I've been doing some electronic tutorials over at allaboutcircuits.com. The idea comes from this lesson: https://www.allaboutcircuits.com/tex...-and-calculus/
At the bottom of the topic there is a circuit showing a neon bulb in parallel with an inductor and the sentence "If current through an inductor is forced to change very rapidly, very high voltages will be produced."
Well I am thinking why not change the bulb for a capacitor and capture the energy? I did some googling and found that the formula for energy of a capacitor is .5CV^2. Also the capacitor voltage would = the breakdown voltage of the inductor which is V = L di/dt. di is large and dt small which gives a large V. The capacitor current is I = C dv/dt. Here again dv is large and dt small which gives a large I.
I'm not sure how this would be done. But why not siphon of some of the energy to repeat the procedure instead of using a battery. The rest could be used to power a load. This should be able to continue indefinitely as the breakdown voltage is much larger than the voltage required to charge up the inductor.
One idea I have is to use 2 capacitors in parallel with the inductor. One with a small capacitance to keep the circuit running and a larger one for the load. The load capacitor would have to use diodes? (I guess) to prevent it emptying back into the inductor. It would also be necessary to employ some switching arangement (transistor?) to obtain the high breakdown voltage.
The idea seems a bit too simple so I wouldn't be surprised if it doesn't work. I've put the topic up at the allaboutcircuits forum and so far nobody has given me a convincing argument why it wont work. Somebody said you can't get out more than you put in. But that's not what the maths (pretty simple stuff. Just what I have mentioned in this post) is telling me.
At the bottom of the topic there is a circuit showing a neon bulb in parallel with an inductor and the sentence "If current through an inductor is forced to change very rapidly, very high voltages will be produced."
Well I am thinking why not change the bulb for a capacitor and capture the energy? I did some googling and found that the formula for energy of a capacitor is .5CV^2. Also the capacitor voltage would = the breakdown voltage of the inductor which is V = L di/dt. di is large and dt small which gives a large V. The capacitor current is I = C dv/dt. Here again dv is large and dt small which gives a large I.
I'm not sure how this would be done. But why not siphon of some of the energy to repeat the procedure instead of using a battery. The rest could be used to power a load. This should be able to continue indefinitely as the breakdown voltage is much larger than the voltage required to charge up the inductor.
One idea I have is to use 2 capacitors in parallel with the inductor. One with a small capacitance to keep the circuit running and a larger one for the load. The load capacitor would have to use diodes? (I guess) to prevent it emptying back into the inductor. It would also be necessary to employ some switching arangement (transistor?) to obtain the high breakdown voltage.
The idea seems a bit too simple so I wouldn't be surprised if it doesn't work. I've put the topic up at the allaboutcircuits forum and so far nobody has given me a convincing argument why it wont work. Somebody said you can't get out more than you put in. But that's not what the maths (pretty simple stuff. Just what I have mentioned in this post) is telling me.
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