Hi Bill,
> Hmm. I base this suggestion on the observation that others had offered
> that the beads nearest the feedpoint of the balun were heating up and
> destroying themselves, while beads away from the feedpoint were fine.
I wonder about that also. Roy Lewallen and a bunch more besides me would
sure like to see one do that first hand!!!
I have had beads in closed areas do that when hung vertically, the upper
beads gets hottest first. When I've turned the balun upside down, hanging
the load BELOW the balun, the beads at the load end run cooler.
> I see now that the suggestion infers that each bead is doing something in
> isolation -- removing current from the sheild and generating heat. But
> that's not true, as you point out.
Yep. It CAN'T be true because things don't work that way. Herr Kirchoff
will come out of his grave and haunt us if we violate his law.
> Since that is the case, why DOES the bead closest to the feedpoint get
> the hottest and fail first?
My own thought is heat rises, and the vertical bead stack gets hottest at
the top first. If you stacked low loss beads at the top, it would certainly
even the heat out!
In my 160 phasing network, I had a string of beads over a flopped coax to
generate 180 degree phase inversion. The beads were horizontal, and spaced
a half inch or so in a random stack. All the similar high loss tangent
beads cracked at the same time. Lower loss tangent beads mixed in did not
get hot at all.
> >I don't understand why that would be true. It is a simple series
circuit,
> >and power dissipation is I^2 R.
> >Since current is equal throughout the string, heat would be proportional
to
> >the ESR (R) of each bead. It always works that way, unless Kirchoff and
Ohm
> >were wrong and their laws are just suggestions.
>
> You're absolutely right. For some reason, I was thinking each bead works
> in isolation.
Only if it has an appreciable shunting electric field between each bead and
some outside path to the source of the common mode excitation. We know that
isn't true!
73 Tom
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