Could someone help me and check to see if my diagram will work. I want to make a simple device which will light an LED once an object is lifted off a mat.

I haven't done this in a while so my diagram may be completely out.

Thanks

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I'm afraid it is - there's no way the LED would light as it's shorted out. Presumably you're wanting the LED to light when the switch opens?, to do this you need an inverter, you can't do it with a diode - easiest way is a simple transistor.

Assuming you're not bothered about current consumption?, there is a nasty crude way - place the switch across the LED, and the LED in series with a resistor from a supply. But this way it takes MORE current when the LED isn't lit.

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PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

Thanks for the quick reply it seems I got a job in IT and not in electronics for a good reason; is there a simple way of creating this circuit without sapping the battery.

 

A single NPN transistor, LED and series resistor in the collector, and a resistor from base to the positve supply, emitter to the negative supply. This configuration turns the transistor ON - so you place the switch from base to emitter, this turns it OFF. Current consumption depends on the base bias resistor.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk