PIV Diode Confusion - Need clearing up

Discussion in 'Homework Help' started by Vicros, Feb 5, 2012.

  1. Vicros

    Thread Starter New Member

    Jan 22, 2012
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    Hey guys

    Just writing my notes out for both half wave and full wave diode rectifiers.
    Just in a bit of a confusion at the moment, Ive found multiply PIV formulas for the both, and i was hoping if someone could clear it up for me.

    Half wave:
    PIV = Vs or PIV = Vs - Vf


    Full wave:
    PIV = 2Vs - Vf or PIV = 2(Vs - Vf)

    where Vs = voltage source
    Vf = voltage the diode conducts at


    I believe that the half wave PIV is equal to the peak of Vs, and that the full wave is approximately twice that for the case of the half wave where PIV = 2Vs - Vf

    but then the other two seem plausible to me as well....

    circuits that i am looking at can be found in the attachment
     
  2. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    I really duno what you are talking about, but all I know is PIV is peak inverse reverse voltage a diode can withstand.

    And they are always indicated in it's data specs. No where else.

    And in a circuit PIV seen is always Vs...or Vrms
     
  3. Vicros

    Thread Starter New Member

    Jan 22, 2012
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    yeah, well the question ask: what is the peak inverse voltage across the diode?

    i thought there was a formula for it, cause there will be no data specs given
     
  4. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    There ain't a formula.

    The PIV is given by the manufacture of the diode.
     
  5. holnis

    Member

    Nov 25, 2011
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    The PIV (Peak Inverse Voltage) rating of a diode is the maximum voltage that you should apply to it in the reverse biased condition. Exceeding this voltage can destroy the diode.

    And as R!f@@ said, The PIV is given by the manufacturer.
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    As Wikipedia notes it's a case of either or .....

    http://en.wikipedia.org/wiki/Peak_inverse_voltage

    PIV is either the manufacturer's specified rating or it is the peak inverse voltage one would anticipate appearing across a reverse biased device in a particular circuit configuration.
     
  7. t_n_k

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    Mar 6, 2009
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    Vicros,

    Not sure about the source and reliability of your attachment.

    In the case of the half wave rectifier with filter capacitor shown on the left the PIV will depend upon the load time constant. If C is sufficiently large and notionally charges to the peak source value [neglecting Vf for the diode] then the diode has a peak inverse voltage of 2*Vs peak. If Vs=12V then the peak value will be ~17V. Hence the diode will see a peak reverse value of about 34V-Vf with the diode drop included. In the absence of the capacitor the PIV would only be about 17V.

    In the case of the full-wave bridge with large filter capacitor / load time constant the peak inverse voltage across a single diode in the bridge will always have a value no greater than the source peak value - again ignoring the diode forward drops. The result will be the same if the capacitor is removed.

    In summary I would regard the PIV value 'equations' given in your pdf attachment with considerable distrust. See if you can work out why.
     
    Last edited: Feb 5, 2012
  8. Vicros

    Thread Starter New Member

    Jan 22, 2012
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    from my sedra/smith text book it says: (for HW)

    "... peak inverse voltage PIV that the diode must be able to withstand without breakdown, determined by the largest reverse voltage that is expected to appear across the diode. in the rectifier circuit (given,pdf) we observe that when Vs is negative the diode will be cut off and Vo will be zero. if follows that the PIV is equal to the peak of vs

    PIV = Vs"

    This is the part im confused at now. it says that it "is the peak of Vs"
    which to me is
    Vp = Vs-Vf
    Vp = 12 root(2) - 0.8 = 16.1V

    But the equation just says PIV = Vs
    which i see as:
    Vs = 12 root(2) = 16.9V ~17V as you said.
     
  9. t_n_k

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    To which circuit topology are the Sedra/Smith comments referring?
     
  10. Vicros

    Thread Starter New Member

    Jan 22, 2012
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    for the half-wave rectifier as seen in the pdf attachment.
     
  11. t_n_k

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    Mar 6, 2009
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    If the capacitor is included then first consider the total circuit at the instant at which the diode just turns off.

    C will be charged to Vs[peak]-VF=16.97-0.8=16.17V.

    The capacitor will then begin discharging into the load resistor RL.

    The AC source reaches a negative maximum one half cycle later or about 10msec after the diode turns off. The capacitor has been notionally selected to limit the ripple voltage to no more than 1V. So the capacitor would have discharged by about 0.5V at the 10msec mark - with recharging of the capacitor starting roughly 20mS after the diode turns off.

    So the additive peak reverse voltage across the diode will 'roughly' be Vs[peak]+Vcap=16.97+16.17-0.5=32.64V

    A more accurate analysis may require one to determine the conduction angle and so forth....
     
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