0
\$\begingroup\$

How do I identify if a component is a photo diode or photo transistor? I have a three lead component removed from an old mouse where it was used to do quadrature encoding of the scroll wheel (a normal side facing IR LED was illuminating other side of the wheel) - presumably it houses two photo detectors side by side to distinguish which direction the wheel was being turned. What tests can I do to determine if it is a photo-transistor or photo-diode and appropriate parameters to drive it? I would like to re-purpose this for a micro-controller project I'm working on.

Only markings on the package is "H6.26" on the back and 'EL' on the top.

Close up of the component - sorry, best I can do with 4am shed lighting

In situ view - placed back from whence it came

Managed to trace the two outer pins directly to separate pins on an unidentifiable micro-controller - (each trace is well below 1 Ohm). As for the middle pin it connects through a 10 Ohm resistor to I don't know where

| improve this question | |
\$\endgroup\$
  • 1
    \$\begingroup\$ Scope each pin and test with a pull-up R \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 14 '18 at 16:58
  • 1
    \$\begingroup\$ How was it connected before you removed it? \$\endgroup\$ – Spehro Pefhany Aug 14 '18 at 17:06
  • \$\begingroup\$ Before removing it was connected to one of two three layer circuit boards with the traces disappearing in among 40 to 50 discreet components (was an old Dell bluetooth optical mouse). \$\endgroup\$ – norlesh Aug 14 '18 at 17:29
  • \$\begingroup\$ please post a closeup picture \$\endgroup\$ – jsotola Aug 14 '18 at 17:32
  • \$\begingroup\$ picture added - have been attempting to trace where the leads reemerge but it's a complete birds nest. \$\endgroup\$ – norlesh Aug 14 '18 at 18:15
0
\$\begingroup\$

Photodiode is a diode, phototransistor is a transistor. So take a DMM and measure diode forward voltage between all pins. If it's a diode, you will find one, if a transistor- two. I wonder what will you do with this information.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thanks, that makes sense now (it's been a long frustrating night/morning so probably didn't think things through as clearly as I should have) 0.59V between 1 and 2, 0.68V between 3 and 2 - I don't suppose it would be particularly important that the two diodes are perfectly matched unless dealing with extremely fast transitions. - I'm using the information to measure rotation of a brushed dc motor. \$\endgroup\$ – norlesh Aug 14 '18 at 19:44
  • \$\begingroup\$ Buy an encoder. What you are doing is amateurish. \$\endgroup\$ – Gregory Kornblum Aug 14 '18 at 19:47
  • 1
    \$\begingroup\$ He is learning* \$\endgroup\$ – gregb212 Aug 14 '18 at 20:04
  • \$\begingroup\$ Not this way. Wasting his time is not effective for learning or for business. \$\endgroup\$ – Gregory Kornblum Aug 14 '18 at 20:05
  • \$\begingroup\$ I'd prefer to waist a night learning than a full week waiting for an encoder in the mail. \$\endgroup\$ – norlesh Aug 15 '18 at 2:16
3
\$\begingroup\$

You've surmised that this is a phototransistor (detector). Since it is quadrature system, it may be a dual phototransistor (left), especially if the illuminating infra-red LED was a single emitter.
Which pin is which is unknown. You may have to try all combinations. A test circuit is shown at right. Monitor the output while illuminating with an infra-red source. Unilluminated, the outputs should remain at +5V. A standard incandescent lightbulb should work as a source. The black plastic casing usually blocks visible light amazingly-well.

schematic

simulate this circuit – Schematic created using CircuitLab

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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