Controlling a garage door with CBUS

Could anyone advise the typical method of controlling a typical garage door opener - is it simply a relay circuit - one or two channel? ie: how to make it rise and how to make it go down?



Thanks....

 

Hi

I would imagine all thats required is a single cbus relay (dry contact) and wire it in parallel with the open / close button on the garage door controller.....Then when required pulse the cbus relay to toggle the garage door status......

HTH
Frank

 

In my experience, MOST electric garage door openers that I have seen or worked with or on have had a simple 2 wire button inside the garage, that when pressed, toggled the door to open or close, depending on its current status.

So Frank's method of simple using a C-Bus relay (of the non-240v variety wired to the NO channel would seem like it would do the job perfectly, as per Frank's instructions.

Regards, Darpa

 

Cheers

Thanks guys for your prompt response! When you say relay of the non 240 variety - do you mean one of the voltage free relays? Could you suggest a relay product?



Cheers

 

All of the CBus relays are voltage free.

 

Perhaps you could use a relay like the 5102RVF, if you dont have any other C-Bus equipment near your garage, or you could use L5504RVFCP if you already have a spare DIN rail in or near your garage.

Either way, you can see the whole range of C-Bus relays here:

C-Bus Relay Units


Darpa

 

Here's a good thread on setting up your garage doors:

http://www.cbusforums.com/forums/showthread.php?t=2322&highlight=garage+door

Cheers

 

So, what you are saying is.
Connect the 2 wires from the supplied 2 wire garage door button to a channel on the L55xxRVFC Relay? (4, 8 or 12 channel)
Then you can program any button on a C-Bus switch to be the button that opens/closes the garage door instead of using the supplied button.
Thus, you do not need a 5504AUX to control the door?

Does a Voltage Free Relay mean that the output side only has the voltage that the unit connected to is uses and not the 240V that the input side has?

ie - the above garage door button, lets say it uses 5 volts.
The Relay has 240V Input, channel 1 connects to a 240V light and thus outputs 240V to the light.
Channel 2 connects to a 12V light and thus outputs 12V to the light.
Channel 3 connects to the garage door button and thus outputs 5V to the button.

Martin

 

Essentially, all that you need the relay to do to open or close your roller door, is that when the C-Bus relay in question's corresponding C-Bus channel is activated, (pulsed on for as short a time as possible), the realy connects the 2 remote wires to your door motor. Your do NOT want 240V on the input side, you simply want the two switching contacts for that particular relay to be "joined" for a moment when that relay's channel is activated.

This means it would be wired slightly differently for each different relay you might use, but if using a changeover relay, there will be NO (Normally Open), NC (Normally closed) and C (Common) terminals. In this configuration, you would connect the 2 remote wires to your door motor to the NO and C terminals, (it doesnt matter which way around they go, as all you need the relay to do is complete the circuit).

If you were using a typical C-Bus relay, and lets say for instance you have an 8 channel relay, and you are using channel 6, then you would put one wire from the door motor remote into the 6A terminal, and the other wire into the 6B terminal.

Please note though, when programming, you dont want the relay to "Latch" on, you only want it to pulse on for a short amount of time (less than 2 seconds), and then switch back off again.


Hope this helps, Darpa

 

Thanks Darpa, I will give that a try soon.

Martin

 

Darpa,

Just to clarify.

When using a RELAY to open/close garage door, you said you DO NOT want 240V on the input side.
So, does this mean that you CANNOT have the ONE relay control both the garage door and lights?
Thus, you would need 2 relays, one connected to 240V input to control lights and the other relay with NO input and controlling the garage door?

 

The way that most garage door openers work, is that you should leave the 240v power supply for the motor always plugged in, not controlled by C-Bus, but just running on a normal power circuit, doesnt really matter if this is a dedicated circuit just for it, or if it is a circuit shared with other things, as long as you dont run too much off that one circuit (as common sense would dictate).

What I was saying about using a C-Bus relay to control your door, is that you do not control the 240v to the unit at all, the ONLY thing you are controlling is the 2-wire remote button that is usually mounted on the wall inside the garage, to allow you to open and close the door without using the remote control.

Also, you should not be mixing different voltage classes on one relay unit, and you can find a great article explaining that Here, on pages 22 to 24 of the document. Basically, the way that the remote button on the wall in the garage works, is that a 2-core wire is run to it from the motor on the ceiling, when the button is pressed, it closes (completes) the circuit. All you need to do is run a 2-core cable from the terminals on the motor that is meant for the button, and run it to a c-bus relay, one that does not have anything above 50v AC or 120v DC connected to it, and simply connect one wire to the "C" terminal, and the other wire to the "NO" terminal.

Then program your C-Bus to simply toggle that relay for a second or two, to simulate you pressing the remote button on the wall that would usually be used with your garage opener, hell, you can even keep the button on the wall if you want it, just run a cable from the motor to both the button on the wall, and the C-Bus relay.


Hope this helps,

Warmest Regards, Darpa

 

I have drawn a basic diagram of how to wire the 2-core cable from the door motor into a changeover relay. And to answer your question that I missed before, no, you should NOT be running the lights and the motor off the same relay. You should use a standard relay for the lights, and a changeover relay for the motor control. There isnt really any inpur or output when wired this way, it simply connects the 2 wires together when the relay is activated, simulating the same thing as pushing the normal NO button on the wall. Thus, there is no need for any voltage to be input into the relay, as there is nothing that needs to be powered, (well, other than the 240v going into the control side of the relay on the left, but that has nothing to do with the outputs on the right). All that the relay does is "joins" the 2 wires together momentarily.

Hope this makes things a little clearer

Best of Luck,

Darpa

 

You don't need a change over relay to control the garage door motor as it will be controlled via dry contacts.

So, channel 1 could be for the garage lights (with a 240v feed into terminal A) and channel 2 could operate the garage door (figure 8 across A & B).

 

I know what you're saying PSC, I just figured if I was going to give advice, I should do it "by the book" and not mix SELV and LV on the same relay, as per the advice in the CIS Tech Support recommendations. But yes, you could control both the lights and door from the same relay, just using 2 channels.

 

OK, in that case you would be better off using a single channel relay for the garage door.

I'm just saying that there is no need to go to the expense of using a change over relay.

 

Lol, sorry Peter, I didnt mean to sound like a prick, I'm sick at the moment and grumpy as hell, so I'm sorry if I sounded like I was having a go at you, because I certainly didnt mean to sound that way.


So in summary for anyone reading this, you have 3 types of relay you can use to control a 2-wire door remote, the best options being a single-channel relay, or a changeover relay, but yes, you can still use a standard relay as well (DIN mounted 4,8,12 channel types).

 

No worries Darpa, no offence taken.

But, if we are going by the book you would need to isolate the DIN rail relay controlling the garage door in a separate enclosure away from the other 240V C-Bus equipment.

 

Ok, Now I think I understand.
I need

1 Relay connected to 240V input to control the lights

1 Relay (preferalbly a single channel relay as it is the cheapest or 5102RVF as 5101R is not Voltage Free) connected to 240V input, then connect the 2 wires for the remote button to that output channel

I just needed to know before I purchase them so that the electrician can wire them up later.

 

Marty, on the relay for the door control, there should be no 240v connected at all on the input side, the relay should only connect the 2 wires from the door motor together. so you would need to wire it so that instead of the relay, when activated, connecting 240v to the two wires to your door motor (this would probably result in a big bang and a dead door motor), the terminals of the relay should simply have one wire from the motor connected to the input, and the other to the output. Looking at the 5102RVF installation guide, its not very clear as to how to do this, you would need to test it with a multi-meter across the terminals to suss out how to connect it. But if you have a relay that can source power from C-Bus when its not connected to mains power, (probably around 18mA), then use that, and let it source its control power from C-Bus instead.

You REALLY REALLY need to make sure that if you have 240v connected to the relay in any way, that its not going to apply that to either of the 2 remote wires to the motor when the relay is "switched on". ALL you want it to do is simply close the circuit between the two remote wires.

Darpa