I have a Linear garage door opener that is powered by a 12vac plug in power supply. Will this unit work on 24v DC? I was suprised to find the internal relay labeled as DC. I am trying to get a DC output as a trigger for a relay/timer circuit module. Any help would be appreciated.

 

That really is not enough information. If the power supply is 12VAC, you should probably stick with it unless you can figure the circuit out. Is the label on the internal relay for the contacts voltage or the coil? They are not necessarily the same.

 

Sorry, I meant it is powered by 24vac not 12vac! I want to use 24vdc.

 

It's risky to answer that question outright. I suspect you are assuming there is a rectifier in the very first stage that supplies the rest of the circuit with DC. What you don't know is the actual DC operating voltage for the rest of the electronics. For example, is it 12V or 5V? In any event, 12V AC will give about 16 to 17 V DC, not 24V DC. One possibility is that there is a voltage regulator in the opener to reduce the rectified AC to 12V. Putting 24V DC into that regulator might cause it to overheat.

What brand of opener is it? For some of the real common ones (e.g., Chamberlain, Sears, etc.), you might find enough of the schematic on the Internet to answer your question with more certainty.

Edit: Written before seeing the above correction.

As an alternative, can you pull the DC you need from the PCB of the door opener and leave the opener's supply intact.

John

 

It's a Linear Delta 3 and my objective is to be able to supply it with 24cDC instead of 24vAC and I don't know if it will work. I am using the receiver to turn on a light via a relay module that uses only DC as a trigger.

It's risky to answer that question outright. I suspect you are assuming there is a rectifier in the very first stage that supplies the rest of the circuit with DC. What you don't know is the actual DC operating voltage for the rest of the electronics. For example, is it 12V or 5V? In any event, 12V AC will give about 16 to 17 V DC, not 24V DC. One possibility is that there is a voltage regulator in the opener to reduce the rectified AC to 12V. Putting 24V DC into that regulator might cause it to overheat.

What brand of opener is it? For some of the real common ones (e.g., Chamberlain, Sears, etc.), you might find enough of the schematic on the Internet to answer your question with more certainty.

Edit: Written before seeing the above correction.

As an alternative, can you pull the DC you need from the PCB of the door opener and leave the opener's supply intact.

John

 

I am using the receiver to turn on a light via a relay module that uses only DC as a trigger.

Sorry, my carelessnes, I missed the brand name. The receiver board almost certainly runs on DC. Since you asked the question, I suspect you also know that rectifiers on the power input are used in some DC devices as protection against using the wrong polarity. In other words, one could put the battery in backwards, and the device either would not be harmed or would continue to work normally.

What we don't know is whether anything in the door opener requires AC or is not rated for DC, like a transformer, some relays, relay contact ratings, switch ratings. While that may be unlikely, it is an unknown.

I would trace the power input to the unit and find out just what happens to the AC input. If there are no AC devices, and you find only a rectifier and filter capacitor, then it might be worth a try. Before I did that, though, I would test some of the voltages to be sure. One other thing to check is to be sure your DC supply can supply enough power (wattage) compared with the AC supply. John