Hi Gary, sounds like you are being hung up by use of series reactance in
the case of the d.e. impedance vs parallel reactance for the hairpin.
I don't have a modeling program for this but doing it the old fashioned
way, the d.e. R component can be transformed from the series form to the
parallel form by R par = R ser times (Q sqr +1). Assuming 22 ohms R, it takes
a
Q of 1.128 to transform to 50 ohms. The Q is still 1.128 in the parallel
form so the parallel equivalent is 50R shunted by 44.3 capacitive reactance.
These numbers agree closely with what you stated.
A Q of 1.128 requires a series X of 22 times 1.128 which equals -24.8
reactance. This is the number provided by shortening the d.e.
The resulting parallel form is R50 in shunt with -44.3 ohms. The shunt
reactance is then canceled by the hairpin inductance.
The hairpin can accurately be modeled as a transmission line. The handbook
formula is XL = Zo times Tan of the length in degrees. Zo is the
characteristic impedance of the line and is defined as 276 times the log of (2
x
spacing) divided by diameter. 1/2" conductors with three inch spacing would be
297.9 ohms characteristic impedance for example.
If we need 44.3 ohms inductive reactance, the required length then is 8.46
degrees. If the frequency is say 14.25 MHz, 360 degrees (one wavelength) is
984/14.25 = 69.05 ft or 828.6 inches. The hair pin length would be
(8.46/360) X 828.6 = 19.47 inches.
If you want the hairpin to be longer, then the Zo has to be less which is
done by reducing the spacing.
By the way, the hairpin can become part of a balun by grounding the short
to the boom and running the coax up one of the hairpin tubes. The coax
shield has to be connected to the same place the hairpin short is connected to
the boom. At the feed point end, the coax center conductor connects to one
side of the d.e. and the shield to the other side. Making this water
resistant might be a challenge.
Hope this makes sense and is helpful.
73,
Gerald K5GW
In a message dated 5/22/2015 12:24:00 P.M. Central Daylight Time,
rxdesign@ssvecnet.com writes:
Ok.... I read the Antenna Handbook... it says to use a chart they have –
reading it at 52 ohms line impedance (coax)... and 22 ohms RL I need 42 ohms
of reactance.
Then shorten the driven element to produce enough capacitive reactance to
cancel out the inductive and transform the impedance up to 52 ohms.
So I enter 42 ohms as a center load in my driven element (6ft/-6ft length
load set to 50% and is 50% ... ) ... then shorten the length of the driven
element. But this doesn’t work. The load never gets transformed.
What I’ve been doing is enter their length/diameter info from their
manual... then adjust the driven element to get the SWR where I need it
(centered) ... then I read the load resistance off the SWR chart and change
the “
alternate’ source impedance to be this value. Finally I add in just enough
hairpin to cancel out any capacitive reactance. This works – I get the same
SWR curves that are published by M2 when I model their antennas... so I know
its ‘right’. But I can’t seem to now NOT use the ‘switch’ to a different
line impedance in order to determine the correct hairpin XL so I can figure
out how to build the hairpin!
any one see what I’m doing wrong please?
This is using EZNEC5
Gary
K9RX
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