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schematic

simulate this circuit – Schematic created using CircuitLab

I am thinking of creating a circuit that lights up a green LED once the garage door is open so I don't need to lean out the car window to see when the garage door is open all the way. A switch will be triggered once the garage door is all the way up. But if I leave the garage door open for more than a couple of seconds, or even minutes, then the LED should stop shining so the battery doesn't drain, and once the circuit is open (garage door is closed) it should 'reset' so that it lights up the next time I open the garage door.

I want to use basic electronic components, and no chips.

I've tried something similar to this:

I've used software to try and emulate but.. I've more or less lost my train of thought.

I've tried this in the circuit designer:

Once I click the switch into closed it lights up, but when I click OPEN and wait, then click to close the circuit it seems the capacitor isn't discharged and led doesn't lit up.

What's the best way to go about this?

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    \$\begingroup\$ You need a fairly high value resistor (10 - 100K?) across the capacitor to discharge it. \$\endgroup\$ – Peter Bennett Jan 3 '17 at 20:29
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    \$\begingroup\$ A 555 timer monostable will do \$\endgroup\$ – Mike Jan 3 '17 at 20:34
  • \$\begingroup\$ @PeterBennett: Thanks that might solve it. The only problem with emulators/simulators is that they don't actually take power consumption in consideration... at least not the ones I know. \$\endgroup\$ – OddBeck Jan 3 '17 at 20:42
  • \$\begingroup\$ I find it easier to lean out the car window ))) \$\endgroup\$ – Nazar Jan 3 '17 at 21:19
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Can you use a SPDT switch for the garage sensor? If yes, how about the circuit below?

It uses a MOSFET. The gate is almost no current so you do not need a ginormous capacitor like you would in your circuit.

When SW1 gets thrown into the up position (garage door fully open), M1's gate immediately goes to 9V because C1 is discharged. This turns on the LED D1.

C1 starts to charge through R1 and the gate drive to M1 slowly drops until finally LED D1 will shut off.

When SW1 is thrown into the down position (garage door begins to close), 9V is disconnected and C1 is discharged through R3. Also, R3 limits the current to prevent arcing which would shorten the life of the switch contact.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I'm not sure, but this was a good solution and I'll even accept this as an answer as I liked it very much and I'll be able to get a SPDT-switch :) Thank you! \$\endgroup\$ – OddBeck Jan 3 '17 at 21:48
  • \$\begingroup\$ You're welcome! Just be careful with the MOSFET. The gate can be damaged by static electricity from touching it. Use proper ESD handling. \$\endgroup\$ – Vince Patron Jan 3 '17 at 22:20

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