@xee,  maybe your right,  I think I got my light bulb measurements backwards.  I did this a while ago.  Good point to check then, I'll follow the advice and go buy a higher power rating resistor.

@grumpy,  I checked it just with the 150 ohm resistor shorted across the 9 volt battery, and it draws around  0.4 amps, and yes things heat up quick, even the battery.  Actually, letting the circuit run for a while (I tried about a minute)  heats the resistor up really good, and I guess the battery warms up a bit too.

gyulasun, thanks for the suggestions

hoppy,  the Vbe voltage swings negative to -10 Volts or so, as can be seen in the scope shots I posted.  I hope nothing will degrade, but it might.  The transistor is certainly operating in a non conventional way.

...in fact, if the battery is still at around 6.25V, as shown on the trace, then 6.25/148.7 = .042A

@EM
any chance your reading of 0.4A is actually 0.04A?

just a thought
sandy
Doc Ringwood's Free Energy site  http://ringcomps.co.uk/doc

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

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

you all seen this long ago im sure...  i sure have   just never bothered to build it

however im sure it works along the same lines.....

almost the same unit em....

http://youtube.com/watch?v=gTAqGKt64WM

the joule thief

so what if you took the output of one of thease units  and used it as your 9v supply.....  or in place of it ...  what will be your return  or your COP


ist

http://youtube.com/watch?v=bRaVI4Solg0

even better vid....

Is your peak to peak voltage really 34 volts?

yes xee,  it is and you can see it in the scope shots I posted.

Look, even if we get picky and say the waveform might not be perfectly sinusoidal (and shape matters), so let's assume an even lower voltage, let's say 15 volts peak amplitude, then

Pout = 0.5 * (15 volts)^2 / 147 = 0.765 watts
Pin = 0.54 watts
COP = 0.765 / 0.54 = 1.40

so we are still way into OU teritory, with 40% more energy output.

@all, I'll report more after I try to close the loop.  I will stop by Radioshack and try and buy a rectifyer bridge.  Hopefully it doesn't eat up to much power.



EM