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.

enter image description here

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?

Thank you


2 Answers 2


It's an Allegro Microsystems A3187L. This is a Hall sensor IC. Search for 'A3187L allegro' and it's the first Google hit. Seems like it's discontinued.

Datasheet Link

There are two sensors. Each one will output a square wave, based on the flux changes it senses as the magnetic disc spins past it. The sensors are arranged to make a 90-degree quadrature waveform, which not only increases the number of pulses per revolution for counting but can be used to detect direction as well.

You probably don't need to fully reverse-engineer this. The sensor outputs will be open drain, and can be used with standard logic with a pull-up. If you have access to an oscilloscope you can measure the number of pulses / rev, and figure out the phasing of the two signals for direction sensing.

Generally, the best way to deal with these in a microcontroller is to wire up each one to an interrupt.

  • \$\begingroup\$ I guess you are right, in that case pins are +VCC, GND and OUTPUT, still I'd like to understand what this circuit does in order to use it. \$\endgroup\$ Commented Aug 12, 2019 at 3:28
  • \$\begingroup\$ @GuillermoRuffino, start by drawing the schematic diagram \$\endgroup\$
    – jsotola
    Commented Aug 12, 2019 at 5:12
  • \$\begingroup\$ And if you haven't got an oscilloscope, have the Arduino continuously read each input and light an LED to match what it read. Then turn the shaft and watch the lights. \$\endgroup\$
    – gbarry
    Commented Aug 12, 2019 at 5:23
  • \$\begingroup\$ Do I really need to make a diagram? Seems simple the PCB traces. \$\endgroup\$ Commented Aug 12, 2019 at 14:59
  • \$\begingroup\$ I don't understand the answer properly. The middle pin is connected to both IC's +VCC then the left and right pins are connected to the GND of the left and right ICs. The output seems to go to a resistor? (3920) I can connect left and right to a microcontroller with pull up, but then the center also to +Ve? (I'm planning using 5v) I don't think that would work, it will be +5V everywhere. \$\endgroup\$ Commented Aug 12, 2019 at 15:05

Well in case somebody has a similar case of hall sensors wired this way and gets confused like me I’ll share what I did to get it working.

enter image description here

I cut some PCB traces and wired ir directly to my MCU I am using 10k pull-ups

Thanks to all who contributed.


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.