General > General Technical Chat
Powering an LED from international line voltage
TimFox:
I have used NE-51s and NE-2s since 1970.
The problem is that over the years electrode metal sputters onto the inside of the glass and darkens the display.
xvr:
I'd peek some circuit from one of PE article. It can solve here.
switchabl:
--- Quote from: soldar on May 03, 2024, 01:41:29 pm ---If all you want is an indication that power is ON then NE-2 neon + resistor is difficult to beat in terms of simplicity and reliability. It is what is still used in switches and power strips. I have quite a few that are half a century old and working like the first day.
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There is nothing wrong with that but there are bidirectional/bipolar LEDs (two antiparallel diodes in a single package) that are basically a more efficient drop-in replacement.
coppice:
--- Quote from: mikeselectricstuff on May 03, 2024, 09:10:09 am ---Capacitive dropper plus some resistance to make the current more consistent- this will not be constant current but may be good enough - does brightness change over supply voltage really matter ?
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More specifically you need enough resistance to limit the capacitor charging current if you turn on near the peak of the mains. Otherwise you can pop the LED. You probably want a reverse diode across the LED, too. Watch out if the product might be supplied from a highly distorted mains, like some UPS solutions give you. The low impedance of the capacitor can put a high current through both LED and resistor with a lot of high frequency harmonics. Quite a few products with cap drop supplies can catch fire with a distorted waveform, as the resistor cooks. Use a large one.
BrokenYugo:
--- Quote from: Gyro on May 03, 2024, 09:17:46 am ---
@OP: I would suggest something like a mains dropper, feeding into a small package, low current bridge rectifier [Edit: or discrete diodes] (flicker removal), then a Zener of, say, 4V7 followed by a small resistor to limit the LED current down from 4V7.
Assuming an efficient LED @ 10mA, a resistive mains dropper would dissipate 750mW at 80V and 2.5W at 250V - much too high in my opinion. A capacitive dropper is the way to go. Assuming a 470nF safety rated capacitor, this would give ~12mA at 80V (Zener essentially dissipating nothing) and 37mA to 250V. At maximum voltage, the dissipation of the Zener, after taking off the 10mA for the LED would be only 127mW, so a 250mW rating would be fine. Don't forget to put a low value series resistor (preferably fusible) of around 100R as the first component between mains live and the capacitor to cope with mains spikes.
EDIT: My capacitor reactance calculation was for 50Hz. At 60Hz, you would get away with a 390nF cap, maybe even 330nF.
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This is exactly what you see inside a cheap LED night light, with the addition of a transistor+LDR to short the current source during the day. They usually use a half wave rectifier and a small low voltage electrolytic for smoothing, but for an indicator I don't find 120hz very flickery. Very proven design.
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