Hopefully I can attract the attention of an electronics whiz here! Anyways, what I've got is a CXA-2011 led. The led likes a little over 40 volts and I plan to run it at 1A, its maximum current, for approximately 3k+lumens. The driver I have is a Recom RACD60-105. It is an AC-DC CC driver. It has an external dimming circuit. On this website : (http://www.recom-international.com/p...ine/RACD60.pdf) it describes adding a zener diode, resistor, and thermistor for a thermal feedback circuit, i.e. temperature at LED is too high, it will lower the current. My question is: what is the formula to select the proper components. Or can someone tell me what resistor and thermistor I would need?
Thanks in advance- as always the genius' on here always set me straight lol
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.
Ok, here we go. So in actuality the LED should draw 46V at 1.0A. With that in mind and using the thermistor you recommended with the 13.4k value at 75deg I figure the other value of resistor to be 63.6kohm. That sound right?
yep, that works out right. And with a 63.4k resistor the current though it when the thermistor is very low is ~0.73 mA. That means it dissipates 33.6 milliwatts of power (0.73 mA * 46V) so a 1/8W resistor is more than capable.
The zener would have to dissipate 10v, with 36v across the 63.4k resistor (V/R= I, 0.57 mA) so 5.7 miliwatts of power would be dissipated (P= V*I) again, almost any device would be suitable.
Now you can re-do the numbers with any other thermistor, or set point, or Vf with the LEDs if anything changes. I'd figure out what temp the divider puts out 10V at, to see what temp the heatsink can get upto with full power, since at 8V its already throttling back the LEDs some.
Now in what it actually works out to wont be exactly the same, since the Vf of the LED will drop as it warms up, meaning it may only be dividing 44V instead of 46V, but that works to our benefit here. Though you might want to consider a 10K pot wired in series with the top resistor and lowering the top resistor ~5k, so you have some adjustment range, though unless youre planning on let it run pretty hot most the time, it should be "good enough"
Right on! I seriously appreciate the help. What do you do for a living that you are this good with electronics?! Here's what I ordered from Digi-key: BC2300-ND1N5240BFSCT-NDCMF63.4KHFCT-ND. That should work right?