Home Automation - Cate 5e and garage door openers

Is it possible to wire a Genie garage door opener with cat5e cable?? My wife and I purchased a new home a little over a year ago and when the home was inspected, the electrician said that the garage was pre-wired for a garage door opener. I just finished installing a Genie Directlift opener and I noticed that the wiring is cat5e...will this work??

I don't know if a Genie opener uses low voltage for the opener or not. Most openers do as far as I know. If yours does, the cat5 would suffice.

A door bell circuit( garage door button is same ) is a ARt 725 Class 2 power limited application. Cat 5 cable is not listed for this application. Besides door bell cable is 18 gage, cat 5 is 24

Class 2 circuits. Class 2 circuits typically include wiring for low-energy (100VA or less), low-voltage (under 30V) loads such as low-voltage lighting, thermostats, PLCs, security systems, and limited-energy voice, intercom, sound, and public address systems. You can also use them for twisted-pair or coaxial local area networks (LAN) [725.41(A)(4)].

Class 2 circuits protect against electrical fires by limiting the power to 100VA for circuits that operate at 30V or less, and 0.5VA for circuits between 30V and 150V. You protect against electric shock by limiting the current of the circuit to 5mA or less for circuits between 30V and 150V [Chapter 9, Table 11].

You can wire Class 2 circuits with Class 2 cable or any of its substitutes permitted by Table 725.61(A), depending on the condition of use.

Table 725.61 permits the use of "CM" cables as a substitute for "CL2" cables. Cat-5 is listed as a "CM" cable. Yes, agree that the conductor size is small, but don't see a code restriction on its use for doorbells

wow, lots of info. At the start you say cat 5 is not acceptable. At the end, you say it is.

lots of stuff in between. good stuff but you ended up in the same place I did.:)

and to overcome the conductor size, parallel the pairs. You can use all 4 pairs and in effect quadruple the conductor size.

from my professional experience, the resistance needed to give closure to the opener unit will vary by unit. Most units will obviously require a very low resistance for contact closure, since a contact closure is a short. When working with this, I actually metered out a 1000 foot roll of CAT5, and its resistance was about 35 ohms, with the last unit I worked with, it took 10 ohms or less to trigger the door opener unit to operate. But this was also a very long cable run. If it is a relatively short cable run, it may work. My suggestion would be to unhook the cable from both ends, and short a pair together, and then put a volt meter on the other end, and see what the resistance reading is in ohms. This will definately tell you if that cable will work or not. From what I have done with these, even if all pairs are paralled together, the resistance still will be too high to trigger the opener unit.

From what I have done with these, even if all pairs are paralled together, the resistance still will be too high to trigger the opener unit.

well, here are some tech specs;

AWG 24 cu .205 mm sq. area 25.67 ohms/1000 ft.
AWG 18 cu .823 mm sq area 6.385 ohms/1000 ft.

so, if one were to combine 4 conductors into 1, one would end up with just about an AWG 18 wire. Based on this and your < 10 ohm requirement, one could run the switch about 783 feet away from the opener before one would exceed the listed max of 10 ohms.

I'm betting we are talking about less than 100 feet (one way). That would calculate out to about 1.3 ohms. In fact, if one were to not combine the conductors, a 100 foot run would result in about 5.14 ohms. So, based on your determination of <10 ohms, using just one pair of a cat 5 cable would be adequate.

Oh, and to be correct, a closure of a contact does not cause it to be a "short". It is a completion of a circuit. The problem with too much resistance in the wire is there will be too much voltage drop in the circuit to cause the circuit to be triggered.

well, here are some tech specs;

AWG 24 cu .205 mm sq. area 25.67 ohms/1000 ft.
AWG 18 cu .823 mm sq area 6.385 ohms/1000 ft.

so, if one were to combine 4 conductors into 1, one would end up with just about an AWG 18 wire. Based on this and your < 10 ohm requirement, one could run the switch about 783 feet away from the opener before one would exceed the listed max of 10 ohms.

I'm betting we are talking about less than 100 feet (one way). That would calculate out to about 1.3 ohms. In fact, if one were to not combine the conductors, a 100 foot run would result in about 5.14 ohms. So, based on your determination of <10 ohms, using just one pair of a cat 5 cable would be adequate.

Oh, and to be correct, a closure of a contact does not cause it to be a "short". It is a completion of a circuit. The problem with too much resistance in the wire is there will be too much voltage drop in the circuit to cause the circuit to be triggered.

these are some great specs for cable. Secondly, in theory, if the cable run is less than 100' this would prove to be right, but for a simple experiment, grab a switch and connect a meter to it. with the meter connected to both leads, measure the resistance of the switch when open, it should measure open. Now, close the switch and read the measurement, it should measure as a short, which theory, will be 10 ohms or less. Finally, do the same experiment with a relay, it will do the same thing. All that a garage door opener door switch is doing in this particular application is providing a switch closure.

Genie openers come with pretty light wire, I'd say no heavier than #24. I installed one in a garage pre-wired with #18 thermostat cable, and it works fine. IMHO Cat5 should work fine.

these are some great specs for cable. Secondly, in theory, if the cable run is less than 100' this would prove to be right, but for a simple experiment, grab a switch and connect a meter to it. with the meter connected to both leads, measure the resistance of the switch when open, it should measure open. Now, close the switch and read the measurement, it should measure as a short, which theory, will be 10 ohms or less. Finally, do the same experiment with a relay, it will do the same thing. All that a garage door opener door switch is doing in this particular application is providing a switch closure.

I know what the switch is doing. I do this for a living.

and why are you tossing in a relay? trying to confuse the situation with superfluous components?

and the specs for the cable are specs, not something I made up. If you want to argue with IEEE, go for it. I tend to accept them at face value.

the switch closure thing. Like I said, why the resistance is important is if it is too great, there will be no voltage to be available to power a relay once it runs through the entire circuit. I gave you calculations that are accurate and as I showed you, even a single pair of #24 wire would not cause too much resistance, in a 100 foot (200 foot total) circuit. Bonding 4 conductors together would provide much less resistance and well below your threshold let alone you using measurements from a 1000 foot roll of cat 5 wire.

and, unless this guy is a millionaire with a 65 car garage, his run is going to be less than 100 feet one way. A typical attached garage is going to be more like 25 feet (even better).

and if you want to argue with me, then be correct and accurate:

with the meter connected to both leads, measure the resistance of the switch when open, it should measure openit will either measure "infinite" resistance or "as an open". It will not measure "open".

it should measure as a short, which theory, will be 10 ohms or less.I don't know what you do for a living but any readable resistance is a complete circuit.

I sure would love to see you hook up a 12.5 kV line with 20 ohms to ground on the dead side. I bet you wouldn't do that more than once.



and by definition (I'll use wiki's since it is simple and accurate):

A short circuit (sometimes abbreviated to short or s/c) in an electrical circuit is one that allows a current to travel along a different path from the one originally intended

First of all, yes, I understand you are providing the correct specs from IEEE. But these are only SPECS. They often need to be taken with a grain of salt becasue they are on paper. The actual specs of cable, or a component, etc. can often vary from spec. Secondly, in the Electronics Engineering, or Electronics Technical world, yes, you are correct with saying that a circuit disconnected would measure infinite, but the electronics term is the circuit will often be referred to as open, hence the fact i mentioned it would measure open. This is the same meaning as being infinite, simply being used interchangeably. Thirdly, by hooking up a 12kv circuit with 20 ohms to ground would be stupid, and dangerous as well. I am an electronics field engineer in the audio video world, and along with that I have spent a lot of time working on tech bench. It takes anywhere between 10kv and 30kv to fire a CRT for a TV. I know what kind of damage that these extremely high voltages can do.

well, out it this way: if the cable does not meet or exceed the specs as determined by the IEEE, then the cable is not acceptable for the purpose so most manufacturers endevour to make sure the sell quality cable.


I know what kind of damage that these extremely high voltages can do.You are the one posting that anything under 10 ohms resistance is considered a complete circuit so inversely, anything over 10 ohms would be an open circuit. I did not say it, you did. And while the voltage in a flyback transformer of a CRT is high, the amperage available is minimal. It would cause little damage to typical electrical (as in non-electronic) components. I was speaking of a high voltage line. That would wake you up a bit.
, you are correct with saying that a circuit disconnected would measure infinite, but the electronics term is the circuit will often be referred to as open, hence the fact i mentioned it would measure open. This is the same meaning as being infinite, simply being used interchangeably. again, you are the one that was mixing your phrase references. I merely stated the obvious.


but, back to the case at hand; by combining the pairs in the cat 5 cable, I will guarantee the system will work. I believe there would be no problem if just using 1 pair by itself but since he has 4 pairs, why not just bond them into 1 pair and remove any possibility of problems?

the limited current draw through the circuit, even at the voltages used, is going to be so minimal there are not going to be any problems with voltage drop.

Cat5 will work and is safe on this opener. I've used it also with on problems. I even used it with my security system on a few runs to control my opener due to the fact I ran out of my regular wire. I have it wired so I can check my security system from my bedroom at night and if the door is opener I can shut it from my bedroom. I haven't had any problems at all.

JimBeer 4U2

Along the same lines;

I have a Genie garage door opener and the button on the wall is broken. It appears to be a doorbell button repurposed by Genie for their doors. Can I replace the button with a garden variety doorbell button from a local hardware store or do I need to get a Genie branded button?

Thanks

Welcome to the forums, Goze.

Are there only two wires to the switch? If so, any basic unlighted doorbell switch will work.

Are there only two wires to the switch? If so, any basic unlighted doorbell switch will work.

The previous owner or installer did it this way:

A Red, White, Black, and Yellow wire come down the wall but only the Black and Yellow plug into the lighted button. Black goes to the screw marked B and the Yellow wire goes to the one marked W. The Red and White are just hanging free not connected to anything. Pretty sloppy.

But at any rate I'll get another button, lit or unlit, and reconnect it the same way and assume it will work.

Thanks