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Old October 16th 05, 12:45 AM
Richard Clark
 
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Default V/I ratio is forced to Z0:was Mythbusters

On Sat, 15 Oct 2005 22:36:11 GMT, Cecil Moore wrote:

these voltages match the voltage drops across R1 and R2
in amplitude.


There is NO R1 and/or R2

This is familiar bait and switch methods so common to your trolling.
Changing the schematics to suit the argument can make any answer come
alive.

Like I said, if you were already familiar with this, you would have
said so and skipped all this stageshowing.

In other words, it is the main Z0 calibration setting


An interpretation spanning as many documents as it takes to assemble
all the words xeroxed to this posting.

If the range of C1 and C2 are great enough, the wattmeter
could be calibrated for 75 ohms rather than 50 ohms.


Notably, there are no C1 and/or C2. Even more, if there were, any
competent tech in this group would be aware that it also would take a
similar shift in value of your R1 and/or R2. In fact, this
overstatement by calling out BOTH Cs needing change begs the chuckle
reflex being stifled.

The question is: Between the "transmitter" terminal and
the "antenna" terminal, what determines the physical
characteristic impedance of the sampling circuit?


It is very lightly loading as a series load by
design and as evidenced by Dave's measurements.


Exactly how much effect does that light loading have on
the primary voltage/current amplitude and phase? Enough
to be detectable if the V/I ratio is not 50 ohms?


[clue: this is in the portion of my original post that you omitted.]

I would normally ask my students "What do you think?" but there is a
danger of that here, with so little obtained in return for so much
said.

Rather, I will offer "Think about it. How much power does it take to
swing a 1mA meter even full scale when there is a 500KOhm calibration
resistor (imagine that, do we assign THAT as the Z?) limiting it."

Something about IČR comes to mind for max power (300W) which works out
to half a watt out of 300 supplied. A simple, back-of-the-napkin
computation suggests less than 1%. A similar study of the actual load
of 82 Ohms needs only be cast back through the NČ turns ratio. Seeing
that it, too, is probably a quarter Watt resistor that we are still
talking about less than 1% consumption.

So, anyway you slice and dice it without actually counting the turns
ratio, the tech at the bench would easily offer the meter injects only
half an ohm or less series resistance into the line.