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Calorimeter response #1



Original poster: "Gary Johnson by way of Terry Fritz <twftesla-at-uswest-dot-net>" <gjohnson-at-ksu.edu>

Paul, Barry, Duncan:

Thanks for the suggestions.  The major influence of humidity on input
impedance was a surprise to me, but I think I am going to be able to use
this feature to explain some of the measurements that have been puzzling me
up to now. I am going to defer answers on humidity until I think I
understand its effects.

The 14 gauge coil is wound on a tube made of 1/8th inch polyethylene sheet.
The 16 gauge and the 22 gauge coils with the same resonant frequencies are
wound on polyethylene barrels claimed from the local salvage yard. The other
three coils are on PVC pipe. Ground consists of three 1 inch copper water
lines spaced a few feet apart and buried a foot or two deep in some very
hydrophilic clay. The lawn around the building is beautiful from all the
water I put on it. Inside the building is basically a dirt floor, covered
with a poly vapor barrier and about 4 inches of asphalt millings (no
concrete, no rebar).

Input impedance is measured two ways. One is the ratio of the rms voltage of
the applied square wave to the rms of the sinusoidal current as measured by
the voltage drop across a 0.02 ohm resistor, with the appropriate factor to
correct for the fundamental component of a square wave. The other is to
apply a sinusoidal voltage, (10 V rms) and measure the sinusoidal current
with a Phillips PM 9355 current probe. The rms values are read from a HP
54654D scope. I have a couple of 50 ohm dummy loads that I use for calibration.

Resonant frequency stays rock solid as humidity changes, so the capacitance
of the coil is not changed by humidity.

The input impedance is not a function of power below breakout. It acts just
like the resistance of a linear RLC model ought to act. Above breakout,
however, the coil changes from constant impedance to constant current. If I
get a 3 inch plume at 300 V and 1 A input, then at 400 V input the plume is
somewhat longer, but the input current is still 1 A. And no, the phase
between voltage and current has not changed. I am still thinking about a
good explanation for that. If anyone has a good explanation, I would
appreciate hearing it.

Gary Johnson