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0 pointspwg2y ago0 comments

> but that’s problematic when the next owner tries to screw in a standard bulb - what happens when it sees 48V DC?

If by "standard" you mean a incandescent tungsten filament bulb, nothing at all.

For a true LED driver power supply, it would be constant current, so the tungsten filament would see 25mA (or whatever the constant current is set for) of DC, and nothing bad would happen (the filament also would not likely illuminate either).

Screwing in an LED bulb with integrated power supply, the external supply will still feed the constant current value, so what happens depends upon the design of the LED bulb's integrated power supply. If 25mA is enough to drive everything, the LED bulb might light up. If 25mA is not enough to drive everything, most likely nothing lights up.

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Dylan168072y ago

48V without a current limit shouldn't be nothing, but you should expect less than 10% brightness.

For constant current, you'd need to drive at least 9 watts so it would be more like 250mA if not higher.

A 1600 lumen LED module might take as much or more current than a 60w incandescent. If your constant current supply can output between 0 volts and input volts, and it's set for a bulb with such a module, it would be able to power an incandescent bulb.

amluto2y ago

I suspect the results would be quite poor. Incandescent filaments increase their resistance when they get hotter, so driving them at constant RMS voltage means that the power will decrease as they heat up, which will give them a degree of stability. At constant current, though, the power will increase with increasing temperature.

(Of course, they’re quite hot and radiative cooling increases like T^4, so this isn’t necessarily a show stopper. But it’s probably not helpful.)

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