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.

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.

That is demonstrably false.

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

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

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?


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.


That is demonstrably false.

Please demonstrate this for us as we wish to learn.

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.

Capacitors are an open circuit at DC and have a frequency dependant
impedance at AC.

Inductors are a short circuit at DC and have a frequency dependant
impedance at AC.

There is no such thing as a transmission line at DC.

Current at DC is constant and does not cause propagation while current
at AC causes a varying electromagnetic field that can propagate.

There is no such thing as a phase angle at DC.

A wire carrying DC current will not induce a voltage into another nearby
wire but a wire carrying AC current will.

More?


--
Jim Pennino

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 7/7/2015 3:05 AM, John S wrote:
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?


You apply 1vdc to a 0.159 microfarad capacitor and you get 0 amps
flowing (open circuit).
You apply 1vac at 1MHz to that same capacitor and you get 1 amp flowing,
with the current leading the voltage by 90 degrees.

You apply 1vdc to a 0.159 microhenry inductor and you get infinite amps
flowing (short circuit).
You apply 1vdc at 1MHz to that same inductor, and you get 1 amp flowing
with the voltage leading the current by 90 degrees.

You place the capacitor and inductor in series.
Fed with DC, you get 0 amps flowing (open circuit).
Fed with 1MHz AC, you get infinite current flowing (short circuit).

You place the capacitor and inductor in parallel.
Fed with DC, you get infinite current flowing (short circuit).
Fed with 1MHz AC you get 0 amps flowing (open circuit).

There is a huge difference between ac and dc!

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

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

On 7/7/2015 10:05 AM, Jerry Stuckle wrote:
On 7/7/2015 3:05 AM, John S wrote:
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?


You apply 1vdc to a 0.159 microfarad capacitor and you get 0 amps
flowing (open circuit).
You apply 1vac at 1MHz to that same capacitor and you get 1 amp flowing,
with the current leading the voltage by 90 degrees.

You apply 1vdc to a 0.159 microhenry inductor and you get infinite amps
flowing (short circuit).
You apply 1vdc at 1MHz to that same inductor, and you get 1 amp flowing
with the voltage leading the current by 90 degrees.

You place the capacitor and inductor in series.
Fed with DC, you get 0 amps flowing (open circuit).
Fed with 1MHz AC, you get infinite current flowing (short circuit).

You place the capacitor and inductor in parallel.
Fed with DC, you get infinite current flowing (short circuit).
Fed with 1MHz AC you get 0 amps flowing (open circuit).

There is a huge difference between ac and dc!


Yes, but the LAWS have not changed. The components have changed. So,
changing the components changes the laws of physics?

Suppose you apply .01Hz AC RMS to the components you specified. What then?

On 7/8/2015 6:09 AM, John S wrote:
On 7/7/2015 10:05 AM, Jerry Stuckle wrote:
On 7/7/2015 3:05 AM, John S wrote:
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?


You apply 1vdc to a 0.159 microfarad capacitor and you get 0 amps
flowing (open circuit).
You apply 1vac at 1MHz to that same capacitor and you get 1 amp flowing,
with the current leading the voltage by 90 degrees.

You apply 1vdc to a 0.159 microhenry inductor and you get infinite amps
flowing (short circuit).
You apply 1vdc at 1MHz to that same inductor, and you get 1 amp flowing
with the voltage leading the current by 90 degrees.

You place the capacitor and inductor in series.
Fed with DC, you get 0 amps flowing (open circuit).
Fed with 1MHz AC, you get infinite current flowing (short circuit).

You place the capacitor and inductor in parallel.
Fed with DC, you get infinite current flowing (short circuit).
Fed with 1MHz AC you get 0 amps flowing (open circuit).

There is a huge difference between ac and dc!


Yes, but the LAWS have not changed. The components have changed. So,
changing the components changes the laws of physics?

Suppose you apply .01Hz AC RMS to the components you specified. What then?

The point is - the rules for AC are different than the rules for DC.

I'm not going to waste my time figuring out the calculations - you can
do that. But the bottom line will be there will be some impedance in
every case. It will be neither zero nor infinity, as it would be in a
DC circuit.

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

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

John S wrote:
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?

What part of "you get a whole new set of laws" was it you failed
to understand?

Here's a clue for you; at DC the reactive components of a length
of wire are irrelevant but at AC they are not.


--
Jim Pennino

On 7/7/2015 1:37 PM, wrote:
John S wrote:
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?

What part of "you get a whole new set of laws" was it you failed
to understand?

Here's a clue for you; at DC the reactive components of a length
of wire are irrelevant but at AC they are not.

So, at 1Hz the law has changed, eh? What new law do I need to use?

John S wrote:

So, at 1Hz the law has changed, eh? What new law do I need to use?

To be pendatic, there is only one set of physical laws that govern
electromagnetics.

However for DC all the complex parts of those laws have no effect and
all the equations can be simplified to remove the complex parts.

In the real, practical world people look upon this as two sets of
laws, one for AC and one for DC.

A good example of this is the transmission line which does not exist
at DC; at DC a transmission line is nothing more than two wires with
some resistance that is totally and only due to the ohmic resistance
of the material that makes up the wires.



--
Jim Pennino