Rectifier Filter alternate Dc or Vr P-P calculation.

Discussion in 'Homework Help' started by Petrucciowns, Jul 3, 2009.

  1. Petrucciowns

    Thread Starter Active Member

    Jun 14, 2009
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    I have a predicament. I'm trying figure out the P-P ripple voltage of a Half-Wave rectifier filter, but the only equation I have is: I / (f * C) yet to find the current I need the DC voltage which I don't have yet.The same goes for the DC voltage. I have the equation: VDC = VOUT P – (VR P-P /2) yet I do not have the ripple voltage yet. So my question is, are there any alternate equations that will allow me to find the ripple voltage without the DC voltage, or the DC voltage without the ripple voltage.

    Thanks.
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Well, if we have the transformer that give as 12Vrms (secondary voltage).
    12Vrms=16.9Vp
    The capacitor (with no load connection) will be charge to
    Vc=16.9V-Vd≈16.2Vdc
    And if we connect C=220uF and RL=1K
    Vpp_ripple=1/(F*R*C)*Vc=1/(50Hz*1K*220uF)*16.2V=0.09*16.2V=1.5Vpp
    Vout_dc≈(16.2+14.7V)/2≈15.4V
    These calculations assume that transformer voltage are constant. They don't change with the load.
    As we know this is not true for real circuits.
     
    Last edited: Jul 4, 2009
  3. Petrucciowns

    Thread Starter Active Member

    Jun 14, 2009
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    Hm the ripple current equation is not seeming to give me the right answer. Could you possibly do an example with these values.
    Vout Peak- 169.3 volts 12 micro farad 15kohms 60hz and a forward biased silicon diode.

    The answer for Vr p-p should be an exact 15 volts p-p


    Also with your DC equation Vout_dc≈(16.2+14.7V)/2≈15.4V

    Where does the value of 14.7 volts come from?

    Thanks.
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Vpp_ripple=1/(60Hz*12uF*15KΩ)*169.3V=92.6m*169.3V=15.7Vpp
    Vdc=(169.3V+(169.3V-15.7V)/2=161V
     
  5. Petrucciowns

    Thread Starter Active Member

    Jun 14, 2009
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    Thank you, is there a similar equation for center tapped full-wave rectifier filters such as:

    60 hz 24.3 v p output 1000 micro farad 100 ohms


    Because I noticed that equation is only for half wave. The answer to the above should be around: 1.94 v p-p


    Thanks.
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Well the equation is exactly the same

    Vpp_{ripple}=(\frac{1 }{F*R*C})*Vin_p

    But the frequency is 120Hz for bridge and center-tapped full-wave rectifier.
    So Vpp_ripple≈1.96V

    And in my previous calculation I forgot about diode and her voltage drop.
     
  7. Petrucciowns

    Thread Starter Active Member

    Jun 14, 2009
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    With the the values in my last post I get a different answer. What am I doing wrong

    1/ (120 x 100 x 1000 micro) x 24.3 = 2.03 Volts

    When it should be 1.96 volts like you have above.
     
  8. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    The answer is simply, you forget about diode voltage drop.
    Capacitor will be charge to Vc=24.3V-0.7V=23.6V
     
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