Inductor turn-off: Can you help me understand the theory behind what happens when the current through an inductor is abruptly stopped? Specifically, I'm confused about why the voltage that appears across the inductor is reversed, relative to the original direction of current flow. As a practical matter I could just drop in a flyback diode to handle the voltage spike, and not worry about it WHY it works, but I really want to understand it better.
Take a look at this diagram from the Wikipedia page about flyback diodes:
http://en.wikipedia.org/wiki/File:FlybackExample.GIFIn figure 1, the switch is closed and current flows through the inductor, eventually reaching a steady state. The upper terminal of the inductor is at a positive voltage relative to the lower terminal.
In figure 2, the switch is opened, cutting off the power supply. According to Wikipedia and every other source, this creates a large NEGATIVE voltage across the inductor, making the upper terminal negative relative to the lower terminal. But if the "goal" of the inductor is to resist changes in current flow, and it wants to keep current flowing in the same direction as it was when the switch was closed, then shouldn't it create a large POSITIVE voltage? That would keep the current moving as it was before. By creating a large negative voltage, the inductor is actually inducing current to flow in the opposite direction as before.
In figure 3, my confusion is compounded. This figure shows current flowing the wrong way through a potential gradient across the inductor. Huh?