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

Discussion in 'Electronic Basics' started by Jean-Marie Vaneskahian, Nov 8, 2004.

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

    My goal is to use simple battery operated combination Smoke / Carbon
    Monoxide detectors in various parts of my house. I want these battery
    operated detectors to trigger a simple contact closure when they
    alarm.

    I noticed that the detectors have a Red LED that lights up when the
    Smoke / Carbon Monoxide detector is triggered. I soldered two small
    wires to the PCB on the back of the detector in "parallel" with the
    Red LED.

    When I connect a voltmeter to the two wires I soldered in "parallel"
    with the Red LED and now run from the back of the PCB of the Smoke /
    Carbon Monoxide detector I get about +0.001V when nothing happens and
    +1.78V when I hit the test button on the Smoke / Carbon Monoxide
    detector. By the way, the Smoke / Carbon Monoxide detector runs on a
    total of 3 AAA 1.5V batteries.

    I have a very basic electronics understanding. One concept that I do
    NOT understand is that of "Ground". I do understand how to "Wire"
    components though.

    My question is this:

    How do I take the 2 wires running from the back of the PCB on the
    Smoke / Carbon Monoxide detector and generate a basic dry "Contact
    Closure" when the detector is triggered?

    I am sure this requires transistors, diodes, resistors, reed switches
    and a separate battery source, but I have no clue how to connect them
    and what types and values to purchase. I buy most of these components
    from Radio Shack (Part Numbers Would Be Great!).

    Please help me put this together. I really appreciate any wisdom on
    this topic. My goal is to protect my family by wiring these battery
    operated Smoke / Carbon Monoxide detectors to my home alarm system
    that uses contact closures.

    Thanks a million in advance,
    Jean-Marie Vaneskahian
     
  2. 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.
     
  3. Bill Bowden

    Bill Bowden Guest

    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
     
  4. 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:
    http://www.terrypin.dial.pipex.com/Images/Smoke1.gif

    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.
     
  5. John Fields

    John Fields Guest

    ---
    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!
     
  6. 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
     
  7. What is the value of the resistor?
     
  8. John Fields

    John Fields Guest

    ---
    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.
     
  9. 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
     
  10. Bill Bowden

    Bill Bowden Guest

    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
     
  11. 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? <g>
     
  12. John Fields

    John Fields Guest

    ---
    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' ;)
     
  13. 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:
    http://www.aaroncake.net/circuits/relaytim.htm
    http://www1.electusdistribution.com.au/images_uploaded/relaydrv.pdf

    Thanks again to all those that helped me!
    Jean-Marie Vaneskahian
     
  14. 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 www.adilink.com
     
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