"Roy Lewallen" wrote in message
...
Richard Fry wrote:
. . .
The source impedance of most transmitters is not published even today.
If
it was, probably we wouldn't be having all of this confusion about it,
and
its effects.


Who's confused? It has no effect.

Roy Lewallen, W7EL

Hi Roy,

Going back to the Slick discussions of last winter, Was that you who made
the statement that you can have 100% re reflection from a transmitter, even
if it has a 50 Ohm output impedance? At first I thought this was all wet,
but after making some low power experiments, I am convinced it is true.

Tam/WB2TT

Tam/WB2TT wrote:
Going back to the Slick discussions of last winter, Was that you who made
the statement that you can have 100% re reflection from a transmitter, even
if it has a 50 Ohm output impedance? At first I thought this was all wet,
but after making some low power experiments, I am convinced it is true.

It is true by definition. All reflected power incident upon a
transmitter is re-reflected, by definition. Never mind that
reflected voltage can cause an over-voltage condition and/or
reflected current can cause an over-current condition *inside*
the transmitter. By definition, any reflected power dissipated
in the source after making a round trip to the load was never
generated to begin with.
--
73, Cecil http://www.qsl.net/w5dxp


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On Mon, 6 Sep 2004 10:39:54 -0400, "Tam/WB2TT" wrote:

Hi Roy,

Going back to the Slick discussions of last winter, Was that you who made
the statement that you can have 100% re reflection from a transmitter, even
if it has a 50 Ohm output impedance? At first I thought this was all wet,
but after making some low power experiments, I am convinced it is true.

Tam/WB2TT

Yep , Tam, it's correct. The internal resistance in Class B and C amps has two
parts, 1) the cathode-to-plate resistance, which is dissipative, and 2) the
non-dissipative resistance established by the V/I ratio within the pi-network
tank circuit--a high resistance at the input and a low resistance at the output.
The V/I ratio also establishes the slope of the load line. Consequently, the
reflected power reaching the network output is not absorbed, but instead adds to
the power delivered by the generator. Although powers are not generally
considered to add, they do in this case, because their respective voltage and
current phasors add. If the reflected voltage and current phasors are not in
phase with those from he source, the only result is that the source is
mismatched to the load and reduces its delivery of power. Readjusting the tuning
and loading controls brings the out-of-phase phasors in phase, establishes a
conjugate match and the source again delivers the maximum available power.

Walt, W2DU

Walter, W2DU wrote:
"The internal resistance in Class B and C amps has two parts, 1) the
cathode-to-plate resistance, which is dissipative, and 2) the
non-dissipative resistance established by the V/I ratio within the
pi-network tank circuit -- a high resistance at the input and a low
resistance at the output."

It`s true that a parallel resonant circuit constructed of ideal
inductance and capacitance has high (infinite) impedance and no loss.
The configuration of the high impedance in series with the load limits
output. But, we use imperfect components and we seek a limiting
impedance equat to the 50-ohm load, not an infinite impedance.

My take on the non-dissipative impedance is that it comes from the
switched-off time of the Class B and C amps. During this time in each
cycle, no current flows through the amp to cause loss. Likewise, there
is no current from the amp (actually it is a switch operating on and off
at a radio frequency) to the load or tank circuit. The tank circuit
cleans up the pulse mess, filling gaps in the RF cycle.

Best regards, Richard Harrison, KB5WZI

On Mon, 6 Sep 2004 17:49:00 -0500, (Richard
Harrison) wrote:

It`s true that a parallel resonant circuit constructed of ideal
inductance and capacitance has high (infinite) impedance and no loss.

Actually, if it were isolated in museum glass case perhaps, but the
loaded Q, by design, is less than 20, often less than 10, and in the
case illustrated by Mendenhall's FM transmitter = 2. Loss comes to it
by nature as a practical necessity of moving power (transmitting) to
the æther.

The configuration of the high impedance in series with the load limits
output. But, we use imperfect components and we seek a limiting
impedance equat to the 50-ohm load, not an infinite impedance.

The impedance transform of coupling that changed from plate resistance
to load resistance is quite capable of reciprocal transformation (at
least in the rigs of transistor persuasion) to allow return power from
load Z to match towards the source Z. This is the natural progression
of linear design.

My take on the non-dissipative impedance is that it comes from the
switched-off time of the Class B and C amps. During this time in each
cycle, no current flows through the amp to cause loss. Likewise, there
is no current from the amp (actually it is a switch operating on and off
at a radio frequency) to the load or tank circuit. The tank circuit
cleans up the pulse mess, filling gaps in the RF cycle.

By the nature of reciprocity, loss must follow from reflected power
returning through it to a known dissipater, the cathode-plate. If
that cathode-plate is available only intermittently during cycles,
reciprocity is fully functioning through the same available flywheel
of this loaded Tank.

All should notice that if this scenario is NOT working as described,
the caloric demands of reflected power would resolve in a high energy
arc along the way (no one has ever experienced this?). This may
displace WHERE the dissipation occurs, but it does not render
reflected power an inert concept.

All such descriptions can be described in a bulk component model, it
is still the same heat (only the remote arcing becomes an enigma some
would sweep under the rug).

73's
Richard Clark, KB7QHC

Richard Clark wrote:
"This may displace WHERE the dissipation occurs, but it does not render
reflected power an inert concept."

No it doesn`t, but it doesn`t require its acceptance back in the
transmitter either.

I`ll repost my earlier posting as near as I can reconstruct it. It seems
to be lost in cyberspace.

Walter Maxwell, W2DU wrote:
"Consequently, the reflected power reaching the network output is not
absorbed, but instead adds to the power delivered by the generator."

My explanation for the above, which is my observation too, is that an
energy wave experiences a phase reversal between the volts and amps
which it will generate after reflection. That fact makes Bird`s
directional coupler in its wattmeter work.

The transmitter`s output isn`t receptive and won`t absorb a wave that
produces out-of-phase volts and amps, so the reflected wave is
re-reflected from the transmitter, placing its amps and volts back
in-phase.

The newly minted RF and the twice reflected RF are similar, both having
their volts and amps in-phase. So, the similar RF constituents merge to
have a go at the reflection point.

Best regards, Richard Harrison, KB5WZI

On Mon, 6 Sep 2004 23:15:24 -0500, (Richard
Harrison) wrote:

The newly minted RF and the twice reflected RF are similar, both having
their volts and amps in-phase.

Hi Richard,

In a reality of 359 other possible phase angles, how does a
transmitter happen to always be "in-phase" to any reflection?

73's
Richard Clark, KB7QHC

Richard Clark wrote:
"In a reality of 359 other possible phase angles, how does a transmitter
happen to always be in-phase to any reflection?"

Connect any generator to any resistor, and current in the resistor is
in-phase with the applied voltage. The Zo of the common transmission
line is a reasonably good resistance. At radio frequencies, Zo is
independent of frequency.

The current in the incident wave is always in-phase with the voltage
applied to a transmission line. The current in the reflected wave is
always 180-degrees out-of-phase with the reflected voltage. It makes no
difference which was inverted by reflection, the volts or tha amps, one,
and only one, of them was flipped upside down. The transmission line can
and does handle the reflected wave.

Standing waves display interference between incident and reflected waves
whiich ideally have in-phase and out-of-phase constituents.

The fact that the Bird wattmeter works is evidence that the theory is
correct at least until a better theory replaces existing theory.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
The newly minted RF and the twice reflected RF are similar, both having
their volts and amps in-phase. So, the similar RF constituents merge to
have a go at the reflection point.

Wonder why we have protection circuitry in transmitters?
The superposed forward voltage and reflected voltage can damage an
unprotected transmitter. The superposed forward current and reflected
current can cause over heating in an unprotected transmitter.

The transmitter sees whatever impedance it sees and that impedance
can be highly reactive. The superposed voltage can be high or low.
The superposed current can be high or low. The phase between the
superposed voltage and superposed current can have lots of values.

Just a for instance - assume the transmitter is putting out 70.7v
in phase with 1.4a at zero deg. The arriving reflected wave is 50v
at 90 deg and 1.0a at -90 deg. The load seen by the transmitter is
86.6v at 35 deg and 1.72a at -35 deg. Over voltage and over current
exist at the transmitter output. The forward power is 100w and the
reflected power is 50w. The net power being delivered to the reactive
"load" seen by the transmitter is 86.6*1.72*cos(70.4) = 50w.

The math model is trying to dictate reality. It is supposed to be
exactly the opposite. There is no magic barrier that automatically
rejects reflected energy from a transmitter. Reflected energy
arriving at the transmitter can drastically alter the impedance
away from the designed-for load impedance. The transmitter sees
one of the transformed impedances that exists on the SWR circle.

Note that the problem disappears in a matched system where reflected
energy is not allowed to reach the transmitter.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil, W5DXP wrote:
"The superposed forward voltage and reflected voltage can damage an
unprotected transmitter."

To do so, they would be in-phase and not out-of-phase. Entirely possible
if the transmission line is the right length.

If the reflected volts are in the same phase as the newly munted volts,
which are larger? With a reflection coefficient of 1, and a lossless
line, the open-circuit value of transmitter volts would face some lower
value of line volts on opposite ends of the internal impedance of the
transmitter. Which way does the current flow?

Theory is that it flows from the higher to the lower. That is, from the
transmitter to the line.

Best regards, Richard Harrison, KB5WZI