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the sum of the EMF and wiring voltage drop must exceed the battery voltage plus 2.5 volts for current to flow. Even if the lowside FET is still turned on, the junction will be shunted by the diode when it is forward biased.
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For further discussions, I will always refer to Vemf as a battery or source and due only to the rotation of the motor. I refer to V_L as the inductor voltage that is caused only by the current rate in the wires.
I still don't agree with the quoted statement and its time we drew some pictures. I will post a few shortly...but I think its the current rate that you are neglecting. As soon as the 12v is switched off, the field created by the current collapses and creates a voltage drop across the inductor that tries to sustain that current. If the low side diode wasnt there it would indeed cause a spark and almost instanteous discharge of the coil. But, the low side diode is there to allow current to flow in the same direction but with a negative rate. So the inductor voltage jumps to the sum of :
V_L = -(i*R + Vemf + V_diode ) and the current now decays at a rate
di/dt = V_L / L until the current goes to zero.
The low side FET that is ON is still conducting current in the same direction as the charge period. (As far as I know, these MOSFETs can conduct current in both directions when turned ON but that is an unknown to me.)