On 14 mayo, 18:06, Cecil Moore wrote:
On May 13, 5:51*pm, Wimpie wrote:

When a source behaves like a 50 Ohms source, it will not rereflect the
reflected power back to the load. In other words the forward power
generated by the source will not change when changing the load. The
reflected power that is absorbed by the source may result in increase
or reduction of power dissipation of the active device. It all depends
on the change of the integral of V*I (for the active device).

Where is the reflected power that is "absorbed by the source"
dissipated? How can a "reduction of power dissipation" be the result?

Hello Cecil,

Ever looked at the plate current indicator during tuning? I think you
did. Here you see that plate loading affects plate current, hence DC
input power to the PA.

Just convert the voltage reflection coefficient (as seen by the PA) to
an impedance and work this back through the Pi-filter to the plate.
Now you can forget the whole reflected power story.

What happens now depends on what the plate sees. Due to internal
feedback (grid capacitances) and/or voltage saturation issues, both
plate dissipation and average plate current (so PA input power) may
change. Same is valid for solid state PAs also. Changing the load
changes device dissipation and DC input power.

I mentioned this in this thread and a similar thread about a year
ago.

your question directly: reduction in power dissipation can occur when
the impedance as seen from the active devices increases. It will go
into voltage saturation, and this leads to a reduction in average (DC)
current and integral of V*I product. V can be plate voltage, I can be
plate current.


The answer is that if the power dissipation of the source decreases
when the reflected power is incident, then destructive interference
energy is being redistributed back toward the load. That destructive
interference energy includes some (or all) of the reflected energy.

If some of the reflected wave re-reflects at the cable-PA transition,
you will notice this very likely at the forward power indicator. It
also shows that in that case the output impedance of the PA isn't 50
Ohms (as is the case with many PAs).


The forward wave is associated with EfwdxHfwd power. The reflected
wave is associated with ErefxHref power. The energy in those two waves
traveling in opposite directions at the speed of light in the medium
must be conserved.

For instance, when the load at the end of an ideal 1/4WL feedline is
changed to a short, the forward power is still the same and the
reflected power is equal to the forward power.

Fully agree, if the source remains a 50 Ohms source under this massive
mismatch.

Yet, the source
dissipation has been reduced to zero even though the reflected power
"is absorbed by the source".

I do generally not agree on this one, please convert your Thevenin
source to a Norton source. Though the source doesn't deliver any power
to the shorted quarter wave line (as its input impedance is infinite),
there is still dissipation in the source.

Practical issue (very related to PAs): when you put your shorted
quarter wave line directly to a PA with a type of L match (as is used
in low voltage solid state PAs), the L match will "convert" this
impedance to a very low value, resulting in an near AC short at the
drain or collector. This will greatly increase the power dissipation
in the BJT or MOSFET.

In the source resistor, the forward
current is equal in magnitude and 180 degrees out of phase with the
reflected current and the net current superposes to zero. That's the
definition of "total destructive interference" and according to the
conservation of energy principle, the energy involved has to go
somewhere - so it is redistributed in the only possible direction -
back toward the load associated with the "total destructive
interference" that occurs in the 50 ohm source resistor.

The main point to this part of the discussion is to realize that the
forward and reflected wave energy is flowing through the 50 ohm source
resistor because the physical reflection coefficient looking back into
the 50 ohm source from a 50 ohm feedline is zero.

--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

For the experimenters:
Make this setup: PA – forward/reflected power indicator – tuner – 50
Ohms load. Have a means to measure the PA input current.

Notice DC input current and forward power when the tuner is not used
(bypassed) and don't touch PA settings anymore. Now activate the tuner
and show slight mismatch to your PA (with your tuner). Notice the PA
input current and forward power reading. You may need to turn of ALC
or other means that automatically change the drive level of the actual
power device (valve or semiconductor).

Very likely both PA input current and forward power reading will
change. You may do this at various power levels and not the
difference.

With kind regards,

Wim
PA3DJS
www.tetech.nl

On May 14, 1:56*pm, Wimpie wrote:
Now you can forget the whole reflected power story.

Which is exactly what most people do - just forget the problem and
hope it will go away. Some have said just to calculate the voltage/
current and the energy (or power) will take care of itself. What this
requires is ignorance of the well-known laws of physics from the field
of optics. When one understands the laws of physics that effect the
boundary conditions for RF waves including the role of interference,
all of the component energies can be easily tracked through a system
in the same manner the track irradiance through an optical system
because they cannot measure the voltages and currents. Method#1 is the
way most RF engineers analyze a problem. Method#2 is the way optical
physicists are forced to solve the problem.

Method#1: Calculate the voltages and currents and superpose them. (One
way is using s-parameter voltage equations.) Calculate power at the
end of this process.

Method#2: Calculate the irradiances (power densities) and merge them
together using the irradiance equations. The irradiance equations are
what one gets when one squares the s-parameter normalized voltage
equations.

When you say "forget reflected power", you are asking people to forget
the ExH power contained in every EM wave. You are asking people to
forget that EM waves must necessarily travel at the speed of light in
the medium, i.e. component EM wave energy cannot stand still in
standing waves. You are asking people to forget the conservation of
energy principle.

Earlier I suggested that we concentrate on a simpler lossless example
that doesn't involve the source impedance at all. It doesn't matter
what the source impedance is. All that matters is that the source is
delivering 70.707 volts to the 50 ohm feedline. Here it is again:

100W--50ohm--+--1/4WL 100ohm--200 ohm load

During steady-state, Pfwd is 100w and Pref is 0 watts on the 50 ohm
line. Pfwd is 112.5w and Pref is 12.5w on the 100 ohm line. Exactly
what happens at the Z0-match point '+' to reverse the flow of that
12.5w reflected power wave? (There are no RF source or DC supply
arguments behind which to hide.) The reflected voltage reflection
coefficient at point '+' is -0.3333, i.e. the negative of the forward
voltage reflection coefficient at the load.

The ExH power (Poynting vector) in the reflected wave from the load is
12.5 watts. It is necessarily traveling toward the source at the speed
of light in the Z0=100 ohm medium. It necessarily contains measurable
real-world energy and momentum. Exactly what happens at point '+' to
reverse the momentum and energy flow of that reflected EM wave? All
the arguments in the world about what happens in the source has not
answered that question and this example doesn't involve the source at
all.

I do generally not agree on this one, please convert your Thevenin
source to a Norton source.

Sorry, that particular source was not my choice but was instead W7EL's
choice in his food-for-thought series. He specifically states he is
NOT talking about any equivalent circuit but is instead talking about
a good approximation to a real-world source with a 50 ohm source
resistance.

I do not see how the above energy flow problem/example can be solved
without an understanding of the principles of interference (covered in
section 4.3 of "Reflections" by Walter Maxwell. If you can adequately
answer the questions without any direct or implied reference to
interference, you will be the first I know of on ham newsgroups to
have done so. Hint: squaring the s-parameter equations yields those
interference power terms. If you choose not to respond to this simpler
example, I will understand your reasons based on "forgetting reflected
power".
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

On 5/15/2011 3:42 PM, Cecil Moore wrote:
On May 14, 1:56 pm, wrote:
Now you can forget the whole reflected power story.

Which is exactly what most people do - just forget the problem and
hope it will go away. Some have said just to calculate the voltage/
current and the energy (or power) will take care of itself. What this
requires is ignorance of the well-known laws of physics from the field
of optics. When one understands the laws of physics that effect the
boundary conditions for RF waves including the role of interference,
all of the component energies can be easily tracked through a system
in the same manner the track irradiance through an optical system
because they cannot measure the voltages and currents. Method#1 is the
way most RF engineers analyze a problem. Method#2 is the way optical
physicists are forced to solve the problem.

Method#1: Calculate the voltages and currents and superpose them. (One
way is using s-parameter voltage equations.) Calculate power at the
end of this process.

Method#2: Calculate the irradiances (power densities) and merge them
together using the irradiance equations. The irradiance equations are
what one gets when one squares the s-parameter normalized voltage
equations.

When you say "forget reflected power", you are asking people to forget
the ExH power contained in every EM wave. You are asking people to
forget that EM waves must necessarily travel at the speed of light in
the medium, i.e. component EM wave energy cannot stand still in
standing waves. You are asking people to forget the conservation of
energy principle.

Earlier I suggested that we concentrate on a simpler lossless example
that doesn't involve the source impedance at all. It doesn't matter
what the source impedance is. All that matters is that the source is
delivering 70.707 volts to the 50 ohm feedline. Here it is again:

100W--50ohm--+--1/4WL 100ohm--200 ohm load

It is a 100 ohm feedline. The impedance at the + point looking toward
the load is 50 ohms. The 100 W source will see the first 50 ohms and
then the impedance at the + point. The total impedance seen by the
source is therefore 100 ohms. A 100 W source will deliver 100V to this
100 ohm combination and 50V of it will appear at the + point. Where does
the 70.707V come from?

John

On May 15, 7:54*pm, John KD5YI wrote:
100W--50ohm--+--1/4WL 100ohm--200 ohm load

It is a 100 ohm feedline. The impedance at the + point looking toward
the load is 50 ohms. The 100 W source will see the first 50 ohms and
then the impedance at the + point. The total impedance seen by the
source is therefore 100 ohms. A 100 W source will deliver 100V to this
100 ohm combination and 50V of it will appear at the + point. Where does
the 70.707V come from?

I said: "All that matters is that the source is delivering 70.707
volts to the 50 ohm feedline." That would be at the source SO-239
terminal to PL-259 50 ohm coax junction. I did *NOT* say there is
70.7v on the Z0=100 ohm feedline. There is indeed 100 volts on the
Z0=100 ohm line.

The source is Z0 matched to 50 ohms and is
*delivering 70.707 volts and 1.414 amps to the 50 ohm coax*.
70.7/1.414=50 ohms. 70.7*1.414=100 watts.

Here is a better diagram:

http://www.w5dxp.com/enfig1.gif

Pfor1=100w, Pref1=0, Z01=50 ohms, Z02=100 ohms and 1/4WL, Load=200
ohms. Everything except the load is lossless.

I apologize if my one line diagram was confusing. It is extremely
difficult to draw ASCII schematics using a proportional font.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

On 5/16/2011 6:54 AM, Cecil Moore wrote:
On May 15, 7:54 pm, John wrote:
100W--50ohm--+--1/4WL 100ohm--200 ohm load

It is a 100 ohm feedline. The impedance at the + point looking toward
the load is 50 ohms. The 100 W source will see the first 50 ohms and
then the impedance at the + point. The total impedance seen by the
source is therefore 100 ohms. A 100 W source will deliver 100V to this
100 ohm combination and 50V of it will appear at the + point. Where does
the 70.707V come from?

I said: "All that matters is that the source is delivering 70.707
volts to the 50 ohm feedline." That would be at the source SO-239
terminal to PL-259 50 ohm coax junction. I did *NOT* say there is
70.7v on the Z0=100 ohm feedline. There is indeed 100 volts on the
Z0=100 ohm line.

The source is Z0 matched to 50 ohms and is
*delivering 70.707 volts and 1.414 amps to the 50 ohm coax*.
70.7/1.414=50 ohms. 70.7*1.414=100 watts.

Here is a better diagram:

http://www.w5dxp.com/enfig1.gif

Pfor1=100w, Pref1=0, Z01=50 ohms, Z02=100 ohms and 1/4WL, Load=200
ohms. Everything except the load is lossless.

I apologize if my one line diagram was confusing. It is extremely
difficult to draw ASCII schematics using a proportional font.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Okay, so your source really looks like a a 100 W generator with no
internal impedance but which feeds a 50 ohm line internally. Since the
200 ohm load appears as 50 ohms when fed with a 100 ohm 1/4WL line,
there is a match and all the power appears at the load. In that case I
agree with your 70.7 volts since 100 W into 50 ohms will result in that
voltage.

But I have no need of reflections. The fact that a 200 ohm load appears
as a 50 ohm load at the input of a 1/4WL 100 ohm line takes care of all
that for me.

73,
John

On May 16, 3:42*pm, John KD5YI wrote:
On 5/16/2011 6:54 AM, Cecil Moore wrote:









On May 15, 7:54 pm, John *wrote:
100W--50ohm--+--1/4WL 100ohm--200 ohm load

It is a 100 ohm feedline. The impedance at the + point looking toward
the load is 50 ohms. The 100 W source will see the first 50 ohms and
then the impedance at the + point. The total impedance seen by the
source is therefore 100 ohms. A 100 W source will deliver 100V to this
100 ohm combination and 50V of it will appear at the + point. Where does
the 70.707V come from?

I said: "All that matters is that the source is delivering 70.707
volts to the 50 ohm feedline." That would be at the source SO-239
terminal to PL-259 50 ohm coax junction. I did *NOT* say there is
70.7v on the Z0=100 ohm feedline. There is indeed 100 volts on the
Z0=100 ohm line.

The source is Z0 matched to 50 ohms and is
*delivering 70.707 volts and 1.414 amps to the 50 ohm coax*.
70.7/1.414=50 ohms. 70.7*1.414=100 watts.

Here is a better diagram:

http://www.w5dxp.com/enfig1.gif

Pfor1=100w, Pref1=0, Z01=50 ohms, Z02=100 ohms and 1/4WL, Load=200
ohms. Everything except the load is lossless.

I apologize if my one line diagram was confusing. It is extremely
difficult to draw ASCII schematics using a proportional font.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Okay, so your source really looks like a a 100 W generator with no
internal impedance but which feeds a 50 ohm line internally. Since the
200 ohm load appears as 50 ohms when fed with a 100 ohm 1/4WL line,
there is a match and all the power appears at the load. In that case I
agree with your 70.7 volts since 100 W into 50 ohms will result in that
voltage.

But I have no need of reflections. The fact that a 200 ohm load appears
as a 50 ohm load at the input of a 1/4WL 100 ohm line takes care of all
that for me.

73,
John


Sorry John, you can't cut it without the reflections. They may be
invisible to you, but they're there doing the work. The 1/4wl line
transformer works only with reflections. If you'll give me your email
address I'll send you a graph showing how the reflections perform the
matching. The actions of the reflections from either end of the
transformer is what performs the matching function. The action of the
reflections is as necessary to the function as breathing is to life.
Believe it!!!

Walt

On May 16, 2:42*pm, John KD5YI wrote:
But I have no need of reflections. The fact that a 200 ohm load appears
as a 50 ohm load at the input of a 1/4WL 100 ohm line takes care of all
that for me.

There's no argument/discussion about that. The question is: Exactly
what happens at the 50 ohm Z0-match to reverse the direction and
momentum of the reflected wave resulting in no reflections on the 50
ohm line? Hint: it is exactly what happens with non-reflective glass
where the 1/4WL thin-film coating cancels the reflections through
destructive interference. Why not alleviate ignorance by taking
advantage of those well understood facts of EM wave physics from the
field of optics?

micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/
waveinteractions/index.html
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

On 5/16/2011 2:55 PM, Cecil Moore wrote:
On May 16, 2:42 pm, John wrote:
But I have no need of reflections. The fact that a 200 ohm load appears
as a 50 ohm load at the input of a 1/4WL 100 ohm line takes care of all
that for me.

There's no argument/discussion about that. The question is: Exactly
what happens at the 50 ohm Z0-match to reverse the direction and
momentum of the reflected wave resulting in no reflections on the 50
ohm line? Hint: it is exactly what happens with non-reflective glass
where the 1/4WL thin-film coating cancels the reflections through
destructive interference. Why not alleviate ignorance by taking
advantage of those well understood facts of EM wave physics from the
field of optics?

micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/
waveinteractions/index.html
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

By the way, what is the impedance of a constant 100W source looking back
into its short piece of internal 50 ohm line?

John