Hey y'all,
I know that I must use a resistor to turn a 12V signal into a 5V signal...but how much resistance. I searched through my books and there isn't anything about a formula to do that. Can anyone else think of something?
Thanks
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Andrew Weitzel
MECP First Class Installer
The circuit you need to build is a voltage divider.
HEREs a nice resource for designing what you need.
yea if you want to burn off over half your power use a voltage divider, you will need to know the load resistance though.
Using resistors to drop 12v to 5v..No big deal..Using resistors to drop 12v to 5v AND have enough current to do anything useful, big deal. So first off what are you trying to give 5v to, and how much current does it draw? If it's just an LED or something, just use the search button I know I explained that one myself at least once in good detail. If something else, let us know what your trying to do and I'm sure you'll get some better responses.
Mike
Thanks for they replies guys,
That resistance calculator is quite handy. I may just save that on our favorites list. So to reply, yes, it is an LED that I am looking to trip off of this....actually two LED's. Now, I want them to work with the alarm but the Black Widow alarm only puts out about 5V from the LED outputs.....It worked fine for the red LED's....but the blue ones would not light up for the life of them. So question here, is if i get a couple of relays, will 5.25V be enough to trip these two relays so that they will work. Also, how durable will a relay be, I have never given them a good load test before and seeing that they would be going every LED blink it makes question their integrity.
If anyone has tried to get more than one LED off of an alarm but the alarm doesn't put out enough voltage....I would be really happy to hear how they got it to work.
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Andrew Weitzel
MECP First Class Installer
dont use a relay, it will draw more current than 10 LEDs. I would suggest using a common bipolar junction transistor. like this...
