Unless you're just up for a masochistic challenge, don't go too crazy on this. I've got something in the works that should be at least somewhat reasonably priced.. To be announced in the next 7-10 days..
I’m new to forums. If this question is not appropriate or posted in the wrong lace, please let me know. I was given some LED driver boards by the principal scientist on a research project I’m working and was told that their use was absolutely essential for a prototype used in our experiment. These are small circular board purchased from DealExtreme by my predecessor. Each board has three AMC7135 drivers and other associated parts.
My question is this…I have to drive a blacklight LED (around 365nm). I have the datasheet for the chip, buy I don’t know how to choose the best match in an LED. What are the critical datapoints I should be matching up between the chip data and the LED data?
Any help that any of you can render will help make life bearable again.
The AMC7135 is a linear current regulator. It's capable of dropping excess voltage from its input to match the Vf of an LED at 350mA (Per AMC7135).
Based upon this, your input voltage must be higher than the LED voltage in order for the AMC7135 to regulate properly. Being a linear regulator, however, it drops the excess voltage as heat, so you don't want an input voltage that is too much higher than the LED Vf.
Pretty much every single specification in the AMC7135 datasheet is relevant. You have to pay attention to temperature rise based on the number of watts you're burning off as heat, absolute maximum input voltage (6V is memory serves), and you'll need to decide how many AMC7135 chips to use to power your LED to the drive level you're trying to attain.
If you have specific questions, I or someone else will be happy to answer them. Asking what the crucial data points are, however, is an impossible question since we don't know your requirements, your application or what is critical to you. What one person considers crucial, you might consider meaningless. The reverse could also be true.