ahhh.... forgive me if I’m wrong – but all of this discussion was relative to
ELEVATED radials – correct? It does not apply to radials on or in the ground –
right? If so then all the ‘stuff’ published up until now is out the window
which I think is highly unlikely. Having 1/2 WL radials on/in the ground ISN’T
the most inefficient ... it might not be a good use of copper if you only have
a few but it isn’t going to reduce the efficiency compared to the same amount
of radials at 1/4 W.
Gary
I've taken the liberty of opening a new thread, which I probably should have
done before.
To Dan, AC6LA,
First let me also thank you for your civil replies. As you stated, "Refreshing
in this internet age".
I think I have a simple comparison to show the problem, using the 1/4 WL
Vertical (36 ohms R radiation) and 1 kW rf.
First, consider antenna operating over the perfect ground plane. All the power
is radiated. P = I sq X R radiation. So, I = 5.27 Amps.
Then, let's replace the perfect ground with an 8' ground rod, which in my soil
is about 100 ohms.
Now the antenna feed impedance is R radiation + R ground, or 36 + 100 = 136
ohms. Then apply our 1 kW. I is now down to 2.71 Amps. Radiated RF power is now
264.7 W & 735.3 W goes into heating the ground. This illustrates two things.
One, why we don't use a ground rod, but more importantly, when R ground
increases, using constant power, the current and efficiency drop.
Let's just see how much power is required if we were to achieve the same ground
current as the perfect ground model. As the I squared term would be identical,
the power required would simply be the ratio of impedances, or P = 1 kW
(136/36) = 3.78 kW!
Now, realizing that the electrical 1/4 WL radials present the lowest R ground
to the antenna, you can see that greatest efficiency is to be had there.
Likewise, with 1/2 electrical WL radials, R ground is the highest, so
efficiency is the worst.
In Rudy's discussion of a constant current into the radials (and that current
MUST be the same into the vertical), the RADIATED power is constant. However
the ground losses vary considerably. As the radial resistance increases when
exceeding 1/4 electrical WL, Power input must be increased to achieve the same
current. Importantly, as the radials are now not being fed at a current
maximum, but elsewhere on the sinusoidal pattern, larger values will be
observed elsewhere along the radials, increasing their losses.
Hope that helps...
Brian K8BHZ
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