On 7/7/2015 3:37 AM, Roger Hayter wrote:
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.




No, the power the transmitter can generate has nothing to do with the
load impedance. The transmitter generates what it can generate.

The amount of power transferred to the load is dependent on the load
impedance, with the maximum power transfer occurring when the impedances
match.

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