Electronic – Lower led bulb intensity with resistor

You know the current through your resistor and the value of your resistor, so you can calculate the power dissipated by the resistor (Ohm's law strikes again):

$$ P = I^2 R $$

So in your case that would be 0.026 W or 26 mW.

However when choosing the power rating, I'd not consider just the normal operating point but the worst case.

So the worst case would be, your power source failing with a higher voltage and the LED failing as a short circuit, in that case the resistor is dissipating the most power.

Well typically you don't consider both happening at the same time, but you might do that calculation as well.

If the LED fails as a short the current will increase and the resistor will dissipate (Ohm's law yet again):

$$ P = \frac{U^2}{R} $$

With your numbers: 167 mW.

The next thing to consider is what is actually meant when a manufacturer tells you "this resistor can handle 250 mW".

This value tells you the resistor will heat up a certain amount if you put 250 mW through it. Say 60 K (I'm making that up right now). So your resistor will heat up quite substantially in case the LED fails. Is your device able to handle that safely? I don't know.

The power dissipation is often also temperature de-rated, so at elevated temperatures your resistor is only allowed to dissipate a lower power. Does this limit your application? I don't know.

If you look at the data sheet for the LED it says that at 20 mA the forward voltage may be between 2.8 V and 3.6 V. Importantly it states that this is at an ambient temperature of 25 degC.

So, how well regulated is the ambient temperature in your experiment? I ask this because you seem to be relying on generating a constant current by using a resistor and a 12 V supply and, if the volt-drop on the UV LED changes (maybe due to temperature or aging) then the current into the device will change and the light output will also change.

Gradual self-heating of the LED cannot be ruled out and neither can drift on the 12 V DC power supply. You should consider a precision constant current supply - there are circuits you can build around simple op-amps that will achieve this.

Now, onto the photodiode - its "gain" will be somewhat sensitive to temperature but less so than for infra red types however, dark current will approximately double for every 10 deg C rise so this might start to affect the accuracy of your experiment if you are using the device in photoconductive mode. If you are using the device in photovoltaic mode there may be temperature effects that make inaccuracies worse compared to photoconductive mode.

Also, what about local ambient lighting conditions changing? In a darkened room I think you should be OK but if there is any "spillage" of the sun through a window, then significant errors could be caused.