-=> On 03-05-98  21:01, Michael Warner said to Russ Wuertz,<=-
-=>"About VSWR...,"<=-
 
Hi, Michael;
 
[Extranious stuff snipped]
 
 MW> ... Seriously, who can give me an explanation,
 MW> (preferably a site or file on the  internet), of Voltage Standing
 MW> waves as they apply to an active transmitter connected to an untuned
 MW> antenna? 
 
This would be a lot easier if I had charts, graphs and drawings to refer
to.   :-}
 
Ok, I guess we can start with an antenna which IS operating at resonance,
then see what happens when the frequency changes.
 
As you probably know, electricity in a conductor does NOT travel at
infinate speed: It travels at the speed of light, 186,000 miles per
second, or a little slower under certain conditions, such as in an
antenna feedline.  Every frequency has a unique wavelength.  For
instance, at 147 MHz, a full wavelength is aproximately 2 Meters, (where
the band gets it's name), or about 76 inches.  So a quarter-wave antenna
at this frequency is aproximately 19 inches long.  Assume that there is
no feedline in this case, that the transmitter is right at the base of
that quarter-wave antenna.
 
Assuming that a single straight-wire antenna is connected directly to the
output of the transmitter, Starting at the instant when the output
voltage of the transmitter crosses zero and starts going positive.
The "wave" of positive voltage travels down the antenna toward the end.
Just as the "wave" reaches the end of the antenna, the voltage at the
transmitter end has reached it's maximum value.  (Say, 100 volts).  Since
the end of the antenna is not terminated, it will reflect back toward the
transmitter, and reach it just as the voltage coming out of the
transmitter again reaches zero, this time going toward negative.  At this
time the voltage at the far end of the antenna has reached maximum.  This
is a quarter-wave antenna at resonance.  In practice there are two
antenna elements of the same length and in opposite directions.  This is a
half-wave dipole.  Under these conditions, the transmitter "sees" a
resistance load of 52 Ohms, and the antenna is said to be in tune, or in
resonance.  So far so good?
 
Now if we change the frequency without changing the antenna, the outgoing
wave will either not reach the end of the antenna, or will have arrived
and been reflected back toward the transmitter before the output of the
transmitter completes it's half cycle, so that when the wavefront reaches
the transmitter, the output will either still be dropping toward zero or
will have already reversed and be going negative.  In either case, the
transmitter will not see a resistance for a load, but will see a
reactive componant: either a capacitor or an inductor and a resistor.
The value of the resistor that the transmitter sees is the radiation
resistance of the antenna - and relates to how much signal is actually
transmitted.  The value of the reactive component will depend on how far
off resonant frequency the antenna is.  A reactive componant will store
energy and feed it back to the source, in this case, the final amplifier
of the transmitter.  If the voltage feeding back into the final is
sufficently high, there may be arcing in the case of a tube amplifier,
and in any case, the final amplifier may be damaged.  (Transistors are
especially prone to damage from high VSWR!)  Also, tube-type amplifiers
have tuning adjustments which will make up for some error in the antenna
resonance, while transistorized amplifiers do not.  Thus it is more
important to have either an accurately tuned antenna or use a matching
network when using a transistorized transmitter than it is with a tube
unit.
 
When you use an antenna matcher, which also contains reactive components,
it is adjusted to cancel out the reactance of the mis-tuned antenna, and
present as close as possible to a 52 Ohm load to the transmitter.
Unfortunately, all that excess power is NOT sent back to the antenna to
be radiated, but it is dissipated in the matching network as heat!  So if
you transmit with an antenna which is WAY off resonance and use a matching
network with a 100 Watt transmitter, you might be radiating 30 watts and
converting 70 watts of heat in the matching network!  So it is MUCH more
efficent to have an antenna which is exactly 1/4 wavelength long at the
operating frequency, but if that is not possible, at least have it close
to resonance, so that your matching network has very little work to do.
 
OK, since we don't want the antenna sitting on the transmitter, but would
like to move it some distance away, now we add a coaxial feedline to the
system.  We shall use a 52-ohm coaxial cable - an insulated center
conductor, with a woven shield around it, protected by a plastic jacket.
This coax, (in this case), has an impedance of 52 ohms, the same as the
antenna, and is the same as the transmitter is designed to have for a
load.  (Feedline impedance is a whole 'nuther lesson.)   :-}
 
The case of a resonant antenna being fed with a coax is exactly the same
as when it is connected directly to the transmitter.  The voltage at all
points along the coax will be the same.  However when we change frequency,
we get a new kettle of fish.  Those reflected waves coming back from the
end of the antenna go zipping back up the feedline to the transmitter at
about 2/3 the speed of light!  The farther off resonance the antenna is,
the more voltage is reflected along the coax.  This will set up a series
of voltage peaks and valleys along the coax, which will also generate heat,
and in severe cases, can actually cause arcing inside the coax!  The ratio
between the voltages at the voltage node, (highest voltage on the coax),
and the voltage null, (the lowest voltage), is the Voltage Standing Wave
Ratio, or VSWR.  What a VSWR bridge does is separate and read the outgoing
and returning voltage waves.  From those two readings, the VSWR is
calculated and displayed on one or two meters or a digital display.
 
Does this answer your question?
 
73 DE KB9QPM
   Ivy
 
 
 
... Double your dipoles. Double your flux.
 
                --
 
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