On 7/4/2015 12:32 PM, Wayne wrote:


"John S" wrote in message ...
On 7/4/2015 10:55 AM, Wayne wrote:


By the way, Wayne...

Are you aware of a companion Excel application for EZNEC called
AutoEZ? You can run many test cases in a few seconds using it. You can
find it on the EZNEC site.

It is how I generated the data I posted.

Thanks, I'll look for that. I run the old wood burning version 3.0.


BTW, Wayne, what are the dimensions of your metal roof? And how high?

"Ian Jackson" wrote in message
...
class-C PAs supposed to be around 66.%?). Also, as much power would be
dissipated in the PA stage as in the load.

I think this is easy to disprove in practice. I have an amp that is
probably class B, but it does not mater about the class. If I adjust it to
an input of 2000 watts from the DC power supply, I get out 1200 watts to a
resistive dummy load. If the above is true, I should have to input 2400
watts to the final stage. Now can someone tell me where the extra 400 watts
are comming from ? This 400 extra watts is not even counting on any loss in
the circuits.

On 7/5/2015 9:23 AM, Ian Jackson wrote:
In message ,
writes
Wayne wrote:


"Jeff Liebermann" wrote in message
...

On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
wrote:
Think of it this way, without the math. On the transmitter side of
the
network, the match is 1:1, with nothing reflected back to the
transmitter.

So you have a signal coming back from the antenna. You have a perfect
matching network, which means nothing is lost in the network. The
feedline is perfect, so there is no loss in it. The only place for
the
signal to go is back to the antenna.

Wikipedia says that if the source is matched to the line, any
reflections that come back are absorbed, not reflected back to the
antenna:

https://en.wikipedia.org/wiki/Impedance_matching
"If the source impedance matches the line, reflections
from the load end will be absorbed at the source end.
If the transmission line is not matched at both ends
reflections from the load will be re-reflected at the
source and re-re-reflected at the load end ad infinitum,
losing energy on each transit of the transmission line."

Well, I looked at that section of the writeup.
And, I have no idea what the hell they are talking about.
Looks like a good section for a knowledgeable person to edit.

If the termination matches the line impedance, there is no reflection.

Both the antenna and the source are terminations.

This is a bit difficult to visualze with an RF transmitter, but is
more easily seen with pulses.

Being essentially a simple soul, that's how I sometimes try to work out
what happening.

You'll be better off if you killfile the troll. You'll get a lot less
bad information and your life will become much easier.


The wikipedia entry is correct as written.

In the real world, the output of an amateur transmitter will seldom
be exactly 50 Ohms unless there is an adjustable network of some
sort.

I've always understood that the resistive part of a TX output impedance
was usually less than 50 ohms.

If a transmitter output impedance WAS 50 ohms, I would have thought that
the efficiency of the output stage could never exceed 50% (and aren't
class-C PAs supposed to be around 66.%?). Also, as much power would be
dissipated in the PA stage as in the load.

Fixed output amateur transmitters are a nominal 50 ohms. It can vary,
but that is due to normal variances in components, and the difference
can be ignored in real life.

But output impedance has little to do with efficiency. A Class C
amplifier can run 90%+ efficiency. It's output may be anything, i.e.
high with tubes but low with solid state. But the output impedance can
be converted to 50 ohms or any other reasonable impedance through a
matching network.

A perfect matching network will have no loss, so everything the
transmitter puts out will go through the matching network. Of course,
nothing is perfect, so there will be some loss. But the amount of loss
in a 1:1 match will not be significant.

Also, the amplifier generates the power; in a perfect world, 100% of
that power is transferred to the load. The transmitter doesn't
dissipate 1/2 of the power and the load the other 1/2. It's not like
having two resistors in a circuit where each will dissipate 1/2 of the
power.

BTW - the resistive part of the impedance is not the same as resistance.
For a simple case - take a series circuit of a capacitor, an inductor
and a 50 ohm resistor. At the resonant frequency, the impedance will
be 50 +j0 (50 ohms from the resistor, capacitive and inductive
reactances cancel). But the DC resistance is infinity. Again, a simple
example, but it shows a point.

--
==================
Remove the "x" from my email address
Jerry, AI0K

==================

On 7/5/2015 10:02 AM, Jerry Stuckle wrote:
On 7/5/2015 9:23 AM, Ian Jackson wrote:
In message ,
writes
Wayne wrote:


"Jeff Liebermann" wrote in message
...

On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
wrote:
Think of it this way, without the math. On the transmitter side of
the
network, the match is 1:1, with nothing reflected back to the
transmitter.

So you have a signal coming back from the antenna. You have a perfect
matching network, which means nothing is lost in the network. The
feedline is perfect, so there is no loss in it. The only place for
the
signal to go is back to the antenna.

Wikipedia says that if the source is matched to the line, any
reflections that come back are absorbed, not reflected back to the
antenna:

https://en.wikipedia.org/wiki/Impedance_matching
"If the source impedance matches the line, reflections
from the load end will be absorbed at the source end.
If the transmission line is not matched at both ends
reflections from the load will be re-reflected at the
source and re-re-reflected at the load end ad infinitum,
losing energy on each transit of the transmission line."

Well, I looked at that section of the writeup.
And, I have no idea what the hell they are talking about.
Looks like a good section for a knowledgeable person to edit.

If the termination matches the line impedance, there is no reflection.

Both the antenna and the source are terminations.

This is a bit difficult to visualze with an RF transmitter, but is
more easily seen with pulses.

Being essentially a simple soul, that's how I sometimes try to work out
what happening.

You'll be better off if you killfile the troll. You'll get a lot less
bad information and your life will become much easier.


The wikipedia entry is correct as written.

In the real world, the output of an amateur transmitter will seldom
be exactly 50 Ohms unless there is an adjustable network of some
sort.

I've always understood that the resistive part of a TX output impedance
was usually less than 50 ohms.

If a transmitter output impedance WAS 50 ohms, I would have thought that
the efficiency of the output stage could never exceed 50% (and aren't
class-C PAs supposed to be around 66.%?). Also, as much power would be
dissipated in the PA stage as in the load.

Fixed output amateur transmitters are a nominal 50 ohms. It can vary,
but that is due to normal variances in components, and the difference
can be ignored in real life.

But output impedance has little to do with efficiency. A Class C
amplifier can run 90%+ efficiency. It's output may be anything, i.e.
high with tubes but low with solid state. But the output impedance can
be converted to 50 ohms or any other reasonable impedance through a
matching network.

A perfect matching network will have no loss, so everything the
transmitter puts out will go through the matching network. Of course,
nothing is perfect, so there will be some loss. But the amount of loss
in a 1:1 match will not be significant.

Also, the amplifier generates the power; in a perfect world, 100% of
that power is transferred to the load. The transmitter doesn't
dissipate 1/2 of the power and the load the other 1/2. It's not like
having two resistors in a circuit where each will dissipate 1/2 of the
power.

BTW - the resistive part of the impedance is not the same as resistance.
For a simple case - take a series circuit of a capacitor, an inductor
and a 50 ohm resistor. At the resonant frequency, the impedance will
be 50 +j0 (50 ohms from the resistor, capacitive and inductive
reactances cancel). But the DC resistance is infinity. Again, a simple
example, but it shows a point.


I encourage all of you to read Walter Maxwell's (W2DU) book "Reflections
III". It will explain everything about this.

Everything you are discussing has been put to bed.

Cheers.

On 7/4/2015 9:43 PM, Jeff Liebermann wrote:
On Sat, 04 Jul 2015 19:33:30 -0400, Jerry Stuckle
wrote:

On 7/4/2015 7:22 PM, Jeff Liebermann wrote:
On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
wrote:
Think of it this way, without the math. On the transmitter side of the
network, the match is 1:1, with nothing reflected back to the transmitter.

So you have a signal coming back from the antenna. You have a perfect
matching network, which means nothing is lost in the network. The
feedline is perfect, so there is no loss in it. The only place for the
signal to go is back to the antenna.

Wikipedia says that if the source is matched to the line, any
reflections that come back are absorbed, not reflected back to the
antenna:

https://en.wikipedia.org/wiki/Impedance_matching
"If the source impedance matches the line, reflections
from the load end will be absorbed at the source end.
If the transmission line is not matched at both ends
reflections from the load will be re-reflected at the
source and re-re-reflected at the load end ad infinitum,
losing energy on each transit of the transmission line."

And you believe everything Wikipedia says? ROFLMAO.
But that also explains your ignorance.

Let's see if I understand you correctly. You claim that with a power
amplifier (source) output impedance that is perfectly matched to the
coax cable, but not necessarily the load (antenna), any reflected
power from the load (antenna) is bounced back to the load (antenna) by
the perfectly matched source (power amp). Is that what you're saying?

Yet, when I have a perfectly matched load (antenna), all the power it
is fed is radiated and nothing is reflected. You can't have it both
ways because the reflected power from the load (antenna), becomes the
incident power going towards the source (power amp). Matched and
mismatched loads do NOT act differently depending on the direction of
travel. If you claim were true, then transmitting into a matched
antenna or dummy load would reflect all the power back towards the
transmitter.

I think this is one of those situations where a casual explanation won't
work. You can use a "casual" explanation when the various
qualifications for a simplification apply. But to do that, the
qualifiers have to be fully understood and no one here is showing what
the qualifiers are much less that they are met. So until we get a real
explanation I will stick with what I recall. In the end, to settle this
we may have to use the math.

I'm sure someone in s.e.d could explain this properly. Some of them may
be purely argumentative, but some really know their stuff. I believe
the description of a conjugate match is the mathematical inverse of the
complex impedance of the antenna "viewed" through the feed line, but I
have to admit I don't really know what that implies or if it is even an
accurate description.

--

Rick

On 7/4/2015 10:37 PM, Jerry Stuckle wrote:

With a perfect matching network and a perfect feedline (which is what we
are discussing), that is true. But I also know that is far beyond your
limited intelligence.

Nothing wrong with it at all - except your limited intelligence can't
understand simple physics. But then that's nothing new.

Jeff, this is the sort of reply that you can expect from Jerry so that I
just don't even respond to him anymore unless... well, at the moment I
can't think of a reason to respond to what he writes.

--

Rick

On 7/4/2015 7:55 PM, Wayne wrote:


"Jeff Liebermann" wrote in message
...

On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
wrote:
Think of it this way, without the math. On the transmitter side of the
network, the match is 1:1, with nothing reflected back to the
transmitter.

So you have a signal coming back from the antenna. You have a perfect
matching network, which means nothing is lost in the network. The
feedline is perfect, so there is no loss in it. The only place for the
signal to go is back to the antenna.

Wikipedia says that if the source is matched to the line, any
reflections that come back are absorbed, not reflected back to the
antenna:

https://en.wikipedia.org/wiki/Impedance_matching
"If the source impedance matches the line, reflections
from the load end will be absorbed at the source end.
If the transmission line is not matched at both ends
reflections from the load will be re-reflected at the
source and re-re-reflected at the load end ad infinitum,
losing energy on each transit of the transmission line."

Well, I looked at that section of the writeup.
And, I have no idea what the hell they are talking about.
Looks like a good section for a knowledgeable person to edit.

What is not clear about your objection. I can't say for sure if it is
accurate, but it seems clear to me saying that at either end if you have
an impedance mismatch some of the wave will be reflected. With
mismatches at both ends the wave will bounce back and forth losing
energy each time, but never dying out completely. That is exactly what
I would expect.

The only case I am aware of that will give total reflection is when the
terminal is open circuit with infinite impedance absorbing *no* signal.

--

Rick

On 7/5/2015 10:48 AM, Ralph Mowery wrote:
"Ian Jackson" wrote in message
...
class-C PAs supposed to be around 66.%?). Also, as much power would be
dissipated in the PA stage as in the load.

I think this is easy to disprove in practice. I have an amp that is
probably class B, but it does not mater about the class. If I adjust it to
an input of 2000 watts from the DC power supply, I get out 1200 watts to a
resistive dummy load. If the above is true, I should have to input 2400
watts to the final stage. Now can someone tell me where the extra 400 watts
are comming from ? This 400 extra watts is not even counting on any loss in
the circuits.

The idea of matched impedance transferring maximum power is one of those
"simplified" descriptions that has preconditions that some people forget
about. It is not a universal truth.

If you have a transmitter output with a fixed impedance you can get
maximum power transferred to the feed line by matching the feed line
impedance to the transmitter output impedance. But if your feed line
impedance is the constant, you get maximum power transfer by minimizing
the transmitter output impedance, meaning zero ohms.

So you could in theory get 1200 watts into your feed line while drawing
only 1200 watts from the power supply.

--

Rick

On 7/5/2015 11:35 AM, John S wrote:

I encourage all of you to read Walter Maxwell's (W2DU) book "Reflections
III". It will explain everything about this.

Everything you are discussing has been put to bed.

Of course it has. We are not inventing anything here, we are trying to
understand it. Does this writing have a Cliff Notes version?

--

Rick

On Sat, 04 Jul 2015 22:37:41 -0400, Jerry Stuckle
wrote:

On 7/4/2015 9:43 PM, Jeff Liebermann wrote:
Let's see if I understand you correctly. You claim that with a power
amplifier (source) output impedance that is perfectly matched to the
coax cable, but not necessarily the load (antenna), any reflected
power from the load (antenna) is bounced back to the load (antenna) by
the perfectly matched source (power amp). Is that what you're saying?

With a perfect matching network and a perfect feedline (which is what we
are discussing), that is true.

Ok, so we agree on that part.

Yet, when I have a perfectly matched load (antenna), all the power it
is fed is radiated and nothing is reflected. You can't have it both
ways because the reflected power from the load (antenna), becomes the
incident power going towards the source (power amp). Matched and
mismatched loads do NOT act differently depending on the direction of
travel. If you claim were true, then transmitting into a matched
antenna or dummy load would reflect all the power back towards the
transmitter.

Nothing wrong with it at all

Ah, but there's plenty wrong with your view. At the load (antenna)
end of the coax, we both agree that with a perfect match, perfect
coax, and perfect load, there's is no reflection. Yet when you look
at the other end of the same coax, the same perfectly matched coax
(S21 back into the PA) suddenly decides to reflect any RF that might
be returned from a load (antenna) mismatch. It seems rather odd that
RF would act differently at opposing ends of the coax cable. In the
forward direction, a matched load either absorbs or radiate. In the
reverse direction, a matched load changes its mind and decides to
reflect? I don't think so.

There's another problem with your view of how VSWR works. If I
transmitted into an open or short circuit load (antenna), all the
forward RF would be reflected back to the source (PA). That would
mean that the PA will need to protect itself from over voltage or over
current using the traditional VSWR protection circuit. Yet, if the PA
were perfectly matched into a perfect coax cable, all that reflected
RF would bounce off the PA and back to the load (antenna). The PA
would not see any of that RF, and there would not be any need for a
VSWR protection circuit. I don't think so.

Personal experience with blowing up finals has demonstrated to my
satisfaction that a perfectly matched PA is quite capable of being
blown up by transmitting into an open or short with no VSWR
protection. Presumably, the damage was done by the reflected RF
(causing over voltage or over current in the PA) which would not be
present in your scheme of things, with a perfectly matched PA and
coax.

Another problem is IMD (intermodulation) products produced in the
power amplifier (PA). This is not a major problem with HF radios, but
is a serious problem with mountain top repeater sites. The antennas
on the towers tend to be rather close together. RF from a nearby
transmitter can couple into adjacent antennas, travel down the coax to
the PA, mix with the transmit signal in the PA, get amplified by the
PA, and get re-radiated by the antenna. The effect is typically
blocked by cavity filters and one-way isolators or circulators. The
problem here is that if the perfectly matched PA really did reflect
anything coming down from the antenna back to the antenna, there would
be no need for such IMD protection. The RF from the adjacent antenna
would simply bounce back towards the antenna and be re-radiated
without any mixing taking place. It would be a wonderful world if it
worked that way, but it obviously does not.

There are other problems, but they require math to explain, which
requires more time than I have available right now.



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