| T O P I C R E V I E W |
| SemosD |
Posted - Nov 12 2015 : 4:33:35 PM
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 |
| 2 L A T E S T R E P L I E S (Newest First) |
| Aaron Cake |
Posted - Nov 15 2015 : 10:36:57 AM
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. |
| audioguru |
Posted - Nov 14 2015 : 10:11:06 PM
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. |
|
|
