I would like to know if this is a sensor and the name of it if it is possible. I post two pictures, sorry about bad quality of them.

In my opinion it works sending different signal if something is between little "crystal sticks" into the dark of the black box.

Would be wonderful to know the schema to connect it to an Arduino, also if is mandatory to add some other component like resistor (I should post another question for this, may be)

Both components are very similar with a few little differences.

Thanks a lot in advance.

enter image description here

enter image description here

  • 5
    \$\begingroup\$ It looks like it might be optically detecting something being in the slot, but that's just a wild guess. \$\endgroup\$ Aug 10, 2017 at 11:46
  • 3
    \$\begingroup\$ I agree with Olin... It looks like a simple opto interrupter for a slotted disk. The odd thing, is that unless there are two boards on each black casing, is that it doesn't look like it is quadrature, so would not be able to determine direction. \$\endgroup\$
    – R Drast
    Aug 10, 2017 at 12:07
  • 1
    \$\begingroup\$ Looks like one is an infrared led (A and K on the PCB) and the other is a photo transistor (C and E on the PCB). \$\endgroup\$
    – HandyHowie
    Aug 10, 2017 at 12:08
  • \$\begingroup\$ @RDrast If the device knows which way it's turning but doesn't know how fast, determining direction might not be necessary. \$\endgroup\$
    – Dampmaskin
    Aug 10, 2017 at 12:10
  • \$\begingroup\$ Is from a robot vacuum, I guess when robot bump against the wall a little piece goes back and is placed between the 'little sticks'. Is not from a mouse, not slotted disk. \$\endgroup\$
    – danihp
    Aug 10, 2017 at 12:28

1 Answer 1


This is a simple transmissive optical sensor. The LED produces (infrared) light, and the phototransistor senses it

Connect the LED (K to ground, A to +5 through 270 ohms or around that). Measure the voltage across the LED- it should be around 1.2V. Don't omit the resistor or you will destroy the LED. You won't be able to see the IR light, but a camera might pick it up. If the voltage is 1.2V then it is almost surely producing light.

Connect the E to ground, C to +5 through 20K. Measure the voltage at C relative to ground- with nothing inserted into the slot. If it is not less than 0.6V or so increase the resistor, up to a maximum of maybe 200K.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ I would have said transmissive, as the transmitter and receiver are not on the same side, but one in front of the other... \$\endgroup\$
    – next-hack
    Aug 24, 2017 at 13:25
  • \$\begingroup\$ @next-hack Thanks, corrected. I missed that that parts were side-emitting (and side-sensitive). \$\endgroup\$ Aug 24, 2017 at 13:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.