# 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?

Oh, I see now, you've substituted the frequency for the period.

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