timer - How do I build a DC time out circuit that stays off until power is toggled?

My girlfriend bought an LED faucet that is actuated when you turn the water on and is powered by a battery pack of 4 AA batteries (2 wire out).

However, when the water is turned off, the slope of the faucet is rarely sufficient to overcome the surface tension and the faucet remains on. How can I build a timeout circuit such that the faucet turns off after being lit for 5-10 minutes and stays off unless water is fully cleared from the faucet and then the faucet is turned on again?

A friend of mine recommended a 555 chip, but after looking into it it seem the faucet would just turn off after 5 min, then turn back on 5 min later and keep cycling indefinitely.

I tried searching this site and came across this question that seem similar, but can't figure out how to deal with the fact that the activation of the circuit would work more like a toggle switch than a push button.

Thanks!

Answer

Yes, a 555 would do the job. The 555 can work in a number of ways - the one you are looking for is called "Monostable". This triggers once on the press of a button and turns itself off after a certain amount of time. Nothing happens until you press the button again.

Replacing the button with a capacitor effectively gives you a single press of the button at power-on time and nothing after that.

Something along the lines of (though I haven't tested it of course):

^{simulate this circuit – Schematic created using CircuitLab}

• R1 and C2 define how long it stays turned on for (around 363 seconds that should be with those values).


• C1 and R2 cause a delay on the trigger rising at power-on (you may want to increase C1 slightly for more initial delay).


• The 555 when activated turns on Q2 which in turn turns on Q1 which allows the power to flow to the LEDs.

By the way, this circuit was inspired by: http://555-timer-circuits.uk/operation/one-shot2.html