Create a Simple Fire Alarm with Voltage Divider and LED Circuit

[quickk]

Hi everyone,

I'm new to circuits, and was thinking about using a voltage divider circuit to make a very basic "fire alarm." I've attached an image of what I was thinking. The top resistor is a thermistor.

I'm having trouble figuring out what this circuit will do. My understanding of the led is that it will not light unless the voltage across it is higher than a set value, say 2 V. If I choose the resistor and thermistor correctly, the voltage across the LED will be below 2 V at room temperature. If I use a hairdryer and increase the temperature, the thermistor's resistance will drop and then at some point the voltage across the LED will exceed its critical value (2 V in my example), and then the LED should turn on.

This is where I am confused. When the LED is on, doens't it act like a short circuit? Then the entire voltage drop should be across the thermistor and once again the LED would turn off (since there is no more voltage drop across it). I know this doesn't make sense, but I can't figure out how to think about this.

Any guidance would be greatly appreciated!

 

[256bits]

The LED on-voltage is what it says. The LED will start to emit light and continue to emit light if the voltage across the LED exceeds the on-voltage. The voltage across the LED varies slightly upward from the on-voltage as the LED conducts. So no short circuit. Just do not exceed the on-voltage by too much or your LED will burn out very quickly.

In your circuit, please note that the point where the LED is attached to the voltage divider will act as a voltage divider for Rth and R1 until the LED turns on.

Then what happens if Rth is heated more its resistance drops and can cunduct more current. Where will that current go - mostly through the LED and it will become brighter and brighter until burnout. It would be best to put a resistance in series with the LED to act as a current limiter.

 

[quickk]

The voltage across the LED varies slightly upward from the on-voltage as the LED conducts.

Thank you for your help. I guess I was thinking in terms of an ideal diode, but in real life the LED's resistance won't go from ∞ to 0 at the critical voltage. Your statement still confuses me a little though: if the LED starts to conduct, shouldn't it's resistance drop? And if it's resistance drops, then so will the voltage across the LED. Why doesn't it turn off again then?

The current limiter resistance is a good idea. Thanks for the tip!

 

[vk6kro]

The voltage across the LED varies by a small amount once it is conducting, so even though its resistance is reducing, the voltage across it remains fairly constant.

It is like this:
http://dl.dropbox.com/u/4222062/LED%20V%20vs%20I.PNG

The actual voltage varies with the type of LED (mainly its colour).

The best thing to do is to get a LED and a potentiometer (say 1000 ohms) and try it with the pot in place of the thermister in your diagram.
Place a 330 ohm resistor in series with the LED, though, to avoid accidents.