GENERATOR- YOU DO THE IN/OUT POWER MATH

[FatChance!!!]

INPUT : 9.57 VDC @ 0.83 A = 0.79431 W

OUTPUT : 19.8 VDC @ 0.22 A = 4.356 W

Your input calculation is wrong.
INPUT: 9.57 VDC @ 0.83 A = 7.9431 W
Efficiency: 4.356 W / 7.9431 W = COP 0.54 = 54% efficiency.

Conclusion:
You can save your money for better things.

[DeepCut]

Hi.

Sorry i typed it wrong !

Should be .083 not .83.

Sorry i do that i lot i gota be more careful when typing.

INPUT: 9.57 VDC @ 0.083 A = 0.79431

I think we should ALL be replicating this since it's piss-easy to construct as well as cheap, and most of us probably have Bedini circuits lying around anyway.


Gary.

[gyulasun]

Gary,  you mentioned your output load is the DC resistance of a coil, correct?

Question: Do you have a full wave bridge (or a single diode) at the output and then you use a puffer capacitor and you load the capacitor with the coil?

Or you do not have a puffer capacitor, maybe you do not have a diode either? I wonder.

EDIT: just noticed you wrote DC output voltage so you surely had the diodes bridge, now the question is the puffer capacitor. 

[DeepCut]

Hi Gyulasun.

I don't have a load on it.

I wnted to know, purely, the watts in vs the watts out without measuring current with a resistor load bla bla so i did what you said before, read the voltage with a DVM and calculated the current because the coil resistance is known.

I couldn't find the post where you previously suggested a puffer cap, am i right in thinking you put it across the bridge output and also what properties should it have (uF and V) ?

(I am in the process of making a new housing for the magnet assembly so that the entire coil is wrapped in the plane of the shaft to maximise induction, getting the wire tomorrow).

*EDIT* I have a 47uf/35v and a 100uf/10v to hand *EDIT*


Thanks.

[gyulasun]

Yes, you put the puffer cap in parallel with the diode bridge DC output, with the correct polarity.
It is ok that you calculated the output current but:

If you did not use a puffer capacitor when you measured your output DC voltage, then you cannot trust on the the measured 19.8V DC output because of the half sinewave series of output. The capacitor would smooth them out, use the 47uF/35V capacitor. A 470 or 1000uF would be much better but you do not have that at hand in at least for 35V WV.

There can be another "problem":  I would suggest watching the input current whether it changes to a higher value than the .083A WHEN you add the 90 Ohm load at the output (the puffer cap is already there too). Here I assume the 9.57V supply voltage remains stable.
IF the input current of .083A changes to a higher value, then note the new value and then check the DC output voltage across the puffer capacitor (which is equivalent with the output where you connect the 90 Ohm load and note the output voltage, then recalculate, ok.