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VSWR doesn't matter? But how about "mismatch loss"?

C

Cecil Moore

Jan 1, 1970
0
Buck said:
Please explain your antenna and radio. I am assuming you have a solid
state rig with an so-239 connector on it for the basic and then you do
what?

Hi Buck, Please check out my associated web page and then
ask me anything that you don't understand.

http://www.w5dxp.com/notuner.htm

Contrary to what you may have been told, you can change
the 50 ohm SWR seen by your transmitter by changing the
length of the 450 ohm ladder-line.
 
R

Roy Lewallen

Jan 1, 1970
0
An understanding of "mismatch loss" doesn't require SWR, reflections,
power waves, "reflected power", "reflected energy", or other real or
imagined complexities other than simple impedances. Here's what it means:

If you have a generator with a fixed output impedance such as a signal
generator, and connect it to a conjugately matched load, the power
dissipated in that load is the most you can get in any load connected to
the generator. For example, if your generator produces 10 volts RMS open
circuit and has a 50 ohm resistive output impedance, it can deliver 0.5
watt to a 50 ohm resistive load. If you connect any other load impedance
to the generator, you'll get less power to the load. You can calculate
exactly how much with simple circuit theory.

"Mismatch loss" is simply a way of expressing the reduction in power you
get due to the load being mismatched, compared to how much you'd get
with a matched load. For example, if you connect a 100 ohm resistor to
the output of the generator, it would dissipate 0.44 watt instead of
0.5, so the mismatch loss is 10 log 0.5/0.44 = 0.51 dB(*). If you
connect a 25 ohm resistor to the output, you also get 0.44 watt in the
load resistor, again a "mismatch loss" of 0.51 dB. These numbers are
calculated using nothing more complicated than simple lumped circuit
principles.

Mismatch loss is a useful concept when connecting fixed-impedance
circuits together, such as in a laboratory environment. But it doesn't
apply to either antennas or to VSWR. All you have to do to reduce the
"mismatch loss" to zero is to insert a tuner or other matching network
between the generator and the load. Presto, the generator sees 50 ohms
resistive, the load dissipates 0.5 watt, and the mismatch loss is zero.

(*) For the 100 ohm example: The circuit consists of a 10 volt
generator, and a 50 ohm resistance (the generator impedance) and 100 ohm
resistance (the load) in series. So the current is V / R = 10 / (50 +
100) = 66.67 mA. The power dissipated in the load is I^2 * R = 0.06667^2
* 100 ~ 0.44 watt. No reflections, VSWR, transmission lines, or bouncing
power waves required.

Roy Lewallen, W7EL
 
D

Dan Bloomquist

Jan 1, 1970
0
Roy said:
An understanding of "mismatch loss" doesn't require SWR, reflections,
power waves, "reflected power", "reflected energy", or other real or
imagined complexities other than simple impedances. Here's what it means:

If you have a generator with a fixed output impedance such as a signal
generator, and connect it to a conjugately matched load, the power
dissipated in that load is the most you can get in any load connected to
the generator. For example, if your generator produces 10 volts RMS open
circuit and has a 50 ohm resistive output impedance, it can deliver 0.5
watt to a 50 ohm resistive load. If you connect any other load impedance
to the generator, you'll get less power to the load. You can calculate
exactly how much with simple circuit theory.

"Mismatch loss" is simply a way of expressing the reduction in power you
get due to the load being mismatched, compared to how much you'd get
with a matched load. For example, if you connect a 100 ohm resistor to
the output of the generator, it would dissipate 0.44 watt instead of
0.5, so the mismatch loss is 10 log 0.5/0.44 = 0.51 dB(*). If you
connect a 25 ohm resistor to the output, you also get 0.44 watt in the
load resistor, again a "mismatch loss" of 0.51 dB. These numbers are
calculated using nothing more complicated than simple lumped circuit
principles.

Mismatch loss is a useful concept when connecting fixed-impedance
circuits together, such as in a laboratory environment. But it doesn't
apply to either antennas or to VSWR. All you have to do to reduce the
"mismatch loss" to zero is to insert a tuner or other matching network
between the generator and the load. Presto, the generator sees 50 ohms
resistive, the load dissipates 0.5 watt, and the mismatch loss is zero.

Bravo. And a great deal simpler to understand than most handwaving on
these threads.
Roy Lewallen, W7EL

Best, Dan.
 
R

Robert

Jan 1, 1970
0
Cecil Moore said:
Seems to me that evolution violates that principle.

Not to mention the Noble Prize Ilya Prigogine won for "Dissipative
Structures" in 1977.

Spontaneous Ordered Structures arising out of disorder. But it takes an
Energy Flow to produce them.

Robert
 
M

mg

Jan 1, 1970
0
What I gleaned from the excellent answers for the original "VSWR
Doesn't Matter?" thread is that high VSWR doesn't really matter in a
lossless transmission line environment between a transmitter's antenna
tuner and the antenna, since any reflected RF energy will simply
continue to "bounce" back and forth between the tuner's output
impedance and the antenna's input impedance until it is, finally,
completely radiated from the antenna without loss.

There's also the case where a perfect VSWR does no good. That happens
when you connect a transmitter to an extremely lossy line and the
signal dissipates before it gets to the load :>
 
B

Buck

Jan 1, 1970
0
Hi Buck, Please check out my associated web page and then
ask me anything that you don't understand.

http://www.w5dxp.com/notuner.htm

Contrary to what you may have been told, you can change
the 50 ohm SWR seen by your transmitter by changing the
length of the 450 ohm ladder-line.

Ok, I have seen that. It isn't that the antenna isn't tuned, it is
tuned with the twin-lead instead of a conventional tuner.

I might have enough twin lead to build one of those. I'll have to
consider it. It has always tweaked my interest.

Thanks
 
C

Cecil Moore

Jan 1, 1970
0
Buck said:
Ok, I have seen that. It isn't that the antenna isn't tuned, it is
tuned with the twin-lead instead of a conventional tuner.

It has a tuned feeder instead of a conventional tuner.
 
B

Buck

Jan 1, 1970
0
It has a tuned feeder instead of a conventional tuner.

I am curious to know, have you measured the power both at the antenna
and the radio to see what loss there might be? I wouldn't expect very
much, personally.

On a G5RV, it is taught that the twin-feedline is also part of the
antenna itself, is that true in your model as well?

thanks
Buck
 
B

Buck

Jan 1, 1970
0
http://www.w5dxp.com/notuner.htm

Cecil,

Looking more at your antenna, I am making several observations. I
posted the URL at the top for other readers to know what we are
talking about.

I kind of wish you would change the name of the antenna and not call
it 'no-tuner' because I feel let down every time I look at it and see
the feed-line tuner. I realize it isn't a conventional tuner. Don't
get me wrong, I think it is a GREAT idea and I like it very much and
would like to try it myself. However, I feel it is a little deceptive
in the name.

I have a few questions about the system. I may have asked some
already in another thread, so please bear with me. This is a better
place to discuss it rather than hijacking another thread.

I see you use 450 ohm ladder line (or window-line as some call it.) I
am wondering if the concept will also work with other impedance
feedline such as 600, 300 or 75 ohm twin-line or even possibly with
coax. ****


**** I just looked over your program. I see that you have it setup to
allow 300 or 450 ohm ladder line. I wonder if you have LLWL available
for 600 and 75 ohm.... I will try to put this in XL. (I don't have a
basic program.) OK, I just found the compiled program.

I am interested in making one, but I only have 300 ohm.

Does the feedline act as part of the antenna? I am sure it will, at
least up to the 'no-tuner' if it acts like a G5RV, but do you know if
the feedline radiates?

Have you measured the power at both sides of the 'no-tuner' to see
what loss there might be? I doubt there would be much considering
that you are using window-line.

I see a couple of the frequencies are above 1.5:1, which I am not
comfortable going over with solid state rigs. Do you think that could
be fine-tuned with the addition of a 1/2 foot section or maybe with
that and a 1/4 foot section?

Would it be safe to assume that I can create a mono-band dipole, maybe
even multi-band - if I am lucky, by fixing the length of the dipole
and the feedline such that the increment gives me the imax at the
balun for the desired frequency(s)?

Finally, you have a 1:1 choke at the feeder. I see that the better
quality coax, the more toroids are needed. Would there be a problem
with using a foot of RG-58 with the fewer toroids and then connect
that to the high-quality line? I suspect that by then the common-mode
currents will be gone and there would be minimum loss in such a short
strand of lower-grade coax for HF. I doubt there would be a
noticeable signal loss.

Does the 'no-tuner' feedline need to be spread out. I see from the
photo that your 16 foot section is one large loop, I figure it must be
close to a four-foot diameter loop.

Is there a similar system that would work with a vertical?

Thank you,

I appreciate your taking time to answer this. I hope many of us can
learn from it.

Buck
N4PGW
 
B

Buck

Jan 1, 1970
0
I am curious to know, have you measured the power both at the antenna
and the radio to see what loss there might be? I wouldn't expect very
much, personally.

On a G5RV, it is taught that the twin-feedline is also part of the
antenna itself, is that true in your model as well?

thanks
Buck


Please refer to my new thread. Thanks
 
B

Buck

Jan 1, 1970
0
After reading your webpage more closely......

I see you use 450 ohm ladder line (or window-line as some call it.) I
am wondering if the concept will also work with other impedance
feedline such as 600, 300 or 75 ohm twin-line or even possibly with
coax. ****
Does the feedline act as part of the antenna? I am sure it will, at
least up to the 'no-tuner' if it acts like a G5RV, but do you know if
the feedline radiates?

That answer would be 'no'. I see that the purpose of the feedline
length is to get the maximum current at the center point of the
dipole.


I see a couple of the frequencies are above 1.5:1, which I am not
comfortable going over with solid state rigs. Do you think that could
be fine-tuned with the addition of a 1/2 foot section or maybe with
that and a 1/4 foot section?

Based on the fact that the antenna is being fed maximum current at the
center, I would guess that this should work.
 
C

Cecil Moore

Jan 1, 1970
0
Buck said:
I see you use 450 ohm ladder line (or window-line as some call it.) I
am wondering if the concept will also work with other impedance
feedline such as 600, 300 or 75 ohm twin-line or even possibly with
coax. ****

Might as well go into some detail here. The feedpoint impedances
encountered in a 130 foot dipole may range from ~50 ohms to
~5000 ohms. In order to limit the maximum SWR the feedline Z0
should be ~SQRT(50)(5000) = ~500 ohms. Thus the choice of 450
ohm line. Given that the feedpoint impedances may range from
~50 to ~5000 ohms, here are the maximum SWRs that may be
expected for the different Z0s. Since the antenna system
is fed at a current maximum point, I have included the
impedance at the current maximum point which needs to be
between 25 ohms and 100 ohms in order to avoid foldback.

Z0 SWRmax Imax Impedance
600 12:1 50 ohms
450 11:1 41 ohms
300 17:1 18 ohms
75 67:1 1 ohm
50 100:1 0.5 ohms

It should be readily apparent why coax is a no-no for this
antenna system. Even 300 ohm twinlead will result in a
50 ohm SWR of 50/18 = 2.8:1, high enough to cause foldback.
I am interested in making one, but I only have 300 ohm.

Let's take 40m as an example. The dipole is one wavelength
and will have a 300 ohm SWR of 16:1 on 7.2 MHz according
to EZNEC. 300/16 = 18.75 ohms at the current maximum
point resulting in a 50 ohm SWR of 2.7:1 enough to cause
foldback without an antenna tuner. Such is life.

I do use 300 ohm line on my 20m rotatable dipole and it
does work 20m-10m but there are probably 10m frequencies
where the 50 SWR is too high.
Does the feedline act as part of the antenna? I am sure it will, at
least up to the 'no-tuner' if it acts like a G5RV, but do you know if
the feedline radiates?

I monitor the current balance in my feedlines. They are
so well balanced that there is no hint of common-mode
current on the coax side of my current/choke/balun.
Balanced currents radiate a negligible amount.
Have you measured the power at both sides of the 'no-tuner' to see
what loss there might be? I doubt there would be much considering
that you are using window-line.

No I haven't measured the losses. I have trusted Owen's
transmission line calculator for that data.
I see a couple of the frequencies are above 1.5:1, which I am not
comfortable going over with solid state rigs. Do you think that could
be fine-tuned with the addition of a 1/2 foot section or maybe with
that and a 1/4 foot section?

The disadvantage of this method is that it lacks one dimension
of tuning necessary to achieve 50 ohms. Unless you do one more
impedance transformation, no amount of fine-tuning the ladder-
line length will get any closer.
Would it be safe to assume that I can create a mono-band dipole, maybe
even multi-band - if I am lucky, by fixing the length of the dipole
and the feedline such that the increment gives me the imax at the
balun for the desired frequency(s)?

One such example is at: http://www.w5dxp.com/HEDZ.htm
This antenna works on 75m and 40m with a fixed length of ladder-
line.
Finally, you have a 1:1 choke at the feeder. I see that the better
quality coax, the more toroids are needed. Would there be a problem
with using a foot of RG-58 with the fewer toroids and then connect
that to the high-quality line?

Just use RG-58 entirely unless you are running high power.
Does the 'no-tuner' feedline need to be spread out. I see from the
photo that your 16 foot section is one large loop, I figure it must be
close to a four-foot diameter loop.

I have my 16 foot loop coiled in a 4 turn spiral around a piece
of fiberglas fishing pole.
Is there a similar system that would work with a vertical?

Verticals are not usually fed with ladder-line so probably
more trouble than it is worth.
 
W

Werty

Jan 1, 1970
0
Yes, and it should hurt , because we are using
English and text to show something that
s/b shown in pictures .

Im E.E. , KC7CC, and more..

i can simply show you with pictures.
It will be instantly clear .

---BTW------
Im doin ARM computers , I will give a free
Op System . It will NOT use English nor
ASCII . No arbitrary definitions ...
No C++ , No Linux , No M$ , no
more "Free Lunch" .....

We use Coax for its isolation from nearby
absorbers .. parallel line is much lower loss
but absorbs into other objects close .
We do not use caps , but stubs .
But they are tuned ( freq dependent ) .
The fast way to follow this , is to draw
a picture , then edit it as you go .
English will only get you a college degree
and a free lunch ( Liberals ) .


__________________________________
 
C

Cecil Moore

Jan 1, 1970
0
Bob said:
... while at the same time this overall trend results in
a very localized DECREASE in entropy (increase in order;
in this case, the evolution of complex systems on Earth).

How about considering the localized W5DXP system? :)
 
B

Bob Myers

Jan 1, 1970
0
Dave Oldridge said:
If the lossless transmission line (obviously no such animal exists) were
tuned with a lossless tuner, then VSWR would not matter at all.

Dave, I'm pretty sure I know what you mean here, but it should
be noted that this isn't entirely true. The line would have to have a
couple
of other characteristics besides being simply "lossless" for VSWR
not to matter at all.

The problem, of course, is that a VSWR of other than 1:1 implies
(by definition!) that the voltages and currents along the length of
the line are not constant; there are cyclic variations in each, with
maxima and minima located at half-wavelength intervals (that's
the whole "standing wave" thing in the first place, right?).
Particularly in high-power situations, it is possible for the maxima
to exceed the ratings of the source or of the line itself.

Bob M. (KC0EW)
 
B

Bob Myers

Jan 1, 1970
0
Cecil Moore said:
Seems to me that evolution violates that principle.

Not at all; "entropy increases" is with respect to the total
entropy of a closed system. But in the case of "evolution,"
the only closed system which makes any sense to consider
is the entire solar system, or at the very least the Sun-Earth
system. It is entirely pemissible for an overall increase in entropy
to occur (i.e., the Sun slowly loses energy to the rest of the
universe) while at the same time this overall trend results in
a very localized DECREASE in entropy (increase in order;
in this case, the evolution of complex systems on Earth).

Bob M.
 
R

RST Engineering

Jan 1, 1970
0
Bill ...

I would bet that most of the "confustion" comes from the conditions people
put on their answers to the question.

Some postulate a "tuner" between generator and load.

Some postulate a specific internal impedance of the generator.

Some postulate a specific length of feeder line (either lossless or
resistive, which is another parameter in and of itself).

Some postulate lots of other stuff, almost all of which is valid in the
context of their answer.

What exactly do we mean when we say that we have a "100 watt transmitter"?
What we are actually saying is that the transmitter will cause a resistor of
a specific value to dissipate 100 watts of energy when tied to the
transmitter output port and the transmitter keyed. Let's not muddy the
waters up by asking if we are talking about peak power, PE power, average
power, or whatnot. Let's just assume an unmodulated carrier putting out a
constant power into the resistor that gets just as hot as when 100 dc watts
(E*I) is pumped into it from a battery.

What value resistor? Whatever the designer/engineer/manufacturer specifies.
32 ohms? Sure. 50 ohms? No problem. 300 ohms? Certainly. Any competent
engineer can give you a specified power into a specified resistive load.

The crux of the question becomes, "What happens if my transmitter is
specified into a 50 ohm resistor and I put a 100 ohm resistor as the load?
How much "loss" do I get (or another way of asking that same question is how
much power is dissipated in the 100 ohm resistor)?"

The answer is that it is impossible to tell without making the measurement.
That may seem like a "wiggle" answer, but the truth of it is that the output
stage design of the transmitter will dictate how it handles a "VSWR load".
In some output stages, the output voltage will increase to the point of
nearly driving 100 watts into the 2:1 VSWR resistor. Some will shut
themselves down with a protection circuit. Some will go into parasitic
oscillation. Some will fry the output devices.

Now increase the magnitude (and probably the sign also) of the problem to
toss in a complex impedance instead of a resistive load and the confusion
factor goes up rapidly. What DOES that output stage do when the load has an
inductive component? Or a capacitive component? And both a resistive
component and an imaginary component that varies with frequency?

The simple answer to your question outside the "laboratory environment"
where everything is nicely matched and the internal impedances are set "just
so" is that there IS NO SINGLE RIGHT ANSWER to this simple question.

And that is most probably the cause of your confusion.

Jim





But here again, I'm probably not seeing the
 
D

Dave Oldridge

Jan 1, 1970
0
Dave, I'm pretty sure I know what you mean here, but it should
be noted that this isn't entirely true. The line would have to have a
couple
of other characteristics besides being simply "lossless" for VSWR
not to matter at all.

Not if there's a lossless tuner.
The problem, of course, is that a VSWR of other than 1:1 implies
(by definition!) that the voltages and currents along the length of
the line are not constant; there are cyclic variations in each, with
maxima and minima located at half-wavelength intervals (that's
the whole "standing wave" thing in the first place, right?).
Particularly in high-power situations, it is possible for the maxima
to exceed the ratings of the source or of the line itself.

In which case it's not truly "lossless" (and after breakdown becomes VERY
lossy).
 
T

Tom Ring

Jan 1, 1970
0
Cecil said:
Seems to me that evolution violates that principle.


Duh, life always violates it locally, but makes the sum total of entropy
higer than it would have been. You aren't the idiot you appear, I hope.

tom
K0TAR
 

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