I have a led tape strip 5630-60D (12v) which one can cut every 3 leds.
Nw the thing is that I need only one led (actually 3 separated which I plan to drive later with a nano and a l298n module)....
The question is which tension and resistance do I need for a single led?
I've made some measurements on a 3led segment trying to understand.... Here below my primitive results:
R between 2&3: 39 Ohm
The current consumed: 55mA
My humble conclusions... The voltage drop of the led is 3.2v and the I of one led is 55mA
Now I guess if I plan to supply 12v-3.2v/55mA = 160Ohm.... Chossing a 180Ohm will give me a 49mA dissipating a lot of power on the resitor... so probably it would be smarter to drive it under 5v with 33Ohm --> 54.5mA
So the real questions are:
- are my calculations right?
- could I drive this led directly from the nano at 5v at 55mA or do I need to reduce the current (to the 40mA from nano spec) if I want to drive directly?
Thank you in advance!
That's too much current. Get yourself a small transistor, drive the transistor base thru a 200 ohm resistor, and use the transistor to sink current thru the LED to Gnd. The arduino can't sink current from a 12V source, while an external transistor can.
Thank you CrossRoads. What about my calculations? If I remove a single led should I use a 180Ohm @ 12v / 33Ohm @ 5v?
They are deisgned to be grouped in 3's....no?
(hence the 12 volt requirements on those type of strips?)
If you are trying to remove and use only 1 led... you dont need 12 volts anymore.. as they are not grouped together to take the 12v any longer.
post a link to the exact led strip you are trying to use/hack a part.
55mA is to much for an Arduino pin. Even 40mA is absolute max so recommended is max 20mA to 30mA per pin. And it also has a total max of 200mA. So get yourself a transistor to drive them
And yeah, voltage drop of 3,2V is about right. So indeed around 33Ohm. Or make it easy and just grab the part with the 39Ohm already on it Will give you 46mA @ exactly 5V but you will barely notice the difference.