> If your coax is the same impedance as your transmitter, but
> different than your load, can it still be a transformer
> though?
It will ALWAYS act as a transformer when the cable's Z does not match the
LOAD Z. The SOURCE device (transmitter) plays NO part in the transformation
that happens. The Z at the source end of the line is a function of only
three things: the load Z (antenna, duplexer, whatever), the cable's
characteristic Z, and the cable's electrical length (in
degrees/wavelengths/radians/whatever).
Cable loss also plays into it, as its effect is to always bring the load Z
closer to the cable's characteristic Z as cable loss goes up, but let's
leave that out for the sake of simplicity.
The source device's impedance does not factor into the equation at all when
it comes to determining what the resultant Z is at the source end of the
line. Any mismatch that occurs at that end of the line (the
source/transmitter end) does NOT affect the VSWR on the line, nor does it
change the Z at the source end of the line.
It's very important to not put the cart before the horse here. While it may
seem to be contrary to instinct, the device that is sourcing the power is
NOT what determinates what happens on the transmission line with respect Z
and VSWR, it's the LOAD mismatch that sets up the standing waves and the
resulting Z's along the length of the line. It is for this reason that your
transmitter CANNOT "detune" your duplexer, and for the same reason, you
can't "tune" (change the Z) of your antenna from the far end of the cable.
> Is it possible to transform a load that isn't 50 ohms to 50 ohms using 50
ohm coax?
NO.
> Yes your right VSWR is the ratio between Vmax and Vmin, node
> and anodes, of the interference pattern caused by standing
> waves. Even still there is a point where the voltage is at a
> minimum on the line. What happens if that point is at the
> transmitters output... does it help keep the heat down in the
> transmitter due to high SWR?
NO. VSWR on a transmission line doesn't directly manifest as "heat" in a
transmitter. The whole notion of high VSWR creating heat in a transmitter
is likely based on a drop in efficiency in SOME transmitters when they are
not properly matched to the feedline. You can have a very high VSWR on the
feedline, and provided the transmitter is matched to the Z at the source end
of the line, the efficiency will not suffer, and there will be nother ill
effects (including "heating") that occur within the transmitter.
A PROPER MATCH DOES NOT OCCUR WHEN THE TRANSMITTER'S SOURCE Z IS THE SAME AS
THE CABLE'S CHARACTERISTIC Z EXCEPT WHEN THE LOAD Z ALSO MATCHES THE CABLE'S
CHARACTERISTIC Z. Or, rewritten, if the load Z (antenna) is not 50+j0, a 50
ohm transmitter delivering power into a 50 ohm transmission line will NEVER
be matched. (Didn't mean to shout, but it's important to understand and
accept that fact.)
Sidebar: VSWR is a conveniently-simplistic scalar value, but it doesn't tell
you anything about the specific impedance at a particular point along the
line, nor what the ratio and phase relationships are between voltage and
current. You can calculate VSWR based on a specific Z, but you can't do the
reverse, except in the case of a true 1:1 VSWR. All Corvettes are Chevys,
but not all Chevys are Corvettes. You get the idea. Many hams like to talk
in terms of VSWR as it's a nice easy number to deal with, and the lower the
number, the better. But that oversimplification seems to also translate
into a relaxation in the attention paid to the theory behind what's really
going on along the transmission line, at the source-to-line interface, and
at the line-to-load interface. When it really comes down to the nuts and
bolts of designing and building matching networks, you don't care about the
VSWR value, you need to know, think, and design based on complex impedances.
Back in the good ol' days when your rig running 6146's had a pi matching
network on its output, you could tune the transmitter into some not-so-good
loads, often in excess of 3:1 VSWR, without any problem. Why? The pi
network did its job, matching the high-Z (and somewhat reactive) output Z of
the tubes to the whatever-Z existed at the end of the antenna feedline. You
could get efficiency just as good at a 3:1 VSWR as you could at 1:1 because
the output of the matching network provided a conjugate match.
--- Jeff
