A reasonably general expression for the RF current capacity of wire or tubing
at a given frequency
can be derived as follows:
First, it will be assumed that the current limitation is due to temperature
rise of the
conductor, and that the very good thermal conductivity of copper will
spread the dissipated heat evenly over the cross-section of the conductor.
Then, the equality P(dissipation) = I(dc)^2* R(dc) = I(rf)^2* R(rf) = constant
can be considered valid.
If power dissipation is considered constant, then the current limit for a given
dissipation can be solved from:
I1(dc)^2* R(dc) = I2(rf)^2* R(rf); where I1 is the DC current limit for a
given temperature rise,
and I2 is the corresponding RF current limit.
Rearranging: I1/I2 = Square root(R(rf)/R(dc)).
The ratio of R(rf)/R(dc) for circular copper wire can be calculated (using
Metric units) from:
R(rf)/R(dc) = 0,0038 * D * square root (f) + 0,26 ; where D is the wire or
tubing diameter in mm, and f is frequency in Hz.
The final expression is then Imax(rf) = Imax(dc)/(square root (0,0038 * D *
square root (f) + 0,26)
A worked example:
A copper wire of 3 mm diameter in free air is rated for about 50 A DC current
for a temperature rise of 45 C.
The same wire would then be able to carry 50/(square root (0,0038 * 3 * square
root (30E6) + 0,26) =
50/7,9 = 6,3 A RF at 30 MHz for the same temperature rise.
This translates to about 1,5 - 2 kW of CCS RF through a Pi-network with a Q of
12.
Hope this helps and 73/
Karl-Arne
SM0AOM
----- Original Message -----
From: "R@contesting.com;Measures" <r@somis.org>
To: "Dan Levin" <djl@andlev.com>
Cc: <amps@contesting.com>
Sent: Friday, April 15, 2005 10:43 AM
Subject: Re: [Amps] Skin Effect and Wire Current Capacity at HF
>
> On Apr 14, 2005, at 10:22 PM, Dan Levin wrote:
>
> > 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.
>
> E. F. Johnson rated its #14 gauge Cu roller coil at 3A @ 30MHz.
> >
> > 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?
> >
> > Thanks!
> >
> > ***dan, K6IF
> >
> > _______________________________________________
> > Amps mailing list
> > Amps@contesting.com
> > http://lists.contesting.com/mailman/listinfo/amps
> >
> >
>
>
>
> Rich Measures, 805.386.3734, AG6K, www.somis.org
>
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