To All from PA0SE
Andy Talbot wrote:
I find that rather impossible to believe - 300m of thick cable being a
dummy
load at 137kHz !
Go back to the fundamental equations and calculate properly rather than
rely
on tables and software used for the wrong purpose .
This may be a better way to estimate the performance....
Inductive reactance of a shorted length of line Xl = Zo.TAN(2 . PI . L
/
vf / Wavelength)
With Wavelength = 2188m, velocity factor = 0.67, Zo = 50
this gives 171 ohms (= 200uH at 137k) you must have slipped a digit
somewhere to get 2mH.
This is the reactance looking into the coax, shorted at the far end and
neglecting any losses.
To get 2mH Xl = 1722 ohms, L = 0.245 wavelength (in air) = 373m of
coax.
Not very much of an increase on 300m and shows how critical the length is
and how fast Xl will change with frequency.
(In fact,since a lot of the numbers above have been rounded and we are
very
close to a shorted quarter wave, a back calculation using the rounded
values
to check gave Xl = 1600 rather than the 1720 ohms used in the forward
calculation - that's how twitchy this technique will be)
For an estimate of losses :
Skin depth of copper at 137kHz is approximately 0.18mm From D =
503
SQRT(Resistivity / Freq / uo)
For Cu Resistivity = 1.7E-8 Ohms / m, and uo (magnetic permeability)
= 1
Diameter of centre conductor = 2.5mm (near enough anyway)
so cross sectional area of conducting path is
0.18mm * 2.5mm = 0.45E-6 m^2
RF Resist = Resisivity * Length / Area = 1.7E-8 * 370m /
0.45E-6m^2 = 14 ohms.
For a quick estimate assume the braid losses are a lot less than the
centre
conductor as they have a much larger surface area, so can be ignored
(although that may not necessaily be the case) and we can also ignore
dielectric losses (a reasonable assumption at these freqs) so Q = Xl / R
= 1722 / 14 = 123
Which is about what I got on my 5mH conventional coil of 1.5mm wire,
300mm
diameter and 400mm long.
In other words, a very expensive, very large and heavy 'coil' - making it
from coax
Andy says "Go back to the fundamental equations and calculate properly". A
good advice but Andy did not follow it himself.
The equation for Xl he used is fine for a lossless transmission line but for
a lossy line the full equation for the input impedance of a transmission
line line of length l, characteristic impedance Zo and terminated by an
impedance Zl should be used:
Zin = Zo x {Zlcosh(gamma) * l + Zosinh(gamma * l)} / { Zlsinh(gamma * l) +
Zocosh(gamma * l)}
I wrote "gamma" in full because my computer refuses to produce the Greek
character for it.
I happily leave it to others to apply this rather formidable equation to
Alan Melia's cable.
I feel not competent to do so but also no need as N6BV's computer program
does that for me as it is based upon that formidable equation.
To check the validity of the computer result I performed a simple test.
When the 984 ft long RG-213 cable is shortcircuited at its end the computer
finds for the impedance (Zin)sc at the input of the cable at 137 kHz:
(Zin)sc = 49.42 + j 172.52 as mentioned before.
When the cable ends in an open circuit the input impedance (Zin)oc is
according to the computer:
(Zin)oc = 2.60 - j 13.72
Now compute the product of these impedances:
(Zin)sc * (Zin)oc = (49.42 + j 172.52) * (2.60 - j 13.72) = 2418 - j 207
According to another basic equation this should be equal to the square of
the characteristic impedance of the cable:
Zo^2 = (50.0 - j 2.3)^2 = 2505 - j 230; pretty near the expected result.
I leave it to experts to judge whether this test is a sufficient proof on
the validity of the computer results.
For me it is enough to stick to my opinion that Alan's cable would be fine
as a dummy load, provided a capacitor with a reactance of -j 172.52 Ohms
(6727 pF) is put in series to tune out the reactance.
73, Dick, PA0SE
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