I might design a box that has a 5000 volt power supply inside. I was thinking that I could have a warning LED (or two independent ones, for redundancy) that indicate that HV is present. It could be inside, shining through an overlay on the top cover, so it's visible with the cover on or off.
So, how do I drive an LED pretty bright, from 5KV down to the officially safe level of 48 volts?
Clearly a resistor won't work very well, with a 100:1 voltage range. Even 1 mA at 5 KV is 5 watts dissipation.
One could charge a cap through a biggish resistor and, when it hits maybe 30 volts, discharge it into the LED and get a flash. Rate would be about proportional to voltage. But it's still just as efficient as a simple resistor:
2.5 volts of LED drop from a 5KV supply is 0.05% efficient no matter how you time slice it.So I need a more efficient 5KV-to-LED circuit (or 48V-to-LED), some sort of switcher. It needs to be low power dissipation, small, and preferably made from standard surface-mount parts. Maybe charge that cap to 30-ish volts through a resistor, but dump it into the LED with some efficiency, to make a bright flash, Still messy.
I've used depletion fets into LEDs as cap-charge indicators. They are cool, constant-current, constant rate discharge into the LED all the way down to about
5 volts, where the LED winks out. But a depletion fet string good for 5KV would get ugly, and efficiency is still 0.05%.I could use a neon relaxation oscillator, but I don't trust neons, and they wouldn't work at 48 volts.
Maybe something with a lithium battery or supercap in it. Ditto messy.