RESONANCE EFFECTS FOR EVERYONE TO SHARE

[overunitydave]

Great post and loved the video. I clicked on the link for the schematic drawing but nothing was there.
Would it be possible for you to upload a picture of the schematic of the circuit you were demonstrating
on youtube here: http://www.youtube.com/watch?feature=endscreen&NR=1&v=lJQvqTpBdiQ  ?

Thank you very much!

[gotoluc]

 The circuit is just a capacitor and coil in series connected to the output of a signal generator.
Luc

[gixo]

It is great to view such works as those here and on previous pages, this resonating circuit board technology.

Thanks to all for the info ...   I would like to establish the following method for updated reference.

The math states obvious error in favor of successfully gathering an output wattage 400% in excess.


Measuring Input
http://s15.postimage.org/jfw6nmt1n/phpa_Mxy0x.gif

Measuring Output
http://s15.postimage.org/50ngg7urv/php_N7d_Ws_H.gif

Diagram
http://www.energeticforum.com/renewable-energy/3609-big-joule-theif-38.html#post213488

This circuit in the preceding results has been modified to have half as many turns on coil 3, a 500ohm 100w rheostat to the Base, and four transistors.

[gyulasun]

Hi gixo,

Nice job,  however I think you considered the power dissipation in the 1.4 Ohm resistor as THE input power,  this 2.64 W is only PART of the actual input power.  The current via this resistor is I=1.935 V/1.4 Ohm=1.38 Amper.  So the input power draw from the 12V is P=12*1.38=16.56 W, and if you deduce the resistor dissipation of 2.64 W you get 16.56-2.64=13.92 W.  Now if you relate this to the 8.42 W dissipated in the output resistor, the efficiency is 8.42/13.96= 0.6    i.e. 60%

(Notice that in the calculation I accepted your measured voltage numbers, in fact the voltage drop across the 1.4 Ohm should be checked with an oscilloscope or a true rms AC meter for getting the real amplitude whether the voltage drop has any AC wave (if you use capacitor filtering at the input across both legs of the 1.4 Ohm with respect to the negative pole of the 12V source, making a low pass RC filter, then the measured voltage drop across this filtered resistor as a DC voltage value is acceptable.)

rgds,  Gyula

[gixo]

Hi gyulasun,

Thanks for your advice; there are a few aspects still in the grey area it seems...  why the two different input results?  The input is 13.96W, I agree, but the output voltage is far in excess of 293.3 without the resistor.  It soared past 500 on capacitive load if memory serves. 

293.3 / 10210 = .0287 A

.0287 * 550 = 15.8 W

What I would like to set up is a configuration in which the output is sent back to the source, perhaps a 20K uF capacitor.  The method by which this may be accomplished is not yet within my full understanding.

I'll try adding a capacitor to the input resistor as you've specified.  Also I may reduce the number of turns on the output coil.