2 ideas, first put a good diode or 2 between the charging cap and the inductor, this way you may be able to keep the inductor and charging cap connected, will simplify automation. (but maybe you won't get the free energy?)
Second, take say 10 capacitors or more, use 5 in parallel as the power cap, and 5 in series as the charging cap, if they are good capacitors and don't lose energy too fast once charged you can reconfigure your charged cap to be your power cap and visa versa.
Ofcoz you can pump up some exciting voltage with a boost forward converter. But there is no need to demonstrate it, since it's use and technology is widely known.
Think critically about the cap size shown here, that blue cap is not 40V 7500µF.
Just do this experiment yourself. It is easy to get some high voltage capacitors and then charge them like this. Then you can see for yourself if it works or not. Thanks!
no chewing gum
eurogoldexchange 4 months ago
I tried it with high voltage cap of 400V. Eff about 94-95% as it should be =(
NoctumDeVir 2 years ago
2 ideas, first put a good diode or 2 between the charging cap and the inductor, this way you may be able to keep the inductor and charging cap connected, will simplify automation. (but maybe you won't get the free energy?)
Second, take say 10 capacitors or more, use 5 in parallel as the power cap, and 5 in series as the charging cap, if they are good capacitors and don't lose energy too fast once charged you can reconfigure your charged cap to be your power cap and visa versa.
aether222 2 years ago
in order for me to see the validity of this being viable is showing all of your wires
legalizeitnow09 2 years ago
Ofcoz you can pump up some exciting voltage with a boost forward converter. But there is no need to demonstrate it, since it's use and technology is widely known.
Think critically about the cap size shown here, that blue cap is not 40V 7500µF.
More like 40V and 75000µF.
Paxmax 2 years ago
Just do this experiment yourself. It is easy to get some high voltage capacitors and then charge them like this. Then you can see for yourself if it works or not. Thanks!
Subspace4d 2 years ago
@ 1:09 he just takes the wrong wire and hooks CAP2 to the source instead of cap.
Folks, there is no magic here.
Paxmax 2 years ago
At 1:09 this happen. C1=7500µF; C2=30µF; U1(1)=25,88V; U1(2)=25,87V ; U2=25.70V
Useing the formula: Energy = C U² /2 you can calculate that.
dE1=7500µF ((25,88V)² - (25,87V)²)/2=1.9milliJoule
dE2=30µF 660.49V² /2=9.9milliJoule
9.9milliJoule - 1.9milliJoule = 8milliJoule
So we made a win of 8milliJoule !!! We made more than we used. Correct baloghcsongor! Magic is really happening! Thank you
Subspace4d 2 years ago
Subspace4d . Yep, you're right about this one, but do this all again with data from 1:09! There's the magic moment.
baloghcsongor 2 years ago
Is this the Ed Grey effect? But where is the catcher electrode wrapping this?
Subspace4d 2 years ago