Why does the PIV in centre tap rectifier is twice of the V source?

  1. In the full wave centre tap rectifier, piv of diode require is twice of the input current of the transformer. I can't understand why is it twice when the source is only 1V-peak.
     
  2. jcsd
  3. NascentOxygen

    Staff: Mentor

    Hi null void. [​IMG]

    Have you drawn the schematic? If so, mark voltages on the secondary terminals, showing CT as ground, upper end +1V, and the lower end as -1V.

    Then proceed to examine the diode inverse voltages from there.
     
  4. This is my just my assumption:

    I consider the voltage drop at the resistance is same as the voltage drop at the reversed biased diode...am i right?

    [​IMG]
     
  5. Hello NV - When the DIode is fully reverse biased - what are the V on ether side of the diode?
     
    1 person likes this.
  6. NascentOxygen

    Staff: Mentor

    CT is the centre tap, and that is grounded here. When one winding on the transformer has a voltage V volts, the other winding has a terminal voltage of -V volts. The conducting diode has no voltage loss across it. Mark these on your diagram, and you'll see what the voltage is on the reverse biased rectifier.

    The load voltage is determined by the conducting diode.
     
  7. Opps accidentally replace the image of previous post. So i have to consider the forward biased diode as a jumper, and point a has voltage of V, and point B has voltage of 0V ?
     
  8. NascentOxygen

    Staff: Mentor

    Between point B and ground is a winding of the transformer! At the instant that A has a voltage of V volts (wrt ground), the voltage at point B has a voltage of -V volts (wrt ground) courtesy of its half of the transformer winding.
     
  9. [​IMG]

    Is that correct to view the circuit independently like in the diagram? How does the ground wire which direct connect to the transformer affect the circuit?
     
  10. sophiecentaur

    sophiecentaur 13,399
    Science Advisor
    Gold Member

    How are you defining the volts from the transformer? Is the "V" you have put on the diagram the volts referenced to Earth (i.e. is the transformer a V-0-V secondary)? It isn't clear from your sketch and it's vital for a proper answer.
    But you have, basically a loop of wire from one end of the transformer to the other (you can ignore the resistor as it just appears in parallel with one of the windings) with a gap in it (the reverse biased diode). The voltage across that gap will be the same as the volts across the ends of transformer winding (=2V for a V-0-V transformer). Simples.
     
    Last edited: May 23, 2013
    1 person likes this.
  11. sorry my V is the peak voltage of the transformer
     
  12. sophiecentaur

    sophiecentaur 13,399
    Science Advisor
    Gold Member

    Which "Peak Voltage"? There are three connections on it. (See my edit on my previous post)
     
  13. in a case of a series circuit with a transformer and a diode only, the voltage across the diode when it is reversebiased is also 2V_peak of the ac supplier source? or the transformer needto be grounded at the middle to create such condition?
     
  14. sophiecentaur

    sophiecentaur 13,399
    Science Advisor
    Gold Member

    If the total PD across the transformer are +V - (-V) =2V then it would have to be 2v.
    This is why I was being picky about your use of the 'unqualified' value V. Do you take my point?
    There is also the issue of Peak and RMS - which is why I slipped in the root 2 factor in my earlier post.
    Grounding the centre tap is of no consequence, aamof, all that counts is the (vector) subtraction of the voltages at each end.
    Note. If you only use one diode, there is a short circuit on the reverse cycle and what you say has no meaning. It's a totally different circuit that's required. You'd need to draw it out and it would be clearer.
     
  15. [​IMG]

    I am not sure if i am really get your point right now, the -V is the only thing bugging right now.
     
  16. NascentOxygen

    Staff: Mentor

    There are 2 identical secondary windings, and at any moment in time they produce identical voltages. So if you say there are V volts across one, then there must also be V volts across the other (regardless of whether there is current being drawn or not).

    So for the situation depicted, with V volts across the lower winding, this places B at -V volts with respect to the other end of that winding (which being connected to earth, we call 0 volts).
     
  17. vk6kro

    vk6kro 4,059
    Science Advisor

    [​IMG]

    As you can see, the center tap of the transformer is always at zero volts because it is grounded.

    So the opposite ends of the transformer winding swing positive and negative relative to this ground connection and produce a peak voltage of 1.414 (√2) times the RMS voltage.

    If you take the moment when the top winding produces a maximum positive voltage and the bottom winding produces a maximum negative voltage, there is a combined maximum reverse voltage across the bottom diode.
    This is when it is most likely to fail and it has to be rated for at least this reverse voltage.
     
    1 person likes this.
  18. this sound like what my lecturer said, he split the ac source into 2:
    [​IMG]

    And if i am not wrong, each of the splited source can produce 0.5 of the V_peak. And is it because the circuit is somehow not completed, the voltage differnce across the the gap is same as the v_peak, like in this diagram:

    [​IMG]
     
  19. NascentOxygen

    Staff: Mentor

    EDIT: oops, I see you have it right after all.

    Because the - of the upper source is earthed, then the upper winding delivers +V volts. Because the + of the lower is earthed (it shares the same centre tap), then the lower winding terminal delivers -V volts.
     
  20. NascentOxygen

    Staff: Mentor

    You have to get it clear in your mind and be consistent, either view it as one long winding producing 2V volts with a centre tap, or view it as two windings, each producing V volts.

    If it's a 36V winding centre-tapped, then each half winding delivers 18V, and the output is 18V full wave rectified. This is the same as viewing it as a pair of 18V windings, with one end of each joined together and earthed (and phasing correct). Your second sketch with one winding showing no voltage is quite wrong.
     
    1 person likes this.
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?

0
Draft saved Draft deleted