[AMRadio] antenna tuners transmision lines and more

John Coleman ARS WA5BXO wa5bxo2005 at pctechref.com
Sun Apr 23 19:58:33 EDT 2006

```	You are correct Gary, it is very confusing to many and I was one
confused guy for many years.  It's not easy to get a grip on things of
this nature.  Invisible radiation and weird parts that have no movement,
makes it all seem like wizardry and magic.  Of course this is what makes
it fascinating.  I'll just add some more to the confusion.

I am by no way a XPERT on this stuff but I have been told that I
have a way with words as long as I can get a spell checker working.  I
have been asked to do some of this writing.  I feel that I should share
this with others and I have chosen this place to do it.  I don't have a
lot of opportunity to go get on the air much any more.  I get stuck here
at home watching kids once in a while and so this is when I type up
these long stories.

So please excuse the long winded transmission here.
I hope some one gets something from it.

FALSE STEAMENT #1 -- A high SWR reading is an indication that a lot of
power is wasted and not being radiated. -----
TRUE STATEMENT --- SWR is the ratio of currents measured at physical
points on a transition line.  It is the ratio of the maximum current on
the line verses the minimum current on the line.  These two physical
points will be 1/4 electrical wavelength apart.  They do not necessarily
have to be at the load end or the source end.  IF the load end is
representative of a pure resistive load then the SWR will be the ratio
of the load resistance to the line characteristic impedance.  If the
load resistance is non reactive and equal to the line characteristic
impendence then the SWR is 1:1 and current will be the same at any point
on the transmission line that you care to measure it except for the
normal loss due to line characteristics.  Even a perfectly matched
load:line such will have slightly less current and voltage at the load
end than at the source end although as some one earlier pointed out, "It
is generally a negligible difference".  It would need to be a very long
line to be significant on 80 or 40 meters.

FALSE STATEMENT #2 --- There is no need for a tuner if the antenna is
resonate and the line is matched.
TRUE STATEMENT -- If the antenna feed point is equal to the line Z and
the transmitter is made to work into this load then there may be no need
for a tuner.  This is an almost impossible task as some one pointed out
earlier, and even if it were to be done it would only be true for a very
small range of frequencies.  QSY would be a compromise.

FALSE STATEMENT #3 --- Tuners waste a lot of power and just make the
transmitter think the antenna is right.
TRUE STATEMENT --- A tuner consists of coils and capacitors neither of
which by mathematical definition consumes energy.  The adjustments of
the coils and capacitors change the phase as well as the voltage to
current ratios of input and output.   The slight amount of energy that
may be consumed by tuners is generally so negligible that it is very
difficult to measure.  In some cases a tuners components maybe made of
poor quality material and too small for the job.  These types of
components will get hot.  Heat is an obvious point of loss.  I had a
small MFJ tuner that was manufactured some years ago. It was just a
small external Pi-Net device and I found it to have a measurable
insertion loss.  It turned out to be the rivets that held the connectors
on the little chassis.  I soldered braid across the connectors to the
chassis and then the loss was then immeasurable.

Modern solid state equipment is designed to work into a 50 ohm non
will cause the rig to put out less RF current and make the automatic
drive level circuitry start pulling back on drive prematurely.  If the
load becomes slightly reactive as well then the RF production will
decrease rapidly.  A tuner is nearly a must for these rigs.

In tube type XMTRs the use of toroidal transformers for the output is
impossible because of the high output Z of tubes.  These rigs used
instead a Pi-NET or link coupled tuned circuitry to do the job of
matching the tube to the low impedance output.  This type of circuitry
could match a relatively wide range of impedances from 25 ohms to
several hundred ohms as well as compensate for some reactance.  Because
of this an external tuner may not have been necessary especially if
confined to one band on one antenna.  A lot of folks put up multiple
antennas one for each band or used a multiband trapped dipole or some
other multiband radiator with a single coaxial down line.  The Pi-Net in
the rig did all the compensation for them.  But with solid state rigs
and no internal tuning it would be an near necessity to have an external
tuner if nothing more than a small PI-Net tuner such as the one I had
from MFJ

Having to do with the conservation of energy laws.  Here are some facts.

1. High quality capacitors (especially air or vacuum type with good
aluminum plates) have little or no measurable loss.  They give almost
100% of the energy they absorb back to the load or source.  They are
adjusted with the inductors so as to send the energy to the load and not
the source.

2. Air inductors are also almost lossless except for a small amount due
to the resistance of the material.  The energy they absorb is stored
magnetically and almost all given back to the load or source.  They also
are adjusted with the capacitors so as to send the energy to the load
and not the source.

3. Antenna systems (including tuners) are made of material that is very
low in resistance to electron flow (or they should be).

With the above facts in mind, consider the following scenario.

1. A transmitter is connected to an antenna system made with quality
components

2. The finals are not dissipating any more heat than they would if
connected to a perfect dummy load.

3. There is no measurable heat dissipated in any of the components of
the antenna system.

Then the energy that is produced from the finals must be being
used by something irregardless of resonance.  The energy must be going
to out into space because nothing is dissipating any heat that we can
measure and it makes no difference what length the antenna is because th
tuner is compensating for the reactance and transforming the current to
voltage ratios as needed to get the energy out.

Most folks mistakenly think of the term radiation resistance as
a fixed value of 73 Ohms.  BUT THIS IS NOT TRUE.  73 Ohms is the
radiation resistance of a center fed 1/2 wave dipole in free space and
by the way increasing the size of the wire has very little effect on it.
A center fed full wave dipole will radiate the same amount of energy but
has a much higher radiation resistance.  It has no greater or less
radiation efficiency than does the 1/2 wave dipole (negligible copper
resistance loss).   It just radiates in a slightly different pattern.

Theoretical, (neglecting copper losses) if all of the energy of
the radiated signal could be recaptured and measured from each of the
two antennas the measured amounts would be equal.

Here is some question that I have never learned the answer to.

I have never seen a value of radiation resistance assigned to a
center fed full wave dipole.  Perhaps it is too difficult to measure?
As Don,K4KYV pointed out, "There has to be some current flow there, else
there would be no power transferred"

I would also like to know the theoretical feed point resistance
of a theoretically infinite length dipole and why a rhombic is
terminated with a 600 ohm resistor instead of, for the sake of argument,
say a 100 Ohm resistor or some other value.

I understand that Rhombic and long wires (10 wave lengths or
more) radiate 90% of there energy before the signal reaches the end of
the wires. And that the terminating resistor is there to lower
reflections that might make the antennas bi-directional.  So could that
mean that 600 ohms is about the Radiation resistance of a infinite
length of wire?

Here is a little tidbit that may not be well known.

Don, K4KYV, once explained to me, the reason for the 300 ohm
feed point of the folded dipole.  It went like this.

There or two wires which must divide there current evenly.

Consider a 100 watt carrier gong into a 1/2 wave dipole.

With the 73 ohm radiation resistance the current at the feed point would
be about 1.17 Amps and the voltage would be 85.4 Volts

If another wire is added to make the antenna into a folded dipole then
each wire would have a current at the center of (1.17 / 2) or .585 Amps.

But since only one wire is fed then in order to get the 100 watt value
the voltage must be double to 170.8 volts.

R = E/I so 170.8 volts divided by .585 Amps equals about 292 Ohms.
Hence the value rounded to 300 Ohms.

It is some what like an impedance ratio of 1:4 for a 2 wire folded
dipole.

In a 3 wire folded dipole the currents will be divided by thirds per
wire.  This will work out to near 600 Ohms for a 3 wire folded dipole.

**********************************************************************

-----Original Message--Edited for space---
[mailto:amradio-bounces at mailman.qth.net] On Behalf Of Gary Schafer
Sent: Sunday, April 23, 2006 11:49 AM

When using an antenna tuner and we tune the reactance out of the circuit
so
we only see a resistive component we can say it is resonant. But what is
really resonant? The antenna is not, the feed line is not, the tuner is
not.

The only thing resonance means in this case is that the capacitive and
inductive reactances at the tuner are equal. It confuses many people.

73
Gary K4FMX

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