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How to replace the audible alarm in a 9v water leak detector with an ledSeptember 19, 2013 12:44 PM   Subscribe

I have a 9v battery powered water leak detector (this one: http://www.glentronics.com/water_alarm.htm ). I'd like to open it up and replace the audible alarm with an led. How do I do that?

I have the basic equipment and I've done a little bit of soldering/desoldering (little jobs like fixing a loose connection or wire) and voltage/resistance measurement with a multimeter. But I've never done an actual project like this. I don't know how to determine what size led to use or what else I might need to do the conversion. How do I do this? Please help. Thanks!
posted by atm to technology (5 answers total) 2 users marked this as a favorite

An assumption: The audible alarm is a piezo oscillator that'll go off with DC. Also, this is a best guess, and I'm generally an electronics duffer. I'll try to document my assumptions as I go.

Get out your volt meter, clip the wire to the "speaker", trigger the alarm and measure the voltage across the two leads which used to go across the speaker. If you have trouble, try measuring this as both DC and AC, but my assumption and hope is that it's DC.

Next assumptions: The "speaker" will only draw as much current as it needs, and there's plenty of current to draw. If it's AC, that may mean you need to use a smaller resistor to get the LED to light up appropriately brightly, but you should at least see light with the resistor you calculate for DC.

Use a "LED calculator" (search Google, there are a gazillion of them) to figure out what resistor you need for your LED. Solder the LED in series iwth the resistor to the two leads. If it doesn't light, you got the resistor backwards. Also note that "close but more Ohms" is a reasonable way to size resistors, if the calculator tells you 220 Ohms, it's okay to start with 330 Ohms.
posted by straw at 1:12 PM on September 19

I agree with straw, except to clarify what I suspect is a typo. If it doesn't light it is the led that is backwards, not the resistor. Resistors generally don't care which way they go, LED's do.
posted by Just this guy, y'know at 1:27 PM on September 19 [1 favorite]

Doh: Yes, Just this guy, y'know is correct: The LED will light one way and block current the other way. The resistor is current-direction agnostic (which also means that it doesn't matter which side of the LED it's on, if that's helpful).
posted by straw at 3:46 PM on September 19

straw: "The LED will light one way and block current the other way."

Don't rely on this if the drive voltage is AC. Typical reverse breakdown V @ usual operating currents are somewhere around 4~5V; 9V AC is likely to blow it in short order.

If you must drive it with AC, use a series diode - even a small signal diode like a 1N4148 - in series with the LED in addition to the resistor.
posted by Pinback at 4:52 PM on September 19

Thanks a lot, you guys. I'm going to get going on this in the next week or so. I'll post a follow-up question if I run into any trouble.
posted by atm at 7:29 PM on September 20

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