In pulse motors there is an additional significant phenomenon that needs to be accounted for.

For simplicity, let's consider an empty air-coil (solenoid) possessing some inductance L

_{MIN}.

**1**) This empty air coil is first energized with current (i

_{MAX}) from some power supply.

**2**) For simplicity of analysis, the charging of the coil's current is stopped at i

_{MAX}, long before the

**point C** on the current waveform is reached and the coil is shorted and a movable soft ferromagnetic element (nonconductive) is introduced a distance away from the coil.

**3**) Now the coil attracts the movable ferromagnetic element towards itself, because current keeps circulating in the coil and magnetic field B-gradient is generated by it.

**4**) As the ferromagnetic element approaches, the inductance of the coil increases and in response to that, the current flowing through its winding decreases.

**5**) When the ferromagnetic element becomes attracted all the way inside the coil, its inductance reaches maximum (L

_{MAX}) and it current falls to minimum (i

_{MIN}). This would happen even in a superconducting coil devoid of resistance.

**6**) The coil is opened and all the energy circulating in it (½L

_{MAX}*i

_{MIN}^{2}) is quickly discharged into a capacitor.

**7**) The movable ferromagnetic element continues to move away from the coil by inertia and unimpeded by attraction from the deenergized coil.

**DISCUSSION**The decrease of current described in pt.4 will be manifested on the scope as flattening of the pulses' tops. This flattening can be confused with reaching the V/R limit of the winding. To distinguish them, a dry run should be made first with the movable ferromagnetic element absent. The results of the dry run can be later subtracted out.

Furthermore, when the coil is discharged into capacitor in pt.6, it should be noted that the smaller this capacitor is the faster the energy in the coil will be discharged into it. The downside of a small capacitor is that the voltage which it becomes charged to, is large.

In other words, the smaller the capacitor - the faster the coil's discharge but the higher the capacitor's final voltage because ½L

_{MAX}*i

_{MIN}^{2} = ½C*V

^{2}, and this transforms to V = i

_{MIN}*SQRT(L

_{MAX} / C)

Also note, that this capacitor and the winding form an LCR circuit that wants to oscillate at the frequency equal to 1/(2π*SQRT(C*L

_{MAX})).

If the goal is to transfer all of the energy from the coil to the capacitor, then this oscillation should be interrupted after ¼ of the cycle, when the current in the coil is zero and the voltage in the capacitor is at its maximum. The energy stored in this capacitor can be transferred back into the power supply or used to energize the coil in the next motor cycle.

Once all of the coil's energy is transferred into the capacitor, the resistance of the winding and the remaining circuit, cannot keep wasting it as heat and you can relax while preparing for the next cycle of your motor.