In article <(E-Mail Removed)>,
Moloney <(E-Mail Removed)> wrote:
-I am trying to detect whether or not my garage door is open. I purchased a
-normally closed contact switch to place on the bottom of the railing and
-bottom of the door. Being new to electronics I only understand the basics,
-but would like to try and use the N/C contact to send some type of signal to
-me that my garage door is open.
Sounds simple enough.
-
-Could I hook this N/C contact to a breadboard and have an LED light up when
-the door is open?
Sure.
- What would I need to detect if the circuit is opened or closed?
You already have the detector, the NC contact. The only problem is how to hook
it up so that the LED lights up when the door is opened. The simplest way is
to utilize the fact that an LED must have a current limiting resistor so that
it won't burn up. So let's start with a simple LED circuit:
V----R----A----LED-----GND
V is a positive voltage source, presumably a DC wall wart. R is the current
limiting resistor (more on that in a bit), A is just a mark point which I'll
get to in a second, just note there's no physical device at point A yet, LED
is the LED (duh?!) and GND is the ground end of the wall wart. If you wire
this on a breadboard, current will flow through the resistor and the LED and
the LED will light. To pick a resistor value divide the voltage by 10
milliamps (.010A) an use a resistor that's close to that value. So
for example if you have a 9 volt wall wart, divide 9 by 0.01 giving 900. So
pick a resistor around 900 ohms.
Now on to the dectector. It's normally closed which means that it's an unbroken
circuit when the door is closed. It wouldn't help to wire the detector between
point A and the LED because then the LED would be on when the door is closed.
But what you can do is use the detector as a bypass around the LED. So if you
wire it between point A and ground, when the door is closed then the current
will flow from point A through the detector and into ground (or vice versa ;-)
bypassing the LED. So the LED will go out because it doesn't have current
running through it. However when the door opens and the detector opens, then
current can no longer flow directly between point A and GND, so it must then
flow through the LED, lighting it.
Hope this helps,
BAJ
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