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I have a vague recollection that while off, a LED can actually receive light and generate a small current (in uA). Is that correct? How does that work? What is the best way to prevent this?

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A LED can work as a (not very good) photodiode, and produce a small current when it receives light. You have to connect it like a photodiode, though, that is reversed polarised. In a normal circuit you won't have trouble with it.

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    \$\begingroup\$ Yes, and in principle, any reverse-biased PN junction is sensitive to light. The more fossilised of us will remember the OC71 germanium transistor which was encapsulated in a small glass envelope painted black. The manufacturers (Mullard) then produced the OCP71 phototransistor in a clear glass envelope which was much more expensive. Enterprising hackers soon realised that you could just scrape the paint off an OC71 and get a phototransistor with identical performance. \$\endgroup\$ – MikeJ-UK Apr 18 '11 at 15:25
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    \$\begingroup\$ Mullard got wise to that, and painted the inside of the glass! \$\endgroup\$ – Leon Heller Apr 18 '11 at 16:39
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    \$\begingroup\$ Why did they use glass? Cheapest to seal? \$\endgroup\$ – stevenvh Apr 18 '11 at 17:14
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    \$\begingroup\$ Does this mean that a LED technically consumes slightly less current when turned on in a very bright environment compared to a dark one? :-) \$\endgroup\$ – Kellenjb Apr 18 '11 at 17:58
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    \$\begingroup\$ Forest Mims used colored LED's specifically to get wavelength-matched photodetection of the sun. opticsinfobase.org/abstract.cfm?URI=ao-31-33-6965 \$\endgroup\$ – Toybuilder Apr 21 '11 at 0:24

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