GENERATOR- YOU DO THE IN/OUT POWER MATH

[Rapadura]

Rectified "overdouble" output is good news...

Try to feedback the output to the input and take a video of it!

[infringer]

indeed it does sound rather interesting...

Surely a video and full detailed information would be of the order weather it is a bad reading or not has yet to be determined but the conscious effort is there and you must continue to post your results.

[DeepCut]

Getting this today :

http://maplin.co.uk/Module.aspx?ModuleNo=2355

If it doesn't kick in then it's time to wind that second coil and perhaps double the output for free.

Bruce,  would another coil, wound as you suggested, induct ? I thought it had to be at 90 deg to the movement of the changing magnetic field ?

@Infringer

Lots more detail eralier on in thread, MagnetMan has his videos posted on the concept.

I am following in his footsteps and ahve a much cruder setup, he will be doing his load-testing this week i think and should have better results but i will post video once it's self-powering, if it ever is lol ;+}


Gary.

[gyulasun]

...
System components :
Drive circuit: Bedini, no JonnyDavro relay. Using 10k, 3-contact linear pot.
Drive coil : Bifilar, single coil.
               Trigger wire : 0.25mm.
               Induction wire : 0.28mm.
Magnet : Neo @ 6,100 gauss.
Full wave bridge rectifier, shop-bought, maximum reverse-current is 10A per diode.
Load resistor : 100k
Resistance of Bedini pot : ~1200
Peak performance result :
Input : 9.43 VDC @ 0.075 A = 0.70725 W
Output (rectified) : 41.5 VDC @ 0.035 A = 1.4525 W

Apparent COP = 2.053729232944503 ...

I ran the test 10 times and all results were similar, these are average figures.

*EDIT* What's an easy way of switching between the PSU and the rectified output ? Can i just connect the output to where the input goes to begin with or will that blow my house up lol ? *EDIT*

Hi DeepCut,

Here is what I would do:

I would not loop directly, FIRST I would load further down the DC voltage at the output of the diode bridge by using much less value resistors than your present 100 kOhm. Try using as low as anything between 100 Ohm to 1 kOhm and see that how this increased load affects the output DC voltage and the current consumption from the input (9.43 VDC @ 0.075 A).

IF and WHEN you have an output voltage across the diode bridge output higher than 10 Volt MEASURED when you use a load resistor of  say 100 Ohm, Than this means your load current is 10V/100 Ohm= 0.1 Amper, hence you have overunity PROVIDED your input conditions 9.43 VDC @ 0.075 A did not change.   

By the way I do not get when you use a 100 kOhm load resistor across the diode bridge output how can 0.035 Amper flow into it if the DC voltage is 41.5V?  Ohm's law should be valid and it calculates as 41.5/100000=0.000415 A   i.e. 0.415 mA and not 35 mA you measured.

Do you have a puffer capacitor across the diode bridge output? It is an electrolytic cap of any value between 470 uF to 2200 uF.  IF you did not have such capacitor when you did the measurements, please start it again with using one. 

Your input circuit as a load at 9.43 VDC @ 0.075 A corresponds to a R=9.43/0.075=125.7 Ohm resistor.  This may change to an even lower value when you use a 100-150 Ohm load resistor at your diode bridge output, you have to check that.  IT is good to use two meters: one ampermeter at the input in series with the 9.43V DC source and one voltage meter across the diode bridge output (where the puffer cap is also connected) , in parallel with the 100-150 Ohm load resistor.

rgds,  Gyula

[DeepCut]

Gyula thanks.

There's a lot i don't know ...

I thought a larger resistor put more load on but it's the other way around is it ? I will try it with a 100 ohm after this post.

The bridge is shop-bought, i'm not sure what it contains but max reverse-current is 10 A per diode.

About the ampere difference, in the Bedini circuit i have a 10k linear pot to 'tune' the circuit with, perhaps this is the reason ? It's set to about 1200k at peak performance.

I have a 470u cap @ 16 V and will try that in addition to the 100 o resistor.

My digi meter just died, it was cheap ... so will only be able to use analogue tonight.

Will report back.


Gary.