We are all told that VSWR doesn't matter when using low loss
transmission lines, since the RF energy will travel from the
transmitter up to the mismatched antenna, where a certain amount of
this RF energy will reflect back towards the transmitter; after which
the RF will then reflect back up to the antenna -- where the energy is
eventually radiated after bouncing back and forth between the
transmitter and antenna. I understand the concept, but what I don't
quite understand is why the reflected RF energy isn't simply absorbed
by the 50 ohm output of the transmitter after the first reflection?
For the RF to bounce back and forth, wouldn't the transmitter's
impedance have to be very, very high (or low) when the reflected RF
energy hit its output stages? I know I'm missing something vital
here...

Thanks!

-Bill

billcalley wrote:

We are all told that VSWR doesn't matter when using low loss
transmission lines, since the RF energy will travel from the
transmitter up to the mismatched antenna, where a certain amount of
this RF energy will reflect back towards the transmitter; after which
the RF will then reflect back up to the antenna -- where the energy is
eventually radiated after bouncing back and forth between the
transmitter and antenna. I understand the concept, but what I don't
quite understand is why the reflected RF energy isn't simply absorbed
by the 50 ohm output of the transmitter after the first reflection?
For the RF to bounce back and forth, wouldn't the transmitter's
impedance have to be very, very high (or low) when the reflected RF
energy hit its output stages? I know I'm missing something vital
here...

That's assuming you use an antenna tuner. The tuner will transform the
transmitter's output impedance* just as it transforms the line. Were
the transmitter output impedance actually at 50 ohms, on the other side
of the tuner it would have the same VSWR as the line when everything was
tuned up.

Having said that, the VSWR _does_ matter somewhat when using low loss
lines, both because the line loss is low but not zero, and the tuner
loss will tend to go up as you correct for higher and higher VSWR.

* I am _not_ going to start the Big Transmitter Output Impedance Debate.
sed denizens -- just don't comment on what a transmitter's "actual"
output impedance may be, lest you start a flame war.

--

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

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

"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html

billcalley wrote:
... I understand the concept, but what I don't
quite understand is why the reflected RF energy isn't simply absorbed
by the 50 ohm output of the transmitter after the first reflection?
For the RF to bounce back and forth, wouldn't the transmitter's
impedance have to be very, very high (or low) when the reflected RF
energy hit its output stages? I know I'm missing something vital
here...

Yep, you are missing the "total destructive interference"
happening toward the source caused by a Z0-match. Here's
an article that might help:

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

A Z0-match eliminates reflected energy from reaching
the source.
--
73, Cecil http://www.w5dxp.com

On Mar 11, 8:08�pm, "billcalley" wrote:
* * We are all told that VSWR doesn't matter when using low loss
transmission lines...

Well, I was never taught that -- which is a good thing because it is
not true.

The most intuitive way to prove this (antenna tuner or not), is to
simply supply more and more power to the load. You will eventually
burn it up.

Now, had the VSWR been better (ie. a better network Return Loss), it
would have taken even more power to achieve the same "charcoal"
results. So VSWR matters.

-mpm

On Mar 12, 1:08 am, "billcalley" wrote:
We are all told that VSWR doesn't matter when using low loss
transmission lines,

No, we are not all told that.

since the RF energy will travel from the
transmitter up to the mismatched antenna, where a certain amount of
this RF energy will reflect back towards the transmitter; after which
the RF will then reflect back up to the antenna -- where the energy is
eventually radiated after bouncing back and forth between the
transmitter and antenna. I understand the concept, but what I don't
quite understand is why the reflected RF energy isn't simply absorbed
by the 50 ohm output of the transmitter after the first reflection?

The active part of the transmitter output isn't 50 ohm.
That would cause half the power to be lost as heat in
the output stage. It's only 50ohm once it becomes a moving
wave in the transmission line.

Bob9

On 11 Mar 2007 20:39:46 -0700, "Bob" wrote:

The active part of the transmitter output isn't 50 ohm.
That would cause half the power to be lost as heat in
the output stage.

Hi Bob,

Well, aside from the initial misunderstanding of how power gets to the
load (much less back, and then to the load again); I will put to you
a question that has NEVER been answered by those who know what the
transmitter output Z ISN'T:
"What Z is it?"

73's
Richard Clark, KB7QHC

Triode or pentode? ;o)

Tim

--
Deep Fryer: A very philosophical monk.
Website @ http://webpages.charter.net/dawill/tmoranwms

"Richard Clark" wrote in message
...
On 11 Mar 2007 20:39:46 -0700, "Bob" wrote:

The active part of the transmitter output isn't 50 ohm.
That would cause half the power to be lost as heat in
the output stage.

Hi Bob,

Well, aside from the initial misunderstanding of how power gets to the
load (much less back, and then to the load again); I will put to you
a question that has NEVER been answered by those who know what the
transmitter output Z ISN'T:
"What Z is it?"

73's
Richard Clark, KB7QHC

In .com, billcalley said:
We are all told that VSWR doesn't matter when using low loss
transmission lines, since the RF energy will travel from the
transmitter up to the mismatched antenna, where a certain amount of
this RF energy will reflect back towards the transmitter; after which
the RF will then reflect back up to the antenna -- where the energy is
eventually radiated after bouncing back and forth between the
transmitter and antenna. I understand the concept, but what I don't
quite understand is why the reflected RF energy isn't simply absorbed
by the 50 ohm output of the transmitter after the first reflection?

Two problems:

1) The transmitter may well have output impedance matching the
characteristic impedance of the transmission line. RF power reflected
back in this case gets converted to heat in the output stage of the
transmitter, in addition to whatever heat the output stage already has to
dissipate.

1a) The reflection may increase requirement of the output
tubes/transistors to both drop voltage and dissipate power. This can be a
problem for many transistors, especially a lot of bipolar ones. It is not
necessarily sufficient to stay within power, current, voltage and thermal
ratings. Many bipolar transistors have reduced capability to safely
dissipate power at voltages that are higher but within their ratings -
sometimes even at voltages as low as 35-50 volts. This problem tends to
be worse with bipolar transistors that are faster and/or better for use
with higher frequencies. The keyphrase here is "forward bias second
breakdown", a problem of uneven current distribution within the die at
higher voltage drop.

2) It appears to me that transmitters can have output stage output
impedance differing from the intended load impedance.
An analog is common practice with audio amplifiers - output impedance is
often ideally as close to zero as possible, as opposed to matching the
load impedance.

If zero output impedance is achieved in an RF output stage, I see a
possible benefit - reflections do not increase output stage heating but
get reflected back towards the antenna. Then again, the impedance of the
input end of the transmission line could be low or significantly reactive
depending on how the load is mismatched and how many wavelengths long the
transmission line is, and that can increase heating of the output stage.
In a few cases transmitted power can also increase.

Not only is increased output stage heating possible and maybe fairly
likely, high VSWR also causes a high chance of the output stage seeing a
partially reactive load. RF bipolar transistors often do not like those
due to increased need to dissipate power with higher voltage drop. As I
said above, RF bipolar transistors are likely to really dislike
simultaneous higher voltage drop and higher power dissipation.

- Don Klipstein )

On Mon, 12 Mar 2007 00:09:26 -0600, "Tim Williams"
wrote:

Triode or pentode? ;o)

For you, cascode, push-pull triodes.

73's
Richard Clark, KB7QHC

*Sigh*

The same misconceptions keep coming up, as they have countless times on
this newsgroup and I'm sure they will for decades or perhaps centuries
to come. After one of the many previous discussions, I wrote a little
tutorial on the topic. Originally in the form of plain text files, I've
combined it into a pdf file for easier viewing. You can find it at
http://eznec.com/misc/Food_for_thought.pdf.

On page 8 you'll find the statement "THE REVERSE POWER IS NOT DISSIPATED
IN OR ABSORBED BY THE SOURCE RESISTANCE". Above it is a chart of several
examples which clearly show that there's no relationship between the
"reverse power" and the source dissipation. The remainder of the
tutorial explains why.

Any theory about "forward" and "reverse" power, what they do, and their
interaction with the source, will have to explain the values in the
example chart on page 8. Does yours?

Roy Lewallen, W7EL

Don Klipstein wrote:
In .com, billcalley said:
We are all told that VSWR doesn't matter when using low loss
transmission lines, since the RF energy will travel from the
transmitter up to the mismatched antenna, where a certain amount of
this RF energy will reflect back towards the transmitter; after which
the RF will then reflect back up to the antenna -- where the energy is
eventually radiated after bouncing back and forth between the
transmitter and antenna. I understand the concept, but what I don't
quite understand is why the reflected RF energy isn't simply absorbed
by the 50 ohm output of the transmitter after the first reflection?

Two problems:

1) The transmitter may well have output impedance matching the
characteristic impedance of the transmission line. RF power reflected
back in this case gets converted to heat in the output stage of the
transmitter, in addition to whatever heat the output stage already has to
dissipate.

1a) The reflection may increase requirement of the output
tubes/transistors to both drop voltage and dissipate power. This can be a
problem for many transistors, especially a lot of bipolar ones. It is not
necessarily sufficient to stay within power, current, voltage and thermal
ratings. Many bipolar transistors have reduced capability to safely
dissipate power at voltages that are higher but within their ratings -
sometimes even at voltages as low as 35-50 volts. This problem tends to
be worse with bipolar transistors that are faster and/or better for use
with higher frequencies. The keyphrase here is "forward bias second
breakdown", a problem of uneven current distribution within the die at
higher voltage drop.

2) It appears to me that transmitters can have output stage output
impedance differing from the intended load impedance.
An analog is common practice with audio amplifiers - output impedance is
often ideally as close to zero as possible, as opposed to matching the
load impedance.

If zero output impedance is achieved in an RF output stage, I see a
possible benefit - reflections do not increase output stage heating but
get reflected back towards the antenna. Then again, the impedance of the
input end of the transmission line could be low or significantly reactive
depending on how the load is mismatched and how many wavelengths long the
transmission line is, and that can increase heating of the output stage.
In a few cases transmitted power can also increase.

Not only is increased output stage heating possible and maybe fairly
likely, high VSWR also causes a high chance of the output stage seeing a
partially reactive load. RF bipolar transistors often do not like those
due to increased need to dissipate power with higher voltage drop. As I
said above, RF bipolar transistors are likely to really dislike
simultaneous higher voltage drop and higher power dissipation.

- Don Klipstein )