Polarized Capacitors

Discussion in 'General Electronics Chat' started by slimnick, Nov 26, 2007.

  1. slimnick

    Thread Starter Member

    Nov 25, 2007
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    i am currently in college and as a project i have built a variable DC power supply. after the rectification process there is a polarized capacitor used to filter. but my question is what is the importance of this capacitor. why cant i just use a NP capacitor like a ceremic. i need an explanation of why one has to be there.
    thankyou for your help:D
     
  2. Distort10n

    Active Member

    Dec 25, 2006
    429
    1
    You answered you own question of 'why' the capacitor should be there. It is like a filter since it helps to reduce the ripple after rectification. Large values, typically electrolytics, are used because a small value capacitors like ceramics are often used for high frequency. They would not smooth out the 'slow' or low frequency ripple. This is how you would look at it in the frequency domain. In the time domain, think of the cap as a charge reservoir. The ceramic would already be charged thus allowing the ripple to continue on its way. The reservior is too small.
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Another way of explaining it is simply that the capacitance per unit volume of polarized caps is much higher than that of nonpolarized ones. Your monetary and board space cost for nonpolarized caps would be much too high. Do a little shopping and see if you can find a 1000 uF nonpolarized cap. If you do, it will still be electrolytic. You might as well use a polarized one.
     
  4. slimnick

    Thread Starter Member

    Nov 25, 2007
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    so would i be right in saying the the only reason why a polarized capacitor as the filtering capacitor instead of a NP capacotr such as cermamic is because NP capacitors do not come in large values, that would be enough to completely get rid of the ripples?
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    That's correct.
     
  6. slimnick

    Thread Starter Member

    Nov 25, 2007
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    thank you very much. i was all over the internet trying to find why this was. atleast i now have closure to this question that i had. the help is very much appreciated
     
  7. cumesoftware

    Senior Member

    Apr 27, 2007
    1,330
    10
    To calculate the capacitance needed, in farads, to achieve a certain ripple, you can use:
    C = I / (2 x Vripple x F)
    C = I / (2 x Vp x (%ripple / 100) x F)

    F is the AC current frequency, Vp is the peak voltage at the secondary, Vripple is the ripple voltage (or the AC component voltage, base to peak), %ripple is the ripple percentage (normally it is used 5% or 10%) and I is the current needed. This formulas are valid for full wave rectification (that is why you multiply the frequency by 2).
     
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