I've lost where this thread started.... But Paul, w9ac,is right.
100 watts heads down the transmission line.
What arrives finds an open, and is 100% reflected back. The
transmission line is warmed. It is warmed again, on the return
trip, further reducing what gets back to the transmitter.
The matching circuit (tuner) works both ways....it matches (if it can)
the load presented to it by the open...transformed by the xmsn line.
The TUNER is warmed, as power flows back into the finals.
You'll find that within your presumed 50 ohm source transmitter,
there is ANOTHER matching circuit, to match the 2 ohm or so source
impedance of the power FET's to 50 ohms. THAT matching circuit
absorbs some power, too, and IT is warmed, as well.
So, depending on the relative efficiencies of each element...and
the length of that coax, you'll have maybe 80 or 90% of the energy
dumped back into the finals.
The Bird 43 is an average reading meter. And IT consumes some power,
as well...although not very much. If it reads 100watts forward, and
a bit less coming back, it's close...but given the open at the far end,
it's not operating within a 50 ohm system, and it's wrong, too.
Incidentally...theory aside, I hope your transmitter sees the reflected
power and shuts down its finals before they melt!
w9ac wrote:
/snip/
I contend this is true when a conjugate (reactance cancellation) exists at
the source. Typically, this is accomplished by the ubiquitous transmatch,
auto-tuner or Pi-network within the transmitter. The degree of re-flection
and combining of the reflected wave with that of the forward wave is a
function of that network's ability to cancel system reactance.
In the case of a fixed, 50-ohm output transceiver (e.g., my Ten-Tec Omni
Six), I am observing that all reflected power is being returned back to, and
absorbed by, my transceiver when the transmitter is feeding an unterminated
line. This is major point of contention on this thread. When transmitting
into an unterminated 100-foot length length of transmission line, my Bird 43
displays 100-watts forward and very close to 100 watts reflected. Under
this condition, my transmitter draws 20-amperes DC at 13.8 volts. This is
not a trivial point. Watt's Law still applies: power is being generated and
dissipated. Notwithstanding any line loss power, the vast majority of that
power must be absorbed in the transmitter's PA. Thoughts on this point?
/snip/
Jim Jarvis N2EA
Keithley Instruments
Essex Vermont
802 872 5830 voice
802 872 5831 fax
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