It depends upon the coil inductance, the coil current, and how fast you want to start and stop the current.How would I go about figuring out how large of a inductive voltage spike I would get when I discharge a coil?
A lot would depend on the voltage you charged it at, not the amperage. With it being an open circuit at the shutdown of charging where is the amps or volts going to go? If the volts are high enough you will get a spark across the air gap, but only if it is high enough to jump your air gap. Voltage jumps the gap not amperage.If I switch the circuit open, what would the slope look like?
I think the current goes instantly to zero here. With the switch open, how could it decay gradually? The voltage spikes, and then decays as the magnetic field dissipates. If the peak voltage is high enough, it can arc across the switch. I am interested, but I do not know how this voltage curve is calculated...When the switch opens, we have a open circuit, but current though the inductor keeps flowing
Excellent. Thank you for this explanation!When the circuit breaks, the inductance energy is pumped into the energy of the parasitic capacitor.
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