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Re: Electrical Properties of Brass



Original poster: "Terry Fritz" <teslalist-at-qwest-dot-net>

Hi Antonio,

If the transition from insulating Al2O3 to pure Al was instant, it would 
indeed be fine.  The problem is the resistivity decreases gradually through 
the layer.  So instead of either a perfect insulator or perfect conductor, 
the currents tend to run through a perfect resistor :-p  I work with 
equipment that runs like 500 amps at 13MHz, so when something goes 
resistive it just explodes.  Silver coated copper pipe with water running 
through it does wonders ;-))  Tarnished silver is actually very conductive 
unlike most oxides (might be a sulfied?).

I think we worked all this out a few years ago and gold over pristine clean 
copper was best for RF currents.  I would think in most cases, the copper 
could be sprayed with polyurethane if corrosion was an issue.

One really has to balance the primary resistive losses against the ~~3 ohm 
resistive loss of a spark gap.  If the primary does not get warm, it is not 
loosing system power...

Of course, coilers will use what they have on hand. but given a choice, "I" 
would avoid aluminum.

Cheers,

         Terry


At 02:16 PM 4/19/2003 -0300, you wrote:
>Tesla list wrote:
> >
> > Original poster: "Terry Fritz" <teslalist-at-qwest-dot-net>
> >
> > Hi,
> >
> > "I" would never use aluminum.  Aluminum forms thick (skin depth thick)
> > resistive oxide layers.  These are terrible for RF currents.
>
>Why would not the current just pass under the oxide layer?
>If this is a problem, then a conductor with an irregular surface
>would be bad, or even a conductor painted with resistive ink would
>be bad. Very strange.
>It doesn't look difficult to compute what is the current distribution
>at the surface of a conductor that has over it a layer with a
>different resistivity, and then see what happens with the losses,
>compared to the case when there is no resistive layer. The first
>thing that I notice is that the skin depth is larger in a more
>resistive material, and so the current is effectively pushed to
>the conductive layer below.
>
>Antonio Carlos M. de Queiroz