Message-ID: <1333251.951933993446.JavaMail.imail@bubbles.excite.com>
Date: Wed, 1 Mar 2000 10:06:33 -0800 (PST)
From: "john sexton" <computernetworks@excite.com>
To: rsgb_lf_group@blacksheep.org
Subject: LF: Re: Low loss inductors
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On Sat, 26 Feb 2000 14:00:10 GMT
"Andy Talbot" wrote:

>For an estimate of losses : 
>
>Skin depth of copper at 137kHz is approximately 0.18mm      From    D = 503

>SQRT(Resistivity / Freq / uo) 
>For Cu    Resistivity = 1.7E-8 Ohms / m,   and uo (magnetic permeability) =
1 
>Diameter of centre conductor = 2.5mm  (near enough anyway) 
>so cross sectional area of conducting path is 
>0.18mm * 2.5mm =  0.45E-6  m^2 
>
>RF Resist   =   Resisivity * Length / Area       =   1.7E-8  *  370m  / 
>0.45E-6m^2    =    14 ohms. 
>
>For a quick estimate assume the braid losses are a lot less than the centre

>conductor as they have a much larger surface area, so can be ignored 
>(that may not necessaily be the case) and we can also ignore 
dielectric losses (a reasonable assumption at these freqs)  so Q =  Xl / R 
>=     1722 / 14 =  123   
>Which is about what I got on my 5mH conventional coil of 1.5mm wire,  
300mm 
>diameter and 400mm long.

You forgot a Pi in the calculation of the cross sectional area. The
circumference is Pi D, not just D. Therefore cross sectional area is approx.
1.41E-6 m^2
and RF Resist = 1.7E-8 * 370m / 1.41E-6m^2 = 4.46 ohms.
So Q = 1722/4.46 = 386, which is somewhat better than your conventional
coil.

John, G4CNN


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