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Shortcircuiting diode bridge

  1. Jul 14, 2012 #1

    Femme_physics

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    I am told that in this circuit (above pic) we shortcircuit D1 and D2, and disconnect D3 and D4... then we are to calculate Vrl(average)

    But the way I see it, if we shortcircuit D1 and D2 the entire circuit is shortcircuit and current only flows like this (below pic)

    http://img195.imageshack.us/img195/8396/d1d2d3.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jul 14, 2012 #2

    lewando

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    When the question asks "shortcircuit D1 and D2", I think it means: "replace D1 with a wire and replace D2 with a wire"
     
  4. Jul 14, 2012 #3

    I like Serena

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    Yep.
    And disconnecting D3 means taking D3 out.
    So there cannot go current anymore through the place where D3 used to be...
     
  5. Jul 14, 2012 #4

    Femme_physics

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    Last edited by a moderator: May 6, 2017
  6. Jul 14, 2012 #5

    I like Serena

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    Yep. The current would flow like that. :)

    What about that block on the right? What does it do?
    Does it draw any current?
    And is the SCR always conductive?
    Doesn't that depend on the voltage at its gate?
     
  7. Jul 14, 2012 #6

    CWatters

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    Yes and no. You have only modelled the positive cycle.

    To model the negative cycle put the diodes back and pretend the other two are shorted/conducting.
     
  8. Jul 14, 2012 #7

    Femme_physics

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    The block to the right is a circuit to control firing angle (At 30 degrees).

    CWatter - There is no negative cycle. The SCR blocks it! All the voltage falls on the SCR and nothing on the load.
     
  9. Jul 14, 2012 #8

    I like Serena

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    Okay.
    Then I guess you are set to calculate the average Vrl.
     
  10. Jul 14, 2012 #9

    CWatters

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    Perhaps you can explain why a bridge rectifier is needed then :-)
     
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