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A problem in Coupling Transformers

  1. Oct 29, 2013 #1
    Hello Guys !!

    While Studing coupling transformers i found this problem
    i tried to solve it
    but i don't know if my solving is right or no !!
    so tell me :)

    The problem :
    27phVRt.png
    eJPQdSJ.png



    My solve is :
    V(out)=nV(in)
    V(out)= 2 * 312 = 624 v
    PIV = V(out) = 624 v
    and if the diode is turned around it will be reversed biased so no current will pass across RL and
    Output voltage will equal to zero ..

    am i right guys or what ??

    Thank you !!
     
  2. jcsd
  3. Oct 29, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    Looks okay. Although whoever set the problem doesn't know his components very well. A 1N4002 diode as a PIV rating of about 100 V if I recall correctly. Expect the magic smoke to escape from the diode very shortly after switching on!

    Also, you have ignored the ~0.7 V forward drop across the diode. Probably okay considering what a small percentage this is of the total voltage.

    Nope. The secondary will present a full AC voltage signal, so if you turn the diode around it will simply conduct when the diode is forward biased on the opposite half-cycles (opposite that is to the half-cycles when it conducted in the original diode orientation). What does that tell you about the resulting potential across the load resistor?
     
  4. Oct 29, 2013 #3
    I think the potential across the load resistor will be the same but with an opposite sign
    i mean that the voltage will take this shape ..am i right ??

    SlNR6jY.png
     
  5. Oct 29, 2013 #4

    gneill

    User Avatar

    Staff: Mentor

    Looks good!
     
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