I find it easier to understand like this:-When current flows through a coil it generates a magnetic field. When this current is suddenly turned off there is nothing to support the field and it starts to collapse. As it collapses the magnetic lines of force cut through the coil. That is how electric generators work, magnetic lines of force cutting through a coil. The way physics works is that the polarity of the voltage this collapsing field produces is opposite in polarity to the polarity that generated the field in the first place.Hence the coil produces a voltage which is opposite or the reverse of the original voltage that produced it. As the field is collapsing it does that a lot faster than the field was originally made to grow and hence the reverse voltage is higher than the voltage used to create the field in the first place.I hope you find this useful.
... With a diode across the inductor the force that is necessary is 0.7V, which is what happens. The alternative path through the transistor involves exceeding the breakdownvoltage of the transistor, which being a lot more than 0.7V, means the current just diverts from the transistorto the diode when the transistor switches off.
Hi,This might help;http://www.douglaskrantz.com/ElecFlybackDiode.htmlTom..
Is it safe to assume that different diodes have different voltage requirements, or are they all about the same 0.7?