Seria interesante saber si la energia extraida es radiante o no sumergiendola en agua y comprobando como la corriente es una pura onda escalar sin electrones por el colapso magnetico de la bobina
It would interested to know if the extracted energy is radiant or no, dipping a bulb in water, & with this checking that there's no electron current, instead of a pure scalar wave of pure potencial due to the coil's magnetic collapse, which is what Bedini, Don Smith. Lindermann & others predicted.
R=E2/power, Thus Bulb is 12x12/6watts=24 Ohms. Coil is 30AWG 0.338496 Ohms per meter. Assume coil dia 75mm x 300 turns thats average of 23.6 circumference which is 70.90 meters of wire , at 0.338 per meter gives 23.99 ohms. Resistors in parallel 23.99 plus 24Ohms = total resistance 12 Ohms. coil bulb , 24 ohms bulb only , just a guess..
Seria interesante saber los amperios que salen de la bateria cuando se conecta la lampara directamente y cuando se conecta la lampara a la salida de la bobina. Ya que la bobina supuestamente tiene una resistencia mucho menor que la lampara seria logico pensar que lo que esta sucediendo es que la bobina extrae la energia mas rapidamente de la bateria transformandola en un campo magnetico y cuando el mosfet abre el circuito el campo se colapsa enviando la corriente extraida hacia la lampara...
Sir I am not trying to load batery, back emf is burned in lamp only` to verify his presence,a connects lamp in direct form to batery, side that its brightness is much smaller than when the oscillator ignites, back emf is burned in the lamp
When you connect amp mitter in series with power suply and the device I think you get a true reading, don't you? Battery never go hot, since I only turned on less then a 5 min. But device in good order still works well. Any other idea what must I do? to het OU?
When just the Bulb was on Amp miter showed .500 m. amp when I turned the timer on bulb turn on very bright but but amp mitter showed 2 Amps. I was really impressed but no OU or access energy here.
Hey Amigo,I built your device to day (july-30-2007) by using N Channel Mosfet IRF 510 device worked just like yours. When I turned on 555 timer, N Channel Mosfet works great, light bulb turns on super right. I further found that, bulb gets even brighter when I removed .1 mf capacitor. I was very excited until I measured the battery and I saw that access energy comes from the battery, no BACK EMF, no Radiant energy! Nothing!
hey amigo, the cuestion is very simply ,put te lamp
direc to batery and look the brigth ,this brightness is equal to 1, when putting lamprara to the circuit the brightness is equal to 10. the question is, where lamp leaves the excess of brightness is clearly a load resisitive,chose by a bridge of diodes and chargea capis its capacity was loaded to voltage of rupture not mattering. where it leaves exes of voltage?
So true. This kind of device draws a lot of current from the battery. A lot more than the light bulb can. The coil then raises the voltage a bit with the fast collapsing magnetic field. No gain.
Seria interesante saber si la energia extraida es radiante o no sumergiendola en agua y comprobando como la corriente es una pura onda escalar sin electrones por el colapso magnetico de la bobina
It would interested to know if the extracted energy is radiant or no, dipping a bulb in water, & with this checking that there's no electron current, instead of a pure scalar wave of pure potencial due to the coil's magnetic collapse, which is what Bedini, Don Smith. Lindermann & others predicted.
artursala2 2 years ago
R=E2/power, Thus Bulb is 12x12/6watts=24 Ohms. Coil is 30AWG 0.338496 Ohms per meter. Assume coil dia 75mm x 300 turns thats average of 23.6 circumference which is 70.90 meters of wire , at 0.338 per meter gives 23.99 ohms. Resistors in parallel 23.99 plus 24Ohms = total resistance 12 Ohms. coil bulb , 24 ohms bulb only , just a guess..
Eltimple 1 year ago
luarsan , tu explicacion es viable ,pero recuerda que al colapsar el campomagnetico tiene que atravesar la
resistencia propia de la bobina, deberia
haber una caida de tencion por la isercion de esta ,pero el campo de retroceso sigue siendo altisimo.
como puede ser????
malagarcha 2 years ago
I am sorry to dissapoint you all here.
This is NO Bemf.
What you do here, is putting a second resistor in parallel with the first one.
That gives a lower resistance. More amps. brighter lamp.
br
Steve
stevie1001H2 3 years ago
don't you see the BLOCKING diod?
armakuni2000 3 years ago
LOL zero point energy ,... ya right XD
there's an old thing called autotransformer (kinda like "inductive kick") go check it out
Szajbave 3 years ago
confunden las cosas ,yo no estoy aqui tratando
de cargar la bateria ,solo verifico que al habrir el circuito se induce una tremenda energia,si conecto la lampara en forma directa
sin el circuito,veo que el brillo es 5,cuando
la conecto al circuito el brillo es 20 ,aun circulando parte de la energia por la bobina,
sin existir transformador elevador alguno
¿como es posible que la lampara brille mas
que cuando es conectada directamente a la bateria.
como usar y capturae esta energia
bansai3 3 years ago
what is the theory behind it?
that the back emf should destroy the battery dipole?
crippled82 4 years ago
Seria interesante saber los amperios que salen de la bateria cuando se conecta la lampara directamente y cuando se conecta la lampara a la salida de la bobina. Ya que la bobina supuestamente tiene una resistencia mucho menor que la lampara seria logico pensar que lo que esta sucediendo es que la bobina extrae la energia mas rapidamente de la bateria transformandola en un campo magnetico y cuando el mosfet abre el circuito el campo se colapsa enviando la corriente extraida hacia la lampara...
luarsan 4 years ago
Sir I am not trying to load batery, back emf is burned in lamp only` to verify his presence,a connects lamp in direct form to batery, side that its brightness is much smaller than when the oscillator ignites, back emf is burned in the lamp
only test presense emf ,burning in the lamp.
bansai3 4 years ago
When you connect amp mitter in series with power suply and the device I think you get a true reading, don't you? Battery never go hot, since I only turned on less then a 5 min. But device in good order still works well. Any other idea what must I do? to het OU?
mksboysal 4 years ago
So the battery is getting hot? Will the battery be emtpy earlier? Because you may be fooled when measureing the amps at the battery.
challenger159 4 years ago
When just the Bulb was on Amp miter showed .500 m. amp when I turned the timer on bulb turn on very bright but but amp mitter showed 2 Amps. I was really impressed but no OU or access energy here.
mksboysal 4 years ago
Hey Amigo,I built your device to day (july-30-2007) by using N Channel Mosfet IRF 510 device worked just like yours. When I turned on 555 timer, N Channel Mosfet works great, light bulb turns on super right. I further found that, bulb gets even brighter when I removed .1 mf capacitor. I was very excited until I measured the battery and I saw that access energy comes from the battery, no BACK EMF, no Radiant energy! Nothing!
mksboysal 4 years ago
hey amigo, the cuestion is very simply ,put te lamp
direc to batery and look the brigth ,this brightness is equal to 1, when putting lamprara to the circuit the brightness is equal to 10. the question is, where lamp leaves the excess of brightness is clearly a load resisitive,chose by a bridge of diodes and chargea capis its capacity was loaded to voltage of rupture not mattering. where it leaves exes of voltage?
bansai3 4 years ago
So true. This kind of device draws a lot of current from the battery. A lot more than the light bulb can. The coil then raises the voltage a bit with the fast collapsing magnetic field. No gain.
kudortamas 4 years ago