I tried today to measure the output current to power my qartz-LED circuit (see
http://www.overunityresearch.com/index.php?topic=718.msg12210#msg12210). The goal is to evaluate the transfer of energy between the generator and the load and to check if there is anomalous results.
I put a 10 Kohm resistance in series with the antenna and measure the HF voltages (for easiness, peak to peak values read from the oscilloscope) :
(1) at the output of the transmitter, there is 48v pp.
(2) after the 10 Kohm resistor (at the connection resistor-antenna), I measured 18v pp.
(3) when the frequency is tuned exactly on the quartz frequency, there is a drop of only 0.3v pp (17.7v pp at the antenna).
It means that around (48-18)/10= 3 mA pp current is permanently provided by the transmitter under voltage 48v, i.e. a power of 134 mW pp = 17 mW.
At the antenna, the voltage being 18v, the maximum available power is therefore 3*18=54 mW pp = 7mW (which is enough to light a LED).
But remember that 7 mW is still dissipated even when the frequency is not accorded with that of the quartz and the LED is not lit, because we have always around 18v at the antenna. I suppose it is absorbed by the environment. When well tuned, the voltage drops by 0.3v pp, meaning that an extra current of 0.3/10=0.03mA pp is consumed, which represents a power of 18*0.03=0.54 mW pp = 0,07 mW which is too weak to light a LED.
It follows that this measurement is not conclusive. The extra power that is provided when I tune the frequency on that of the quartz, is not enough to light a LED. But we can suppose that, when the frequency is well tuned, a part of the energy which was previously wasted in the environment, is now redirected to our resonant circuit. It is likely but I don't know how to be sure of that. An idea?
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Topic: Dr. Stiffler SEC circuits (Read 93731 times)
