VSWR doesn't matter?

[billcalley]

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

 

[Tim Wescott]

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

 

[Cecil Moore]

... 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.

 

[mpm]

    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

 

[Bob]

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

 

[Richard Clark]

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

 

[Tim Williams]

Triode or pentode? ;o)

Tim

 

[Don Klipstein]

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 ([email protected])

 

[Richard Clark]

Triode or pentode? ;o)

For you, cascode, push-pull triodes.

73's
Richard Clark, KB7QHC

 

[Roy Lewallen]

*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

 

[Jan Panteltje]

On a sunny day (Mon, 12 Mar 2007 07:36:25 +0000 (UTC)) it happened
[email protected] (Don Klipstein) wrote in
<[email protected]>:

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 ([email protected])

All true.
Also normally, there is a pi type filter (to prevent harmonics), between
amplifier and antenna.
This filter _WILL_ match the antenna to the output impedance of the
transmitter, so _even_ if the transmitter output impedance is very
very low (low voltage high current output stage for example), the reflected
power will be nicely converted to match the transmitter, and heat up the
output amp, with its possible destruction as result.

 

[Ulrich Bangert]

Bill,

an exellent treatment on this question has been published in QEX December 94
under the title "Where does the power go?"

73 de Ulrich, DF6JB

 

[Cecil Moore]

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?"

It really doesn't matter except for overall
efficiency. A 10 ohm source outputting 100 volts
into a local load of R +/- jX sources the same
amount of power as a 100 ohm source outputting
100 volts into the same local load.

 

[Cecil Moore]

If zero output impedance is achieved in an RF output stage, I see a
possible benefit -

Yep, 100% efficiency would be quite a benefit.

 

[Cecil Moore]

*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?

Mine does. All of your values can be understood by looking
at the destructive and constructive interference and applying
the irradiance (power density) equations from the field of
optics. You see, optical engineers and physicists don't have
the luxury of measuring voltage and current in their EM waves.
All they can measure is power density and interference and
thus their entire body of knowledge of EM waves rests upon
measurements of those quantities. Those power density and
interference theories and equations are directly applicable
and 100% compatible with RF theories and equations. Any
analysis based on power density and interference will yield
identical results to the ones you reported in your "food for
thought" article which includes the following false statement:

"While the nature of the voltage and current waves when
encountering an impedance discontinuity is well understood,
we're lacking a model of what happens to this "reverse power"
we've calculated."

We are not lacking a model of what happens to this
"reverse power" we've calculated. The model is explained
fully in "Optics", by Hecht. When one has standing waves of
light in free space, it is hard to hide the details under
the transmission line rug.

In general, it is just as easy, and sometimes easier, to deal
with the energy values and then calculate voltage and current
as it is to start with voltage and current and then calculate
the power.

All this is explained in my WorldRadio article at:

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

The great majority of amateur antenna systems are Z0-matched.
For such systems, an energy analysis is definitely easier to
perform than a voltage analysis. Here's an example:

100W------50 ohm---+---Z0>50 ohms-----load
Pfor1=100W--> Pfor2-->
<--Pref1=0W <--Pref2

The power reflection coefficient is 0.5 at point '+'.
The power reflection coefficient is 0.5 at the load.

What are the values of Pfor2 and Pref2? What is the physics
equation governing what happens to Pref2 at point '+'?

 

[Cecil Moore]

Also normally, there is a pi type filter (to prevent harmonics), between
amplifier and antenna.
This filter _WILL_ match the antenna to the output impedance of the
transmitter, so _even_ if the transmitter output impedance is very
very low (low voltage high current output stage for example), the reflected
power will be nicely converted to match the transmitter, and heat up the
output amp, with its possible destruction as result.

Some gurus will say that it's the voltage and/or current
that is destroying the final, not the reflected energy.
They have yet to explain how those dangerous voltages
and/or currents can exist without assistance from the
ExH joules/second in the reflected energy wave. Depending
upon phase, the E in the ExH reflected wave is what causes
the overvoltage due to SWR. The H in the ExH reflected
wave is what causes the overcurrent due to SWR.

The impedance seen by the source is

Z = (Vfor+Vref)/(Ifor+Iref)

Where '+' indicates phasor (vector) addition.

The above equation also gives the impedance anywhere
along the transmission line and anywhere along a
standing-wave antenna.

 

[Cecil Moore]

an exellent treatment on this question has been published in QEX December 94
under the title "Where does the power go?"

Unfortunately, that article doesn't explain where the
power does go. A much better treatment of the subject
is in "Optics", by Hecht. To understand where the
power does go, one must understand destructive and
constructive interference. Please see my transmission
line example in another posting.

The energy content of a transmission line during
steady-state is always exactly enough to support
the forward traveling wave and the reverse traveling
wave without which there would be no standing wave.

 

[Don Klipstein]

Yep, 100% efficiency would be quite a benefit.

There are audio amplifiers with output impedance around .1 ohm, driving
8 ohm speakers, and having efficiency nowhere near 80/81. The theoretical
limit for efficiency of a class B amp driving a resistive load with a
sinewave is 78.54%.

- Don Klipstein ([email protected])

 

[Cecil Moore]

There are audio amplifiers with output impedance around .1 ohm, driving
8 ohm speakers, and having efficiency nowhere near 80/81. The theoretical
limit for efficiency of a class B amp driving a resistive load with a
sinewave is 78.54%.

Of course, that was a tongue-in-cheek posting.
But if you could design a Thevenin equivalent
source with a 0.1 ohm source impedance, wouldn't
the efficiency calculate out to be pretty high?

 

[billcalley]

Thanks guys -- some really great posts here -- it will take me quite a
while to digest all this!!

Thanks Again,

-Bill