Growdad54
Member
Mass is not a factor.
Thickness has a negative effect on thermal management. If you look at the equation for heat transfer by conduction one coefficient is distance. Distance being the distance the heat flux must travel which in the formula is expressed as L (length) See blue circles below.
View attachment 4045999
And here is the formula for the above. Notice that L is in the denominator meaning as L increases the conduction qx decreases. Where qx is the heat transfer from the LED, through base heatsink to the air.
View attachment 4046001
Calculating the heat transfer with your heatsink is complicated. I have attached a PDF with the formulas needed to calculate the heat transfer. If you have the heatsink and CoBs in hand, just try it and monitor the temperature.
In the CXA 3070 datasheet there is a graph on page 2 with the max case temperatures based on current. The max case temperature would be 90°C for operating at 100W.
Cree also states the max temperature of the Light Emitting Surface being 135°C.
http://www.cree.com/led-components/media/documents/ds-CXA3070.pdf If you have an infrared thermometer gun, you can monitor the LES temperature.
There is more information (a must read) in the CX Family LED Design Guide
http://www.cree.com/led-components/media/documents/CXA_design_guide.pdf
You can use the heatsink calculator provided by USA Heatsink:
https://www.heatsinkcalculator.com/calculator_heatsinkusa.html
You can also create a free use account on https://www.heatsinkcalculator.com
View attachment 4046358
I found from this calculator the heatsink you specified will dissipate about 100 Watts with a heatsource temperature of 100°C.
The best practice is to have an effective passive heatsink without the exhaust air flow. Then if the exhaust fan fails the LEDs will not be damaged. So shoot for a case temperature of 90°C with out the fan then use the fan as well.
Given all that it is highly unlikely you can operate four of these CoBs at 100W. You would need substantial thermal management to dissipate 400 watts. Cree has a reference design where they use passive heatsinks for 550 Watts. I estimated the cost of those heatsinks to be over $200.
http://www.cree.com/led-components/media/documents/HorticultureReferenceDesign.pdf
Mass is not a factor.
Thickness has a negative effect on thermal management. If you look at the equation for heat transfer by conduction one coefficient is distance. Distance being the distance the heat flux must travel which in the formula is expressed as L (length) See blue circles below.
View attachment 4045999
And here is the formula for the above. Notice that L is in the denominator meaning as L increases the conduction qx decreases. Where qx is the heat transfer from the LED, through base heatsink to the air.
View attachment 4046001
Calculating the heat transfer with your heatsink is complicated. I have attached a PDF with the formulas needed to calculate the heat transfer. If you have the heatsink and CoBs in hand, just try it and monitor the temperature.
In the CXA 3070 datasheet there is a graph on page 2 with the max case temperatures based on current. The max case temperature would be 90°C for operating at 100W.
Cree also states the max temperature of the Light Emitting Surface being 135°C.
http://www.cree.com/led-components/media/documents/ds-CXA3070.pdf If you have an infrared thermometer gun, you can monitor the LES temperature.
There is more information (a must read) in the CX F
a lot of good info,thanks guys. It's a little over my head, but still appreciated lol.I was kidding, saying it was going to get very hot, much hotter than 90°C
BTW 90°C was the operating limit for case temperature at 100W (2.76 Amps).
I did see previous post. See attached conductionConvectionHeatsink.pdf and
Thermal Management of Light Sources Based on SMD LEDs.pdf
Mass is relevant in fluid convection and conduction. Heatsinks are about conduction heat flow to convection and radiation surfaces. No matter how much mass is involved in the conduction heat flow it does no good if there the surface convection and radiation are ineffective.
The very basic thermal conductivity formula. Where q = thermal flux, k=thermal resistance, A = area, ∆T=temperature difference heatsource to heatsink, L=length. No mass, just thermal resistance.
View attachment 4046503
And simplifies to just temperature delta and thermal resistance.
View attachment 4046516
And the example for 250°C delta T, Rth =1.7 W/m K, L=0.15 m
Notice there is NO mass in the equation.
View attachment 4046518
Mass will slow things down from getting too hot. If you put a 400 watts of CoB on a 1 kg horizontal plate of copper, eventually you will end up with a very hot plate of copper.
Recently I was running some thermal experiments using a bar of copper 1.5" x 24" x 0.125" ( #2 in this image) with a 20 Watt CoB and it over heated. Problem was while there was plenty of mass there was very little convection (i.e. convection heat transfer coefficient) and very little radiation (poor emissivity and little free convection buoyancy-driven flow). Furthermore heat transfer rate depend strongly on whether the surface is hot or cold and on whether it is facing upward or downward. Free convection (passive) is very ineffective when the hot surface facing downward and the cold surface facing upward.
The entire bar got very hot because the convection and radiation were very ineffective. Great conduction does no good if there is no place to dump the thermal energy.