RadioBanter - The power explanation

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Owen Duffy

February 28th 07 08:35 PM

The power explanation

Breaking out of the previous thread to explore the "power explanation" in
a steady state situation:

The scenario for discussion is a transmitter connected to a half wave of
600 ohm lossless transmission line connected to an antenna with a
feedpoint impedance of 70+j0.

The transmitter is rated for 100W output, 100W is developed in the 70 ohm
load, the VSWR on the transmission line is 8.6, the "forward power"
(meaning Vf^2/Zo) on the transmission line is 267W, the "reflected
power" (meaning Vr^2/Zo) on the transmission line is 167W, the DC input
power to the transmitter is 200W.

The questions a

Is there any internal inconsistency in the scenario characterisation, if
so, identify / explain?

What is the heat dissipated in the transmitter (and why)?

What part of the "reflected power" of 167W is dissipated in the
transmitter (and why)?

Owen

Dave	February 28th 07 11:01 PM

The power explanation

"Owen Duffy" wrote in message
...
Breaking out of the previous thread to explore the "power explanation" in
a steady state situation:

The scenario for discussion is a transmitter connected to a half wave of
600 ohm lossless transmission line connected to an antenna with a
feedpoint impedance of 70+j0.

The transmitter is rated for 100W output, 100W is developed in the 70 ohm
load, the VSWR on the transmission line is 8.6, the "forward power"
(meaning Vf^2/Zo) on the transmission line is 267W, the "reflected
power" (meaning Vr^2/Zo) on the transmission line is 167W, the DC input
power to the transmitter is 200W.

The questions a

Is there any internal inconsistency in the scenario characterisation, if
so, identify / explain?

i think the best answer is... it depends.
it depends on what the radio looks like to the reflected power.
for instance, if the tx output network included a circulator and dummy load,
then the reflected power would be lost and there would be no way to get 100w
into the load.
however, because of the conditions given it would appear you have assumed a
perfect re-reflection of the reflected power at the tx end of the line so
everything balances perfectly.

What is the heat dissipated in the transmitter (and why)?

200w dc in, 100w rf into load, lossless line, sounds like 100w dissipated in
the tx to me.

What part of the "reflected power" of 167W is dissipated in the
transmitter (and why)?

none. it is all re-reflected back toward the load.

Cecil Moore

March 1st 07 12:52 AM

The power explanation

Cecil Moore

March 1st 07 01:26 AM

The power explanation

Owen Duffy wrote:
Breaking out of the previous thread to explore the "power explanation" in
a steady state situation:

The scenario for discussion is a transmitter connected to a half wave of
600 ohm lossless transmission line connected to an antenna with a
feedpoint impedance of 70+j0.

The transmitter is rated for 100W output, 100W is developed in the 70 ohm
load, the VSWR on the transmission line is 8.6, the "forward power"
(meaning Vf^2/Zo) on the transmission line is 267W, the "reflected
power" (meaning Vr^2/Zo) on the transmission line is 167W, the DC input
power to the transmitter is 200W.

What part of the "reflected power" of 167W is dissipated in the
transmitter (and why)?

There is a conceptual exercise that I use on such problems.

Source---1WL 70 ohm line---+---1/2WL 600 ohm line---70 ohm load
Pfor1=100w-- Pfor2=267w-- 100w
--Pref1=0 --Pref2=167w

Now the picture becomes perfectly clear. The insertion of 1WL
70 ohm line doesn't affect the steady-state condition. A 70
ohm Z0-match is achieved at '+' so none of the 167w is dissipated
in the transmitter. What happens at point '+' toward the source
is total destructive interference, i.e. no reflected energy. What
happens at point '+' toward the load is total constructive
interference. Reference "Optics", by Hecht, 4th edition, page 388.

rho at '+' is 0.791. That makes rho^2 = 0.626.

The amount of Pfor1 that makes it through the impedance discontinuity
at '+' is Pfor1(1-rho^2) = 100w(0.374) = 37.4 watts - call that P1

The amount of Pref2 that is reflected from the impedance discontinuity
at '+' is Pref2(rho^2) = 167w(0.626) = 104.5 watts - call that P2

The power equation says that Pfor2 = P1 + P2 + 2*SQRT(P1*P2)cos(A)

Since it is a Z0-match point A=0 so cos(A)=1. Therefore,

Pfor2 = 37.4w + 104.5w + 2*SQRT(37.4w*104.5w) = 267 watts

What do you know? The power equation works perfectly. Pfor2 didn't
even need to be given.
--
73, Cecil http://www.w5dxp.com

Richard Clark

March 1st 07 01:51 AM

The power explanation

On Wed, 28 Feb 2007 20:35:27 GMT, Owen Duffy wrote:

Breaking out of the previous thread to explore the "power explanation" in
a steady state situation:

The scenario for discussion is a transmitter connected to a half wave of
600 ohm lossless transmission line connected to an antenna with a
feedpoint impedance of 70+j0.

The transmitter is rated for 100W output, 100W is developed in the 70 ohm
load, the VSWR on the transmission line is 8.6, the "forward power"
(meaning Vf^2/Zo) on the transmission line is 267W, the "reflected
power" (meaning Vr^2/Zo) on the transmission line is 167W, the DC input
power to the transmitter is 200W.

The questions a

Is there any internal inconsistency in the scenario characterisation, if
so, identify / explain?

Hi Owen,

After a fashion, yes. You don't have enough information. The lesson
of past correspondence reveals the source resistance is not a benign
element that can be discarded in the examination of these problems.

Explain? I thought I had by example, so to continue by simple
analysis we have to ask: "What is the source R?"

If it is 70 Ohms, it sees a 70 Ohm load and an Impedance Match. In
effect there is no other way for a transmitter that is capable of 100W
being able to source 100W to a load, is there?

If it is 50 Ohms, it immediately sees a mismatch at the connector -
you did do your lumped equivalent of the transmission line system,
didn't you? Hence a transmitter limited to supplying 100W could not
supply 100W to a mismatched load.

If it is 600 Ohms, we again see a massive mismatch, further discussion
is unnecessary as being repetitive.

What is the heat dissipated in the transmitter (and why)?

I will skip that exercise as obvious pending discussion below.

What part of the "reflected power" of 167W is dissipated in the
transmitter (and why)?

Presuming the 70 Ohm source as it is the only one capable of
supporting the 100W application of power to the 70 Ohm load (by your
definition and limitations) we shall proceed to the determination of
those pesky powers with absolute finality.

The load voltage is 84 volts (to sufficient accuracy). The voltage at
the source port is 84 volts (to sufficient accuracy). The line has
been defined to be lossless. The same potential at the same angle is
returned to the source through 360 degrees of travel and no loss
(analysis allows us to treat the load as a generator). That signal
applied to the transmitter finds there is no potential difference, and
thus there is no current flow. This variously describes either a
short or an open; and either way it represents an infinite mismatch -
ALL the reflected power is re-reflected to the load.

Voila, another statistical curiosity!

Tuners do this more generally and we use tuners to re-reflect the
power, and to not suffer the transmitter the additional heat burden of
reflected power arriving at the finals.

If we repeat this exercise with the other source resistances, we stand
to appreciate a paradox in the initial condition of:
the DC input power to the transmitter is 200W.
forcing us to resolve the additional heat burden (roughly 50W)
exceeding the reflected power's ability to contribute that much
(especially when all the reflected power is re-reflected); or worse,
the source going into clipping. A non-linear source is not a pleasant
thing to behold.

73's
Richard Clark, KB7QHC

Richard Clark

March 1st 07 08:06 PM

The power explanation

On Wed, 28 Feb 2007 17:51:30 -0800, Richard Clark
wrote:

The load voltage is 84 volts (to sufficient accuracy). The voltage at
the source port is 84 volts (to sufficient accuracy). The line has
been defined to be lossless. The same potential at the same angle is
returned to the source through 360 degrees of travel and no loss
(analysis allows us to treat the load as a generator). That signal
applied to the transmitter finds there is no potential difference, and
thus there is no current flow. This variously describes either a
short or an open; and either way it represents an infinite mismatch -
ALL the reflected power is re-reflected to the load.

Hi All,

Well, in lieu of Owen providing his own solution, much less
fulminating against the BS provided by yours truly in the quote above
(how seductive is it to the rest of you?). The problem here arises
out of a mixture of models, but this topology is not my choice and it
is inclined to invite such problems.

We can look at the details provided to find a 600 Ohm line terminated
at both ends by 70 Ohm resistors (again, I am forced to the
presumption of the source resistance, but given the constraints it is
the only value available). The energy within the resonant line is
ringing (one reason why a mismatched line it is called resonant is
because of circulating energy - the allusion to ringing).

If we deposit ourselves in the line, within the fields of energy, it
would be like inhabiting a hall with two partial mirrors, one at each
end. To other descriptions, these partial mirrors are leaky
interfaces. Some of the energy is passed through, some is reflected.
The "some" can be rendered into an absolute through knowledge of the
degree of mismatch - it is already provided within the original post
as being 8.6:1 (to a sufficient accuracy). In this regard, the
resonant line has a poor Q, however "poor" is in the eye of the
beholder because a flat line has the poorest of Qs at 1 (or worse).

The topology, and the question, invited me to respond with both the
lumped equivalent (this topology is in the classic expectation of
repetition of the load, cast back identically through a lossless line
to its input) and a Thevenin analysis (200W of power applied to a
source that could only supply 100W). The topology here has the
classic expectation of 50% efficiency that only a 70 Ohm source can
provide. Given that there is special constraint of a very strongly
specified line length, that too invites an analysis (which then
corrupts the lumped equivalent analysis if they are not separated).

So, we have three avenues commingled here in pursuit of an answer as
to explaining the power. Let us return to that hall of mirrors and
traverse its length (and unfortunately it must violate another
expectation of steady state, but this topology, much less expectations
galore, was not my choice - but I wander into any discussion with so
many invitations).

The first pass of energy sees the partial mirror of the load
interface. Some energy passes, some reflects. That which passes
cannot supply the expected potential of 84 volts (to a sufficient
accuracy); that remains to be justified by further analysis. The
reflected energy approaches an IDENTICAL mirror at the source end. The
reflected energy is coherent to the source (this is enforced by the
topology) behind that mirror. However, and revealing my deception
above, the reflected energy is not the same amplitude. There are two
sources of energy now embracing the source resistance of 70 Ohms.
There is a potential difference across that source resistance that is
different from that of what would be a presumed steady state 84 volts
(to a sufficient accuracy). Others can provide that computation if
they wish as the magnitude is one of degree (pun intended).

We have COOLED the source resistance! But we have not provided the
full load (anyone want to bet we cooled it by an equal degree? - pun
intended again). And the 200W application must be accounted for
somewhere (this inclusion is, of course, a sham of moving between
topologies and expectations that are so inviting).

And so on, and so on, ad infinitum through the successive waves of
reflection and dissipation for this statistical curiosity for me to
once again conclude:
ALL the reflected power is re-reflected to the load.
which is to say, energy (through the corruption of terms that has been
generally allowed).

73's
Richard Clark, KB7QHC

Owen Duffy

March 1st 07 08:45 PM

The power explanation

I did not intend this to be difficult, or incomplete. I think it is
solveable without recourse to knowledge or speculation about source
impedance, or changing the problem by inserting additional lines, or
injecting photons, or whatever.

My answers are inline:

Owen Duffy wrote in
:

Breaking out of the previous thread to explore the "power explanation"
in a steady state situation:

The scenario for discussion is a transmitter connected to a half wave
of 600 ohm lossless transmission line connected to an antenna with a
feedpoint impedance of 70+j0.

The transmitter is rated for 100W output, 100W is developed in the 70
ohm load, the VSWR on the transmission line is 8.6, the "forward
power" (meaning Vf^2/Zo) on the transmission line is 267W, the
"reflected power" (meaning Vr^2/Zo) on the transmission line is 167W,
the DC input power to the transmitter is 200W.

The questions a

Is there any internal inconsistency in the scenario characterisation,
if so, identify / explain?

No, I cannot see any.

What is the heat dissipated in the transmitter (and why)?

Approximately 100W. It is reasonable to assume that all DC power into the
transmitter is converted to RF power output and heat. The heat then is
approximately DC power input minus the RF output power. Since the power
in the load is stated to be 100W, and the line is stated to be lossless,
the power out of the transmitter must be 100W, and given the stated DC
input power of 200W, the transmitter heat dissipation is approximately
100W.

What part of the "reflected power" of 167W is dissipated in the
transmitter (and why)?

None. The notion of the "reflected power" is part of describing the field
setup / energy storage / energy flow on the transmission line. It follows
from trying to reconcile the restriction the Vf/If=Zo, and Vr/Ir=Zo with
the V/I ratio of the circuit attached to the transmission line.

The problem defined the standing wave effects on the line. The
transmitter sees a load of 70+j0 due to the antenna and half wave
lossless line, and must be delivering 100W to that apparent load so as to
achieve the stated antenna load power.

Any notion that reflected power *must* flow back to the source and is
dissipated as heat will not lead to the correct solution of this problem.

As an exercise, think of a generator that has a Thevenin equivalent of
some voltage V and a series impedance of R+j0, connected to a half wave
of lossless transmission line where Zo=R. To give a numerical example,
lets make V=100 and R=50, so Vr=50 and "reflected power"=50. How much of
the "reflected power" is dissipated in the generator. In this case, the
generator dissipates less heat than were it terminated in 50 ohms.

When I said that the reflected power explanation is not a good
explanation, I did not say it was necessarily invalid, but it is my
opinion that it provides an incomplete explanation that folk with limited
knowledge and supported by ham radio folklore like to use to explain
things that are not necessarily explained by that explanation (if you are
still with me), for example why a transmitter (sometimes or necessarily)
gets hotter when the antenna VSWR is high.

Owen

Richard Clark

March 1st 07 11:00 PM

The power explanation

On Thu, 01 Mar 2007 20:45:44 GMT, Owen Duffy wrote:

Any notion that reflected power *must* flow back to the source and is
dissipated as heat will not lead to the correct solution of this problem.

Hi Owen,

Why must it flow back, when the generator itself presents a huge
mismatch to the line? The laws of reflection work at both ends of the
line and to force the presumption that initial reflected power would
flow through this discontinuity without reflection is a strained
expectation. However, I see by the alteration that follows, that we
now have a fully matching source:

As an exercise, think of a generator that has a Thevenin equivalent of
some voltage V and a series impedance of R+j0, connected to a half wave
of lossless transmission line where Zo=R. To give a numerical example,
lets make V=100 and R=50, so Vr=50 and "reflected power"=50. How much of
the "reflected power" is dissipated in the generator. In this case, the
generator dissipates less heat than were it terminated in 50 ohms.

"reflected power"= 50 what? Watts? So be it. Your intent may not to
be incomplete nor difficult, but what is wrong with specifying the
load? Do we get a line length?

Let's see, I start with your termination of 50 Ohms to find 1 Amp
flowing through both source and load resistances. 50 Watts each. Now
I move to a load that will reflect 50 Watts - the former load's
complete contribution. I presume by this you mean either an open or a
short at the load end. As you left it unspecified, and it being my
choice, I can show that the source resistor will get hotter than a $5
pistol. I will, however, be complete and discuss how the source
resistance dissipates the reverse power for both short and open.

To cut to the chase, and not knowing the length of the line, we have a
spectrum of choices, but we may as well force the situation with
another resonant line any number of lossless halfwaves.

It is obvious that the two powers combine constructively or
destructively. One is with the energy reflecting a zero voltage
component and all the current (or now twice the current through the
source resistance), the other is with the energy reflecting an
identical voltage component, and none of the current flows. As I've
offered, the spectrum of possible answers yields to the same analysis,
this was simpler.

Both times the reflected power had to be absorbed by the source
resistance. It does not mean that contribution has to always resolve
to more heat (haven't I already demonstrated this in this thread?).

I will at this point re quote Chipman to roughly this scenario (being
more general, he didn't specify the reflection).

"At the signal source end of the line ... none of the power
reflected by the terminal load impedance is re-reflected on
returning to the input end of the line."
The ellipsis reveals that the source Z matches the line Z.

73's
Richard Clark, KB7QHC

Owen Duffy

March 2nd 07 12:31 AM

The power explanation

Richard Clark wrote in
:

On Thu, 01 Mar 2007 20:45:44 GMT, Owen Duffy wrote:

Any notion that reflected power *must* flow back to the source and is
dissipated as heat will not lead to the correct solution of this
problem.

Hi Owen,

Why must it flow back, when the generator itself presents a huge
mismatch to the line? The laws of reflection work at both ends of the
line and to force the presumption that initial reflected power would
flow through this discontinuity without reflection is a strained
expectation. However, I see by the alteration that follows, that we
now have a fully matching source:

As an exercise, think of a generator that has a Thevenin equivalent of
some voltage V and a series impedance of R+j0, connected to a half wave
of lossless transmission line where Zo=R. To give a numerical example,
lets make V=100 and R=50, so Vr=50 and "reflected power"=50. How much
of
the "reflected power" is dissipated in the generator. In this case, the
generator dissipates less heat than were it terminated in 50 ohms.

"reflected power"= 50 what? Watts? So be it. Your intent may not to
be incomplete nor difficult, but what is wrong with specifying the
load? Do we get a line length?

The line is stated as a half wave of lossless line, why do you need to
know more about its length?

The load you are looking for isn't there, the line ends in an open
circuit. Sorry for the confusion, I should have been explicit that there
was no load, just o/c.

The misunderstandings will frustrate your analysis, so try again.

Owen

Jim - NN7K

March 2nd 07 01:25 AM

The power explanation

And, in fact, the "Reflected power" is
"Re-reflected" from the Source, Back to
the Load, if memory serves (minus loss's
accumulated from the first "Reflection"
of power, if memory serves! Jim NN7K

Owen Duffy wrote:
Richard Clark wrote in
:

On Thu, 01 Mar 2007 20:45:44 GMT, Owen Duffy wrote:

Any notion that reflected power *must* flow back to the source and is
dissipated as heat will not lead to the correct solution of this
problem.

Owen Duffy

March 2nd 07 03:42 AM

The power explanation

Jim - NN7K wrote in news:C2LFh.5953$re4.1319
@newssvr12.news.prodigy.net:

And, in fact, the "Reflected power" is
"Re-reflected" from the Source, Back to
the Load, if memory serves (minus loss's
accumulated from the first "Reflection"
of power, if memory serves! Jim NN7K

All this consideration of re-reflected, re-re-reflected, re-re-re-re-
reflected energy is something that is appropriate to analysing how the
steady state is established, and yes I know it takes an infinite time,
but it establishes subtantially quite quickly. The transient converges to
the steady state.

To find the steady state solution, take a short cut, bypass the transient
analysis and jump straight to the converged situation. In the steady
state... The complex ratio of Vf to Vr at the load end is entirely
determined by the constraints of a transmission line of Zo and the
passive load. The complex ratio of Vf to Vr at the line input can be
determined from the load end conditions by applying the line transmission
formulas with the complex propagation constant. Knowing Vf and Vr at the
input of the line and Zo, the complex V/I ratio (the equivalent load)
that the loaded line input will enforce can be calculated. The power
delivered by the source can be then found by finding the voltage or
current it will supply into the equivalent load.

This might seem long winded, but it is a sight easier than solving some
thousands of iterations of reflection, re-reflection and so on.

Owen

Owen Duffy

March 2nd 07 04:33 AM

The power explanation

Owen Duffy wrote in
:

As an exercise, think of a generator that has a Thevenin equivalent of
some voltage V and a series impedance of R+j0, connected to a half
wave of lossless transmission line where Zo=R. To give a numerical
example, lets make V=100 and R=50, so Vr=50 and "reflected power"=50.
How much of the "reflected power" is dissipated in the generator. In
this case, the generator dissipates less heat than were it terminated
in 50 ohms.

Ok, the solution:

Lets examine the matched load scenario for a start, the 100V generator
with 50 ohms internal resistance and a matched 50 ohm load. The current
is 100/(50+50) or 1A, the power in the load is 1^2*50 or 50W, the power
dissipated in the source is 1^2*50 or 50W.

No lets look at the scenario with the o/c half wave lossless line
attached to the generator. In the steady state, current from the
generator is zero, dissipation in the generator is 0^2*50 or 0, voltage
at the generator terminals and at the o/c (load end) of the line is 100V.
At the o/c load end, the complex reflection coefficient is 1, so Vf=50V,
Vr=50V and "reflected power"=50^2/50 or 50W.

But, wait a minute, there is 50W of "reflected power" on the line, the
line is matched to the source, and there is zero dissipation in the
source, less than when it has a matched load.

Don't take anything above to mean that I represent that a simple linear
model is a good representation of a transmitter PA.

This simple example that shows that existence of "reflected power" on a
transmission line does not necessarily result in some or all of the
"reflected power" being dissipated in the generator.

I will leave it to Cecil to take to confuse this simple example with some
photon based complication.

Owen

Richard Clark

March 2nd 07 05:52 AM

The power explanation

On Fri, 02 Mar 2007 00:31:55 GMT, Owen Duffy wrote:

The misunderstandings will frustrate your analysis, so try again.

Hi Owen,

Is it noteworthy that I found a blistering hot resistor for exactly
the conditions you set forth? Was I balanced in my reply to note the
alternative did the opposite? Even with my gaff of missing the
halfwave description, was the discussion incomplete in noting there
being a spectrum of responses?

Given I come to exactly the same analysis, same solution, same
conclusion (with more explanation, perhaps in that I do demonstrate
the reflected energy is absorbed/dissipated/what-have-you in the
source resistor) as your posting timestamped 2033 hours my time - what
exactly do I need to try again? Did the intervening 4 hours between
this post and the second one of yours find some catharsis?

Could you give me a dope slap instead of a hint about this frustration
I seem to be suffering? Like, should I take an aspirin or a syringe
of morphine?

73's
Richard Clark, KB7QHC

Owen Duffy

March 2nd 07 06:16 AM

The power explanation

Richard Clark wrote in
:

On Fri, 02 Mar 2007 00:31:55 GMT, Owen Duffy wrote:

The misunderstandings will frustrate your analysis, so try again.

Hi Owen,

Is it noteworthy that I found a blistering hot resistor for exactly
the conditions you set forth? Was I balanced in my reply to note the
alternative did the opposite? Even with my gaff of missing the
halfwave description, was the discussion incomplete in noting there
being a spectrum of responses?

Richard, I must admit I didn't read the rest of your post when you stated
that you didn't know the line length and requested that info.

I have now read it.

I make the comment that just because the situation exists where the Volts
and Current from the source are the same as the Volts and Current into
the line (they have to be don't they), that does not imply matching in
the Jacobi Maximum Power Transfer Theoram sense.

Owen

Richard Clark

March 2nd 07 06:50 AM

The power explanation

On Fri, 02 Mar 2007 06:16:15 GMT, Owen Duffy wrote:

that does not imply matching in
the Jacobi Maximum Power Transfer Theoram sense.

-um, OK-

I find negative propositions a bit oblique, what DOES it imply?

73's
Richard Clark, KB7QHC

Owen Duffy

March 2nd 07 07:04 AM

The power explanation

Richard Clark wrote in
:

On Fri, 02 Mar 2007 06:16:15 GMT, Owen Duffy wrote:

that does not imply matching in
the Jacobi Maximum Power Transfer Theoram sense.

-um, OK-

I find negative propositions a bit oblique, what DOES it imply?

I was responding to your words:

"I will at this point re quote Chipman to roughly this scenario (being
more general, he didn't specify the reflection).

"At the signal source end of the line ... none of the power
reflected by the terminal load impedance is re-reflected on
returning to the input end of the line."
The ellipsis reveals that the source Z matches the line Z."

I disagree that the conditions that exist in the steady state at the
source end of the line imply in the general case, matching in the Jacobi
Maximum Power Transfer Theoram sense.

Owen

Cecil Moore

March 2nd 07 01:21 PM

The power explanation

Owen Duffy wrote:
To find the steady state solution, take a short cut, bypass the transient
analysis and jump straight to the converged situation.

That certainly works but in the process, many people
have ignored and/or forgotten the total energy content
of a transmission line with reflections is greater
than the net energy being transferred to the load.

A one microsecond long transmission line transferring
100 watts to a matched load (Pfor=100w, Pref=0w)
contains 100 microjoules of energy during steady-state.

A one microsecond long transmission line transferring
100 watts to a mismatched load (Pfor=200w, Pref=100w)
contains 300 microjoules of energy during steady-state.

The transmission line always contains exactly the right
amount of energy to support the forward and reflected
waves. The most logical conclusion is that forward and
reflected waves continue to exist during steady-state
and those EM waves cannot stand still.

Forward and reflected waves during steady-state are
the building blocks of the standing wave which couldn't
exist without them. Forward and reflected waves contain
energy levels which can be easily calculated. The extra
energy, more than the matched flat case, is what causes
the extra losses due to SWR.
--
73, Cecil http://www.w5dxp.com

Denny

March 2nd 07 01:27 PM

The power explanation

Ya know, all the bits sent on to the great bit bucket in the sky
during this 'discussion' could have been avoided by simply reading
Walt Maxwell's "Reflections" until you understand it...
denny / k8do

Cecil Moore

March 2nd 07 01:47 PM

The power explanation

Owen Duffy wrote:
This simple example that shows that existence of "reflected power" on a
transmission line does not necessarily result in some or all of the
"reflected power" being dissipated in the generator.

I will leave it to Cecil to take to confuse this simple example with some
photon based complication.

No photons necessary, Owen. You are using a Thevenin
equivalent source. Paraphrasing Ramo et.al of "Fields
and Waves ..." fame: No valid conclusions can be
automatically drawn from the calculation of power
dissipation inside a Thevenin equivalent source.
(Sorry, I don't have the book with me for the exact
quote.)

For a lot of real-world sources, double the voltage
with zero current output would be very bad news.
--
73, Cecil http://www.w5dxp.com

Cecil Moore

March 2nd 07 02:26 PM

The power explanation

Cecil Moore wrote:
Owen Duffy wrote:
This simple example that shows that existence of "reflected power" on
a transmission line does not necessarily result in some or all of the
"reflected power" being dissipated in the generator.

I will leave it to Cecil to take to confuse this simple example with
some photon based complication.

No photons necessary, Owen. You are using a Thevenin
equivalent source.

What is the dissipation in the generator using a Norton
source?
--
73, Cecil http://www.w5dxp.com

Richard Harrison

March 2nd 07 04:25 PM

The power explanation

Cecil, W5DXP wroote:
"What is the dissipation in the generator using a Norton source?"

According to page 76 of Terman`s 1955 opus:
"Alternatively, a load impedance may be matched to a source of power in
such a way as to make the power delivered to the load a maximum (the
available power of the power source). This is accomplished by making the
load impedance the conjugate of the generator impedance as defined by
Thevenin`s theorem."

On page 75 Terman labels a Norton diagram: "Equivalent Arrangement".

Equivalence means the open-circuit voltage and short-circuit current are
the same whichever diagram represents the power source.

The power dissipated in the source under matched conditions depends not
on the diagramatic representation, but upon how much of the internal
resistance of the source behaves as a resistor does, and how much is
"dissipationless resistance". It`s real, but makes no heat. If it were
fictional, final amplifiers would be limited to 50% efficiency. We all
know many R-F amplifiers have discontinuous input power
which allows efficiencies much in excess of 50%.

Best regards, Richard Harrison. KB5WZI

Richard Clark

March 2nd 07 05:51 PM

The power explanation

On Fri, 02 Mar 2007 07:04:32 GMT, Owen Duffy wrote:

I find negative propositions a bit oblique, what DOES it imply?

I was responding to your words:

"I will at this point re quote Chipman to roughly this scenario (being
more general, he didn't specify the reflection).

"At the signal source end of the line ... none of the power
reflected by the terminal load impedance is re-reflected on
returning to the input end of the line."
The ellipsis reveals that the source Z matches the line Z."

I disagree that the conditions that exist in the steady state at the
source end of the line imply in the general case, matching in the Jacobi
Maximum Power Transfer Theoram sense.

Hi Owen,

This is still oblique. Are you speaking of the conditions you created
(for either instance in this thread)? Or that Chipman created to
illustrate his figure? Or through a loose reference to me? Or
perhaps it is compounded in the missing preceding comma for a
parenthetical "in the general case?"

If the last serves, and to the point of obliquity, when did the JMPTT
show up and who are you disagreeing with?

73's
Richard Clark, KB7QHC

Richard Clark

March 2nd 07 05:54 PM

The power explanation

On 2 Mar 2007 05:27:13 -0800, "Denny" wrote:

Ya know, all the bits sent on to the great bit bucket in the sky
during this 'discussion' could have been avoided by simply reading
Walt Maxwell's "Reflections" until you understand it...
denny / k8do

Hi Denny,

I am to presume you are adding bits to the bit bucket then? Your own
answer to any of Owen's examples would have trumped hoary advice.

73's
Richard Clark, KB7QHC

Owen Duffy

March 2nd 07 10:12 PM

The power explanation

Cecil Moore wrote in news:fuWFh.3131$M65.1761
@newssvr21.news.prodigy.net:

Cecil Moore wrote:
Owen Duffy wrote:
This simple example that shows that existence of "reflected power" on
a transmission line does not necessarily result in some or all of the
"reflected power" being dissipated in the generator.

I will leave it to Cecil to take to confuse this simple example with
some photon based complication.

No photons necessary, Owen. You are using a Thevenin
equivalent source.

What is the dissipation in the generator using a Norton
source?

The existence of cases that show that dissipation in the source does not
necessarily increase due to VSWR on the transmission line does not
support the assertion that "VSWR causes reflected power that is
dissipated in the source". One sound case is enough to disprove the
generality.

Sure, transforming the source to a Norton equivalent would produce an
answer, and in this case a different answer for what happens inside the
generator. That Norton equivalent source with half wave s/c line will
also produce zero dissipation in the source.

Cecil, it appears your motive is to create confusion to divert attention
from the cases that are inconsistent with the assertion that "VSWR causes
reflected power that is dissipated in the source".

I have no difficulty with the statement "a transmitter is usually
specified to work over a limited range of load impedances (often
specified as a maximum VSWR at the transmitter terminals), the user
should expect it works properly over that range and should understand
that operation outside of that range may expose it to voltages or
currents (consequent heat), that may cause permanent damage". This advice
can be given to a six hour ham without telling them any lies, but
imparting the knowledge that they need to operate safely.

The explanation expressed / supported by some here that "we use ATUs to
cause total re-reflection of power "reflected" from the antenna so
protecting the PA" is a nonsense explanation of how the ATU protects the
PA from the effects of a poor load.

Owen

Cecil Moore

March 3rd 07 12:08 AM

The power explanation

Owen Duffy wrote:
Cecil, it appears your motive is to create confusion to divert attention
from the cases that are inconsistent with the assertion that "VSWR causes
reflected power that is dissipated in the source".

Appearances can be deceiving. My motive is to uncover facts
and I agree with virtually everything you have said.

My personal opinion is that from 0% to 100% of reflected
power can be dissipated in the source depending upon the
relative phase of the incident reflected wave and the
configuration of the source. It can be argued that if the
source sees an infinite or zero impedance, then all of the
source power is reflected at the source output. This, of
course, would be a same-cycle reflection, something that
also occurs at the mismatched load.

By convention, any power same-cycle reflected at the source
output was never generated to start with - one of the original
copouts.

Since the great majority of amateur transmitters are
looking into a Z0-match resulting in total destructive
interference in the direction of the source, IMO, this
subject is pretty much moot. That's why I poke fun at it.

All one has to do to calculate the reflected power
dissipated in the source is to understand the constructive
and destructive interference occurring at the source
output terminal. This is easier said than done.
--
73, Cecil http://www.w5dxp.com

Ian White GM3SEK

March 3rd 07 12:10 AM

The power explanation

Owen Duffy wrote:
Richard Clark wrote in
:

On Fri, 02 Mar 2007 06:16:15 GMT, Owen Duffy wrote:

that does not imply matching in
the Jacobi Maximum Power Transfer Theoram sense.

-um, OK-

I find negative propositions a bit oblique, what DOES it imply?

I was responding to your words:

"I will at this point re quote Chipman to roughly this scenario (being
more general, he didn't specify the reflection).

"At the signal source end of the line ... none of the power
reflected by the terminal load impedance is re-reflected on
returning to the input end of the line."
The ellipsis reveals that the source Z matches the line Z."

The confusion arises out of Richard's misleading quotation, which is out
of context and includes an inaccurate final sentence ("The ellipsis
reveals...") added by Richard himself.

The context in Chipman's book is specifically about the scale of
"reflection loss" as found alongside a Smith chart. Reflection loss is
the power delivered into a mismatched (reflecting) load impedance,
relative to the power that would have been delivered into a matched
(non-reflecting) load.

Chipman points out that "The concept is directly applicable only to
transmission-line circuits of the form shown in Fig 9-26, in which the
source impedance is equal to the characteristic impedance of the line."

In more detailed analysis, he explains why the definition of reflection
loss only holds good if there is no further re-reflection at the source,
so the "reflection loss" scale on the Smith chart can only be used in
cases where the source impedance Zs is equal to Zo of the line.

Richard's comment that "The ellipsis reveals that the source Z matches
the line Z" is misleading, for that isn't at all what Chipman was
saying. Chipman makes it perfectly clear that this section is dealing
with a special case, which only applies if Zs has deliberately been made
equal to Zo. Most of the rest of the book deals with the more general
case where Zs is NOT necessarily equal to Zo.

However, that special case often does apply to signal generators and
similar test equipment. If the RF output comes through an attenuator
that has a design impedance of Zo, and if the attenuation is large
enough, this creates a good approximation to a source of impedance Zo.
Then the reflection loss concept becomes valid.

--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Richard Clark

March 3rd 07 04:37 AM

The power explanation

On Sat, 3 Mar 2007 00:10:21 +0000, Ian White GM3SEK
wrote:

The confusion arises out of Richard's misleading quotation,

blah blah blah blah blah

Hi Ian,

I have restrained the afterburners until 1200 GMT (seeing as you are
undoubtedly in bed) so that you can fully read the entire side thread
and discover the complete botch you've made in an attempt to weakly
persuade the group that a transmitter could never absorb the energy of
a reflected wave.

That was your point for this chain of accusation, wasn't it, or were
the accusations a solitary pleasure? (I hope you opt for the first,
or other explanation.) I don't expect any apologies so that I can
enjoy a dish that is best eaten cold (re-heated actually). ;-)

73's
Richard Clark, KB7QHC

Richard Clark

March 3rd 07 09:04 AM

The power explanation

On Fri, 02 Mar 2007 20:37:22 -0800, Richard Clark
wrote:

until 1200 GMT

"Don't talk to me about a man's being able to talk sense;
everyone can talk sense. Can he talk nonsense?"

walt	March 7th 07 02:34 PM

The power explanation

On Mar 3, 4:37 am, Richard Clark wrote:
On Sat, 3 Mar 2007 00:10:21 +0000, Ian White GM3SEK

wrote:
The confusion arises out of Richard's misleading quotation,

blah blah blah blah blah

Hi Ian,

I have restrained the afterburners until 1200 GMT (seeing as you are
undoubtedly in bed) so that you can fully read the entire side thread
and discover the complete botch you've made in an attempt to weakly
persuade the group that a transmitter could never absorb the energy of
a reflected wave.

That was your point for this chain of accusation, wasn't it, or were
the accusations a solitary pleasure? (I hope you opt for the first,
or other explanation.) I don't expect any apologies so that I can
enjoy a dish that is best eaten cold (re-heated actually). ;-)

73's
Richard Clark, KB7QHC

I'm in a hotel in Jax, FL, unable to use my routine connections to
rraa, so I'm learning my way around in unknown territory here, and
hope to make a post that will appear.

I've been reading this thread with interest, but the discussions
appear to be only academically related. On the other hand, I've made
measurements that prove the sailent points of these academic
discussions. These measurements were made since those reported in
Reflections 2, and will appear in Reflections 3. However, they are
available on my web page at www.w2du.com. Go to 'Preview of Chapters
from Reflections 3' and click on Chapter 19A.

You may want to disregard the first portion of the chapter, which is
an epilog to Bruene's fiction concerning the conjugate match. The
pertinent portion here is that which reports in detail the step-by-
step procedure in measuring the output impedance of a Kenwood TS-830S
transceiver feeding a reactive-impedance load.

With a careful review of these steps I'm sure you'll find empirical
proof of the academics appearing in the previous posts.

Walt, W2DU

Richard Clark

March 7th 07 04:52 PM

The power explanation

Richard Harrison

March 7th 07 06:08 PM

The power explanation

Owen Duffy wrote:
"Breaking out of the previous thread to explore the "power explanation"
in a steady state situation:"

All OK as I see it.

Bird tells us that if you have significant standing waves, reflected
power is 10% or more of the forward power, and the ratio of reflected
power to forward power is then easily determined on the Bird Thruline
Wattmeter. Ratio of the reflected power to forward power is easily
converted to VSWR.

Bird supplies, charts, slide rules, and a formula for this conversion.

Bird confirms: "Power delivered to and dissipated in a load is given by:

Watts into load = Watts forward - Watts reflected."

Owen Duffy told us 100W is developed in 70 ohm load and the DC input
power of the transmitter is 200W.

Obviously 100W is dissipated in the transmitter and the efficiency is
50%.

Best regards, Richard Harrison, KB5WZI

Owen Duffy

March 7th 07 09:35 PM

The power explanation

"walt" wrote in news:1173278086.390893.310040@
30g2000cwc.googlegroups.com:

On Mar 3, 4:37 am, Richard Clark wrote:
On Sat, 3 Mar 2007 00:10:21 +0000, Ian White GM3SEK
....
I've been reading this thread with interest, but the discussions
appear to be only academically related. On the other hand, I've made
measurements that prove the sailent points of these academic
discussions. These measurements were made since those reported in
Reflections 2, and will appear in Reflections 3. However, they are
available on my web page at www.w2du.com. Go to 'Preview of Chapters
from Reflections 3' and click on Chapter 19A.

You may want to disregard the first portion of the chapter, which is
an epilog to Bruene's fiction concerning the conjugate match. The
pertinent portion here is that which reports in detail the step-by-
step procedure in measuring the output impedance of a Kenwood TS-830S
transceiver feeding a reactive-impedance load.

With a careful review of these steps I'm sure you'll find empirical
proof of the academics appearing in the previous posts.

Hi Walt,

I have read your document with interest, and will reread it a couple of
times yet.

The measurements are interesting, and on a first read, appear consistent
and in agreement with how things work as I understand it.

However, I don't believe any of your measurents actually reveal the
source impedance. You have shown the impact of the changed load on the
transmitter. You have demonstrated admirably the transformation of the
two external loads to the load seen by the plate(s), and you have shown
what the transmitter looks like from the antenna socket with a resistor
in place of the valves.

So, IMHO, the promise "The output source resistance of the amplifier in
this condition will later be shown to be 50 ohms" is not fulfilled.

It is my view that the statement "Because the amplifier was adjusted to
deliver the maximum available power of 100 watts prior to the resistance
measurement, resistance RLP looking into the plate (upstream from the
network terminals) is also approximately 1400 ohms" is not proven.

If we were to view the plate as a generator, and the pi network as a
lossless flexible variable impedance transformer, and you were to adjust
the pi network for maximum power transfer, that would imply that the
impedance loading the generator was the complex conjugate of its
equivalent series resistance... IF and ONLY IF the generator can be
accurately represented by an equivalent series circuit of a fixed voltage
generator and fixed equivalent series impedance.

The question is can the plate (or the whole transmitter for that matter)
be accurately replaced by an equivalent series circuit of a fixed voltage
generator and fixed equivalent series resistance (independent of load).

How would such a simple model deal with the case of a transmitter that at
maximum power output is close to voltage saturation (ie cannot develop
more output voltage) and close to current saturation (ie cannot develop
more output current)? These non-linear behaviours close to operating
point are not captured in a simple linear equivalent circuit.

Owen

Richard Clark

March 7th 07 10:06 PM

The power explanation

On Wed, 07 Mar 2007 21:35:14 GMT, Owen Duffy wrote:

The measurements are interesting, and on a first read, appear consistent
and in agreement with how things work as I understand it.

However, I don't believe any of your measurents actually reveal the
source impedance.

Hi Owen,

Seems like the band has struck up another waltz.

For what it is worth, Walt would like:
Hi Richard,

Without my home computer I'm relatively helpless with no address library
available. I'd like to contact both Denny and Owen, but their email addresses
appearing in the Google venue are shortened--a big help.

So would you please give me both Denny's and Owen's email addresses.

Walt

It appears neither Denny nor Owen reveal their email here, so the only
alternative is for this appeal.

73's
Richard Clark, KB7QHC

Richard Clark

March 7th 07 10:09 PM

The power explanation

On 7 Mar 2007 06:34:46 -0800, "walt" wrote:

I'm in a hotel in Jax, FL, unable to use my routine connections to
rraa, so I'm learning my way around in unknown territory here, and
hope to make a post that will appear.

I've been reading this thread with interest, but the discussions
appear to be only academically related. On the other hand, I've made
measurements that prove the sailent points of these academic
discussions. These measurements were made since those reported in
Reflections 2, and will appear in Reflections 3. However, they are
available on my web page at www.w2du.com. Go to 'Preview of Chapters
from Reflections 3' and click on Chapter 19A.

You may want to disregard the first portion of the chapter, which is
an epilog to Bruene's fiction concerning the conjugate match. The
pertinent portion here is that which reports in detail the step-by-
step procedure in measuring the output impedance of a Kenwood TS-830S
transceiver feeding a reactive-impedance load.

With a careful review of these steps I'm sure you'll find empirical
proof of the academics appearing in the previous posts.

Walt, W2DU

Another relay:

Hi All,

I have now been able to access this thread from my hotel room In Jax,
FL. I've read the posts with interest, but they appear to be only
discussions in academia. On the other hand, I have made measurements
that prove the salient points being discussed here.

My measurements include those made since those reported in Reflections
2, and are going to be available in Reflections 3. However, they are
presently available from my web page at www.w2du.com. Go to 'Preview
Chapters from Reflections 3' and select Chapter 19A. You may disregard
the first portion, which is intended as an epilog to Bruene's fiction
concerning the conjugate match. The pertinent portion here is a
detailed step-by-step description of the measurement procedure I
erformed on a Kenwood TS-830S transceiver with a reactive impedance
for its load.

A careful review of these steps will add empirical proof to the
academic enlightenment already evident in the previous posts.

Walt, W2DU

73's
Richard Clark, KB7QHC

walt	March 7th 07 10:26 PM

The power explanation

On Mar 7, 9:35 pm, Owen Duffy wrote:
"walt" wrote in news:1173278086.390893.310040@
30g2000cwc.googlegroups.com:

On Mar 3, 4:37 am, Richard Clark wrote:
On Sat, 3 Mar 2007 00:10:21 +0000, Ian White GM3SEK
...
I've been reading this thread with interest, but the discussions
appear to be only academically related. On the other hand, I've made
measurements that prove the sailent points of these academic
discussions. These measurements were made since those reported in
Reflections 2, and will appear in Reflections 3. However, they are
available on my web page atwww.w2du.com. Go to 'Preview of Chapters
from Reflections 3' and click on Chapter 19A.

You may want to disregard the first portion of the chapter, which is
an epilog to Bruene's fiction concerning the conjugate match. The
pertinent portion here is that which reports in detail the step-by-
step procedure in measuring the output impedance of a Kenwood TS-830S
transceiver feeding a reactive-impedance load.

With a careful review of these steps I'm sure you'll find empirical
proof of the academics appearing in the previous posts.

Hi Walt,

I have read your document with interest, and will reread it a couple of
times yet.

The measurements are interesting, and on a first read, appear consistent
and in agreement with how things work as I understand it.

However, I don't believe any of your measurents actually reveal the
source impedance. You have shown the impact of the changed load on the
transmitter. You have demonstrated admirably the transformation of the
two external loads to the load seen by the plate(s), and you have shown
what the transmitter looks like from the antenna socket with a resistor
in place of the valves.

So, IMHO, the promise "The output source resistance of the amplifier in
this condition will later be shown to be 50 ohms" is not fulfilled.

It is my view that the statement "Because the amplifier was adjusted to
deliver the maximum available power of 100 watts prior to the resistance
measurement, resistance RLP looking into the plate (upstream from the
network terminals) is also approximately 1400 ohms" is not proven.

If we were to view the plate as a generator, and the pi network as a
lossless flexible variable impedance transformer, and you were to adjust
the pi network for maximum power transfer, that would imply that the
impedance loading the generator was the complex conjugate of its
equivalent series resistance... IF and ONLY IF the generator can be
accurately represented by an equivalent series circuit of a fixed voltage
generator and fixed equivalent series impedance.

The question is can the plate (or the whole transmitter for that matter)
be accurately replaced by an equivalent series circuit of a fixed voltage
generator and fixed equivalent series resistance (independent of load).

How would such a simple model deal with the case of a transmitter that at
maximum power output is close to voltage saturation (ie cannot develop
more output voltage) and close to current saturation (ie cannot develop
more output current)? These non-linear behaviours close to operating
point are not captured in a simple linear equivalent circuit.

Owen- Hide quoted text -

- Show quoted text -

Hi Owen,

Thank you for the insightful response.

First, let me say that although the average source resistance at the
plates appears to be 1400 ohms in the case I described, and IMHO I
believe it is, I'm not in the position of stating that is as a fact.
What I do claim as a fact is that when the transmitter is loaded to
deliver all available power to its load, the OUTPUT source resistance
(or impedance) at the output terminals is the conjugate of its load.
I'm differentiating between the conditions at the input of the pi-
network and those at the output, because the energy storage effect of
the network Q isolates the output from the input, such that the
conditions at the output can be represented by an equivalent Thevenin
generator. At the output terminals the conditions appearing at the
input are irrelevant, such as the shape and duration of the voltage
applied to the pi-network, as long as the energy storage Q is
sufficient to support a constant voltage-current relationship (linear)
at the output for whatever load is absorbing all the available power
from the network.

Thus, when all available power is delivered into a 50-ohm load the
source resistance at the output terminals is 50 ohms. Please also
review the later portion of Chapter 19, also available on my web page.
On those pages I report the results of measurements using the load-
variation method, which also shows the output source resistance to
equal the load resistance when the amp is delivering all its available
power.

Walt

Richard Harrison

March 7th 07 11:17 PM

The power explanation

Owen Duffy wrote:
"It is my view thatthe statement "Because the amplifier was adjusted to
deliver the maximum available power of 100 watts prior to the resistance
measurement, resistance RLP looking into the plate (upstream from the
network terminals) is also approximately 1400 ohms" is not proven.."

The maximum power transfer theorem is a classic. It has been proven
countless times. It is explained by Terman in his 1955 opus on page 76.
It requires linearity in the area examined. A tuned circuit is often the
linearizer in an r-f amplifier. It eliminates harmonics of the
fundamental frequency so that Ohm`s law prevails across the output
terminals of the device. Thus, it is linear. It makes no difference that
ahead of the linearizer the power source is being pulsed so long as the
pulses don`t appear in the output. We don`t have pulses in the outputs
of our radios feeding our antennas.

When you have a conjugate match, the resistive parts of the source and
load are equal by definition.

Best regards, Richard Harrison, KB5WZI

Owen Duffy

March 7th 07 11:22 PM

The power explanation

"walt" wrote in
ups.com:

Walt,
....
First, let me say that although the average source resistance at the
plates appears to be 1400 ohms in the case I described, and IMHO I
believe it is, I'm not in the position of stating that is as a fact.

Ok, I think we are agreed that the measurements haven't directly
supported that belief.

What I do claim as a fact is that when the transmitter is loaded to
deliver all available power to its load, the OUTPUT source resistance
(or impedance) at the output terminals is the conjugate of its load.

If it were a linear source and you delivered *maximum* (as opposed to
*all*) to the load, I agree that the load impedance is the complex
conjugate of the source impedance. That is essentially the Jacobi Maximum
Power Transfer Theoram.

The question is whether it is a sufficiently linear source to use that
model.

I'm differentiating between the conditions at the input of the pi-
network and those at the output, because the energy storage effect of
the network Q isolates the output from the input, such that the
conditions at the output can be represented by an equivalent Thevenin
generator. At the output terminals the conditions appearing at the
input are irrelevant, such as the shape and duration of the voltage
applied to the pi-network, as long as the energy storage Q is
sufficient to support a constant voltage-current relationship (linear)
at the output for whatever load is absorbing all the available power
from the network.

Thus, when all available power is delivered into a 50-ohm load the
source resistance at the output terminals is 50 ohms. Please also
review the later portion of Chapter 19, also available on my web page.
On those pages I report the results of measurements using the load-
variation method, which also shows the output source resistance to
equal the load resistance when the amp is delivering all its available
power.

Walt, I have just re-read that section and note your measurements which
explored the delta V and delta I for small load variation (delta R) where
delta R is always negative, and calculated results.

Your results are interesting.

I have seen others report quite different results, and have found
differently myself on rough measurements, but I note your comments on the
sensitivity of the calculated Rs to tuning/matching which might reveal
why other tests disagree.

(It only takes one sound repeatable experiment that shows that the source
impedance is not the conjugate of the load to disprove the generality.)

On a practical note, the sensitivity discussed above does mean that if
your assertion about matching is true, it is unlikely that transmitters
are exactly matched.

My measurements have been on transistor PAs with broadband transformer
coupling to the load. The transmitters have had a lowpass filter with a
break point well above operating frequency between the transistors and
load. It is a different configuration, and although my measurements were
rough, they indicated different apparent source impedance at different
drive levels which questions the linear model for large signal operation,
especially for modes with varying amplitude such as SSB telephony.

Owen

Cecil Moore

March 7th 07 11:33 PM

The power explanation

Owen Duffy wrote:
The question is can the plate (or the whole transmitter for that matter)
be accurately replaced by an equivalent series circuit of a fixed voltage
generator and fixed equivalent series resistance (independent of load).

Can a real-world dynamic source impedance be
independent of load?
--
73, Cecil http://www.w5dxp.com

Dave	March 8th 07 01:27 AM

The power explanation

Owen Duffy wrote:

SNIPPED

The question is can the plate (or the whole transmitter for that matter)
be accurately replaced by an equivalent series circuit of a fixed voltage
generator and fixed equivalent series resistance (independent of load).

SNIPPED

The answer is NO!!

Power Amplifiers are not linear devices! [Regardless of manufacturer's claims].

My AL-80B puts out ~850 watts with ~1300 watts input. IF a conjugate match
existed the output would be ~650 watts. The Zo of the 3-500 varies depending on
the phase of the pulse signal. The amplifier provides a pulse of energy into the
tuned circuit, tank circuit, tuner, antenna, where the reactances maintain the
current/voltage under resonant conditions.

My understanding is that reflected energy is also coupled into the tuned
circuits. The active device, power amplifier, may be cutoff, saturated, or
somewhere on the active load line. The energy in the tuned circuit will increase
the Vmax and/or Imax depending on the circuit QL. This increase in stored energy
provides the extra stress on the active device. If the QL 10, nominal design
value, then 90% of the reflected energy is re-reflected back towards the load.
The missing 10% produces heat in the tuned circuit.

Devices that operate from saturation to cutoff are by definition NON-LINEAR.

The tuned circuits store both forward and reflected energy. Ultimately all the
power is radiated, either as a rf field or as heat.

Richard Clark

March 8th 07 02:57 AM

The power explanation

On Wed, 07 Mar 2007 20:27:10 -0500, Dave wrote:

Ultimately all the power is radiated, either as a rf field or as heat.

Hi Dave,

Well, having said that, it was hardly worth constraining the rest of
the discussion with linearity, SWR, efficiency, pulses, Q, cut-off,
saturated, tube or transistor.

73's
Richard Clark, KB7QHC

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