Joel Koltner wrote:
Hi Jim,

Thanks for the thoughts; I hadn't thought of many of the additional loss
mechanisms you mention.

"Jim Lux" wrote in message
...
In the subject case here, think of this: say you had a 2cm diameter
copper bar and you run 100 Amps of DC through it. The current is
distributed evenly, as is the power dissipation. Now run 1 MHz RF
through that same bar. The skin depth is about .065 mm, so virtually
ALL the RF current is contained within a layer less than 1/3 mm thick.
That's a very different heat and thermal distribution (sort of like
the difference between putting that thick steak in the 200F oven and
throwing it on the blazing hot grill).

If you're just looking at surface temperature (i.e., with a thermal
camera), will it take more or power at 1MHz to obtain a given surface
temperature increase than at DC?

At DC, since you're heating up the entire bar, and the only way for the
heat to go is up "out" to the surface... I'm thinking... less power is
needed for a given rise?

That would certainly then overestimate antenna efficiency.


And one would need to be careful about when you've reached thermal
equilibrium (if ever)