COB efficiency vs heat output

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Shugglet

Well-Known Member
So Ive been doing some little experiments with my new COBs before my driver for final light gets here.

In doing so I came up with some rather interesting data points that left me questioning some of the info around here.

I am using water cooling to cool 2 Vero 29 Bs at 700ma. This comes out to ~68watts or 232btu.

I always here people talking about using heatsinks not rated for the actual draw but the "efficiency" or par watts.

Yet, when I ran the set up listed above for 6 minutes it raised a gallon of water by 3*. ( 8 pints to a gallon means 24btu for every 6 min or 240 btu/hr.)

This falls right in line with actual wattage, not par watts. So what gives? Am I doing the math wrong or should all watts be considered when figuring for heatsinks?
 
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churchhaze

Well-Known Member
Shugglet said:
So Ive been doing some little experiments with my new COBs before my driver for final light gets here.

In doing so I came up with some rather interesting data points that left me questioning some of the info around here.

I am using water cooling to cool 2 Vero 29 Bs at 700ma. This comes out to ~68watts or 232btu.

I always here people talking about using heatsinks not rated for the actual draw but the "efficiency" or par watts.

Yet, when I ran the set up listed above for 6 minutes it raised a gallon of water by 3*. ( 8 pints to a gallon means 24btu for every 6 min or 240 btu/hr.)

This falls right in line with actual wattage, not par watts. So what gives? Am I doing the math wrong or should all watts be considered when figuring for heatsinks?
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Calculating how much heatsink you need is nontrivial despite what many members here believe. You can get a rough idea how well it will conduct heat by its surface area, but developing a formula that accurately models any heat sink is masters thesis level material. (sure you can use a simulator) My point is that the calculations based on surface area are going to be somewhat off. There's a little bit of guess work involved. This is why I think copying a working design is a good idea.
 
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Shugglet

Well-Known Member
This isn't so much about effectiveNess of heatsinks but the general idea that COBs put out less heat than the wattage draw would imply.

Anyone know what the 90cri 3000k vero 29B efficiency is ? Mainly curious from 500ma to 1a. Thanks
 
MeGaKiLlErMaN

MeGaKiLlErMaN

Well-Known Member
Someone has heat coming off of their cobs??? Lol its not much at all. If you push more than 50Ws per led I suggest active cooling... its more cost efficient.
 
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churchhaze

Well-Known Member
Shugglet said:
This isn't so much about effectiveNess of heatsinks but the general idea that COBs put out less heat than the wattage draw would imply.

Anyone know what the 90cri 3000k vero 29B efficiency is ? Mainly curious from 500ma to 1a. Thanks
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They don't put out less heat than the wattage draw. Some of power that's dissipated is emitted toward the plant as radiant heat (light) while the rest is conducted away through the heat sink. The rate of total heat radiated and conducted will be equal to the power dissipated. You were asking about how big of a heat sink you need, not about the heat the lamp puts out.
 
alesh

alesh

Well-Known Member
Shugglet said:
So Ive been doing some little experiments with my new COBs before my driver for final light gets here.

In doing so I came up with some rather interesting data points that left me questioning some of the info around here.

I am using water cooling to cool 2 Vero 29 Bs at 700ma. This comes out to ~68watts or 232btu.

I always here people talking about using heatsinks not rated for the actual draw but the "efficiency" or par watts.

Yet, when I ran the set up listed above for 6 minutes it raised a gallon of water by 3*. ( 8 pints to a gallon means 24btu for every 6 min or 240 btu/hr.)

This falls right in line with actual wattage, not par watts. So what gives? Am I doing the math wrong or should all watts be considered when figuring for heatsinks?
Click to expand...
Guess there's an error on the input side.
 
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Shugglet

Well-Known Member
wietefras said:
Interesting test. Can you dim the drivers and if so does the water heat up more slowly then?
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Yeah using HLG 185H 1400. Tested them at 1400ma and 700ma. Each test resulted in the water temp being raised in line with what the actual wattage would suggest.

alesh said:
Guess there's an error on the input side.
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I would assume as much. And I did realize I forgot to add the ~8w pump. Which would equate to roughly a ~10-12% gain... any idea on rough efficiency figures I could expect from the Vero 29 B model?
 
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Shugglet

Well-Known Member
alesh said:
Close to 50% at around 1.4A.
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See, this is what I was expecting, but assuming 50% efficient means only 50% of the watts get dumped as heat into the heatsink, something isnt right. Even with error margins I cant see my numbers being so far off.

Granted I dont think Ive ever seen anyone actually test heat vs efficiency, so who knows.
 
PhotonFUD

PhotonFUD

Well-Known Member
Yeah, your maths are off. For example, you are forgetting to calculate the time involved to raise the temperature.

Also, once an equilibrium is reached with the environment, the heat transfer from source is consumed maintaining the equilibrium. How many environmental factors impact that?

50% efficiency should mean 50% for light and 50% for heat waste. Or, 75% efficiency means 75% light and 25% heat. In my opinion, improving efficiency produces exponential gains.
 
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Shugglet

Well-Known Member
PhotonFUD said:
Yeah, your maths are off. For example, you are forgetting to calculate the time involved to raise the temperature.

Also, once an equilibrium is reached with the environment, the heat transfer from source is consumed maintaining the equilibrium. How many environmental factors impact that?

50% efficiency should mean 50% for light and 50% for heat waste. Or, 75% efficiency means 75% light and 25% heat. In my opinion, improving efficiency produces exponential gains.
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I certainly did consider the time it took to raise the temp... and equilibrium wasnt in question because the starting temp on the water was 47*f.
 
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Shugglet

Well-Known Member
Well, looks like I definitely screwed up somewhere because I got my driver finally ( HLG 320 1050) and ran some tests with all 6 cobs hooked up and I came up with an efficiency of about 56% at 525ma. Regardless of errors, this is much more in line with what should be expected so Im happy my initial sentiments here were unfounded.
 
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Shugglet

Well-Known Member
So I hooked up my multimeter tonight and got some interesting results...

The driver I got is the Meanwell HLG 320H C 1050A .

Amps ranged from 410ma to 1210ma, with a voltage range of 288 to 301. Which gives me a range of roughly 118watts to 364watts.

I didnt run the tests very long at all to get the readings so not exactly sure what theyll settle in at.
 
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