This circuit board is soldered at the back of a dc motor The motor has been removed and a magnetic disc is attached to the shaft and spins over these smd components. They seem to say A87L but I couldn’t find any info.
It has solder pads for 4 wires but there are only 3 used.
Question is: "How can I tell if the shaft is rotating with this encoder?"
I expected it had a +VCC, GND and two Outputs, but this has only three wires. I understand quadrature encoders, I can count steps up and down. But not sure how this one works.
I'd like to write an arduino like program or wire some i2c pulse counter to it.
The motor is intended to be driven by 12V DC, in both directions as it's taken from a linear actuator of the back door of a car.
The only BIAS visible in the image is connecting the isolated left pin on this side with the bias there, connecting the center pin of the left IC.
Also the component numbered 4302 makes no sense to me, unless this is a multilayered PCB (which does not seem to be) it's just there.
Thanks, and sorry for not putting the question clear.
UPDATE: Here is what I did. I supplied +12V to the middle pin (this will give +12V to the VCC pin according to the schematics) Now I only had the other pin left so I connected it to the long pin of a LED, then the other pin of the led in series with GND. LED lit up. Now using the disc I can see how the LED lits a two distinguishable brightness levels. Now, what would be the involved maths to calculate the required components to have a pin pull up down at 5V so I can connect it to a 5v Arduino? I don't have oscilloscopes. Just multimeters.
Another (easier?) option I guess would be to remove the 392 Ohm resistors and wire the output of the ICs to the arduinos PIN? And power the device continuously with nothing in between? Will this work too right?