On Tue, 10 May 2011 16:27:00 -0700 (PDT), walt wrote:

Richard, the point I missed until rereading your post, is that there
is total re-reflection of the reflected wave at either the output of
the correctly-adjusted pi-network, or at the output of the antenna
tuner, not the same mismatch as between the line and the load.

Walt

Hi Walt,

The mismatch, of the reflection wave approaching the transmitter, is
seen at the antenna/tuner connection. Hence we have a reflection
there back to the load.

In regard to solving the transmission line problem where the
mismatched load acts as the source of energy (found in the
reflection); then that energy must travel to meet a matching load
where the load (our transmitter) complete takes it in, or a mismatch
where some energy passes into the load (our transmitter) and some is
reflected according to the conventional mechanics of mis-matching.

I take this step given the evidence of physics and reflection at even
the most specular of surfaces. There is some portion of the energy
that exists within a quarter wave behind/beneath the surface of a
reflector (I am speaking of light, but lower frequencies doesn't alter
the situation). That is to say, there is no infinitesimally thin
boundary through which energy does not pass - even for a perfect
reflector.

In antennas, we call this zone the reactive field.

Hence, the reflection seeing the mismatch of the tuner/tuned output,
finds its way beyond the connector, back into the internals, passing
through transform circuitry.

73's
Richard Clark, KB7QHC

On May 10, 8:40*pm, Richard Clark wrote:
On Tue, 10 May 2011 17:14:56 -0700 (PDT), Richard Fry
wrote:

On May 10, 7:00*pm, Richard Clark wrote:
Another way of putting it is saying the ringing line is holding 10W
until it is dissipated. *So far, only the antenna qualifies as
dissipation.

Apparently you have never had to repair the components of a
transmitter PA, an output network, or a transmission line that arced
over and/or melted down due to load reflections (coherent, or not).

Not more than a quarter million Watts, fur shure. *Does it show?

Seriously, if that was your only rant, based on that snippet, like
Wim, you have just indicted your own witness (experience?).



RF

73's
Richard Clark, KB7QHC

To Richard C:

Richard, I must totally disagree that the same degree of mismatch
occurs at the tx output as at the dissipative load, the antenna. As I
have said many times, with the non-dissipative output source impedance
of the RF amp (tubes) the reflection at the junction of the line input
and the pi-network output is total. In other words, all of the
reflected power incident on the output of the pi-network is TOTALLY re-
reflected. Can't be anything less!

To Richard F. And I disagree that the load mismatched at 2:1 absorbs
88.8889% of the forward power. It is 90% !!! We are discussing steady-
state values, in which arrival of the first forward wave is
irrelevant. If you are so confident that the number is 88.8889%,
please derive the conditions that yield that number.

I never thought I'd be having this discussion.

Walt

On Tue, 10 May 2011 19:35:55 -0700 (PDT), walt wrote:

To Richard C:

Richard, I must totally disagree that the same degree of mismatch
occurs at the tx output as at the dissipative load, the antenna. As I
have said many times, with the non-dissipative output source impedance
of the RF amp (tubes) the reflection at the junction of the line input
and the pi-network output is total. In other words, all of the
reflected power incident on the output of the pi-network is TOTALLY re-
reflected. Can't be anything less!

Hi Walt,

There's an active load behind that 2:1 mismatch seen by the reflected
power returning from the load. That is my part of the discussion.

However, if you are going to neglect that interpretation, then there
must be some explanation for TOTAL reflection. Fortunately there is,
it is called an open, or a short.

Now, we come to a philosophical quandry of the "Conjugate Match." If
we tune to a load, and we guarantee that the tune is the conjugate of
the load, then at any point along the transmission line in between
there exists looking at either end a conjugate of the opposite end.

In other words, putting a network analyzer into the line looking at
the load it sees R +jX, and looking back at the source it sees R -jX.
This is conjugation. Moving the analyzer along the length of the line
will transform R and X, but they will always track to a solution of
the conjugate.

You've been arguing this for years.

And, as the load is neither a short, nor an open (which is a special
solution of conjugation); then it stands to reason it is something
other which necessarily demands a solution in the form of:
R ±jX
from the source, where R has some non-zero, and non-infinite value.
For trivial antennas, that value of R falls somewhere between the 10s
of Ohms and the 100s of Ohms. For trivial, retail transmitters
available to Hams, similar values of R abound but most often fall
closest to 50 Ohms, or they would have a hard time getting optimal
power into the 50 Ohm pipe. Vendors are not going to shoot themselves
in the foot before that race.

A TOTAL reflection is available courtesy of the "active" part of this
special load. The "active" load (the transmitter sinking the
reflected power from the dissipative load) can offer one of two
mechanisms to provide the necessary short or open. When the source
wave coming out meets the returning wave coming in, and they are phase
coherent at 0 degree difference, then there is no potential difference
through the connector - no current flow. No current flow = open
circuit. Open circuit satisfies TOTAL reflection mechanics (but
violates conjugation). If they are phase coherent, but 180 degrees
different, then there is maximum current = short circuit. Short
circuit satisfies TOTAL reflection mechanics (but violates
conjugation).

However, in either of these scenarios, both powers (actually energies)
must be equal. Obviously that isn't the case with a nominal 100W
source looking at a 10W reflection.

Hence the analysis draws out slightly more along the lines I've
already posted and won't repeat here.

However, for the sake of argument, the condition of tune (conjugation)
and the state of coherency (a slippery slope, but here we are),
combine through a ringing in the line to create a virtual TOTAL
reflection at the connector of the Tuner/tuned amplifier. So, Walt, I
am not denying your case, but in the earlier posting I do preserve
conjugation in the explanation.

No transmitter, even with the best of tuning attempting to match will
forgive a short or open as load however. And, yes Richard, I have
experienced such conditions with 10s of W to 100s of kW and not all of
them accidental.

Another issue of moving sensors along the transmission line between
high mismatches is that there will be considerable error in the
determination of power - I presume this is where Jim's bookkeeping
explanation will lead, but that is still hanging on the line (no pun
there). I have experience the difficulty of those conditions too when
making precision power measurements.

73's
Richard Clark, KB7QHC

On May 10, 9:35 pm, walt wrote:
If you are so confident that the number is 88.8889%,
please derive the conditions that yield that number.

à = R(load) - R(source) / R(load) + R(source)

SWR = (1 + |Ã|) / (1 - |Ã|)

Power Accepted by the Load = Incident Power * (1 - Ã^2)

For a 50 ohm source connected to a 100 ohm load:

à = (100 - 50) / (100 + 50) = 50 / 150 = 0.333333...

SWR = (1 + 0.333333) / (1 - 0.333333) = 1.333333 / 0.666666 = 2:1

Load Power = Incident Power * (1 - 0.333333^2) = Incident Power *
0.888889 (or 88.8889%)

The answers are the same for a 50 ohm source with a 25 ohm load.

RF

On 11/05/2011 12:21, Richard Fry wrote:
On May 10, 9:35 pm, wrote:9.542
If you are so confident that the number is 88.8889%,
please derive the conditions that yield that number.

à = R(load) - R(source) / R(load) + R(source)

SWR = (1 + |Ã|) / (1 - |Ã|)

Power Accepted by the Load = Incident Power * (1 - Ã^2)

For a 50 ohm source connected to a 100 ohm load:

à = (100 - 50) / (100 + 50) = 50 / 150 = 0.333333...

SWR = (1 + 0.333333) / (1 - 0.333333) = 1.333333 / 0.666666 = 2:1

Load Power = Incident Power * (1 - 0.333333^2) = Incident Power *
0.888889 (or 88.8889%)

The answers are the same for a 50 ohm source with a 25 ohm load.

RF

Or expressed another way a 2:1 vswr equates to a return loss of 9.542dB
and so a mismatch loss of -0.512dB which equates to 0.888 or 88.8% .

Jeff

On May 10, 5:34*pm, Richard Fry wrote:
Let us start at T=0. *The transmitter is keyed, and instantaneously
delivers 100W to the input of a lossless transmission line. *That 100W
travels to the end of the transmission line and encounters a load
connected there having a 2:1 mismatch to the Zo of the transmission
line.

That load will dissipate about 88.889W, and reflect about 11.111W back
to the source.

Likewise, the load will dissipate only 88.889% of the RE-reflected
power from the source for the same reason it dissipated only 88.889%
of the power in the original incident wave.

So even if the RE-reflection from the source is lossless, and phase-
coherent with the original incident power at the load, how does this
yield the 100W of load dissipation that you posit?

It's what happens while going from T0 to steady-state.

Pload = 88.889w -- 98.8w -- ... -- 100w
Pfor = 100w -- 111.11w -- ... -- 112.5w
Pref = 11.11w -- 12.3w -- ... -- 12.5w
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

On May 10, 9:35*pm, walt wrote:
Richard, I must totally disagree that the same degree of mismatch
occurs at the tx output as at the dissipative load, the antenna.

May I suggest that we simplify the example by removing the tuner from
the discussion?

100w--+----1/4WL 100 ohm feedline----200 ohm load

Same conditions as before. The voltage reflection coefficient at the
load is 0.3333. The power reflection coefficient at the load is
0.1111. A 50 ohm Z0-match exists at point '+' without a tuner. No
reflected energy reaches the source.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

On 10 mayo, 23:37, walt wrote:

[deleted]

Now Wim, your math is very challenging: You state 100w delivered by
the source, but at one point you also state 111w is delivered to the
100-load--at another point you state that 100w is delivered to the 100-
ohm load. Which is it?

You also state that voltage out of the source is 212.1v--sum ting wong
here. 212.1v across 50 ohms yields 899.73w, and 212.1v across 100 ohm
yields 449.86w. These power values are nowhere near the values
appearing in you statements.

In my calculations, with 100w the voltage across a 50-ohm line is
70.71v, and across a 100-ohm line the voltage is 100v.

With a 50-ohm line terminated in 100 ohms, the vrc is 0.3333 as you
stated, thus the power-reflection coefficient is 0.1111. This means,
as I've stated continually, that with 100w delivered by the source,
the power reflected at the 2:1 mismatch is 11.111w, which when added
to the 100w supplied by the source, makes the forward power 111.11w.
Now with 11.111w of power reflected at the mismatch, this leaves 100w
delivered to the 100-ohm load, not 111w. Please tell me where the 111w
came from. You also haven't told us why you calculate 130w forward and
15w reflected. Someone else may have made these calculations, but it
was you I asked for an explanation, because you repeated those
calculations.

When using the correct physics and math in this example, how can the
results *be so different?

To close, let me present the procedure I use to calculate the total
forward power--I guess it is related to Ohm's Law:

With the power-reflection coefficient as prc, the forward power PF = 1/
(1 - prc).

So Wim, can you clarify the confusion you appear to have made?

Walt

Hello Walt,

The source (that is the PA in this case) produces an EMF (as mentioned
before) of 212.1Vrms, Source impedance is 100 Ohms (so forget the
"conjugated match" thing for this case).

When terminated according to the numbers above the socket ("100W into
50 Ohms"), this results in 100W ( Vout = 212.1*50/(50+100) = 70.7V,
just a voltage divider consisting of 100 Ohms and 50 Ohms ).

70.7V into 50 Ohms makes 100W.

Now we remove the 50 Ohms load and create a mismatch (referenced to 50
Ohms), by connecting a 100 Ohms load (VSWR = 100/50 = 2)

Now the output voltage will be:

Vout = 212.1*100/(100 +100) = 106V


106V into 100 Ohms makes 112W (forgive me the one Watt difference).

This all without any transmission line theory.


Now the forward and reflected power balance (this is exactly the same
as in my previous posting, except for using 112W instead of 111W):

100 Ohms equals |rc| = 0.3333 (50 Ohms reference).

Hence "reflected power" = 0.333^2*"incident power" = 0.111*"incident
power".

"net delivered power" = "incident power" – "reflected power" = 112W.

"net delivered power" = (1-0.11111)*"incident power" = 112W.

"incident power"= 126W, "reflected power" = 14 W

As far as I can see, there is nothing new with respect to the previous
calculation, except for the truncation/rounding.

Similar reasoning can be applied to my class-E PA. It is designed for
500W into 4.5 Ohm, but the output impedance is below 1 Ohms. If you
try to achieve conjugate match, you fry the active devices. Same is
valid for most audio amplifiers, they may mention: "80W into 8 Ohm",
but that does not mean that Zout = 8 Ohms.

Awaiting your comment,



Wim
PA3DJS
www.tetech.nl

On 11 mayo, 02:05, walt wrote:
On May 10, 7:28*pm, Richard Fry wrote:



On May 10, 5:48*pm, Richard Clark wrote:

The transmission line is bounded at either end by identical mismatches
in conjugation. *There is a "ringing" in the line, which at only one
end there is dissipation (neglecting loss and damage to the tube for
exceeding maximum tolerances, of course).

If, as you have posted, dissipation occurs only at one end of the
transmission line -- which presumably for the most useful benefit
would be at the end of the line opposite the source, e.g., the load --
and the incident power generated and delivered to that line by the
final r-f stage in the transmitter is a constant, then how could
"damage to the tube" occur?

Just to note that as a field engineer for RCA Broadcast for many years
I have had to troubleshoot and repair failures that occurred in
transmission lines, transmitters, and output networks that resulted
from reflections -- either from within the transmission line alone, or
together with the load connected at its far end.

RF

Yes Richard, I'm familiar with your work at RCA Broadcast Div.
However, in my posts I'm concerned only with tube rigs used in the
Amateur Service. In these rigs reflected power doesn't cause
overheating, or other damage to the tubes.

To Walt: (Steady state) Load conditions can be specified in many forms
(complex impedance, complex reflection coefficient, VSWR with phase,
etc), but depending on the actual integral of the Vp*Ip product, the
plate's dissipation may decrease or increase (due to a load change).

So changing the load (without retuning as you mentioned below) may put
your tubes at risk. Your qualification below seems to contradict your
statement above.

But I'll qualify that
statement--if the pi-network is originally resonated into, say, a
dummy load, and is then switched to a mismatched line without
retuning, the reactance appearing at the input of the mismatched line
detunes the network and the result is the same is if the network was
left off resonance initially. As you well know, a mis-tuned pi-network
results in excessive plate current, which, if high enough will damage
the tube. Consequently, it's the mis-tuning that causes the damage,
not the reflections per se.

Incidentally Richard, I was also an electrical engineer with RCA, from
1949 thru 1980, first at the RCA Labs in Princeton and beginning in
1958 with Astro.

Walt


Wim
PA3DJS
www.tetech.nl

On May 11, 10:57*am, Wimpie wrote:
On 11 mayo, 02:05, walt wrote:









On May 10, 7:28*pm, Richard Fry wrote:

On May 10, 5:48*pm, Richard Clark wrote:

The transmission line is bounded at either end by identical mismatches
in conjugation. *There is a "ringing" in the line, which at only one
end there is dissipation (neglecting loss and damage to the tube for
exceeding maximum tolerances, of course).

If, as you have posted, dissipation occurs only at one end of the
transmission line -- which presumably for the most useful benefit
would be at the end of the line opposite the source, e.g., the load --
and the incident power generated and delivered to that line by the
final r-f stage in the transmitter is a constant, then how could
"damage to the tube" occur?

Just to note that as a field engineer for RCA Broadcast for many years
I have had to troubleshoot and repair failures that occurred in
transmission lines, transmitters, and output networks that resulted
from reflections -- either from within the transmission line alone, or
together with the load connected at its far end.

RF

Yes Richard, I'm familiar with your work at RCA Broadcast Div.
However, in my posts I'm concerned only with tube rigs used in the
Amateur Service. In these rigs reflected power doesn't cause
overheating, or other damage to the tubes.

To Walt: (Steady state) Load conditions can be specified in many forms
(complex impedance, complex reflection coefficient, VSWR with phase,
etc), but depending on the actual integral of the Vp*Ip product, the
plate's dissipation may decrease or increase (due to a load change).

So changing the load (without retuning as you mentioned below) may put
your tubes at risk. *Your qualification below seems to contradict your
statement above.

But I'll qualify that
statement--if the pi-network is originally resonated into, say, a
dummy load, and is then switched to a mismatched line without
retuning, the reactance appearing at the input of the mismatched line
detunes the network and the result is the same is if the network was
left off resonance initially. As you well know, a mis-tuned pi-network
results in excessive plate current, which, if high enough will damage
the tube. Consequently, it's the mis-tuning that causes the damage,
not the reflections per se.

Incidentally Richard, I was also an electrical engineer with RCA, from
1949 thru 1980, first at the RCA Labs in Princeton and beginning in
1958 with Astro.

Walt

Wim
PA3DJSwww.tetech.nl


To Richard C.: It seems to me that you're overlooking the fact that
the source resistance appearing
at the output terminals of the pi-network is non-dissipative, and thus
cannot absorb the energy in the reflected wave incident on it. The
resistance is E/I and therefore non-dissipative, has a real value and
is thus a part of the conjugation. The result is total re-reflection,
no transfer of energy into the amp. I thought I'd made this very clear
over the years.

To Richard F. My humble apology for my moronic error. My bad!!! I
can't believe I was so stupid as to not know where the 0.888889 came
from. Thanks for spelling it out for me. I got totally sidetracked
with the 100/111,11 = 0.90, and didn't see any farther.

Wim: I'm working on your response--it's finally beginning to make
sense--gimme a few more minutes.

Walt