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What value of capacitor to drop the ripple %

Discussion in 'General Electronics Discussion' started by jalapino4, Feb 26, 2013.

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  1. jalapino4

    jalapino4

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    0
    Feb 26, 2013
    Here in my company we try to get all rectifier box under 5% or below. My co-worker is doing the calibration and all he know is getting the input and list the box pass or fail. Now we got 2 box that have high rectifier ripple. Its a rectifier box that handle up to 500 Amp using 10 Volt. When calculate 50% Amp with load to get under 5% using formula Vac/Vdc *100%. He receive one box come up to 30% and the other is 12% ripple. How do we calculate what capacitor size to use to drop the ripple to 5% or below? Can you help me out and explain in detail. Thank you, this will help me alot and give me a lot of knowledge of what to tell him.
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    11,007
    2,506
    Nov 17, 2011
    Look here or Google rectifier ripple calculation.
     
  3. BobK

    BobK

    7,682
    1,687
    Jan 5, 2010
    So the OP has 10V 500A which is a load resistance of 0.02R and wants 5% ripple which is 0.5V.

    So we have:

    Vr = Vp / (Rl * C) * dt

    or

    C = Vp / (Rl * Vr) * dt

    or

    C = 10 / (0.02 * 0.5) * 0.01 = 10F

    I knew it would be big, but that is enormous!

    Bob
     
  4. jalapino4

    jalapino4

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    Feb 26, 2013
    thank you for your fast reply, What is a dt? So you mean whatever ripple i want to achieve, eg 3% the voltage would be .5v? Do i connect the capacitor parallel with the load?

    For your information, it a big unit. I think its a 3 phase coming into the unit and output 10v and can handle 500amp max. We have some that can handle up to 2,000 Amp. It a rectifier that use for plating into a chemical. Just to give you an idea what we are working.

    So the formula should work as a standard to all of the rectifier.
     
  5. BobK

    BobK

    7,682
    1,687
    Jan 5, 2010
    The ripple voltage is a percentage of output voltage. 3% of 10V would be 0.3V.

    dt is delta time. The time for one cycle. Assuming 50Hz this is 1/100th of a second, assuming full wave rectification.

    BTW: 10F is a HUGE capacitance, in the range of supercapacitors. It might be very difficult and expensive to find a 10F 10V capacitor.

    Bob
     
  6. jalapino4

    jalapino4

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    0
    Feb 26, 2013
    How do you know if it 50Hz or 60Hz. Someone told that all US is 60Hz, is that right? If it 60Hz is it still 1/100th?
    What if it a half wave rectifier, any different in the formula?
    One of my co-worker suggest that SCR in the unit might be short, causing the high ripple and it not using it full voltage potential. Do you think it might effect the ripple?
     
  7. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    In #4 we got some critical information. The supply is three phase and with full wave rectification there will be six pulses per cycle. The voltage will never drop to zero like a single phase supply, even so, I do not see a capacitor doing anything significant.

    It may be that a rectifier is faulty if the ripple is too high.

    Look up Wikipedia 'Rectifier'
     
    Last edited: Feb 26, 2013
  8. jalapino4

    jalapino4

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    0
    Feb 26, 2013
    It might be one of the SCR is fail, causing the voltage to be not accurate. I think this might work. What if i put an actual load into the tank and test the result from there. It might give me a lower reading on the ripple. I will let you guy know that outcome. thanks
     
  9. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    You could connect the output with to a load of say a 12V headlight bulb and look at the number of pulses with an oscilloscope.

    It appears to be more than a rectifier, with SCRs there must be a control circuit.

    Information is creeping out bit by bit.
     
  10. woodchips

    woodchips

    43
    0
    Feb 8, 2013
    As has been mentioned, 3 phase will reduce the ripple to cos30 or 0.866 with no smoothing at all. Presumbly at 500A you are using a dual phase transformer with 6+6 rectifiers reducing the ripple to cos15 or 0.966 which gives you your 5% using no caps at all. A choke would also help, if needed.

    Bob
     
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