I understand that the bigger resistor is the lower current will
but I miss some explanation about how to choose the resistor's
value. You have to make sure you only have Vf available to the
LED, and need to lose any other voltage from the supply, across
the resistor. You know a few things: the voltage supplied Vs,
the voltage the LED can take Vf, and the current I. You need to
lose the difference in the voltages across the resistor, call
that Vr: so Vr = Vs - Vf Apply Ohm's Law to the resistor, where
it's R we need to find: R = Vr / I = (Vs - Vf) / I. I don't know
Vs in your example, let's say it's 9v for example... then we
need to lose 9 - 2.6 = 6.4v across the resistor so R = 6.4 / .02
= 320. So to summarise, you need to apply Ohm's Law to the
resistor, not the LED, and the resistor needs to drop the
surplus voltage.