Then the essence of this of this is what effect does a base loading inductor
have on the current profile on the radiator. My answer is none.
It's too bad we can't have attachments here so I'll try to describe it with
words. I "built" a model of a vertical in AutoEZ and used EZNEC Pro/2+ V 7.0
running the NEC 5 engine to do the computations. I used 7.25 MHz with a height
of 1/8 wavelength at that frequency (16.55 feet), diameter of 0.1 inch and 33
segments. I used perfect ground and conductor loss of zero. The current source
was 1.0A.
The feedpoint Z without any loading is 6.15 -j444.6 (SWR = 651). Using the
current plotting feature of AutoEZ, the current profile v. segment is a nearly
linear, negative slope line from (almost) 1.0A down to (almost) zero. (Almost
because current is reported along the length of a segment, not at the ends) I
used the snapshot feature to capture this plot.
Next I added a series inductor at the feedpoint of 1.48 +j444.6 (Q=300). The
calculated feedpoint Z=7.64 j0 (SWR = 6.55). The current profile is nearly
identical as before. The snapshot completely overlays the newly calculated plot.
My conclusion: A non-radiating, high Q inductor at the feedpoint of an
electrically short monopole affects the feedpoint Z but has no effect on the
current distribution on the radiator, which is determined entirely by its
geometry. A load placed anywhere on the radiator other than the feedpoint is a
different story.
Wes N7WS
On 7/3/2025 4:19 PM, David Gilbert wrote:
The loss is the same (probably), but the current profile at the antenna is
different (definitely).
Dave AB7E
On 7/3/2025 4:10 PM, Wes wrote:
I'll confess some difficulty (probably my failing) in understanding your
point(s). To try and simplify the situation to something my pea brain may
understand, let's use a toroid at either location. Now tell me what the
difference is.
Wes N7WS
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