On Mon, 24 May 2004 16:40:44 -0500, "Steve Nosko"
wrote:
This is an interesting twist, Tam. I think if this were the case, then
there would be MORE power dissipated in the Tx

Hi Steve,

And why would that be?

73's
Richard Clark, KB7QHC

Dave Shrader wrote:
Cecil Moore wrote:

snip

For what it is worth, I believe that the first homo sapien
originated about a quarter of a million years ago and was
a female with dominant genes.
--
73, Cecil http://www.qsl.net/w5dxp


My wife has me convinced that ALL women have the dominant genes!!!!

I prefer women without jeans, thankyouvermuch!

- Mike KB3EIA -

Richard Clark wrote:

wrote:
You asked - I answered.

Did you? What was the original question?

I don't know. I wasn't around 7 billion years ago.



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Richard Clark wrote,

On Mon, 24 May 2004 15:28:40 -0500, Cecil Moore
wrote:
I am sure that our solar system and homo sapiens didn't exist
when the original question was asked.
So the question is which came first, the homo or the solar system?


Homos before Helios, or Helios before homos? That's a profound
question which I'll have to think about over my after-dinner port. On
the surface it looks about as meaningful as "turkeys from Turlock,"
but first impressions are sometimes deceiving.
73,
Tom Donaly, KA6RUH

Cecil wrote,

Richard Clark wrote:

wrote:
I am sure that our solar system and homo sapiens didn't exist
when the original question was asked.

So the question is which came first, the homo or the solar system?

For what it is worth, I believe that the first homo sapien
originated about a quarter of a million years ago and was
a female with dominant genes.
--
73, Cecil http://www.qsl.net/w5dxp



Yes, she married Org Orgluk and they had a bunch of halfwit kids who sired
the human race.
73,
Tom Donaly, KA6RU

Richard Clark wrote:

On Mon, 24 May 2004 14:06:28 -0500, Cecil Moore
wrote:

What was the original question? [accredited stock response]

When totally ignorant, divert the issue as long as possible.
Why am I not surprised?

Are you sure this is the original question? [accredited stock
response]

Richard

Your earlier comment on things was quite accurate. This thread is
hilarious!

tom
K0TAR

On Mon, 24 May 2004 18:37:09 -0500, Cecil Moore
wrote:
I don't know. I wasn't around 7 billion years ago.
We were beginning to wonder.

"Steve Nosko" wrote in message
...

"Tam/WB2TT" wrote in message
...

"Richard Fry" wrote in message
.........................
...............................
Concept below

However this is not an accurate model of a transmitter.

For an example, take an old Heathkit DX-100 generating a measured 180
watts
of CW RF into a matched 50 ohm load. To do this, it does NOT also
dissipate
180 watts of RF into some "virtual" internal RF load in the DX-100.
In
fact, the PAs and power supply in the DX-100 could not produce a total
RF
output power of 360 watts without exceeding their ratings.

The dissipation in the PA is essentially related only the DC to RF
conversion efficiency of the PA, which in this case probably is about
75%,
max (Class C). So a PA input power of about 240 watts DC is required
to
produce 180 watts of RF output power. The other 60 watts of plate
input
power is converted to heat by the PA tube anodes.

The entire RF output generated by the PA stage is applied virtually
100%
to
the output connector. How much of that is absorbed by the load
connected
there is a function of load SWR and system losses.

- RF
There is a Motorola ap note that agrees with what Richard is saying. To
paraphrase it, if the the DX100 had an output impedance of 50 Ohms, then
the
overall efficiency would be 37.5%.

Unfortunately I can't read all the digressions in the thread. I skim
by
author...

This is an interesting twist, Tam. I think if this were the case,
then
there would be MORE power dissipated in the Tx than Mr. Fry is saying -
making the situation worse. By that, I mean, getting further from what is
going on. I think this goes in the wrong direction. I believe the flaw
is
believing that the Rs=RL must exist for the transmitter.

That is what I am saying. The efficiency goes from 75% to 37.5%; so, there
is more power dissipated in the TX.

Tam/WB2TT
--
Steve N, K,9;d, c. i My email has no u's.

Richard Clark wrote:

On Sun, 23 May 2004 02:07:13 GMT, "Henry Kolesnik"
wrote:


I know that any power not dissipated by an antenna is reflected back to the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an open
is required to reflect power and I'm searching for which it is, an open or a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or analogy
and some math wouldn't hurt.
tnx
Hank WD5JFR



Hi Hank,

At last count there are 130 responses to this post, this is #131, and
the question still hasn't been answered.

When energy or power is transmitted in any medium where the wavelength
and the length of the transmission medium are significant percentages of
one another some energy/power is reflected at any discontinuity in the
transmission medium. The reflected energy/power may be re-reflected if a
discontinuity exists in the backward path.

The simplest example that we can all understand is the common case of
the echo!! H E L L O ! .... HELLO ! .... hello ! .... etc. The
energy/power is re-reflected many times until we can't hear it. But is
is still re-reflecting at sub-audible levels until 100% dissipation
occurs. As long as the discontinuities exist the echoes exist!

DD, W1MCE


What you describe as reflection and re-reflection occurs between the
mismatched antenna and the tuner that has been adjusted to minimize
power returned to the transmitter. The sole function of the tuner is
to keep this power from being dissipated by the transmitter (common
experience of arcing, denoting a voltage reflection, or thermal
runaway, denoting a current reflection). The "virtual" reflection
(offered by the tuner) is generally know as the complex conjugate of
the remote load, seen at the near end of the line through which it is
returning. This means that the line transforms the phase and
amplitude of the reflection, and the tuner's job is to invert that
relationship to counteract it, and return it to the antenna.

There are both wave descriptions of this process, and lumped circuit
equivalents. Both work, and both describe the same process from
different points of view. One does not negate the other's validity
(unless, of course, you attempt to mix the points of view and demand
consistency in terms - a frequent rhetorical trap here).

There will no doubt be a flurry of denials to this simple example with
contortions of logic to match. As for the math, you will find it by
the reams, once you've been overwhelmed with the arcana of hyperbolic
descriptions of a novel physics that have to proceed its proof.

Keep your eye on how your literal points in your question go abandoned
with these arcane theories.

73's
Richard Clark, KB7QHC

On Tue, 25 May 2004 01:45:19 GMT, Dave Shrader
wrote:
At last count there are 130 responses to this post, this is #131, and
the question still hasn't been answered.

Hi Dave,

Well, actually, it has been answered sufficiently, apparently it
hasn't been understood - quite a gulf between those two positions. As
such, your response doesn't necessarily constitute an answer either,
that is, until the gulf is spanned.

73's
Richard Clark, KB7QHC

Richard Clark wrote:

On Tue, 25 May 2004 01:45:19 GMT, Dave Shrader
wrote:

At last count there are 130 responses to this post, this is #131, and
the question still hasn't been answered.


Hi Dave,

Well, actually, it has been answered sufficiently, apparently it
hasn't been understood - quite a gulf between those two positions. As
such, your response doesn't necessarily constitute an answer either,
that is, until the gulf is spanned.

I've gotten some education from it, and had some fun too!

- Mike KB3EIA -

Richard

You say my question has been answered but I haven't seen the answer, too
old, too stupid, whatever but I still would like to understand. So, would
you be kind enough to give me a better understanding of the "mechanism" in
the transmitter final that can dissipate and transform yet it can't
dissipate a reflection because there's some kind of one way device that acts
like a checkvalve or diode that reflects.
tnx


--
73
Hank WD5JFR
"Richard Clark" wrote in message
...
On Tue, 25 May 2004 01:45:19 GMT, Dave Shrader
wrote:
At last count there are 130 responses to this post, this is #131, and
the question still hasn't been answered.

Hi Dave,

Well, actually, it has been answered sufficiently, apparently it
hasn't been understood - quite a gulf between those two positions. As
such, your response doesn't necessarily constitute an answer either,
that is, until the gulf is spanned.

73's
Richard Clark, KB7QHC

On Tue, 25 May 2004 02:59:52 GMT, "Henry Kolesnik"
wrote:

Richard

You say my question has been answered but I haven't seen the answer, too
old, too stupid, whatever but I still would like to understand. So, would
you be kind enough to give me a better understanding of the "mechanism" in
the transmitter final that can dissipate and transform yet it can't
dissipate a reflection because there's some kind of one way device that acts
like a checkvalve or diode that reflects.
tnx

Hi Hank,

There is no answer as you phrase the question because the transmitter
does dissipate power reflected to it. I can easily imagine you may
find no understanding as I go hyperbolic in the following lines. :-)

It can also reflect (some or all of) what it does not absorb, but only
if it does not match the line connected to it. The degree to which it
absorbs/reflects is determined by the ratio of its Z to the line/load
at the antenna terminal. In this case (bear with me), the source Z is
NOW found at the load's mismatch, transformed through the line. This
means matching works both ways (often the singlemost ignored aspect of
obvious reciprocity in any of these arguments). Such circuit analysis
is called superposition. This is merely an academic way of saying you
look at the problem from both points of view where you reverse roles
(source becomes load, and load becomes source). This is a common
practice learned in first quarter circuit analysis when Kirchhoff's
laws are developed (Thevenin/Norton equivalent circuits also emerge).

For example (a strained one at that), if you had a 500 Ohm Source
characteristic Z, you will never launch much power into a 50 Ohm
system because it would immediately hit a reflective interface at the
antenna connector. This means that a 500 Ohm source, when confronted
by power going towards it from a 50 Ohm system will reflect most of
that power (but how did we get this power into the system in the first
place - Karma?) This, by design, will never happen in any transistor
ham rig built in the last quarter century.

So, this is why you see such a vacuum of response when you (the
general readership "you") ask:
"What is the source Z if it is not 50 Ohms?"
Silence guarantees they either don't know (a fatal admission for egos)
or if they offered a value, we could all balance the checkbook and
that would end the game being played. Hence we are treated to all
these suppositions of shorts and opens or magic reflectors hidden
beneath the hood. The frequent toss-off comment of "no one knows" is
called projection, a psychological salve for the ego meaning if I
don't know, then certainly no one else does either - or they are
wrong.

The answer is the transmitter source Z is 50 Ohms at rated power. If
a watt of power is chooglin' down the line toward it, that 50 Ohms is
going to dissipate into a watt worth of calories. This can be argued
with wave mechanics, or lumped circuit equivalents - doesn't matter
because it's all the same calories. Modern rigs can tolerate this
watt through limited feedback and level controlling circuitry - if you
paid more than a kilobuck for your rig that is. For the rest of us,
it's a spin of the wheel and you take your chance. 1 watt hardly
amounts to much, but are you that lucky that it is ONLY 1 watt? Are
we to suppose those unfortunate souls who blasted their rig
transmitting into a mismatch lost it only because they lacked enough
magic pixie dust? I will bet no rig was lost to a cold snap with
frosty finals.

I've already answered about where the heat can be found, so we will
conserve bandwidth.

=== WARNING to the logic impaired, the following is a supposition ===

Now, lets simply accept those answers that require the magic pixie
dust of total reflection from the transmitter. Fine, the mismatched
antenna reflects some power, this power returns to the transmitter,
the transmitter simple routes it ALL back to the antenna,
round-and-round until the antenna finally radiates it. If this were
true, what do we need tuners for? Take a survey of everyone who
chants this mantra of the rig reflecting, and ask if they have a tuner
in the line. Is it a paper weight holding down their license? Is it
a line stretcher because they needed another foot extension between
the antenna lead and their rig? Dare I point out the utter failure of
their faith?

I will pause to allow those answers to emerge and enjoy the thrashing
dawn of a new era of metaphysics. Ooooohhhhhhmmmmms Laaaaaaaaaaw!

73's
Richard Clark, KB7QHC

Henry Kolesnik wrote:
You say my question has been answered but I haven't seen the answer, too
old, too stupid, whatever but I still would like to understand. So, would
you be kind enough to give me a better understanding of the "mechanism" in
the transmitter final that can dissipate and transform yet it can't
dissipate a reflection because there's some kind of one way device that acts
like a checkvalve or diode that reflects.

By definition, reflected energy dissipated in the source was never
generated in the first place.
--
73, Cecil http://www.qsl.net/w5dxp



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"Cecil Moore" wrote in message
...
Henry Kolesnik wrote:
You say my question has been answered but I haven't seen the answer, too
old, too stupid, whatever but I still would like to understand. So,
would
you be kind enough to give me a better understanding of the "mechanism"
in
the transmitter final that can dissipate and transform yet it can't
dissipate a reflection because there's some kind of one way device that
acts
like a checkvalve or diode that reflects.

By definition, reflected energy dissipated in the source was never
generated in the first place.
--
73, Cecil http://www.qsl.net/w5dxp


That's not a very good definition.
Would you say that a rock, thrown vertically, never was thrown just because
it returned to hit you on the head?

Ed
wb6wsn

Post below

REALITY CHECK: If tx source Z really was 50 ohms, a tx connected to a 50
ohm load would lose 1/2 of the RF power it generates to that internal Z.
And if that is true, then the heat produced by the tx, and the AC power
consumed by the tx would be proof of that.

However careful measurement of heat production and AC power consumption of
broadcast txs (for example) does NOT support the belief that the source Z of
a tx designed for 50 ohm loads IS 50 ohms.

Following is the output of an old DOS program I wrote years ago to show
operating parameters for a 25kW FM broadcast tx manufactured by my employer,
when operating at any RF power and line voltage in its rated range. These
values closely track the average of factory test data accumulated over 100s
of delivered units.

In this example, if the tx had a 50 ohm source Z it would have to generate
50kW of RF in order to deliver 25kW to a 50 ohm load. In the first place,
it doesn't have a big enough power supply or a big enough PA to do that.
But even if it could, the input power and heat production would be almost
twice the real values shown below.

The same physics applies to ham transmitters, which should be verifiable by
an unbiased evaluation of comparable operating parameters.

- RF

* FM TRANSMITTER AC LOAD *

Transmitter name (HT-??FM): ? 25
TPO in kW: ? 25
AC line voltage: ? 220

RESULT:

Total Tx = 103.7 amps/phase, 39.5 kVA total.
H.V. Power Supply Current = 92.7 amps/phase
Total AC/RF Efficiency = 66.5 %

PA Air = 400 cu. feet/min.
Tx Cabinet Flushing Air = 335 cu. feet/min.
Power Supply Air = 250 cu. feet/min.
Total Air = 985 cu. feet/min.

Total Heat = 42810 BTU/Hr.
Heat Rise, total tx system = 39 deg. F.
Air Conditioner Load = 3.6 tons

____________

"Richard Clark"
"What is the source Z if it is not 50 Ohms?"
(clippage)
The answer is the transmitter source Z is 50 Ohms at rated power.

Richard
This is in response to your answer of last night. Before going to bed I got
out the book REFLECTIONS II by Walt Maxwell W2DU. I'm typing verbatim from
page 2-2 and 23-1 for those who don't have the book.page 2-2 "Contrary to
what many believe, it is not true that when a transmitter delivers power
into a line with reflections, a returning wave sees an internal generator
resistance as a dissipative load. Nor is the reflected wave converted to
heat and, while at the same time damaging the final amplifier....the
reflected power is entirely conserved...." from page 23-1 "One of the most
serious misconceptions concerned reflected power reaching the tubes in the
RF amplifier of the transmitter. The prevalent, but erroneous thinking was
that the reflected power enters the amplifier, causing tube overheating and
destruction. However, I dispelled this misconception in the above mentioned
publications, using wave-mechanics treatment, discussed here in greater
detail, by showing that when the pi-network tank is tuned to resonance, a
virtual short circuit to rearward traveling waves is created at the input of
the network. Consequently, instead of the reflected power reaching the
tubes of the amplifier, it is totally re-reflected toward the load by the
virtual short circuit appearing only to waves at the network input".
I'm guessing it's a virtual short because the pi-network is resonant but
what happens if it is a bit off. Also what happens in a transistor final
with no pi?

73 Hank WD5JFR


"Richard Clark" wrote in message
...
On Tue, 25 May 2004 02:59:52 GMT, "Henry Kolesnik"
wrote:

Richard

You say my question has been answered but I haven't seen the answer, too
old, too stupid, whatever but I still would like to understand. So,
would
you be kind enough to give me a better understanding of the "mechanism"
in
the transmitter final that can dissipate and transform yet it can't
dissipate a reflection because there's some kind of one way device that
acts
like a checkvalve or diode that reflects.
tnx

Hi Hank,

There is no answer as you phrase the question because the transmitter
does dissipate power reflected to it. I can easily imagine you may
find no understanding as I go hyperbolic in the following lines. :-)

It can also reflect (some or all of) what it does not absorb, but only
if it does not match the line connected to it. The degree to which it
absorbs/reflects is determined by the ratio of its Z to the line/load
at the antenna terminal. In this case (bear with me), the source Z is
NOW found at the load's mismatch, transformed through the line. This
means matching works both ways (often the singlemost ignored aspect of
obvious reciprocity in any of these arguments). Such circuit analysis
is called superposition. This is merely an academic way of saying you
look at the problem from both points of view where you reverse roles
(source becomes load, and load becomes source). This is a common
practice learned in first quarter circuit analysis when Kirchhoff's
laws are developed (Thevenin/Norton equivalent circuits also emerge).

For example (a strained one at that), if you had a 500 Ohm Source
characteristic Z, you will never launch much power into a 50 Ohm
system because it would immediately hit a reflective interface at the
antenna connector. This means that a 500 Ohm source, when confronted
by power going towards it from a 50 Ohm system will reflect most of
that power (but how did we get this power into the system in the first
place - Karma?) This, by design, will never happen in any transistor
ham rig built in the last quarter century.

So, this is why you see such a vacuum of response when you (the
general readership "you") ask:
"What is the source Z if it is not 50 Ohms?"
Silence guarantees they either don't know (a fatal admission for egos)
or if they offered a value, we could all balance the checkbook and
that would end the game being played. Hence we are treated to all
these suppositions of shorts and opens or magic reflectors hidden
beneath the hood. The frequent toss-off comment of "no one knows" is
called projection, a psychological salve for the ego meaning if I
don't know, then certainly no one else does either - or they are
wrong.

The answer is the transmitter source Z is 50 Ohms at rated power. If
a watt of power is chooglin' down the line toward it, that 50 Ohms is
going to dissipate into a watt worth of calories. This can be argued
with wave mechanics, or lumped circuit equivalents - doesn't matter
because it's all the same calories. Modern rigs can tolerate this
watt through limited feedback and level controlling circuitry - if you
paid more than a kilobuck for your rig that is. For the rest of us,
it's a spin of the wheel and you take your chance. 1 watt hardly
amounts to much, but are you that lucky that it is ONLY 1 watt? Are
we to suppose those unfortunate souls who blasted their rig
transmitting into a mismatch lost it only because they lacked enough
magic pixie dust? I will bet no rig was lost to a cold snap with
frosty finals.

I've already answered about where the heat can be found, so we will
conserve bandwidth.

=== WARNING to the logic impaired, the following is a supposition ===

Now, lets simply accept those answers that require the magic pixie
dust of total reflection from the transmitter. Fine, the mismatched
antenna reflects some power, this power returns to the transmitter,
the transmitter simple routes it ALL back to the antenna,
round-and-round until the antenna finally radiates it. If this were
true, what do we need tuners for? Take a survey of everyone who
chants this mantra of the rig reflecting, and ask if they have a tuner
in the line. Is it a paper weight holding down their license? Is it
a line stretcher because they needed another foot extension between
the antenna lead and their rig? Dare I point out the utter failure of
their faith?

I will pause to allow those answers to emerge and enjoy the thrashing
dawn of a new era of metaphysics. Ooooohhhhhhmmmmms Laaaaaaaaaaw!

73's
Richard Clark, KB7QHC

"Henry Kolesnik" wrote in message
.. .
Richard
This is in response to your answer of last night. Before going to bed I
got
out the book REFLECTIONS II by Walt Maxwell W2DU. I'm typing verbatim
from
page 2-2 and 23-1 for those who don't have the book.page 2-2 "Contrary to
what many believe, it is not true that when a transmitter delivers power
into a line with reflections, a returning wave sees an internal generator
resistance as a dissipative load. Nor is the reflected wave converted to
heat and, while at the same time damaging the final amplifier....the
reflected power is entirely conserved...." from page 23-1 "One of the
most
serious misconceptions concerned reflected power reaching the tubes in the
RF amplifier of the transmitter. The prevalent, but erroneous thinking
was
that the reflected power enters the amplifier, causing tube overheating
and
destruction. However, I dispelled this misconception in the above
mentioned
publications, using wave-mechanics treatment, discussed here in greater
detail, by showing that when the pi-network tank is tuned to resonance, a
virtual short circuit to rearward traveling waves is created at the input
of
the network. Consequently, instead of the reflected power reaching the
tubes of the amplifier, it is totally re-reflected toward the load by the
virtual short circuit appearing only to waves at the network input".
I'm guessing it's a virtual short because the pi-network is resonant but
what happens if it is a bit off. Also what happens in a transistor final
with no pi?

73 Hank WD5JFR
Henry,
Here is an example of what you just said. Take a sine wave source, and
connect it to a 1/4 wave section of shorted transmission line through a
series resistor R. The reflected wave will reach this resistor 1/2 cycle
later, and will be in phase with the source. For a lossless transmission
line, there will be *0 Volts across the resistor*. There will be 0 current
through the resistor, and the reflected wave will be re reflected for all
values of R, including R=Z0, because the reflected wave will not "know" what
R is. You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Tam/WB2TT

Ed Price wrote:
"Cecil Moore" wrote:
By definition, reflected energy dissipated in the source was never
generated in the first place.

That's not a very good definition.
Would you say that a rock, thrown vertically, never was thrown just because
it returned to hit you on the head?

Nope, I wouldn't. But that seems to be what some people
are saying. I objected to that definition about 15 years
ago. I was asked to prove otherwise and was unable to
do so. The generated power is *defined* as the *net*
steady-state output power of the source.

That's why many of my examples involve signal generators
equipped with a circulator and load, outputting a constant
forward voltage in phase with a constant forward current.
--
73, Cecil http://www.qsl.net/w5dxp



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Richard Fry wrote:

Post below

REALITY CHECK: If tx source Z really was 50 ohms, a tx connected to a 50
ohm load would lose 1/2 of the RF power it generates to that internal Z.
And if that is true, then the heat produced by the tx, and the AC power
consumed by the tx would be proof of that.

That describes a Class-A amplifier.

However careful measurement of heat production and AC power consumption of
broadcast txs (for example) does NOT support the belief that the source Z of
a tx designed for 50 ohm loads IS 50 ohms.

With other classes of amps, the amplifier's individual component signals
are non-linear.
--
73, Cecil http://www.qsl.net/w5dxp




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"Cecil Moore" wrote in message
...
That's a steady-state shortcut which assumes pure sine waves that
don't exist in reality. Please don't confuse steady-state shortcuts
with reality. Noise and modulation cause the "steady-state complex
impedance" not to be steady-state at all. Many will say it's close
enough, but one cannot understand reflections by assuming an un-
varying steady-state.

In a TV system with ghosting due to reflections, the unvarying steady-
state condition doesn't exist. In fact, when you assume steady-state
conditions, you eliminate ghosting, at least in your own mind. In
reality, steady-state doesn't really exist because of random noise
and unpredictable modulation.

The more you talk about 'steady-state-, the more apparent it becomes that
you don't really understand what it means.

73, Jim AC6XG

Henry Kolesnik wrote:

Richard
This is in response to your answer of last night. Before going to bed I got
out the book REFLECTIONS II by Walt Maxwell W2DU. I'm typing verbatim from
page 2-2 and 23-1 for those who don't have the book.page 2-2 "Contrary to
what many believe, it is not true that when a transmitter delivers power
into a line with reflections, a returning wave sees an internal generator
resistance as a dissipative load. Nor is the reflected wave converted to
heat and, while at the same time damaging the final amplifier....the
reflected power is entirely conserved...." from page 23-1 "One of the most
serious misconceptions concerned reflected power reaching the tubes in the
RF amplifier of the transmitter. The prevalent, but erroneous thinking was
that the reflected power enters the amplifier, causing tube overheating and
destruction. However, I dispelled this misconception in the above mentioned
publications, using wave-mechanics treatment, discussed here in greater
detail, by showing that when the pi-network tank is tuned to resonance, a
virtual short circuit to rearward traveling waves is created at the input of
the network. Consequently, instead of the reflected power reaching the
tubes of the amplifier, it is totally re-reflected toward the load by the
virtual short circuit appearing only to waves at the network input".

Please note that when this was written, there were no solid-state transmitters
with fixed 50 ohm outputs. Walt was talking specifically about tube transmitters
with built-in adjustable pi-net antenna tuner circuits, like my old Globe Scout.
The closest present-day transmitters to match the above statements would be
something like my IC-756PRO with a built-in auto-tuner.

I'm guessing it's a virtual short because the pi-network is resonant but
what happens if it is a bit off.

If it is a bit off, reflected voltage and current flow into the finals
and superpose with the source voltage and current. Since the reflected
voltage and current are 180 degrees out of phase, the superposition
usually causes the load on the amp to become reactive. Even if the
voltages are in phase or the currents are in phase, the load on the
amp becomes something other than the designed-for load. The result
can be over-voltage or over-current or both. That's not good.

Also what happens in a transistor final with no pi?

Note: most transistor finals have a fixed pi-network designed to
operate into a fixed 50 ohm load. The results would be the same as
the detuned example above. Such commercial finals are protected
from over-voltage and over-current by decreasing the source power.
--
73, Cecil http://www.qsl.net/w5dxp



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On Tue, 25 May 2004 13:50:09 GMT, "Henry Kolesnik"
wrote:

Richard
This is in response to your answer of last night. Before going to bed I got
out the book REFLECTIONS II by Walt Maxwell W2DU. I'm typing verbatim from

Very familiar stuff which I have held Walt accountable for. We have
had considerable correspondence to this point and he is holding my
argument in suspension, pending his bench work confirmation of my
simple experiments that I have posted here to this effect.

I'm guessing it's a virtual short because the pi-network is resonant but
what happens if it is a bit off. Also what happens in a transistor final
with no pi?

What you describe is consistent with Walt's language, you deserve
points for following his logic. As for your question about the
transistor final with no pi network. That was handled quite
explicitly by me for the thread "Loop Antenna for Class-E amplifier"
in correspondence with Toni, ea3fya and Marc Battyani. If you will
note, this thread was entirely devoid of ego so puffed up in these
issues, and we had the rare opportunity of a third, independent
observer who also sat at the bench and followed the instructions and
comments I offered with actual construction and testing to POSITIVE
results. You may also note the absolute vacuum of other's comments
imploring that his data was chimerical, illusions of a meter. In fact
it was one of those rare threads with a question much like yours, with
the same simple observations offered by me, confirmed and ending in
very much less than 600 postings of supposition and superstition.

However, back to Walt's treatise and your reading of it. I again
simply ask: "Do you have a tuner in line each day you fire up your
rig? Why, if you have faith in your logic, do you need it?"

As I've pointed out, there's the recent thread, and any number of
simple tests and experiments that I have offered here. These etudés
are typically ignored in favor of vacuous theory-spinning - NONE of
which answer the simple question: "What is the Source Z if it is not
50 Ohms?"

73's
Richard Clark, KB7QHC

Before it came down my 20 meter antenna had a SWR approaching 20:1 and I was
running a Collins 30S-1 and no tuner!. I did that for several years and
would be still doing if the roofers hadn't destroyed ~130 ft. dipole. One
of these days after my back improves I'll put another one up.

--
73
Hank WD5JFR
"Richard Clark" wrote in message
...
On Tue, 25 May 2004 13:50:09 GMT, "Henry Kolesnik"
wrote:

Richard
This is in response to your answer of last night. Before going to bed I
got
out the book REFLECTIONS II by Walt Maxwell W2DU. I'm typing verbatim
from

Very familiar stuff which I have held Walt accountable for. We have
had considerable correspondence to this point and he is holding my
argument in suspension, pending his bench work confirmation of my
simple experiments that I have posted here to this effect.

I'm guessing it's a virtual short because the pi-network is resonant but
what happens if it is a bit off. Also what happens in a transistor final
with no pi?

What you describe is consistent with Walt's language, you deserve
points for following his logic. As for your question about the
transistor final with no pi network. That was handled quite
explicitly by me for the thread "Loop Antenna for Class-E amplifier"
in correspondence with Toni, ea3fya and Marc Battyani. If you will
note, this thread was entirely devoid of ego so puffed up in these
issues, and we had the rare opportunity of a third, independent
observer who also sat at the bench and followed the instructions and
comments I offered with actual construction and testing to POSITIVE
results. You may also note the absolute vacuum of other's comments
imploring that his data was chimerical, illusions of a meter. In fact
it was one of those rare threads with a question much like yours, with
the same simple observations offered by me, confirmed and ending in
very much less than 600 postings of supposition and superstition.

However, back to Walt's treatise and your reading of it. I again
simply ask: "Do you have a tuner in line each day you fire up your
rig? Why, if you have faith in your logic, do you need it?"

As I've pointed out, there's the recent thread, and any number of
simple tests and experiments that I have offered here. These etudés
are typically ignored in favor of vacuous theory-spinning - NONE of
which answer the simple question: "What is the Source Z if it is not
50 Ohms?"

73's
Richard Clark, KB7QHC

On Tue, 25 May 2004 07:27:03 -0500, "Richard Fry"
wrote:
However careful measurement of heat production and AC power consumption of
broadcast txs (for example) does NOT support the belief that the source Z of
a tx designed for 50 ohm loads IS 50 ohms.

Hi Richard,

This quote above bears directly to one of my observations. You define
by negatives, and that is a shoddy technical discussion worthy of the
creationist argument.

Tell us what it IS instead. A simple complex number is all that is
asked, derived from actual measurement. I've done this numerous
times, and not simply for milliwatt signal generators and have
returned these readings to the client with certificates of
traceability to Primary Standards Labs.

Barring any explicit answer that yields this complex number; then that
answer is simply conjecture and illusion that is commonplace in this
discussion. Draping it with programs and formula doesn't give
substance to the shape of that dream.

73's
Richard Clark, KB7QHC

Tam/WB2TT wrote:
You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance? Hint: it's the net current
that is zero at the input of a 1/4WL stub. The forward current
and reflected current are equal in magnitude and opposite in
phase. Their individual magnitudes can be very large.
--
73, Cecil http://www.qsl.net/w5dxp





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On Tue, 25 May 2004 16:12:05 GMT, "Henry Kolesnik"
wrote:
Before it came down my 20 meter antenna had a SWR approaching 20:1 and I was
running a Collins 30S-1 and no tuner!. I did that for several years and
would be still doing if the roofers hadn't destroyed ~130 ft. dipole. One
of these days after my back improves I'll put another one up.

Hi Hank,

I used to teach this Collins equipment in the Navy. I presume yours
had a finals tuning circuit still in it? Hard to imagine it
otherwise.

Tubes are different from transistors only by approaching the Source Z
with an inverted ratio of transformation. Ever hear of plate
resistance? It is literal resistance. Ever see a plate glow when
under the stress of hi SWR? It is literal heat. The same heat is
generated irrespective of it being explained by wave mechanics or
lumped equivalent circuits. Choose the model you are comfortable
with, and then tackle the SWR if you care about efficiency.

If anything, tube sets prove the problem of reflected power through
your ability to directly observe the heat generated and experience the
cost of new finals tubes through their degraded life span. This stuff
is all rote teaching; and my students were given practical tests to
troubleshoot, tune, and repair against such scenarios.

I've had hands-on experience with this topic both academically and at
several benches - the sophistries that deny these points are amusing,
but remain amateur scribblings.

73's
Richard Clark, KB7QHC

The source Z of a typical tx having a single-ended Class C PA is just a few
ohms. The exact value depends on circuit design and tuning adjustments.
Heavy loading of the output tuning network of the PA reduces the return loss
looking back into the tx, and lighter loading improves it.

Please read my post using the new subject line "Tx Source Impedance & Load
Reflections" that I started for this a few minutes ago. The paper I refer
to and quote there (not mine) is accepted in the broadcast industry as a
benchmark on this subject.

- RF
____________

"Richard Clark"
Hi Richard,

This quote above bears directly to one of my observations. You define
by negatives, and that is a shoddy technical discussion worthy of the
creationist argument.

Tell us what it IS instead.

On Tue, 25 May 2004 11:42:42 -0500, "Richard Fry"
wrote:

The source Z of a typical tx having a single-ended Class C PA is just a few
ohms. The exact value depends on circuit design and tuning adjustments.
Heavy loading of the output tuning network of the PA reduces the return loss
looking back into the tx, and lighter loading improves it.

Please read my post using the new subject line "Tx Source Impedance & Load
Reflections" that I started for this a few minutes ago. The paper I refer
to and quote there (not mine) is accepted in the broadcast industry as a
benchmark on this subject.

- RF

OK you cannot give us a specific answer derived from actual
measurement.

73's
Richard Clark, KB7QHC

Henry Kolesnik wrote:
Before it came down my 20 meter antenna had a SWR approaching 20:1 and I was
running a Collins 30S-1 and no tuner!.

If the 30S-1 doesn't have a built-in tuner, what are those knobs
and meters for?
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil wrote,

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance?


In the case of a lossless, 1/4WL stub, (no such thing, except on this
newsgroup) in the steady state you can disconnect the transmitter and
the currents will still be there. Current doesn't flow, Cecil, charge flows.
Current is just the rate of flow of charge, dQ/dt. In a 1/4WL stub, charge,
and the fields associated with it, are in a state of oscillation. Over time,
their average movement in space is zero. It's too bad you have to think of
current
as like the water in a big river that has to flow from one place to another
in order to exist. (That's wrong, too. Current in a river is the rate of flow
of
water past a point, not the water itself.) Sloppy semantics sure screw up
understanding.
73,
Tom Donaly, KA6RUH

Good points, my 30S-1 has load and tune still in it and the final a
3CX1500B is ceramic and can't be seen but it never lost it's Eimac stencil.
However my Collins 30L-1 was a different story. Once or twice I had 811As
with holes melted in their plate structure. They continued to work well and
I always thought this was from overdriving but now I'm not so sure because
it ran at the same high swr as the 30S-1. I used the 30L-1 in summer
because the air conditioner couldn't keep up with the heat generated by the
30S-1 so it was my winter amp.
I notice that Walt Maxwell's Reflections II was published by World Radio
whereas Reflections was by the ARRL. There is some disagreement on
"conjugate match' and probably other things between ARRL and Walt. So now I
have to wonder if the final tube is dissipative or non-dissipative for
reflections.

Looks like my new thread isn't working but I'm posting this to it just in
case since I don't really know the customary protocol on a long thread that
sometime digresses.

--
73
Hank WD5JFR
"Richard Clark" wrote in message
...
On Tue, 25 May 2004 16:12:05 GMT, "Henry Kolesnik"
wrote:
Before it came down my 20 meter antenna had a SWR approaching 20:1 and I
was
running a Collins 30S-1 and no tuner!. I did that for several years and
would be still doing if the roofers hadn't destroyed ~130 ft. dipole.
One
of these days after my back improves I'll put another one up.

Hi Hank,

I used to teach this Collins equipment in the Navy. I presume yours
had a finals tuning circuit still in it? Hard to imagine it
otherwise.

Tubes are different from transistors only by approaching the Source Z
with an inverted ratio of transformation. Ever hear of plate
resistance? It is literal resistance. Ever see a plate glow when
under the stress of hi SWR? It is literal heat. The same heat is
generated irrespective of it being explained by wave mechanics or
lumped equivalent circuits. Choose the model you are comfortable
with, and then tackle the SWR if you care about efficiency.

If anything, tube sets prove the problem of reflected power through
your ability to directly observe the heat generated and experience the
cost of new finals tubes through their degraded life span. This stuff
is all rote teaching; and my students were given practical tests to
troubleshoot, tune, and repair against such scenarios.

I've had hands-on experience with this topic both academically and at
several benches - the sophistries that deny these points are amusing,
but remain amateur scribblings.

73's
Richard Clark, KB7QHC

Tdonaly wrote:
In the case of a lossless, 1/4WL stub, (no such thing, except on this
newsgroup) in the steady state you can disconnect the transmitter and
the currents will still be there.

Since that configuration doesn't exist in reality, only God
can cause what you are asserting. Why am I not surprised that
you need a supernatural being to prove your arguments?

Current doesn't flow, Cecil, charge flows.

:-) :-) Having to resort to trivial arguments is the sure sign
of a loser. :-) :-)
--
73, Cecil http://www.qsl.net/w5dxp



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On Tue, 25 May 2004 19:43:42 GMT, "Henry Kolesnik"
wrote:

Good points, my 30S-1 has load and tune still in it and the final a
3CX1500B is ceramic and can't be seen but it never lost it's Eimac stencil.
However my Collins 30L-1 was a different story. Once or twice I had 811As
with holes melted in their plate structure. They continued to work well and
I always thought this was from overdriving but now I'm not so sure because
it ran at the same high swr as the 30S-1. I used the 30L-1 in summer
because the air conditioner couldn't keep up with the heat generated by the
30S-1 so it was my winter amp.
I notice that Walt Maxwell's Reflections II was published by World Radio
whereas Reflections was by the ARRL. There is some disagreement on
"conjugate match' and probably other things between ARRL and Walt. So now I
have to wonder if the final tube is dissipative or non-dissipative for
reflections.

Looks like my new thread isn't working but I'm posting this to it just in
case since I don't really know the customary protocol on a long thread that
sometime digresses.


Hi Hank,

Those pinholes (or larger) that you found in the plates occurred by a
very fascinating display of plasma. If you are really putting the
hurts to it, you will observe a football shaped plasma between the
plate and the cathode where the point terminates (and penetrates) in
the enlarging ruby glow of the plate. This is visible in some tubes,
not others, depending of course on tube structure. I've had students
point out tubes still operating with the glass envelope completely
melted (or jelled rather) and collapsed, hugging the structure!

Hence you have confirmed the dissipation of return power directly -
all mumbling aside from theorists who describe it as virtual. ;-)

As for Walt's publishing house shift, I don't recall him having any
technical issue, but I let that go as he found another outlet and got
to say what he considered was important. I proofed a number of his
appendices and we corresponded quite often about our differences. To
his credit, although I dispute his stand, he maintains an open mind on
the subject and offers he will look into my experiments that reveal
this. He has already measured the value of 50 Ohms, the point of
departure between us is whether this same value absorbs power.

So, you see, you get numbers from those that sit down to the bench and
do it, you get pixie dust from the rest.

And as for new thread or old, this topic has run to 600+ postings in
years past; so opening it as a new topic is fine to allow news readers
(like mine) from pushing the indentations out to the event horizon.
This post is not so deeply nested though.

73's
Richard Clark, KB7QHC

The30L-1 811A holes were dime size and I wish I knew then what I know now
because I would have made some comparisons. I would have set the drive to
get the tubes red hot on my antenna with SWR 20 and then using same drive
and Ip see what happened loading into a 50 ohm dummy and running. Perhaps
someone has done this and can report the results.
tnx

--
73
Hank WD5JFR
"Richard Clark" wrote in message
...
On Tue, 25 May 2004 19:43:42 GMT, "Henry Kolesnik"
wrote:

Good points, my 30S-1 has load and tune still in it and the final a
3CX1500B is ceramic and can't be seen but it never lost it's Eimac
stencil.
However my Collins 30L-1 was a different story. Once or twice I had
811As
with holes melted in their plate structure. They continued to work well
and
I always thought this was from overdriving but now I'm not so sure
because
it ran at the same high swr as the 30S-1. I used the 30L-1 in summer
because the air conditioner couldn't keep up with the heat generated by
the
30S-1 so it was my winter amp.
I notice that Walt Maxwell's Reflections II was published by World Radio
whereas Reflections was by the ARRL. There is some disagreement on
"conjugate match' and probably other things between ARRL and Walt. So
now I
have to wonder if the final tube is dissipative or non-dissipative for
reflections.

Looks like my new thread isn't working but I'm posting this to it just in
case since I don't really know the customary protocol on a long thread
that
sometime digresses.


Hi Hank,

Those pinholes (or larger) that you found in the plates occurred by a
very fascinating display of plasma. If you are really putting the
hurts to it, you will observe a football shaped plasma between the
plate and the cathode where the point terminates (and penetrates) in
the enlarging ruby glow of the plate. This is visible in some tubes,
not others, depending of course on tube structure. I've had students
point out tubes still operating with the glass envelope completely
melted (or jelled rather) and collapsed, hugging the structure!

Hence you have confirmed the dissipation of return power directly -
all mumbling aside from theorists who describe it as virtual. ;-)

As for Walt's publishing house shift, I don't recall him having any
technical issue, but I let that go as he found another outlet and got
to say what he considered was important. I proofed a number of his
appendices and we corresponded quite often about our differences. To
his credit, although I dispute his stand, he maintains an open mind on
the subject and offers he will look into my experiments that reveal
this. He has already measured the value of 50 Ohms, the point of
departure between us is whether this same value absorbs power.

So, you see, you get numbers from those that sit down to the bench and
do it, you get pixie dust from the rest.

And as for new thread or old, this topic has run to 600+ postings in
years past; so opening it as a new topic is fine to allow news readers
(like mine) from pushing the indentations out to the event horizon.
This post is not so deeply nested though.

73's
Richard Clark, KB7QHC

On Tue, 25 May 2004 22:32:47 GMT, "Henry Kolesnik"
wrote:

The30L-1 811A holes were dime size and I wish I knew then what I know now
because I would have made some comparisons. I would have set the drive to
get the tubes red hot on my antenna with SWR 20 and then using same drive
and Ip see what happened loading into a 50 ohm dummy and running. Perhaps
someone has done this and can report the results.
tnx


Hi Hank,

A little follow-on, if you will. I presume you tuned into the load
first before switching to the antenna. How did you determine the SWR?
Were you driving a twin lead transmission line going to your
mismatched antenna, or a coax?

73's
Richard Clark, KB7QHC

Ed Price wrote:
By definition, reflected energy dissipated in the source was never
generated in the first place.
--
73, Cecil http://www.qsl.net/w5dxp

That's not a very good definition.
Would you say that a rock, thrown vertically, never was thrown just because
it returned to hit you on the head?

Ed
wb6wsn

Electromagnetic waves can cancel, but rocks can't. Could that maybe
make a difference?

73, Jim AC6XG

On Tue, 25 May 2004 16:43:37 -0700, Jim Kelley
wrote:

Would you say that a rock, thrown vertically, never was thrown just because
it returned to hit you on the head?

Ed
wb6wsn

Electromagnetic waves can cancel, but rocks can't. Could that maybe
make a difference?

Hi Jim,

This is like mistaking electrons and charge displacement. Sound like
Ed described two rocks hitting with an inelastic collision which
results in power dissipation.

73's
Richard Clark, KB7QHC

Richard Clark wrote:

On Tue, 25 May 2004 07:27:03 -0500, "Richard Fry"
wrote:

REALITY CHECK: If tx source Z really was 50 ohms, a tx connected to a 50
ohm load would lose 1/2 of the RF power it generates to that internal Z.


Dear Richard,

This is the time-honored misreading and misapplication of Thevenin's
theorem posed by Edison to confound investors in Westinghouse's AC
generation plants. Look at the ratio of DC to AC power plants
constructed in the past century to find the poor accuracy of your
reading.

73's
Richard Clark, KB7QHC

Boy, I'm gonna have to try and find my old EE5xx Non-Linear Transistor
Design notebook. I doubt I could translate it into words for this
section of the thread, but it is incredibly applicable.

I took it not knowing what it really was - RF amplifiers, tank circuits
and matching circuits of every flavor, and how much DC power was used
and RF delivered at the primary freq and all harmonics. He also covered
doublers, triplers, phase locked loops (from scratch), the whole 9
yards. I learned more in that one class than in any other 5 I took.
The professor was fantastic and very very tough. It was also the most
demanding course I ever took.

tom
K0TAR

I used 450 ohm ladder line and a balun along with a cheap SWR meter.

--
73
Hank WD5JFR
"Richard Clark" wrote in message
...
On Tue, 25 May 2004 22:32:47 GMT, "Henry Kolesnik"
wrote:

The30L-1 811A holes were dime size and I wish I knew then what I know
now
because I would have made some comparisons. I would have set the drive
to
get the tubes red hot on my antenna with SWR 20 and then using same
drive
and Ip see what happened loading into a 50 ohm dummy and running.
Perhaps
someone has done this and can report the results.
tnx


Hi Hank,

A little follow-on, if you will. I presume you tuned into the load
first before switching to the antenna. How did you determine the SWR?
Were you driving a twin lead transmission line going to your
mismatched antenna, or a coax?

73's
Richard Clark, KB7QHC