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

B

billcalley

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.

But then why does the concept of "mismatch loss" exist in
reference to antennas? I have quickly calculated that if a
transmitter outputs 100 watts, and the TX antenna has an impedance
that will cause a VSWR of 10:1 -- using lossless transmission line --
that the mismatch loss in this "lossless" system would be 4.81dB!
(Reflected power 66.9 watts, RL -1.74).

Since mismatch loss is the "amount of power lost due to
reflection", and is as if an "attenuator with a value of the mismatch
loss where placed in series with the transmission line", then I would
think that VSWR would *definitely* matter, and not just for highly
lossy lines either. But here again, I'm probably not seeing the
entire picture here. What am I missing??

Confused!

-Bill
 
C

Cecil Moore

Jan 1, 1970
0
billcalley said:
Since mismatch loss is the "amount of power lost due to
reflection", and is as if an "attenuator with a value of the mismatch
loss where placed in series with the transmission line", then I would
think that VSWR would *definitely* matter, and not just for highly
lossy lines either. But here again, I'm probably not seeing the
entire picture here. What am I missing??

If the system is Z0-matched, e.g. antenna tuner, there is
a mismatch gain at the tuner that offsets the mismatch
loss at the load so, in a lossless system, nothing is
lost. Wave cancellation toward the source is balanced
by constructive interference toward the load.
 
B

billcalley

Jan 1, 1970
0
If the system is Z0-matched, e.g. antenna tuner, there is
a mismatch gain at the tuner that offsets the mismatch
loss at the load so, in a lossless system, nothing is
lost. Wave cancellation toward the source is balanced
by constructive interference toward the load.

Now my head *really* hurts! This is a VERY confusing subject, to
say the least. (And I also thought antenna tuners actually had a
*loss* due to their limited Q...I think I'm going to change careers
now and just become a pet groomer; or perhaps simply give up
completely and work at Radio Shack).

-Bill
 
C

Cecil Moore

Jan 1, 1970
0
billcalley said:
Now my head *really* hurts! This is a VERY confusing subject, to
say the least. (And I also thought antenna tuners actually had a
*loss* due to their limited Q...I think I'm going to change careers
now and just become a pet groomer; or perhaps simply give up
completely and work at Radio Shack).

Real-world antenna tuners do have a loss but we previously
specified a lossless system. Of course, real world tuners
and transmission lines suffer losses but we all just live
with those losses while striving to minimize them. The point
is that an antenna tuner reflects most of the reflected
energy back toward the load thus accomplishing a mismatch
gain that offsets some, if not most, of the mismatch loss.
High SWR transmission lines are indeed lossier than flat
matched transmission lines of the same material.
 
C

Charles Schuler

Jan 1, 1970
0
billcalley said:
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.

That's basically true but ignores the stress that can be placed on the
output circuit of the transmitter. Why do you think VSWR shut-down circuits
are so popular? One can see rather high voltages or currents that are
potentially damaging to the transistors and capacitors.

Now, since "lossless" is an abstraction and all materials have voltage and
current limits, just make this easy on yourself and always strive for a VSWR
of 2:1 or less. It simply works better and is more reliable.
 
C

Cecil Moore

Jan 1, 1970
0
Charles said:
Now, since "lossless" is an abstraction and all materials have voltage and
current limits, just make this easy on yourself and always strive for a VSWR
of 2:1 or less. It simply works better and is more reliable.

But renders many all-HF-band dipoles useless. :) I regularly
run up to an 18:1 SWR on my 450 ohm ladder-line. Owen's
transmission line calculator says I'm losing about 0.8 dB
in 100' of line under those conditions on 40m. IMO, it's
a small price to pay for all-HF-band operation.
 
C

Charles Schuler

Jan 1, 1970
0
Cecil Moore said:
But renders many all-HF-band dipoles useless. :) I regularly
run up to an 18:1 SWR on my 450 ohm ladder-line. Owen's
transmission line calculator says I'm losing about 0.8 dB
in 100' of line under those conditions on 40m. IMO, it's
a small price to pay for all-HF-band operation.

How many Ham transmitters have a balanced output?

How are you feeding a balanced line?

If you are using an antenna tuner with unbalanced in (50 ohms) and balanced
out (variable impedance), you should be OK in most situations.
 
J

Jimmie D

Jan 1, 1970
0
Charles Schuler said:
That's basically true but ignores the stress that can be placed on the
output circuit of the transmitter. Why do you think VSWR shut-down
circuits are so popular? One can see rather high voltages or currents
that are potentially damaging to the transistors and capacitors.

Now, since "lossless" is an abstraction and all materials have voltage and
current limits, just make this easy on yourself and always strive for a
VSWR of 2:1 or less. It simply works better and is more reliable.
If a tuner is placed directly after the TX and properly adjusted the TX
will always see a 50 ohm load and the shutdown circuit will always be
happpy. Again as long as the TX sees a match there is no unusual stress
placed on it. Remember that before the invention of coax cable SWR was
rarely considered. Instead the tx was tuned for proper established
operational parametrs and all was right with the world.

1:1 SWR CAN MEAN YOUR COAX IS FULL OF WATER.

Jimmie
 
C

Cecil Moore

Jan 1, 1970
0
Charles said:
How are you feeding a balanced line?
http://www.w5dxp.com/notuner.htm

If you are using an antenna tuner with unbalanced in (50 ohms) and balanced
out (variable impedance), you should be OK in most situations.

No tuner! I don't like tuner losses. The feedpoint impedance
is always between 35 ohms and 85 ohms resistive. My choke
has an impedance in the thousands of ohms.
 
T

Tim Wescott

Jan 1, 1970
0
billcalley said:
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.

But then why does the concept of "mismatch loss" exist in
reference to antennas? I have quickly calculated that if a
transmitter outputs 100 watts, and the TX antenna has an impedance
that will cause a VSWR of 10:1 -- using lossless transmission line --
that the mismatch loss in this "lossless" system would be 4.81dB!
(Reflected power 66.9 watts, RL -1.74).

Since mismatch loss is the "amount of power lost due to
reflection", and is as if an "attenuator with a value of the mismatch
loss where placed in series with the transmission line", then I would
think that VSWR would *definitely* matter, and not just for highly
lossy lines either. But here again, I'm probably not seeing the
entire picture here. What am I missing??
Your "amount of power lost due to reflection" statement would be true if
the line were connected to something resistive at the line's
characteristic impedance. With a properly tuned tuner, that's not the
case -- instead, the impedance looking into the tuner will also reflect
power, and in a way that makes it all work out so that the power all
ends up being radiated, which is what you wanted in the first place.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Posting from Google? See http://cfaj.freeshell.org/google/

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html
 
T

Tim Wescott

Jan 1, 1970
0
billcalley said:
Now my head *really* hurts! This is a VERY confusing subject, to
say the least. (And I also thought antenna tuners actually had a
*loss* due to their limited Q...I think I'm going to change careers
now and just become a pet groomer; or perhaps simply give up
completely and work at Radio Shack).
If you start considering loss in the tuner and the line then yes, a
greater mismatch between the antenna and the line will result in more
lost power (and more component heating in the tuner). You really want
to leave that subject be until you understand the properties of a
lossless system.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Posting from Google? See http://cfaj.freeshell.org/google/

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html
 
D

Dave Oldridge

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.

Well, there will be tuner losses, depending on how good the tuner's
components are.
But then why does the concept of "mismatch loss" exist in
reference to antennas? I have quickly calculated that if a
transmitter outputs 100 watts, and the TX antenna has an impedance
that will cause a VSWR of 10:1 -- using lossless transmission line --
that the mismatch loss in this "lossless" system would be 4.81dB!
(Reflected power 66.9 watts, RL -1.74).
Since mismatch loss is the "amount of power lost due to
reflection", and is as if an "attenuator with a value of the mismatch
loss where placed in series with the transmission line", then I would
think that VSWR would *definitely* matter, and not just for highly
lossy lines either. But here again, I'm probably not seeing the
entire picture here. What am I missing??

Confused!


Yes, you're confused.

If the lossless transmission line (obviously no such animal exists) were
tuned with a lossless tuner, then VSWR would not matter at all.

The loss due to mismatch in any real system will depend upon frequency,
VSWR, type of feedline, length of feedline, and the quality of the tuning
circuits used to match the system to the transmitter.

Let's take your example. 100 watt transmitter into, let's say 100ft. of
feedline at 10:1 VSWR and assume tuner losses are negligble (they often
aren't). Here are the losses for some different kinds of 50 ohm coax at
10mhz:

Belden 8237 2.19db
Belden 9913 1.63db
Belden 9258 3.19db
Belden 8240 3.71db
Belden 9201 3.83db

So, what's obvious here is that different coaxes have different losses at
high SWR. Why is that? Because as power is reflected back and forth in
a transmission line, the losses accumulate. So line that is very low-
loss to begin with will be less affected by high SWR than line that has
moderate to high losses when flat.

If 10 percent of the power in a line is lost travelling from the
transmitter to the antenna, and if the antenna only radiates half that
power, sending the rest back down the line, then 45 percent of the
transmitter power is radiated immediately, while 45 percent is reflected.
But only 40.5 percent reaches the tuner or transmitter. If ALL of that
is re-reflected, then only 36.45 percent of the power is available at the
second reflection to the antenna. The antenna will radiate 18.225
percent of the transmitter's power at this point, making the total 63.225
percent of the transmitter's output. Another 18.225 percent will be
reflected again and, of that 16.4025 percent of the transmit power will
live to be re-reflected from the tuner and 14.76225 percent will arrive
at the antenna on the next bounce. Of that we can expect 7.381125
percent of the transmitter's total power to end up radiated while an
equal amount starts its way back to the tuner. Anyway, it becomes a
pretty simple bit of limit math to predict exactly how much will be
radiated and how much lost in the coax under these conditions.
 
R

Richard Clark

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.

Hi Bill,

As a short description, that is adequate.
But then why does the concept of "mismatch loss" exist in
reference to antennas?

"Mismatch loss" is a system description, not an antenna description.
So your reference is wrong.
I have quickly calculated that if a
transmitter outputs 100 watts, and the TX antenna has an impedance
that will cause a VSWR of 10:1 -- using lossless transmission line --
that the mismatch loss in this "lossless" system would be 4.81dB!
(Reflected power 66.9 watts, RL -1.74).

Two things wrong here:
1. You say nothing of a tuner inline;
B. Your math is wrong either way.
Since mismatch loss is the "amount of power lost due to
reflection", and is as if an "attenuator with a value of the mismatch
loss where placed in series with the transmission line", then I would
think that VSWR would *definitely* matter, and not just for highly
lossy lines either.

It does matter if you lack a tuner (in more ways that one). Most
discussion of "mismatch loss" omits such matters as tuners as it is a
separable issue. Combining these topics raises your chance of
confusion.

73's
Richard Clark, KB7QHC
 
E

ehsjr

Jan 1, 1970
0
billcalley said:
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.

But then why does the concept of "mismatch loss" exist in
reference to antennas? I have quickly calculated that if a
transmitter outputs 100 watts, and the TX antenna has an impedance
that will cause a VSWR of 10:1 -- using lossless transmission line --
that the mismatch loss in this "lossless" system would be 4.81dB!
(Reflected power 66.9 watts, RL -1.74).

Since mismatch loss is the "amount of power lost due to
reflection", and is as if an "attenuator with a value of the mismatch
loss where placed in series with the transmission line", then I would
think that VSWR would *definitely* matter, and not just for highly
lossy lines either. But here again, I'm probably not seeing the
entire picture here. What am I missing??

Confused!

-Bill

I think the confusion (which *always* seems to arise)
comes from the mix of the concept with the real world.
The concept claims that the system is lossless, so the
power bounces around until it eventually exits the "system"
via the antenna. Real world, the system is lossy, so
with all the bouncing around some of the power fails to
leave the system via RF radiation and instead leaves
the system via IR radiation. More heat, less RF.
Tastes great, less filling. :)

Ed
 
J

Joerg

Jan 1, 1970
0
Tim said:
Your "amount of power lost due to reflection" statement would be true if
the line were connected to something resistive at the line's
characteristic impedance. With a properly tuned tuner, that's not the
case -- instead, the impedance looking into the tuner will also reflect
power, and in a way that makes it all work out so that the power all
ends up being radiated, which is what you wanted in the first place.

Unless your coax decides that it had enough of all this current. I had
that happen once. The SWR had gone up and I pressed on. Then a muffled
boom outside and it became 100% reflective. Now I had to get on a ladder
and painstakingly scrape all the molten gunk off the stucco.
 
R

Rich Grise

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.

But then why does the concept of "mismatch loss" exist in
reference to antennas? I have quickly calculated that if a
transmitter outputs 100 watts, and the TX antenna has an impedance
that will cause a VSWR of 10:1 -- using lossless transmission line --
that the mismatch loss in this "lossless" system would be 4.81dB!
(Reflected power 66.9 watts, RL -1.74).

Since mismatch loss is the "amount of power lost due to
reflection", and is as if an "attenuator with a value of the mismatch
loss where placed in series with the transmission line", then I would
think that VSWR would *definitely* matter, and not just for highly
lossy lines either. But here again, I'm probably not seeing the
entire picture here. What am I missing??

Confused!

It truly wouldn't matter if there were no such thing as resistance
and so on. The whole circuit could be tuned, with the transmission line
a part of it, and all of the power would go out the antenna.

Unfortunately, that's not the way reality works, more's the pity.

Every time those "standing waves" bounce back and forth, they warm
up the transmission line, the connectors, the transmitter tank, the
transmitter itself, etc, etc, and Entropy is conserved. ;-)

Hope This Helps!
Rich
 
R

Richard Clark

Jan 1, 1970
0
I have quickly calculated that if a
transmitter outputs 100 watts, and the TX antenna has an impedance
that will cause a VSWR of 10:1 -- using lossless transmission line --
that the mismatch loss in this "lossless" system would be 4.81dB!
(Reflected power 66.9 watts, RL -1.74).
[/QUOTE]

Taking this at face value, yes, the "mismatch loss" is 4.8dB.

73's
Richard Clark, KB7QHC
 
B

bw

Jan 1, 1970
0
Every time those "standing waves" bounce back and forth, they warm
up the transmission line, the connectors, the transmitter tank, the
transmitter itself, etc, etc, and Entropy is conserved. ;-)

Humor noted, but Entropy is not conserved.

Entropy increases
 
B

Buck

Jan 1, 1970
0
No tuner! I don't like tuner losses. The feedpoint impedance
is always between 35 ohms and 85 ohms resistive. My choke
has an impedance in the thousands of ohms.

Cecil,

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?

Thanks
 

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