On 7/6/2015 4:50 PM, Ian Jackson wrote:
In message , rickman writes

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.

Also when it is a zero impedance (short circuit).

I'm trying to picture this.

In the case of an open circuit a matched driver drives the transmission
line to 50% of the driving voltage. The wave reaches the open
termination and is reflected with the same polarity resulting in a
return wave that reaches 100% of the driving voltage.

In the same vein, if the wave hits the short circuit the reflected wave
will be the opposite polarity making the reflected wave 0% of the
driving voltage resulting in the short circuit eventually showing to the
drive circuit.

--

Rick

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.

Also when it is a zero impedance (short circuit).

I'm trying to picture this.

In the case of an open circuit a matched driver drives the transmission
line to 50% of the driving voltage. The wave reaches the open
termination and is reflected with the same polarity resulting in a
return wave that reaches 100% of the driving voltage.

In the same vein, if the wave hits the short circuit the reflected wave
will be the opposite polarity making the reflected wave 0% of the
driving voltage resulting in the short circuit eventually showing to the
drive circuit.

Yup. You can see this happen on a time-domain reflectometer (an
o'scope and a pulse generator will do).

(although, to pick nits, I'd clarify your latter paragraph to read
"will be of the opposite polarity, making the sum of the forward and
reflected wave 0% of the driving voltage...")

Jerry Stuckle wrote:

On 7/6/2015 12:41 PM, John S wrote:
On 7/6/2015 11:01 AM, Jerry Stuckle wrote:
On 7/6/2015 4:20 AM, Ian Jackson wrote:
In message , rickman
writes



How about we quit with the speculation and come up with some numbers?

Here is a simulation of a 50 ohm load with a 50 ohm matched series
output impedance and a voltage source of 200 VAC peak. Power into the
load is 100 W.

http://arius.com/sims/Matched%20Load%20Power.png

Same exact circuit with the series impedance of just 1 ohm, power into
the load is 385 W.

http://arius.com/sims/UnMatched%20Load%20Power.png

I'd say that is pretty clear evidence that matched loads are not the
way to maximize power transfer when the load impedance is fixed and
the output impedance is controllable.

Quite simply, if your prime objective is to get maximum power out of a
power (energy?) source, the source having an internal resistance is a
BAD THING. You don't design the source to have an internal resistance
equal to its intended load resistance. No one designs lead-acid
batteries that way (do they?), so why RF transmitters?

While theoretically you can extract the maximum power available from the
source when the load resistance equals the source resistance, you can
only do so provided that the heat you generate in the source does not
cause the source to malfunction (in the worst case, blow up).

Because DC power transfer is not the same as AC power transfer.


Why not? Does something happen to the laws of physics with AC?


Yup, AC has reactance. DC does not. Big difference.


If what you say is correct, then it wouldn't matter what antenna
impedance I had, as long as it matches the transmission line. VSWR
would be immaterial.


There is no VSWR nor ISWR if the load matches the line.


Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there.


That is demonstrably false.

Please demonstrate this for us as we wish to learn.



OK, take your amateur transmitter. Connect it through a 1:1 balun to
300 ohm feedline. Connect that to a 300 ohm antenna.

The VSWR would be near to 1:1. Of course, you would need a 300 ohm
meter to measure it. It would be the same whether or not you connected
your amateur transmitter




According to you, you should get full power output at the antenna. In
reality, you will get a 6:1 SWR and about 49% of the power at the
antenna, minus transmission line loss (assuming, of course, your
transmitter hasn't cut it's power back).


You probably won't get full power, because the transmitter would have to
produce 2.4 times it's usual output voltage to achieve it. But the
system would be gratifyingly low in SWR and transmission line loss.


--
Roger Hayter

On 7/6/2015 7:59 PM, Roger Hayter wrote:
Jerry Stuckle wrote:

On 7/6/2015 12:41 PM, John S wrote:
On 7/6/2015 11:01 AM, Jerry Stuckle wrote:
On 7/6/2015 4:20 AM, Ian Jackson wrote:
In message , rickman
writes



How about we quit with the speculation and come up with some numbers?

Here is a simulation of a 50 ohm load with a 50 ohm matched series
output impedance and a voltage source of 200 VAC peak. Power into the
load is 100 W.

http://arius.com/sims/Matched%20Load%20Power.png

Same exact circuit with the series impedance of just 1 ohm, power into
the load is 385 W.

http://arius.com/sims/UnMatched%20Load%20Power.png

I'd say that is pretty clear evidence that matched loads are not the
way to maximize power transfer when the load impedance is fixed and
the output impedance is controllable.

Quite simply, if your prime objective is to get maximum power out of a
power (energy?) source, the source having an internal resistance is a
BAD THING. You don't design the source to have an internal resistance
equal to its intended load resistance. No one designs lead-acid
batteries that way (do they?), so why RF transmitters?

While theoretically you can extract the maximum power available from the
source when the load resistance equals the source resistance, you can
only do so provided that the heat you generate in the source does not
cause the source to malfunction (in the worst case, blow up).

Because DC power transfer is not the same as AC power transfer.


Why not? Does something happen to the laws of physics with AC?


Yup, AC has reactance. DC does not. Big difference.


If what you say is correct, then it wouldn't matter what antenna
impedance I had, as long as it matches the transmission line. VSWR
would be immaterial.


There is no VSWR nor ISWR if the load matches the line.


Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there.


That is demonstrably false.

Please demonstrate this for us as we wish to learn.



OK, take your amateur transmitter. Connect it through a 1:1 balun to
300 ohm feedline. Connect that to a 300 ohm antenna.

The VSWR would be near to 1:1. Of course, you would need a 300 ohm
meter to measure it. It would be the same whether or not you connected
your amateur transmitter


That's right. It's a 1:1 SWR on the transmission line. And it matches
your requirements.




According to you, you should get full power output at the antenna. In
reality, you will get a 6:1 SWR and about 49% of the power at the
antenna, minus transmission line loss (assuming, of course, your
transmitter hasn't cut it's power back).


You probably won't get full power, because the transmitter would have to
produce 2.4 times it's usual output voltage to achieve it. But the
system would be gratifyingly low in SWR and transmission line loss.



Not according to you. Since the transmitter has a comparatively low
impedance, it should dissipate very little power and virtually all of
the power should go to the antenna. After all, you do have a 1:1 VSWR
on the transmission line/antenna, fed by a low impedance transmitter.

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

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

On Mon, 6 Jul 2015 20:33:29 +0100, Ian Jackson
wrote:

In message , Jeff Liebermann
writes



One problem with the thermocouple ammeter. It's slow.

The other problem with the thermocouple ammeter is that it's easy to
burn out the thermocouple.

Yep. All mine were guaranteed to be fried when I obtained them for
free. However, they can be repaired. I've only done one successfully
so far:
http://www.tuberadio.com/robinson/Thermocouples/

But if you've been unfortunate to buy a duffer, don't totally despair.
With a simple internal rewire, at least you'll have a usable* 500uA or
1mA FSD moving coil meter.
*Convert to diode type?

That's the most common solution:
http://k4che.com/GO-9%20Transmitter/RF%20Ammeters/RF%20Ammeters.htm
I had problems with RF leakage being rectified by the diode resulting
in inaccurate readings. At least the typical WWII ammeter is well
shielded. Calibrating the scale is impossible but replacing the meter
scale is easy:
http://www.tonnesoftware.com/meter2.html

--
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 7/6/2015 1:03 PM, wrote:
John S wrote:
On 7/6/2015 11:01 AM, Jerry Stuckle wrote:
On 7/6/2015 4:20 AM, Ian Jackson wrote:
In message , rickman writes



How about we quit with the speculation and come up with some numbers?

Here is a simulation of a 50 ohm load with a 50 ohm matched series
output impedance and a voltage source of 200 VAC peak. Power into the
load is 100 W.

http://arius.com/sims/Matched%20Load%20Power.png

Same exact circuit with the series impedance of just 1 ohm, power into
the load is 385 W.

http://arius.com/sims/UnMatched%20Load%20Power.png

I'd say that is pretty clear evidence that matched loads are not the
way to maximize power transfer when the load impedance is fixed and
the output impedance is controllable.

Quite simply, if your prime objective is to get maximum power out of a
power (energy?) source, the source having an internal resistance is a
BAD THING. You don't design the source to have an internal resistance
equal to its intended load resistance. No one designs lead-acid
batteries that way (do they?), so why RF transmitters?

While theoretically you can extract the maximum power available from the
source when the load resistance equals the source resistance, you can
only do so provided that the heat you generate in the source does not
cause the source to malfunction (in the worst case, blow up).

Because DC power transfer is not the same as AC power transfer.


Why not? Does something happen to the laws of physics with AC?

Yes, quite a lot, you get a whole new set of laws.

If you apply 1vDC to a 1 ohm resistor, you get 1A of current. If you
apply 1vAC RMS (at any frequency) to a 1 ohm resistor, you get 1A of
current. How does the AC change the law?

On 7/6/2015 12:02 PM, rickman wrote:
On 7/6/2015 12:50 PM, John S wrote:
On 7/5/2015 11:39 PM, rickman wrote:
On 7/5/2015 4:45 PM, Jerry Stuckle wrote:
On 7/5/2015 3:58 PM, Roger Hayter wrote:
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.

The resistive part of the impedance is exactly the same as a
resistance
as far as the frequency you are using is concerned. And if the
amplifier output impedance *and* the feeder input resistance were
*both*
matched to 50 ohms resistive then 50% of the power generated (after
circuit losses due of inefficiency of generation) would be
dissipated in
the transmitter. It would, at the working frequency, be *exactly*
like
having two equal resistors in the circuit each taking half the
generated
power. So the amplifier has a much lower output impedance than
50ohms
and no attempt is made to match it to 50 ohms.


OK, so then please explain how I can have a Class C amplifier with 1KW
DC input and a 50 ohm output, 50 ohm coax and a matching network at the
antenna can show 900 watt (actually about 870 watts due to feedline
loss)? According to your statement, that is impossible. I should not
be able get more than 450W or so at the antenna.

It is true that if the transmitter is thought of as a fixed voltage
generator in series with, say, a 5 ohm resistor then the maximum power
transfer in theory would occur with a 5 ohm load. But to achieve this
the output power would have to be 100 times higher (ten times the
voltage) and half of it would be dissipated in the PA. Not the best
way
to run things, it is better to have a voltage generator chosen to give
the right power with the load being much bigger than its generator
resistance.


So why do all fixed-tuning amateur transmitters have a nominal 50 ohm
output instead of 1 or two ohms? And why do commercial radio stations
spend tens of thousands of dollars ensuring impedance is matched
throughout the system?

The maximum power transfer at equal impedance theorem only applies if
you started with a *fixed* output voltage generator. We don't; we
start with a load impedance (50 ohm resistive), then we decide what
power output we want, and we choose the voltage to be generated
accordingly. (Thank you for giving me the opportunity to think about
this!)


Actually, it doesn't matter if it's a fixed or a variable output
voltage
- maximum power transfer always occurs when there is an impedance
match.

How about we quit with the speculation and come up with some numbers?

Here is a simulation of a 50 ohm load with a 50 ohm matched series
output impedance and a voltage source of 200 VAC peak. Power into the
load is 100 W.

http://arius.com/sims/Matched%20Load%20Power.png

Same exact circuit with the series impedance of just 1 ohm, power into
the load is 385 W.

http://arius.com/sims/UnMatched%20Load%20Power.png

I'd say that is pretty clear evidence that matched loads are not the way
to maximize power transfer when the load impedance is fixed and the
output impedance is controllable.


There can be a lossless resistive part of source impedance according to
the IEEE (and most every other well educated EEs). After all, a
transmission line has a resistance but it's loss resistance is much
lower.

Can you provide a reference to any of this?


IEEE Standard Dictionary of Electrical and Electron Terms, IEEE Std
100-1972. For a complete treatment on the topic, see Reflections III by
Walter Maxwell, W2DU, Appendix 10.

Jerry Stuckle wrote:

On 7/6/2015 7:59 PM, Roger Hayter wrote:
Jerry Stuckle wrote:

On 7/6/2015 12:41 PM, John S wrote:
On 7/6/2015 11:01 AM, Jerry Stuckle wrote:
On 7/6/2015 4:20 AM, Ian Jackson wrote:
In message , rickman
writes



How about we quit with the speculation and come up with some numbers?

Here is a simulation of a 50 ohm load with a 50 ohm matched series
output impedance and a voltage source of 200 VAC peak. Power into the
load is 100 W.

http://arius.com/sims/Matched%20Load%20Power.png

Same exact circuit with the series impedance of just 1 ohm, power into
the load is 385 W.

http://arius.com/sims/UnMatched%20Load%20Power.png

I'd say that is pretty clear evidence that matched loads are not the
way to maximize power transfer when the load impedance is fixed and
the output impedance is controllable.

Quite simply, if your prime objective is to get maximum power out of a
power (energy?) source, the source having an internal resistance is a
BAD THING. You don't design the source to have an internal resistance
equal to its intended load resistance. No one designs lead-acid
batteries that way (do they?), so why RF transmitters?

While theoretically you can extract the maximum power available from the
source when the load resistance equals the source resistance, you can
only do so provided that the heat you generate in the source does not
cause the source to malfunction (in the worst case, blow up).

Because DC power transfer is not the same as AC power transfer.


Why not? Does something happen to the laws of physics with AC?


Yup, AC has reactance. DC does not. Big difference.


If what you say is correct, then it wouldn't matter what antenna
impedance I had, as long as it matches the transmission line. VSWR
would be immaterial.


There is no VSWR nor ISWR if the load matches the line.


Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there.


That is demonstrably false.

Please demonstrate this for us as we wish to learn.



OK, take your amateur transmitter. Connect it through a 1:1 balun to
300 ohm feedline. Connect that to a 300 ohm antenna.

The VSWR would be near to 1:1. Of course, you would need a 300 ohm
meter to measure it. It would be the same whether or not you connected
your amateur transmitter


That's right. It's a 1:1 SWR on the transmission line. And it matches
your requirements.




According to you, you should get full power output at the antenna. In
reality, you will get a 6:1 SWR and about 49% of the power at the
antenna, minus transmission line loss (assuming, of course, your
transmitter hasn't cut it's power back).


You probably won't get full power, because the transmitter would have to
produce 2.4 times it's usual output voltage to achieve it. But the
system would be gratifyingly low in SWR and transmission line loss.



Not according to you. Since the transmitter has a comparatively low
impedance, it should dissipate very little power and virtually all of
the power should go to the antenna. After all, you do have a 1:1 VSWR
on the transmission line/antenna, fed by a low impedance transmitter.

True. I agree. But the power it can generate depends on its load
impedance. Consider a very high or even open circuit load. It can put
little or no power into it, while having its normal voltage output.



--
Roger Hayter

In message , Jerry Stuckle
writes

So why don't manufacturers design transmitters with 1 ohm output
impedance, Rick?

They probably would - if they could (at least for some applications).
That would then enable you to step up the TX output voltage (using a
transformer), so that you could drive more power into a higher (eg 50
ohm) load.

But of course, the overall output impedance would then become
correspondingly higher. You would also be drawing correspondingly more
current from the original 1 ohm source, and if you used too high a
step-up, you would risk exceeding the permitted internal power
dissipation (and other performance parameters).

So yes, you are getting more power output when you match* the source
impedance to the load - but it doesn't necessarily mean you always can
(or should) go the whole hog.
*Or, at least, partially match.



--
Ian

In message , Jerry Stuckle
writes


Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there.

If there's no reflection, there can be no standing wave. So, being
pedantic, there's no such thing as an SWR of 1:1!



--
Ian