[Amps] Re: [Amps] Re: [Amps] Re: [Amps] Bird® 43 Ma nual

[Paul Christensen]

> We have to remember that when a transmission line is involved and there are mismatches, there is impedance transformation going
on. Any
> time the load is not matched to the line the line no longer presents the same impedance at the watt meter or the transmitter that
it
> did before.
> What I am saying is that I don't think the bird meter will read true power unless it sees a 50 ohm resistive load.  So what is
seen on
> the meter is not necessarily what is going on in the transmission line.

Provided that the Bird directional coupler matches the characteristic impedance of the line, the metered results should be
reasonably accurate.  Interpreting the displayed results is another story.

> If a coax line is unterminated the bird meter reads 100 watts forward and 100 watts reflected as you saw but is there really 100
watts
> there?

I'm not too sure it matters.  What does matter is that whatever power really is generated, all of it (less line loss power) is being
self-absorbed in the transmitter's solid-state PA.  (For those folks just jumping in, we're discussing transmitters with a fixed
50-ohm source impedance.)

If there is zero current and some finite voltage then is there really any power actually present? The transmission line is no
> longer 50 ohms at the watt meter because of the impedance transformation.  (sorry about the additional questions but I don't have
a
> definitive answer)

Perhaps not at that precise instant in time.  Recall that power is defined as energy divided by time (e.g., 1 Joule per second = 1
watt).  The characteristic impedance of the line never changes as a function of impedance or SWR.  It is held constant by the
conductor sizing and spacing.

 > What I was saying about an open quarter wave line is that an open circuit at the end of a quarter wave line will be transformed
into a
> short at the other end due to impedance transformation of the line. If this is hooked to the watt meter then the watt meter would
be
> looking into a short. Would it still read 100 watts? The transmitter probably could not put out an infinite amount of current at
zero
> voltage to get a true 100 watts at that point.

That point is an instant in time.  If we are looking at voltage or current at any given instant snapshot in time, then by
definition, there is no power.  I do not know if the directional coupler will indicate any differently just because we made the open
termination at the end of a quart-wave line.  Any takers?

> If the reflected power were to get absorbed by the transmitter finals then an open wire transmission line would never work but in
a few
> instances where it was matched to the antenna.

An open-wire line or a relatively loss-less coaxial line behave similarly.  The reason that an open-wire balanced line works in most
of our applications is because a *conjugate* network is placed somewhere on the transmission line (typically at the transmitter with
an auto-tuner or transmatch), or a *matching* network is placed at the antenna terminals.  In your example above, the same rules
apply.

Consider an RF amplifier (gotta' stay somewhat on topic here...heh) with a fixed, balanced 450-ohm source impedance terminated into
100-feet of 450-ohm balanced open line.  If the opposite end of the open-wire line is unterminated, all generated power is returned
back to, and absorbed by, the transmitter.  Again, I ask the question:  In this example where else can the power go?

In reality the reflected power on that type of line is often very high if you were to
> measure it. The reason that it does work is that most all of the reflected power gets re- reflected back to the antenna and is
absorbed
> by the antenna.

Agreed, but for the right reasons.

-Paul, W9AC