Can anyone explain why I haven't blown my 1W LEDs on my testing platform?
12V in, to LM317 set as constant current @ 100mA.
3 series connected LEDs, 2.1Vf each, for total LED Vf of 6.3V
3V loss in the 317 in CC mode leaving 3V unaccounted for.
Why are my LEDs still alive? What I've been taught is if you overvolt OR overcurrent an LED you will kill it.
Not to mention that according to all the rules I've been taught, voltage drop through a circuit MUST equal total supply. So where is the missing 3V?
If an LED is working, it is impossible to get too much forward voltage across it if you keep the current to less than its rating.
You don't quote the full specifications of the LEDs, but they are probably something like this:-
https://www.avagotech.com/docs/AV02-1640EN
Now those quote a Vf of between 1.7 and 2.3 volts at 350 mA. The graph also shows that the typical voltage is around 2.2 V at 350 mA, dropping to 1.8 V at 100 mA.
However the voltage rating of an LED is not what you should supply it with. You should supply it with current, and the voltage is what the circuit has to put up with.
Most electrical devices are the other way round. They are rated at a voltage, and the current that they take is what the power supply has to allow for. For instance, you plug a heater or a mobile phone charger into the same socket in the wall. Both are rated at mains voltage, the heater takes lots more current.
LEDs are connected to circuits like yours that supply constant current. That is why the LEDs are in series, forcing them all to take the same current. The LM317 with the resistor will supply the same current whatever the load voltage is. If you short out all the LEDs, the current will still be the same.
The voltage of the LEDs is useful for calculating the supply voltage you need. The minimum voltage across the LM317 plus the resistor is about 3 V, leaving about 9 V, so you could probably run 4 LEDs in series. If you put 5 LEDs, there would be less than 2 V across the LM317 and resistor, so the current would be less.
The heat generated by the LM317 depends on the voltage drop, so if you did short out all the LEDs, the LM317 would get a lot hotter. The minimum LED voltage should be used to calculate the maximum voltage drop on the LM317, and therefore the amount of heat it will produce.