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Re: [TenTec] was OT: Indoor Antenna: re B&W type terminated dipoles

To: Discussion of Ten-Tec Equipment <tentec@contesting.com>
Subject: Re: [TenTec] was OT: Indoor Antenna: re B&W type terminated dipoles
From: Ken Brown <ken.d.brown@hawaiiantel.net>
Reply-to: ken.d.brown@hawaiiantel.net, Discussion of Ten-Tec Equipment <tentec@contesting.com>
Date: Tue, 07 Dec 2010 20:08:09 -1000
List-post: <mailto:tentec@contesting.com>


"Efficiency" is a term well understood by most professional engineers as: Efficiency = Pout/Pin
Yep,
If we are talking specifically about an antenna's efficiency - not an antenna *system* - the expression becomes: Efficiency = (Radiated Power)/(Power input) = (Power input - Losses)/(Power input)
Just be sure to separate the radiated power at the RF frequency that the transmitter produces from the power radiated as infrared wavelengths (heat), and you've got it right. Which is what you are saying when you say:
The only losses in a simple antenna are I^2*R losses in the element(s). If we take a resonant 20m half-wave dipole constructed from #14 wire, its efficiency is about 98.5% (-0.06dB); if we operate the same antenna on a different band - say 17m - the efficiency improves to 98.9% (-0.05dB).


The efficiency of a Double Bazooka will be lower than that of a half-wave dipole because of extra losses in its coax stubs.
Absolutely. There are a lot of people who think that poor dummy loads (ones that radiate a lot of RF) are really efficient, because they have a low SWR over a wide band. Like for instance 75 and 80 meters. Some of us know better than to think the SWR numbers are the golden standard by which we judge antennas.

Of course there are various ways of defining efficiency. Some would claim that it is more efficient use of the operators time if he/she never has to readjust a matching network when tuning from one end of the band to the other. But that is not antenna efficiency.

It all depends how you draw the lines. A physicist or thermodynamic engineer could say that all antennas are 100% efficient, because all of the power delivered to them is radiated, even if some of it is radiated as heat. As radio engineers we don't think of it that way.

You could also claim that an antenna low to the ground, or close to other objects, so that the RF field generates a lot of heat in the ground or the other nearby objects, is just as efficient as a similar antenna in free space, because the heat is not due to antenna losses, but rather due to what happens to the energy AFTER that antenna radiates it efficiently as an RF field. In this case the installation as a whole is inefficient, even if the antenna is not where the inefficiency lies.

DE N6KB

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