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Calculating current ripple?

  1. Sep 1, 2010 #1
    Hi,

    I am trying to, roughly, calculate the current ripple associated with a 12-pulse power supply. I have managed to workout an (idealised) voltage ripple from the rectified signal and I heard that there is a way to work out the current ripple from that.

    I'm a physicist not an electrical engineer so I'm sorry if I have neglected some important information or have confused my definitions but if someone can help that will be great.

    Rob
     
  2. jcsd
  3. Sep 1, 2010 #2

    berkeman

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    Staff: Mentor

    Could you give more details about the power supply? You call it "12-pulse", but I don't know what that means. Do you mean it is a DC-DC converter power supply? If so, what topology is it?
     
  4. Sep 1, 2010 #3
    You will find a definition/explanation of pulse numbers (including 12 pulse) near the bottom of this page.

    http://en.wikipedia.org/wiki/Inverter_(electrical)

    The ripple depends upon the current draw ie the load as well as the supply. It is only partly a function of the supply configuration.
     
    Last edited by a moderator: Sep 1, 2010
  5. Sep 1, 2010 #4
    If your output goes directly from a bridge rectifier to a capacitor bank without a current limiting device, you will have very high current surges for a small % of output voltage waveform. You cannot calculate the rms current or peak current until you have a specific circuit design.

    Bob S
     
  6. Sep 1, 2010 #5

    berkeman

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    Oh, he's asking about an inverter. Got it, thanks Studiot.
     
  7. Sep 1, 2010 #6

    vk6kro

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    Because this system uses overlapping input pulses fed to rectifiers, the output is fairly good DC without any filtering.

    I just did a simulation of this and got about 3.3 % ripple with sinewave inputs 30° apart.


    EDIT:
    The dip in input voltage happens half way between input peaks or at 15 degrees after a maximum.
    So the voltage of either waveform at this point would be at sin (75°) times the maximum voltage.
    This is (1 - 0.9659) (ie sin (90°) - sin (75°)) or 0.03407 times the maximum voltage. So about 3.4 % p-p ripple.
    This agrees OK with the simulation, allowing for diode drops.
     
    Last edited: Sep 1, 2010
  8. Sep 3, 2010 #7
    Ok thanks for all the replies,

    I'm glad that you put this in as it was the way I had been calculating my voltage ripple and so great to have it confirmed. My problem, however, was in using this value to determine a percentage current ripple.
    I think now that It is not such a simple relationship that I was hoping for. From all the other posts and other research I have done it seems that I have to include the characteristics of the circuit I am investigating to determine the current ripple (ie the resistances, inductances and capacitances within the circuit).

    I am now using similink (the Matlab electric engineering function) to model my circuits and then determine the current ripple that the program detects. so basically the higher the inductances the smaller the current ripple.
    Does this sound like the right thing to do? I apologise again if I haven't explained myself well enough.

    Rob
     
  9. Sep 3, 2010 #8

    vk6kro

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    This system would not be very much affected by load, unlike capacitor filtering methods.

    Because the inputs overlap, there is already excellent ripple suppression so this would be a suitable system for very high powered designs.

    The penalty, of course, is that you need some formidable hardware to achieve this. Two 3 phase transformers and some massive rectifiers.

    However applications like electroplating and aluminium production by electrolysis would possibly use such systems.
     
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