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Author Topic: Determining maximum power for homebrew antenna  (Read 3392 times)
N3WWN
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Posts: 9


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« on: April 05, 2001, 01:37:23 PM »

Hi!  I was wondering if anyone could give me a little info on homebrew antennas.  

I love building antennas!  I like the idea of homebrew equipment, but at this point in my life, I am not comfortable building complex equipment..  Antenna building is fairly simple with measurements and tuning probably being the most complex aspects.

Now, on to my questions:

How can I determine a safe usable maximum power for a homebrew antenna?

What are the factors which influence the max power that an antenna can handle?  I know wire gauge is a major factor, but I don't know what sizes will handle what amount of power.

Thanks and 73,

-Rich - N3WWN
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WB6BYU
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Posts: 13334




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« Reply #1 on: April 05, 2001, 04:20:15 PM »

There are several factors which limit the power an antenna can
handle.  For a wire dipole, you can assume a 50 ohm feedpoint and
calculate the current as a function of the power using Ohm's law.
For example, 100 watts results in around 1.4 amps on the wire.
The ARRL handbook has a table of the maximum current carrying
capacity for different wire sizes - I don't have it handy, but use the
value for free air.

Since most wire antennas will have a radiation resistance higher
than 50 ohms, the wire size you calculate for a dipole will be on the
safe side for longwires, quads, etc.  The only exceptions would be
some wire yagis and small loops (and other high-Q antennas),
which can have a  much lower radiation resistance.  These
antennas need larger conductors for reasonable efficiency, and this
will probably limit the use of small wire for these antennas anyway.

The next thing to consider is the voltage gradiant at the ends of
the antenna.  The higher the power (and the higher the Q of the
antenna) the higher the voltage at the ends.  There are two
conditions to be concerned with.  The first is the insulation between
the end of the wire and surrounding conductors.  For most antennas
at the 100 watt level, only a small insulator is needed, and, in most
cases, a length of synthetic rope will suffice.  But relying on the
wire insulation alone may no be enough.  The other condition is
corona discharge from sharp ends ons the antenna - generally this
is not a problem at amateur power levels, but it can happen,
particularly at high altitudes and with sharp ends on a high-Q
antena.  The solution is to use rounded ends or plates on the ends
of the elements to spread out the charge.

One remaining possible power limitation is the dielectric strength
of the insulating materials used to build the antenna.  I've heard
stories of loading coils wound on PVC pipe which melt at high
power, for example.  (This is highly dependent on the additives
in the specific piece of pipe being used.)

You can estimate the voltage across the feedpoint using the
feedpoint impedance and ohm's law.  Again, presuming 100 watts
and a 2000 ohm feedpoint impedance (such as a center fed full
wave dipole) this is several hundred volts, which only requires a
small insulator.

Antenna traps containing a capacitor will have a power limit
which is dependent on the voltage rating of the capacitor (and, in
some cases, the maximum current that the capacitor can pass.)

But if you are just interested in experimenting with wire antenas,
and you aren't trying anything fancy, then the truth is that you can
use quite small wire.  I've heard of folks running 100 watts into wire
as small as #36, but one problem with wire that small is that the
birds can't see it in time to avoid it, so it often doesn't last long.
(I suspect that #36 is not rated to carry the calculated 1.4 amps,
but it might work on SSB where the duty cycle is very low.)  
Somewhere around #28 is a good practical limit for experimenting,
since the wire becomes more difficult to work with if it gets any
smaller.
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N3WWN
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Posts: 9


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« Reply #2 on: April 19, 2001, 01:25:50 PM »

Thanks for the info!

I have since obtained a copy of the ARRL Handbook (circa 1988, but the info is still valid).  There is a table which states the maximum continuous amps that different wire gauges can handle.

How would I apply this type of table for items such as spade lugs and such?

Thanks and 73!

-Rich - N3WWN
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WB6BYU
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Posts: 13334




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« Reply #3 on: April 23, 2001, 12:43:28 PM »

Most lugs these days are designed to crimp onto a cable, and the
insulation on the crimp section is color coded to indicate the range
of wire sizes for which the crimp is designed.  For example, I think
blue is something like #22 to #16 wire.  A properly designed lug
should handle the maximum current for at least the mid-sized wire
in it's size range, if not the largest size.  

When in doubt, find the narrowest portion of the lug which needs to
carry current, and measure the cross-sectional area of the lug at
that point.  A good start is that the lug should handle as much
current as a wire of equivalent area.  There are, of course, other
effects to consider, such as the material from which the lug is made
and whether (like a ground lug) it is connected to a large thermal
mass which can help to dissipate the heat.  But this should be a
reasonable first estimate.

Many lugs (such as the spade lugs found in many car electrical
systems) also have a current rating, typically 5 or 10 amps or more.
The equivalent power rating in an antenna will depend on the
current distribution at the point where the lug is located.  In a long
wire, or the outer half of a dipole, the current is fairly low, but the
same amount of power in a small loop antenna will create much
higher currents.
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WT8Y
Member

Posts: 65




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« Reply #4 on: April 30, 2001, 08:15:01 AM »

To summarize, with the average current levels in amateur service, if the wire will support itself you don't have much to worry about...consider 100 watts/50 ohms= 1.4 amps X duty cycle SSB (30%) or CW (50%) = .42 amps.average current SSB or .7 amps avg. current CW..even at KW powers 18/16ga wire will handle it with ease.
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