Guy Olinger, K2AV k2av@contesting.com
Wed, 9 Feb 2000 20:00:34 -0500

```This problem is usually modeled as a SERIES resistance. In a series
circuit the same current flows through each resistor, call it I. The
power dissipated in each branch is I squared times R. Since the I
squared is the same in each branch, the ratios of power dissipated in
each branch to the total power are the same as the resistance ratios to
the total resistance.

So if the ground resistance is 35 ohms, the radiation resistance is 12
ohms and the ohmic loss is 3 ohms, then the radiation efficiency is
12/(35+12+3) = 24%. (This is a common set of real life figures.)

On a 100 watt signal only 24 watts gets radiated. That's 6 dB down from
a near lossless antenna (e.g. 2" copper pipe 1/4 wave radiator with
radials in a salt water marsh on 40 meters).

[ Oh yes, and our 24% (in)efficient, 6 dB down antenna has a perfect 1:1
swr at resonance, and since the major resistance in the equation
(ground) is very broadband, it will have a low swr across the whole
band. That's what really counts....  Sorry, just couldn't resist. ]

- - . . .   . . . - -     .   . . .     - - .   . - . .

73, Guy
k2av@contesting.com
Apex, NC, USA

----- Original Message -----
From: EUGENE SMAR <SPELUNK.SUENO@prodigy.net>
To: <towertalk@contesting.com>
Sent: Friday, February 04, 2000 9:31 PM

>
> TowerTalkians:
>
>      Here's a question the answer to which is mainly of academic,
rather
> than practical, interest.  But I'll ask it anyway:
>
>      In several treatises on RF ground fields for vertical antennas
> especially, (see any antenna handbook and
the term
> "antenna efficiency" has been defined as the ratio of radiation
resistance
> (the "good" antenna resistance) to the total feedpoint resistance
> plus ground plus Ohmic loss resistances).   We all, then, go on to
calculate
> the efficiencies of our resultant vertical antenna systems as the
product of
> this "efficiency" and the power input at the feedpoint of the antenna.
>
>      But is this the correct procedure?  As I imagine this Radiation
> Resistance, I see three "resistors" in series:  the ground and Ohmic
losses
> plus this Radiation Resistance - a voltage divider circuit.  Taking
the
> above ratio looks like I'm calculating the voltage drop across that
one
> Radiation Resistance value in a three-valued voltage divider circuit.
>
>      If this is truly what is being modeled when we say "Radiation
> Resistance", ought we not to calculate the POWER efficiency of the
actual
> antenna as this ratio SQUARED?   If we are truly to use the ratio
as-is (not
> squared), then this value can be used HOW?  Efficiency of WHAT?  Not,
in my
> estimation,  the net power delivered into that Radiation Resistance
and sent
> on its way to ZL8.
>
>      Looking at a simple example, if I calculate this ratio for my
> hypothetical 80 meter vertical, I might come up with a figure of 1/3.
This
> is 33.3% "efficiency" according to the above definition.  But I'm
arguing
> that I ought to SQUARE this percentage figure (33.3% X 33.3%) to come
up
> with a POWER efficiency of 11.1%.  I say this  because the power is
applied
> to all three "resistors" in my antenna model but gets radiated by only
1/3
> of the total resistance for 1/9 of the applied power.
>
>      In a practical antenna, we would all work toward lowering the
ground
it is
> when we're done.  But I hope someone can explain what the term
"efficiency"
> in all of these articles really means and what can we do with the
number.
>
> 73 de
>
>
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