Electrical How to use capacitors in a Solenoid circuit

[lekh2003]

The current in an inductor continues to flow in the same direction when you disconnect it. It is the voltage that reverses, causing the “flyback” voltage spike that can damage semiconductors.
With metal contacts in a relay, a high reverse voltage arc will form as the contacts separate. That negative voltage makes di/dt = V / L very high so the current flow rapidly reduces.
The energy that was stored in the inductor's magnetic field appears as heat in the arc. E = ½·L·i²

Ah, so if I'm understanding correctly, the solenoid still has a pullback affect on the projectile, but the energy from that pullback is transferred to heat energy in the relay module?

If that were the case, it explains why the relay modules heat up during the experiment.

 

[Baluncore]

If that were the case, it explains why the relay modules heat up during the experiment.

The two relays heat due to the resistance of the coil. One relay is also heated by the arc between the opening contacts. If you take the cover of that relay and operate it in the dark, you should see that arc.

A "flyback" or "freewheeling" diode across the gun coil would drop a low reverse voltage, (≈1 volt), so the current continues to flow for a longer time, and the projectile will be pulled back.

The arc at the opening relay contacts can be timed to rapidly kill the magnetic field as the projectile enters the centre of the coil. That greatly reduces pull back of the projectile.

A capacitor discharge through a diode to the gun coil produces a single half sine wave pulse, so it does not cause pull back of the projectile, and it does not produce a flyback pulse that might destroy a semiconductor switch.

 

[lekh2003]

The two relays heat due to the resistance of the coil. One relay is also heated by the arc between the opening contacts. If you take the cover of that relay and operate it in the dark, you should see that arc.

A "flyback" or "freewheeling" diode across the gun coil would drop a low reverse voltage, (≈1 volt), so the current continues to flow for a longer time, and the projectile will be pulled back.

The arc at the opening relay contacts can be timed to rapidly kill the magnetic field as the projectile enters the centre of the coil. That greatly reduces pull back of the projectile.

A capacitor discharge through a diode to the gun coil produces a single half sine wave pulse, so it does not cause pull back of the projectile, and it does not produce a flyback pulse that might destroy a semiconductor switch.

Thank you very much for the explanation. It makes sense.

The fact that a capacitor would have been a better choice is also useful information. I can point out that it exists as a weakness of my experimental setup.