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Door Monitor

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BigBill

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I am looking to build a wired door monitor. I am a newbie to this.
I am thinking about something like this:
Normally Closed Magnetic switch attached to the door so that when the door is open it completes the circuit and lights a flashing red LED.
I was hoping this could be done using a single AA battery. I have searched the web and this forum but I guess this is too simplistic for anyone to have written it down.
Anyone have a diagram of what the circuit would need to look like and any parts I would need?
Thanks,
Bill
 
Do you mean a normally open switch and when door open's completes the cct flashing an led indicator.:)
 
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Most reed relay switches close when the magnet is nearby.

So door closed relay closed.

You need to look at the reed relay driving the base of a PNP transistor via a 1k:eek:hm: resistor, which in its turn switches the power on to the LED.

Door open reed relay open, no drive to PNP transistor and LED is on.
You need to use a pull down resistor at the base circuit. About 2k2:eek:hm: is ok.
 
RODALCO said:
Most reed relay switches close when the magnet is nearby.

So door closed relay closed.

You need to look at the reed relay driving the base of a PNP transistor via a 1k:eek:hm: resistor, which in its turn switches the power on to the LED.

Door open reed relay open, no drive to PNP transistor and LED is on.
You need to use a pull down resistor at the base circuit. About 2k2:eek:hm: is ok.

I think I am looking for the opposite so when the magnet is there the circuit is open, as soon as the door moves the magnet away the circuit closes and completes the curcuit.

I am very new so if anyone knows where a simple diagram that I could buy the parts from and build that would be great!
 
It doesn't have to be a reed switch unless it has to be hidden or inconspicuous.
Ordinary contacts would do just as well.
Arrange them so that the door opens them when its shut,
when it opens you can wire up a bell or a lamp as you choose.
Similar to a car door working an interior light.

John :)
 
As BlueRoom suggested, you can get both NC (normally closed) and NO (Normally Open) door switches. The most common one, used in alarm systems, is the one where the switch is closed (On) when the door is closed. You want the opposite type. If you are confused by this, buy one that has 3 terminals. That way it will have both NC and NO contacts:
**broken link removed**
Then just put the flashing LED, switch, and battery (You'll need 3V or more for a flashing LED) in a series loop.
 
If you're using it it for security, you're better off using a NO reed switch that closes when door is closed, as rodalco suggested, then if a wire is cut from switch it would indicate door open condition.
 
Thanks! I am just using it to clue me in if someone leaves the garage door open so an NC sounds like it would work. If the magnet is missing the circuit closes and lights up the flashing LED. I have ordered these parts from allelectronics to make it.
Anyone have idea's on how long I should expect the two alkaline batteries to be able to run a single T-1 3/4 flasher?

I am hoping its a long time..


Quantity: 2 Item: LED-4 Price: $ 0.45
- RED FLASHER LED, T-1 3/4
Quantity: 1 Item: BH-32 Price: $ 0.75
- BATTERY HOLDER 2 AA CELLS
Quantity: 1 Item: MS-7 Price: $ 4.50
- SPDT N.O./N.C. MAGNET SWITCH
Quantity: 50 Item: WRB-18 Price: $ 0.22
- 18 GAUGE BLACK/RED ZIP CORD
Quantity: 1 Item: 1591-MSBK Price: $ 3.20
- BLACK PLASTIC CASE, 3.3" X 2.2" X 1.04"
 
Well according to the allelectronics site, the LED should draw an average of about 10ma. The typical alkaline cell should be good for about 2800mah so it should last for around 280hrs of flashing. Read 280hrs of the door being open. Your mileage may vary slightly.
 
kchriste said:
Well according to the allelectronics site, the LED should draw an average of about 10ma. The typical alkaline cell should be good for about 2800mah so it should last for around 280hrs of flashing. Read 280hrs of the door being open. Your mileage may vary slightly.

Wow, I didn't know how that related to one another. Thanks!!

-Bill
 
BigBill

If drawing programmes weren't such a mission on a computer I draw you a schema.

Basically your circuit will be as follows.

1.5 Volts supply.

Positive to reed switch, then to 1000 ohm resistor, then to 620 ohm resistor then to negative. Currentflow about 0.9 mA.
At the point wher you join the 1000 and 620 ohm resistors , drive the base from a PNP transistor.
The PNP transistor can switch the LED ON or OFF.

How it works.

Door closed, switch closed, LED OFF. The PNP transistor has a positive bias at the base and is turned off.
Door open, switch open the positive bias has gone and the 620 ohm resistor pulls the base of the PNP to negative which turns the LED on.

Hope it helps,

Regards, Raymond
 
I hope this attachement works.

I'm not sure if your 1.5 volt battery voltage is high enough for the LED.

If you go for 3 volts, double the resistor values.

Regards, Raymond
 

Attachments

  • door circuit.GIF
    door circuit.GIF
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RODALCO said:
If you go for 3 volts, double the resistor values.
If BigBill uses a flashing LED he doesn't need a resistor. The circuit you posted has a high standby current and will drain the battery unnecessarily. A 1.5V battery will not light a LED. He'll need a 3V supply:
 

Attachments

  • door.GIF
    door.GIF
    3.2 KB · Views: 129
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what's wrong with just using the reed switch ?
 
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