Mysterious Resonant Circuit

[Bruce_TPU]

{snip}


I've found an excellent way to smooth battery RF voltage fluctuations to where they are below 0.4 volts,  and the following modified circuit illustrates the modification.    (series L + C, when tuned have inpedance = 0 )

Also, I now have an almost perfect sinewave at the output, of 17.5 Volts Amplitude across the 150 ohm load resistor.

Here's the power calculations:

POWER IN  =  ( 8.2 Volts DC )  x  ( 0.1 Amp DC) = 0.82 watts

POWER OUT = ( 17.5 Volts AC) ^2 / ( 150 ohms)  / 2 = 1.02 watts

EXCESS POWER =  (1.02 - 0.82) / 0.82 = 0.24 = 24 % extra energy

COP = 1.02 / 0.82 = 1.24


{snip}

[Grumpy]

Try putting a resistor accross your 9v bat of about 80 ohms and if the bat gets hot - something ain't right.

[nul-points]

thanks, Gyula - i managed to wake up in my follow-on email 
(with the example of a battery voltage at 6.25V giving 0.042A thro' 148.7R)

i'm very interested to hear if EM's inductor-based circuit can be confirmed as showing the same level of COP as my switched-cap test circuit (1.2 approx)

we both have very simple (and very different) circuits but both have ferrite-cored coils, so i'm wondering if it's the coils in our circuits which are tapping the 'free energy' source?

all the best
sandy
Doc Ringwood's Free Energy site  http://ringcomps.co.uk/doc

[hartiberlin]

Hi EMdevices,
good work,
but check your output resistor,
as what the others said, 150 Ohms at 9 Volts should give:
9Volts / 150 Ohm= 60 mA.

Well, if you can scale up your circuit and it produces at least 3 Watts
of free energy, then you can apply for the overunity prize.

Regards, Stefan.

[EMdevices]

xee had me worried about the resistor changing it's value when heated up,  so I meaused its resistance by itself while I heated it up,   with a magnifyer lens in direct sunlight,  hot hot hot....  and guess what,  the value bearly changed from 147.5 to perhaps 147.7, or some insignificant amount like that.

well,  good news,  I got better results now:

Power Input = ( 6.0 Volts ) x ( 0.09 Amps) = 0.54 watts

Power Output = 0.5  *  (17 Volts_peak )^2  / (148 ohms) = 0.976 watts

COP = 0.976 / 0.54 = 1.80

So, nearly 80 % more energy out.   And I have been rounding the numbers up or down to be conservative, for example the voltage seemed to be more like 17.5 Volts, so I used 17.0 volts, and the amperage reading was changing between 0.08 and 0.09 amps, so it's probably 0.085 so I used 0.09 amps to be safe.

For those scrutenizing the number I might have made a mistake of writing 0.4 instead of 0.04,  also, the battery I'm using is depleating as I'm experimenting with it.  It came from the fire alarm after it started beeping that the battery was low, so then I took it out and have been using it for a few months now with different experiments, so you can see why it's voltage is droping so low.  The circuit seems to be more efficient as the input voltage drops.  I still get around 17 volts output which is amazing.

EM