[Amps] Re: BirdŽ 43 Manual

[Mike]

My point is that power dissipation in the source will vary
depending on the length of that unterminated line since the
actual impedance presented to the final amplifier could be
anything on the rim of the Smith Chart ranging from a dead
short to an open circuit (I am assuming a very low-loss line).
In other words, which case would you expect to result in a
higher level of dissipation in the final amplifier, a shorting
plug placed on the RF output connector, or an open circuit?
The same goes for more sane levels of VSWR. A 2:1 VSWR,
for instance, can be represent any value of impedance to the
final amplifier lying on a 2:1 VSWR circle on the Smith
Chart. This can range from 25 ohms resistive up to 100 ohms
resistive with a bunch of other complex impedances in
between.

73 de Mike, W4EF..........................


----- Original Message -----
From: "Paul Christensen" <w9ac@arrl.net>
To: "Mike" <W4EF@dellroy.com>; "Gary Schafer" <garyschafer@attbi.com>
Cc: "AMPS" <amps@contesting.com>
Sent: Thursday, March 28, 2002 3:41 AM
Subject: Re: BirdŽ 43 Manual


> Hi Mike,
>
> > The reflected wave combines with the incident wave at the source
> > such that the source sees a mismatched load.
>
> 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?
>
> -Paul, W9AC
>
>
>
>