thingmaker3 -> Cboot calc.

Discussion in 'The Projects Forum' started by HellTriX, Jul 5, 2008.

  1. HellTriX

    Thread Starter Active Member

    Apr 11, 2008
    83
    1
    This question is not just for thingmaker3 if anyone else has the answers I seek.

    So I tried the calculation its much more simple then the one I was going to use from the application guide, but I'm getting a fairly large value.

    So lets do one.
    Cboot must be larger than: (Qg + Dmax*Tsw*Idrv ) / ΔV

    (.00000017 + .75*20000*.2) / 2
    is equal to 1500.000000085. This would appear to be 1500 farads?
    What am I doing wrong?

    Or, if I don't convert to actual values I could do:
    (170nC + 75% * 20khz * 200mA) / 2v
    and get a value of 235 which I could assume means 235 picofarads?

    So a tally of the issues I'm having:
    1) Did I do any of the calculations right?
    2) Do I use the average current of my driver or peak pulse current?
    3) What happens when you get to duty cycles of 1% and 99%?
    4) A 2V ripple? Meaning I'm driving with 12v and will allow it to sag to 10v.


    Thanks
    TriX
     
  2. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    Why are you using a six hour switching period?:eek:

    Oh, I see now, you've substituted the frequency for the period.:cool:
     
  3. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    The formula, by the way, comes from Christophe Basso, author of Switch Mode Power Supplies: SPICE Simulations and Practical Designs.
     
  4. HellTriX

    Thread Starter Active Member

    Apr 11, 2008
    83
    1

    Mmk so I should use switching period instead?

    The switching period is 0.00002 for 20khz? (I'm not sure on this).

    When I do (.00000017 + .75*.00005*.2) / 2
    I come up with .000003835 or 3.8uF.
    Which seems closer to reality.
     
  5. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    Period is the multiplicative reciprocal of frequency. Divide one by frequency to get period. 1/20000 = 0.00005
     
  6. HellTriX

    Thread Starter Active Member

    Apr 11, 2008
    83
    1
    Ewps, yes, I appeared to get the correct answer that I used in my calculation but typoed the first one to .00002

    Guess my application of math was good, but now that I know its the inverse multiple, that will help for the memorization.

    tnx
     
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