What Bernard says - it needs a significant heatsink, and clearly you have not read the datasheet which is available on that product page. The datasheet specifically mentions thermal design !
If you look at table 2 (Electrical and Thermal Characteristics), you'll see that the nominal current rating is 700mA - so neither of the drivers you've tried is "correct". 1050mA is the maximum continuous current (table 3, Absolute Maximum Ratings) - you should NOT be designing to drive at that level as it gives you zero headroom in your design.
Also from table 3 you'll see that the MAXIMUM junction temperature is 135˚C, and you should be designing for MUCH less than that. You need (if stuff I learned <cough> decades ago at uni is still correct in my memory) to be adding up all the thermal resistances (that from the junction to rear heatsink surface of the board is given in the datasheet) to work out the temperature difference from junction to ambient, add in ambient, and check that the result is within limits.
So returning again to table 2, at 700mA and Tc=85˚C (which is max operating point from table 3) the max forward voltage of the 5 LED unit is 55V. So that's a max of 0.7 * 55 = 38.5W. Table 2 gives thermal resistance from junction to heatsink surface as 1.2 K/W (degree/watt) - so at 38.5W that gives 46.2˚C between heatsink face and junction. With max juntion temp of 135˚C, that means you must keep the rear face of the board below 88.8˚C.
Having arrived at this point, you can now set your max ambient temperature, and choose a heatsink that will do that - with some headroom. And don't forget when speccing the heatsink that you also need to include the resistance between the board and the heatsink - a good thermal pad/compound works wonders, and check whether the board is electrically isolated (some aren't, in which case you may need electrical isolation which typically adds to the thermal resistance.
The heatsink Bernard shows wouldn't work as it's only about half the length needed - but it's the sort of style you'll need. The specs for that are 0.69K/W - but watch out for the effect of orientation on performance (check the spec sheet). But if we ignore thermal resistance between board and heatsink, we can add the 0.69 to 1.2 to get a combined 1.89K/W, and at 38.5W that gives a max ambient of 135 - (1.89 * 38.5) = 62˚C. Sounds high, but if inside a closed box, you need to consider that the inside of the box can get quite warm.
If you decide to run the LEDs at 1A (or more), which according to the specs for the chips themselves - datasheet at https://www.lumileds.com/uploads/354/DS103-pdf - you can, then you need to use the higher power dissipation in calculations, and also derate the junction temperature. See figure 1 on page 5 of the chip datasheet.
As you can see, this is not a trivial design to do properly. It needs at least a basic understanding of thermal management and how to calculate operating conditions. Hopefully the above will have given you enough clues to find the information to do it successfully.
If you look at table 2 (Electrical and Thermal Characteristics), you'll see that the nominal current rating is 700mA - so neither of the drivers you've tried is "correct". 1050mA is the maximum continuous current (table 3, Absolute Maximum Ratings) - you should NOT be designing to drive at that level as it gives you zero headroom in your design.
Also from table 3 you'll see that the MAXIMUM junction temperature is 135˚C, and you should be designing for MUCH less than that. You need (if stuff I learned <cough> decades ago at uni is still correct in my memory) to be adding up all the thermal resistances (that from the junction to rear heatsink surface of the board is given in the datasheet) to work out the temperature difference from junction to ambient, add in ambient, and check that the result is within limits.
So returning again to table 2, at 700mA and Tc=85˚C (which is max operating point from table 3) the max forward voltage of the 5 LED unit is 55V. So that's a max of 0.7 * 55 = 38.5W. Table 2 gives thermal resistance from junction to heatsink surface as 1.2 K/W (degree/watt) - so at 38.5W that gives 46.2˚C between heatsink face and junction. With max juntion temp of 135˚C, that means you must keep the rear face of the board below 88.8˚C.
Having arrived at this point, you can now set your max ambient temperature, and choose a heatsink that will do that - with some headroom. And don't forget when speccing the heatsink that you also need to include the resistance between the board and the heatsink - a good thermal pad/compound works wonders, and check whether the board is electrically isolated (some aren't, in which case you may need electrical isolation which typically adds to the thermal resistance.
The heatsink Bernard shows wouldn't work as it's only about half the length needed - but it's the sort of style you'll need. The specs for that are 0.69K/W - but watch out for the effect of orientation on performance (check the spec sheet). But if we ignore thermal resistance between board and heatsink, we can add the 0.69 to 1.2 to get a combined 1.89K/W, and at 38.5W that gives a max ambient of 135 - (1.89 * 38.5) = 62˚C. Sounds high, but if inside a closed box, you need to consider that the inside of the box can get quite warm.
If you decide to run the LEDs at 1A (or more), which according to the specs for the chips themselves - datasheet at https://www.lumileds.com/uploads/354/DS103-pdf - you can, then you need to use the higher power dissipation in calculations, and also derate the junction temperature. See figure 1 on page 5 of the chip datasheet.
As you can see, this is not a trivial design to do properly. It needs at least a basic understanding of thermal management and how to calculate operating conditions. Hopefully the above will have given you enough clues to find the information to do it successfully.
