I Need To Trigger a Relay When an LED Lights Up On a Smoke Detector

Ground is just a reference point from which all voltage measurments are made. Usually ground is the negative side of the battery or power supply. So, if you measure 1.78 volts from the negative side of the batteries to the LED, it means the negative side of the LED is probably grounded or connectd to the barrery (-).

If this is the case, you might be able to drive a small relay with a transistor and resistor and diode. Something like below might work if the relay coil current is less than 50mA. There also should be a diode in parallel with the relay coil (not shown). The cathode of the diode goes to the top (+4.5). You can also use a higher voltage if needed, for a 12 volt relay for example.

+4.5 | | Relay | C 510 |/ +---//----B| NPN (2N3904) | | LED E | | | | GND GND

-Bill

This sounds like a fun project.

One thing to consider is that you probably won't be able to power the relay open using the current through the LED. If you are willing to use a separate battery, however, then there is an easy way to do this.

Define the ground to be the more negative side of the LED. (Note: you can define ground to be anything. It's just the point you measure other voltages against)

Then, attach one side of a 4.7k resistor to the positive side of the LED, and the other side to the base of an NPN transistor. Attach the emitter of the NPN to the negative side of the LED. Now, when the LED turns on, it'll allow current to flow through the transistor.

Hook the negative terminal of a 9V battery to the emitter of the transistor, and the positive side to one side of the coil for the relay. Hook the other coil of the relay to the collector of the NPN transistor.

Now, when the LED is on, the transistor will allow current to flow, so current will flow in a circuit through the relay coil, the transistor and the 9V battery. This should cause the relay to open.

--
Regards,
   Robert Monsen

"Your Highness, I have no need of this hypothesis."
     - Pierre Laplace (1749-1827), to Napoleon,
        on why his works on celestial mechanics make no mention of God.

I understood this perfectly!!! Thanks a million I will try this and let you know if it worked! I guess I just need to use a 9V relay.

Jean-Marie

What is the value of the resistor?

--
I'd wind a reed relay and substitute it for the LED or, if I needed to
keep the LED for some reason, substitute it for at least _part_ of the
LED's current limiting resistor. Before I could proceed, though,  I'd
need to know, as a bare minimum, the amount of current flowing through
the LED, the value of the current-limiting resistor in series with it,
and whether the alarm system is looking for a closing or an opening
contact closure as the triggering event.

If you can reverse-engineer all of the LED driving circuitry and post
what you find, so much the better!

--- A way to do it without having to modify the CO/smoke detectors (which, if you did, would surely void their warrany and perhaps give your insurance company grounds to reject a claim, if you're concerned about that kind of thing) would be to use an electret microphone to detect the raucous audio output from the alarm and eventually activate a relay or a little wireless transmitter which activates a receiver at the alarm. I'm pretty sure there's some X10 stuff out there which you could easily adapt for the task, or if you want, we can help you roll your own.

-- John Fields

Pity there are no such 'extension terminals' available on your units; (I'm guessing that some models now offer that convenience?)

Anyway, it should be easy enough to do what you want using the direct wiring approach that you propose, i.e. to end up with a circuit that closes a normally-open switch which can use trigger your house alarm. (Or opens a normally-closed switch; you need to check which your alarm requires.)

Rather than wire across the LED, it would simplify matters if you could find an accessible connection point which would be 0V when inactive and 3.5 to 4.5V when active. So I'd make that the first step. The other connection point should be permanently 0V, i.e. the negative battery terminal, or a point directly connected to it.

If you can find such a pair of connections, then you can use a circuit like the one at the top here:

If any of the smoke detectors is activated, the relay is also activated and the required switching of the house alarm occurs.

If you *can't* find such a 'full range' connection point, before modifying that suggested circuit it's a good idea to establish which of various circuit switching arrangements your smoke detector adopts. I've illustrated 4 examples in the lower diagram, showing the active and inactive states for each.

You've measured voltage across the LED, so these distinctions won't have been apparent. If instead you connect the negative lead of your DMM to the negative terminal of the battery (or any convenient wire directly connected to it), any measurements you make with the positive DMM lead will be *relative to 0V. So your DMM will give the same results as I've shown on the diagram. A little careful experiment will now quickly establish how your detector is working. Alternatively, you could remove the battery and measure resistances instead. If it comes to that, report back and we'll see if any changes are needed.

--
Terry Pinnell
Hobbyist, West Sussex, UK

Thanks so much Terry! Your diagrams are great! I will look for a trigger point that gives me the full 4.5V as you reccomended. I really appreciate your time and expertise.

Just to give you an idea of what I am doing. I have a house full of hardwired 120V AC smoke detectors that are all wired to each other and are also wired to my alarm system via a relay.

I am adding Carbon Monoxide detectors to certain parts of the house. I have these small wireless Caddx Alarm system transmitters that detect a contact closure that have these simple screw down terminals that were designed to wirelessly monitor doors and windows. I thought I could use these same wireless Caddx Alarm system transmitters to detect the status of the Carbon Monoxide detectors based on the RED Alarm LED.

Since all I need is a simple contact closure a reed relay that operates at a low voltage should work.

I will take your advice and map out how the Carbon Monoxide detector works using the DMM. Thanks a million!!!

Jean-Marie

It's labeled 510 which is 510 ohms. But I'm assuming the LED is running near 20mA so the resistor only bleeds off a couple mA and doesn't upset things too much. If the LED draws much less than 20mA, you may need a larger resistor and a extra transistor.

-Bill

--- If the voltage across the LED is 1.78V and the drop across what's driving it is 0.3V, then if the LED current is 20 mA and the supply voltage is 4.5V, the current limiting resistor for the LED should be something like

E 4.5 - (1.8 + 0.3) R = --- = ------------------ = 120 ohms I 0.02

Now, if you wanted to use a reed relay to do the switching you could choose something like a Hamlin MITI-3V1 reed switch and wind it with enough turns to make sure it closed.

Assuming a worst case of 15AT and, assuming end-of-life battery voltage of 3V, then you'd have about 7.5mA to play with and you'd need

AT 15 n = ---- = -------- = 2000 turns A 0.0075

The switch has a length of 0.275" and a diameter of 0.071", so you'd probably need a bobbin to hold all the wire; say with an inside length of 0.3" and a diameter 0.1".

If you used #40 wire for the coil, with a diameter of 0.003145" (and you rounded up to 0.004", just because...) you could get

0.275 n = --------- ~ 68 turns 0.004

per layer. Since you need 2000 turns, that would be 29.4 layers and, rounding up again that would give us 30 layers.

If we started with a bobbin diameter of 0.1" and added 0.004" for the wire, the first layer would have a diameter of 0.104". Adding 0.004" for each additional layer would result in a diameter of 0.220 for the

30th layer, resulting in a mean length of turn of 0.509" for the winding, and with 2000 turns that would be 1018" or, about 85 feet of wire. # 40 wire has a resistance of 1.05 ohms per foot, so that would be about 90 ohms.

If the LED's current limiting resistor was 120 ohms, then, you could replace it with a 30 ohm resistor and wire the coil in series with it giving you 20 mA into the LED and a definite metallic contact closure when the LED went on with no additional circuitry. If you wanted to, you could wind the bobbin with enough wire to get you to 120 ohms, which would eliminate the resistor altogether. If you decide to go this route, it wouldn't hurt to put a diode (something like a 1N4148 in parallel with the coil, with the diode's cathode connected to the most positive end of the coil.

Thinking about it, since you probably won't be able to level-wind the coil you're going to wind up putting more than 85' on it for 2000 turns. Maybe, even, 114.28' ;)

-- John Fields

OK, let us know how you get on.

I didn't see any response to the question about warranty invalidation that others raised? If that *is* an issue, you might consider the alternative LDR suggestion. To check its feasibility, get an ORP12 or similar, connect it in series with say a 100 ohm resistor (to protect it against too much current in case of inadvertent intense illumination), connect a DMM set to resistance across the combination, and experiment. With the LDR carefully positioned inside the case, close to the rear of the LED, see if you can get a significant variation (say 2 to 1 or better) between inactive and active states. And of course there must be no significant resistance change due to

*external* illumination; in practice I suspect that might be hard to achieve for a CO detector positioned close to bright indoor lighting or a sun-facing window.

Alternatively, assuming the CO detectors have an audible alarm, John's sound detector approach is another option, although with similar practical difficulties.

If warranty is *not* an issue, the direct wiring approach is by far the simplest solution.

BTW, where do you live - a thatched cottage next door to a steel foundry?

--
Terry Pinnell
Hobbyist, West Sussex, UK

Mission Accomplished! I have a CO Detector Wirelessly Sending Out Alarm Status

I finally got the battery operated Carbon Monoxide detector to also send a trigger to the house alarm system!

I want to thank everyone who pointed me in the right directoion and also those that came up with novel ways of making this work.

Here is the setup:

I have already in place in my single family home hardwired interconnected smoke alarms that are also hardwired to my monitored alarm system. What I do not have is any Carbon Monoxide detectors. I found very nice battery operated Carbon Monoxide detectors at "Home Depot" that would work great as standalone units. I wanted to have these nice battery operated Carbon Monoxide detectors also fault independent zones on my monitored alarm system.

My alarm system has wireless contact closure transmitters. If the wireless transmitters detect a contact closure on the screw terminals it sends a zone faulted alarm to the alarm system.

The battery operated Carbon Monoxide detectors have two LEDs on the face, one that blinks every 30 seconds to indicate normal operation and one that is RED that ONLY turns on in an alarm condition.

I took the PCB out of the battery operated Carbon Monoxide detector (it snaps right out) and soldered 2 ? 24 Gauge, 12 inch wires onto the anode and cathode of the red LED on the back side of the PCB.

With my voltmeter I saw that the voltage across the red LED when the battery operated Carbon Monoxide detector when into an alarm condition was 1.7 ? 1.9 Volts. The LED would also flash rapidly because the voltage would go on and off. In other words the LED would stay fully lit if it always received the 1.7V and would only turn off when no voltage was present.

Here is what I did to get contact closure from the voltage across the red LED.

With the help of many people here:

1 ? I ran the two small wires I had soldered to the battery operated Carbon Monoxide detector to a breadboard.

2 ? I used an external 9V battery run to a voltage regulator that dropped the Voltage to 5V out

3 ? From there I went to a NPN transistor and resistor on the base leg.

4- between the emitter and collector I want to another NPN transistor that was the switch for a 555 timer circuit

5 ? The 555 Timer then has wired small 5V reed relay with diodes across the coil. I also used a 1 mega ohm resistor so the 555 would hold the relay closed for about 16 seconds.

Now when the battery operated Carbon Monoxide detector goes off and the red LED flashes, the reed relay closes for 16 seconds and does not care about the flashing nature of the red LED, it stays latched for 16 seconds as soon as the first 1.7V hits the NPN transistor.

I was amazed that all this really worked!!!! Now the hard part.

I want to put all this circuitry in little project boxed but I have never soldered to a circuit board. There are lots of junction points and I worried that I may burn many of the components trying to solder all this together. Any advice?

Here are the 2 circuits I put togher to make all this work:

Thanks again to all those that helped me! Jean-Marie Vaneskahian snipped-for-privacy@vaneskahian.com

Jean-Marie Vaneskahian has brought this to us :

Doing that modification will void the UL certification of the device and lets your insurance company off the hook if ever you have a claim. There are smoke detectors available to interface with your alarm system.. Look at

--
This is an automatic signature of MesNews.
Site : http://www.mesnews.net