System to light 2 LED's when garage door is open

[stume]

I'd like to assemble a system to use LED's to indicate when my garage door is open, but I want one different from the ones that are for sale, so I think I'll need to make one.

I'd like each of two locations to have a single LED that lights if and only if the garage door isn't 100% closed. It seems like it might be simple to design for someone who knows what they're doing, so that's why I'm here asking.

I'm comfortable soldering and following directions (but unfortunately I don't know how to read a wiring diagram). I don't want it to be wireless. I can run wires to the necessary places (garage door area and both locations where I'd like the LED's). All three places are within a 15 foot wire path from each other.

I have two garage doors, so I was going to do build a twin system for the other garage door, but I thought I should mention this in case a system for both doors could be designed and be cheaper on parts.

Thanks very much for your help

 

[KeepItSimpleStupid]

Depends on what 100% really is. What is the slop range? 1/4"? Contact or non contact technologies?

A microswitch is a typical contact technology.
An alarm magnetic switch is yet another.

A photo interrupter (something blocking an opening) is yet another
Reflective object sensing (place reflective tape at sensor)

Hall effect is a solid state version of the reed relay/magnet sensor.

Take a look here: Slotted Switches Product Training Module PTM

 

[stume]

You're right, "100%" was a bad choice of words. I should've just said "closed". It's a normal hinged-panel garage door, like this: garagedoorguide.files.wordpress.com/2009/09/garage_door_parts_diagram.jpg

I know that different technologies have pros and cons and prices vary for each choice. I'm hoping it can be done for $25 in parts. I don't know which technology is best to use for this project. I'm depending on the experience, judgment, and expertise of the experts here to recommend what to use, and to know what a layman like myself has access to purchase at Radio Shack, Home Depot, online, etc.. Thanks everybody.

 

[KeepItSimpleStupid]

If you can find a way to mount a sensor such as this: OPTEK TECHNOLOGY|OPB911L55|OPTO SWITCH, LOGIC OUTPUT | Newark.com we're there. This one has a wide gap.

For the lowest parts count, Inverted Open collector is preferred, otherwise we have to add a few parts. If you can find inverted Open collector, then 1 resistor per sensor, 1 LED per sensor and a power supply is needed which you can get from a USB wall charger. The power supply can be shared.

You need something to block the slot when the door is closed.

 

[Reloadron]

Given the options I would run with one of KeepItSimple's first suggestions and use a everyday SPDT magnetic reed switch. Should be easily installed and positioned where you want. Switch on structure and magnet on door. Something like the 6300 series of these. The bottom of the link includes the actuator (magnet) part numbers. Switches like this can be had just about anywhere, places like any home improvement store. Hardware stores and the like. No electronics, maybe a 5 volt power supply and a few LEDs and resistors. Oh yeah some wire.

Ron

 

[MikeMl]

Another idea:

 

[KeepItSimpleStupid]

The only reasons why I picked the slotted switch was:
1. Contactless
2. Probably more sensitive to actual position. Change the "flag" size and you change the +-position it responds to.

The magnetic alarm switches is what I was thinking about, although this Magnetic Alarm Contacts | Burglar Alarm Door and Window Contacts link shows a contact made just for garage doors. It fails to say if it's normally open or normally closed.

 

[KeepItSimpleStupid]

Better yet: **broken link removed**

 

[Reloadron]

Yeah, what I linked to was SPDT so you have a choice. They come in all flavors. I don't know, roller micro switch or magnetic reed switch? I guess whatever fits best and is more available?

Ron

 

[stume]

It seems like this is becoming more complicated than I thought it would, and maybe it's due to the $25 budget leaving too many options. Truthfully, I'd like it to be as cheap as possible (to save money, and to make component selection easier).

Also, as information, I'd say that the door closes to within 1/4 of an inch the same position every time.

Thanks everyone.

 

[KeepItSimpleStupid]

This isn’t a bad project to illustrate the finer points of design.

W can start out with simple requirements aside from the mechanical one and assume the simplest case:
1. A power source
2. A switch that is closed when the door is closed
3. A LED for each door

We know that this has to be a class II system, so power has to be limited (100 W or so). Voltages have to less than 24 VAC/DC to not have to use conduit etc.

LED’s can operate on AC or DC with additional components. There are inefficiences with using a DC regulated power supply. You can get 4W 120 V switching power supplies. An isolated supply needs to be used for safety.

One could consider the possibility of using a Bi-color LED to indicate OPEN/Closed based on color. Red- OPEN, Green – Closed. Green being associated with OK rather than it’s safe to enter the garage. One could decide that a green LED for the door is closed is an OK color.

Wiring: Doorbell, CAT3 or CAT5 (Ethernet cable) or quad telephone wiring would probably be appropriate. If this were alarm wiring then Red Teflon fire cable would be appropriate. Plenum rated wiring, would not likely have to be used, but wire entry may have to be sealed if it has to penetrate a fire wall of the home. Solid is best used within the walls. It’s easier to fish and won’t look sloppy when tacked on the ceiling.

If you use solid wire, you have to typically transition from solid to stranded in some places and provide a mechanism of strain reliefing the connections.

The LED’s: You can buy them in fixed voltages and with aestic looks or you cab use a clip like mount for standard LED’s. Typical LED’s used for this application drop about 2.1 V and the design createria is to usually operate at 20 mA.

A standard blank outlet plate can be drilled and dry transfer lettering be applied to give a very professional appearance.

If you use a low voltage box, there is typically no way to strain relief the wiring, so I’m advocating a typical electrical box or a low voltage box with a cord type strain relief. An underwriter’s knot and a grommet may suffice for a strain relief mechanism.

There needs to be a way to strain relief the wires to the LED. One way is to use a bezel type LED and place a small piece of bent metal behind it so that the wires to the LED can be strain reliefed. Stranded wiring, matching the colors of the wire used or even the polarity of the LED such as RED and Green which are quad shield colors. Small wirenuts can be used to make the connectons. Personally, I like DIN terminals with wire protectors. These are typically hard to find and may have long lead times.

At the garage door, you may or may not have to provide signal conditioning, but we are assuming not, since I have assumed a simple switch. So a box of some sort needs to make the transition from the garage door switch to the premise wiring. Agin, this could be a standard electrical box with grommets or a real cord connector. Cord connectors can be had for various internal range of diameters and use standard electrical conduit fittings. They can be had in plastic and Aluminum. Again, wire nuts can be used to make the connections. If the difference of wire gauges is large, then other options could be considered. Another way is to just mount a plastic box with a metal cover nearby and use grommets. Some of the wires could be strain reliefed by say a wire staple.

It’s best to run all the wires as a home run to the controller which needs to be near a power source. Wires can be run through joists and holes drilled in the center of the joists. Cable clips can be used to tack the wires in place, For home wiring, I tend to use expensive RIP ties, because wires can be easily added or subtracted from the route. You don’t want to pinch the wire and sharp bends should be avoided.

The location could be the ceiling near the light since that has a power source, but troubleshooting might be awkward there. Let’s assume that location for now.

A box has to be picked that will fit and this is where you can end up with troubleshooting issues, A professional approach might use disconnectable terminal strips on a PC board or use terminal strips and a header connecter to the main board.

We might chose a PC board system or we could use DIN rail and DIN terminals. The latter would increase the cost significanty, let’s go with pluggable terminals. We can change the size to avoid the wrong insertion. In our simple case.
5 terminals for the sensor (Run a 4 conductor cable).
Possible sensor power, sensor gnd, sensor signal, sensor GND
In our simple case, 2 conductors would be used
Two such connectors.
4 terminals for the LED (2 such connectors)
Lots of options here. Let’s assume one 4 conductor cable to LED’s
3 terminals for power.
One might become a shield
Assume a wall Wart

You have to make a decision as to whether to allow swapping of LED’s and sensors. In any even, they can be swapped with the screws on the connectors and it would be easy with the right pluggable headers.

When creating the contents of the box I’d probably go for a 12 V unregulated wall wart.

Now, I might change the design slightly and put a 1N4001 diode in series with the LED for reverse polarity protection, Soldered and heat shrinked at the box, so if I hook the LED u backwards nothing bad happens.

The LM334 is a nice maximum 10 mA current regulator. If 10 mA is enough for the LED’s we could use this. I have used it and I like the method. We can use other techniques too.

Using a 5V or 12 V power regulated supply poses other issues. Without some sort of regulation or transient suppression, the design may not be reliable, so the design should have a fuse for the power and at a minmum have transient protection. The fuse could be internal or external. You could place a power LED on the box to indicate power. The fuse could be external or internal. Assume internal.

So, the PC board contains a few components (a current limiter, transient protection, a fuse and essentially the point to point wiring, so that the system is clean and easy to troubleshoot. Standoffs would be required to mount the board.

This illustrates the design process and some of the tradeoffs. I like systems that are reliable and easy to troubleshoot. At work, in a research environment, I liked things to be modular and reuseable,

This illustrates that a very simple project has lots of hidden costs. Anyone building projects realized that the power supply, real estate (PC board) and the case contributes largely to the costs. Connectors, while a necessary evil contributes negatively to reliability

Here is a datasheet for the LM334. https://www.electro-tech-online.com/custompdfs/2011/01/LM334.pdf So your only real components are: The current regulator, 2 resistors, a transient voltage protector, optional diodes for reverse polarity protection ( 2 for the LEDs and 1 for the power supply and a fuse)

I hope you find this info useful. Typical vendors are Electronic Components Distributor | DigiKey Corp. | US Home Page, www.jameco.com , Mouser Electronics - Electronic Component Distributor and US - Electronic Components Distributor | Newark.com

The system is simple. It's just the details that kill you. If you don't have an electrical outlet nearby that just adds more expense.

The simple version is a switch, a regulated power supply, a resistor for each LED. I'll end with "If you don't have time to do it right, when will you have time to do it over". Now there is always the option of having someone do most of it for you. Say, make the main box and the LED plate.

 

[stume]

Oh boy. A few things I noticed.

2. A switch that is closed when the door is closed

Does switch being closed mean an open circuit? I'd want no voltage (open circuit) when the door is closed, and a closed circuit when the door is open (so the LED's will be lit). Forgive me if I don't understand the terminology, as I'm sure was obvious since my first post.

3. A LED for each door

This isn't quite what I'm trying for. There are two locations where I can't see my garage doors. At each location I'd like to have one LED for for each door that would light if the corresponding door is open.

One could consider the possibility of using a Bi-color LED to indicate OPEN/Closed based on color.

As I was saying, I only require an indication of an open door.

Wiring: Doorbell, CAT3 or CAT5 (Ethernet cable) or quad telephone wiring....Solid is best used within the walls. It’s easier to fish and won’t look sloppy when tacked on the ceiling.

No wiring will be in the walls. Also, any wiring will look fine for this job.

... A standard blank outlet plate can be drilled and dry transfer lettering be applied to give a very professional appearance.

This is only in my basement. I will do a neat job, so it will be neat and orderly, but besides that, appearance is not a concern.

Most of the rest is beyond me. I am only asking for a simple plan on how to use some cheap little switches to light some LED's when a door is open. I'm trying to take the advice of your screen name.

 

[KeepItSimpleStupid]

No big deal, but it does help a bit.

As you said, the circuit is very simple. So, I may have misunderstood. Light when the door is open.

In the sense of a simple magnetic switch with a common a NO (Normally Open) and NC (Normally Closed). What this says, that with the magnet not nearby, there is a connection between NC and C. When the magnet is close by or activated, then there is a connection between NO (Normally Open) and C.

The switch is the issue. The commercial garage door switch makes the most sense, but it doesn't mean that other types of switches can be used. Switches don't have to have both the normally open and normally closed contacts.

LED's are what I'll call current driven and there are two specs that are important: Vf, If and Imax. They have typical operating currents and it affects brightness and lifetime. Let's use 5V and 10mA or .010 Amps

If you assume a fixed power supply. Let's use 5V for now, the resistor in units of ohms should be <= (5-2.1)/.010. The power rating of the resistor should be greater than I*R. 1/4, 1/2 and 1W are common resistors.

When you add a diode in series, you need to take into account another 0.6 Volts.

This is the simple resistor method.

If you use the LM334, you have to add another 3 Volts and use the resistor selection process in the data sheet.
so 3+0.6+2.1 is greater than 5V, so I picked 12V for the supply voltage.

An unregulated power supply will work, but at some point the circuit might fail because of a power line transient. That's the reason for using the LM334 instead of a fixed power supply and a resistor.

Usually a neat solution is better than a messy one. I had a chemical engineer ask me why the wires have to be neat in a panel? I said it helps troubleshooting and accidently pulling a wire off.

"closed" may have a different meaning to some people and that's a critical parameter. Each switch used will have a range that it will work.

The power supply can be shared. Having two locations for the LED's also increases the cost.

Reliability adds some cost, but in the long run it would be worth it.

Now, you could do the following and use standard electrical boxes, grommets and underwriter's knots. The fuse could be an inline automotive fuse. The regulator can be encapsulated into a piece of heatshrink with two wires. All conections can be made with small wirenuts.

Each system would be separate except for the power supply. That would have a fuse and LED and even one of the current regulators before it goes to a junction box for each system (sensor, power, led). Inside that box would be the current regulator made from some wire and heat shrink.

Near the LED, would be a diode for reverse polarity protection.

Did we make it simpler?

You might have $25 in just in the boxes, wires, wire nuts, heat shrink and wire clips. Now add a wall wart from Radio Shack. Estimate $10 and the two switches for $30.

Very little soldering:
The diodes in series with the LEDs.
The resistor, LM334 and 2 wires.
I'd use clear heat shrink.

Then there is postage if you have to order stuff.

 

[Reloadron]

Given a choice and in keeping it simple I would just go with something like the attached. The power supply is a 6 VDC Stancore STA 3560B everyday wall wart. If you want to spend a few bucks more get a 5 volt regulated wall wart. The manufacturer Stancore is merely an example as there are dozens out there and they are cheap. The magnetic reed switch is another everyday generic item. I used a NTE 54-627 SPDT version having N/O and N/C contacts. An example can be found here. The LEDs are also generic everyday LEDs. The series resistors were chosen based on the LEDs.

If you have two doors and want two LEDs for each just duplicate all but the wall wart power supply.

If you choose to go with a regulated wall wart power supply Radio Shack (and any parts house) sell LEDs for voltages like 5 and 12 volts with built in series resistors. Making that easy.

Magnetic reed switches like this come in all flavors and foe many of them can be had at a Lowes or Home Depot (also a good hardware store) or any place dealing with alarm systems. Make sure the mountable magnet is included. Mount the switch and magnet wherever you choose on the door and frame. Wherever they fit is fine.

<EDIT> As to the wire? Everyday telephone cable would do fine. </EDIT>

Anyway, in the interest of simplicity I would do it something like the attached.

Just My Take
Ron

 

[stume]

KISS and Reloadron, thanks for all your help.

Reloadron, I must say this is exactly what I was hoping for. Thanks so much!

Since I have several wall warts left over from various discarded devices, could I use one of those (cut off the silver tip)? If so, what would I change in the design to adjust to be able use one of my wall warts? In other words, are there formulas I can apply on my own to adjust the specs in the design to use one of my wall warts? Or do I need to post the specs on my wall warts and let you decide?

 

[KeepItSimpleStupid]

I actually gave you the formulas. I also gave you a reason not to use 5V. If you have a thunderstorm and it breaks. Well it's broken.

I think it would be best to post the specs and we are going to have to be able to determine the polarity. If you have a multimeter, that would be great. If not, there are other ways.

Way back when, my company bought a few thermocouple scanners. We used these with systems that had 1000 W quartz lights that blinked during their temperature control. The scanner's didn't last long. It turned out the mfr had designed it with a regulated power supply, but they were not stuffed. I added a $2.00 component and we had no more failures.

I have designed systems that have to work, or there would be loss of life. A system was put to the test and it worked. I'm the guy that also gets angry when I have to keep fixing the same problem over and over again.
So, this is where I come from.

 

[Reloadron]

KISS and Reloadron, thanks for all your help.

Reloadron, I must say this is exactly what I was hoping for. Thanks so much!

Since I have several wall warts left over from various discarded devices, could I use one of those (cut off the silver tip)? If so, what would I change in the design to adjust to be able use one of my wall warts? In other words, are there formulas I can apply on my own to adjust the specs in the design to use one of my wall warts? Or do I need to post the specs on my wall warts and let you decide?

As to a wall wart you just need enough current for the LEDs so 200 mA of current should be more than enough. I also agree with KISS in that you really should prefer low voltage. Personally anything in the 5 to 12 volt range should be fine. Just make sure it has a DC output as some do output AC depending on their original application. Yes, if you post the name plate data off of a few you have we could make suggestions. Also, per KISS if you have a voltmeter that would help. The tip could be cut off or a mating socket found (maybe Radio Shack). Most wall warts just fold under a short but in the interest of safety I would look for a 200 mA or maybe 1/4 Amp (250 mA) fuse. Pigtail or whatever type.

KISS did post the formulas for the LEDs but what you have is this. The LED should have specifications that include the Vfwd (Forward Voltage) and Ifwd (Forward Current) so things look like this:

Vsupply - Vled / Ifwd = Resistance needed in series with the LED.

Just keep it safe and simple. Hell, you could configure it on a bench sans doors and test it.

Lacking a voltmeter if you get polarity confused place both leads in a piece of potato close to each other. One lead should turn green where it enters the potato, that is (+).

Ron