How to figure out how many LEDs a driver can run
 

I've got some ldd-1000h drivers and I'm just wondering how to figure out how to calculate how many LEDs I can run (3w or 10w) or a combination of the two

I don't really understand all of this but it might help

http://www.meanwelldirect.co.uk/publ...27/r1627_3.pdf

Add up the voltage of the led string to be a bit lower than the rated output (about 80% utilization of total output voltage).

The supplied voltage to the driver must be at least 2-2.5V higher than the total maximum voltage drop of your string of LEDs that is being run by that driver. Look up the spec sheets for your LEDs and they'll typically have a voltage versus current plot. Figure out the voltage drop at 1000 mA for each LED and then add them all up. Since you can supply up to 52 VDC to the LDD-1000H, you can get quite a few on there (basically a LDD-1000H can handle up to ~50W of LEDs). Make sure your LEDs can handle 1000 mA - some pop after 700 mA.

The closer you get your supply voltage to the voltage drop of the string, the more efficient the driver will run. If your string voltage drop is 24V, run your power supply at 26V instead of 30V (the drivers will run hotter because they have to sink more excess voltage). Most power supplies can be adjust +/- a few volts to help with this.

For example take this Cree XT-E LED:
http://www.cree.com/~/media/Files/Cr...g/XLampXTE.pdf

If you scroll down to page 7 you see a graph. The LED itself can handle 1500mA, but with the LDD-1000 you'll only be able to drive it at a maximum 1000mA (the limit of the driver). Which in itself is not bad, as you shorten the lifespan of the LED if you run them maxed out.
So on the graph, @ 1000mA the vF (or forward voltage) is somewhere around 3.2V. In theory, divide the max voltage your driver can handle by the vF of your LED's. So:
52V / 3.2= 16.25 LED's.
That would max it out to the limit so you could drop one and run 15 LED's fairly safely. (if your were looking solely at the driver, but we're not!)
Now to figure out what you need for a power supply you need to add up all the vF's of the LED's per string. Using the above example, if you have 15 LED's running at a vF of 3.2V you need to supply the driver with at least 48V's. And like Scubasteve mentioned this is where the voltage needs to be 2-2.5V higher than what the string needs. So what I do is add 1 phantom LED to my total vF's on a string. So:
15 leds * 3.2vF = 48V
48V + 3.2V (the phantom LED) = 51.2V's That's a fair bit of voltage. You'll see many people strongly advise DIY'ers to remain under 48V's for safety sake. And besides the price of power supplies start to really creep up as the voltage increases. The choice is ultimately yours though.
But let's say for this example we didn't want to need a 51.2V power supply. Then we may need to remove a few LED's off this string. Or as this example progresses, we may decide to change the configuration.
Another consideration we need to make besides the voltage requirements is the Amperage. This is somewhat easier. While in figuring out the voltage we added the required vF's of each LED in the string. For the Amp's we add the mA of how many strings we run.
We know the driver runs at 1000mA. So for each string we need to provide 1 amp. So again taking the example we started out with, we realized it might not be the wisest idea to run 15 LED's on a single LDD-1000 at 51.2V's. This is where planning comes in a bit and we need to ask how many LED's do we WANT to run. For this example let's say we originally wanted to run 16 (because it's easy for our example). We decided at the outset that because running 16 LED's brought us the the max of the driver capabilities we dropped one and ran 15. But let's say we try running 2 parallel strings of 8 LED's each. Now this needs some clarification: We are talking about 2 drivers running parallel, not 1 driver running 2 parallel strings of LED's. This is much safer for the LED's and doesn't require fusing the strings and a whole bunch more math.
So using our previous method's of 2 strings running 8 LED's at 3.2vF:
8 LED's * 3.2vF = 25.6V's.
25.6V + 3.2V (again, the phantom LED) = 28.8V
We would need to supply each of the 2 drivers with 28.8 V's. Very easily done and really safe voltage to work with.
But now we have 2 strings. Each string is going to require 1A so 2 strings is going to require 2A. But we don't want to make the power supply run maxed out either so add 20% to the amp requirement to give you a bit of head room. So:
2A + 20% = 2.4A.
So for a 2 strings of 8 - 3W, 3.2vF LED's, running off of 2 LDD-1000 drivers you'd need a power supply capable of supplying 28.8V @ 2.4A.

Obviously that would be a very lightly run system and you couple afford to add a few LED's per string, but that gives you the idea of my methodology for arriving at some numbers.
PLEASE: someone correct me if I'm wrong :)

Explained it better than I did!

Very helpful thank you


Can you mix LEDs that have different forward voltage if I compensate for it in my math?

Say running a string with x number of LEDs with a forward voltage of 2.6v max and y number of LEDs that are 3.2v max. Forward current is the same for both tho (750ma 3w LEDs)

(Looking to run them on ldd-700 drivers)

Or will I have to run them on separate drivers

Yes, you can mix. Since the LEDs are in series, the current through them all is the same and the forward voltages just add up. As long as you maximum voltage drop is a couple of volts less than the power supply, you're golden.