On 7 jun, 01:51, walt wrote:
On Jun 6, 6:36*pm, Wimpie wrote:



On 6 jun, 19:00, Richard Clark wrote:

On Sun, 6 Jun 2010 07:42:31 -0700 (PDT), Wimpie
wrote:

However a PA is not a 50 Ohms source

Hi Wimpie,

You say this like others, with the air of "knowing." *However, when I
ask in response of those who "know" what the PA is NOT, what IS it?

Give me the Z value of your transmitter. *Specify all initial
conditions.

We have had lengthy correspondence with Walt Maxwell's very rigorously
measured Kenwood TS830s that demonstrates a Z of 50 Ohms, or nearly
that as is practicable (say +/- 20%); and yet your voice was missing
from this discussion with evidence to the contrary.

73's
Richard Clark, KB7QHC

Hello Richard,

Walt did respond and did a solid statement regarding the amplifier
with matching section to obtain maximum output (so I know the
conditions) *This is the point where the tube/transistor is at the
edge of current/voltage saturation.

At that operating point the output impedance is 50 Ohms (for small
load variations). When you change the load significantly (or change
drive level), current or voltage saturation will dominate, hence the
output impedance is no longer 50 Ohms. * This is also the reason that
PA intermodulation may occur in close spaced transmitters where some
power from transmitter A enters the amplifier of transmitter B and
vice versa. This also proves that there is no linear 50 Ohms output
impedance.

Best regards,

Wim
PA3DJSwww.tetech.nl

Hello Walt,

I think the problem is in the initial conditions that we use as
starting point. Your starting point is matched to maximum power output
in all cases, also under bad antenna VSWR. My starting point was a
fixed tuned amplifier, made for 50 Ohms and experiencing large
mismatch (50W reflected, 100W forward).

Sorry *Wim, I can't agree with some of your statements in your last
post. Concerning maximum output, I will agree that saturation will
occur when the minimum of the peak AC plate voltage equals the peak AC
grid voltage. This condition occurs when the tube is delivering its
total maximum possible power. However, when the grid drive level is
less than that which brings the plate-voltage minimum down to the grid-
voltage level, saturation does not occur.

In that case the final tube will more or less behave like a current
source (LF appraoch, there will be feedback via capacitance), so the
impedance will change from the value for maximum power.

There are two saturation conditions: voltage saturation, this lowers
the plate impedance, current saturation (below maximum voltage swing),
this increases the plate impedance (especially for pentodes).

We start with a maximim power matched situation (so ZL = Zout) and
don't touch the tuning.
When you now change the load (for example VSWR =2), you will probably
have a strong voltage saturation condition or strong current
saturation condition, so voltage source or current source behavior
dictates, but not both. Therefore the output impedance at the plate
will rise (current saturation) or decrease (voltage saturation). For
triodes the effect is less then for pentodes as the transition for
current saturation to voltage saturation is relatively smooth for
triodes.

In addition, when the pi-network has been adjusted to deliver all the
available power at some drive level less than the maximum possible
power, the source resistance at the output of the pi-network will be
exactly equal to its load resistance, not somewhat higher or lower.
This follows from the Maximum Power Transfer Theorem. This is not
speculation, but proof determined by data from many, many
measurements.

Fully agree with this, but is this always the case under practical
circumstance?

When you change one of the paramers (drive level or VSWR, but not the
matching), you will divert from the optimum setting, hence impedance
will change. You can try the VSWR dependence by measuring the forward
power under varying VSWR. I am sure it will change (keeping drive
level and pi filter matching the same).

Most solid state amplifiers do not have the possibility of matching
when you change VSWR, so with bad VSWR, you will be in the current or
voltage saturating range of the active device and your output
impedance will change.

Virtually all high efficient amplifiers work in voltage saturated mode
and are therefore not operated at maximum available power, therefore
their output impedance doens't match the expected load.

Except for CW, most SSB amplifiers run in the current saturated mode
(so below maximum voltage swing), so they do not work at their maximum
power point given the instantaneous drive level (or you need to
modulate your supply also).

In real world the situation is more complex because of phase changes
in VSWR and parasitic feedback that may change the gain of the active
device (worst case spoken, you may get instability). I discovered
(when I was younger) that with a class C fixed tuned amplifier, small
changes in VSWR resulted in change of forward power, but significant
change in DC supply current (and heatsink temperature). So what
happens with the reflected power depends on many factors.


Walt, W2DU

Wim
PA3DJS
www.tetech.nl

On Jun 6, 9:22*am, "JC" wrote:
Basic question (at least for me) for a very poor antenna matching :
-100 w reach the antenna and 50 w are radiated.
- 50 w are "reflected", what is their fate ?
Are they definitely lost for radiation and just heat the line, the final.....

What happens to the 50 joules/second of reflected energy depends upon
the phasing between the source wave and the reflected wave at the
source impedance. What most amateurs don't understand is there are
two mechanisms that can redistribute reflected energy back toward the
antenna. Those mechanisms are a re-reflection based on the physical
reflection coefficient (what RF engineers understand) and wave
interaction resulting in constructive/destructive interference (what
most RF engineers don't seem to understand) because, unlike optical
physicists, have not been forced to follow the energy flow.

If the reflected wave arrives 180 degrees out of phase with the source
wave, the two waves undergo destructive interference and all of the
reflected power is redistributed as constructive interference energy
back toward the antenna. This is what happens at the Z0-match
established at the input of an antenna tuner.

Thus your conditions of 100w forward power and 50w reflected power
could be accomplished with a 50w matched source.

If the reflected wave arrives 90 degrees out of phase with the source
wave, there is zero interference and the reflected power is dissipated
in the source resistor (in a source with a source resistor).

If the reflected wave arrives in phase with the source wave, all of
the reflected power and more than 1/2 of the source power can be
dissipated in the source resistor.

Such knowledge is old hat for optical physicists who don't have the
luxury of measuring voltages in light waves. They rely on a power
density (irradiance) equation.

Ptot = P1 + P2 + 2*SQRT(P1*P2)*cos(A)

where A is the phase angle between the electric fields of two waves.
The last term is called the "interference term" and that value is what
most amateurs are missing in their energy analysis. If the sign of the
interference term is negative, the interference is destructive. If the
sign of the interference term is positive, the interference is
constructive.
--
73, Cecil, w5dxp.com

On Jun 6, 3:13*pm, Cecil Moore wrote:
On Jun 6, 9:22*am, "JC" wrote:

Basic question (at least for me) for a very poor antenna matching :
-100 w reach the antenna and 50 w are radiated.
- 50 w are "reflected", what is their fate ?
Are they definitely lost for radiation and just heat the line, the final....

What happens to the 50 joules/second of reflected energy depends upon
the phasing between the source wave and the reflected wave at the
source impedance. *What most amateurs don't understand is there are
two mechanisms that can redistribute reflected energy back toward the
antenna. Those mechanisms are a re-reflection based on the physical
reflection coefficient (what RF engineers understand) and wave
interaction resulting in constructive/destructive interference (what
most RF engineers don't seem to understand) because, unlike optical
physicists, have not been forced to follow the energy flow.

If the reflected wave arrives 180 degrees out of phase with the source
wave, the two waves undergo destructive interference and all of the
reflected power is redistributed as constructive interference energy
back toward the antenna. This is what happens at the Z0-match
established at the input of an antenna tuner.

Thus your conditions of 100w forward power and 50w reflected power
could be accomplished with a 50w matched source.

If the reflected wave arrives 90 degrees out of phase with the source
wave, there is zero interference and the reflected power is dissipated
in the source resistor (in a source with a source resistor).

If the reflected wave arrives in phase with the source wave, all of
the reflected power and more than 1/2 of the source power can be
dissipated in the source resistor.

Such knowledge is old hat for optical physicists who don't have the
luxury of measuring voltages in light waves. They rely on a power
density (irradiance) equation.

Ptot = P1 + P2 + 2*SQRT(P1*P2)*cos(A)

where A is the phase angle between the electric fields of two waves.
The last term is called the "interference term" and that value is what
most amateurs are missing in their energy analysis. If the sign of the
interference term is negative, the interference is destructive. If the
sign of the interference term is positive, the interference is
constructive.
--
73, Cecil, w5dxp.com

as in the other thread, what is the mechanism of that 'interaction'
between waves? i contend there can be no 'interaction' between
forward and reflected waves if the device is linear. so in an ideal
case of a voltage or current source and ideal source resistance there
is no interaction, it is reflected by and/or absorbed in the source
depending on the impedance of the line and source. if the source is
not linear then you would have to calculate the effect of the sum of
the voltages or currents at the source to determine the effect.

On Sun, 6 Jun 2010 11:01:20 -0700 (PDT), K1TTT wrote:

as in the other thread, what is the mechanism of that 'interaction'
between waves? i contend there can be no 'interaction' between
forward and reflected waves if the device is linear. so in an ideal
case of a voltage or current source and ideal source resistance there
is no interaction, it is reflected by and/or absorbed in the source
depending on the impedance of the line and source.

There is no mixing as in multiplication of waveforms. Perhaps I can
offer a simple analogy. Instead of two AC waveforms (forward and
reflected), use a DC equivalent. Start with two DIFFERENT batteries.
Connect the two negative ends together and declare that to be ground.
Connect a resistor between the positive terminals. The two voltages
most certainly "interact" across the resistor, resulting in the
current and power being proportional to the difference between the two
battery voltages. Nothing in this crude example is non-linear, so
there's no need for mixing in order to get interaction.

Similarly, the coax cable acts much in the same way. The two
batteries are replaced by the incident and reflected signals. At any
time, or position on the transmission line, the model can be frozen
and the instantaneous voltages and currents be calculated.

if the source is
not linear then you would have to calculate the effect of the sum of
the voltages or currents at the source to determine the effect.

If the source (or load) is non-linear, then the waveforms seen on the
transmission line will be distorted. This is unlikely because we
usually don't install diodes in antennas, or build HF amplifiers with
substantial non-linearities (i.e. distortion).



--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

On Jun 6, 8:55*pm, Jeff Liebermann wrote:
On Sun, 6 Jun 2010 11:01:20 -0700 (PDT), K1TTT wrote:
as in the other thread, what is the mechanism of that 'interaction'
between waves? *i contend there can be no 'interaction' between
forward and reflected waves if the device is linear. *so in an ideal
case of a voltage or current source and ideal source resistance there
is no interaction, it is reflected by and/or absorbed in the source
depending on the impedance of the line and source. *

There is no mixing as in multiplication of waveforms. *Perhaps I can
offer a simple analogy. *Instead of two AC waveforms (forward and
reflected), use a DC equivalent. *Start with two DIFFERENT batteries.
Connect the two negative ends together and declare that to be ground.
Connect a resistor between the positive terminals. *The two voltages
most certainly "interact" across the resistor, resulting in the
current and power being proportional to the difference between the two
battery voltages. *Nothing in this crude example is non-linear, so
there's no need for mixing in order to get interaction.

it is very misleading to try to make a lumped circuit analogy out of a
transmission line problem. the first step in any circuits 101 course
would be to simplify the two batteries into one then solve for the
single simple current across the resistor... no waves, no back and
forth, no interaction between batteries, just a single current and
voltage.


Similarly, the coax cable acts much in the same way. *The two
batteries are replaced by the incident and reflected signals. *At any
time, or position on the transmission line, the model can be frozen
and the instantaneous voltages and currents be calculated. *

they can be, but what does it prove? i want to freeze it just as the
first wave gets to the load, there is no reflected wave yet, so what
good is that? in order to use our common equations there are many
unstated but necessary assumptions. the most restrictive of which is
that we normally only solve for the sinusoidal steady state waves...
this requires that a long time(in terms of the length of the line) has
passed since the source was energized, that it is a single frequency
pure sine wave, and that nothing is changing in time in the load or
line characteristics. given all that there is no need for
instantaneous values, sure they can be calculated or measured, but
they are of no use in most cases.


if the source is
not linear then you would have to calculate the effect of the sum of
the voltages or currents at the source to determine the effect.

If the source (or load) is non-linear, then the waveforms seen on the
transmission line will be distorted. *This is unlikely because we
usually don't install diodes in antennas, or build HF amplifiers with
substantial non-linearities (i.e. distortion).

and obviously if either one is non-linear then all the simple
equations can be abandoned and a more complete analysis must be done.


--
Jeff Liebermann * *
150 Felker St #D * *http://www.LearnByDestroying.com
Santa Cruz CA 95060http://802.11junk.com
Skype: JeffLiebermann * * AE6KS * *831-336-2558

On Jun 6, 4:55*pm, Jeff Liebermann wrote:
On Sun, 6 Jun 2010 11:01:20 -0700 (PDT), K1TTT wrote:
as in the other thread, what is the mechanism of that 'interaction'
between waves? *i contend there can be no 'interaction' between
forward and reflected waves if the device is linear. *so in an ideal
case of a voltage or current source and ideal source resistance there
is no interaction, it is reflected by and/or absorbed in the source
depending on the impedance of the line and source. *

There is no mixing as in multiplication of waveforms. *Perhaps I can
offer a simple analogy. *Instead of two AC waveforms (forward and
reflected), use a DC equivalent. *Start with two DIFFERENT batteries.
Connect the two negative ends together and declare that to be ground.
Connect a resistor between the positive terminals. *The two voltages
most certainly "interact" across the resistor, resulting in the
current and power being proportional to the difference between the two
battery voltages. *Nothing in this crude example is non-linear, so
there's no need for mixing in order to get interaction.

Similarly, the coax cable acts much in the same way. *The two
batteries are replaced by the incident and reflected signals. *At any
time, or position on the transmission line, the model can be frozen
and the instantaneous voltages and currents be calculated. *

if the source is
not linear then you would have to calculate the effect of the sum of
the voltages or currents at the source to determine the effect.

If the source (or load) is non-linear, then the waveforms seen on the
transmission line will be distorted. *This is unlikely because we
usually don't install diodes in antennas, or build HF amplifiers with
substantial non-linearities (i.e. distortion).

--
Jeff Liebermann * *
150 Felker St #D * *http://www.LearnByDestroying.com
Santa Cruz CA 95060http://802.11junk.com
Skype: JeffLiebermann * * AE6KS * *831-336-2558

Looks like I should step in here, as the answer to this question is
the main theme in the book Reflections--Transmission Lines and
Antennas, the first edition published in 1990, the second in 2001, and
the third in just this past month of May, released at Dayton.

The notion that ANY reflected power enters the source, such as an RF
power amp using tubes and a pi-network, is FALSE!!! The output source
resistance of these amps is non-dissipative, and totally re-reflects
all reflected power from a mismatched antenna. The same is true when
using an antenna tuner. When correctly adjusted the antenna tuner
totally reflects all reflected power, resulting in a conjugate match
at the antenna-coax mismatch, canceling all reactances in the system
to zero, thus tuning the non-resonant antenna to resonance. This
action if fundamental, and has been a misunderstood myth for
centuries.

For proof of the above statements I invite you to read Chapter 23 of
Reflections, which you can find on my web page at www.w2du.com. Click
on 'Read Chapters from Reflections 2' and then click on Chapter 23.

In addition, Chapter 19 gives more insight, and the addition to
Chapter 19 can be found by clicking on 'Preview Chapters from
Reflections 3'. The addition shows measured data proving that the
output source impedance of the RF amp is the conjugate of the complex
load impedance when the pi-network is adjusted to deliver all the
available power at a given level of grid drive.

Furthermore, a completely revised edition of Chapter 23 and the total
Chapter 19 appear in Reflections 3, which is now available from CQ
Magazine.

Walt Maxwell, W2DU

On Jun 6, 5:27*pm, walt wrote:
On Jun 6, 4:55*pm, Jeff Liebermann wrote:



On Sun, 6 Jun 2010 11:01:20 -0700 (PDT), K1TTT wrote:
as in the other thread, what is the mechanism of that 'interaction'
between waves? *i contend there can be no 'interaction' between
forward and reflected waves if the device is linear. *so in an ideal
case of a voltage or current source and ideal source resistance there
is no interaction, it is reflected by and/or absorbed in the source
depending on the impedance of the line and source. *

There is no mixing as in multiplication of waveforms. *Perhaps I can
offer a simple analogy. *Instead of two AC waveforms (forward and
reflected), use a DC equivalent. *Start with two DIFFERENT batteries.
Connect the two negative ends together and declare that to be ground.
Connect a resistor between the positive terminals. *The two voltages
most certainly "interact" across the resistor, resulting in the
current and power being proportional to the difference between the two
battery voltages. *Nothing in this crude example is non-linear, so
there's no need for mixing in order to get interaction.

Similarly, the coax cable acts much in the same way. *The two
batteries are replaced by the incident and reflected signals. *At any
time, or position on the transmission line, the model can be frozen
and the instantaneous voltages and currents be calculated. *

if the source is
not linear then you would have to calculate the effect of the sum of
the voltages or currents at the source to determine the effect.

If the source (or load) is non-linear, then the waveforms seen on the
transmission line will be distorted. *This is unlikely because we
usually don't install diodes in antennas, or build HF amplifiers with
substantial non-linearities (i.e. distortion).

--
Jeff Liebermann * *
150 Felker St #D * *http://www.LearnByDestroying.com
Santa Cruz CA 95060http://802.11junk.com
Skype: JeffLiebermann * * AE6KS * *831-336-2558

Looks like I should step in here, as the answer to this question is
the main theme in the book Reflections--Transmission Lines and
Antennas, the first edition published in 1990, the second in 2001, and
the third in just this past month of May, released at Dayton.

The notion that ANY reflected power enters the source, such as an RF
power amp using tubes and a pi-network, is FALSE!!! The output source
resistance of these amps is non-dissipative, and totally re-reflects
all reflected power from a mismatched antenna. The same is true when
using an antenna tuner. When correctly adjusted the antenna tuner
totally reflects all reflected power, resulting in a conjugate match
at the antenna-coax mismatch, canceling all reactances in the system
to zero, thus tuning the non-resonant antenna to resonance. This
action if fundamental, and has been a misunderstood myth for
centuries.

For proof of the above statements I invite you to read Chapter 23 of
Reflections, which you can find on my web page atwww.w2du.com. Click
on 'Read Chapters from Reflections 2' and then click on Chapter 23.

In addition, Chapter 19 gives more insight, and the addition to
Chapter 19 can be found by clicking on 'Preview Chapters from
Reflections 3'. The addition shows measured data proving that the
output source impedance of the RF amp is the conjugate of the complex
load impedance when the pi-network is adjusted to deliver all the
available power at a given level of grid drive.

Furthermore, a completely revised edition of Chapter 23 and the total
Chapter 19 appear in Reflections 3, which is now available from CQ
Magazine.

Walt Maxwell, W2DU

Forgot to mention that the output of the RF power amp is LINEAR, even
though the input is non-linear. The reason is that the the pi-network
tank circuit is not only an impedance transformer, it's an energy-
storage device that isolates the output from the input. The linearity
of the output is indicated by the sinusoidal shape of the output wave,
and that the voltage and current are in phase when the load impedance
is resistive.

Walt, W2DU

On Jun 6, 1:01*pm, K1TTT wrote:
as in the other thread, what is the mechanism of that 'interaction'
between waves? *i contend there can be no 'interaction' between
forward and reflected waves if the device is linear.

David, you are preaching to the choir. I explained before that there
is NO interaction between forward and reflected waves because they are
traveling in different directions. The only time that coherent,
collimated waves can interact is when they are traveling in the same
direction in a transmission line.

Keith's argument requires that forward waves and reflected waves
interact. I say they cannot interact in a constant Z0 environment. You
say they cannot interact. We are on the same side of this argument.

The mechanism of the interaction of two coherent, collimated waves
traveling in the same direction is that the superposition process is
irreversible. The source photons and the reflected photons are
indistinguishable.
--
73, Cecil, w5dxp.com

On Jun 7, 5:00*pm, Cecil Moore wrote:
On Jun 6, 1:01*pm, K1TTT wrote:

as in the other thread, what is the mechanism of that 'interaction'
between waves? *i contend there can be no 'interaction' between
forward and reflected waves if the device is linear.

David, you are preaching to the choir. I explained before that there
is NO interaction between forward and reflected waves because they are
traveling in different directions. The only time that coherent,
collimated waves can interact is when they are traveling in the same
direction in a transmission line.

Keith's argument requires that forward waves and reflected waves
interact. I say they cannot interact in a constant Z0 environment. You
say they cannot interact. We are on the same side of this argument.

The mechanism of the interaction of two coherent, collimated waves
traveling in the same direction is that the superposition process is
irreversible. The source photons and the reflected photons are
indistinguishable.
--
73, Cecil, w5dxp.com

This is what YOU said:

What happens to the 50 joules/second of reflected energy depends upon
the phasing between the source wave and the reflected wave at the
source impedance. What most amateurs don't understand is there are
two mechanisms that can redistribute reflected energy back toward the
antenna. Those mechanisms are a re-reflection based on the physical
reflection coefficient (what RF engineers understand) and wave
interaction resulting in constructive/destructive interference (what
most RF engineers don't seem to understand) because, unlike optical
physicists, have not been forced to follow the energy flow.

There is no interaction between the waves, they may be superimposed,
and maybe their photons are indistinguishable, but there is no
'interaction'. interaction implies that one wave affects the other,
energy is transferred or fields are affected, such things occur in non-
linear media, but not in linear ones.

On Jun 7, 4:21*pm, K1TTT wrote:
There is no interaction between the waves, they may be superimposed,
and maybe their photons are indistinguishable, but there is no
'interaction'. *interaction implies that one wave affects the other,
energy is transferred or fields are affected, such things occur in non-
linear media, but not in linear ones.

Note that I was not talking about forward and reflected waves in a
constant Z0 transmission line. I am talking about the four wavefront
components that are generated at an impedance discontinuity. It has
been proven in experimentally that those waves indeed interact. In
fact, the transfer of destructive interference energy from wave
cancellation to the areas that permit constructive interference is
obviously interaction since the canceled waves disappear in their
original direction of travel. But, as the FSU web page says, they are
not annihilated - their energy components simply change direction.

How can you possibly argue that wave cancellation doesn't require wave
interaction? Those two waves completely disappear in the direction of
destructive interference. Dr. Best argued that those two waves don't
interact and continue propagating (completely devoid of energy)
forever in the direction of original travel. I asked him to prove that
his phantom waves exist but he could not. Is there such a proof
available?

At a 1/4WL thin-film coating on non-reflective glass, when the
internal reflected wave arrives with equal magnitude and 180 degrees
out of phase with the external reflection, wave cancellation occurs.
That is an *obvious* effect that one wave has on the other. Wave
cancellation is an obvious interaction. In the s-parameter equation:

b1 = s11*a1 + s12*a2 = 0

the s11*a1 wavefront has obviously interacted with the s12*a2
wavefront to accomplish wave cancellation.
--
73, Cecil, w5dxp.com