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Old November 18th 06, 09:03 PM posted to rec.radio.amateur.antenna
Owen Duffy Owen Duffy is offline
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First recorded activity by RadioBanter: Jul 2006
Posts: 168
Default Operation of bootstrap coax traps

On Sat, 18 Nov 2006 16:01:55 GMT, Cecil Moore
wrote:

Dave wrote:
a lot of the performance probably depends on the frequency vs length of the
coax. if the frequency is low enough that the delay in the coax is small
then you basically have a capacitor (the inner conductor to the inside of
the shield), in parallel with an inductor (the outside of the shield). it
gets a bit uglier because of capacitance between turns on the outside, but
that is likely much smaller than the internal capacitance. so you end up
with a basic parallel resonant circuit. at high frequencies where the
length of coax is longer than a small fraction of a wavelength that probably
falls apart and would take much more complex analysis to figure out.


This thing is much easier analyzed as a distributed network
transmission line. At the resonant frequency where I2 is 180
degrees out of phase with I1, assuming |I1|=|I2|, common-mode
currents are eliminated so there are none available at 'B'.
On frequencies where I2 is not 180 degrees out of phase with
I1, common-mode currents flow on the outside braid of the trap
and thus continue to flow down the antenna wire at point 'B'.


I designed a trap using VE6YP's calculator. The trap was designed for
resonance at 7MHz, used Belden 8262 (RG58C/U) on a 50mm dia former. I
assumed Q of the inductor is proportional to f^0.5, and that Q at 1MHz
was 60, which gives a Q of around 160 at 7MHz. I think the Q
assumptions are realistic considering the effect of the braided
conductor and proximity effect of the close space turns.

Analysis of that trap using my model suggests that at resonance (which
is 2% lower than predicted by the calculator) |I1||I2|, |V2| is a
little less than |V1|. The coax is not a 1:1 impedance transformer or
current transformer in any way, shape or form. If one examines the
standing wave on the coax, at trap resonance, a current minimum and a
voltage maximum occur the A end of the coax.


The key to understanding the operation of this trap is to
realize that current ceases to flow on the outside of the
trap braid at the trap's designed-for frequency.


I suggest that at resonance (if that is what you meant), impedance is
a maximum, I1 becomes small, |I2||I1|, and the current flowing on
the outside of the coax (I1+I2) is not zero.

Owen
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