The Mathematics is pretty simple according to ohms law.
Lets say your LED is dropping 1.6v at 20mA and you have a 12v supply.
12 - 1.6 = 10.4v dropping across the resistor.
10.4 / .02 = 520 ohm resistor.
The same at 9V would be:
9 - 1.6 = 7.4v
7.4 / .02 = 370 ohm Resistor
What you have to watch out for is the power dissipation of the resister
when
using higher power supplies. So taking the 12V example:
we multipy the voltage drop by the current which would be:
10.4 * .02 = 208 mW
So you would probably consider using a 1/4 W resistor. Higher values
will of
course require higher rated resisters and perhaps heatsinks.
I think this is right, hopefully my Ohms law is not too rusty. The
moral of the
story is as has already been said, go for a higher value resistor and work
down
!
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