EMRFD Message Archive 7233
Message Date From Subject 7233 2012-02-10 08:24:11 Jim Miller Tapping a tank?
puzzled for a while but I think I'm finally getting it figured out.
The 40m match was a straight forward L-match and I could replicate the
values with TLW.
I attempted to do the same for the 160 and 80 meter sections before
realizing that those configurations were quite different. For those the
author chose to resonate the capacitive reactance of the antenna with a
suitable inductor which was then tapped near its bottom end,
autotransformer-style, to match to 50ohms.
My question: is the tap point calculable in general or is it a matter of
trial and error with an antenna analyzer? Basically "you know it's down
there somewhere, go find it."
I read thru EMRFD and the earlier RFD and couldn't find much on this topic.
Ditto the ARRL Antenna Book and Handbook.
tnx and 73
jim ab3cv
[Non-text portions of this message have been removed]7234 2012-02-10 16:43:06 William Carver Re: Tapping a tank?
a first approximation of the tap point by pretending the coil is an
ideal autotransformer: multiply the total number of turns by the square
root of antenna resistance divided by 50 ohms. That is, if the antenna
resistance was 5000 ohms, the square root of 5000/50 = 10, s you'd tap
up 10% on the coil.
If the coil is a toroid with fairly good coupling between turns, this is
fairly accurate. I would a coil on a T200-6 core whose reactance would
cancel the slight capacitance of my close-to-a-halfwave vertical, then
putting the tap at the point indicated by the calculation above.
If you don't know the antenna resistance then it becomes a matter of cut
and try with whatever instrumentation you have. Toughest would be an
SWR bridge, 'cause then you don't know if you're too high or too low and
have to diddle around more to figure out where the tap goes.
W7AAZ7235 2012-02-11 09:00:36 John Marshall Re: Tapping a tank?
First, we need to find the Q needed to match 10.688 to 50 ohms:
Q = SQRT( 1 + 50 / 10.688 ) = 2.3829
Now that we know the Q, the reactances are:
XLshunt = 50 / Q = j20.983 and XCseries = 10.688 * Q = -j25.468
So, at 1.9 MHz the shunt inductor is 20.983 / ( 2 * pi * 1.9e6 ) = 1.7576 uH.
Now we need an inductive reactance to cancel most of the antenna's -j508.59 and leave a result of -j25.468:
XLseries = -j25.468 - ( -j508.59 )= j483.12,
and at 1.9 MHz the series inductor is 483.12 / ( 2 * pi * 1.9e6 ) = 40.469 uH.
Of course, that's just an example and I took the liberty of ignoring the shunt capacitance used in the article to tune the network. Including that makes the calculation more complicated. I would do it by converting the antenna's impedance to admittance, do the same with the tuning capacitance, combine them, then convert back to impedance and calculate the L-network as above.
If you'd rather not go through the mental exercise, the calculati7236 2012-02-11 09:22:38 Jim Miller Re: Tapping a tank?
that was very helpful!
thanks!
jim ab3cv
[Non-text portions of this message have been removed]7237 2012-02-11 09:28:12 dan edwards Re: Tapping a tank?
As a crude method to 'gitter done', the easy thing to do is first resonate the antenna, slightly above the desired freq. then, add the shunt inductor, and tweak for best match. and perhaps re-tune the series L to center the frequency in the range of interest.
to me, the amount of inductance across the feedpoint seems, intuitively, remarkably small....like 'wow, this is almost a short, the XL is so small.....'
but it works nicely...FWIW....73 w5xz, dan