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I have the two components, which are in the picture below.

I have a couple of questions. The first one is, how do I make sure that the IR LED detector is working? I tried searching for specs, but the searches for "IR sensor", "IR detector", "phototransistor", and "photodiode" kept throwing inconclusive information, at least for me (I don't have a lot experience in the field).

The second question is, how do I actually use the phototransistor (which I believe is what the one below is)? I tried making a circuit like so:

But I couldn't get the red LED to turn on, even when I made the emitter and detector touch each other.

I have jumped from site to site, but I can't seem to get it right. Could you help me?

Thanks.

LED emitter and detector, right?

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    \$\begingroup\$ Can you get the emitter to work? (Hint: Use your camera to detect it.) \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Nov 26, 2013 at 20:21
  • \$\begingroup\$ Note that if the current is very low, the light coming from the LED will be quite dim. You may need to shield it from outside light in order to tell if it's lit. \$\endgroup\$
    – Scott Winder
    Commented Nov 26, 2013 at 20:23
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    \$\begingroup\$ Your photodiode will work better (block current when no light is input) if it is turned around the other way. \$\endgroup\$
    – The Photon
    Commented Nov 26, 2013 at 20:23
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    \$\begingroup\$ Also with 33kOhm at R1 you're not going to get more than about 10 uA through the LED. You probably want about 100x more to get a reasonable light output. \$\endgroup\$
    – The Photon
    Commented Nov 26, 2013 at 20:24
  • \$\begingroup\$ @ThePhoton So the black component is indeed a photodiode or a phototransistor? \$\endgroup\$
    – OFRBG
    Commented Nov 26, 2013 at 20:26

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Putting a photodiode and a LED in series won't do anything useful. Photodiodes work on much lower currents than LEDs do.

There are two ways to use a photodiode, in forward mode like a solar cell, and in reverse mode where the light causes leakage. Generally the second method is simpler because the current is directly proportional to the light level (of the wavelength of light that the photodiode can see).

Try this:

Q1 and Q2 amplify the tiny leakage current from photodiode D1 to eventually drive the LED D2 to about 10 mA. The LED will light with very little current into the base of Q1. Putting 1 finger on the 5 V supply and the other on the base of Q1 should easily light the LED. To decrease the sensitivity, make R1 smaller.

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  • \$\begingroup\$ Sorry about the trouble. I will indeed use this answer in the future, but it seems that the component is a phototransistor: steren.com.mx/doctosMX/PT1302BC2.pdf I couldn't indicate that in the diagram since there is no symbol for the component. \$\endgroup\$
    – OFRBG
    Commented Nov 26, 2013 at 21:33
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    \$\begingroup\$ @Fiir: A phototransistor has more gain than a photodiode, but otherwise they look pretty much the same to the circuit. Get rid of D1 and R1 in the circuit above, and replace Q1 with the phototransistor. The values as shown will give you maximum sensitivity for this circuit. To reduce sensitivity, reduce R4. \$\endgroup\$
    – Olin Lathrop
    Commented Nov 26, 2013 at 22:41

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