Smoothing capacitor

Discussion in 'Homework Help' started by cdennis414, Mar 1, 2013.

  1. cdennis414

    Thread Starter New Member

    Dec 19, 2012
    14
    1
    Hi everyone
    I have to design a power supply for my assignment but i struggle to find a right equation for smoothing capacitor... There are a few different equation i have found but they all giving me different result.
    The most common equation i came across is C=I/f*V and f is equal 100 Hz.
    But the one i have used is from a book. I have attached a pictures from the book. What i can't understand is why in they calculations the frequency timed by 2. Its already a 100 (supply frequency by 2)???
    Thanks.
     
  2. t06afre

    AAC Fanatic!

    May 11, 2009
    5,939
    1,222
    The frequency is the number of occurrences of a repeating event per unit time. I took this drawing from Wiki to illustrate. After the rectifier the waveform has changed and frequency has doubled, compared to the AC input frequency. It is that simple
    [​IMG]
     
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  3. cdennis414

    Thread Starter New Member

    Dec 19, 2012
    14
    1
    Ok I understand that but if you look at the calculation from the book the 50 Hz original AC frequency been doubled which makes a 100Hz and then doubled again. The part i cant understand why they doubled it twice. Sorry if i didn't explain it properly. I attached the first page from the book. As i understand the original frequency is 50Hz.
    Thanks.
     
  4. t06afre

    AAC Fanatic!

    May 11, 2009
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    The number 2 has nothing to do with the frequency.
    The formula is for half-wave rectification:
    Vripple=I/fC
    And for full-wave rectification
    Vripple=I/2fC
    Remember that the frequency for half-wave rectification will be the AC voltage frequency. So the ripple in a full-wave rectification. Will be four times less compared to half-wave rectification
     
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  5. cdennis414

    Thread Starter New Member

    Dec 19, 2012
    14
    1
    Thanks! I get it now.
     
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