Good day all,
I've been going through many tutorials to determine the value of the resistor i should use (using Ohm law, etc.) but I wanted to make sure I got it right.
Here is the datasheet of the LED i want to use
According to my calculation, the resistor value is around 100 (Using a 5V power source), but since this is my first project I would like to make sure I got it right,
Thanks for taking the time for reading this and helping.
The formula is:
( V[power source] - V[LED forward voltage drop] ) / I[desired current]
So if your power source is 5V and LED Vf is 2V and you want, say, 5mA passing through it:
(5-3)/.005 = 400 ohms
nordi:
...but I wanted to make sure I got it right.
Just make sure you don't get near the absolute max rating of the LED.
For tiny LEDs like that it might be 10-15mA. See the datasheet.
Or just use less than 10mA.
Those tiny LEDs are still very bright with 1mA.
Just use 1k for (5-2)*1k= 3mA.
10k (0.3mA) might still be too bright.
Leo..
The two different colours of LED need to be treated differently due to the different forward voltages:
2V orange and 3.3V blue.
For 20mA at 5V that means 150 ohms for orange, 82 ohms for blue. Scale up for lower currents.
You'll need enough pcb copper to dissipate the heat, note, if running them at full power, because they
are so small. Another reason to scale down the current.
MK1888:
The formula is:( V[power source] - V[LED forward voltage drop] ) / I[desired current]
So if your power source is 5V and LED Vf is 2V and you want, say, 5mA passing through it:
(5-3)/.005 = 400 ohms
Thank you very much, now I understand even better how to use the formula !
MarkT:
The two different colours of LED need to be treated differently due to the different forward voltages:
2V orange and 3.3V blue.For 20mA at 5V that means 150 ohms for orange, 82 ohms for blue. Scale up for lower currents.
You'll need enough pcb copper to dissipate the heat, note, if running them at full power, because they
are so small. Another reason to scale down the current.
Please correct me if I'm wrong but my idea was to use one resistor for both colors (using a value strong enough for both):
I have an atmega328p-au (SMD), pin 1 is linked to the blue color, pin 2 to orange color and my idea was to have both pin 1 and 2 connected to the resistor, and said resistor is connected to both the blue and orange lead of the LED.
Since both colors are not on at the same time it would allow me to save space and components.
[EDIT]
Your schematic didn't show when I first typed my reply.
You have shorted the pins together, you will damage the outputs. One high, the other low, lots of current between them. Use 2 resistors.
[EDIT EDIT]
You might be OK if you swap the LEDs and the resistor; put the LEDs on the pins, join the cathodes and have 1 resistor to 0V.
What bothers me is: what is it you don't understand about basic electrical circuits that made you think that might work?
