Topband: 160 metre vertical with 'top loading'
Michael Tope
W4EF at dellroy.com
Wed Apr 27 06:26:45 PDT 2011
On 4/26/2011 1:49 AM, Tom Boucher wrote:
> I take your point Yuri, but my simplistic way of looking at the current decreasing along a straight quarter wave of wire is due to the current flow through it's distributed capacitance to ground. Maybe that is wrong and I should go back and look at my transmission line theory.
A matched low-loss transmission line has distributed capacitance to
ground (i.e. x number of pf/foot), but yet it does not have an
appreciable current taper. Only when you introduce an open circuit at
the end does the large current taper appear (i.e. the standing wave).
Similarly it's the open circuit at the tip of the radiator that forces
the current taper. In fact if you imagine a two-wire transmission line
where the wires start out parallel and then are slowly spread apart at
an ever increasing included angle, you'll see that at first you have (at
very small included angles) an impedance transformer (step up to higher
Zo), then as you get to larger included angles (i.e. ~90deg) you have
the equivalent of an inverted-v dipole (well not really inverted unless
you imagine the feedline coming downward toward the ground, but you get
the idea). At an included angle of 180degrees you have a flat-top
dipole. Thus, there is a continuum between the transmission line
paradigm and the antenna paradigm, which can be helpful when trying to
abstract what goes on the antenna element, but you have to be careful
how far you push the transmission line analogy.
> Surely the fact that it is an electrical quarter wave is due to the straight wire being part of one very large coil turn and therefore having an inductance, combined with that distributed capacitance causing resonance at one particular frequency? If that is so, then isn't my theory of the current decay being due to flow through distributed capacitance along it's length still correct?
Consider that the inductance of the wire is also distributed between
those infinitesimal capacitors to ground. This is analogous much more to
a transmission line than a lumped element resonant circuit. The ratio of
the distributed L to distributed C does change along the length of the
wire (i.e. the characteristic Zo of this putative transmission line is
not constant), but is still more like a transmission line than a lumped
circuit. Of course, as I said before you have to be careful how far you
push the transmission line analogy. It is, after all, an antenna :-)
> AA7JV said that the argument against current at both ends of the loading coil being the same, is that you cannot have high current and high voltage at the top of the coil because that represents much higher power than you are sticking into the base of the antenna. That's not true because we are talking about VA not Watts, i.e. there is a phase difference between V and I.
Yes, in fact if you use the resonant series LC circuit analogy, as you
move from source voltage Vs toward the load, the voltage rises across
the series inductor in proportion to I * XL and then tapers back down
across the series capacitor in proportion to I * XC until you reach the
load where the voltage is I*RL which is equal to the source voltage Vs.
There is no current taper in this circuit. The peak voltage occurs at
the common node between the inductor and capacitor and is Vs +jI*XL.
That is the voltage step-up that George was referring to. It does not
necessitate a current taper because of the 90 deg phase shift between
the inductor voltage and the current through it.
73, Mike W4EF........................
> 73
> Tom G3OLB
>
>
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