smoothing capacitors

Discussion in 'Homework Help' started by gunmetol, May 4, 2015.

  1. gunmetol

    Thread Starter New Member

    Jan 29, 2015
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    Hi,
    i have been tasked to design a power suply 3 - 20 volts but my smoothing capacitor seems to be huge could somebody please have a look at my calculations
    its a full wave rectification system
    output = 0.4 amps
    ripple voltage = 300 mv
    s0
    0.4/(300*10-3)*100= 13.3 mf

    is this correct ?
    thanks
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    C = Q/V = (I* t)/V = (0.4A * 10ms)/0.3V = 13.3mF
    But remember that this equation assume constant current discharge. In real life you will need a smaller cap.
     
  3. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    I do it this way: Q = I*t =C*ΔV

    Fullwave at 50Hz, means t = 10ms (t=8.33ms at 60Hz)

    Transposing, C = I*t/ΔV = 0.4*0.01/0.3 = 0.01333F = 133,000uF

    This is why LM78xx were invented. You can let the ripple be much higher and still get regulation... If you are using a LM78xx, then the constant-current discharge assumption of the filter capacitor is correct.
     
  4. gunmetol

    Thread Starter New Member

    Jan 29, 2015
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    thank you for the help
    i am using a LM 317k chip what would be the calculation of the smoothing capacitor
     
  5. dl324

    Distinguished Member

    Mar 30, 2015
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    What's the secondary voltage and current rating?
     
  6. gunmetol

    Thread Starter New Member

    Jan 29, 2015
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    23 v RMS
    0.4 amps or 8 wats @ 20 volts
     
  7. dl324

    Distinguished Member

    Mar 30, 2015
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    You're not going to get 0.4A out of it, but you can cut way back on the filter cap because your going to have more than enough headroom for the LM317 to stay in regulation at 20V. You could allow volts of ripple...

    What kind of transformer is it? 23V RMS is a strange number. Used to seeing multiples of 6.3...
     
  8. gunmetol

    Thread Starter New Member

    Jan 29, 2015
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    I have currently gotten 0.4 amps at 20 volts out of the lm317 on a simulation software but it know to be not very relabael . could you explain why you would think that i would not be possible ?

    i have cut it back to about 40 % of this but i need to prove this in mathematical form.

    sorry i should have been more precise with the voltage its 23.4 volts because of the rectifier removing 1.6 volts as its a 25 volt transformer

    regards
     
  9. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Read the data sheet for the LM317. Look up the Dropout voltage. The sag in the filter capacitor voltage can go down to (20V + Vdo),

    The peak filter capacitor voltage is 1.414*Vrms - drop across winding resistance - drop across two Si rectifiers. The discharge of the filter capacitor is ~10ms @ 0.4A constant-current.

    It looks to me that ~350uF is big enough to keep the ripple above the dropout. I'll let you prove it...
     
  10. dl324

    Distinguished Member

    Mar 30, 2015
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    Not sure why you'd use a simulator to design a power supply. Kind of like using a calculator to add 200 + 200...
    This is from the National Semi Voltage Regulator Handbook (it's available on the WEB).
     
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