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To All from PA0SE
At the risk of boring you even more I report the result of another
experiment I made on my loading coil.
To avoid disasters I replaced the transmitter by an old Rohde &
Schwarz valved power signal generator type SMLR, tuned to 137kHz.
The voltage at the top end of the loading coil was measured by an
even older Philips valve voltmeter type GM3765.
The voltage at the bottom end of the coil was indicated on an
oscilloscope. The current in the earth wire was passed through
a 0.78 ohm resistor; the voltage over the resistor being indicated
by the second trace on the scope.
Test A.
The system was tuned for maximum voltage at the top end of the coil
(= voltage on the aerial) and the generator output adjusted to make
this 50V.
Voltage at bottom end of the coil:
0.636V
Current in earth
wire:
13mA
Phase
angle:
0°
Power into
system:
8.3mW
Test B.
The aerial is disconnected. The vacuum capacitor in parallel with
the the coil is increased to replace the 370pF of the aerial. The system is
tuned for maximum voltage at the top end of the coil and the generator output
adjusted to make this 50V again.
Voltage at bottom end of the coil:
5.6V
Current in earth
wire:
1.8mA
Phase
angle:
68° capacitive
Power into
system:
3.8mW (= U * I * cos68°)
Current, phase angle and power are approximate values because
the deflection on the scope indicating the current was small and the
trace widened by the coil picking up signals from
broadcast stations.
C. Conclusion
In test B current flows into the bottom end of the coil and
none into the aerial. The circuit in which this current flows must be via
capacitance between coil and earth and via the earth wire back to the
generator.
Because in test A and B the voltage over the coil is roughly the
same that current must also be flowing when the aerial is connected. That
should explain the difference in current in earth and aerial wire that
started this long discussion.
The real part of the 1.8mA earth current in test B is
1.8mA * cos68° = 0.67mA, which is about 5% of the
13mA (real) current in test A.
That is the same percentage I noted as difference between currents
at top and bottom end of the coil using full transmitter power (2.0A at top and
2.1A at bottom end).
Jim, M0BMU, has explained it all very well in his e-mail of May
14.
Is this the end of my reports? I hope for you it is
...
73, Dick, PA0SE
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