I realize this topic is old but I have researched this subject long before seeing this topic, and others could benefit from that, could save them a lot of time...
- First you must understand the context for manufacturer spec sheets, then you must understand that Chinese/generic spec sheets are often not only vague but often just plain wrong, IF you even get a spec sheet instead of someone just posting a specs
picture or copy/pasting specs they saw somewhere else with no idea if it's really the same LED.
Anyway, onto specifics. Your LED as specified is arranged in a 3 parallel strings of 3 dies in series, or just 3 x 3 for short. This means the minimum voltage to get a series of 3 to illuminate at all is a little below 8V, but to get remotely close to rated output it's around 10.6V. So, how bright it is at 5V = not lit at all.
Since you want only 200 lumens, which I think is not bright enough - that you should instead design for the thermal capacity of the light enclosure and available space for your choice of heatsink while enclosed, which is probably around 4W unless you want extended/continuous running, your LED forward voltage (Vf) is lower than 10.6V, probably closer to 10.0V but this is ONLY a ballpark estimation. To get valid numbers you must use a multimeter to measure voltage @ current flow then calculate out how high each can go before you exceed the thermal capacity of your heatsink and enclosure.
You can use a resistor, or series of them to arrive at the target voltage, but the problem is that your dome light will be powered by roughly 12.6V when the vehicle is off, then closer to 14.4V when the engine (and alternator) is running. Since you have such a low 200lm target brightness, you should be okay because even such a large jump in power supply voltage will still be in the context of a chip designed for up to 1050mA or so. HOWEVER, it begs the question of why you are even bothering? 200 lumens is probably not much brighter than the original bulb was. You'd be doing the work for a trivial gain, just a little colder color temperature light and a very very minor efficiency increase considering it has a BIG battery and roughly 100A alternator recharging system. Efficiency just doesn't seem important unless this is a cargo light in a large vehicle that you're using to camp in and that light might be on for hours at a time.
Anyway, you can plug the numbers into an LED resistor calculator. Do a web search for those terms using the values I mentioned above until you measure different values.
12.6V power - 10.0V LED Vf = 2.6V difference
2.6V difference / 0.25A (about what it would take to get 200lm) = 10.4 ohm resistor, 0.65W dissipated so a minimum of 1W resistor rating and better 2W to 3W.
Punching the numbers again for 14.4V alternator powered circuit, 4.2V difference (Vf rises with current increase in an LED) / 10.4 ohm resistor = 400mA, about at the 4W power level I suggested a few paragraphs above. 2W you could keep cool with a metal plate heatsink or a small pentium 1 era CPU heatsink. 4W is closer to Pentium 2 era heatsink, about 1.5" x 1.5" x 1" tall. If you simply don't have room for that size heatsink you can use smaller if you never run the light for more than dozens of seconds, or reduce drive current to what the heatsink you use will allow. Test it outside the vehicle, if it gets too hot to touch it will be hotter still enclosed in the vehicle overhead light housing.
SO, the above would work. However, you can get ready made 10W constant current drivers designed to take the vehicle battery and alternator input and output regulated drive current for about $1 each on Ebay! They are designed to output about 900mA (so they claim, but actually closer to 833mA). The majority of these $1, 10W drivers use a pair of current sensing resistors in parallel, one 0.2ohm and one 0.3ohm resistor.
Since you don't want full brightness, what you could do is get one of those $1 constant current drivers, then remove the 0.2 ohm resistor to lower the output current. The math to determine current is 0.1 / (resistance) = drive current. 0.1/0.3ohm = 333mA drive current. That would get you reasonably close to your 200lm target, a bit brighter but I think after the lens losses you will be happier with it a bit brighter and it will stay nearly constant brightness in both engine running and engine off mode.
Anyway, that is my recommendation. Buy the $1 regulator and I've linked an example below, then either rip off or desolder the 0.2Ohm resistor leaving only the 0.3ohm resistor. That will be almost as easy to wire as the resistors would, and easier in the end because you don't have to worry about thermally insulating the resistors (they could get hot enough to melt plastic if touching it directly). All the LED driver board needs is heatshrink tubing around it and some sellers even include a piece of tubing for that.
http://www.ebay.com/itm/1PCS-10W-In...Driver-Power-For-High-Power-LED-/400954019570
There are many different sellers, just make sure it's a 10W driver, that the input range is similar at 12V, the output range is similar at being lower rather than higher than the input range (so a buck, not boost driver), and the current is claimed to be in the ballpark of 700mA to 1050mA - better to look at pictures to see it's the same and has the parallel 0.2ohm and 0.3ohm resistors as mentioned above. If you were to want it even brighter you could remove the 0.3ohm resistor instead of the 0.2ohm, but I would not run it with both resistors at ~ 833-900mA because that's too hot for a standard plastic vehicle overhead light enclosure once you take away the original design elements of an incan bulb and no active electronics.
. So what voltage would make the chip glow at around 200 lumens? And what would be the simplest way to power it? Resistor? If i'd need a full constant-current LED driver then I'll forget it and probably look at other options like a board with an array of LEDs.
