Re: [TowerTalk] ground losses (vertical antennas)

[Jim Lux <jimlux@earthlink.net>]

At 04:23 PM 6/12/2004 -0700, Al Williams wrote:

> ----- Original Message -----
> From: "Tom Rauch" <w8ji@contesting.com>
> 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". 

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