Is Maxwell wrong?
Cecil Moore
repairs and I suffer from the PC/UNIX translation on the DELPHI system.
JB>I'm not sure what you mean by a "Z0 match." In my Dec, 1994 QEX article
JB>example (Fig 4), the only "match" is of the load seen by the
JB>transmitter to the *design* load impedance.
Hi again, Jon. I'm not being contrary just for the fun of it. I
am trying to explode a myth. I know I'm right but am having
trouble putting it into common words. A "Z0 match" is defined in
the ARRL Antenna Book. The Z0 match is where the conjugate match
equals the Z0 of the system. Z0 is a special conjugate match and
for the average solid-state transmitter it is 50+j0. There is
always a change in impedances at a Z0 match. The Z0 match is what
gives a virtual open circuit or short circuit to the reflected
waves. The example you sent to me of 50 ohm coax into a
half-wavelength of ladder-line into a 50 ohm load is an example of
a Z0 match. At the coax/ladder-line junction the reflected waves
are 100% re-reflected by a virtual _open_ circuit. If the
ladder-line is a quarter-wave and the load is 1800 ohms for 300
ohm ladder-line, the Z0 match is achieved and is a virtual _short_
circuit. That's how a quarter-wave transmission line transformer
works. In either case, a virtual open or virtual short circuit is
necessary for a Z0 match. I intend to prove that if a Z0 match
does not exist anywhere in the system, then reflected waves invade
the transmitter and cause I^2R heating losses. There is nothing that
can reflect waves 100% except an open or short circuit. (Thanks, Gary)
JB>As long as you continue to try to figure out what happens to the
JB>reflected power in isolation from the rest of the system, you'll keep
JB>running in circles.
Quantum physics says that as long as you _don't_ isolate the
reflected waves from the rest of the system, _you'll_ keep running
in circles. The rule that reflections are 100% re-reflected is a
rule-of-thumb that works as far as the mathematics go but it is not
what happens in reality. Someone in the past couldn't look inside a
transmitter to see what what was going on and decided that the math
works if one just says that the reflected waves are 100% re-reflected.
Well, we can now look inside a transmitter. I have done that with a
wafer probe and I have seen the truth. Reflected waves are not 100%
reflected unless they see a Z0 match. If they do not see a Z0 match,
they maintain their identity and are dissipated as heat in the
transmitter. Jon, you, yourself said in your article that you didn't
know exactly what was going on inside a transmitter.
JB>The simple fact is, all of the power applied to the effective load
JB>impedance seen by the transmitter gets to the load (if the transmission
JB>line is lossless, of course). Dance around that all you want, but it's
JB>really all you need to consider in order to perform an effective system
JB>analysis.
I agree with you when you say that is all you need to consider in
order to perform an effective system _analysis_... no argument.
What we are arguing about is what actually happens. Jon, please
be open-minded here. We are about to explode a myth and prove
Maxwell wrong. He says, "Reflected power (waves) does not flow
back into the transmitter and cause dissipation and other damage."
I can prove that to be incorrect. Here is my statement based on
the principles of quantum physics which have never been proven
wrong.
**********************************************************************
* *
* In a transmitting system where a Z0 match is not achieved anywhere *
* in the entire system, some reflected waves will penetrate the *
* transmitter, causing mismatches and/or heating. *
* *
**********************************************************************
Remember when I'm famous, you read it here first.
The mathematical effect will be identical to the assumption that
all reflected waves are re-reflected by the transmitter because
of the conservation of energy principle. The following conditions
result in the same energy balance.
Transmitter previously generating four watts into a 50 ohm load.
Now we read four watts forward and two watts reflected.
(1) Four watts generated, two watts transmitted, two watts 100%
reflected, two watts dissipated in the final. (your view, and Gary's
and Maxwell's and, therefore, almost everyone else's)
(2) Four watts generated, three watts transmitted, one watt re-reflected,
one watt of reflections absorbed and dissipated as heat in the transmitter.
(my view of what's possible, I think you say this is impossible)
(3) Four watts generated, four watts transmitted, nothing re-reflected,
two watts of reflections absorbed and dissipated as heat in the network
analyzer. (nobody can argue with this one)
The equations in your QEX article are trivial and are true at any point
in a conjugate-matched system if one is not allowed to make measurements
farther down the transmission line. i.e. pick any point on the trans-
mission line and don't allow any measurements any closer to the
transmitter and one can conclude that, at that point, 100% of the
reflected power is re-reflected. The math all works out just fine.
However, anyone with the capacity for measuring anything closer to
the transmitter will conclude that you are deluding yourself.
I am absolutely amazed that no one has discovered this contradiction
before. It means that few, if any, quantum physicists are hams.
Jon, still no hard feelings, just a search for the facts.
73, Cecil, KG7BK, OOTC
Gary and Jean Peach
across the antenna terminals and press the PTT ?
I suspect neither as you both know what will happen.
Now if a large sum of money was offered, if you would
do it and you where allowed to interpose a short length
of transmission line, ( less than one wavelength) between
the short and the transmitter, you are allowed to choose
the length, What length would you choose?
(a) lambda/4
(b) lambda/2
(c) lambda/x where x is a number
(d) some other length d is included as it is
traditional to have
4 answers
you may choose d but then you must supply the exact
length to three decimal places.
Velocity factor is 0.87935
Gary Peach G7SLL
Zack Lau (KH6CP)
: Which of you guys is prepared to put a short piece of wire
: across the antenna terminals and press the PTT ?
: I suspect neither as you both know what will happen.
Why not, isn't this a good test of a final amplifier circuit
design? I believe Solid State Design actually suggests
a similar experiment somewhere (with a transmatch to simulate
a variety of impedances).
With a good design, the current drawn will jump up and down
but stay within tolerable limits. No damage will result.
With a bad design, you can get excessive voltage or current.
Both can destroy parts.
In one of my 40 meter transceivers, the final will oscillate
into a short, but I don't consider it a problem worth fixing
since I doubt anyone will hear a radiated signal.
--
Zack Lau KH6CP/1 2 way QRP WAS
8 States on 10 GHz
Internet: zl...@arrl.org 10 grids on 2304 MHz
cmo...@sedona.intel.com
Gary and Jean Peach <wise...@cix.compulink.co.uk> wrote:
>Which of you guys is prepared to put a short piece of wire
>across the antenna terminals and press the PTT ?
Hi Gary, way back last year under "Where does the power go?" I told
about my antenna blowing down and shorting at the feedpoint. I didn't
know it, tried to tune up, and burned up my 1625s.
> (a) lambda/4
> (b) lambda/2
> (c) lambda/x where x is a number
> (d) some other length d is included as it is
I guess I would have to choose (d) and make it 100 miles. That should
protect the transmitter unless you specify lossless line and then
nothing will protect it except itself. A short at the end of any
length of lossless line leaves no real load so all the transmitted
power will be reflected and all you have is a choice of phase. I
suppose a .7Vmax*.7Imax might be the lesser evil.
What we are arguing over here is whether the Output Return Loss (S22)
of an RF amplifier really exists. I say yes, others say no, there is
no output return loss. Wonder why we measure it and specify it if
there's no such thing?
For those not familiar with the S22 parameter, it is measured by
connecting a signal generator to the _output_ of an RF amplifier
and measuring the _output_ VSWR. If by its nature of being an RF
amplifier it re-reflects all reflected energy, then why measure
and specify this parameter?
--
73, Cecil, KG7BK, OOTC (All my own personal fuzzy logic, not Intel's)
Jon Bloom (KE3Z)
[snip]
: What we are arguing over here is whether the Output Return Loss (S22)
: of an RF amplifier really exists. I say yes, others say no, there is
: no output return loss. Wonder why we measure it and specify it if
: there's no such thing?
: For those not familiar with the S22 parameter, it is measured by
: connecting a signal generator to the _output_ of an RF amplifier
: and measuring the _output_ VSWR. If by its nature of being an RF
: amplifier it re-reflects all reflected energy, then why measure
: and specify this parameter?
Actually, it measures the output reflection coefficient, by comparing
the amplitude and phase of the returned signal to that of a signal
applied to the output, i.e.: Vret/Vapp, where both values are complex.
Hmmm... let's see what would happen if you did measure the S22 of an
operating transmitter, ignoring for the moment the smoke pouring out of
your S-parameter measurement set. To properly make the measurement, you
need to have the test set apply a signal that is at exactly the same
frequency as the transmitter signal and at an amplitude equal to the
(presumed) reflected signal from the antenna--remember, we're trying to
simulate an actual system here. Now the test set sees a "returned"
signal at an amplitude greater than the applied signal, and with some
you-pick-it phase difference. The calculated reflection coefficient
thus has a magnitude greater than 1. Clearly, the transmitter is not
acting merely as a reflective load, and *you can't treat it as though
it is just a reflective load*. Again, as I keep saying, you have to
consider the effect of the transmitter's signal on the system.
Now change the experiment. Offset the test-set applied frequency from
the transmitter frequency. (We'll assume the test set ignores any
signal at a frequency other than the applied signal.) Do you now get an
S22 with a magnitude less than 1? You bet, because the transmitter
signal is no longer part of the measurement. (By the way, this is
essentially what Warren Bruene, W5OLY, did in his November, 1991 QST
article to measure the output impedance of a transmitter.)
What this demonstrates is that to consider what happens to the
reflected wave, you *must* include the transmitter source in the
analysis.
--
Jon Bloom KE3Z jbl...@arrl.org
Tom Bruhns
(Some "Z0 match" stuff deleted...since I don't have the context
from which that started, I won't try to say anything about it,
except that it looks rather strange... but the following I
gotta comment on:)
: There is nothing that can reflect waves 100% except an open or short
: circuit.
Huh? Any non-dissipative load will reflect 100%; there can be no power
dissipated in such a load, so all must return. That includes all pure-
reactive loads as well as shorts and opens.
I'm trying to find a diplomatic way of commenting about the rest of the
posting...maybe just a simple statement of the facts as I see them is
sufficient:
You can analyze a transmitter load as either just a (possibly complex)
impedance across its antenna terminals, or as a line with forward and
reverse waves on it. Done properly, either analysis will yield the same
result. If you want to consider "forward and reverse waves" inside the
transmitter, you better understand exactly how those waves are propagating
and what impedance the transmission medium is; that all seems rather silly
to me.
To say a transmitter generates a certain amount of power, some of which is
dissipated because it comes back in a reverse wave, is one way of looking
at it, but it's liable to lead you to some strange results if you fail to
consider the phase of the reverse wave. If I have your "4 watt
transmitter" -- one that delivers 4 watts to a 50+j0 ohm load -- and
connect it to a 50 ohm line which is terminated in a load which dissipates
2 watts when the transmitter is on, I _CAN_NOT_ say that the transmitter is
still generating 4 watts, and dissipating 2 watts of reflected wave! That
is because the load on the output stage of the transmitter is no longer
what it was for the 4 watts being generated. And unless I know the design
of that transmitter AND the effective load, I won't know whether that
transmitter output is more or less efficient than with a 50 ohm load. The
effective load will depend on the PHASE as well as the amplitude of any
reflected waves. And it's because that phase can be anything that I'm
QUITE sure that it's inappropriate to talk about a "4 watt transmitter"
in such simplistic terms. The "4 watts" is NOT independent of the load.
73, K7ITM
William P. Osborne
Well I can't watch this without comment any longer so here goes. Cecil and
Jon are both right they are just working different problems.
In Jon's case:
he is assuming a lossless matching network between the generator and the
load which is connected with lossless line (remember the Pi network he
keeps bringing up). Consider a 50 ohm generator putting out 1 watt into a
50 ohm line, then put a tuner in that line and connect the far side of the
tuner to a 100 ohm load through 50 ohm lossless line. Now the tuner is
lossless and the lines are lossless so one watt must be going to the load
but if you measure the forward power between the tuner and the load it
will be 1 + p^2 where p is the reflection coefficient of the load and the
reflected power will be p^2. All of the reflected power appears to stay in
the line in this cases because the tuner is providing a conjugate match.
The forward and the reflected waves are really the result of summing up an
infinite number of relected waves from both ends.
Now in Cecil's case:
There is no tuner and S22 does exist and since there is no tuner the
reflected wave does enter the generator and some or all of it is obsorbed
in the real part of the source impedance. In this cases there is much more
loss in the system due to the load mismatch since if the generator has a 50
ohm source Z then all of the reflected power is terminated in the
generator, where as in Jon's case with the matching network only
the loss in the coax is important, i.e., no power is lost due to the
reflection itself.
Guess what Maxwell is right after all!!
William Osborne, AA5ZQ wosb...@nmsu.edu
The Klipsch Dept. of Electrical & Computer Engineering
New Mexico State University
cmo...@sedona.intel.com
Jon Bloom (KE3Z) <jbl...@arrl.org> wrote:
>Clearly, the transmitter is not
>acting merely as a reflective load, and *you can't treat it as though
>it is just a reflective load*. Again, as I keep saying, you have to
>consider the effect of the transmitter's signal on the system.
Jon, I feel somewhat like Copernicus standing in the presence of the
Pope telling him that the earth is not the center of the universe.
I trust that you are more open-minded than the Pope was and will
not put me under house arrest. :-)
If you can prove that any transmitter, under all circumstances of
mis-match, re-reflects the reflected waves, please just present it.
In the meantime, consider this experiment. Let's say we have a
transmitter putting out 10w into 50 ohm coax with a 50 ohm load
at the end. SWR=1:1. The 50 ohm load is in a signal generator/
network analyzer and is designed to absorb all forward power so
there will not be any reflections of the transmitted waves. Phase-
lock the signal generator (SG) to the transmitted signal so the waves
sent back down the coax by the SG are the same frequency as the
xmtr. The xmtr cannot tell the SG signal from a reflected wave.
Now vary the amplitude and phase of the SG signal. What you have
stated is that no matter what the amplitude or phase of the SG
signal seen by the xmtr, the xmtr will follow the amplitude and
phase so that the reflected waves are totally re-reflected. You
say all the measurements made at the xmtr prove it. I say you can
make those exact same measurements at the output of an antenna
tuner or anywhere up and down a xmission line. You are relying on
measurements being impossible inside the transmitter. It might have
been true for tubes, but it is certainly not true for transistors.
Now would you say in the above experiment if one increases the
SG power to 11w, the xmtr will continue to re-reflect all the
reflected waves? If it can't handle that one, it also cannot
handle out-of-phase reflected waves. Just because the forward
waves and the reflected waves add vectorally at the output of
the xmtr doesn't prove that the reflected waves have changed
direction. I believe they will 100% change direction only at a
Z0 match.
>What this demonstrates is that to consider what happens to the
>reflected wave, you *must* include the transmitter source in the
>analysis.
IMHO, including the transmitter source in the analysis is just the
smoke and mirrors you need to demonstrate that a transmitter is
really a magic device capable of perfectly matching any amplitude
and phase of a reflected wave. If I can prove what I am saying, will
you publish it in QEX?
cmo...@sedona.intel.com
>Cecil Moore (cecil...@delphi.com) wrote:
>
>: circuit.
>
>Huh? Any non-dissipative load will reflect 100%; there can be no power
>dissipated in such a load, so all must return. That includes all pure-
>reactive loads as well as shorts and opens.
Hi Tom, slightly out of context. I was talking about reflected waves.
At a Z0 match, there is a virtual open or virtual short. I admit that
a virtual non-dissipative impedance would re-reflect reflected waves but
I don't know how to create one.
>You can analyze a transmitter load as either just a (possibly complex)
>impedance across its antenna terminals, or as a line with forward and
>reverse waves on it. Done properly, either analysis will yield the same
>result. If you want to consider "forward and reverse waves" inside the
>transmitter, you better understand exactly how those waves are propagating
>and what impedance the transmission medium is; that all seems rather silly
>to me.
I have said over and over, I am not disagreeing with conventional analysis
methods and because of the convervation of energy, any reflected energy
reabsorbed by the transmitter CAN BE CONSIDERED to be 100% re-reflected
and merely subtracting from the transmitted power. The math works and that's
great. I am not arguing with our methods. I _am_ arguing with the myth that
a transmitter is a magical device capable of re-reflecting anything thrown
at it no matter what the amplitude or phase.
>And it's because that phase can be anything that I'm
>QUITE sure that it's inappropriate to talk about a "4 watt transmitter"
>in such simplistic terms. The "4 watts" is NOT independent of the load.
In an earlier thought experiment, I asked that a linear KW transmitter
be considered to be transmitting the 4 watts. Within reasonable limits
(50% reflected power is reasonable) it seems impossible to me that
a KW linear amplifier would decide not to transmit those two watts of
reflected power. If the reflected waves are in phase with the transmitted
waves, the forward power will increase to 6 watts and a Z0 match will
have been achieved. If the reflected waves are 90 degrees out of phase
with the transmitted waves, some of the reflected energy will be re-
absorbed by the transmitter which is so big it will never notice the
difference.
Gary and Jean Peach
exactly my point only modern circuits would be safe
doing this as they have some protective design
precisely because it has been recognised what will
happen if an unfavourable impedance is applied
73 Gary G7SLL
incidently I have designed protected circuits even
at DC for stalled motors precisely because I recognise
the maximum power transfer occurs when load
resistance equals source resistance. Protection
designs seek to limit power dissipation in the source
by current or voltage limiting they do not bring about
a match.
Gary and Jean Peach
Hi Cecil
Fine please see my answer to your protagonist.
This topic does interest me because of the related
topic of high and low pass filters. the are often used
between Tx and antenna. A load to dissipate the rejected
frequency is never provided hence it returns from whence
it came heating the final and reducing the efficiency of the
filter. And radiating all around the shack and else where
from the now mismatched feeder( at the unwanted frequency)
between the Tx and the filter. how many filters do you see
with resistive elements included within the design, nearly
always passive elements only.
Consider a hypothetical transmitter output power 1KW
with second harmonic 10dB down (OK so its only hypothetical)
now a low pass filter is inserted in the line and 100W of
second harmonic no longer goes up the line to the Antenna.
so where does it go with the type of low pass that is
currently (forgive the pun) generally available?
[ Tx ]=======[ filter ]====== low frequency ===== antenna
...........................II
......................[ filter ]====== high frequency ===== load
Not on your life.
Now as to the existence of S22, I think there is a crossed
wire here. If S22 exists then it had better have power handling
capability.
Consider a 50 Ohm Tx, that is when a 50 Ohm resistive load is
connected across its output terminals exactly half of the power
generated is dissipated in that 50 Ohm load (the sort of power
that boils water, remember Watts is joules per second a heat
unit, and if things go wrong it is the heat that does the damage)
now if the internal resistance (not impedance) of the PA is
sufficiently physically large to dissipate the other half of the
power then things are fine if not then get bigger tubes next time.
Now if a load other than 50 OHM resistive is used to terminate
the Tx then less than half the power is dissipated in the load
in consequence more than half is dissipated in the Tx, if no safety
device is fitted to the PA to limit it's power and it is not physically
big enough to dissipate the extra power from the mismatch then
poof, try bigger tubes next time, (for tubes also read transistors
and heat sinks).
In conclusion S22 exists (you measured it, If it has a VSWR then
it can be expressed as a complex impedance of the form R + jx )
the question is, CAN IT HANDLE THE POWER ?
When calculating that power use half the power generated plus
that which comes back from a bad load and any that is not at the
required frequency that has come back because the load doesn't
look the same at all frequencies.
Happy New Year
73 Gary G7SLL
cmo...@sedona.intel.com
>(By the way, this is
>essentially what Warren Bruene, W5OLY, did in his November, 1991 QST
>article to measure the output impedance of a transmitter.)
Thanks Jon, for referencing W5OLY's article. It proves my point.
He measured an SWR of 2:1 looking back into the transmitter output
proving that some of the power traveling toward the transmitter
is absorbed unless a Z0 match exists in the system. I think we agree
that the f1 reflections do not see the same impedance as f2 sees.
(The transmitter is transmitting on f1 and f2 is the test frequency
10 khz away from f1).
Warren comes to an erroneous conclusion when he says, "Therefore, I
have shown that a conjugate match does not exist in these correctly
tuned power amplifiers." It is explained in the ARRL Antenna Book,
15t edition, page 25-2. "The phase relationship of 180 degrees between
the resultant voltage and current waves creates a _VIRTUAL OPEN CIRCUIT_
to waves traveling toward the source..."
Because the transmitter is properly tuned, a conjugate match does indeed
exist for f1, just not for f2, the test frequency. The f1 forward waves
have created a virtual open circuit for the f1 reflected waves because
of the conjugate match at f1 (which Warren thinks doesn't exist). Detune
the transmitter so there is no f1 conjugate match, and f1 reflections
will be absorbed by the transmitter just like the f2 waves.
CMo...@sedona.intel.com
>There is no tuner and S22 does exist and since there is no tuner the
>reflected wave does enter the generator and some or all of it is obsorbed
>in the real part of the source impedance.
>Guess what Maxwell is right after all!!
Sorry Bill, your above statement contradicts Walter Maxwell. I agree with
you that the reflected wave can enter the generator and some be absorbed
in the source impedance. Maxwell says that it is impossible. If the tuner
is deliberatly detuned for purposes of experimentation, then the conjugate
match disappears and a real S22 does exist for the reflected waves.
W5OLY accidentally proved Maxwell wrong, he just didn't realize it
because he seems not to understand virtual opens and shorts.
Presumably, Maxwell didn't understand it either. In W5OLY's experiment
there actually was a conjugate match at the transmitter frequency 1
but not at the signal generator frequency 2. W5OLY proved that if a
transmitter is not conjugately matched, then some of the reflected
energy is dissipated in the xmtr source impedance, contrary to what
Maxwell states to be a fact. W5OLY's mistake was ignoring the effect
of the forward wave and assuming that the reflected waves saw the
same impedance that the signal generator saw. The signal generator
f2 saw 25 ohms, the f1 reflected waves saw a virtual open circuit.
This is what I have been saying all the time. The reflected energy
is 100% re-flected only if there is a conjugate match. If there is
not a conjugate match, i.e. a mis-match, then some reflected energy
is absorbed by the transmitter. In addition, the power generated by
the transmitter may or may not be reduced by the mis-match. Energy
is conserved so that the energy in the absorbed reflected waves CAN
MATHEMATICALLY BE CONSIDERED to be re-reflected and the transmitter
power reduced by that same amount. It works for analysis but it is not
what actually happens (conservation of momentum). The only thing that
can cause reflected waves to be 100% re-reflected is a (virtual) open
circuit, short circuit, (or pure reactance). Transmitters are not
intellegent or magic!
William P. Osborne
CMo...@sedona.intel.com <CMo...@sedona.intel.com> wrote:
>In article <46147.w...@nmsu.edu> <wosb...@nmsu.edu> writes:
>
>>There is no tuner and S22 does exist and since there is no tuner the
>>reflected wave does enter the generator and some or all of it is obsorbed
>>in the real part of the source impedance.
>>Guess what Maxwell is right after all!!
>
>Sorry Bill, your above statement contradicts Walter Maxwell. I agree with
>you that the reflected wave can enter the generator and some be absorbed
>in the source impedance. Maxwell says that it is impossible. If the tuner
>is deliberatly detuned for purposes of experimentation, then the conjugate
>match disappears and a real S22 does exist for the reflected waves.
Cecil: This does not in anyway contradict Maxwell's equations (I would
never be so bold)!
Can you please state clearly which of Maxwell's equations
are violated by the generator absorbing part or all of the reflected wave?
Gary and Jean Peach
I found the reference that I was looking for, it is
S. Silver Microwave Antenna Theory and Design
page 20, 21 section 2.4
--- Thevenin's Theorem and the Maximum-power Theorem ----
================================================
......................................
.................. by Thevenin's theorem the system is equivalent
to a generator of
emf VgZ12 / (Z11+Zg)
and internal impedance
Z12 - (Z12)^2 / (Z11+Zg)
feeding the load impedance ZL directly.
.
It follows then that Maximum power transfer will be achieved
with a load that is the complex conjugate of the internal
impedance of the effective generator:
ZL = Z*22 - (Z*12)^2 / (Z*11+Z*G)
.................................................................
[end quote]
transposing that we get
Z*22 = ZL + (Z*12)^2 / (Z*11+Z*G)
There is a nice diagram to illustrate it which would take
all night to do in ASCII, so please look up the book
Vol 12 in the MIT Radiation Laboratory Series
The other References are W.L. Everitt
R.W. King, H.R. Mimno, A.H.Wing
Transmission Lines, Antennas, and Wave Guides
McGraw-Hill New York, 1945, Chap 1
Of course the relationship is well derived and evolves from work by
Rayleigh, Everitt, Guillemin, Kirchhoff et.al
Maxwell in this case does not actually get a mention.
I think he can hardly take the blame for the lack of understanding
in this discussion. And therefore can not be wrong.
Gary Peach G7SLL
NEWBURY
ENGLAND
cmo...@sedona.intel.com
> frequency is never provided hence it returns from whence
Hi again Gary, Maxwell says that you are simply overcoupled or
undercoupled.
I agree with you. There seems to be some vested interest in perpetuating
myths within ham radio. The professional RF designers know the truth.
If they believed what a lot of hams believe, they would lose their jobs
and naturally change from professional to amateur. Hey, I think I'm
on to something! :-)
The maximum power transfer theorum assumes a generator and a load.
What we may be dealing with here is a maximum energy conversion
efficiency point that may or may not correspond to the maximum
power transfer point. Which would you rather have, an amplifier
that was 50% efficient or an amplifier that was 80% efficient?
If you are designing for maximum (RF power/DC input power) the
characteristic output impedance of the transmitter may not be the
load impedance. W5OLY's experiment seems to confirm that.
>S22 exists (you measured it, If it has a VSWR then
>it can be expressed as a complex impedance of the form R + jx )
>the question is, CAN IT HANDLE THE POWER ?
Ah Gary, you know we are not talking about practical stuff here. The
question is not can it handle the power. For the sake of this discussion,
nobody cares if it blows up or not. The question is, do any reflected
waves make it into the final tube of a mis-matched amplifier. I say it
does unless that final tube is magical and "knows" how to violate the
quantum physics law of conservation of momentum.
Most sane folk believe that reflected waves are theoretically 100% re-
reflected at a Z0 conjugate match (defined in the ARRL Antenna Book).
Most don't seem to realize that in order for all reflected waves to
be re-reflected in a mis-matched system, a conjugate match must exist
at the reflection point and therefore there cannot be such a thing as
a mis-matched system... but there is.
A transmitter that adjusts its output to perfectly match the phase of
the incoming reflected waves so that 100% will be re-reflected, has
achieved a conjugate match. Either the voltage is in phase and the
current out of phase or the voltage is out of phase and the current
in phase. If there is a conjugate match at any junction in the system,
then there is a conjugate match at every junction in the system, according
to the conjugate matching theorum. Someone has realized this, and that
is why we are being told to forget conjugate matches... that they are not
important and may not even exist.
If one believes that (1) all the reflected waves are re-reflected by any
transmitter whether matched or not, and one believes (2) the conjugate
matching theorum, then one arrives at a logical contradiction. One of
those concepts is not true. Some say (2) is not true. I say (1) is not
true. Maxwell says both are true and lives with the contradiction... or
else he believes in magic, like some other people posting here.
cmo...@sedona.intel.com
William P. Osborne <wosb...@nmsu.edu> wrote:
>On 6 Jan 1995 15:26:50 GMT,
>CMo...@sedona.intel.com <CMo...@sedona.intel.com> wrote:
>
>>In article <46147.w...@nmsu.edu> <wosb...@nmsu.edu> writes:
>>>reflected wave does enter the generator and some or all of it is obsorbed
>>>in the real part of the source impedance.
>>
>
>never be so bold)!
>Can you please state clearly which of Maxwell's equations
>are violated by the generator absorbing part or all of the reflected wave?
Hi Bill, note that I said Walter Maxwell, not James Maxwell. In
"Reflections", Walter Maxwell says, "Reflected power does not flow
back into the transmitter and cause dissipation and other damage."
He says it doesn't, you say it does. Sounds like a contradiction
to me.
cmo...@sedona.intel.com
>
> It follows then that Maximum power transfer will be achieved
> with a load that is the complex conjugate of the internal
> impedance of the effective generator:
Hi again Gary, You have strayed a long way from the original subject.
Do you think the power company dissipates 50% of their power in their
generators? Why does 75% of my DC input power make it to the load?
There has not been any argument with your quotes. This thread started with
a quote from Walter Maxwell's book, "Reflections". He says, "Reflected
power does not flow back into the transmitter and cause dissipation
and other damage." I say that is true only in conjugate-matched systems.
Maxwell says it happens whether there is a conjugate match or not...
hence the title of this thread.
We are not talking about maximum power transfer. We are not talking
about conjugate matches. We are talking about mis-matches and non-
maximum power transfers... what happens when things are not optimum?
There is no argument as to what happens when things are optimum. It
has been studied to death. What hasn't been sufficiently studied is
what happens when there is no conjugate match.
Let's have your opinion. Is it possible to re-reflect all the reflected
waves without achieving a conjugate match? I say if you set up a
mechanism of any kind, short of magic, at the output of a transmitting
tube that will re-reflect 100% of the reflected waves, that mechanism
can be nothing but a conjugate match.
Gary and Jean Peach
Thank you for keeping it going Cecil
> . Which would you rather have, an amplifier
>that was 50% efficient or an amplifier that was 80% efficient?
>
>If you are designing for maximum (RF power/DC input power) the
>characteristic output impedance of the transmitter may not be the
>load impedance.
no matter what the conversion efficiency Cecil the maximum that
you will be able to transfer from the transmitter to the load will be
half only, that is one of the fundamental laws of nature, whether
it is balls colliding or what ever. This has nothing what ever to do
with the efficiency of the design.
>The
>question is not can it handle the power. For the sake of this discussion,
>nobody cares if it blows up or not. The question is, do any reflected
>waves make it into the final tube of a mis-matched amplifier. I say it
>does unless that final tube is magical and "knows" how to violate the
>quantum physics law of conservation of momentum.
Surely Cecil the blowing up of the final is it's self the experimental
evidence of the presence of the additional power, or as I prefer to
think of it the power that never got away, (a bit like putting a blanket
over a radiator on a system without a thermostat and watching the
boiler blow). If you admit to the fact that it is possible that the final
may suffer damage, then that is tantamount to admitting the
existence of R22, (please note I used R cause that is what boils
the kettle). It's called the smoke test, some day there will be a
device for measuring the density and duration of the smoke then
we can convert it to joules, and thus derive the answer by direct
measurement, as instead of going inside to get it it will have come
out to meet us.
>.................................... "knows" how to violate the
>quantum physics law of conservation of momentum.
Best illustrated with one of those executive toys consisting
of 5 large ball bearings each independently suspended on
a piece of fine fishing line. The ball on the right is raised
a little way and released, the balls in the middle hardly move
but the ball on the left moves up almost as much as you
raised the ball on the right, it falls the balls in the middle
barely move the ball on the right rises almost as much as
the ball on the left fell from, the cycle repeats until eventually
the motion is in perceptible and finally dies out. Now where
did the original excitation energy that your hand put in go. Why
did all of the initial energy put in on the right dissipate in the
ball on the left. If this had been snooker (pool) balls lined up
touching then only the end ball would end up with most of the
energy of the struck ball. Now in one case we have reflection
and in the other case not, the difference between the two cases
is that in the first case the energy is converted into potential
energy by the increase in height of the end ball, and in the second
case there is no storage or conversion of energy. The energy
within the transmission system under discussion is either converted
to heat as in case 1 adiabatically or it is not converted as in the
absence of
R22 so where now does it go? if R22 does not exist then what
a) takes it's place if the energy is reflected b) some new undiscovered
mechanism which is 100% efficient converts the energy to some
other form of which we are unaware c) the energy is annihilated and
disappears from the universe entirely, thus adding to the
General Entropy isentropicly.
>Most sane folk believe that reflected waves are theoretically 100% re-
>reflected at a Z0 conjugate match (defined in the ARRL Antenna Book).
I am unqualified to comment upon most sane peoples beliefs, it
would be difficult to define a sane person, let alone identify one
it is all relative and my expertise does not lie in that direction.
A Wave does not know if it is reflected or not
so it behaves exactly like any other wave when met by some
discontinuity what ever the cause. a wave approaching the
transmitter from the transmission line will see the transmitter
as a load and behave accordingly. What throws the perception
into disarray is the presence of the new wave being generated
there will simply be vector addition of the two waves by voltage
and current sums and the new forward wave is born. This scenario
suggests that the transmitter efficiency is changed and a different
level of power is transferred to the load.
>Most don't seem to realise that in order for all reflected waves to
>be re-reflected in a mis-matched system, a conjugate match must exist
>at the reflection point and therefore there cannot be such a thing as
>a mis-matched system... but there is.
I think the former deals with this, the transmitter efficiency changes.
I choose not to answer the next para as I can not envisage
a magical transmitter that knows it has to change.
Your final para seems to point the finger at poor Maxwell
whereas the power transfer theorem is derived from another
authority please see my post referring to
Silver MIT Radiation Series McGraw-Hill sec 2.4
--... ...-- Gary G7SLL My opinions are the company
cmo...@sedona.intel.com
> you will be able to transfer from the transmitter to the load will be
> half only, that is one of the fundamental laws of nature,
Hi again Gary, this is interesting and I admit I don't understand how
you can put 100 watts of DC power into an amplifier and get 75 watts
delivered to the load when only half the power can possibly be transferred.
But that is not the subject of this thread. You keep trying to change the
subject... why? If you want to talk about the power transfer theorum,
please start another thread.
>If you admit to the fact that it is possible that the final
>may suffer damage, then that is tantamount to admitting the
>existence of R22,
You know, I think you are on my side against Walter Maxwell. If R22 (S22)
exists, then some reflected power will be dissipated in the final amp
unless there is a Z0 (Zg) conjugate match. Maxwell says no reflected power
will ever, under any circumstances, be dissipated in the final amp.
Seems you and I agree that he is wrong.
> Best illustrated with one of those executive toys consisting
> of 5 large ball bearings each independently suspended on
> a piece of fine fishing line.
This is uncanny. This is exactly the experiment I suggested to someone
else to prove that out-of-phase reflected energy does not get re-reflected.
> if R22 does not exist then what
> a) takes it's place if the energy is reflected b) some new undiscovered
> mechanism which is 100% efficient converts the energy to some
> other form of which we are unaware c) the energy is annihilated and
> disappears from the universe entirely, thus adding to the
> General Entropy isentropicly.
My point exactly. What other mechanism besides a conjugate match can
re-reflect the reflected waves? The Z0 conjugate match results in an
effective open circuit or short circuit for the reflected waves.
> there will simply be vector addition of the two waves by voltage
> and current sums and the new forward wave is born.
This is the crux of the problem. Vector addition works anywhere up
and down the transmission line without re-reflection. At a re-reflection
point there is a special in-phase vector addition that changes the
direction (momentum) of the reflections. Is there something besides
a Z0 conjugate match that can accomplish that feat?
> Your final para seems to point the finger at poor Maxwell
> whereas the power transfer theorem is derived from another
> authority please see my post referring to
Dang it Gary, look at the subject of the thread. I am not arguing
with the power transfer theorum. I am arguing with a statement in
Walter Maxwell's book, "Reflections". Do you have some ulterior
motive for trying to change the subject?
Now please, all you RF experts out there on the information superhighway,
besides a Z0 (Zg) conjugate match, what mechanism can 100% reverse the
momentum of reflected waves? If that mechanism exists, please tell me
about it. If that mechanism does not exist, Walter Maxwell (and his
unquestioning followers) are wrong.
mor...@ipers1.e-technik.uni-stuttgart.de
Just as a matter of curiosity, Who is Maxwell, and why is he wrong?
73, Moritz DL5UH
Charles T. Rauch
discussion. Let me say this, I just finished testing a group of
RF power amplifiers and I 1.) tried to confirm Bruene's
measurements as in Nov 1991 QST and with identical setup CAN NOT
confirm his measurements, 14 different amps all measured around
50 ohm source imp. I devised a different test method and had same
results, all amps measured 50 ohm source when properly tuned into
50 ohm load. Maxwell appears to be exactly correct, and the guy
who started the mess (Bruene) questioning him is absolutely
wrong. 73 Tom W8JI
--
Tom W8JI
Roy W Lewallen
interesting conversation last night with Tom Rauch, W8JI. Tom designs
RF amplifiers for a living, and has access to a wide variety of amplifier
types. He came up with a clever, and it looks to me, valid, way of measuring
amplifier output Z. He put a 1/8 wavelength line from the amplifier to a
30 dB power attenuator. To the other side of the attenuator he connected
another transmitter, running 200 watts output at a frequency a few kHz away
from the amplifier under test. He put probes on both ends of the transmission
line to measure the voltages. I'm sorry but I don't recall how he sensed
the currents. By looking at the voltages and currents at the frequency of
the second transmitter with a frequency-selective detector, he could infer
the output Z of the amplifier under test. This method doesn't require seeing
very small changes in large values, like other methods might.
The result of measuring 14 amplifiers of various types (including an old
DX-100 and a couple of solid-state ham rigs) is that virtually all of them
were very close to 50 ohms when the amplifier was adjusted for maximum
output into 50 ohms. He said that this was in conflict with the measurements
which Bruene had made.
Without following the discussion closely, I've had the feeling that
amplifier output impedance is a key issue, so thought I'd add this
information.
73,
Roy Lewallen
roy.le...@tek.com
Charles T. Rauch
Bruene. In every amplifier tested using two seperate methods all
amplifiers produced maximum power when their operating impedance
was equal to the conjugate of the line impedance. This included
class C amplifier with 73% eff in plate input to load output
power. Generator models do not work with non-linear circuits, and
dc rules do not aplly to reactances. Impedance and resistance
are not interchangable. 15 different amps tested 2 differnt ways
and all the results confirm that conjugate match is required, and
yes the eff can be much better than 50% in ac circuit.
--
Tom W8JI
Steve (Stephen) G. Miller
> British Physicist
> a Clerk of Penicuick
> Born 13 November 1831 in Edinburgh Scotland
> Educated at the Edinburgh Academy, University of Edinburgh
> Trinity , Cambridge
>
> Among many other Scientific topics he described the behaviour
> of waves mathematically, working upon 'Faraday's lines
> of force' and predicted Electromagnetic Waves before they had
> been generated by man, it was Maxwell's theoretical work which
> stimulated Heinrich Hertz to undertake practical experiments,
> thus producing these waves.
> IMHO Maxwell can not be wrong in this case as the topic under
> discussion is based upon work of others viz Thevenin, Everitt
> King, Mimno and Wing. So as he was not present at the scene
> of crime, the jury will most probably return a verdict of not guilty.
> Doesn't it make you want to CURL up and DIV.
>
> 73 Gary G7SLL
>
Sorry Gary, this is not the Maxwell being disputed. If James Clerk Maxwell
is wrong then someone should tell my old college professors ;). This
thread concerns Walter Maxwell who published a book in the ARRL library.
BTW, several of the arguments in this thread are incorrect, Caveat Emptor!
--
+=============================================================+
| Steve (space cowboy) Miller | WD8IXE |
| Email: mill...@dmapub.dma.org | Ridin' the aethereal waves |
+=============================================================+
Dave Kelley
How did you determine the output impedances of the amplifiers?
73, Dave NB4J
==========================================================
Dave Kelley Electrical Engineering Dept.
email: da...@yoda.psu.edu Pennsylvania State University
cmo...@sedona.intel.com
Gary and Jean Peach <wise...@cix.compulink.co.uk> wrote:
>
> MAXWELL James Clerk (1831 - 1879)
> British Physicist
> IMHO Maxwell can not be wrong in this case as the topic under
> discussion is based upon work of others viz Thevenin, Everitt
Hi Gary, you are apparently not reading the replies. The Maxwell in
question is Walter Maxwell, author of "Reflections" not James Clerk
Maxwell. The title of the thread should have been, "Is Walter Maxwell
wrong." I appologize for the confusion.
W8JI Tom
energy is reflected under all conditions, he said if the network was
adjusted for maximum power transfer there would be a conjugate match and
all reflected power would be re-reflected towards load in phase with
initial wave. This statement is exactly correct, and it is very simple to
measure the output port of an amplifier under actual operation and prove
it.
cmo...@sedona.intel.com
> Which view is physically correct? Really neither is...
>The concept must represent reality right? Well, no.
Gary, honest to God, I didn't do the above on purpose. I was just
trying to trim down your posting when the above resulted. You are
talking metaphysics and I don't think any of us can prove we exist,
Ayn Rand said, "Existence exists and A is A" (borrowed from Aristotle).
>Analyzing transmitter output tube dissipation should be done by treating the
>feedline/antenna combination as a load of the impedance observed at the
>connector on the back of the transmitter.
I have done exactly that in a later thread named "reflected energy". I
shared my insights with one of the resident Internet antenna Gurus and
he was so upset, he broke off communications with me.
>It is the inablility to handle a strange load impedance that causes
>overheating. Forward and Reflected transmission line power are
>meaningless in this analysis.
While approaching forward/reflected energy from a quantum physics
basis, I performed a transform and guess what, I was staring at
Kirchhoff's equation. It proves that reflected energy can enter the
transmitter and cause damage. Conceptually, visualize the moment when
the generator voltage goes to zero (twice per cycle, 14,000,000 times
per second at 7 MHz). There is nothing to withstand the onslaught of
external energy, be it a charge on a capacitor, the field stored in
a coil, or the energy in a reflected wave. Quanta packets of energy
move back into the transmitter and can cause damage. In our ham
language, it means a capacitor stores some energy while it is being
generated and dumps it back when the generator voltage is zero. There
is only one thing known to man that can stop it --- a conjugate match.
If an equal amount of energy has been stored in a coil, it can
neutralize the effect of the capacitor --- that's what resonance and
conjugate matches are all about.
Now we are faced with a choice. We can religiously believe in Walter
Maxwell, a great man, but a limited mortal nevertheless, or we can
revise our concepts in much the same way we were able to explain the
progression in the orbit of Mercury with Einstein's theory of relativity
but not with Newton's laws of gravitation. Was Newton a great man? ---
yes! Was he ever wrong? --- unfortunately, his theory didn't explain
everything that needed explaining.
It is way past time to lay to rest the myth that reflected energy cannot
cause damage to a transmitter. IMHO, a transmitter is not magic. Some
of the posters to this newsgroup think it is. Can magic be proven? Now
that is a real metaphysical question that I think I will simply avoid.
Cecil Moore
>BTW, several of the arguments in this thread are incorrect, Caveat Emptor!
>
IMHO, the incorrect arguments are based on one-dimensional thinking.
DC, RMS, Power, Rg, and RL are one-dimensional. The QEX article,
"Where Does the Power Go?" is one-dimensional. I don't recall seeing
a "J" anywhere. All the examples are either in phase or 180 degrees
out of phase which is simply the other direction in that one-dimension.
Waves are two-dimensional. Two waves, 90 degrees out of phase, can
pass without interaction at a transmitter output terminal. Would
you agree that I have proven Maxwell to be incorrect when he said,
"Reflected power does not flow back into the transmitter and cause
dissipation and other damage."? IMHO, this statement is true if and
only if there is a conjugate match - which there isn't in my Thevenin
generator example.
73, Cecil, KG7BK
Gary Coffman
>
>Just as a matter of curiosity, Who is Maxwell, and why is he wrong?
James Clerk Maxwell or Walter Maxwell?
Neither one is wrong.
Gary
--
Gary Coffman KE4ZV | You make it, | gatech!wa4mei!ke4zv!gary
Destructive Testing Systems | we break it. | emory!kd4nc!ke4zv!gary
534 Shannon Way | Guaranteed! | ga...@ke4zv.atl.ga.us
Lawrenceville, GA 30244 | |
Gary Wescom
>From: cmo...@sedona.intel.com
>Subject: Re: Is Maxwell wrong?
>Date: 9 Jan 95 13:51:53 GMT
>Now please, all you RF experts out there on the information superhighway,
>besides a Z0 (Zg) conjugate match, what mechanism can 100% reverse the
>momentum of reflected waves? If that mechanism exists, please tell me
>about it. If that mechanism does not exist, Walter Maxwell (and his
>unquestioning followers) are wrong.
Sure, other things besides a conjugate match can have 100% reflection. If the
driving source is either a pure current source or a pure voltage source, 100 %
re-reflection will occur. This is actually seen in situations in which coax
cables are driven by very low impedance buffers. Series 75 Ohm or so
resistors are usually employed at the output of video buffers to absorb any
reflected voltage on a 75 video feed system. Of course, this does mean that
about half the available output power is usually absorbed by the series
resistor.
True, this may be a bit off of the reflected power/tube output resistance
question but I think it answers your specific question.
Gary Wescom N6CH
gr...@pge.com
W8JI Tom
describe a non-linear impedance. Look at the rules. A more correct model
is a horsepower limited generator system. This is a non-linear impedance,
and the efficiency can reach nearly 100%. Maximum power is transferred
when the load equals the source impedance. Think about that. The only
amplifier that has a linear impedance is a class A amp, and it follows the
50% rule. A class C amp can be 99.999% eff and still follow the conjugate
match rule because it is impedance limited, not Rs limited. If we just
quit using the wrong rules we would be fine. The continued use of Thevenin
models with non linear impedances or power limited sources violates rules
of the basic theorems. Does everyone agree that maximum power transfers
when a conjugate match occurs? Does everyone agree that a class C
amplifier can have 90+% eff? Before we can discuss the transmission line
we had better agree on the source!
N7WS
error so I'm reposting here with corrections.
I have read with interest the many comments in this forum regarding the
question of whether the reflected power in a mismatched system is or is
not dissipated in the final amplifier of a transmitter and whether
conjugate matching is taking place. The following are some of my thoughts
on the subject:
1. An amplifier (or any other two port device) DOES have an S22. By
definition, S22 is a measure of the signal out of port two due to a signal
applied to port two. The value may indeed be zero, but S22 exists.
2. When looking into the input of a transmission line of unknown
impedance, there will be unique value of impedance R+/-jX at each
frequency. A measurement of this impedance (not SWR) does not indicate how
long the transmission line is, what terminates the line, what the line
impedance is or what the SWR is.
3. In a linear system, maximum available power is transferred to a load,
R+jX, when the generator impedance is R-jX, or conjugately matched. This
is not necessarily the condition for greatest efficiency.
4. For a class A amplifier to generate maximum power, there is an optimum
(plate, collector, or drain) load impedance (ZL) which facilitates this.
If there is no phase shift in the active device to compensate for, this
optimum load impedance will be real (non-reactive).
5. The presence of feedback (positive or negative) in the active device
may require a load impedance (ZL) that is complex.
6. Any output network that can transform the load seen at the amplifier
output port (Zo) to ZL will cause the device to deliver maximum output
power.
7. Under the foregoing conditions, the active device doesn't have a clue
as to whether the transmission line is matched (no standing waves) and it
doesn't care either.
8. If the output network is properly tuned, Maxwell is correct when he
says the mismatch at the line input is absorbed by the amplifier output
circuit. The effect of mismatch is only a deviation from the design loaded
Q. This may increase the circulating current in the output network and
result in lower efficiency for that reason, but there is no "reflected
power" seen by the active device. The only power dissipated by the device
is that dissipated in the dynamic plate, collector or drain resistance.
9. To improve efficiency, the conduction angle of the active device can be
reduced and the ZL adjusted for greater efficiency rather that maximum
power transfer. In this case, there is no conjugate match, however, this
does not mean that reflected power in the transmission line somehow
increases device dissipation. If the output circuit can still be tuned to
present the desired ZL to the device in the presence of mismatch at the
output, then the active device continues to operate as desired. The device
still does not know or care that there is reflected power in the
transmission line. If the active device was disconnected and replaced with
a resistance equal to ZL then a measurement at any point in the system
would show a conjugate match.
10. Warren Bruene’s (W5OLY) November 1991 QST article describes a
procedure for measuring the output return loss of an operating amplifier.
Based on his return loss measurements, he purports to be able to know what
the dynamic source resistance is. As my previous discussion states, I
don’t believe it makes any difference whether we know this or not. By
design there is no conjugate match and no necessity to prove that it does
or does not exist. There is a flaw in Bruene’s conclusions, however. He
makes a scalar measurement and determines SWR and then states flatly that
he knows what the real part of a complex impedance is. He knows no such
thing. What he knows, assuming we accept his methodology, is the magnitude
of a complex reflection coefficient vector. He does not know the phase of
the vector. Indeed, mapping the length of this vector onto a Smith chart
and rotating it through all possible phase angles draws the familiar SWR
circle. There are an infinite number of impedances on this circle that
satisfy the SWR but are a long way from the real axis. Bruene may argue
that because the system is resonant, there is no imaginary component to
measure. I don’t believe that this is necessarily the case. Just as the
real source resistance does not equal the load resistance for maximum
efficiency, it is possible that the imaginary part acts the same way,
especially in a dynamic case and in the presence of feedback. If this is
true, then the output network is actually operating off-resonance and the
measurement must consider the imaginary component.
11. I have some other problems with the Bruene article. The Kenwood TS-930
solid-state amplifier is untuned and shouldn’t be considered in a
discussion that is trying to dispel myths about tuning for conjugate
matches. In fig. 1, the length of transmission line is supposed to
simulate the phase delay or electrical length of the output network.
Bruene can’t have it both ways; if there is a phase shift between the
output port and the ideal transformer, then he has made my case that a
scalar measurement cannot determine the value of Rs. If the system is
truly resonant, then the separate transmission line is superfluous.
Furthermore, if the line section is removed (as I think it should be),
network losses should be included as either series or parallel loss
resistance in the network. When this is done, the argument expressed in
end note 4 goes away. The argument is faulty anyway. For example, if all
the network losses were represented by a loss resistance in parallel with
the primary (tube) side of the ideal transformer, then the error is such
that Rs is actually higher than predicted by the measurement. If the SWR
is represented by Rs/Rl, then the loss makes the SWR measure better than
it is. However, in the opposite case (Rl/Rs), the SWR measures worse than
it is. Conversely, if the losses are simulated with a resistance in series
with the ideal transformer primary, then the errors are in the opposite
direction, Therefore, losses just add uncertainty to the measurement.
12. If we continue to insist on knowing what the device S22 is, then the
most straightforward (although time consuming) method is this: Use a
system similar to Bruene’s except use a vector network analyzer, or
equivalent to measure the operating impedance at the amplifier output.
Turn off the amplifier, remove the tube, transistor, or FET from the
circuit. Replace the device with an adjustable RLC network. Without making
any tuning changes, adjust this network until the amplifier output
impedance measures the same as before. Remove the RLC and measure its
impedance. This is the operating impedance of the active device. If you
want to know whether this is a conjugate match, measure the impedance
looking back into the output network (from the tube side). If you
SIMPLY-GOTTA-ABSOLUTELY-MUST-KNOW-THE-ANSWERS, please arrange to donate
(permanently) an Alpha 76 to the undersigned and I promise to make the
measurements.
13. One further experiment. We’ll use my new Alpha. We will drive a 50 Ohm
low-loss transmission line 1 wavelength long. At the end of the line, we
connect a dual, high-directivity, 30 dB directional coupler. The coupler
through-line output is terminated in a 50 Ohm, 40 dB attenuator. The two
coupled ports are terminated in 50 Ohm, 10 dB attenuators. The outputs of
the attenuators are connected to HP8481A power sensors and HP438 power
meters. Power readings of 100 mW correspond to 1000 W at each port. We now
apply RF drive to the amplifier. We alternately adjust the output tuning
for maximum output while increasing the drive until the output power meter
shows 100 mW output. The reflected power meter shows zero while the
forward power meter shows 100 mW. We note DC plate voltage and current and
compute the product to arrive at the DC input power. Knowing the DC input
and the RF output, we compute the power-added efficiency. We now open the
connection between the directional coupler and the 40 dB attenuator and
insert a series low-loss capacitor with a reactance equal to -35 Ohm at
the operating frequency. The load impedance is now 50-j35 and the line SWR
is 2.0. We also add a series low-loss inductor with a reactance of 35 Ohm
at the operating frequency between the amplifier output and the line
input. Without any adjustment of the transmitter, we now measure a forward
power of 112.5 mW and a reflected power of 12.5 mW. The output power meter
continues to indicate 100 mW. The DC input power is unchanged as is the
efficiency. There are 125 W of reflected power in the line, but the
transmitter doesn’t know it. We now remove the 35 Ohm inductor. Keeping
the RF drive the same as before, we readjust the output network for
maximum output. We again measure 100 mW at the load, 112.5 mW forward and
12.5 mW reflected and the same DC current and efficiency. All we have
effectively done is retune the output network to include the 35 Ohm
inductive reactance necessary to maximize the transmission line current.
The tubes still don’t know a thing about the 2.0 SWR. The circulating
current in the output network may have increased, causing slightly higher
network losses, but the tube efficiency and dissipation are unchanged. To
demonstrate that the reflected power occurres only in the transmission
line, let the line length be zero. The load impedance, Zo, is still
50-j35. Can we still match this with the output network? Of course. Is
there a load mismatch? Yes. Is there any "reflected power"? No there is
not. Are there any negative consequences to the amplifier? No.
Respectfully submitted:
Wes Stewart, N7WS
n7...@aol.com
N7WS
One last post (I hope)..
Rather than repeat a lot of what has been posted in this and other
threads, I'll just make some general comments in response. The guilty will
know who they are.
1. It has been said that reflected power is not real. I will answer this
in a word.... RADAR. A radar system works by generating an RF signal,
coupling it to the impedance of free space (120*pi Ohm or ~377 Ohm) where
it propagates until it encounters a mismatch which reflects some of the
REAL energy back to the receiver. (Sound familiar?) At both the
transmitter and receiver, the impedance transforming network used is
commonly known as an antenna. Its function is to match the impedance of
the transmission line (coax or waveguide) to 377 Ohm. A target reflects
some of the incident energy because the impedance of its surface
is_not_377 Ohm and is therefore a mismatch to the incident (forward) wave.
If the reflected energy is not real, then it cannot be received. If you
don't believe this, the next time Smokey pulls you over for speeding, just
tell him his radar was giving him an imaginary reading. Or tell it to the
F-15 pilot that turned on his fire control radar in the hangar and burned
out the receiver..
BTW, the shielding around your transmitter works because of impedance
mismatch and reflection as does your low-pass filter. The parasitic
elements in your Yagi-Uda antenna do their thing by reflecting incident
energy. In the transmitting case, the driven element delivers energy to
the parasitic elements, which because the real part of their load
impedance is ~zero have no choice but to re-radiate it. If you doubt this,
open the center of your parasites and put in some resistive terminations
and check the gain (loss actually).
2. Several of the arguments here are using the term source resistance as
if it is something you can measure with an ohmmeter. This is nonsense. In
the vacuum tube case, the dynamic plate resistance is determined, not with
an ohmmeter but by calculation. With a fixed grid-cathode bias, the plate
voltage is varied and the change in plate current is measured. (OK, some
will call this an ohmmeter) Plotting this function (over a limited range)
yields a slope which could be replicated by replacing the tube with a
resistor. The value of this resistor is the dynamic plate resistance. It
is a mathematically derived number. It is also non-linear. The plate
current certainly goes to zero when the voltage is zero, but it doesn't go
to infinity with infinite plate voltage. All the arguments using ideal
voltage or current sources with linear resistors are flawed.
3. Many of the posts in this and other threads use hypothetical loads that
are simple, ideal reactances in an attempt to analyze something or the
other that is occurring in an instant of time. Guys, we are using
transmitters that have tank circuits. While they may have a net reactance
value that is capacitive or inductive at some point, they still have the
ability to store energy in both electric and magnetic fields. Because some
hypothetical source delivered a half-sine current pulse to the tank and
then shut off, doesn't mean all that energy has to be immediately returned
to the source (which has just gone into a very high impedance state, BTW).
This energy can actually circulate in the tank for a while until it is
dissipated in circuit losses. Just because the tank is off-resonance and
looks capacitive or inductive doesn't mean there is no circulating
current. It just means the current isn't as high as it would be at
resonance. If you can’t fathom this, then consider the response of a
narrow-band i-f or audio filter to noise impulses. The noise pulses aren’t
occurring at the resonant frequency of the filter, but they sure can cause
the filter to ring long after the impulse is gone.
Respectfully submitted IMHO
Wes Stewart, N7WS
n7...@aol.com
cmo...@sedona.intel.com
W8JI Tom <w8j...@aol.com> wrote:
>You model still is using a Thevenin source to model a power limited
>(circuit with source impedance limiting power available). It will continue
>to come out the same as you continue to use the wrong model. Instead
>plaese substitute a current limited voltage source in series with the Rs
>and set the impedance higher than the Rg. This will be closer but will not
>include the standing dissipation in the PA device. Try the same sample but
>limit the Eg to 25 volts, the I max to 1 amp and the Rs to one ohm. Now
>you are closer.
Hi again Tom, pick whatever model you prefer. Make it as complex as you
want. Then answer this question: What keeps external energy out of the
transmitter when the single "current limited voltage source" goes to
zero which it does twice per cycle? Unless you include a variable
reactance internal in your model, any energy lurking at the transmitter
terminals when the generator goes to zero, will invade the transmitter.
If you include a variable reactance internal in your model, you have
invented a magical conjugate match that is not there in reality.
cmo...@sedona.intel.com
W8JI Tom <w8j...@aol.com> wrote:
>Gee I'm glad it's a small % :-). Anyway I agree with you on most of this.
>Instead of clarifying the little bit that needed clarification in
>Maxwell's text, Bruene set us all back and continues to set us back
>further every time he gets his pen out.
Hi again Tom, don't be too hard on W5OLY. He inadvertently debunked the
myth in his latest CQ article. He said, "When the generator is turned
off... all reflected power will be dissipated in Rs (signal generator)
on its first trip back." Considering that the generator is turned off
14,000,000 times per second at 7 MHz, that's a lot of reflected power
dissipated in Rs.
Walter Maxwell just was not careful enough and started this myth. He
clearly states, "When a _matched_ RF power amplifier is actively
supplying power when the reflected wave returns, the reflected wave
encounters total reflection..." Note the word _matched_. The implication
is clearly that something else happens with mismatches. He neglected to
say _matched_ later when he said, "Reflected power does not flow back
into the transmitter and cause dissipation and other damage."
Why didn't he preface that statement with, "In a matched system?"
He probably thought nobody in their right mind would assume otherwise.
Here's the mother of all mismatches. Assume a solid-state transmitter
designed to work with a 50 ohm load. Connect a 5/8 wavelength of
50 ohm lossless coax and a 300 ohm load. No matter what the poor
transmitter magically accomplishes, the darned old reflections maintain
their 90 degree relationship to the transmitted wave and thus sneak
straight to the final, unopposed.
Cecil Moore
> It looks to me like the camp which is saying the reverse power is absorbed
> in the source has a bit of a problem with this one.
Hi Roy, don't know how closely you have been following the thread. Nobody has
said that reflected power will not be re-reflected if the voltage and current
vectors are in phase. In your example, the voltage and current vectors are
in phase so all reflections will be re-reflected.
Please set up an example where the reflected voltage/current vectors are 90
degrees out of phase with the transmitter terminal voltage/current vectors
and see what happens.
Your example is one-dimensional. Please don't neglect the other 358 degrees.
These simplistic examples have the voltage/current vectors in phase at all
the measurement points so the forward and reflected waves _will_ respond
as predicted. Try 5/8 wavelength instead of 1/2 wavelength and see what
happens.
Again, no one has said that reflected waves are not re-reflected. What I have
said is that the part of the reflected wave that is out of phase with the
transmitted wave will not be reflected. The part that is in phase with the
transmitted wave will be re-reflected. I hope that is 100% clear to everyone.
A reflected wave arriving 45 degrees out of phase with the transmitted wave
will be half re-reflected. A reflected wave arriving in-phase with the
transmitted wave will be 100% re-reflected. A reflected wave arriving 90
degrees out of phase with the transmitted wave will totally dissipate in
Rg if Rg=50. If Rg is not equal to 50 ohms, some will dissipate and some
will be reflected at that point.
73, Cecil, KG7BK, OOTC
Cecil Moore
> 50 ohms ____________________________
> ----/\/\/\--/___________________________/--
> | |
> | |
> source O /
> 2 V RMS | \ (resistor)
> | / 25 ohms
> | \
> |__________________________________________|
Hi again Roy, I forgot something. Your above example is not a conjugate
match, yet it re-reflects all of the reflected waves. May I suggest a
name for this condition? How 'bout we call it a "partial conjugate
match"? That is the condition where the reflected waves are in phase
with the forward waves but are not of a magnitude t result in a true
conjugate match. The partial conjugate match is what has had the experts
fooled for lo these many years.
Hasn't anyone except me read "Flatland"? Remember what happens when one-
dimensional creatures meet the two-dimensional creature for the first
time? Well, all you one-dimensional reflected wave thinkers... welcome
to two dimensions. What are we going to do with three dimensions?
73, Cecil, KG7BK, OOTC (ain't life a kick?)
John Mallick
the voltage and the current on the transmission line are determined
by the solution of the transmission line equations; there are two
solutions, which we interpret as forward and reverse travelling waves.
To determine the coefficients in front of each (their magnitude and
phase) we need to look at the "boundary conditions" imposed upon the
transmission line at both ends. At the load end, we have the load
impedance determining the ratio of V to I. At the transmitter end,
we have V and I determined by whatever impedance the transmission line
terminated in its load presents to the transmitter. I can use the
Thevenin equivalent for the transmitter to determine that V and I.
If you actually write out the solution for V and I in the sinusoidal
steady state case, you'll find that for any passive load the solution
in terms of forward and reflected waves can be re-arranged into the
form of an amplitude-modulated *forward travelling wave*. (In the
case of a purely reactive load or a short/open, this degenerates into
a pure standing wave). The V and I peaks/troughs are 90 degrees out
of space phase and represent points of excess magnetic or electric
energy storage. At all points along the transmission line, the time
average power is flowing toward the load.
Sure, you can calculate the reflection coefficient at the transmitter
end; just use the Thevenin impedance of the generator. But thinking
about the reflected power as partially being absorbed by the generator
impedance is not useful in the sinusoidal steady state.
Just my $0.02 to the thread...
73, John WA1HNL
PS...think about the case of a DC voltage source connect through
a 50 ohm resistor to a 50 ohm cable which is open-circuited at the
load. Here, you can resolve the solution into forward and reverse
travelling waves with non-zero power, but the net power flow is
zero. And the "reflected wave" power does not get absorbed by
the resistor, since it's power dissipation is zero.
--
.... .. .... .. .... .. .... .. .... .. .... ..
John A. Mallick WA1HNL E-mail: mal...@crd.ge.com
GE Corporate Research and Development Phone: (518)-387-7667 (W)
Schenectady, NY 12301 FAX: (518)-387-7592 (W)
.... .. .... .. .... .. .... .. .... .. .... ..
"Work like hell. Tell everyone everything you know. Close a deal
with a handshake. And have fun." --- "Doc" Edgerton
Tom Bruhns
..
: How 'bout we agree on the following definitions?
: Z-match - The point in a conjugately-matched transmitting system
..
: partial Z-match - The point in a transmitting system where the
..
: So let's state an axiom, to be proved by W8JITom and me.
: Unless there is a Z-match or a partial Z-match in the transmitting
: system, at least some reflected energy will enter the final amplifier
: and cause dissipation and/or damage. There is no magic barrier across
: which reflections cannot cross. The barrier is not magic and is achieved
: by either a full Z-match or a partial Z-match.
Independent of the proof of this, I'm trying to understand the practical
significance. From the standpoint of designing equipment, I must
understand the operation of all components, and for a transmitter, that
means understanding what happens at different phases of reflected waves, or
equivalently, a variety of different impedance loads with various reactive
and resistive parts. Is this "Z-match, partial Z-match" concept something
that can help me?
BTW, did anyone here ever carefully define "conjugate match" for this
thread? I'm a little bothered by some of the things tossed around. For
example, if I have a line of impedance Z0 = 50+j5 ohms, then for no
reflected waves, what impedance should the line be terminated in?
(I know the answer to this, but what are the implications with respect to
this "conjugate match" that's been tossed about here so lightly?)
73, K7ITM
Gary and Jean Peach
Cecil Wrote
> "Gary and Jean Peach" <wise...@cix.compulink.co.uk> writes:
>
> > When Zo = R +jX and x = 0 , Zo = R which I have used
> > throughout for reasons of KISS.
>
> Hi again Gary, I am in agreement with the above and have never disagreed
> with it. It is a smokescreen to change the subject. My question is:
> What happens when Z0 is not R+jX? What happens when X does not equal
> zero. What happens when Z0 does not equal R? The "Keep It Simple Stupid"
> is really keeping it Stupid.
>
> 73, Cecil, KG7BK, OOTC
you sure try hard to alienate your allies Cecil
The real water heating part of the energy still heats water
it doesn't vanish just because it has changed phase, the
(voltage and current)
vector shifts around. The magnitude of the reflected wave
is always less than the magnitude of the generator, in
consequence the vector sum can never become Zero,
even at the most unfavourable ( or favourable depending
upon your perspective) phase shift.
My stupid answer to that is: it doesn't shift by 45 degrees.
The new transmitted wave is as we agreed in the first
place the vector sum of the new generator wave and
the old wave reflected.
I found a new toy to show what happens, my 4 year
old grandson showed me the answer. He has a large
plastic helix, the type sold to the kids to climb down
stairs. The thickness of the turns is very small compared
to the diameter so it has a very low attenuation.
If you hold an end in each hand flat against the palms,
and turn both palms up.
The spring sits like an inverted 'U' or an 'n' in front of you.
now if you hold one hand still and pump the other hand
gently up and down you can easily see the compression
wave travel along the axis of the spring. What do you know
when it gets to the stationary hand it comes back to the
moving hand you can time the movements to reinforce
the wave or mess it up have fun. But what happens
when you stop pumping ?
Well the wave propagates back and forth for a bit and then
it doesn't quite reach the end.
Next it travels toward the distant end and it doesn't quite
reach that either.
This goes on for a bit travelling less and less.
You know what ?
The last bit of the spring to move is right in the centre.
If that doesn't mean reflection from both ends,
tell me about it.
The wave dies out towards the middle because it is
DISSIPATED at *BOTH* ENDS of the transmission
medium.
And that's KISS for KIDS
73
Gary Peach G7SLL Jean Peach
Wisdom to perceive, Lace the Ultimate thread
Grace to acknowledge
John Boyer
: cmo...@sedona.intel.com wrote:
The complex conjugate of 50+j5 is 50 -j5.
Cecil Moore
>and resistive parts. Is this "Z-match, partial Z-match" concept something
>that can help me?
transmitters. Do you believe there is a magic boundary across which
nothing is allowed to cross? There are examples of matched systems that
re-reflect all the reflected waves. There are examples of mismatched
systems that re-reflect all the reflected waves. Are those all the
cases? The answer is no and the partial Z-match explains why a mis-matched
system can still re-reflect all reflected waves. Try putting down on
paper the vectors in the latest posting under "reflected waves" and if
you can, you will understand.
>(I know the answer to this, but what are the implications with respect to
>this "conjugate match" that's been tossed about here so lightly?)
effect on reflected waves. It takes an impedance change to have any
effect. I don't think you can make a measurement at a single point on
a transmission and prove a conjugate match. The question is, does a
conjugate match result in a Z-match?
Cecil Moore
>Roy Lewallen's post goes to the heart of the matter. The behavior of
>the voltage and the current on the transmission line are determined
>by the solution of the transmission line equations; there are two
when the signals are out of phase?
>end; just use the Thevenin impedance of the generator. But thinking
>about the reflected power as partially being absorbed by the generator
>impedance is not useful in the sinusoidal steady state.
Don't you think it is useful to know the truth behing your rules-of-
thumb?
Gary Coffman
>"Gary and Jean Peach" <wise...@cix.compulink.co.uk> writes:
>
>> When Zo = R +jX and x = 0 , Zo = R which I have used
>> throughout for reasons of KISS.
>
>Hi again Gary, I am in agreement with the above and have never disagreed
>with it. It is a smokescreen to change the subject. My question is:
>What happens when Z0 is not R+jX? What happens when X does not equal
>zero. What happens when Z0 does not equal R? The "Keep It Simple Stupid"
>is really keeping it Stupid.
What happens? Why the same thing happens of course, only now we have
to invoke sin(phi) and cos(phi) in the equations. That makes it messy
trig, but doesn't change the result one whit because when we simplify
the equations, sin(phi)*cos(phi) always equals zero.
>Will somebody just please answer this question: By what mechanism, short
>of magic, does a transmitter re-reflect a reflected wave that is 90
>degrees out of phase with the transmitted wave? For a transmitter to
>be able to re-reflect all reflected waves no matter what their phase,
>it must be able to shift its transmitter wave phase by plus or minus
>45 degrees. If it does that, it will have achieved a conjugate match.
>If it achieves a conjugate match, I agree that all the reflected energy
>will be re-reflected. How does a transmitter shift its transmitted
>wave phase by 45 degrees?
There's no magic Cecil, it's just that there's no real load in the
transmitter to dissipate the waves, so they get absorbed in the
circulating tank currents, vector summing at whatever phase they
happen to be at that instant, and the energy goes back out on the
next cycle.
Depending on whether the vectors aid or oppose, the load line of
the tube will be shifted up or down from where it would be for a
conjugate matched case. Given that the drive level doesn't
change, tube efficiency will then go up or down, and the tube
may run hotter or cooler than it would in the conjugate matched
case. But this isn't "reflected power" being dissipated in the
tube "load resistance", it's just good old DC power from the
power supply.
Cecil Moore
>
> you sure try hard to alienate your allies Cecil
don't learn anything. I only learn when I'm wrong and I learn a lot. :-)
73 and good luck, Cecil, KG7BK, OOTC
Gary Coffman
>In my opinion this is a thread where technically astute and curious people
>have gathered. This is a longer posting than I ever intended to make, but
>from the comments on this thread it is very necessary. I hope everyone
>forgives me for using up their bits. I would like to add that anyone that
>thinks the efficiency can't be more than 50% with a conjugate match is
>mistaken.Maximum power transfer always occurs with a conjugate match.
>Please go back and read the rules on Thevenin's theorem.
>
>How to model an Vacuum Tube (or any other) RF Power amplifier: The
>amplifying
>device in a vacuum tube RF power amplifier can be viewed as a variable
>resistance in series with a source voltage. The resistance is controlled
>by
>varying the resistance with a control voltage applied to a grid in the
>device. The controlled resistance has unique characteristics that vary
>with
>the tube type, the most important are limited current and voltage
>capabilities and increasing non-linearity at low voltages and high
>currents.
I've deleted the rest of Tom's excellent post because I only want
to amplify :-) a few points that Tom doesn't state as clearly as I'd
like.
A tube is a current controlling valve (the Brit name really fits
better than ours). It is *not* a resistor or voltage or current
source, but within certain careful limits, it can be *modeled* as
a variable resistor in series with the output network and the power
supply (which is the real voltage and current source). Where the
reflected power folks are going wrong is in assuming it's a real
resistor when it is really stressed electric fields in a vacuum that
*simulates* the same action as a resistor under certain conditions.
One of the things it is *not* is a dissipative load. It's a
transconductance, or more properly at RF, a transadmittance.
The loadline of a tube is a convienent mathematical fiction to
describe the behavior of the transadmittance working with a
given voltage and current, IE it's E/I for the particular anode
voltages and currents experienced under a particular drive condition.
This mathematical fiction is useful because it shows what load impedance
the tube *needs to see* for maximum power transfer at a given power
supply voltage and input drive level. It is not a power dissipating
termination in the reverse direction. A transmitter is not a load
in this fashion.
The only dissipative element is the loss resistance of the tube and
associated circuitry. This is very small compared to the transadmittance
of the operating tube, with which it is in series. Hence it will not offer
a match to a signal being fed back into the output of the operating amplifier.
In fact, in any reasonable transmitter, the ratio of transadmittance to loss
resistances is such that the tube looks like a virtual open circuit to any
signal trying to make it's way back through the matching network to the tube
and be dissipated there.
The tube offers virtually no loading to the output network. Any signal
trying to re-enter the transmitter will vector sum with the circulating
currents in the tank, and be absorbed into the next outgoing cycle of
energy, IE it is in effect completely re-reflected.
Note well that feeding a signal which is *not* coherent with the
transmitted signal back into the amplifier won't behave the same
way. It's vector is rotating at a different rate from that of the
outgoing wave, and any measurements made in this way will lead to
erroneous conclusions.
Philip Munts
>>We are not talking about maximum power transfer. We are not talking
>>about conjugate matches. We are talking about mis-matches and non-
>>maximum power transfers... what happens when things are not optimum?
>>There is no argument as to what happens when things are optimum. It
>>has been studied to death. What hasn't been sufficiently studied is
>>what happens when there is no conjugate match.
>>
>>Let's have your opinion. Is it possible to re-reflect all the reflected
>>waves without achieving a conjugate match? I say if you set up a
>>mechanism of any kind, short of magic, at the output of a transmitting
>>tube that will re-reflect 100% of the reflected waves, that mechanism
>>can be nothing but a conjugate match.
Consider the following experiment:
Suppose you design an amplifier to deliver power to a 50 ohm resistive load,
50+j0 ohms.
You connect some impedance that is not 50 ohms through some
transmission line that is not 50 ohms either. You observe the voltage at the
output terminals of the transmitter with an oscilloscope.
You get out your trusty Smith chart and figure out what your non-50 ohm
transmission line transforms your non-50 ohm load impedance to. Call it R+jX.
Now I come along and replace your load and transmission line with a resistor
of R ohms in series with a reactance (either a capacitor or inductor) of X
ohms. I also observe the voltage at the output terminals of the transmitter
with an oscilloscope.
Question: Do we both observe the EXACT same waveform?
I submit that it DOES NOT MATTER what is inside the transmitter box, you
and I will ALWAYS observe the EXACT same waveform at the output terminals.
Furthermore, I submit that at constant frequency, you CANNOT distinguish
my lumped constant load from your transmission line system.
Clearly, my lumped constant impedance does not reflect any power "back into
the amplifier" and I would posit that your transmission line system doesn't
either. QED.
Whatever output matching network the transmitter has transforms the load to
a complex impedance "seen" by the output tube or transistor. This may
well be non-optimal or even destructive, but it isn't because power is
reflected into the transmitter output terminals.
N7WS
writes:
I agree and said pretty much the same thing in my post "Re:Re:Re:Maxwell"
elsewhere in this forum.
Wes Stewart, N7WS
n7...@aol.com
cmo...@sedona.intel.com
| o--------+
| Broad Band Transmitter | | V=10v peak to peak
| with fixed 50 ohm output | __|__
| No Low Pass Filter | _____ -j50
| Untuned Linear Final | |
| o--------+
+-------------------------------------+
It's obvious that a lot of us have failed to communciate. I appologize
for that. In an attempt to establish communications, I've tried to make
this example as simple as possible.
True or False:
(1) The voltage out of the transmitter _is_ the voltage across the
capacitor.
(2) While the transmitter voltage is driving toward maximum, energy is
stored in the capacitor.
(3) After the transmitter voltage has reached its maximum and falls
to zero, the capacitor discharges its energy back into the
transmitter where it is dissipated.
(4) The amount of power delivered back into the transmitter by
the capacitor charge is Vmax^2/pi*XL.
If anyone needs proof, please email me or reference "Capacitance" in
the ARRL Handbook.
(5) If transmitter Vmax=10v then the power delivered from the capacitor
to the transmitter during the discharge time is 0.63662 watts.
(6) The energy delivered back to the transmitter from the capacitor is
not DC although the original source of its energy was the DC supply.
(7) If a wavelength of transmission line is put between the transmitter
and the capacitor, the capacitor discharges its stored energy into
reflected energy waves.
(8) For a continuous wave output from the transmitter, the transmitter
will not see steady-state conditions any differently with or without
the transmission line.
(9) RF energy from reflections dissipates inside the transmitter.
This is an out-of-phase mis-matched system and I rest my case about
only that single kind of system. This is the only kind of system I
have ever been talking about. I am _not_ saying anything about
(1) in-phase, mis-matched systems (100% re-reflection because of phase)
(2) in-phase, matched systems (no reflections to worry about)
(3) out-of-phase matched systems. (Z-match = 100% reflection)
Please stop giving these last three configurations as examples. I am
_not_ talking about or trying to prove anything about them.
Cecil Moore
>>Clearly, my lumped constant impedance does not reflect any power "back
>into
>the amplifier" and I would posit that your transmission line system
>doesn't
>either. QED.
>
does not discharge anywhere after it has been charged? Are you saying
the energy stored in an inductance in your lumped constant doesn't
colapse back into the system? I guess I'm going to have to disagree
with that. I Either that or you are going to have to show me that AC
diode that you have invented.
73, Cecil, KG7BK, OOTC
Cecil Moore
>Depending on whether the vectors aid or oppose, the load line of
>the tube will be shifted up or down from where it would be for a
>conjugate matched case. Given that the drive level doesn't
a little different approach. If someone offered you a million
dollars to come up with a transmitter that dissipated out-of-phase
reflected waves without interfering with the transmitted forward
wave, could you do it? The only rule is that you cannot design in
anything designed to dissipate the reflected waves. The final
must be an ordinary design, no circulators.
What I would do is design a 10kw linear amplifier and choose the
operating voltage and current to give a 50 ohm output impedance.
I would then transmit 10 watts into a 291 ohm load 5/8 wave-
lengths down the 50 ohm transmission line. 5 watts will be re-
flected. I can prove by measurements at the collectors that the
reflected wave energy is being dissipated in the transistors.
If I can do that and if you can do that, then all reflected waves
are not always 100% re-reflected.
If Walter Maxwell said that transmitter is impossible to build,
then he was wrong.
73, Cecil, KG7BK, OOTC
cmo...@sedona.intel.com
are similar to fig 11, page 24-6, 15th edition, ARRL Antenna Book.
+--------------------+ VF=100v at 90 degrees
| |<==1/2 wavelength==>| VR=50v at 90 degrees
| o--------------------x-----o---/\/\/\/---+
| Transmitter | 50 ohm lossless coax 291 ohms |
| 50 ohm fixed o--------------------x-----o-------------+
| | 1/8 wavelength===>| |<===
+--------------------+
Please plot the vectors for the forward voltage and the reflected
voltage. We need one reference point so lets set up the transmitter
so the forward voltage at the load is 100v at 90 degrees. Half the
voltage is reflected in phase so the reflected voltage at the load is
50v at 90 degrees. 1/8 wavelength toward the generator at point 'x', the
forward voltage is 100v at 135 degrees and the reflected voltage is 50v
at 45 degrees. 1/2 wavelength toward the generator from 'x' just before
the transmitter, the forward voltage is 100v at 315 degrees and the
reflected voltage is 50v at 225 degrees. Assuming the transmitter
re-reflects 100% of the reflected voltage, the resultant _new forward
voltage_ must be 111.8v at 288.44 degrees which is the vector sum of
the forward voltage and reflected voltage at the transmitter. Makes
sense if forward power equals transmitted power plus reflected power.
But doesn't the forward voltage 1/2 wavelength down the lossless
transmission line have to equal the forward voltage at the transmitter?
Try to get this model to settle down to steady-state. I haven't said
anything about current vectors, but they follow the same basic pattern
with the reflected current 180 degrees out of phase with the reflected
voltage. Forward current is in phase with forward voltage.
Remember, we set the transmitter output so the forward voltage is 100v
at 90 degrees at the load and the _forward_ voltage at the transmitter
must be 111.8v if all the reflected energy is re-reflected at the
transmitter. How can these two voltages have different amplitudes in
a lossless transmission line?
Would somebody please show me the error in my logic? I will be eternally
grateful, appologize, and go back to hamming.
Gary Coffman
>Gary Coffman <ga...@ke4zv.atl.ga.us> writes:
>
>>No. Two voltage waves will *vector sum* at all points. That's the
>>basis for "voltage standing waves" on a line. Two current waves
>>will also sum at all points along the line. What *doesn't* sum is
>>power, since there isn't any on a pure "J" lossless line due to the
>>90 degree phase shift between current and voltage. Power can only
>>express itself across a real resistance.
>
>Hi Gary, it is a trivial statement to say vectors sum at all points.
>Of course they do. But the forward energy vectors are moving in the
>opposite direction to the reflected energy vectors. That means that
>an instant after you have summed them and looked at them with a scope,
>one moves to the right and one moves to the left having no effect on
>each other.
Yep, the principle of superposition holds for voltages and currents
on the line.
>The reason that power doesn't sum is they are moving in
>opposite directions. The forward voltage and current vectors are in
>phase and the reflected voltage and current vectors are 180 degrees
>out of phase. Thus they are both real not "vars". Please read the
>ARRL Antenna Book for an explaination. If there is no conjugate match
>in the system, the reflected power will definitely express itself
>across a real resistance, the generator internal resistance.
This is where you go astray. There is no power on a lossless line
since it's a pure reactance. Only VARs. Voltage and current are
90 degrees out of phase at all points along the line, *in either
direction*. There's no need to look at the Antenna Book, look
at chapter 2 of the Handbook for this simple relationship.
A transmission line is best modeled as a *transformer*. And
like any other good transformer, it has negligble internal
real components to it's characteristic impedance. The only
real impedance components appear at the terminations after
being transformed by the purely reactive transformer from
one end to the other.
Now it's true that the reflected energy will express itself across
the generator internal resistance, but that's in *series* with the
generator load line, IE equal to it's transadmittance. The load line
can't dissipate power, it's not a real resistor. The circuit loss
resistance *can* dissipate power, but it is tiny in comparison to the
load line value. Since the reflected energy has nowhere else to go,
it circulates in the output tank circuit until it can go back
out the output port toward the load on the next cycle. This happens
regardless of whether the amplifer is "matched" to the line or
not.
>>No, you haven't proven that, even disregarding the fact that a real
>>transmitter isn't a Thevenin generator. The net of current flow is
>
>Ok Gary, I challenge you to find anything wrong with my example. Maxwell
>didn't say anything about his theories not working with linear amplifiers.
>With a Vg of 100sinWt, an Rg=50, an RL=50, and an XL=-j100, how can you
>possibly keep the capacitor voltage from appearing across Rg when Vg=0
>which it does twice each cycle. The only way would be for the generator
>to be capable of internally producing +j100 and I believe that to be
>magic.
There's no magic. Your fault is continuing to assume that Rg is a
real resistance. It isn't. It's the series value of the very low
Rloss (tube contact resistance, wire resistance, etc), and the
very high transadmittance of the active device. Since the trans-
admittance is *not* a resistor, it can't dissipate any power.
As far as the returning wave is concerned, it presents a negligble
load on the output tank circuit. The only significant load on
the output tank is the actual load for the transmitter, and that's
where all the circulating energy, neglecting circuit losses, finally
dissipates.
Gary Coffman
>Gary Coffman <ga...@ke4zv.atl.ga.us> writes:
>
>>Depending on whether the vectors aid or oppose, the load line of
>>the tube will be shifted up or down from where it would be for a
>>conjugate matched case. Given that the drive level doesn't
>
>Hi Gary, maybe you don't understand what I'm saying so let's take
>a little different approach. If someone offered you a million
>dollars to come up with a transmitter that dissipated out-of-phase
>reflected waves without interfering with the transmitted forward
>wave, could you do it? The only rule is that you cannot design in
>anything designed to dissipate the reflected waves. The final
>must be an ordinary design, no circulators.
I couldn't do it. That's because an ordinary design doesn't have any
impedances with significant real parts that can dissipate the reflected
wave. That's why real transmitters always re-reflect any energy fed back
into their outputs (after circulating it in their tank circuits). The
only way I could do it would be to put a dissipative resistance in the
circuit, IE a Norton or Thevenin model, and you've specifically disallowed
that. Good thing too, since real transmitters aren't Norton or Thevenin
sources with a real valued Rs equal to their desired output impedance.
>What I would do is design a 10kw linear amplifier and choose the
>operating voltage and current to give a 50 ohm output impedance.
>I would then transmit 10 watts into a 291 ohm load 5/8 wave-
>lengths down the 50 ohm transmission line. 5 watts will be re-
>flected. I can prove by measurements at the collectors that the
>reflected wave energy is being dissipated in the transistors.
Can you really? I think not unless you've somehow given the DC
power supply a dynamic AC response at the frequency of interest.
It's the power supply that's the voltage source, not the active
device, and it's the input signal that modulates the device
transadmittance (which is not a real dissipative resistor),
not output terminal voltages.
Gary Coffman
>Hi again:
>
>One last post (I hope)..
>
>Rather than repeat a lot of what has been posted in this and other
>threads, I'll just make some general comments in response. The guilty will
>know who they are.
>
>1. It has been said that reflected power is not real. I will answer this
>in a word.... RADAR. A radar system works by generating an RF signal,
>coupling it to the impedance of free space (120*pi Ohm or ~377 Ohm) where
>it propagates until it encounters a mismatch which reflects some of the
>REAL energy back to the receiver. (Sound familiar?) At both the
>transmitter and receiver, the impedance transforming network used is
>commonly known as an antenna. Its function is to match the impedance of
>the transmission line (coax or waveguide) to 377 Ohm. A target reflects
>some of the incident energy because the impedance of its surface
>is_not_377 Ohm and is therefore a mismatch to the incident (forward) wave.
>If the reflected energy is not real, then it cannot be received. If you
>don't believe this, the next time Smokey pulls you over for speeding, just
>tell him his radar was giving him an imaginary reading. Or tell it to the
>F-15 pilot that turned on his fire control radar in the hangar and burned
>out the receiver..
Sorry, but you are confusing *energy* with *power*. They are not the
same thing. Reflected energy across a pure reactance, like a lossless
transmission line or free space, is not power (watts), instead it is
energy in VARs (volt-amperes-reactive). It's said not to be *real*
because it's across an impedance with no real part, IE the R+jX has
no R for the line, or for free space. There's no real part to the
impedance that can dissipate power, and energy across a pure reactance
has it's current and voltage vectors at 90 degrees to each other.
Since sin(phi)E*cos(phi)I equals zero, there is no power there. But
there is certainly energy there in the form of VARs, and if it is
expressed across an impedance with a real part, it'll become power
and be dissipated.
The rest of your argument is relatively sound, and is deleted to
save space.
cmo...@sedona.intel.com
Then would you do it for ten million dollars? :-)
>That's because an ordinary design doesn't have any impedances with
>significant real parts that can dissipate the reflected wave.
Take a look at the schematic of an MAR-n and tell me again that it
"doesn't have any impedances with significant real parts". I don't
know enough about tubes to argue impedances with you but transistors
certainly have a real component of impedance that an out-of-phase
reflected wave would see if the transistor was running in class-A.
You can prove that by supplying a separate DC voltage to the collector
while it is transmitting. The transistor will get hotter. Out-of-
phase energy is just like that DC voltage except changing with time.
Why doesn't the transmitter re-reflect the extra DC voltage? Because
it is completely out-of-phase!
>That's why real transmitters always re-reflect any energy fed back
>into their outputs (after circulating it in their tank circuits).
I agree for matched circuits and in-phase mis-matched circuits. What
magic allows an out-of-phase wave to circulate in a tank circuit?
A transmitter does not produce two signals so it must shift its
phase by up to 45 degrees to meet the reflections head on. That's
how it achieves the virtual open or short. If the class-A transmitter
does not somehow match phase with the reflected wave, some reflected
wave energy will be dissipated in the transmitter, and I don't believe
in magic.
>It's the power supply that's the voltage source...
Obviously, unless one believes in perpetual motion. But the reflected
waves are not DC even though their energy came from the DC supply.
Have you drawn the vector diagram I suggested? If the vector sum of
the forward and reverse waves somewhere in the transmitter are not
equal to the forward power vector 1/2 wavelength away, then the
reflected energy has not been 100% re-reflected. I think this
measurement can be done on a bench. Remove the tank circuit and the
low pass filter and feed DC power to the class-A final through an RF
choke. Now all the smoke and mirrors are gone. Generate a small amount
of power into a 5/8 wavelength of 50 ohm heliax feeding a 291 ohm load.
IF the sum of the forward and reflected vectors is equal to just the
forward vectors 1/2 wavelength away, I'll appologize and shut up.
Forrest Gehrke
Re: your capacitor load example.
My mail reader hashed up your circuit showing
a -j50 load to a fixed tuned 50 ohm output
circuit. So I can't show it.
CC> It's obvious that a lot of us have failed to communciate. I
CC> appologize for that. In an attempt to establish communications, I've
CC> tried to make this example as simple as possible.
Cecil, I have been following this thread for awhile and have
seen you complain of posters changing the subject. But
now you are doing it! Regardless of whether Phillip Munts or
N7WS are correct in saying substitution of a lumped
constant load identical to a transmission line load involving
a non 1:1 SWR would make no difference, they had a resistance
in their load. They are maintaining that the same, i.e.
*all*, the RF power would be dumped into that resistance and
that the active device would not know the difference.
There is no load resistance in your example, and again,
to use the substitution idea in reverse--how would it look
any different from an open circuited transmission line
reflecting back the same reactance of your capacitor?
No one has proposed either of these examples.
This is getting ridiculous. Here is lacking anything
to dissipate power. What would you hope to measure
under such a circumstance and how would you correlate
that with the case of a resistive component as part
of a reactive load?
Prove their idea wrong if you can; don't change the subject.
* RM 1.3 02583 * Irony: A tortoise has a drag coefficient of .03.
John Mallick
|> John Mallick <mal...@crd.ge.com> writes:
|>
|> >Roy Lewallen's post goes to the heart of the matter. The behavior of
|> >the voltage and the current on the transmission line are determined
|> >by the solution of the transmission line equations; there are two
|>
|> Roy is a smart fellow but all his signals were in phase. What happens
|> when the signals are out of phase?
It doesn't matter; if you write the solutions in terms of complex functions,
then the solution is algebraic. You just take the real part at the end
and that's the answer. See "Engineering Electromagnetics" by DC Cheng.
|>
|> >end; just use the Thevenin impedance of the generator. But thinking
|> >about the reflected power as partially being absorbed by the generator
|> >impedance is not useful in the sinusoidal steady state.
|>
|> All of the RF designers that I have talked to say that's what happens.
|> Don't you think it is useful to know the truth behing your rules-of-
|> thumb?
It's not "rules of thumb"...it's how the problem is actually solved...
Cecil Moore
>>>waves without achieving a conjugate match? I say if you set up a
>>>mechanism of any kind, short of magic, at the output of a transmitting
>>>tube that will re-reflect 100% of the reflected waves, that mechanism
>>>can be nothing but a conjugate match.
an example where there wasn't a conjugate match only something like
a partial conjugate match and that's why I prefer the term "partial
Z-match" for that condition. I do believe that a reflected wave is
never re-reflected unless it sees a virtual open or short and that
happens when the reflected wave encounters an in-phase forward wave.
>Clearly, my lumped constant impedance does not reflect any power "back into
>the amplifier" and I would posit that your transmission line system doesn't
>either. QED.
the amplifier. The capacitor dumps its charge and an inductor dumps its
field. Either one results in a current in the load and in the transmitter.
Do you think a transmitter is an AC diode passing AC in one direction
and not the other. Quick, get a patent on that thing. I want to install
one in my transmittion line close to my antenna.
>Whatever output matching network the transmitter has transforms the load to
>a complex impedance "seen" by the output tube or transistor. This may
>well be non-optimal or even destructive, but it isn't because power is
>reflected into the transmitter output terminals.
There is no where else for it to go. It divides across the load resistor
and the transmitter impedance, whatever it is. Otherwise, the capacitor
would not discharge and it does. Please see "Capacitance" in the ARRL
Handbook for the power equation from a discharging capacitor. What do
you think "out of phase" means anyway?
73, Cecil, KG7BK, OOTC
Roy W Lewallen
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From: aa...@raider.raider.net (Bill W. Cox [AA4NU])
Subject: The E word ...
To: ro...@tekgp4.CSE.TEK.COM
Date: Thu, 19 Jan 1995 10:17:06 -0600 (CST)
Cc: roy.le...@email.HUB.TEK.COM
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GM ! Saw two diff addresses for you, so tried both !
Anything new going on with ELNEC ? I am still using version 2.??
Have gone from a 8088/mono to a 80386/387/SVGA and thght I'd see if anything
was out or in the works ...
Have a FB week there !
73 Bill
* hope this note was ok via this method, if not please advise ...
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Date: Fri, 20 Jan 1995 13:21:44 -0500 (EST)
From: Morbius <mor...@gate.net>
To: NEC-List <nec-...@ee.ubc.ca>
Subject: Mininec ADA181681 software
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Status: R
Does anyone know if the software in the NTIS package for MININEC numbered
ADA181681 is any good and if it runs on an IBM PC class machine?
And, does anyone have or know where I can get the MININEC software
published by Artech? Its currently out of print.
Regards, Vic... N4QK
W8JI Tom
source with Rs. The common use of R to describe an impedance or the use of
ohms to define a reactance probably causes people to equate a source
impedance to an inductance, a capacitance or a resistance....but this is
untrue. For example while RF transistor output Smith charts plainly state
the values given are the conjugate of the optimum load impedance, people
tend to equate the value to a resistance. It can be + or - J zero and the
source must still be considered a non-dissipating impedance. If we want to
analyze the source, the model needs to be very complex.
The R value in the Smith chart is not the Rs of the device. And we don't
match Rs to anything anyway.
In a PA stage or in a generator if Rs is significant compared to the
impedance (nothing to do with XL or Xc, remember that) the device will
have very poor efficiency. If Rs dominates or even approaches the source
impedance the stage would not produce maximum efficiency at or close to
the point where maximum output power occurs (in an amplifier without
feedback). This is almost never the case, because Rs is insignificant in
well designed devices.
Besides mixing impedance and resistance we are also mixing models. If we
want to use reflected power (wave theory) as a model, we have to stay with
the rules. Reflected waves can never have any phase shift other than 180
degrees between the voltage and current in the reflected wave. When using
wave theory in the model, the rule is the current and voltage in the
reflected wave are always 180 degrees out of phase. If they aren't, the
model is incorrect for wave theory analysis.
If we want to shift the current and voltage to other phase relationships
we have to treat the load as a reactance and resistance (impedance), and
we have to use reactance and resistance all the way through the system.
Then the whole system becomes a impedance matching problem. We can't mix
models and use the mess that results to condem a person (Maxwell) that
used the model correctly. After weeks of research I believe Maxwell is
exactly correct. The small doubt I had was because I wasn't comfortable
with wave theory. 73 Tom
Roy W Lewallen
> . . .
>Sorry, but you are confusing *energy* with *power*. They are not the
>same thing. Reflected energy across a pure reactance, like a lossless
>transmission line or free space, is not power (watts), instead it is
>energy in VARs (volt-amperes-reactive). It's said not to be *real*
>because it's across an impedance with no real part, IE the R+jX has
>no R for the line, or for free space. There's no real part to the
>impedance that can dissipate power, and energy across a pure reactance
>has it's current and voltage vectors at 90 degrees to each other.
>Since sin(phi)E*cos(phi)I equals zero, there is no power there. But
>there is certainly energy there in the form of VARs, and if it is
>expressed across an impedance with a real part, it'll become power
>and be dissipated.
Power and energy aren't the same, but the difference isn't as stated
above. Power is simply the time rate of transferring or transforming
energy (a direct quote from the IEEE Standard Dictionary of Electrical
and Electronics Terms). That is, power is to energy as acceleration
is to velocity, velocity is to distance, or current is to charge. The
kilowatt is a unit of power; the kilowatt-hour, energy. The fundamental
MKS unit of energy, the joule, is a watt-second.
VARs, or "reactive power", represents power being stored but not dissipated
or otherwise used. The IEEE dictionary states that "In a passive network,
reactive power represents the alternating exchange of stored energy
(inductive or capacitive) between two areas." Sorry; no energy, no power.
Roy Lewallen, W7EL
roy.le...@tek.com
N7WS
(Gary Coffman)writes in response to my post which follows:
>1. It has been said that reflected power is not real. I will answer this
>in a word.... RADAR. A radar system works by generating an RF signal,
>coupling it to the impedance of free space (120*pi Ohm or ~377 Ohm) where
>it propagates until it encounters a mismatch which reflects some of the
>REAL energy back to the receiver. (Sound familiar?) At both the
>transmitter and receiver, the impedance transforming network used is
>commonly known as an antenna. Its function is to match the impedance of
>the transmission line (coax or waveguide) to 377 Ohm. A target reflects
>some of the incident energy because the impedance of its surface
>is_not_377 Ohm and is therefore a mismatch to the incident (forward)
wave.
>If the reflected energy is not real, then it cannot be received. If you
>don't believe this, the next time Smokey pulls you over for speeding,
just
>tell him his radar was giving him an imaginary reading. Or tell it to
the
>F-15 pilot that turned on his fire control radar in the hangar and burned
>out the receiver.. N7WS
::::Sorry, but you are confusing *energy* with *power*. They are not the
same thing. Reflected energy across a pure reactance, like a lossless
transmission line or free space, is not power (watts), instead it is
energy in VARs (volt-amperes-reactive). It's said not to be *real*
because it's across an impedance with no real part, IE the R+jX has
no R for the line, or for free space. There's no real part to the
impedance that can dissipate power, and energy across a pure reactance
has it's current and voltage vectors at 90 degrees to each other.
Since sin(phi)E*cos(phi)I equals zero, there is no power there. But
there is certainly energy there in the form of VARs, and if it is
expressed across an impedance with a real part, it'll become power
and be dissipated.:::
:::The rest of your argument is relatively sound, and is deleted to
save space.:::
To which I reply, get real. (pun intended)
When the power, energy or whatever you want to call it leaves the lossless
free-space transmission medium, it re-enters the REAL medium of the input
of the detector (radar receiver, power meter, or whatever). It most
certainly can be measured, detected, processed, and argued about ad
infinitum...
Nevertheless, the example cited is trying to convince the non-belivers in
reflected *power*, that reflected power exists. If you want to argue
semantics, that's another story. In other words, even though the
"transmission line" is lossless, the loads on the ends are not. As I said,
tell it to Smokey. :))
This has gone on so long, I forgot what the rest of my argument was. I
think I'm going to delete this thread and go to bed.
73
Wes Stewart, N7WS
n7...@aol.com
cmo...@sedona.intel.com
Gary Coffman <ga...@ke4zv.atl.ga.us> wrote:
>
>There's no magic Cecil, it's just that there's no real load in the
>transmitter to dissipate the waves, so they get absorbed in the
>circulating tank currents, vector summing at whatever phase they
>happen to be at that instant, and the energy goes back out on the
>next cycle.
Hi Gary, the magic is that you do not believe in the conservation
of energy principle. I was going to run a bench test to prove this
but I will just state it here and see if anyone disagrees. And yes,
I am shouting. Assuming a single generator system:
A REFLECTED WAVE CANNOT BE RE-REFLECTED AT THE POINT WHERE IT IS 90
DEGREES OUT OF PHASE WITH THE FORWARD WAVE.
The reason is quite simple. It doesn't matter whether you add or subtract
the generated and reflected waves, the result is of a greater magnitude
than the forward wave. The new larger forward wave will again be
reflected at a greater magnitude and there is no steady-state solution.
This is an infinite series that diverges to infinity and violates the
conservation of energy principle.
In an in-phase system, the reflected waves are re-reflected and subtract
arithmetically from the generator power to give the forward power. The
transcient conditions form an infinite series that converges to a
steady-state value.
Now Gary, if reflected waves 90 degrees out of phase with the forward
wave cannot be reflected, where does the power go?
Gary and Jean Peach
with my interjections in between
> Article: 6730 of rec.radio.amateur.antenna
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>
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> From: gr...@pge.com (Gary Wescom)
> Newsgroups: rec.radio.amateur.antenna
> Subject: Re: Is Maxwell wrong?
> Message-ID: <grw1.121...@pge.com>
> Date: 10 Jan 95 19:50:59 GMT
> References: <1995Jan3.1...@arrl.org>
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> In article <D1yL1...@cix.compulink.co.uk>
wise...@cix.compulink.co.uk
> > ("Gary and Jean Peach") writes:
> >From: wise...@cix.compulink.co.uk ("Gary and Jean Peach")
> >Subject: Re: Is Maxwell wrong?
> >Date: 6 Jan 95 00:51:09 GMT
>
>
> >the maximum power transfer occurs when load
> >resistance equals source resistance.
>
> Hi,
>
> How about if I jump into this discussion. The subject of this thread
is
> interesting but I think the information presented so far is somewhat
off the
> mark. The above statement is a good example. It is true in some
situations
> and not in others and blindly applying it to transmitter output
circuits is
> definately an error.
>
> First, about power transfer and source resistance. If one considers
our
> electric utility power generation and distribution system, perhaps a
little
> perspective can be gained. Electric power generators are built to have
very
> low internal resistance. This is necessary to minimize internal
heating. A
> 1000 Megawatt generator is roughly the volume of a two car garage. An
object
> that small is not capable of internally dissipating 1000 Megawatts for
more
> than a fraction of a second without melting down. The internal
resistance of
> one of these generators actually dissipate only a small fraction of a
percent
> of its output power.
>
> Now, where does this source and load resistance being equal thing come
in?
> Very simply, the maximimum theoretical output power that could be
produced by
> that 1000 Megawatt generator would be into a load that has a resistance
equal
> to the internal resistance of the generator. That load is not even of
> curiosity value to power plant operators since it would represent an
output
> power from the generator probably on the order of 1,000,000 Megawatts
which
> is WELL outside the range of power that its turbine could apply.
>
If the turbine is driven from a head of water and you require 10^12
Watts
that head will have the potential energy of 2 x 10^12 Watts at least.
Consider for a moment what happens when there is an additional demand
for power, Industry is starting up for the day.
Another generator is brought on to line. Do you just close the switch ?
In the old days I would connect a lamp between the live terminal of the
generators and only when I had nudged the phase of the off line
generator
to extinguish the lamp did I close the switch, to parallel the two
generators
instead of serialise them. What would have happened had I not done
this ?
Perhaps you had forgotten this now that this procedure is performed by
a little
black box.
> The bottom line here: There is a large category of power generation
> situations in which the the generators source resistance is
intentionally
> different than load resistance. This also applies to generators of RF
power
> such as our ham band transmitters.
>
The machine that rotates to produce electrical energy is a transducer,
that
is a devise that transforms energy from one state to another. Its
efficiency
as a transducer can be >90% but the transfer of energy from the
primary source to the place of use is maximum 50%
> Tube type transmitter output stages, at least those without negative
> feedback, have very high output impedances..............................
NFB reduces the O/P impedance.
>.............................................................. They
operate more like current
> sources than voltage sources and thus have very high equivalent source
> resistance.
..........................................................................
...............
What about the impedance transformation that occurs between the 1 or 2
turn coupling loop and the tank coil, or the Auto transformer effect
of
taking the power from a few turns up the tank, or the tap pair of
unequal
value capacitors?
>....................... When we adjust their TUNE and LOAD controls for
proper
> operation, we are adjusting transformation of feed line load resistance
to
> make it appear to the tubes as some other, higher value that allows the
tubes
> to operate properly. This setting of the TUNE and LOAD controls does
not
> transform the output resistance of the tubes to 50 Ohms!
If you have answered my question above then you will know the answer to
this.
>
> Were we to take the "source resistance must equal load resistance for
maximum
> power transfer" in the wrong way, we would have to assume that the
maximum
> efficiency we could ever achieve in a transmitter would be 50%. Since
we
> regularly use tube amplifiers, including linear amplifiers, with
efficiencies
> much higher than this, having source resistances higher than the load
while
> achieving efficient power transfer is still possible.
>
50% Which it is
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Remember your old mechanical formula mh = 1/2 mv^2
> OK, next subject: Reflected Power
>
> Forward versus Reflected Power is one of those concepts like RF
radiation
> itself. For essentially everything we do with radio, we consider RF
> radiation to be waves of some sort. The RF roaring off the end of a
KT34XA
> beam is the same stuff as the sunlight gleaming off of its elements.
We
> think of the sunlight a bombardment of photons but the RF coming off
the
> antenna as waves. Which view is physically correct? Really neither
is. We
> use views as tools to describe real observed physical effects though,
and
> quite successfully.
If you where a surfer instead of a cybersurfer you would know a lot more
about waves, and how to get the energy out of them, how large and small
waves combine, how reflected waves from the beach cause waves to break
before reaching the beach, how waves that still contain energy will
work away
at a beach modifying its angle until the energy remaining in the waves
is
insufficient to further change the beach, and it settles for a time to
an angle that
exactly matches the in coming wave give or take a little. Nature's own
Negative
Feedback. There where some excellent films made by the Air Ministry
(RAF)
during WWII of waves in specially constructed tanks with a commentary
explaining
the three velocities associated with waves, the cameras where moved at
these
three different velocities tracking the waves, at first it was difficult
to see what was
being demonstrated but several viewings made all clear. It is now 40
years since I
saw those films, if they still exist I would very much like to see them
again. Or perhaps
NASA or MIT or some body has made some better ones, if they have not it
is time
those old war time cameras have been improved upon and better lighting
is available.
>
> Forward Versus Reflected Power is concept that we apply as a tool to
analyze
> the transmission of AC power. (Remember that the output of an Alpha 87
at HF
> is AC like the stuff available in a AC wall socket execpt at a somewhat
> higher frequency) This concept provides us with a mental model of
operation
> of transmission lines that is really handy. We are able to picture our
RF
> moving through our RF feed lines, winging its merry way from
transmitter to
> antenna and some of it bouncing back again. We buy meters that are
> calibrated to indicate values called forward and reflected power. The
> concept must represent reality right? Well, no.
Had you been able to play with a slotted line you would realise that
this is
more than a concept. You could construct your own Lecher line and make
a
crude carriage to play with some waves. Try 2m or 70cm so that the
construction
is not too critical. If you are running a bit of power into the line a
pea lamp with a
short loop will illuminate at intervals along the line. Try a neon
with a short piece
of wire on one electrode, dangle it into the space between the lecher
line with a
mismatch on the line observe the neon lights some distance from where
the pea
lamp is brightest. If you have a florescent tube place it upon the
line and observe
the pattern of light.
If you want to get really technical put a meter rule beside the line
and make some
measurements, you sound like the sort that will get more fun from this
than from one
of those rag chew QSO's that go on for hours and never reach a
conclusion HI
I have done these experiments, on up to 40GHz always with a sense of
wonder,
But the engineering gets difficult up there as the position of the
probe in the
waveguide slot need to be controlled to 1/200000 mm which is beyond
feasibility
so another device is used a rotary standing wave meter, which further
bears out
the existence of waves, the transitions from peaks to troughs are
smooth continuous
changes not steps that you would expect if photons where the mechanism.
>
> It sure is attractive to think of the concept as real. We describe
> situtaions that must surely prove that the concept is truth. Think of
an
> infinitely long transmission line that we transmit a single CW "dit"
into.
> Well, heck, any fool can see that there is a pulse of RF power moving
down
> the transmission line. That must be real forward power. If we cut the
> transmission line somewhere ahead of the pulse, it would be reflected
back
> and that must be real reflected power.
>
Even better don't use RF just consider what happens if you apply a DC step
transition to a line with a mismatch at the end. Special lumped component
lines are constructed that use this principle to produce high energy
pulses
to pulse a MAGNETRON in a RADAR set. These bits of line masquerade under
the name pulse forming network PFN, or delay line.
Consider the following experiment
BATERY SWITCH RESISTOR SCOPE PROBE PFN MISMATCH
2 * yVolts...........................................______________
yVolts................................._____l
0Volts _______________l
Note that the voltage at the scope probe doubles at a time equal to
twice the
time taken for the voltage rise at switch closure takes to propagate
down the
PFN or pseudo line. If you try this be careful it bights.
> What is really happening occurs at a sub-atomic level. Electromagnetic
> fields interact with subatomic particals within the materials that make
up
> the our transmission lines. Our physicists have mathematical models to
> describe these interactions. Unfortunately, the more accurate model is
> usually too cumbersome for our every day use and generally does not
provide
> any more insight into transmitter, feedline, or antenna operation. A
notable
> exception to this is the NEC/Mininec antenna analysis software we run
on our
> PCs.
>
> The thing to remember is that the concept of Forward versus Reflected
Power
> is a tool. It is a very useful tool but only a tool. It can be the
wrong
> tool for some jobs. Analyzing transmitter output tube dissipation
problems
> is one of the places where it is not the appropriate tool.
I don't want to labour the point but waves real, and our perception of
their
behaviour though imperfect is the tool. The wave analysis only
describes
the behaviour at the transmitter output terminal, I mean the socket on
the back
having performed that function a lumped equivalent network may be
substituted
for the calculation of what happens to the transmitter when terminated
in that
kind of load. The result will be the same, the problem is to produce a
lumped
equivalent with just the characteristics of the line and load, since
you already
have it why bother, to make a substitution?
>
> Analyzing transmitter output tube dissipation should be done by
treating the
> feedline/antenna combination as a load of the impedance observed at the
> connector on the back of the transmitter. You can then see we have an
output
> circuit consisting of tubes, a matching network, and a load of some
sorts.
> The matching network is supposed to transform the load impedance to
something
> that the tubes need.
>
> Now, what happens with high feedline SWR? The load presented at the
> connector at the back of the transmitter is other than pure 50 Ohms
> resistive. If it is not too far off, a little retuning of the TUNE and
LOAD
> controls should make the tubes happy again and the only problem we
might see
> is perhaps a little higher heating or voltage stress in the output
circuit.
AH! so where does this extra energy come from to cause this heating, or
put
another way why does this energy that is causing the heating in the
tube not
leave the transmitter.
Heat does not of itself flow from a cooler body to a hotter body.
> If the SWR is really bad, there may not be sufficient adjustment range
in the
> output variables and the tubes will not be presented with the proper
load and
> overheat. Even if the output circuits can be adjusted to handle the
high
> SWR, heating or voltage stress may damage those components.
There you are maximum power transfer is 50%
>
> It is the inablility to handle a strange load impedance that causes
> overheating. Forward and Reflected transmission line power are
meaningless
> in this analysis.
>
You may change your mind after performing the experiments that I have
suggested.
> AnyHoo, enough rambling fer now.
>
> Gary Wescom N6CH
> gr...@pge.com
>
See you on the ice 73
William P. Osborne
cmo...@sedona.intel.com <cmo...@sedona.intel.com> wrote:
Cecil, I have a question for you. Let's assume you have a transmitter
driving a 50 ohm line of any length terminated in a quarterwave of 70.7 ohm
line which is loaded with 100 ohm resistive load. Since the 50 ohm line is
matched there is no reflected wave on it and we assume the system to be
lossless where does the reflected power in the 1/4 wave line go? Seems
that if it gets to the transmitter it needs to do so via the 50 ohm line or
maybe it is "MAGIC"!
William Osborne, AA5ZQ wosb...@nmsu.edu
The Klipsch Dept. of Electrical & Computer Engineering
New Mexico State University
Gary Coffman
>Gary Coffman <ga...@ke4zv.atl.ga.us> writes:
>
>>way street.
>
>Hi Gary, and that's why we bias a class-A amplifier at the half-on
>current point. Are you implying that a class-A amplifier does not emit
>close to a sine wave RF on CW?
No, not at all. Of course it emits close to a sine wave, but currents
*never* flow backwards in the tube to do so. They are always from cathode
to plate, *always*. At the low (current) point of the wave, only a little
current flows forward in the tube (a fraction above cutoff), and at the peak
current much more current flows forward in the tube, but the current always
flows *forward* from cathode to plate. And there is *never* a negative
voltage at the tube plate, IE the sine wave generated is not symmetric
around zero volts, it has a DC bias on it. Only when, or if, we AC couple
the signal via a capacitor or transformer does the wave become symmetric
around zero volts. A tube is not a variable resistor, thinking of it as
one leads to misconceptions such as you have continually exhibited.
>>The net charge goes the only way it can, toward the load. Some of the
>>charge may circulate in the transmitter's output tank for a cycle first,
>>but it ultimately can only be dissipated in a real resistance, and that's
>>the load.
>
>You are again changing the subject back to matched systems.
No I'm not. It doesn't matter if the transmitter is matched or not.
In fact, in the matched case the question is moot since there is
no "reflection".
>What if there
>is no output tank and what if there's no real resistance in the load?
>If you believe there's no power in a perfectly reactive transmission
>line and load, try grabbing it at the load. Question: How does the
>transmitter know you grabbed it? If there were no power in the perfectly
>reactive system, you would not be able to affect the phase by grabbing
>the load.
If I'm a perfect insulator, I *won't* affect it at all. Of course I'm
*not* a perfect insulator, so now I'm supplying some *real* R value,
and it shocks the hell out of me. No problems with that. There is never
power across a pure reactance, the power factor is *zero*.
Laurence Gene Battin
What I want to know regarding this thread is:
if those who claim that reflected power can damage the tube, why I have never
heard of any dip meters being blown up via this mechanism?
After all, a dip meter is a transmitter which is commonly coupled in all
sorts of ways into totally unknown loads. Surely, in all the millions of times
that dip meters have been used, there must have been plenty of times that
one was coupled at just the right way into just the right load to produce
a tube-destoying reflected wave? Anyhow, I have never heard of this
happening, and I certainly don't recall seeing any warnings about this
in the countless article about dip meters I've read.
Am I way off base here?
I'm certainly no expert, but perhaps the experts can comment on _this_
possiblity?
--
Gene Battin, N9XAM
bat...@iucf.indiana.edu
Cecil Moore
>if those who claim that reflected power can damage the tube, why I
>have never heard of any dip meters being blown up via this mechanism?
not in conflict with the laws of physics. If the generator is
capable of dissipating the amount of extra power in the reflected
waves, it will not blow up. Thus a kw amp generating 10 watts
into a 50 ohm load will not blow up if the load is disconnected.
A pair of 1625s running 150 watts will blow up if the load is
disconnected. It happened to me 40 years ago. The reason
solid-state rigs don't blow up is they are so well protected. The
reason a dip meter doesn't blow up is that it is designed to
dissipate the mismatch with no problem, like the kw amp running 10
watts.
The question here is, when the kw amp running 10w into a 50 ohm
load is disconnected, does it continue to generate RF? Some say
it stops generating RF except for line losses because reflected
waves hit a brick wall at a transmitter and are always 100%
re-reflected if there is a forward wave present. They say
reflections are trapped in the transmission line and cannot get
out as long as the forward energy is greater than the reflected
energy, no matter what the phasing. They say the tube burning up
is a DC effect, not an RF effect.
I think it's possible that reflections out-of-phase with the
generated wave can enter a mis-matched transmitter and dissipate
inside. Others say that the generator is always smart enough to
match phase with the reflections and keep them out. All I know is
that my pair of 1625s sure burned up when the antenna was
accidentally shorted.
If the answer to my question is so simple, are there any
references with photographs of waveforms proving that reflected
energy is always 100% re-reflected? All I've seen are vector
additions identical to what occur every half-wavelength on the
transmission line where the reflected waves are not re-reflected.
73, Cecil, KG7BK, OOTC
Gary Coffman
Your heart's in the right place, Gene, but you're overlooking a difference
between transmitters and dip meters. Dip meters aren't usually operated
near device limits, transmitters often are. While reflected "power" per
se bothers neither of them, the loss of efficiency due to an improper
presented load impedance *may* damage a transmitter.
cmo...@sedona.intel.com
Gary Coffman <ga...@ke4zv.atl.ga.us> wrote:
>Cecil, why do you think that a wave cannot be reflected at the point
>where it is 90 degrees out of phase with a forward wave?
Sorry Gary, I meant to say 90 degrees out of phase with the _GENERATED_
wave. How does a wave get reflected by something 90 degrees out of
phase?
>The tube looks like a virtual *open circuit* to the wave attempting to
>enter it's output port. It's a *perfect* mismatch (aside from some minor
>Rloss components in the wiring and output network parts).
For weeks I have been asking you for some proof and you just keep
giving me the same old bromide - "it's a perfect mismatch - it's a
perfect mismatch." I simply do not believe that you are omniscient.
Prove what you say is true and I will shut up.
S22 exists. The only way a transmitter can re-reflect all the reflected
waves is for the generator wave to have a component equal to and in phase
with the reflected wave. If it doesn't, the reflected wave will see
S22, not a virtual open circuit.
cmo...@sedona.intel.com
Gary Coffman <ga...@ke4zv.atl.ga.us> wrote:
>
>Oh, let me take a shot, Cecil. Assume that we have a single solitary
>pulse from the transmitter into the system Bill described. Now tell
>us how the pulse cancels it's own reflection.
Careful Gary, you are going to prove my point. A pulse does not cancel
its own reflection. The reflected pulse will see S22 and be at least
partially dissipated inside the generator.
cmo...@sedona.intel.com
Roy W Lewallen <ro...@tekgp4.cse.tek.com> wrote:
>It looks to me like everything adds up. Please refresh my memory: What's
>the mystery or problem with this particular situation, again?
Hi Roy, thanks for the calculations. The mystery for me is, does any
of the charge on the capacitor dissipate in Rs? The charge on the
capacitor comes from the generator. It builds up to a maximum voltage
and then discharges back into the system. It is a series system. Some
say the generator has a magic shield against the entry of energy from
the outside. I say, at least in a linear transmitter, some of the
energy from the collapsing voltage on the capacitor dissipates in Rs.
If it doesn't, how does the capacitor get discharged?
cmo...@sedona.intel.com
Gary Coffman <ga...@ke4zv.atl.ga.us> wrote:
>The tube looks like a virtual *open circuit* to the wave attempting to
>enter it's output port. It's a *perfect* mismatch (aside from some minor
>Rloss components in the wiring and output network parts).
I have some questions about Gary Coffman's "virtual *open circuit*"
that re-reflects 100% of the reflected energy no matter what
degree of mis-match exists at the transmitter. My references
indicate that any wave encountering an open circuit keeps the
voltage in phase and reverses the phase of the current..
In the following vector diagrams "f" means forward, "r" means reflected
and "t" means total. Assume Vf=40v, Vr=20v, If=100mA, and Ir=50mA.
Vf Vt It If Vf Vt If It
| / \ | | / | /
| / \ | | / | /
| / \ | | / | /
|/ \| |/ |/
+----Vr Ir----+ +----Vr +----Ir
Before Re-Reflection After Re-Reflection
Vt=44.7v at 63.4 deg Vt=44.7v at 63.4 deg
It=111.8mA at 116.6 deg It=111.8mA at 63.4 deg
Pf=4w, Pr=1w Pf=4w, Pr=1w
Vt*It*cos(53.2)=3w Vt*It*cos(0)=5w
The "Before Reflection" vectors seem to match reality since the
total (load) power equals the forward power minus the reflected
power.
The "After Reflection" vectors seem to contain some magic. The
total power has increased to 5w and the generator has somehow
managed to bring its voltage and current into phase thus creating
a conjugate match. I can see how this is possible with a matching
circuit but the generator power equals the forward power plus the
reflected power. And how can a class-A amplifier with no tank
circuit and no matching circuit create a conjugate match?
William P. Osborne
Cecil Moore <cecil...@delphi.com> wrote:
>"William P. Osborne" <wosb...@nmsu.edu> writes:
>
>>Cecil, I have a question for you. Let's assume you have a transmitter
>>driving a 50 ohm line of any length terminated in a quarterwave of 70.7 ohm
>>line which is loaded with 100 ohm resistive load. Since the 50 ohm line is
>>matched there is no reflected wave on it and we assume the system to be
>>lossless where does the reflected power in the 1/4 wave line go? Seems
>
>Hi Bill, as I told you over email, I am just believing the ARRL Antenna
>Book. It says that there is an initial reflection at the 50/70.7 step-up
>point where the reflected voltage is in phase and the reflected current
>is out of phase with the forward wave. The rest of the forward wave
>continues on to the load where another reflection occurs at the 70.7/100
>step-up point. The forward voltage at the load is 180 degrees out of
>phase with the forward voltage at the 50/70.7 point so the reflections
>at the load will be 180 degrees out of phase with the reflections at
>the 50/70.7 point. When the reflections from the load encounter the
>reflections at the 50/70.7 point, it results in a virtual open or short
>for the reflections from the load at the 50/70.7 point and the reflections
>are cancelled at that point. Please take it from there and explain why
>that concept is wrong.
I promise all watching that this is the last time I take up the nets BW on
this subject. Cecil, you missed the point I was tring to make with the
example above. So I'll try again, first the explaination from the
handbook given above is lacking in it's treatment of the multiple
reflections which occur in this case. But, that does not change the point,
which is that the reflected power stays in the 1/4 section which your
quote from the handbook agrees with and I assume that you agree since you
picked the quote. Now for the case of a 50 ohm source nothing changes if I
choose to make the 50 ohm line have a length of zero. Now the 70.7 ohm
line has forward and reflected power of it and it is mismatched on both
ends but NONE of the reflected power goes into transmitter! I think this
disproves your orginal statement (which I agreed with before I studied the
problem) that reflected power is dissipated (at least some of it) in the
source.
The overall problem with this whole discussion is that power only exist
when it is dissipated in a resistive load of some sort, i.e., if you can
boil water with it. In the case of reflected power you can only boil water
if you can take it out of the line which means seperating the forward and
reflected waves. You can do this with an isolator (many VHF/UHF
transmitters use isolators) but when you do the problem is changed since
now the transmitter always sees a 50 ohm impedance and the reflected power
is absorbed in the isolator (heating it up). You cannot take the
reflected power out of the line without changing the problem since the
load absorbes all of the power put into the line (in our lossless case)
if you try to take out the reflected power then the generator must put
more into the line but the generator will only put more power into the
line if we change the Z seen at the generator output which changes the
problem. I am talking about a voltage generator with a series resistor
for a source resistor; I am not saying what the output of a nonlinear
tuned amplifier looks like when you look back into it, but as Roy said
the theory should work with a voltage source anyway.
Phil Hodge
Gary Coffman writes:
> In article <ZK24idP.c...@delphi.com> Cecil Moore <cecil...@delphi.com> writes:
> >the 50/70.7 point. When the reflections from the load encounter the
> >reflections at the 50/70.7 point, it results in a virtual open or short
> >for the reflections from the load at the 50/70.7 point and the reflections
> >are cancelled at that point. Please take it from there and explain why
> >that concept is wrong.
>
> Oh, let me take a shot, Cecil. Assume that we have a single solitary
> pulse from the transmitter into the system Bill described. Now tell
> us how the pulse cancels it's own reflection.
Cecil is right. That's how a low-reflection coating on glass works. The
wave reflected from the 50/70.7 point and the wave reflected from the 70.5/100
point are 180 degrees out of phase and nearly the same in amplitude, so they
add to nearly zero. Gary's question is a red herring; if you actually had
a wave pulse that short, the reflected waves would be separated, so of course
they wouldn't cancel, but so what?
Phil
John Derry
NO!
73 de K9CUN, Jack
Gary and Jean Peach
So your transmitter has infinite Q ha! Ha! HA!
Gary Coffman Wrote
>
> In point of fact, *any* heated cathode vacuum tube *is* a diode. Current
> can only flow from cathode to anode because only the cathode is an
*emitter*
> of electrons. A tube conducts more or less current depending on the
potential
> applied to the grid, the applied plate voltage, and the impedances
around
> the loop, but current can never flow from plate to cathode. It's a one
> way street.
>
An alternating voltage containing energy does not have to
change about a mean value of zero volts absolute.
Are you prepared to fire up a high power transmitter and
let it run all day with no load on the output connection ?
> The net charge goes the only way it can, toward the load. Some of the
> charge may circulate in the transmitter's output tank for a cycle first,
> but it ultimately can only be dissipated in a real resistance, and
that's
> the load.
>
> Gary
> --
> Gary Coffman KE4ZV | You make it, |
gatech!wa4mei!ke4zv!gary
> Destructive Testing Systems | we break it. |
emory!kd4nc!ke4zv!gary
> 534 Shannon Way | Guaranteed! | ga...@ke4zv.atl.ga.us
> Lawrenceville, GA 30244 | |
If that where true you could start up a Tank circuit disconnect
load and tube and it would run for ever. Perpetual Motion
perpetual-motion machine
perpetual-motion machine, a machine, considered impossible to build, that
would be able to operate continuously and supply useful work without
needing a continuous supply of heat or fuel. A perpetual-motion machine
of the first kind, which would produce more ENERGY in the form of work
than is supplied to it in the form of heat, violates the first law of
THERMODYNAMICS. A perpetual-motion machine of the second kind, which
would continuously supply work without a flow of heat from a warmer body
to a cooler body, violates the second law of thermodynamics.
The Concise Columbia Encyclopedia is licensed from Columbia University
Press. Copyright © 1991 by Columbia University Press. All rights reserved.
Gary and Jean Peach
cmo...@sedona.intel.com
Phil Hodge <ho...@bowline.stsci.edu> wrote:
>Cecil is right. That's how a low-reflection coating on glass works.
Thanks Phil, those first three words haven't appeared here lately. See
what you think of this.
+------------------+ +--/\/\/\/--+
| Solid-State | | RL=50 |
| Linear Xmtr, o-------------------------------------+ |
| 50 ohm Output | 1/2 wavelength 50 ohm lossless coax |
| Design Impe- o-------------------------------------+ |
| dance, No Tuner | | XL=-j100 |
+------------------+ +----| |---+
| switch |
+-----/ ----+
Sometime ago, I was playing with equations from quantum
physics when I found some energy flowing to the load and back
to the transmitter to be dissipated. It has taken me many weeks to
realize what is happening in reality. Assume the switch is closed and
there are no reflections. Then open the switch at the zero-crossing
point of the forward wave. Reflections begin but don't reach the
transmitter for 1/2f (assume 50uS at 10MHz).
On the upswing of the forward voltage, a certain amount of energy is
stored in the capacitor. On the downswing of the forward voltage, that
energy stored in the capacitor is returned to the system as power and
there are equations for calculating the magnitude of that power. Some
of that power is dissipated in RL and some is dissipated 50uS later in
the transmitter since it is a series circuit. The same thing happens
without the transmission line. The power that has exited the
transmitter and re-entered the transmitter simply becomes part of the
_average_ power dissipation in the transmitter. It can be ignored when
discussing steady-state average power as if it had not made the round
trip to the load and back. However, the energy returned to the
transmitter by the capacitor at the end of the transmission line is
100uS "older" than the presently generated wave because it _did_ make
the round trip and returned as energy in the reflected wave to be
dissipated in the transmitter. Note that this dissipation did not
occur until we closed the switch and the reflections began. This
dissipation will happen anytime the generator voltage and current are
out of phase, i.e. we have a non-zero current when we have a zero
voltage - indicating an instantaneous power dissipation in Rg that is
not coming from the generator so it must be coming from outside.
From a quantum physics, quarter cycle perspective, Maxwell is wrong,
every even 25uS period in the example above. The reason that it is not
obvious to everybody is that steady-state RMS power is equivalent to
DC, i.e. average power. Take a look at the instantaneous power values
and kiss a myth goodby. If the capacitor does not dump its power back
into the transmitter the conservation of energy principle will have
been violated and the capacitor cannot charge for the next cycle.
--
---------------------------------------------------------------
Intel has shipped over 100,000,000 MCS-96 Microcontrollers
---------------------------------------------------------------
Cecil_...@ccm.ch.intel.com MCS-96 Architectural Engineer
Intel Corp., CH3-46, 5000 W. Chandler Blvd., Chandler, AZ 85226
Opinions expressed on Internet are my own personal fuzzy logic.
I am not an official spokesman for Intel Corp.
Gary Coffman
>No wonder you break it Gary
>
> So your transmitter has infinite Q ha! Ha! HA!
No, but it can have a very high unloaded Q. In the ideal lossless network
case, it does have an infinite Q.
> Gary Coffman Wrote
>> In point of fact, *any* heated cathode vacuum tube *is* a diode. Current
>> can only flow from cathode to anode because only the cathode is an
>*emitter*
>> of electrons. A tube conducts more or less current depending on the
>potential
>> applied to the grid, the applied plate voltage, and the impedances
>around
>> the loop, but current can never flow from plate to cathode. It's a one
>> way street.
>>
>
> An alternating voltage containing energy does not have to
> change about a mean value of zero volts absolute.
An alternating voltage does represent *potential* energy. Depending on
current magnitude and phase relationships, it may be proportional to
power as well. Currents in a tube, however, never flow from plate to
cathode. So net power flow is always forward. If we ignore the DC component,
we can say in an operational sense that AC power can flow in either direction
through a tube, but we can't say that in a precise physical sense because
the current loop topology won't allow that to be true.
> Are you prepared to fire up a high power transmitter and
> let it run all day with no load on the output connection ?
Sure, I'll just reduce the grid drive to zero. In fact that's exactly
how we ran the hot spare in our broadcast plant.
>> The net charge goes the only way it can, toward the load. Some of the
>> charge may circulate in the transmitter's output tank for a cycle first,
>> but it ultimately can only be dissipated in a real resistance, and
>that's
>> the load.
> If that where true you could start up a Tank circuit disconnect
> load and tube and it would run for ever. Perpetual Motion
Yes, in a sense that's true. In fact, that can happen in a superconducting
loop. Currents have circulated for years in such lossless networks in
actual experimental setups. Now in non-superconducting tanks, Q is finite,
and the currents will decay over time due to being dissipated in the
loss resistances of the tank. However, the unloaded Q of a tank can be
quite high, over a 1,000 in the case of good cavities. When the load
is connected, the loaded Q drops to 10 or less. That means that of the
current circulating in the tank, 100 times more of it will be dissipated
in the load than in tank losses. Now 99% is close enough to perfect for
practical work.
The key thing is that the *tube* doesn't load the circuit very much
at all. It's a series combination of a very high value load line (which
can't dissipate power) and a very low value set of loss resistances. It
hardly affects the tank Q at all. *And* with grid drive, it's *supplying*
current to the tank, not dissipating it. The only major place for power
to go is to the termination load.
Gary and Jean Peach
Gene N9XAM wrote
>
> What I want to know regarding this thread is:
> if those who claim that reflected power can damage the tube, why I have
never
> heard of any dip meters being blown up via this mechanism?
You will most probably find that the power generated
by the device is very small compared to it's design power
handling ability.
Transmitters that blow up are usually being run at or near
to their design limits.
This becomes more likely as the power out increases,
for practical reasons.
If a dip oscillator is running 10 mW the device may easily
be rated at 500mW or 1/2 Watt so even if it was possible
to reflect all of that back into the device it would only
represent 1/50 th (2%) of the devices rated power handling
73 de
cmo...@sedona.intel.com
William P. Osborne <wosb...@nmsu.edu> wrote:
>which is that the reflected power stays in the 1/4 section which your
>quote from the handbook agrees with and I assume that you agree since you
>picked the quote. Now for the case of a 50 ohm source nothing changes if I
>choose to make the 50 ohm line have a length of zero. Now the 70.7 ohm
>line has forward and reflected power of it and it is mismatched on both
>ends but NONE of the reflected power goes into transmitter! I think this
>disproves your orginal statement (which I agreed with before I studied the
>problem) that reflected power is dissipated (at least some of it) in the
>source.
Hi again Bill, This is a Z0-matched case - NO reflected power is
dissipated in the source in a Z0-matched system, NO reflected power
is dissipated in the source in a mis-matched in-phase system, and NO
reflected power is dissipated in the source in a flat-matched system.
We hams always want a flat-matched or a Z0-matched system because that
is where maximum efficiency occurs. I am not arguing anything about
those kinds of systems. The common characteristic of all of those
systems is that the generator voltage and the generator current are in
phase. I am NOT talking about any system where the generator voltage
and current are in phase.
If the generator voltage and current are out of phase, there must be
some energy source, external to the generator, causing the current when
there is no generator voltage. All the energy originates in the
generator but a reactive element stores energy when the generator is
strong and returns the same energy when the generator is weak - and it
happens even when the reactive element is at the end of a transmission
line.
I'm sorry that we don't seem to be communicating. One of us is looking
at the forest and the other is looking at one particular leaf on one
particular tree.
Cecil Moore
>> Gary Coffman Wrote
>>> In point of fact, *any* heated cathode vacuum tube *is* a diode. Current
>>> can only flow from cathode to anode because only the cathode is an
>>*emitter*
current (holes) flows from anode to cathode. :-) However, if you
bias the tube like you bias an opamp, i.e. two power supplies with
the plate of a class-A amplifier at zero volts DC, you can obtain
a direct output without a capacitor to block the DC.
>we can say in an operational sense that AC power can flow in either direction
>through a tube, but we can't say that in a precise physical sense because
>the current loop topology won't allow that to be true.
>
clearer. In a class-A amplifier we definitely can say that AC can flow
in either direction through a tube. Set up a class-A amplifier as above,
put a 1.5v battery on the plate in either direction. A linear amplifier
will measure approximately the same current for both possible polarities
of the battery voltage indicating a reverse path to ground for AC that
is close to linear as long as the AC peak-to-peak voltage doesn't get
outside the linearity range of the tube.
We use biased-on diodes to pass RF in many designs. As long as the RF
voltage is not high enough to cause non-linearity in the diode, it
passes both the positive and negative excursions just fine.
Anytime the generator voltage is at zero crossing and the generator
current is not zero, there is energy being delivered to the generator
from the reactive components in the system and these reactive
components can be effecting the generator phase all the way from the
load. (dang, make that affecting - I hate this editor)
73, Cecil, KG7BK, OOTC
Gary and Jean Peach
Gary Coffman Wrote in etc.
> From: ga...@ke4zv.atl.ga.us (Gary Coffman)
> Subject: Re: Is Maxwell wrong?
> Message-ID: <1995Jan28....@ke4zv.atl.ga.us>
>
------snip-------
> Your heart's in the right place, Gene, but you're overlooking a
difference
> between transmitters and dip meters. Dip meters aren't usually operated
> near device limits, transmitters often are. While reflected "power" per
> se bothers neither of them, the loss of efficiency due to an improper
> presented load impedance *may* damage a transmitter.
>
> --
> Gary Coffman KE4ZV | You make it, |
gatech!wa4mei!ke4zv!gary
> Destructive Testing Systems | we break it. |
emory!kd4nc!ke4zv!gary
> 534 Shannon Way | Guaranteed! | ga...@ke4zv.atl.ga.us
> Lawrenceville, GA 30244 | |
So an incorrectly terminated transmitter dissipates more than
a correctly terminated transmitter.
I also call it a loss of efficiency.
All that DC power goes into the tube instead of some of it
going into the load.
You may haggle about the mechanism or the name of the
extra energy in the transmitter under mismatch, but you
do concede that extra energy is there and under certain
circumstances can do damage. And would you be prepared
to concede that should a mismatch exist and the transmitter
does not blow up because it is adequately over rated, that
extra energy exists in the transmitter.
73
Gary Coffman
>From: cmo...@sedona.intel.com
>Date: 27 Jan 1995 00:55:19 GMT
>Message-Id: <3g9g9n$n...@chnews.intel.com>
>
>In article <6e.40...@cencore.com>,
>Forrest Gehrke <forrest...@cencore.com> wrote:
>>
>>This is getting ridiculous. Here is lacking anything
>>to dissipate power. What would you hope to measure
>>under such a circumstance and how would you correlate
>>that with the case of a resistive component as part
>>of a reactive load?
>
>Hi Forrest, one way to test the consistancy of a theory is to
>take it to the limit. If the theory won't handle 0 ohm, it
>probably won't handle 0.001 ohm either - or 50 ohms. If you add
>0.001 ohm to the circuit, does all the energy avoid the
>transmitter and rush to the 0.001 ohm? I don't think the capacitor
>"knows" it's not supposed to discharge into the transmitter.
Oh, dear, Cecil, have you ever heard of asymptotic functions?
I was going to post the following in answer to one of Gary
Peach's postings, but I think it might do more good here.
In article <D3Cs6...@cix.compulink.co.uk> wise...@cix.compulink.co.uk ("Gary and Jean Peach") writes:
>So an incorrectly terminated transmitter dissipates more than
> a correctly terminated transmitter.
Not necessarily, though it can. It depends on the magnitude of the
incorrect termination, and some other factors. IE a RL greater than
the match value will cause *less* current to flow in the transmitter
loop, and less power to be dissipated in the Rloss of the tube. For
a RL less than the match value, more power will be dissipated in the
tube.
> I also call it a loss of efficiency.
> All that DC power goes into the tube instead of some of it
> going into the load.
Hopefully not. There shouldn't be any DC going to the load.
> You may haggle about the mechanism or the name of the
> extra energy in the transmitter under mismatch, but you
> do concede that extra energy is there and under certain
> circumstances can do damage. And would you be prepared
> to concede that should a mismatch exist and the transmitter
> does not blow up because it is adequately over rated, that
> extra energy exists in the transmitter.
No, not at all. There is no "extra" energy. Efficiency is Po/Pi*100.
If Pi is constant, there can be no extra energy, just a different
apportionment of the available energy in different ratios between
losses and load. The actual situation is more complex because the
power supply is typically not infinite in its ability to deliver
voltage or current in arbitrary ways. A clearer conceptual model
of the tube circuit may be useful here.
A tube is not a source of any kind. A useful model for a tube is
a resistive analog of a transconductance in series with a loss
resistance. Like this:
(fig 1)
T
cathode x-----/\/\/\/\ Rloss
^------------/\/\/\/\-------x plate
Grid
We can picture the transconductance as a reostat with the slider
position controlled by the grid voltage. We must remember, however,
that this isn't an actual resistor. It can't dissipate power. The
British call tubes valves, and that's a good analogy. The transconductance
"valves" more or less current between cathode and plate depending on the
grid voltage. We can use the reostat model as long as we understand that
it only partially models what's happening. The transconductance is the
dominant "resistance" in the tube, usually equivalent to several thousand
ohms. (Nit: transconductance and my reostat are really reciprocals since
conductance is the reciprocal of resistance, but that doesn't alter the
analogy, just makes the numbers whole rather than fractions.)
Rloss is partially real resistances; contact resistance and bulk component
resistance. At DC, these resistances are usually negligible. At RF, these
resistances become more substantial due to skin effect, growing from a
fraction of an ohm to several ohms.
There is another component to Rloss that isn't a resistor, but is
proportional to current and voltage through the tube. This is the
Ke loss from electrons slamming into the plate. The electron carries
charge to the plate, but it also carries a certain amount of kinetic
energy. Electrons have mass and are travelling at a high velocity.
When the electron strikes the plate, it delivers its charge and it
delivers kinetic energy in the form of heat. Ke=.5*m*v^2. The mass
delivered is proportional to forward current, and the velocity is
proportional to the cathode to plate voltage. This kinetic energy
heating dominates the Rloss term in many tubes.
Tube heating depends mainly on how much current passes through the
tube, and on how much accelerating potential is present between the
cathode and plate. This is only indirectly related to matching between
the tube "resistance" and load resistance in that it is the combination
of the two that sets the loop current and the voltage drops around the
loop. All power comes from the power supply, of course, and it has it's
own internal resistances and current and voltage limits.
Now let's draw in a more complete circuit.
(fig 2)
T
x-----/\/\/\/\ Rloss
| ^------------/\/\/\/\-------x
| Grid |
| Rload |
|-----(-battery+)---/\/\/\/\/\---------|
In an ideal tube, T can vary from zero to infinity in response to
changes in grid potential. In real tubes, T can't quite reach zero
due to filament saturation, but it can reach infinity at cutoff.
Ok, so in the limit cases, with T=0, maximum power will be transfered
to Rload when Rload=Rloss, and stage efficiency will be 50%. Note
that Rloss is from fractions of an ohm to a few ohms. At the other
limit, where T=infinity, there is no power transfer through either
Rloss or Rload. At the center of the linear range, T will be several
thousand ohms in series with Rloss. Now maximum power will be transfered
to Rload when T+Rloss=Rload, again a matched case. But, now power can
only be dissipated across Rloss and Rload because T isn't a real resistor.
So, with the same currents flowing in both (they're in series), the
power dissipated in the load and tube loss resistances is I^2*Rload and
I^2*Rloss. Since Rload is much larger than Rloss, the stage will show an
efficiency greater than 50%.
Ok, that should explain why matched stages can exceed 50% efficiency.
Now lets look at mismatches at Rload. First lets look at the limit cases.
With a perfect mismatch caused by Rload being open, no current flows around
the loop, and no power is dissipated in the tube. With a perfect mismatch
caused by Rload being a dead short, maximum current flows around the loop,
and all power is dissipated by Rloss. Two perfect mismatches, two very
different behaviors.
Now lets draw the circuit again, with some changes.
(fig 3)
T
x-----/\/\/\/\ Rloss Tank
| ^------/\/\/\/\-------|--)
| Grid | )
| C = ) L
|-----(-battery+)----------------|--)
This begins to look like a transmitter. Static bias is applied to
the grid so that T+Rloss will put the stage at an idling current
Iidle of battery/(T+Rloss). Now the power dissipated in the stage
is Iidle^2*Rloss. Note that this is a small fraction of the power
that would be dissipated if T were a real resistor, Iidle^2*(T+Rloss).
Since this is DC, the tank looks like a short and doesn't enter into
the situation.
Ok, lets feed RF grid drive in at a frequency where the tank is
resonant. The drive level is set so that the tube reaches neither
cutoff nor saturation, IE it remains in the linear range. Now the
signal current, Irf, is set by (battery+Vtank)/(T+Rloss+Ztank).
Since the tank is resonant, Ztank will be very high, in the ideal
case it will be infinite, in the real world it will be determined
by the unloaded Q of the tank. With a reasonable unloaded Q, it
will be on the order of tens of thousands of ohms. Vtank will equal
battery on positive peaks, so positive peak plate voltage will be
2*battery. On negative peaks, plate voltage will be near zero. This
averages to just battery. AC current flow will be very low through
the tube, just enough to make up for the circulating current losses
caused by the finite Q. The current through the loop will be
Itotal=Iidle+Irf, and power dissipated by the stage will be
Itotal^2*(Rloss+Rcopper) with Rcopper being the tank loss resistance,
a small number, and of course Rloss is also a small number, so little
heating occurs.
Now lets couple a load into the circuit.
(fig 4)
T
x-----/\/\/\/\ Rloss Tank
| ^------/\/\/\/\-------|--) (-----x
| Grid | ) ( / Rload
| = ) ( \
|-----(-battery+)----------------|--) (-----/
We'll assume a coupling factor of 1 and a transformer ratio
of 100:1 for the link so that Rload will appear in parallel
with Ztank as 100*Rload. If Rload were 50 ohms, we'd have
5,000 ohms appearing across Ztank. We'll assume this gives
a loaded Q of 10 for a Ztank on the order of 4500 ohms.
(Note that this is near a very common *loadline* value
for a tube RF amplifier, so it offers a match to T+Rloss.)
Now this brings us back to fig 2. Maximum power will be
transfered when T+Rloss=Ztank. The AC current Irf through
the loop is battery/(T+Rloss+Ztank). But RF power dissipated
in the stage is still Irf^2*(Rloss+Rcopper). Since both
of those are small numbers, stage efficiency remains high,
and most of the power is coupled out to Rload by the link.
(Note: I've been implicitly assuming that battery offers
zero ohms impedance to RF. That's justified if we assume
low reactance bypass capacitors.)
Ok, finally, we're ready to introduce a wave trying to get back
in the transmitter's output port (we'll let some third party
introduce the wave). It will couple through the link, and induce
a voltage, Vback, across the parallel combination of Ztank+Rcopper
and T+Rloss. Since Ztank+Rcopper=T+Rloss, current will divide
equally into each. We'll call them Ihalfback. Now the current
induced through Ztank just kicks the tank and couples back out to
Rload, minus a small Ihalfback^2*Rcopper loss. The current that
couples across T+Rloss also kicks the tank, (remember, the battery
is transparent to RF), minus the small Ihalfback^2*Rloss of the tube,
and goes back out to the load as well.
Because the real resistances in the transmitter output stage are small,
transmitters absorb very little of a wave trying to get back in, and
reflect the vast majority of it right back out the outport to the only
real load available. An *ideal* transmitter would have no Rloss or
Rcopper, and would reflect 100% of the backwave. All this is *regardless*
of whether a transmission line is present on which waves can stand.
The only effect of standing waves is to offer a different, and perhaps
complex, impedance to the transmitter output than would be the case with
the load connected directly to the outport. If that impedance, transformed
by the output network, is higher than the design tube loadline, the tube
will draw less current, run cooler, and not deliver as much power to
the load, but stage efficiency will go *up*. If the impedance presented
is lower than the tube loadline, the tube will draw more current *from
the power supply*, run hotter, deliver less power to the load, and
stage efficiency will go *down*.
Gary
Gary Coffman
In article <196_950...@trisoft.com> Cecil.Moore.@f1.n328.z1.fidonet.org (Cecil Moore ) writes:
>"William P. Osborne" <wosb...@nmsu.edu> writes:
>
>>Cecil, I have a question for you. Let's assume you have a transmitter
>>driving a 50 ohm line of any length terminated in a quarterwave of 70.7 ohm
>>line which is loaded with 100 ohm resistive load. Since the 50 ohm line is
>>matched there is no reflected wave on it and we assume the system to be
>>lossless where does the reflected power in the 1/4 wave line go? Seems
>
>Hi Bill, as I told you over email, I am just believing the ARRL Antenna
>Book. It says that there is an initial reflection at the 50/70.7 step-up
>point where the reflected voltage is in phase and the reflected current
>is out of phase with the forward wave.
So, you're saying that there *is* a reflected wave on the initial
50 ohm section? Must get absorbed in the transmitter's output
circuit, huh?
>The rest of the forward wave
>continues on to the load where another reflection occurs at the 70.7/100
>step-up point. The forward voltage at the load is 180 degrees out of
>phase with the forward voltage at the 50/70.7 point so the reflections
>at the load will be 180 degrees out of phase with the reflections at
>the 50/70.7 point.
Huh? The forward voltage at the load will be 90 degrees out of phase
with the forward voltage at the 50/70.7 point. After all, they've
traveled a quarterwave, which by definition is a 90 degree shift.
Perhaps you meant that when the reflections get *back* to the 50/70.7
point they'll have traveled 180 degrees.
>When the reflections from the load encounter the
>reflections at the 50/70.7 point, it results in a virtual open or short
>for the reflections from the load at the 50/70.7 point and the reflections
>are cancelled at that point. Please take it from there and explain why
>that concept is wrong.
So where does the *power* go? Thin air? You keep saying there's power
in them thar backwaves, where does it go?
Gary
PS The book's not wrong, but your explanation of why makes no sense.
Cecil Moore
A B
GC>So, you're saying that there *is* a reflected wave on the initial
GC>50 ohm section? Must get absorbed in the transmitter's output
GC>circuit, huh?
Hi Gary, I'm assuming that you don't understand what I said rather than
Just being mean. When you key the CW transmitter, the initial wave front
will encounter an impedance change at the 50-71ohm point A and there will
be reflections at that point. The remainder of the initial wave front
keeps going to the load where another reflection takes place at the
71-100ohm point B. The reflections from point B return to point A and
are 180 deg out of phase with the initial reflections at point A. As
we build up to steady-state, the reflections from the load _cancel_ the
reflections at point A so there are no longer any reflections in the 50
ohm coax. The first reflection from the load is smaller than the first
reflection at point A but it subtracts from the reflection at point A
giving more forward power to the load and therefore more reflections
from the load. Cycle by cycle, the reflections in the 50ohm coax
decrease and the reflections in the 71ohm coax increase until there are
no longer any reflections in the 50ohm coax and all reflections are
in the 71ohm coax. The steady-state reflections see a virtual short at
the 50-71ohm junction. Until steady-state conditions are achieved, the
reflections in the 50 ohm coax can indeed be absorbed in the transmitter's
output circuit. W5OLY, in his latest CQ article, admits that reflected
energy will be dissipated in the transmitter under transcient conditions.
CM>The forward voltage at the load is 180 degrees out of
CM>phase with the forward voltage at the 50/70.7 point so the reflections
CM>at the load will be 180 degrees out of phase with the reflections at
CM>the 50/70.7 point.
GC>Huh? The forward voltage at the load will be 90 degrees out of phase
GC>with the forward voltage at the 50/70.7 point.
You are right, of course. Don't know why I typed 180 there. It's not
my first dumb typing mistake. I guess I was thinking about the half-
wave matching section at which I was also looking.
GC>So where does the *power* go? Thin air? You keep saying there's power
GC>in them thar backwaves, where does it go?
One more time, IN A STEADY-STATE MATCHED SYSTEM, ALL REFLECTED WAVES
ARE RE-REFLECTED. I HAVE NEVER, EVER DISAGREED WITH THAT. The above
system is a matched system and the power in the reflected waves adds
to the power in the forward wave. If you think there's no power in
"them thar backwaves" measure the forward power in the 50ohm coax and
then measure the forward power in the 71ohm coax. Wonder of wonders,
there _is_ power in reflected waves and you can measure it. In the
above MATCHED system, the re-reflections become indistinguishable from
the forward wave so I would say some of the energy in the re-reflections
is actually dissipated in the 100 ohm load while an equal amount of the
newly generated energy is reflected at the load. Too bad we can't "color"
the different energies. Then it would be easy to prove what I am saying.
IMO, It would take some sort of magic for the re-reflected waves to 100%
combine with the forward waves and then "decide" to 100% divorce
themselves to be re-re-reflected at the load. Come on, Gary, admit it.
Only the scalar values have to add up. How does the load "know" to
reject the energy that has been previously reflected and dissipate only
the energy that has been newly generated? If we, with all our
sophisticated technology, cannot separate combined generated waves from
the reflected waves that joined them in phase, how in the world does a
dumb resistor do it?
You didn't address the contradiction in your earlier "virtual open
circuit" statement. It's understandable that you like to talk concerning
things about which you correct. I would rather talk concerning the things
about which you are incorrect. :-)
73, Cecil, KG7BK, OOTC
Jon Bloom (KE3Z)
: In article <3gj98d$3...@bowline.stsci.edu>,
: Phil Hodge <ho...@bowline.stsci.edu> wrote:
: >Cecil is right. That's how a low-reflection coating on glass works.
: Thanks Phil, those first three words haven't appeared here lately. See
: what you think of this.
: +------------------+ +--/\/\/\/--+
: | Solid-State | | RL=50 |
: | Linear Xmtr, o-------------------------------------+ |
: | 50 ohm Output | 1/2 wavelength 50 ohm lossless coax |
: | Design Impe- o-------------------------------------+ |
: | dance, No Tuner | | XL=-j100 |
: +------------------+ +----| |---+
: | switch |
: +-----/ ----+
: Sometime ago, I was playing with equations from quantum
: physics when I found some energy flowing to the load and back
: to the transmitter to be dissipated. It has taken me many weeks to
: realize what is happening in reality. Assume the switch is closed and
: there are no reflections. Then open the switch at the zero-crossing
: point of the forward wave. Reflections begin but don't reach the
: transmitter for 1/2f (assume 50uS at 10MHz).
OK, let's look at this example. When the switch is opened, the
reflection coefficient at the load (pL) changes from 0+j0 to 0.5-j0.5.
If we call the forward voltage at the load Vf, the reflected voltage at
the load is Vr=Vf(pL)=Vf(0.5-j0.5). Since you've chosen a half-wave line
length, this is also the reflected voltage at the transmitter end of
the line.
If the transmitter's S22 impedance is 50 ohms, there is no
re-reflection of the signal returning from the load. Thus the forward
voltage is not changed. So the *net* voltage at the transmitter end of
the line is simply the sum of our original forward voltage plus the
reflected voltage: Vt = Vf + Vf(0.5-j0.5) = Vf(1.5-j0.5). In other
words, the voltage at the output has risen in amplitude and it has
changed phase.
Now, what effect does this have on the transmitter? It does *not*
necessarily mean that more power is dissipated in the transmitter! To
find out what effect it does have, we need to know something about the
transmitter's internal arrangement. As an example, let's say the
transmitter is a Thevenin source: an ideal voltage source in series
with a 50-ohm resistance. Note that I'm not saying we're *modeling* the
transmitter as a Thevenin source, I'm saying it actually *is* such a
circuit. (I'll deal with a different case shortly.)
Originally, before we opened the switch, the voltage from the
transmitter's voltage source was evenly divided between the internal
50-ohm resistance and the external 50-ohm load. Vf was then one half of
the source voltage: Vf=Vs/2. This is also the voltage dropped across
the transmitter's internal 50-ohm resistance, making the power
dissipated in the transmitter: (Vs/2)^2/50=Vs^2/200.
Now, with the switch opened and the reflected voltage returned to the
transmitter, the voltage at the output is as calculated above,
Vf(1.5-j0.5)=(Vs/2)(1.5-j0.5). That means the voltage across the
transmitter's internal resistance is the difference between the source
voltage and the output voltage: Vs-(Vs/2)(1.5-j0.5)=Vs(-0.25+j0.25).
Now the power dissipated in the generator is (Vs(-0.25+j0.25))^2/50.
That works out to (Vs^2/8)/50=Vs^2/400. Contrast this to the power that
was dissipated before we opened the switch. There is *less* power being
dissipated in the generator now!
Now let's calculate the powers in the system. The forward power is
always going to be Vf^2/Zo, and the reflected power is Vr^2/Zo. The
load power is going to be determined by the voltage that appears at the
load end of the transmission line, which is Vf+Vr. So the load power is
(Vf+Vr)^2/ZL. In the switch-closed case, the forward power is
(Vs/2)^2/50=Vs^2/200, and the reflected power is 0^2/50 = 0. The load
power is (Vf+0)^2/50=(Vs/2)^2/50=Vs^2/200. No surprise here: the
forward power equals the load power. In the switch-opened case, the
forward power is unchanged, at Vs^2/200, and the reflected power is:
|((Vs/2)(0.5-j0.5))^2/50|=Vs^2/400. The load power in this case is
|(Vs/2+(Vs/2)(0.5-j0.5))^2/(50-j100)=Vs^2/179 (approximately). The real
part of this can be found by multiplying Vs^2/179 by the load power
factor, cos(tan^-1(-100/50))=0.4472. The result is that the real part
of the load power--the power dissipated in the 50-ohm resistor--is
Vs^2/400.
Let's compare the two cases. In the switch-closed case, the generator
dissipates Vs^2/200 watts and the load dissipates Vs^2/200 watts. The
generator is operating at 50% efficiency. In the switch-open case, the
generator dissipates Vs^2/400 watts and the load dissipates Vs^2/400
watts. The generator is *still* operating at 50% efficiency. It's just
delivering less power overall.
I mentioned that the transmitter's internal configuration impacts this
discussion. Let's change the transmitter to a Norton generator. This
consists of a current source in parallel with the 50-ohm resistor, to
which the load is connected, again in parallel. To make the transmitter
operate in exactly the same way on the load as with our earlier
configuration, we need to set the current source to get the same
open-circuit voltage (with no load connected) as we got with our
earlier scheme. Thus the current source must supply a current of
(Vs/50) amps. Now consider what happens when the reflected signal
causes the voltage to rise at the line input. There is a greater
voltage across the 50-ohm internal resistance. Therefore, the
transmitter power dissipation goes up, instead of down, as it did
previously. The load is still dissipating Vs^2/400 watts, but the
transmitter is dissipating |(Vf+Vr)^2/50|=|(Vs/2+(Vs/2)(0.5-j0.5))/50|=
Vs^2/80 watts. The system is operating at 16.7% efficiency. Not too
good!
Lesson 1: You have to know what is inside the transmitter before you
can say what will happen to its power dissipation because of the
reflected signal.
Lesson 2: You can't treat the transmitter as a purely dissipative load.
You *must* consider its internal source as part of the system. In the
first case, using a Thevenin generator, we found that the reflected
signal actually caused the transmitter dissipation to decrease because
the terminal voltage was closer to the internal source voltage.
Lesson 3: Since it is not necessarily the case that a reflected signal
will cause the transmitter power dissipation to rise, it's silly to
worry about reflections per se. All you need to do is ensure that your
transmitter is seeing the load impedance it was designed for. In some
cases, it may be seeing the desired impedance purely *because* of these
reflections--as in the case of a quarter-wave matching section. In
others, unwanted reflections may cause the load impedance to be
incorrect. It is not necessarily true, however, that this will cause
the transmitter to overheat or otherwise damage itself. As we saw in
the example using the Thevenin generator, the transmitter may operate
just as efficiently as before, but output less than the design power.
: On the upswing of the forward voltage, a certain amount of energy is
: stored in the capacitor. On the downswing of the forward voltage, that
: energy stored in the capacitor is returned to the system as power and
: there are equations for calculating the magnitude of that power. Some
: of that power is dissipated in RL and some is dissipated 50uS later in
: the transmitter since it is a series circuit. The same thing happens
Except that, as we've seen, the presence of the source signal inside
the transmitter may actually cause the transmitter's dissipation to
*decrease*.
: dissipated in the transmitter. Note that this dissipation did not
: occur until we closed the switch and the reflections began. This
: dissipation will happen anytime the generator voltage and current are
: out of phase, i.e. we have a non-zero current when we have a zero
: voltage - indicating an instantaneous power dissipation in Rg that is
: not coming from the generator so it must be coming from outside.
Power is the rate of doing work. Thus "instantaneous power" is a
conundrum. For example, the voltage across a pure reactance is 90
degrees out of phase with the current through it. Therefore, according
to your logic, the reactance must dissipate "instantaneous" power when
the current is at the peak of the sine wave and the voltage is at zero.
But you know that a pure reactance dissipates no power, right? That's
because the power dissipated in any circuit is the *time average* of
the voltage times the current. If the voltage across the circuit is
sin(2*pi*f) and the current through it is cos(2*pi*f), like in a
reactance, integrating the product sin(2*pi*f)cos(2*pi*f) over the
period of a cycle gives 0. You have to calculate this time average to
determine what, if any, power is dissipated in a circuit. Or, you can
use RMS values of voltage and current, taking phase into account, as
I've done above. Since RMS values are time-average equivalents, using
them (if done properly) is the same as taking the time average of the
instantaneous values. You can draw *no* conclusion about the power
dissipated in a circuit by considering only instantaneous values!
: From a quantum physics, quarter cycle perspective, Maxwell is wrong,
: every even 25uS period in the example above. The reason that it is not
: obvious to everybody is that steady-state RMS power is equivalent to
: DC, i.e. average power. Take a look at the instantaneous power values
: and kiss a myth goodby. If the capacitor does not dump its power back
: into the transmitter the conservation of energy principle will have
: been violated and the capacitor cannot charge for the next cycle.
Your physics is flawed. The myth here is that "instantaneous power" is
meaningful. The capacitor dumps its energy back into the line. What
affect this has on the transmitter cannot be determined without doing a
full analysis. Thinking that the energy (not power, by the way) that is
traveling from the load toward the transmitter must be dissipated as
power in the transmitter is naive.
(Sorry for the delay in responding to this post. We're experiencing a
several-day delay in incoming news. Also, sorry for trying to cram
equations into a medium not well suited to them. Couldn't be helped.)
--
Jon Bloom KE3Z jbl...@arrl.org
Gary and Jean Peach
Gary Coffman wrote
> No, but it can have a very high unloaded Q. In the ideal lossless
network
> case, it does have an infinite Q.
Zg = Rg + jxg Rg = 0 Zg = xg
Therefore Zg does not Equal Zo
Maximum power transfer does not take place.
Not the same transmitter that we started with.
>
> > Gary Coffman Wrote
> >> In point of fact, *any* heated cathode vacuum tube *is* a diode.
Current
> >> can only flow from cathode to anode because only the cathode is an
> >*emitter*
> >> of electrons. A tube conducts more or less current depending on the
> >potential
> >> applied to the grid, the applied plate voltage, and the impedances
> >around
> >> the loop, but current can never flow from plate to cathode. It's a
one
> >> way street.
AGREED that has never been in dispute.
> >>
> >
> > An alternating voltage containing energy does not have to
> > change about a mean value of zero volts absolute.
>
..............................................................!!!!!!!!!!!!
!!!!!!!!!!!!!!!
> An alternating voltage does represent *potential* energy. Depending on
> current magnitude and phase relationships, it may be proportional to
> power as well.
OK I was trying to keep it simple and discuss what happens at
zero reactance and a power factor of one.
There are so many variables that if some agreement can be
reached about the special case when Zg = R, then the next
step of the discussion could proceed upon a sound basis.
The AC energy is * real kinetic energy* you can't store it in a
battery,without rectifying it, to convert it into potential energy.
That is why power engineers pump water back up hill with it
off peak.
BTW, you can transformer couple it out of the circuit and boil
water with it.
>....................... Currents in a tube, however, never flow from
plate to
> cathode. So net power flow is always forward. If we ignore the DC
component,
> we can say in an operational sense that AC power can flow in either
direction
> through a tube, but we can't say that in a precise physical sense
because
> the current loop topology won't allow that to be true.
Just subtract the mean current from all instantaneous
current values. It's a bit like sitting in a stationary train
reading, when the train pulls out, for a moment you think
its the platform that is moving.
I believe that a man called Einstein had a name for it.
The DC current of varying magnitude represents real energy
that can be transformer coupled out of the circuit to perform
real work such as boil water. Our power standard.
Watts = Joules per Second. A Thermal Unit.
I have often done this, into a water load.
The only real way to measure power.
Bolometers are about the next best thing.
>
> > Are you prepared to fire up a high power transmitter and
> > let it run all day with no load on the output connection ?
>
> Sure, I'll just reduce the grid drive to zero. In fact that's exactly
> how we ran the hot spare in our broadcast plant.
Hardly a transmitter just an expensive space heater.
We ran our Klystron Amps the same way on Apollo etc.
but the drive was not applied till the STD BYE TX was
looking into a good load. It only became a transmitter
at that time. 5W in 20 kW out. 0W in 0Watt out.
We only did this as the EHT took some time to run up to
the operating value. Seconds can mean lives in that game.
>
> >> The net charge goes the only way it can, toward the load. Some of the
> >> charge may circulate in the transmitter's output tank for a cycle
first,
> >> but it ultimately can only be dissipated in a real resistance, and
> >that's
> >> the load.
Consider for a moment that a reflected wave from a
transmission line, terminated in a load with VSWR greater
than 20:1, arrives back at the transmitter almost in
phase, the voltages vector sum.
Vpeak is almost twice V ht
If your tube is rated OK you probably will not get flash over.
The line has stored some energy V*I*t Joules
where
V is the RMS applied voltage,
I is the RMS applied current
t is the trip time for the wave.
>
> > If that where true you could start up a Tank circuit disconnect
> > load and tube and it would run for ever. Perpetual Motion
>
> Yes, in a sense that's true. In fact, that can happen in a
superconducting
> loop. Currents have circulated for years in such lossless networks in
> actual experimental setups.
Yes I've played with Masers I know about that. If you turn
the fridge off the currents stop so they are being driven
by some complex thermal property. They are not perpetual
motion machines.
Incidently tetrodes have a negative resistance portion to
their curve. That too is not perpetual motion.
>................................... Now in non-superconducting tanks, Q
is finite,
> and the currents will decay over time due to being dissipated in the
> loss resistances of the tank. However, the unloaded Q of a tank can be
> quite high, over a 1,000 in the case of good cavities. When the load
> is connected, the loaded Q drops to 10 or less. That means that of the
> current circulating in the tank, 100 times more of it will be dissipated
> in the load than in tank losses. Now 99% is close enough to perfect for
> practical work.
Q = ( ( L / C )^ 1/2 ) / R for lumped circuits
We polished cavities to Q greater than 18,000 for Pound
stabilisers used to control the output frequency of
reflex klystrons. In the days before synthesisers etc..
But that was a lot of very hard work with metal polish and
dry gas filling. We never reached infinity, the OFHC Cu
refused to have zero resistance.
However Zg = R + jX, The R we are discussing is a
different animal. I am thinking of the general set of all
transmitters, Triodes, pentodes, Beam tetrodes,
transistors, magnetrons, klystrons, TWT, BWO, HF, LF,
AF, VHF,UHF, Microwave, etc..
( haven't played with Lasers yet but sure would like to.)
Try a thought experiment. We terminate our transmitter
in a number of resistors in turn, starting at a low value and
each time applying a resistor of a slightly greater value.
Each resistor is immersed in an identical volume of water
contained in a Dewar. We ensure that the temperature is
at some reference value before we apply power.
We time a temperature rise (say 10 degrees) in the water.
It is found that the time is shortest at some value of
terminating resistance.
We note the value of that particular resistor.
That is the value of Rg. Not a direct measurement I grant.
There is within the transmitter a distributed real resistance
equal to this value. When it burns out you'll find it, or the
weakest link.
> The key thing is that the *tube* doesn't load the circuit very much
> at all. It's a series combination of a very high value load line (which
> can't dissipate power) and a very low value set of loss resistances. It
> hardly affects the tank Q at all. *And* with grid drive, it's
*supplying*
> current to the tank, not dissipating it. The only major place for power
> to go is to the termination load.
>
What causes the change of current in the tank coil
in the first instance if there is no couple between
tank and tube?
Like towing a car with a slack rope, it's fine up hill
but downhill the guy in the rear car better apply the
breaks or your rear end is going to suffer.
If he hits the break up hill the rope will probably part.
Which is fine until the extra energy that has been
stored from an earlier time comes back from the
transmission line.
Now I would be the first to admit that I haven't figured
out how to put an Ohm meter on Rg. The only physical
evidence for it is a scrap yard of burnt out transmitters
and PA tubes etc., apocryphal tails of narrow escapes.
cmo...@sedona.intel.com
Gary Coffman <ga...@ke4zv.atl.ga.us> wrote:
GC> What's faulty about your reason is that you are implicitly
GC> assuming there is no *discontinuity* at the point of reflection
GC> for the reverse wave. What I have been telling you over and over
GC> is that this is a false assumption for a transmitter. The tube
GC> looks like a virtual *open circuit* to the wave attempting to
GC> enter it's output port. It's a *perfect* mismatch (aside from
GC> some minor Rloss components in the wiring and output network
GC> parts).
What happens to reflected power 90 deg out of phase with the forward
wave when it is re-reflected by a "virtual open circuit"?
Assume Vforward=40v, Vreflected=20v, Iforward=100mA, Ireflected=50mA
Forward Power = 4w, Reflected Power = 1w
+---->Vr Ir<----+ +---->Vr +---->Ir
Before Re-Reflection After Re-Reflection by a
"virtual open circuit"
If we subtract the re-reflected vectors from the forward vectors, we
get the generator vectors.
Vf Vg If Ig
| / | /
| / | /
| / | /
|/ |/
Vr----+ Ir----+
Assuming generator power equals forward power minus reflected power,
Vg=Vf-Vr and IG=If-Ir
Vt=44.7v at 63.4 deg
It=111.8mA at 63.4 deg
Vg*Ig*cos(0)=5 watts with Vg and Ig in phase.
Before the re-reflection, total power equals 3 watts which is the 4
watts forward minus the 1 watt reflected and agrees with reality.
After the re-reflection by the "virtual open circuit" and assuming that
forward power equals the generator power plus the reflected power, the
generator power has to be 5 watts.
This points up the difficulty in re-reflecting reflected waves at the
_point_ where they are 90 degrees out of phase with the forward wave.
If we do re-reflect them, they are still 90 degrees out of phase with
the forward wave and therefore CANNOT add in phase with the forward
wave. The generator has to generate one watt to kill the reflections
and then generate all of the forward power. Thus 1 + 4 = 5 watts from
the generator.
If that seems like magic, it is. We have taken a severe mismatch, and
by re-reflecting the reflected waves with a "virtual open circuit", we
have brought the generator voltage and current into phase and thus have
achieved a perfect match. We don't need an antenna tuner. "And if
you'll buy that, I've got some ocean front property in Arizona..."
The results of a "virtual short circuit" are the same. This proves that
there cannot be a virtual open or short circuit at the generator when
the reflected waves are 90 degrees out of phase with the forward waves.
The virtual open or short circuit explanation is obviously wrong. Want
to try some other explanation?
Anthony R. Gold
Laurence.G...@f1.n328.z1.fidonet.org
"Laurence Gene Battin" writes:
> I'm certainly no expert, but perhaps the experts can comment on _this_
> possiblity?
The pair of 4CX250B's in my dip meter need replacing frequently. I think
you've solved my problem. Thanks a million :-)
--
Tony - G3SKR / AA2PM / tg...@microvst.demon.co.uk
N7WS
> > possiblity?
>
> The pair of 4CX250B's in my dip meter need replacing frequently.<<
I'll bet you're using a series regulated screen supply and when you get
the mismatched condition, you enter the negative screen current region.
The screen voltatage soars and the screen dissipation rating is exceeded.
Try a shunt regulated supply. I know it will be tough to carry up to the
top of the tower, but you gotta save those tubes:))
Wes Stewart, N7WS
n7...@aol.com
Gary and Jean Peach
It takes a nut to crack a wallnut. (brick wall that is):->
>
> > I'm certainly no expert, but perhaps the experts can comment on _this_
> > possiblity?
>
> The pair of 4CX250B's in my dip meter need replacing frequently. I
think
> you've solved my problem. Thanks a million :-)
Gary Peach G7SLL Jean Peach
C. C. (Clay) Wynn, N4AOX
(Jon Bloom (KE3Z)) writes:
>Power is the rate of doing work. Thus "instantaneous power" is a
>conundrum. For example, the voltage across a pure reactance is 90
>degrees out of phase with the current through it.
>Your physics is flawed. The myth here is that "instantaneous power" is
>meaningful. The capacitor dumps its energy back into the line. What
>affect this has on the transmitter cannot be determined without doing a
>full analysis. Thinking that the energy (not power, by the way) that is
>traveling from the load toward the transmitter must be dissipated as
>power in the transmitter is naive.
Touche' on the power definition. I think some of the confusion comes from
the PEP concept. This was suppose to be so easy?? What is the standard
time increment for computing the PEP of a two-tone modulated AM signal?
Careful, you are not allowed to use instantaneous power values.
In the time domain, calculus definition of power, I think the time imcrement
can be arbitrary. I have blown up the active devices in amplifiers in much
less than one or two cycles. While your power observations are illuminating,
I am still not sure how this proves Cecil's thesis is wrong.
73,
C. C. (Clay) Wynn, N4AOX
w...@ornl.gov
=========================================================================
= ...- .. ...- .- - . .-.. . --. .-. .- .--. .... -.-- =
=========================================================================
Gary Coffman
>In article <1995Feb6.2...@arrl.org> jbl...@arrl.org
>(Jon Bloom (KE3Z)) writes:
>
>>Power is the rate of doing work. Thus "instantaneous power" is a
>>conundrum.
>>Your physics is flawed. The myth here is that "instantaneous power" is
>>meaningful. The capacitor dumps its energy back into the line. What
>>affect this has on the transmitter cannot be determined without doing a
>>full analysis. Thinking that the energy (not power, by the way) that is
>>traveling from the load toward the transmitter must be dissipated as
>>power in the transmitter is naive.
>
>Touche' on the power definition. I think some of the confusion comes from
>the PEP concept. This was suppose to be so easy?? What is the standard
>time increment for computing the PEP of a two-tone modulated AM signal?
>Careful, you are not allowed to use instantaneous power values.
Indeed you aren't allowed to use instantaneous power values since such
things are undefined by requiring division by zero seconds. The FCC has
an explicit standard for PEP. It is the *average* power contained in the
*one cycle* of the RF waveform nearest the *peak* of the modulating envelope,
treated as if it were the average power over one second.
They give an example method that they will find acceptable. Take the peak
voltage, as observed on an oscilloscope or other peak reading device, multiply
it by 0.707, square the result, and divide by the actual load resistance
presented to the transmitter output port.
In fact, they treat PEP as if it were the average power that would be
radiated if the highest one cycle of RF were to occur for all cycles
instead of just one. So the FCC PEP definition yields a different
value than the *true* power in that one cycle. The true power would
be the power calculated above divided by the operating frequency.
>In the time domain, calculus definition of power, I think the time imcrement
>can be arbitrary. I have blown up the active devices in amplifiers in much
>less than one or two cycles. While your power observations are illuminating,
>I am still not sure how this proves Cecil's thesis is wrong.
Remembering that power is the time rate of doing work, and that the
watt is the amount of work done in one second. As the measurement
interval becomes less than one second, the number of watts must
decline by the reciprocal of the time interval of measurement.
If you've blown devices in one or two cycles, it hasn't been because
of exceeding the power dissipation rating of the device. With
transistors, however, it certainly can be because the peak *voltage*
punched a junction. That's an entirely separate issue.
Gary
Gary Coffman
>
>You didn't address the contradiction in your earlier "virtual open
>circuit" statement. It's understandable that you like to talk concerning
>things about which you correct. I would rather talk concerning the things
>about which you are incorrect. :-)
I didn't ignore the "contradiction" since there isn't one. I fully
explained that the "load" you see looking back into a transmitter's
output port consists of a very high transadmittance in series with
a very low Rloss component. Since a transadmittance isn't a resistor,
it can dissipate no power, and since the series Rloss is small, it can't
dissipate much power either (same current through both since they are
in series). Thus, the "load" presented by the device looks like a
virtual open circuit to the wave coming in the out port. Since an
open circuit yields a total reflection, the power goes back out
toward the load.
Cecil Moore
OK, I can accept that you think that I'm silly to worry about
reflections per se. But that is what this thread is all about -
reflections per se. It must have seemed silly for Einstein to worry
about light per se, but he did anyway while the classical physicists
laughed, at least for awhile.
CM>>On the upswing of the forward voltage, a certain amount of energy
CM>>is stored in the capacitor. On the downswing of the forward
CM>>voltage, that energy stored in the capacitor is returned to the
CM>>system as power and there are equations for calculating the
CM>>magnitude of that power. Some of that power is dissipated in RL
CM>>and some is dissipated 50uS later in the transmitter since it is a
CM>>series circuit.
JB>Except that, as we've seen, the presence of the source signal
JB>inside the transmitter may actually cause the transmitter's
JB>dissipation to *decrease*.
I have never said otherwise. The question is, does _any_ of the
xmtr's dissipation come _directly_ from the reflected waves? I
think I have proved that it does. Average power remains the same,
of course, but _some_ of the dissipation inside the transmitter
comes from _some_ energy that has been outside the transmitter and
returned from the capacitor or reflected waves.
JB>You can draw *no* conclusion about the power dissipated in a
JB>circuit by considering only instantaneous values!
I think I made it perfectly clear that I was talking about 1/4 of
a cycle. Do you think power dissipated over 1/4 of a cycle is
meaningless? Do you think 1/4 of a cycle is instantaneous?
JB>The capacitor dumps its energy back into the line. What affect
JB>this has on the transmitter cannot be determined without doing a
JB>full analysis. Thinking that the energy (not power, by the way)
JB>that is traveling from the load toward the transmitter must be
JB>dissipated as power in the transmitter is naive.
It is naive not to believe Kirchhoff's laws. Move the RL+XL back
to the transmitter and you will find that for a sine wave to exist
on the capacitor, it must be charged and discharged through the
same impedance. Part of that impedance is the internal impedance
of the transmitter, whatever it is. If the transmitter is not
100% efficient, part of the impedance will be resistive and energy
will move from the capacitor back into the transmitter anytime the
instantaneous voltage on the capacitor is of an absolute magnitude
greater than the generator voltage. If you disagree with that,
you disagree with Kirchhoff. If the capacitor is at the end of a
transmission line, the energy that re-enters the transmitter is
contained in the capacitor image in the reflected waves and IT
DOES NOT CHANGE THE _AVERAGE_ POWER DISSIPATION! What we are
arguing about is whether some energy leaves the transmitter and
then returns to the transmitter to be dissipated. I and some RF
designers say it does and I think you and a number of others say
it doesn't.
I have never disagreed with your average power calculations so I
don't know why you keep offering them as proof of anything. The
question is: does the capacitive load on a transmitter dump energy
back into the transmitter. If it does, then there is no magic
barrier through which reflected energy cannot pass. If it
doesn't, please draw the capacitor waveform for me under those
conditions where the capacitive charge does not discharge back into
the transmitter.
Also, in a system where 1/2 of the forward wave is reflected at
the load, the most likely probability is that 1/2 of the newly
generated energy is reflected along with 1/2 of the re-reflected
energy. I wish you guys would quit saying that all the reflected
energy stays in the transmission line. It would take some magic
for the quanta of reflected energy to somehow stick together and
decide to be reflected each time it encountered the load. I know
a lot of physicists who would be interested in obtaining whatever
glue you are using to make that happen.
Also, anyone who thinks reflections cannot occur in 3 inches of
coax has never designed a high-speed computer bus. Why do you
think very short computer buses still require terminators?
Standing waves may need somewhere to stand but reflections don't.
Please enlighten me as to the exact length coax has to be for
reflections to suddenly start occuring.
73, Cecil, KG7BK, Old, Old Timer's Club (40+years a ham)