1. I have to confess I do not understand what you explained. I thought that the inductor "wants" to hold on the current and increases voltage. The diode should then open quench this voltage. Is there something else going on ?
2. Thank you. I will try to find that book and learn more. Thank you again.
Yes, the inductor opposes changes in the current flowing through it and releases the energy stored in its magnetic field to oppose that change of current. All real inductors have a finite resistance so the the voltage across the inductor is composed of two components: the voltage across the inductive component and the voltage drop across the resistive component of the inductor.
Consider the situation where you have the inductor connected at the top end to your 24V supply and the bottom end is connected through the switch to ground. In the steady state a current will be flowing through the inductor determined by its resistance and the supply voltage, I = V/R. The current flowing through the inductor will have stored a certain amount of energy (E=1/2 LI
2 ) in its magnetic field but the voltage across the inductive component is zero because the current in not changing.
Upon opening the switch the current the inductor current will start to reduce but the inductor will convert some of that stored magnetic energy to produce a voltage across itself, V=L di/dt, where di/dt is the rate of change of current. Since the current change is negative the voltage across the inductor is negative and opposes the 24V drop across the resistive component and the voltage at the bottom end of the inductor will start to rise. The more rapid the decrease in current the greater the negative voltage across the inductor and if the current decreases rapidly enough the bottom end of the inductor can actually produce a spike "fly-back" voltage which exceeds the 24V supply by a considerable margin, maybe hundreds of volts.
The rapidly increasing voltage across the opening switch creates an arc which is sustained by the stored energy in the inductor until such time as the switch contacts are too far apart for the the inductive spike voltage or all the stored energy in the inductor is dissipated in the switch arc and the resistive losses in the inductor. Thus the flyback voltage can produce a considerably larger arc and cause more damage to the switch contacts than what might be expected from the 24V supply alone.
The addition of the diode across the inductor helps dissipate more of the stored energy in the resistance of the inductor rather than in the switch arc. As the voltage across the inductor flies back the diode will start conducting when the voltage at the bottom of the inductor reaches the 24V supply voltage plus 0.7V for the diode. When the diode starts conducting the current flows in a circular loop through it and the inductor until all the remaining stored energy is dissipated in the resistance of the inductor.
The effect of the diode is to clamp the maximum voltage across the opening switch to 24.7V rather than the hundreds of volts that might be seen in the case of an unclamped inductor. There is still a switch arc produced during the period while the voltage across the switch is increasing from zero to 24.7V but this arc is more quickly extinguished than in the unprotected case. To fully extinguish this remaining smaller arc requires a more complex resistor-capacitor-diode network across the contacts as detailed in the Ott book.