On Wed, 29 Jun 2005 17:28:01 -0400, "Tam/WB2TT"
wrote:

It is the directional coupler that is balanced for a particular value of Z0.

Tam/WB2TT
*****

Correct

james

Frank Gilliland wrote, among other things, "The point is that the error
is insignificant when the directional coupler is much shorter than the
wavelength."

Certainly "directional couplers" for HF may be built at essentially
zero length, and ideally would have exactly zero length, monitoring the
current and voltage at a single point on a line. Then SWR or
reflection coefficient magnitude or even complex reflection coefficient
may be calculated under the assumption we know the desired reference
impedance. But if the equipment combines the voltage and current
samples in the wrong ratio, you will get the WRONG answer. Even if the
coupler looks like a perfect 50 ohms impedance section of transmission
line (with some attenuation), the error _in_measurement_output_ can be
significant indeed. Just because the coupler looks like a 50 ohm line
to the line it's hooked to doesn't mean it will read zero reflection
when IT's presented with a 50 ohm load.

And by the way, not everyone who measures and cares very much about SWR
(or reflection coefficient) cares a whit about field strength. Not all
loads are antennas.

Indeed, as Reg says, we might do better in amateur applications to
consider the SWR meter as an indicator of the degree to which we're
presenting a transmitter with the desired load. That's really what
we're using it for, most of the time. It may ALSO be interesting to
know the field strength, but please be aware that a transmitter's
distortion products may be significantly higher if it's presented the
wrong load impedance, even though the power output may be increased.
Field strength alone is not acceptable to me as a means to adjust an
antenna load to a transmitter, or as a way to adjust the operating
point of the transmitter.

Cheers,
Tom

james wrote:
On Wed, 29 Jun 2005 11:07:17 -0400, "Fred W4JLE"
wrote:


What is the reason a 2:1 SWR can cause such havoc?

How can I avoid this catastrophic condition?

I feed my dipoles with 450 Ohm ladder line, but the last 20 feet or so is 50
Ohm coax, I guess that makes it work ok. I haven't blown up my finals yet.

Lions and tigers and bears Oh my...

*****

Actually can happen if you push the finals to where there is
insufficeint margin to the maximum heat dissapation. Tubes are a bit
more forgiving. Transistor inadequately heatsinked and overdriven,
typical CB usage, often have little of no margin for heat dissapation.

If the transmitter has a refelction coefficient of zero and the load
say .3, then that reflected power from the load is dissapated as heat
in the output circuits and any final transistors or tubes. Now if the
radio has a reflection coefficient other than zero that will lessen
the heat dissapation on the transimiiter. Now you get load and source
reflections convoluting within the transmission line.

You ought to model a 400 Mhz square wave with source and load
refelctions coefficients other than zero. It can get ugly


james



Consider the MRF 140, a 150 Watt 2.0 - 150.0 Mhz N-Channel
linear RF power fet. From the technical data sheet: "100% Tested
For Load Mismatch At All Phase Angles With 30:1 VSWR." You'd have
a tough time damaging this device with a mere 2:1 VSWR.
How do load and source reflections convolute within the
transmission line? That's a new one on me. My old dictionary
defines 'convolute' as "Rolled or folded together with one part
over another; twisted; coiled." The rest of the post is pretty
fanciful, too. A trip to the library would do wonders.
73,
Tom Donaly, KA6RUH

james wrote:

On Wed, 29 Jun 2005 16:42:49 -0500, Cecil Moore
wrote:


Sometimes yes, sometimes no. If the reflected current arrives out
of phase with the forward current, then the final dissipation can
actually be *reduced* by the mismatch.

*****

Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

QED

james


Cecil was talking about current, not power. You can't add
power the way you can voltage and current. If you could, you
could build a very nice perpetual motion machine just by using the
reflections in a transmission line to add power so that the output
was greater than the input.
73,
Tom Donaly, KA6RUH

The rig has no way of detecting any alleged "reflected power". It can't
tell the difference between a feedline with a lot of "reflected power",
a feedline with no "reflected power", and a plain resistor. It behaves
exactly the same in all cases, provided only that the impedance that
each provides to it is the same.

Anyone not convinced of this should put a couple or more dummy loads in
series or parallel, make up a few lengths of transmission line of
various impedances, and see for himself.

Roy Lewallen, W7EL

james wrote:
On Wed, 29 Jun 2005 16:42:49 -0500, Cecil Moore
wrote:


Sometimes yes, sometimes no. If the reflected current arrives out
of phase with the forward current, then the final dissipation can
actually be *reduced* by the mismatch.

*****

Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

QED

james

On Wed, 29 Jun 2005 13:56:53 -0700, Frank Gilliland
wrote:

I'm sure you have studied the lumped-constant equivalent of a
transmission line.....
****

And Maxwell's equations

james

On Wed, 29 Jun 2005 14:24:44 -0700, Frank Gilliland
wrote:

And #4 is exactly why #1 is incorrect: the 'characteristic' impedance
of a coax is constant, but it's 'input' impedance varies according to
load mismatch at the other end. If it wasn't for this fact, a tuner at
the radio end would be useless. But the point here is that if the SWR
meter is left floating with the coax shield (both of which should be
RF grounded) then the measurement can be darn near anything.
*****

What I will agree with is that the impedance seen at the input to the
coax is a reflection of the load impeadance as transformed, altered if
some don't like the word transformed, by the length of the coax. I
have no problem with that. Still this impedance is highly dependant on
the load and its reflection coefficient.

Correct the characteristic impedance of the coax never changes and
that was what I intended with #1.


james

On Wed, 29 Jun 2005 16:37:10 -0500, Cecil Moore
wrote in :

Frank Gilliland wrote:
Point is that they are usually calibrated for Z0=50 ohms
and are in error when used in Z0 environments differing
from Z0=50 ohms, e.g. Z0=75 ohms.

The point is that the error is insignificant when the directional
coupler is much shorter than the wavelength.

Nope, that's not the point at all. It is true that a 50 ohm
SWR meter designed for HF may not work on 70 cm but the error
I'm talking about is the calibration error in a 50 ohm SWR meter
designed for HF and used on HF in, for instance, a Z0 = 450 ohm
environment instead of its calibrated-for 50 ohm environment....


There lies our misperceptions; I was not referring to using an HF SWR
meter designed for coax and plugging it into 450 ohm ladder line.






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On Wed, 29 Jun 2005 13:56:53 -0700, Frank Gilliland
wrote:

Well, let's put it this way: the radio and antenna don't connect to
the dielectric of a coax.
*****

No it does not directly.

I know this concept is not easy to see but at the begining of the
coax, one can then consider the energy that travels down the coax as a
TEM wave. It is inside the dielectric where the E and H fields of the
traveling wave can be measured and found.

In transmission lines it is by far easier to think of E and H fields
within the the transmission line. Once that concept is mastered then
the rest is rather easy.

When the wave reaches the end, you have the final induced currents.
You can take a dipole and look at it as if the legs were an extension
of the transmission line. This can better be seen if you consider a
dipole and it is fed with open twin lead. The leads of the dipole then
are an extention of the twin lead except they are now at 90 degrees to
the transmission line.

Current is high when the magnetic field is high. This is so because
the induced current is controlled by the density of the magnetic
field. The E field is high when magnetic field is low. The E field
does not require current but voltage. On a center fed dipole the
impedance is low and the corresponding currents are high. The E field
off teh antenna is also low. As you progress a quarter wave from the
feed point in either direction the H field increases and the E field
decreases. With increasing H field the RF currents induced in the
antenna are high. Thus Ohm's law hald true. Z = I^2*R. Where R is the
radiation resistance of the antenna. The ends of a center fed dipole
are high impedance.

james

On Wed, 29 Jun 2005 22:58:51 GMT, "Tom Donaly"
wrote:

james wrote:

On Wed, 29 Jun 2005 16:42:49 -0500, Cecil Moore
wrote:


Sometimes yes, sometimes no. If the reflected current arrives out
of phase with the forward current, then the final dissipation can
actually be *reduced* by the mismatch.

*****

Power is power. Phase is not a problem. Take the mafnitude of the
transmitted power and teh magnitude of the reflected power. The
results are phaseless. The magnitudes add linearly.

QED

james


Cecil was talking about current, not power. You can't add
power the way you can voltage and current. If you could, you
could build a very nice perpetual motion machine just by using the
reflections in a transmission line to add power so that the output
was greater than the input.
73,
Tom Donaly, KA6RUH
******

Tom

The problem is that current is not reflected back from the load, power
is. Thus the you can add magnitudes of power.


james