Electronic – How to use an LED as a photodetector

Increasing heat from power dissipation causes a failure of the LED die.

The change in colour, e.g. red and green LEDs going yellow at high currents, is probably because the die is actually glowing hot, i.e. near failure. Note that the red LED has a fall in wavelength, but the green one has an increase in wavelength.

White LEDs going blue could be explained by the yellow-emitting phosphor in the LED being less effective at high currents. White LEDs are often constructed from a blue LED coated with a special phosphor which emits yellow light when blue light hits it creating a fairly even white light. So perhaps you are seeing more blue than the yellow phosphor can convert.

You can use another technique that uses just circuit observation. For a simple circuit like this, I think it's easier than Thevenin.

The LED has a characteristic forward voltage, Vf. It varies with current, but the resistance in this circuit will keep the drop across the LED very close to the reported value in the datasheet. Let's use Vf = 2V for convenience.

Since the LED is parallel to R1, the voltage across R1 is also Vf. That must mean the drop across R2 is what's left over: $$V_{R2} = 5V-V_f = 5V-2V=3V$$

Using Ohm's Law on R2: $$I_{R2}=V_{R2}/100=30ma$$

Since R2 is in series with the source, that 30ma is the total current going into the circuit and is being split between R1 and the LED. We can figure out how much is going through R1 using Ohm's again.

$$I_{R1} = V_f/75= 2V/75 = 26.67ma$$

There's only 30ma going through the whole circuit and 26.67ma of it is going through the resistor. Therefore, only 3.3ma is left to go through the LED:

$$I_{LED} = 30ma - 26.67ma = 3.3ma$$