LED replacement for 48" fluorescents

You mean those things with a capacitor on the AC side of a bridge rectifier? UGH!

What I'm using is an isolated, current-regulated supply. It gives a very flicker-free output, which I really like.

Jon

While you could use the inductive reactance of a choke as in fluorescent lamps, but this requires quite a large inductance, due to the low current of most LEDs. Of course, if you put about 100 "1 W" or "3 W" LEDs in series, you might be able to run them with a fluorescent ballast :-)

So you are talking about the series capacitor. This of course works well as long as the input voltage is a clean sinus waveform without harmonics (from electronic or motor loads) or lightning induced voltage peaks.

Think what happens to the capacitive reactance when such "unexpected" voltages occur. You might need quite a lot additional protection against these.

Yes, this is a very cheap, dirty solution, but with some large capacitors on the output end, the light produced is very stable.

Typically, a large value, high wattage "inrush" resistor and a varactor prevents damage due to voltage spikes. In addition, most of these LED lamps are attached to a Edison or GU10 wall socket, and so if the circuit fails, the switch is flicked off, the lamp is replaced. Same idea is applied to the el cheapo wall warts -- once they are cooked, get a new one. While a power supply with isolation is certainly one's best choice, they are heavy, and certainly cannot be held to a wall socket with a Edison or GU10 fixture.

only on 240V AC 120V wouldn't be enough.

just a single extra capacitor is all

R1 C1 ---\/\/--||---(~) (+)--+------ BR1 === C2 --------------(~) (-)--+-----