Selfrunning cold electricity circuit from Dr.Stiffler

[mudwump]

I feel like a little kid waiting for Santa Claus only this time its the mailman. I really need to get some experimental time in before I start trying to recommend ideas. This will be so much fun.

My intentions are to build a full wave diode bridge rectifier on the output of the SEC. The output of this would go to a dual capacitor circuit that puts one capacitor in charge mode and the other in service mode providing power to the SEC. I am still waiting parts, so I am still in design phase. I am not sure at this point, if I will put the capacitors charge/service circuit on a timer, switch manually, or base the switch on voltage. Any ideas?

@mudwump
Unless you are thinking of something I can not visualize, how are you connecting this bridge so that it floats? The AV Plug is at a high impedance above ground, so what is the other leg for the bridge?

This has been tried by a number of people, but the minute you complete the circuit back to any common reference SEC is DEAD! I can not say this enough although this is most likely the 10kth time.

Now, I may just not see what you want to do, can you post to us a drawing? by hand is great, then I can see if I am missing something here.

Keep up the work..........

[DrStiffler]

@SEC Board Owners
A few people have addressed the voltage across the Neon as measured to ground, what? The AV Plug is floating (download instruction pdf), you can not measure the voltage across the Neon to anywhere except across the Neon and that will not even work...........

The Neons used have a fire voltage of 95V and an average current of 1.9mA. Rest assured that the voltage is at least 95V, but you are not going to hang a meter across it and measure it. When you connect from the AV Plug with a meter to ground or the +V you are not seeing what you think you are.

[plengo]

@Stefan,

I said 13watts from the input power supply that feeds the SEC after running for about 8 hours or so. That's the number I see on the computer monitor because I use a PowerPro from Western Radio to measure the amount of power I am using.

Today the battery is above 12.56v and so far I used 44 watts of input power to the SEC. The amount of watts I will get from this exercise from the battery will only be known when I do a load test on the battery also using another computerized device CBAII (also from Western Radio).

So I measure my cost of power from my power supply to achieve an output power from the battery. IF and only IF I got more power OUT from the battery than of what I used from my power supply INTO to the bettary I will have COP > 1. Cross fingers.

Fausto.

[plengo]

@Dr.Stiffler

I watched your last video three times now and still have not figured out where you are connecting the one end of that large parasitic coil (grey duct tape covered unit with the metal bolt). I fully understand the replacement of the tunable base coil with the choke, but where is this coil connected???

Thanks and great work.....

Thanks Doc. Yea, you can not figure it out because it is NOT connected to anything. I just put it there close to the board so that it pickup some of that radiant energy in a very decoupled way. It is the signal from the coil that is driving my experiment.

I just connected one end only to the probe. The other end of the probe is also not connected to anything. And just for completness, that is a bifilar coil, 400 turns awg 23 and 26 with a galvanized bolt of 1 1/2" long and 1/4 diameter.

Fausto.

[hartiberlin]

@Stefan,

I said 13watts from the input power supply that feeds the SEC after running for about 8 hours or so. That's the number I see on the computer monitor because I use a PowerPro from Western Radio to measure the amount of power I am using.

Hi Fausto,
okay, I see, so it must be Watthours and not Watts.
if you counts Watts x hours you get an energy value, not a Power value.

As you used about 2 Watts all the time, I calculate
2 Watts x 8 hours = 16 Watthours,
maybe it was more pulsed, so it was not all the time 100 mA integrated over all the time.

Today the battery is above 12.56v and so far I used 44 watts of input power to the SEC. The amount of watts I will get from this exercise from the battery will only be known when I do a load test on the battery also using another computerized device CBAII (also from Western Radio).

Okay, again 44 Watthours..

So I measure my cost of power from my power supply to achieve an output power from the battery. IF and only IF I got more power OUT from the battery than of what I used from my power supply INTO to the bettary I will have COP > 1. Cross fingers.

Fausto.

Great, I hope you can prove it.
Good luck !

Regards, Stefan.