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## Re: [TowerTalk] ground losses (vertical antennas)

 To: "Al Williams" , Re: [TowerTalk] ground losses (vertical antennas) Jim Lux Fri, 11 Jun 2004 17:06:05 -0700
 At 04:23 PM 6/12/2004 -0700, Al Williams wrote: ```----- Original Message ----- From: "Tom Rauch" snip > The ground losses in my large F-250 Super > Cab long bed truck with the antenna mounted....``` ```I am having difficulty accepting the concept of ground losses that Tom (and many others in articles and books) refer to.``` ```On the one hand, the vertical is often depicted sticking up out of the ground and having dotted rays of antenna current (field?) flowing out from it on their way to ground and back through ground to the source connection. Even though the surface path through ground is humongous, I can rationalize that there is still resistance and as is often mentioned some radiated power is heating the earth!``` ```On the other hand, what bothers me is that current flowing back through ground must be equal to the current flowing into the antenna and thus radiated. It seems the radiated current from the antenna is split. Part goes into space and part goes into ground. If the ground resistance is reduced it would seem that more of the antenna current would go into ground and less into space?????``` help! k7puc A couple things to think about: 1) The current flowing in the ground radiates too. That's what the "image" is all about, so ground current, per se, isn't bad. 2) Consider the whole "circuit" as a loop... up the vertical, through the space, to the ground, back through the ground to the feedpoint. Each and every one of those pieces has some loss (although the resistive loss of the "space" part is quite low!). This has NOTHING to do with the radiation resistance, which is essentially in series, and is primarily (solely?) determined by the physical shape of the "loop". It is technically incorrect, but useful, to think of it as a special case of a system where you have a big electrode suspended over the ground (forming a capacitor), connected to the feed point by an inductor. The ground resistance is in series with the "other" terminal of the capacitor (i.e. the ground). As a circuit, it follows all the usual rules... the current in the loop is proportional to the voltage applied divided by the total resistance. The power dissipated in each "resistance" is proporational to the resistance, etc. Let's consider two cases, where the physical structure is basically the same, so the radiation resistance is the same, assumed here to be, say, 20 ohms. Case 1: Real good ground, real good antenna conductors. Say the total "loss resistance" is 5 ohms. The total resistance in the system is 25 ohms. If you put 100 Watts in, 80 watts will be radiated and 20 watts will go to heating up things. Case 2: Crummy ground, antenna made of leftover christmas tinsel and wet wood, so the resistance is, say, 80 ohms. The total system resistance is 80+20 or 100 ohms. Put in the same 100 watts and the power divides: 20 watts gets radiated, 80 watts to heat. A very different situation.. Note that the feed point impedance is different in these two cases! You'd need some sort of matching network to make your 100Watt source have an output impedance of 25 or 100 ohms, as the case may be. The current's also different. In the first case, the current is 2 amps, in the second it's 1 amp. You can use these currents to get at the losses a different way: 2 amps (squared)*5 ohms = 20 Watts (Case 1). In a REAL system, it would be hard to separate out the losses from "radiated power absorbed in the earth" and "ground return currents" because they are really all one and the same in the near field. It's the same as losses in any other part of the antenna. The current in some part of an antenna is partly there because of conduction (from an adjacent part of the antenna) and partly because it's induced by currents in some other part of the antenna. When you get out into the far field, then you can talk about "ground losses" due to absorbing the radiated energy. There is yet another aspect, and that is that most "real" RF sources aren't really constant power, constant impedance devices, particularly if there's some reactance in the system. This is particularly true of receivers, where the actual sensing mechanism might be a high impedance voltage sensor (whip feeding a FET) or a low impedance current sensor (loop feeding a BJT). In a receiver, the goal is usually to optimize the signal to noise ratio, and that would seem to imply that you want a conjugate match (i.e. maximize the transfer of power from the antenna to the receiver) but it might well be that using a different load impedance (the receiver input impedance) might reduce the internal noise power more than the reduction of signal power from the "mismatch". _______________________________________________ See: http://www.mscomputer.com for "Self Supporting Towers", "Wireless Weather Stations", and lot's more. Call Toll Free, 1-800-333-9041 with any questions and ask for Sherman, W2FLA. ```_______________________________________________ TowerTalk mailing list TowerTalk@contesting.com http://lists.contesting.com/mailman/listinfo/towertalk```
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