Cesare,
Thank you for your timely proximity equation! I calculate the spacing
required to keep the proximity loss at 10% for two #12 ga (2.0 mm) copper
conductors to be 0.5 inches. The spacing required for #12 Litz wire would be
almost 0 because the sq root terms at the right side, based on Rrf/Rdc
(mostly skin effect), would be close to 0.
Would appreciate a check of my math from anyone.
Bill A
-----Original Message-----
From: cesare tagliabue [mailto:[email protected]]
Sent: Monday, March 18, 2002 1:06 PM
To: [email protected]
Subject: LF: R: Loop Conductors and proximity effect Rdc Rac
Hello John
I have found the Mesny's expression on an old university stenciled
book where only the data were given, but some time ago, for purpose of
include it in a program, I have elaborated an empiric expression that
approximates sufficiently these data. That is:
q = ( 1 + .95 d / D )^2 -
.22
Best 73 to all Cesare
Cesare Tagliabue I 5 TGC
WW-Loc JN53PS
e-mail: [email protected] <mailto:[email protected]>
url: http://www.dadacasa.com/i5tgc <http://www.dadacasa.com/i5tgc>
-----Messaggio originale-----
Da: john sexton < [email protected]
<mailto:[email protected]> >
A: [email protected] <mailto:[email protected]> <
[email protected] <mailto:[email protected]> >
Data: lunedì 18 marzo 2002 17.46
Oggetto: LF: Loop Conductors and proximity effect Rdc Rac
Hi Paul,
From the interesting formula given by Cesare, the short answer to your
question appears to be 1/2 inch.
I tried to find out more about Mesny's expression from the Internet, but it
looks like you have to buy the book.
I wonder whether Cesare can be persuaded to tell us the formula that relates
q to D/d? At a guess it appears to be logarithmic.
The second formula can be simplified to
Reff/Rrf = 1 + (q-1)*sqrt(1-Rdc/Rrf)
When D/d = 9, q = 1 so the second term disappears and Reff = Rrf.
73, John, G4CNN
_____
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