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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.

circuit

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.

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    \$\begingroup\$ It's not clear what you are trying to achieve and your circuit doesn't seem to lift the fog on your intentions. State what you are trying to achieve and don't assume that a partial circuit or some attempt at a circuit is going to help on this occasion. \$\endgroup\$ – Andy aka Feb 24 '17 at 9:19
  • \$\begingroup\$ 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. \$\endgroup\$ – TWiz Feb 24 '17 at 10:38
  • \$\begingroup\$ 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. \$\endgroup\$ – TWiz Feb 24 '17 at 10:44
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As the first approximation the transformer effect can be ignored due the loose magnetic connection between L1 and L2.

Open the switch -> the current from L1 starts to collect voltage to the C. That voltage makes a gradually growing current to L2. It finally sinks the C empty, but there's still a growing current through L1 and L2 if nothing controls the input voltage. In practice it soon develops to a short circuit. This happens already when the switch is on, if the input voltage is a constant DC without another control somewhere.

If your input voltage comes from a charged capacitor, this circuit is NOT impossible. L2 really can have it's max current at the right moment and the resonance can return a substantial part of the energy.

This circuit is quite easy to simulate if you know the inductances. You can assume max. few % coupling between L1 and L2 and see its effect. The effect of the projectile needs experiments or some ultra high cost software that needs a phd to input all parameters properly.

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  • \$\begingroup\$ I don't think the coupling between the inductors can be ignored? Looking at similar adjacent coils the coupling coefficient should be at least 0.2 or 0.3 or so. I know the effect will be dependant on that but would like to know if this effect is something that can be achieved, I could always overlap the coils a little to increase the coupling it was worthwhile. And yeah the input is from a capacitor, that circuit is just cut down to what is needed for an example. \$\endgroup\$ – TWiz Feb 24 '17 at 21:29
  • \$\begingroup\$ @TWiz 20...30% k is substantial. The coupling is useful because the increasing current in L2 reduces the current in L1 which you probably want, if you want to shoot. Do the simulation. It helps you to get something ready for experiments. Only a genius can handle this complex circuit ok also quantifically by the intuition. \$\endgroup\$ – user287001 Feb 24 '17 at 21:48
  • \$\begingroup\$ Thanks, yeah I was expecting to have to simulate it, but if it was sure to fail I would not bother since the model where there is no connection between stages is much simpler. I won't use this to start with but will build it so that it can be adapted to this configuration. \$\endgroup\$ – TWiz Feb 24 '17 at 22:21

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