He is absolutely right! The 47u is fully charged to 5V. Once the transistor turns on (which must be fast), one side of the cap gets forced to ground. Since the cap will resist instantaneous voltage change, it will cause the other side of the cap to be 5V lower, i.e. -5V (since the 100 ohm resistor prevents a direct connection to ground on the inductor / coil side). For a short time, the voltage across the coil will be creater than 5V. The only things that I see that prevent this from being a straighforward solution is:
1) The 5V control / supply signal for the coil must be able to be quickly turned on / off & supply enough current to not be "browned out".
2) The turn on time of the relay must be fairly small, like 100 micro seconds. Increasing the cap could increase the negative supply.
3) The turn on voltage must be as low as 8V. This circuit will not produce exactly -5V, but closer to -4V on the bottom of the coil. Therefore, to have some margin, the coil should energize down to around 8V.
Otherwise, this is a wonderful circuit and I will keep this one in my personal library! Tensi, you get a star for sticking to your guns and backing up what you claimed!!!
Excellent thread!
BTW, I was able to verify the theory by running a simple simulation with PSpice...