Yesterday I was trying to figure out how many ohms of resistor should I be using for my LED, and found this post right here in the forum.
The formula states that one should subtract the voltage drop of the LED from Vcc, and then divide it by the target current (A), in the example, it was (5-3.4)/0.005 = 320 ohms resistor should be used.
My question is this: why the resistance is calculated not with whole 5V but only fraction of it? Suppose the resistance of entire circuit would be at the lowest level of 320 ohms if LED is 0 ohms, and if I used Ohm's Law on this circuit: 5/320 = 0.015625 (A), which is way off the target current I was hoping for.
It's like a mystery for me, and I hope someone could really explain it to me.
EDIT: Another way to put it: If one imagine a blackbox payload consumes 5V, in order to supply 5mA through the circuit, the blackbox needs to be 5/0.005 = 1000 ohms. And I know the blackbox is composed of 2 components: the LED and resistor, and the sum of their resistance must be 1000 ohms. Let R of LED is x, and R of resistor is y, they have the relationship x+y = 1000. That means whatever resistor I put in it, the R of LED automatically changes to a fitting value so their sum is 1000, is that possible?
Doesn't one need to know the resistance of LED, or the LED's resistance is really dynamic?