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Old February 15th 06, 11:44 PM posted to rec.radio.amateur.antenna
Cecil Moore
 
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Default 300 ohm folded dipole from ARRL Handbook, early 1990's

John, N9JG wrote:
The 1986 handbook has the details for this type of antenna on page 33-14,
under the title of "Simple Antennas for HF Portable Operation." For 7.15
MHz, the value of the capacitor is 152 pF, and the length of the matching
stub is 6'11-1/2". An open-end stub, made from twin lead, of length 20'-1/2"
can be substituted for the capacitor.


It might be interesting to some to explain how/why this works.
On a feedline with reflections, there are purely resistive
current maximum points existing every half-wavelength up and
down the feedline. Since the resistance at the current maximum
point is often in the ballpark of 50 ohms, this is often
a logical point at which to connect the transmitter/tuner. That's
why I feed my 130 foot dipole at a current maximum point on each
HF band.

However, for a folded dipole with a 300 ohm resonant impedance
fed with 300 ohm feedline, there are no reflections and therefore
no current maximum points because there are no standing waves.
So the trick is to cause a 50 ohm current maximum point to occur
by causing reflections and standing waves on a short piece of
series matching section. A parallel capacitor can often accomplish
this function. Obviously, a capacitive stub can do the same thing.

Note: the following calculations were done with a paper Smith Chart
and therefore suffer from some inaccuracies.

A parallel capacitive reactance of -j150 ohms will shift the 300+j0
ohm impedance to 60-j120 ohms. Some may want to refresh their
memories on the parallel/series and series/parallel impedance
equations.

60-j120 ohms will cause an SWR of ~6:1 from the point where the
capacitor is installed on a flat 300 ohm feedline to the necessary
1:1 choke/balun. If we divide the 300 ohm Z0 by the SWR, we will
obtain the resistance at the 6:1 SWR current maximum point. That
value is 300/6 = 50 ohms and it exists ~0.062 wavelengths
farther along than the 60-j120 point where the cap is located.

The cap needs to cause an SWR of 9:1 if flat 450 ohm line is used
and 12:1 if flat 600 ohm line in used. Needless to say, the Z0 of
the line determines the value of capacitive reactance that is
necessary to accomplish that function.
--
73, Cecil http://www.qsl.net/w5dxp