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How can we find the emitter and collector of an LDR. I am using it in the following circuit but I don't know how to identify which terminal of the LDR is emitter or collector.

http://i.stack.imgur.com/zo1id.jpg

Here it is mentioned as Photo-transistor but it is basically 2 leg transistor which is hard to find in the market so I decided to use LDR.

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    \$\begingroup\$ @W5VO Maybe he means his local market? The fact something is available on Digikey doesn't mean that it's accessible enough to use. \$\endgroup\$ – AndrejaKo Jun 22 '12 at 4:23
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    \$\begingroup\$ @W5VO I doubt it. Also the question isn't about phototransistors. It's about LDRs. \$\endgroup\$ – AndrejaKo Jun 22 '12 at 4:27
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    \$\begingroup\$ Hang on a minute, I'm totally puzzled now - in your previous question you give the part number as ST1KL3B which is definitely a phototransistor and what the answers were based upon. Now it's an LDR? If so what part number is it? LDRs are different to Phototransistors. \$\endgroup\$ – Oli Glaser Jun 22 '12 at 5:27
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    \$\begingroup\$ @AndrejaKo - Any phototransistor should suffice, as you can just leave the base floating on a 3-pin phototransistor and it will function the same as a 2-pin. \$\endgroup\$ – Oli Glaser Jun 22 '12 at 6:06
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    \$\begingroup\$ @Oli Glaser I'm aware of that, but looking at the last sentence of the post, looks like OP isn't. \$\endgroup\$ – AndrejaKo Jun 22 '12 at 6:07
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You don't. Light dependent resistor is a resistor which means that it doesn't have a collector and emitter. It instead has two terminals that are used just like on regular resistor.

You can actually see that from the schematic symbol:

LDR symbol from Wikipedia

Both leads are the same and it doesn't matter in which direction current goes through it.

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  • \$\begingroup\$ I think the OP was overthinking his phototransistor. This looks like a transistor that is not polarized, i.e., it doesnt mattter what end is connected to what, this will still act like a transistor. (Interesting fact, transistor = trans + resistor = changeable resistor) \$\endgroup\$ – CyberMen Jun 22 '12 at 12:53

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