Go Down

Topic: Can't figure out this sensor (Read 958 times) previous topic - next topic


Hi guys,

I've disassembled a Brother printer and I found a sensor that determines whether the tray is open/close.
It looks like a pretty straight-forward sensor with 3 pins, so I tried to figure out how to use it although I don't have the datasheet, but couldn't :(
I tried all combinations of 5v, GND and input pin. None worked.

I'm attaching two images of the sensor. You'll notice 2 little "bumps" with a small distance between them. When something comes in between, the sensor should detect that the tray is "closed", otherwise it's "open".



Looks like an opto sensor. If you wire it up to power and ground you could have killed it, I suspect it needs a seriese resistor in the power to limit the LED current.


Looks like an optical sensor to me. One side will be an IR emitter - you can identify this using the diode test function on your multimeter (but if you have put 5V across it without, you may have blown it). The other side will be a phototransistor. You need to connect the diode through a series resistor to +5V so that it draws about 10mA, and the phototransistor between ground and a digital input pin, with the internal pullup resistor enabled.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.


It's a OPTICAL INTERRUPTER something  like this one H21A1


I forgot to mention that I a beginner :)
If I understand correctly, what you're suggesting is to connect:
5v -> resistor (10K?) -> pin1
input -> pin2 (pullup resistor enabled)
ground -> pin3

Is that correct? I'm not sure of the order of the pins, is there a standard?

Go Up