Is there a "hack" to bypass those sensors? Possibly a resistor, capacitor or something....temporarily of course until the new Genie sensors arrive?

 

I am not familiar with your sensors, but, if they only have two wires running to them, there are only two possibilities for bypass.

Cut the wires, and see if the door functions normally. If not, wire the wires together. One or the other ought to work.

:gespann:

 

...no, there is a 3rd possibility. There could be a pulse generated by the 2 sensors and an "open" or "short" won't let the door operate properly. That's why I asked if anyone knew of a way to bypass the sensors. Key word "knew" but thanks anyway for the response.

 

Excuse me, but I still don't understand if there are only two wires running to the sensor, how you could possible have more than two options. But obviously, you know something that I do not.

Please post back with what you learn, and we might all learn something.

 

Since this is a technical question concerning Genie sensors you may have to call the Genie techs at 1-(800) 354-3643. This sounds like a question only they can answer since they designed it.

Good luck and post back

 

Depending on your layout, I know alot of guys that just mount the sensors in the attic or just above the opener itself facing each other and wire up as needed.

I know they'e a pain, but ya always gotta think about the little ones.

 

I will do that when the new ones get here, but for now I want to bypass the broken ones.

 

Let us know if Genie told you anything, but I would be surprised if they did. I'm working on something where I need to 86 the sensors (linear actuator for cabinets). Your right; the electronics are smarter than an open or shorted circuit. I'm working off the assumption that it is a voltage divider circuit where the sensors introduce a resistance into the circuit which consumes a certain percentage of the voltage and the IC detects the pull down. If this is correct it is just a matter of finding the correct resistance and introducing it into the line. The problem is that you can not use an ohm meter to detect it because you need a live circuit. I'm thinking a voltage drop test, and then dividing that figure by amperage (ammeter in series). I believe source voltage is 20 volts DC on that side of the circuit.