If you want to put non-DC voltage across an LED & its series resistor (for example, the always non-negative but "bouncy" output of an unsmoothed rectifier bridge), do you use the RMS or peak voltage (& therefore current) to work out the value of the series resistor?
--
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You read the diode data sheet carefully. I'm pretty sure that the important parameter is heating due to current flow in the diode -- but the heating happens at several time constants. Moreover, if I remember correctly and at least in older LEDs, there was a significant issue with high current densities tending to drag aluminum nuclii along with them (or perhaps opposite the flow), creating a wearout mechanism.
If it's just thermal, and if it's fast enough that it all gets averaged out, and if the pulsed current rating of the device isn't exceeded, then the important parameter is average current (which is _not_ proportional to average voltage), because the voltage across a forward-biased LED is more or less constant, so the heating is more or less proportional to current.
'course, if you're current limiting with a smallish resistor, you should make sure that doesn't burn up either.
-- Tim Wescott Wescott Design Services http://www.wescottdesign.com
It's complicated. Spice it!
The average LED current is what will determine the visual brightness.
The RMS current heats the resistor.
-- John Larkin Highland Technology, Inc jlarkin att highlandtechnology dott com http://www.highlandtechnology.com
"Adam Funk"
** Neither - the average DC voltage as read on a normal meter what you need. The rated current of a LED is the average value, which equates to the DC voltage across the series resistor divided by its value.The most reliable way is to first calculate a resistor value using the difference between the expected LED voltage and the DC supply voltage. Then fit such a resistor and test the DC voltage across it in normal use.
Amend the resistor value if necessary to get the average current level needed.
... Phil
That's a good starting place, often the led will be too bright at rated current anyway.
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Thanks very much (& to everyone else who replied) for the interesting comments.
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