Re: Need Help With Capacitor

• To: tesla-at-grendel.objinc-dot-com
• Subject: Re: Need Help With Capacitor
• From: jim.fosse-at-genie-dot-com
• Date: Tue, 6 Feb 96 03:42:00 UTC 0000
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```>>>
Since my garage is about 20 degrees below freezing, I thought I would make
some caps indoors.  I was thinking of using wide straps instead of the
6 gauge wire that I used in my other cap.  The question is how wide is
wide enough?  I was leaning towards a few 2" straps placed along
the length of the conductor, thereby reducing inductance.  The conductor
is about 10 or 12 mils thick aluminum.
<<<
Chip ,
The inductance of a straight wire in free space is:

L = 0.00508 * b * ( (ln(2b/a)-0.75)

L is in microhenries
a is wire diameter in inches -- replace 0.00508 with 0.0002 for mm.
b is wire lenght in inches.
ln is the natural log or 2.303 times the common log (base 10 log).

#6 solid wire is .162" in diameter. let's use 1'=12" as the length
L= 0.00508 * 12 * ((ln(2*12/.162)-0.75) = 0.259uH. At
200Khz  Z=2*pi*f*l= 325 miliohms.

The inductance of a flat strip in free space is:

L = 0.00508 * b * (ln (2b/(h+w)) + 0.5 + 0.2235(w+h)/b)

L is in microhenries
b = length in inches
w = width in inches
h = thickness in inches

using 1'=12" as the length, 2" as width, and 10mils thick:
L= 0.00508 * 12 * (ln(2*12/(0.010+2)) + 0.5 + 0.2235(0.010+2)/12)
L= .184uH At 200Khz Z= 231 miliohms.

Note that these formulie don't take into account skin effect, and
the material must be non-magnetic.

I'm supprised, I had expected the inductance of the flat strip to be
MUCH lower!

jim

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