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Re: [TowerTalk] Q of coil Al vs Cu - Re: OT: Inductor Calculator

To: jimlux <jimlux@earthlink.net>, towertalk@contesting.com
Subject: Re: [TowerTalk] Q of coil Al vs Cu - Re: OT: Inductor Calculator
From: "Richard (Rick) Karlquist" <richard@karlquist.com>
Date: Thu, 28 Feb 2019 14:50:05 -0800
List-post: <mailto:towertalk@contesting.com>
On 2/28/2019 1:05 PM, jimlux wrote:

Aluminum resistivity is 2.65, Copper is 1.68
Skin depth at 7MHz for AL is 30.97 micron, for Cu is 24.66 micron

So, resistance of a coil in Al vs Cu is = 2.65/1.68 * 24.66/30.97 = 1.26

It is simpler to just take the square root of 2.65/1.68 =
sqrt 1.58 = 1.26. It is not necessary to calculate the skin depth ratio,
which is also 1.26, not coincidentally.


Q of a Al coil would be 80% of the Q of a Copper coil.

So far correct.


Assuming the current in the coil is the same, the resistive power dissipation in the Aluminum will be 60% higher.

Correction.

P = I^2*R, hence the aluminum coil dissipates 26% more heat than the
copper, not 60%  If you past DC through the coil, that dissipation
will be 58% higher than if copper is used.

How hot the coil gets is not just 26% more, but depends on the
delta T, and the emmissivity of the respective coils.  The
aluminum should ideally be black anodized, not shiny.

Rick N6RK
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