What ultimately kills the LED if it is overdriven is the heat it can't get rid of. As usual it is (forward voltage * current). As I is a function of Vf and vice versa, it is sufficient to make sure that the current won't exceed the maximum permissible value, in this case 50mA. Here this can be accomplished by using a suitable resistor (ignoring variations of temperature in the LED).
Surely heat is directly related to current, so if you're putting the same current through each it doesn't matter?
I'm relatively new to this whole circuit design thing, but I'm pretty sure that current isn't the only thing that matters.
"For a standard LED of 5mm diameter, Figure 1 shows the forward voltage (VF) vs. forward current (IF). Note that the voltage drop across an LED increases with forward current."So I think what the graph is showing is how the forward voltage changes as I put more current through the LED.With that in mind, I think that means if I reduce the current I'm trying to put through the LED, the forward voltage will drop. So if these LEDs can have a max forward voltage of 5.3v at 50mA, then if I reduce that to say 40mA, I will have less of a chance of getting a batch of LEDs that simply won't work at 5v.
To use a water analogy, let's say you have two pipes going to a pool, and the diameter of the pipe is the current, and the pressure which the water is under is the voltage. If both pipes are 3" in diameter, but one has twice the pressure, then you can see there's a lot more energy there and it will fill your pool twice as fast.So to go back to the LED example, the led with the lower forward voltage is using up more of the voltage inside it to generate light and heat, so it should get a little hotter than the one with the higher forward voltage.
so where does the voltage go once it hits the LED if not into generating light and heat?
where does the voltage go
But I would still like to understand what that graph is trying to tell me.
If the graph were showing me how much current I could put through the LED if it had different forward voltages, then wouldn't the current be inversely proportional to the forward voltage, rather than proportional to it?
So I think what the graph is showing is how the forward voltage changes as I put more current through the LED.
I don't understand how the LED can lower the voltage without actually using up power.