On Wed, 11 Jan 2006 00:13:14 GMT, Cecil Moore wrote:

Owen Duffy wrote:
Is this the antenna described at http://www.qsl.net/w5dxp/G5RV.HTM ?

Nope, that's just an off-the-shelf vanilla G5RV.

In that article, ...

Forget that article which only shows why the *standard* G5RV is
a fairly well matched antenna on 80m and 40m.

But... in that article which recommends 28' of 300 ohm ladder line you
say "To improve the 75m SWR, try installing a 1000pF capacitor (mica
or doorknob) in parallel across the ladder line at the ladder line to
coax junction. Remove the capacitor for all other bands."

IMHO, just considering in isolation what is shown on that page there
is something inconsistent about the Smith chart, the impedances,
lengths, and assertions about the SWR improvement.

....

What I have done on my G5RV is find the point where the SWR circle
intersects the 1/50 conductance circle in the capacitive reactance
region on 3.8 MHz and install a 1000 pf parallel cap there. My series
section line is 22.5 ft. of Wireman #554 at that point. The 50 ohm
SWR is reduced from about 5:1 to 1.3:1 on 3.8 MHz.

This implies you are trying to "tune out" the shunt capacitive
reactance at a point on the line where the shunt resistive component
is 50... but you need the opposite sign of reactance reactance (so
that the susceptances subtract), you need an inductive reactance in
that case.

If "your G5RV" has a feedpoint impedance of 36-j324 (that seems
reasonable), your 22.5 ft. of Wireman #554 will transform that to
21.53-j53.33, and the VSWR in 50 ohm line connected at that point
would be 5.3.

A shunt capacitance CANNOT improve the 50 ohm VSWR at that point

The effect of the shunt 1000pF capacitance is to change the impedance
at the junction to around 3.6-j25, which would cause a VSWR of around
17 in the 50 ohm line.

However:

If the ladder line was around 31' in length, then the Z at that point
would be around 21+j25 (equivalent to 50 ohms R in parallel with +43
ohms X), and a shunt 1000pF (~ -42 ohms X) capacitor would give nearly
perfect VSWR on the 50 ohm line.

In summary, in a general sense, if you want to use a shunt capacitor
as you propose, you need to find length of line such that the
admittance at that point is 1/50-jB (negative susceptance is
inductive), and the correct shunt capacitor has a reactance of 1/B.

Flawed explanation aside, the only way that 22.5' works is if your
feedpoint Z is quite different to 36-j324.

Owen
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