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Re: Please help with capacitor



Original poster: Esondrmn-at-aol-dot-com 

In a message dated 1/5/04 4:15:57 PM Pacific Standard Time, 
tesla-at-pupman-dot-com writes:


>Original poster: "Philip Brinkman" <peeceebee-at-mindspring-dot-com>
>
>I'm making a new flat plate capacitor for use with my 15,000 volt 60 ma
>NST. I'm using aluminum foil plates 7.5"x9" (67.5 sq. in.) with 30 mil
>polyethelyne (6 layers of 5 mil each) between each plate. Using the formula
>C=kA/d (k=2 ) I get 3.25 sq. meters or 74 plates, does this sound right? My
>last capacitor had only 30 plates. I'm aiming for .013 microfarads.. is
>this right? Using different programs I get different values of capacitor to
>use.


Philip,

I have always used the formula: C (pf) = .225 K A / d  with A in square 
inches and d also in inches

So, with your numbers I get about 1,012.5 pf per plate set (one 
capacitor).  You are looking for 13,000 pf so you wold need 12.85 caps or 
13 dielectric layers and 14 plates.  That is a big difference from your 
numbers.  Maybe my memory for the formula is incorrect?

Ed Sonderman