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p75213
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This post is inspired by a lesson on the allaboutcircuits education website. In chapter 15 of Direct Current there is a heading "Inductors and Calculus" - https://www.allaboutcircuits.com/textbook/direct-current/chpt-15/inductors-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? The formula for energy of a capacitor is .5CV^2. Therefore the higher the voltage the more energy and the formula for voltage in an inductor is V = L*di/dt. At the instant the switch is turned of di would be at a maximum and dt a minimum giving a large V.
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. Timing would also be an issue with the larger capacitor taking longer to charge.
This just seems too simple so I'm wondering why it won't work?
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? The formula for energy of a capacitor is .5CV^2. Therefore the higher the voltage the more energy and the formula for voltage in an inductor is V = L*di/dt. At the instant the switch is turned of di would be at a maximum and dt a minimum giving a large V.
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. Timing would also be an issue with the larger capacitor taking longer to charge.
This just seems too simple so I'm wondering why it won't work?