[TowerTalk] Radiation Resistance
Guy Olinger, K2AV
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. ]
- - . . . . . . - - . . . . - - . . - . .
Apex, NC, USA
----- Original Message -----
From: EUGENE SMAR <SPELUNK.SUENO@prodigy.net>
Sent: Friday, February 04, 2000 9:31 PM
Subject: [TowerTalk] Radiation Resistance
> Here's a question the answer to which is mainly of academic,
> 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
> http://www.bencher.com/pdf_download.html#tech_notes as examples),
> "antenna efficiency" has been defined as the ratio of radiation
> (the "good" antenna resistance) to the total feedpoint resistance
> plus ground plus Ohmic loss resistances). We all, then, go on to
> the efficiencies of our resultant vertical antenna systems as the
> 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
> plus this Radiation Resistance - a voltage divider circuit. Taking
> above ratio looks like I'm calculating the voltage drop across that
> 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
> antenna as this ratio SQUARED? If we are truly to use the ratio
> squared), then this value can be used HOW? Efficiency of WHAT? Not,
> estimation, the net power delivered into that Radiation Resistance
> 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.
> is 33.3% "efficiency" according to the above definition. But I'm
> that I ought to SQUARE this percentage figure (33.3% X 33.3%) to come
> with a POWER efficiency of 11.1%. I say this because the power is
> to all three "resistors" in my antenna model but gets radiated by only
> of the total resistance for 1/9 of the applied power.
> In a practical antenna, we would all work toward lowering the
> resistance component by adding radials and the efficiency is whatever
> when we're done. But I hope someone can explain what the term
> in all of these articles really means and what can we do with the
> 73 de
> Gene Smar AD3F
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