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LF: The case of the missing amps

To: "LF-Group" <[email protected]>
Subject: LF: The case of the missing amps
From: "Dick Rollema" <[email protected]>
Date: Tue, 15 May 2001 13:14:38 +0200
Reply-to: [email protected]
Sender: <[email protected]>
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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