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 Looking for a liitle help - cct for 12V LED setup
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SemosD
New Member

Bermuda
1 Posts

Posted - Nov 12 2015 :  4:33:35 PM  Show Profile  Visit SemosD's Homepage  Reply with Quote
I have a LED 12V and I would like to have it on, dim at first and when a switch is push the LED will get brighter, what formula would I use to calculate the size of the resister is series? SemosD

audioguru
Nobel Prize Winner

Canada
4218 Posts

Posted - Nov 14 2015 :  10:11:06 PM  Show Profile  Reply with Quote
You look at the forward voltage of the LED on its datasheet, subtract it from the 12V and see its typical current on the datasheet. The bright resistor value is the subtracted voltage divided by the current.
Add another resistor in series with the first resistor and LED that has a value maybe 10 times the other resistor that the switch shorts.

BUT you said you have an LED 12V which I thought was that you had an LED AND 12V. Do you mean you have a 12V LED? Then maybe you have a few LEDs in series or maybe you have an LED that already has a series resistor. Try adding a 4.7k resistor in series to see if it is dim enough.
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Aaron Cake
Administrator

Canada
6718 Posts

Posted - Nov 15 2015 :  10:36:57 AM  Show Profile  Visit Aaron Cake's Homepage  Send Aaron Cake an ICQ Message  Send Aaron Cake a Yahoo! Message  Reply with Quote
How complicated do you want to get? Because if it is for a tail light application, as I suspect, then I am doing something very similar:

http://www.aaroncake.net/76cosmo/taillights

But yeah, as mentioned, you can look up the datasheet and it will give you a brightness vs. current and voltage chart. You can then use Ohm's law (Google "LED calculator") to figure out a series resistor, or just try a few and see what you like.

Then put the resistor in series with the LED, and use either a relay (probably best for your application if it is automotive) to short the resistor (bright) and unshort it (dim) or a MOSFET to accomplish the same.
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