The unit takes a 0-10v signal to dim it, they have a divider where the thermistor is the lower leg a fixed resistor for the upper leg, and the zener clamps the voltage to 10v.
In they datasheet they suggested a 100k@25c thermistor
http://search.digikey.com/us/en/products/NTCLE203E3104FB0/BC2300-ND/769410 here is one that I'll use as the example.
In the datasheet (
http://www.vishay.com/docs/29048/ntcle203.pdf ) on page 7 it lists values if the thermistor at various temperatures, look on the column for the 100k thermistor.
Now use the resistor divider formula (Vout=Vin*(R2/ (R1+R2) ) to pick an upper resistor(R1) that will give you a voltage of 8V from 40v in at 75C or so. Looking at the chart in the recom datasheet, 8v will give you ~93% (assuming their RACD60-1100 trace is actually the 1050, since there is no 1100 in the datasheet...) As the thermistor gets warmer, it will get lower in resistance and the value will drop more.
Once you pick a top resistor value, you can pick your component power rating. The top resistor will have 40V across it worst case (with that 100k resistor dropping to 5k @100C, its a very real possiblity) So P=(V/R)*V will give you your power dissipated in the resistor, make sure you get one that can handle a decent bit more than that number. I.e. if it dissipates .24W a 1/4W resistor will handle it, but be burning hot the whole time. A 1/2W may be a better choice.
To pick the zener, use ohms law (V=IR) and plug it 30V and your resistance of your upper resistor to figure out how much current from the upper resistor the zener will have flowing though it. 10V * Current = wattage, again, dont push it too close to the thermal rating, as it will be operating this way whenever the driver isnt in thermal protection.
Again, just as a quick recap, youre making a voltage divider, with the top resistor being the thermistor, and youre trying to divide the 40V that powers the LED down to 8V or so at 75C, so as it gets hotter it drops the current even more. The zener wont affect the circuit untill the voltage until its cooled off and the output would normally be higher than 10V.