim hardy

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    i'm not quite sure what you mean by " and loaded the windings 180 ° out of phase "

    but a centertap with full wave bridge gives you two supplies, one positive and one negative
    provided you use the centertap as circuit common. I think that's what you are proposing ???

    I think that'd be equivalent to the diagram labeled " Full Wave Bridge Resistive Load" just above and right. Reason is each half of the transformer winding conducts all the time
    Compare that one to your proposed circuit - in your scheme, if positive and negative currents are equal then there's none through centertap so circuits are equivalent and your current available is 0.9 not 1.27, mostly because transformer doesn't get that rest between half cycles.
     
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