> From: Peter Chadwick <Peter.Chadwick@gpsemi.com>
> The arguments about folding raising radiation resistance or not are
> interesting. If it doesn't alter the radiation resistance, what is the
> radiation resistance (as opposed to feed impedance) of a folded dipole?
You can find that in Electromagnetic Waves and Radiating Systems,
Jordan-Balmain, on page 403 in "Antenna Fundamentals".
The radiation resistance of a thin freespace folded dipole of any
number of wires, using the IRE standard definition, is about 73
ohms. You can also find this in early commercial texts dealing with
the folded dipole.
> 75 ohms? It's interesting that Lamont V. Blake in his book 'Antennas'
> (Wiley, 1966) says that 'If there is no ohmic loss in the antenna -
> that is, if all the input power is radiated - then the radiation
> resistance referred to the feed point is equal to the resistive
> component of the antenna input impedance'.
You'll probably go blind reading all the esoteric definitions, uses,
and mis-uses of radiation resistance. I spent years researching the
topic, and I can quote many "definitions". Three of the most common
are:
1.) The terminal resistance at any point the user chooses in the
radiating system in a lossless system.
2.) The resistance at the current loop in a system in a lossless
system.
3.) The total power radiated as EM energy divided by the square of
effective current causing that radiation taken at the current maximum
of the radiator.
Number three is the IRE method as defined in the late 1940's.
No matter which definition you use, the results are all the same.
Folding the element does NOT increase efficiency nor does it reduce
ground currents.
The error most people make is they take the loss resistance (like it
is a resistor they can hold in their hands), and use that loss
resistance in comparisons against a value of radiation resistance
taken at another point or a different way in the system. Such misuse
is like measuring the loss resistance at one end of a transformer,
and comparing the UN-transformed loss resistance to a TRANSFORMED
impedance on another separate winding.
Efficiency, radiation resistance, or loss resistance can not be
determined without knowing the intensity of the radiated field in any
practical SW, HF, or MW frequency system. When measuring the terminal
impedance of any antenna, while fussing around with the radiator
or ground system, we really have almost no idea what we are doing. I
can make a poor efficiency 65 foot vertical antenna that has a
combined base and radial resistance of two ohms, or two hundred ohms.
The base impedance tells me NOTHING, except what the base impedance
is.
> Certainly raising the feed impedance can help reduce matching network
> losses by reducing the large currents.
If you use a nichrome coil or a conventional inductor operating at
very high loaded Q's that may be true, but most conventional
matching networks operate at low loaded Q values and are so
efficient even modest components produce nearly 100 percent
efficiency.
Do this test.
1.) Turn a 100 watt table lamp on for five minutes, then quickly shut
off the power and touch the bulb. This will give you an idea
what 100 watts of lost power "feels like" in that amount of surface
area.
2.) Now do the same with your transmitter into your worrisome
antenna matching system. Shut off the transmitter and quickly touch
the base matching components.
I suspect most of us won't be burning our fingers on our matching
systems. Even one dB of loss would quickly destroy a matching
transformer or matching network at legal power levels.
Finally, let's consider what you REALLY do when you add a folded
element, and not use any "buzz-words".
Assume you have a 1/4 wl monopole over a perfect ground, and have one
ampere of radiator current. The current flowing into the ground
terminal HAS to equal the current in the base of the monopole (if the
feedline is not radiating). You have one ampere flowing into the
ground terminal, and one ampere in the radiator. Let's assume the
power is 30 watts. The base impedance is 30/1*1, or 30 ohms. The
feedline voltage is 30 volts and the current is one ampere.
Now add a second radiator very near the first one of the very same
size, connecting them like a "folded unipole". Feed them with the
same power. The current equally divides between the two wires, and
they each now carry 1/2 ampere. The impedance of any ONE wire is now
30 watts/ .5*.5, or 30 over .25...... meaning the impedance of
any ONE wire at the base is 120 ohms! The current flowing into the
ground is one half ampere from each wire, for a total of one ampere!
The feedline current is one half ampere, and because the termination
impedance of the feedline is four times higher the voltage is twice
as high. The feedline power remains 30 watts, the ground current
remains one ampere, the NET current in the radiator (divided
between two conductors) is still one ampere, the total power radiated
is still 30 watts.
The only thing that changed is the feedline current is now 1/2 ampere
at 60 volts, which is STILL 30 watts.
We could do the same thing with an L network, or a simple
transformer. But make no mistake about it, no matter what method we
use ground loss would remain EXACTLY the same.
Adding ground loss into this equation makes no difference at all.
Since ground CURRENT remains the same, ground loss remains the same
in either system. Any thoughts to the contrary are absolutely
incorrect.
73, Tom W8JI
-
---
FAQ on WWW: http://www.contesting.com/topband.html
Submissions: topband@contesting.com
Administrative requests: topband-REQUEST@contesting.com
Problems: owner-topband@contesting.com
|