Cupboard alarm

Discussion in 'The Projects Forum' started by roll cast, May 13, 2011.

  1. roll cast

    roll cast Thread Starter New Member

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    [​IMG]

    Here is an alarm circuit I have designed to be incorporated in a metal cabinet, can't use a reed switch as the cabinet is steel.

    Quick talk through

    Op amp and the 2 pot. dividers compares the light levels inside and outside the circuit, when the one is close to the one on the right it will activate the op amp output, this will need to be set up in situ using variable resistors

    When this occurs it will activate the thyristor which acts like a latching switch, this allows current to flow to the 555 timer which will let some leds and a buzzer to work in pulses.

    This can be reset by closing the door to darken the inside of the cabinet and then turning the circuit on again
  2. Bill_Marsden

    Bill_Marsden Moderator Staff Member

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    The reed switch is glued to the steel cabinet, the magnet (I'm assuming a tiny super strength type) is mounted with a nonferrous stand (plastic or wood) where the field from the magnet goes to the metal case through the reed switch with the door closed. It should work fine.

    You should consider paralleling the LEDs with a resistor and the alarm, your alarm is going to be starved for power, as will the LEDs. Also, paralleling LEDs like that is not a good idea, as there is a lot of variation between LEDs. If you try to split current three ways it will not flow equally between the three. Resistors are cheap, and will insure each LED gets the current you want.

    A tutorial on LEDs and more...

    LEDs, 555s, Flashers, and Light Chasers

    Bill's Index
    Last edited: May 13, 2011
  3. roll cast

    roll cast Thread Starter New Member

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    The cabinet has to be made entirely from steel sheeting as I have to do this from my project, I send in my form stating what I was going to make and out of what too soon and forgot about the steel causing problems

    Understand the first part but not the second
  4. Bill_Marsden

    Bill_Marsden Moderator Staff Member

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    LEDs are not simple light bulbs. They drop a voltage, called Vf, and it is treated as a constant. However, many cases they don't drop quite the same voltage. What happens is one LED will hog the current. If you are running them close to the limit of their current (not a problem on your schematic) then one will be overdriven while the others are starved. The tutorial explains this in detail.

    Instead, give each LED its own resistor, or put them in series.

    [​IMG] [​IMG]
  5. ErnieM

    ErnieM AAC Fanatic!

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    I did something similar for a fridge door ajar alarm, but there is can be assumed the light is always bright (a good bulb was a given in the exercise).

    5 parts total, and it consumes micro amps when it is not detecting anything. My breadboard has run for the last 2 years off the original coin battery.

    [​IMG]

    Of course, it helps when one of the parts is a micro controller with a built-in analog to digital converter.

    Normally (in the dark) the processor is asleep and draws minimal current. When the door opens the resistance of LD1 drops greatly causing a change of state in the digital input. That change wakes up the processor, which beeps till it sees dark again. Then back to sleep, rinse, wash, repeat.

    Nothing really practicable here, I just wanted some excuse to play with a 5 pin computer.
  6. roll cast

    roll cast Thread Starter New Member

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    Thanks Ernie and Bill

    E. I would love to use a pic chip but spec won't allow it

    B. I get it now, same as tolerance in resistors then, will rejig my circ. to accomodate that.
  7. SgtWookie

    SgtWookie Expert

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    The 270 Ohm resistors are far too small, and the 100uF cap is far too large.

    If you tried to build the circuit with those values, you would wind up burning up the 555 timer. Pin 7 is only rated for up to 15mA current before the saturation voltage gets out of hand, causing high power dissipation in the timer. 9v/270 Ohms = 33.3mA current.

    I use a "rule of thumb" of a minimum of 100 Ohms per volt of Vcc for the resistor between pin 7 and Vcc; 9x100=900 Ohms; 910 Ohms is the closest standard value. This will keep the current sunk from Vcc to 10mA or less.

    You could use a pair of 2.7k resistors and a 10uF cap; a pair of 27k resistors and a 1uF cap, or a pair of 270k Ohm resistors and a 0.1uF (100nF) cap and get the same timing.
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