I am designing a coil gun and would like to dump some energy from each coil into the next coil before (preferably) dumping its remaining energy back into its storage capacitor.
If I have two coupled coils, with current only flowing in one, and then switch the other one in series with it (common mode) will the current rapidly drop to nearly half (since the magnetic field is already partially established through both coils) or will something else happen?
Producing a voltage spike on the MOSFET that cannot be easily supressed would be quite undesirable, I think the capacitor on the MOSFET is still necessary as the magnetic field still needs to shift a fair bit since the coils are not co-wound. If the current can be brought near equal very quickly this will be very beneficial as the magnetic equilibrium point for the slug will be shifted past the first coil very rapidly and without using the capacitor for the second coil at all.
Here is a basic partial diagram of the concept, the diode is just to prevent a current flowing backwards in the second coil while the first coil charges. The MOSFET would be turned off to switch the second coil in. The coils would be two cylindrical coils placed next to each other coaxially.
EDIT: MOSFET replaced with switch for clarity
Ok so I have clarified the circuit a little (sorry I just use eagle since I have it open for another project). So what I would like to know is if the switch is closed and there is a fair magnetic field established in L1 what will happen when you open the switch? Will the current in L2 increase very rapidly and roughly proportionally to a decrease in L1 current, or will L2 charge at a pace similar to as if they were not coupled while L1 tries to push its existing current through the small snubber capacitor and destroy the MOSFET.
I am pretty sure it is the former, and I feel like it would be very rapid, but I would like to know from someone who properly knows these things roughly how two adjacent air core inductors will behave in this situation.