My two cents, also from a non-expert:
I think Doug is right about the reflected power not necessarily being lost
(i.e., being available to be radiated), but it depends on the impedance
match between the finals and the antenna/feedline. My understanding is that
if the impedances are matched, then the amp delivers all of its power to
the feedline/antenna. If the impedances are not matched, then it does not.
If the plate output impedance is not properly matched to the load
impedance, by using the amp's output network (and possibly an external
antenna tuner), then the reflected power will be dissipated by the plate
in the form of heat. This heat represents power that is not getting to the
feedline/antenna and hence will not be radiated. In this case, you do have
to subtract the reflected power from the forward power to get the actual
power delivered to the antenna.
When the amp's output network is properly tuned to the load, then the
finals do not see any reflected power, and should deliver all of their
available power into the output network. Normally, you can't see this on
the meter because forward and reflected power are typically measured after
the output network. If the output network is properly designed and tuned,
then I believe that it will also deliver all of the power into the
feedline. When the initial impedance mismatch exceeds the capabilities of
the output network, then an external tuner can be used to effect a match.
Again, in theory, 100% of the available plate output power should be
delivered to the feedline/antenna when the impedances are matched.
However, there is one caveat to that theory: heat dissipation by the output
or tuning network. Take a look at the latest QST for information on how
some popular antenna tuners dissipate a significant percentage of output
power as heat when called upon to match tough loads. The percentage seems
to be very dependent on tuner design and build. This leads me to believe
that it is also possible to lose power through heat dissipation in the
amp's output network if it is not properly designed or if it is used
outside of its operating parameters (does anyone know if this can actually
happen?) In both cases, this heat dissipation represents power that is not
getting to the feedline/antenna. However, this applies to unusually
difficult amtching situations and I have no idea how you would measure the
heat dissipation and subtract its effect from the power reaching the
antenna/feedline. Therefore, let's ignore this caveat.
I think we can assume that in the case of a properly matching a typical
load with a well-designed output and matching network, all of the power
gets to the feedline. I know absolutely nothing about how a mismatch
affects feedline losses, but would love to hear about it from someone on
this reflector. Supposedly, it's a myth that coax feedlines radiate when
there's a mismatch, so power won't be lost there. Do they ever get hot?
If all of the power gets to the antenna, then that raises the whole issue
of antenna radiation efficiency. Let's say that you have an 80 meter
dipole, and use your antenna tuner to match the amp at 160 meters. In
theory, with a proper match all of the power gets to the load. But we know
that under these circumstances the antenna will be a lousy radiator. What
happens to the power? My guess is that it heats up the antenna or coax (or
both). Anybody know?
To address the original question about the FCC power limit of 1500 watts
PEP, I'm sure that the limit applies to the power output by the transmitter
when it is properly matched to the load (which represents the maximum power
that can be dumped by the transmitter into the feedline.) In this case,
when you measure 2500 watts forward power and 1000 watts reflected power,
you are breaking the rules. You would have to scale back the output power
to 1500 watts to be operating legally. In other words, what happens after
the power leaves your transmitter is your problem. If the antenna and/or
feedline waste power, too bad. At least you will not exceed the limit. If
the antenna is a lousy radiator, that doesn't mean you get to pump in more
power. The good news is that if the antenna is a great radiator and
concentrates most of the power in one direction, effectively giving you
many multiples of the transmitter power in gain, you don't have to reduce
the output power!
Dick, WC1M
----------
> From: Doug.Hall. <dhall@jps.com>
> To: amps@contesting.com
> Subject: Re: [AMPS] Lets chew on this
To: <amps@contesting.com>
> Date: Wednesday, March 05, 1997 11:00 AM
>
> Tom Rauch (W8JI) <W8JItom@worldnet.att.net> wrote:
> >
> >If you have 2500 watts forward, and 1000 watts reflected, it is 1500
> >watts output.
> >
> >73 Tom
>
> No wonder you have such a big signal :-)
>
> Seriously, I'll be the first to admit that I'm no expert here, but
this
> seems totally incorrect to me. What happens to the 1000 watts of
reflected
> power? Does it come back into the amp and just sit there? Of course not!
> It's reflected back up the feedline to the load, right? So it seems to me
> the only way SWR (and hence reflected power) can reduce the power
delivered
> to your antenna is if you have excessive feedline losses. I'm reasonably
> certain that the scenario described above does NOT satisfy the 1500 watt
> output rule. Under your logic I could run 5 KW to my extended double zepp
> fed with ladder line on 40m and still be within the legal limit, right?
> 'Cause the VSWR on the ladder line is quite high, with considerable
> reflected power.
> Again, I'm not the expert here, so could you (Tom) or somebody please
> enlighten me on this particular subject?
> 73,
> Doug, KF4KL
>
>
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