UK Fire Service

Discussion in 'The Projects Forum' started by keithdixon, Nov 22, 2006.

  1. keithdixon

    Thread Starter New Member

    Nov 22, 2006
    4
    0
    I work for the UK Fire service.

    We have a problem with our lockers opening (Due to not being closed correctly) whilst on route to a call.

    What we need is a way in which to monitor the lockers of the fire truck.

    I have in my mind 8 reed switches (One for each locker) connected to a control box

    Within the control box there is an eight way nand gate providing an output via buzzer. (When any locker is open)

    in addition a set of LED's displaying which is open and which is closed.

    Sounds easy?

    if so you are the one for me.

    Apart from the sensors I have no idea how to wire this baby up.

    What resistors do i use, do i need diodes etc etc etc.

    I have ordered a number of Texas 8-input NAND gate's but do not have a clue what next

    PLEASE HELP!!!

    regards


    Keith Dixon
     
  2. Gadget

    Distinguished Member

    Jan 10, 2006
    613
    0
    Are the reed switches' contact's Closed or Open when a door is open..?

    Do you want to use the NAND gates, or would you be happy with a simpler option using the reed switches, LED's resistors and a buffer transistor..?
     
  3. keithdixon

    Thread Starter New Member

    Nov 22, 2006
    4
    0
    My plan was to have +ve running through the reed switches, back to the control box. Thus when doors are closed the circuit would be complete (+ve)

    I am not aware of using just reed switches, LED's resistors and a buffer transistor for the solution.

    My first thoughts were that i needed nand gates to determine if all were closed

    I am happy to look at either solution. As mentioned i have ordered the nand gates. If this option is a lot more complexed i am happy to ditch them and go for the 2nd option, if it is as complexed i could go with the nand's

    regards


    keith dixon
     
  4. richbrune

    Senior Member

    Oct 28, 2005
    106
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    What Gadget is saying is, if you have Normally closed reed switches, you don't need the logic gate or any transistors. Normally closed, as I am refering to, is one that opens the connection when a magnet is in it's proximity, but closes and allows current to go through it when the magnet is not close to the reed switch. If you build a small box with a light for each door, you don't need to know anything except that one or more of the red lights is on, and to close that (or those) corresponding box(es). I'll try to get to my scanner later, and send a diagram. I've got a vague recollection that vehicles in the U.K. have the positive lead connected to the chassis of the vehicle, rather than the negative lead, as here in the U.S.---I'll try to remember the appropriate footnotes--
    Rich
     
  5. richbrune

    Senior Member

    Oct 28, 2005
    106
    0
    Here's a sketch of the circuit I think he means. Please see the notes below reguarding the fuse. Use 16ga shielded, stranded wire with two conductors, red and black, jacketed. Use the appropriate butt splice connectors for 16ga, avoid wiring where exaust heat or other strains may damage wires. The LED's act as SECONDARY fuses in this circuit, so if an LED burns out-check to see if there's a bare wire that's touching the chasssis, before replacing the LED. (on Negative ground systems). If the chassis of the vehicle is connected to the Positive lead, we need to redesign. Find an appropriate circuit to branch off of, or use an inline Fuse from the battery to the LED box, placing the inline fuse holder as close to the battery as possible (NOT SHOWN IN DRAWING). Use a 1 amp fuse or less. The proximity switches used are not really that typical, but I have found them from time to time. Most proximity switches for door alarms are designed to open the circuit when the magnet is removed (door is open).
    Hope this helps.
    Rich
     
  6. Gadget

    Distinguished Member

    Jan 10, 2006
    613
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    Can't see Richbrune's sketch, but Here was what I was thinking assuming the switches are Normally Open when the doors are closed
    As you can see, when a door opens, the reed switch closes and lights the LED. Now the junction of the resistor and LED is biased at the forward voltage of the LED (1.7v for red, 2.2v for green). The 1N4148's isolate each LED and bias the transistor on if an LED is lit. Depending on the transistor used, you may need to reduce the resistor in series with the base to drive it into saturation.....

    [​IMG]
     
  7. n9352527

    AAC Fanatic!

    Oct 14, 2005
    1,198
    4
    If you connect the 1N4148 between the SW and the R instead, you could connect a 12V buzzer directly to the diodes common and save on the transistor and resistors.
     
  8. Gadget

    Distinguished Member

    Jan 10, 2006
    613
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    Very true. I got so tied up with using a buffer that I neglected the obvious.
     
  9. keithdixon

    Thread Starter New Member

    Nov 22, 2006
    4
    0
    Thanks for the responses,

    I have edited the diagram with the alterations.

    Could you check over for me?

    The fire trucks are 24v negative earth

    regards


    keith Dixon
     
  10. n9352527

    AAC Fanatic!

    Oct 14, 2005
    1,198
    4
    The diodes are correct, but you don't need the resistor, and you have to connect the other side of the buzzer to ground instead of positive supply. Otherwise, looks good.
     
  11. keithdixon

    Thread Starter New Member

    Nov 22, 2006
    4
    0
    thanks for that,

    I have ammended again,

    could you advise upon resistor & diode types for use with std 5mm red led's fed from 24v power

    rgs

    keith dixon
     
  12. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Hi,

    For 10 ma current through the LED's use 2.4K (you may have it as 2R4) resistors. For real attention-getters, use blue LED's. They are just amazingly bright and abnormal enough in appearance to be noticed. Current is low enough that any old diodes will do, as long as it's not a Schottkey type - they have very low reverse voltage ratings. 1N4148 will do nicely, as will any with a rating of 50 volts and up, and 50 ma and up.
     
  13. wireaddict

    Senior Member

    Nov 1, 2006
    133
    0
    Just a few more comments. First, the "buzzer" should be a Mallory "sonalert" or other electronic annunciator; 1N4148s probably can't handle the current of a magnetic/mechanical buzzer. Whatever you use, it must be able to operate on 24 V or you'll need to add some voltage dropping resistance in series with it. And note the polarity on the electronic annunciator. Also, since the operating voltage is 24 V, the series resistors for the LEDs need to all be 2K, 0.5 W instead of 1K to drop the required voltage and handle the additional E^2/R. Should work great for you.
     
  14. wireaddict

    Senior Member

    Nov 1, 2006
    133
    0
    Sorry, Beenthere, somehow I missed your comments about the resistor sizing.

    Dave
     
  15. Gadget

    Distinguished Member

    Jan 10, 2006
    613
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    2k2 resistors are easier to find, and should be just fine and dandy.
    If you use a piezo sounder, then 1n4148's are fine, and most piezos work from 3 - 30 volts so no inline resistor required.
    A higher current diode for the mech buzzers would be the 1n4001 (up to 1n4007).... (1 amp)
    Looks like you've got things well in hand.
     
  16. Spoggles

    Well-Known Member

    Dec 2, 2005
    67
    0
    Hello:

    One thing you may want to consider is that in security and fire alarm protection, parallel circuits are generally to be avoided as it is difficult to impossible to monitor the integrity of the wiring between the sensors.

    If you use reed switches that are closed when the magnet is applied, in series, you will also monitor the integrity of the system (supervise it). If it is not too important to tell WHAT door is at fault, then this is the way I would do it.

    regards
    Spoggles
     
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