----- Original Message -----
From: "Dan Levin" <djl@andlev.com>
> I'm trying to design a coil that will be used in a 1500 watt filter at HF.
> I know that I need to allow for frequency related skin effect that will
> reduce the area of the wire or tubing carrying the current.
>
> First, I check the ARRL Handbook and the Antenna Book. No formula and no
> table showing suggested wire sizes for a given frequency and power level.
>
> Next, I determine that the skin depth at 28 mHz is .0005", and I calculate
> the copper volume that the current is traveling through at one skin depth
> in
> wires and tubes of various sizes. I get results that seem unreasonable -
> using a current density of 2500 amps / square inch I get a current
> capacity
> for #14 wire at 28 mHz of under .5 amps and for 1/2" tubing of just a
> couple
> of amps. Yes, I realize that the skin depth is one standard deviation,
> but
> even allowing for that the capacities seem very low.
>
> Can someone point me at either a formula or a table that will tell me or
> allow me to calculate what the current capacity of a wire or tube is at a
> given HF frequency? Or, I suppose to put it differently since the
> capacity
> is related to the acceptable temperature rise - something that will tell
> me
> the copper loss in a piece of wire or tubing at a given frequency?
>
The amount of heating you will see in your inductor will depend on
the ratio of the loaded vs. unloaded Q of the inductor. In circuits like
TVI low-pass filters, the loaded Q is very low (something like 0.5),
so the amount of heating will be relatively low compared to a high
Q circuit operating at the same power level for the same value of
unloaded Q. If you look at the inductors in a 1500 Watt TVI low-pass
filter you will see that they are pretty wimpy in comparison to a tank
inductor in a 1500 watt HF amplifier. The difference is the loaded Q.
The loaded Q in a HF amp tank circuit is normally around 10. This
is why the upper HF section a 1500 watt amplifier tank coil will be
made out of 1/4 copper tubing, whereas the center conductor of
RG8X coax is puny by comparison even though it can carry 1500
watts at 28 MHz with only modest warming.
As per the SM0AOM, I think all the equations you need are in
Terman.
73 de Mike, W4EF...................
> Thanks!
>
> ***dan, K6IF
>
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