How does a garage door motor wall control on a two wire circuit determine between door control and light control?

Question

I am trying to recreate the garage door wall switch (making a custom assembly that has two buttons to toggle the door and toggle the light). The first thing that I am attempting to understand is the voltage being passed from the motor, to the wall switch and then back to the motor.

While looking at different articles and other places on the internet I discovered that the COM port on the motor passed power down to the wall controller and when the door button was pressed closed the circuit for a brief amount of time to make the motor move the door. To control the light, I thought that the same voltage was passed down through the same wire but instead a lower voltage was passed back up (making the switch more of a voltage divider). There is only one wire going back into the motor meaning it has to be something of a varying voltage to control the door and light.

When using a Fluke Multimeter I was getting a jumping reading from the wire connected to the COM port on the back of the wall control around 0.750 VAC. (I didn't like that it wasn't giving me a constant reading, originally I thought the meter was broke but put it into a standard outlet and got a solid 120 volts). When pressing the door control the voltage on the wire that went back up to the motor read 0.750 VAC. I figured that jumping the circuit would toggle the door, which it did. This made me think that the door opening was a simple instance of closing the circuit for about half a second but made me unsure of what the true voltage was.

The second challenge was how to tell the motor to turn the light on/off through that same wire. A few articles I read stated that the wall control acted as a voltage divider but not by how much or if the motor needed said amount of voltage for how long. When measuring with the light switch pressed down I got readings jumping around 0.350 VAC, but once again nothing concrete.

Here are my questions from all of this, and I am mainly asking those who have tried something like this before or know the common circuitry of these products.

1. How many volts are sent through the COM port on the motor to the wall controller?

2. How many volts are returned to the motor from the controller to open the door?

3. How many volts are returned to the motor from the controller to toggle the light?

I have a Genie Pro Max system, but from what I can tell and the fact that manufactures sell universal door switches I think most systems operate around the same circuitry.

Thank you in advance for your help.

Answer 1

I haven't played with a Genie unit, but I spent some time with an oscilloscope and a Chamberlain/Lift Master unit several years ago. I found it has two modes of operation. One is the simple contact closure -- just short the two wires together and the motor will operate. The other mode is digital. The two wires carry power to the wall control, but they also carry a serial data communication protocol between the motor and the control. The only way to control the advanced features, such as turning the light on and off, is to use one of the wall controls with the digital interface.

As I recall, I also found that the two modes can't be used together. For example, you can't put the fancy wall control in one place in the garage and put a simple doorbell type switch in another place. When the simple switch shorts the wires the door operator goes into "dumb switch" mode and the fancy wall control ceases working (until the door operator is power cycled).

I couldn't say for sure whether your Genie works the same as that Chamberlain did, but I would not be surprised if it does. In any case, you wouldn't get far with discovering this nor reverse-engineering it armed with only a volt meter, unfortunately.

Answer 2

I opened the wall control on my older non digital chamberlain/liftmaster (orange/red program button) to see how they do it.

There is only a single 1k ohm resistor on the board in series with the led and it is used for current limiting to the led.

The door control button is a dead short across the two wires. The LED on the keypad goes out and the voltage drops to zero.

For the light and lock push buttons they use two capacitors. A 1uf for the light and a 22uf cap for the lock/vacation button.

When the switch for light or lock function is pushed it does not short the wires but puts one or the other capacitor across the two wires.

There is no circuit to discharge the cap when you release those buttons so I would assume that the cap forms an oscillator circuit and the door opener determines which function is selected from the different oscillator frequencies for each button.

-A dead short is open/close door.

-A high frequency oscillation signal (1uf cap across wires) is identified as the light push.

-A low frequency oscillation signal (22uf cap across wires) is the lock/vacation button push.

I did not put a scope across the wires to confirm but that also answers why the original posters fluke meter does not provide a steady reading. Hope that helps.

Answer 3

Methinks you are tilting at windmills here.

Modern garage door openers are tiny computers with a set series of programmed responses, not dumb robots with simple voltage controls. If you want an overhead light controlled by a wall switch, you'll get much better results installing an overhead light and a wall switch for it than trying to hack your garage door opener into turning on its light without opening or closing the garage door.

Among other reasons, if you do manage to hack your opener and it then goes on to malfunction and hurt/kill someone, (as they still do from time to time, which is one reason that they have become so highly computerized) the manufacturer will be off the hook and you will be on the hook for responsibility because you have modified the opener.

Answer 4

Open/Close is a BUTTON that shorts the two wires - 0 Ohms. Light On/Off is a BUTTON that connects the two wires through a 200 Ohm resistor. Lock-out is a SWITCH that connects the two wires through an 82.5 Ohm resistor. The only other component on the board is the LED which uses one of the resistors as a current limiter.

Answer 5

Not a good answer. Listen to the other posters. It is likely a resistor placed in line with the light button. My Genie has three buttons. One for the door motor, one for the light and a push and hold button to lock the door. I suspect that the 'microprocessor' in the main unit is looking for one of three signals. A momentary dead short (motor control), a long dead short (lock control) and a momentary resistance, 10K ohms for instance (light control). Also, the programmed functions of the microprocessor also monitor your garage door block sensors to provide crush protection, so as long as you aren't trying to modify the block sensor operation, there should be no issue with anyone getting hurt...or killed. :)

Answer 6

In my case, LiftMaster/Chamberlain model with red erase button, it has written erase instead of the usual program.

• The two wires have an 82 Hz signal (no buttons pushed).
• The signal changes its duty rate, due to the capacitor when pushing the light on/off button.
• When the garage door open and closes it is only a short circuit between the two wires (actually pretty insecure in my opinion). I can confirm with my oscilloscope that the signal only goes to 0v and also measured condutivity, it is only a short, I also tested this by shorting the wires and the garage door open/closes.

Some images:

• The first one is the pure measured signal, no pressed buttons, the second one is when pushing the button for on/off of the light in the garage door opener.