Hi there,
I'm planning on using 3 shf485P led's with 4x rechargable 1,2V (2000 mAh or more) batteries.
When i do the resistor math R= (Vs-Vl)/I = (4,8-4.5)/0.07 = 4.3 Ohm
As I'm no wizzard to electronics I would like to know if my calcs are right or not.
But I've also read that you should also have to calculate 2V for the resistor. Is this correct or did I misunderstood the text.
Hope you can help me figure it out.
Regards,
The equation that you're using is correct: R = (VS - VL) / I
VS = supply voltage
VL = LED voltage (usually 2V, but 4V for blue and white LEDs)
I = LED current (e.g. 20mA), this must be less than the maximum permitted
According to your math above, this is what it says:
Supply voltage = 4.8V
LED Total voltage = 4.5V (1.5v each LED)
LED current = 70mA or 0.070A
So R=(4.8-4.5)/.07 = 4.28ohm
So you need 1 4.3 Ohms resistor
Here is the reference page for you:
http://www.kpsec.freeuk.com/components/led.htm#calculate
You do not need to calculate any draw from the resistor. That does not calculate into the equation, and will not cause you any problems.
Now, you can go with any resistor above the 4.3 Ohm one. Meaning, that you can put a 20 Ohm resistor in there and they will still work fine. They may be a tiny bit dimmer, but you would never notice it. You just don't want to put a 1 Ohm resistor in there, because the LED's will draw too much power and burn out very quickly.
I hope this answers your questions, and good luck!