I have a LiftMaster commercial door operator that uses a through beam photo eye for a door safety or LiftMaster Commercial Protection System CPS-U . I have a spare sensor pair, 001D8857-1 “Receiving Eye” and 001D8857-2 “Sending Eye” (CPS-U consists of these two). I would like to use this spare pair for another application but I have not been able to find any technical information on this sensor including the most basic stuff like operating voltage. The logic board is located 14’ off the floor for the sensor that’s in use and is not readily accessible for me to take readings.

What’s most puzzling to me is how this sensor works: both the sender and receiver each only have two wires and they are wired in parallel. Thus, there are only two wires going to the logic board. I would think at the minimum, 3 wires would be needed, two for power and one for the signal. See page 3 of the attached pdf.

Anybody have any info or specs on the CPS-U or how, in general, this sensor could work with only two wires going to the logic board?

Thanks in advance.

 

Guessing here...
It may work similarly to the Dallas 1-wire system. The data line is pulled high by a resistor, and power is stored in a capacitor in the sensor. The sensor can then pull the data line low to send a signal back to the controller.

 

I had to Google "Dallas 1 wire" and I'm not sure I follow all that I read. It would seem like there'd be a lot of tricky timing to get that to work because both the sender (LED?) and receiver (photo-transistor?) are wired in parallel and go back to the logic board on the same two wires (see attached pdf, page 3). If that were the case, there's probably a bunch of circuitry in both the sender and receiver that would make using this in my other application way more complicated than I'm capable of dealing with. However, I have no better thoughts! Thanks for your response.