Ripple Voltage

Discussion in 'General Electronics Chat' started by david mendes, Mar 23, 2006.

  1. david mendes

    Thread Starter New Member

    Mar 23, 2006
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    0
    Hello
    Does anyone know how to calculate the Ripple Voltage on a half-wave rectifier circuit and on a full-wave?
    I found the following formulas but when I compare them with what I get on the oscilloscope, the difference is huge!

    Ripple (half-wave)= 4,5 x I(mA)/C( μ F)

    Ripple (full-wave)= 1,7 x I(mA)/C( μ F)

    Cheers

    David Mendes
     
  2. Papabravo

    Expert

    Feb 24, 2006
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    The problem could be the constant or it could be the units that you are mixing. That is mA and uF. The approximations are derived from the following equation
    Code ( (Unknown Language)):
    1.  
    2. dV/dT = - I/C
    3.  
    4. where
    5.  
    6. dV/dT is the voltage sag from one rectified peak to the next. dT will be 8.33 milliseconds for the full wave case and 16.67 milliseconds for the half wave case.
    7.  
    8. I is the current in Amperes
    9.  
    10. C is the capacitence in Farads.
    11.  
    12. In a less tha rigorous fashion this leads to
    13.  
    14. dV = dT*(I/C)
    15.  
    So using 0.00833 or 0.01667 for dT and I in amperes and C in Farads do you get more consistent results?
     
  3. david mendes

    Thread Starter New Member

    Mar 23, 2006
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    >So using 0.00833 or 0.01667 for dT and I in amperes and C in Farads do you get more consistent results?

    Where do the 8.33msec and the 16.67 msec come from?

    Cheers
    david
     
  4. Papabravo

    Expert

    Feb 24, 2006
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    8.33 msec is the time from one peak to the next peak of a full wave rectified 60 Hz. sine wave, and 16.67 msec is the time from one peak to the next peak of a half wave rectified 60 Hz. sine wave.

    Naturally if you are outside North America your sine waves will be 50 Hz. and not 60 Hz., and if you are on an airplane they will be 400 Hz.
     
  5. gladeeader

    New Member

    Oct 26, 2008
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    PapaBravo, is the 'I' in that equation based on the RMS voltage or the peak voltage? Is the 'I', Irms or Ipeak?
     
  6. Papabravo

    Expert

    Feb 24, 2006
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    In a somewhat ironic fashion for a thread which died 2 1/2 years ago it is neither. It is instantaneous.
     
  7. gladeeader

    New Member

    Oct 26, 2008
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    I was a little bit worried about the age of this thread, but i figured i'd try it anyway. Also thank you for replying. I'm just getting into some of my electrical engineering classes and this helped me out a lot. and a couple of my peers as well. thank you very much for sharing your knowledge so freely PapaBravo
     
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