Laser BEAM heat and Heatsink Coloring..

[Ilotrophic]

Ok so basically I have 2 questions I want to knock out in 1 thread.

1.) It's obvious that a laser DOT produces heat if powerful enough. So for experimental sake we'll use variable 1 as a >1000mW 445 laser.
So if you were to capture the laser DOT with a thermal imaging camera it would obviously produce a signature on the camera.
What I am basically asking is does the beam produce heat in the air? Meaning will a thermal imaging camera pick up a heating signature from the beam and not just the dot?


2.) I notice a vast majority of commercial lasers out for sale have there housing units and heat sinks are painted black. My question now is are they painted black just for the looks or does it act as a catalyst to draw heat out of the heat sink to the surface?


Any info is appreciated.
-Cheers

 

[Blade]

very interesting! i never really thought about this... but

1) i do not think the beam would produce heat in the air, meaning the camera wont pick up on it. simply because the heat is produced by the object absorbing mass amounts of concentraited radiation. almost the same way a black balloon pops when a powerful enough green laser is pointed at it, and a green balloon wont. this is because the green balloon reflects green light, so the laser radiation isnt absorbed... in order for the beam to produce heat in mid air, it would have to be powerful enough to excite the air molecules enough for them to give off energy. and that would be insanely powerful!:eg: EDIT: the beam is only visible because the light is bright enough to be reflected off of the molecules in the air / dust...

2) well, the heatsink works by convection. so i dont think the color has anything to do with effectiveness. as stated above, the object (in this case, the heatsink) would have to be absorbing some sort of radiation for the color to matter. but it is only absorbing heat directly from the surface of the diode through heat transfer. because of this, it it is the material of the heatsink that matters most.

somebody could probably explain this better than i can. im not the best at explaining things. also, correct me if im wrong

 

[GoolGaul]

my votes:

1) Very interesting question... I think there will be a slight measurable signature caused be a 1000mw laser.

2) the color has nothing to do with it's heat conductivity.

 

[Benm]

On point 1 i'd agree: perhaps a very tiny amount of laser power is absorbed by air, but it would not be enough to be visible on FLIR and such.

There is, however, a reason that heatsinks are often black: it improves radiative cooling, provided that black in the visible spectrum is also black in the infrared. Convection is often a much larger effect than radiation, but having both at an optimum still helps.

Interestingly, radiator paint (for central heating) is often white in the visible spectrum, but black in the infrared to make the radiators more efficient, and also a source of radiative heat you can feel from a distance.

 

[Cyparagon]

radiator paint (for central heating) is often white in the visible spectrum, but black in the infrared

All paint is "black" in mid-IR, though. Hell, pretty much everything but metal is "black".

A laser dot to a thermal camera is like a blu-ray on a glow board:

 

[rocket689]

^^^^ That's pretty cool, Cyparagon.
I wish I had a thermal imaging camera to play with.

That and a high speed camera to watch stuff blow up s-l-o-w-l-y

 

[Ilotrophic]

All paint is "black" in mid-IR, though. Hell, pretty much everything but metal is "black".

A laser dot to a thermal camera is like a blu-ray on a glow board:

Yes yes, I got it now.
Thank you

And thank everyone for there answers
-Cheers.

 

[BShanahan14rulz]

1: if the beam power density is high enough, I have a feeling you might see bits of dust being burnt along the beam, that ought to show on an IR camera, no? But I don't think any of our lasers would be visible.

2: The black color often comes from the process of anodizing. Anodizing creates a porous layer of aluminum oxide (?) on the surface of the aluminum, and with a heatsink, surface area is one of the key factors in heat removal from the sink. It also has the properties of an electric insulator, is harder than the aluminum under it, and protects against corrosion.

 

[Bluefan]

A black body radiator is black for a reason. Black doesn't only absorb all light, it radiates it also. That's why black heatsinks work better, not because of increased conduction but because of the heat they radiate.
Black body - Wikipedia, the free encyclopedia

Surface area is important, but anodizing doesn't help that. The boundary layer of air that is at the surface of the heatsink stands still, and this layer is bigger than the microscopic size of a porous layer. That's why usually passive heatsinks have bigger fins than the many small ones on active heatsinks, because active heatsink (with a fan) have a higher flow and a smaller boundary layer of air.

 

[BShanahan14rulz]

So how come many heatsinks are anodized black instead of painted black? Wouldn't it be cheaper to paint it black, especially if black paint is "blacker" than the dyes used in anodizing? Maybe is the special paint really expensive? I have noticed that most passives are anodized while they don't really bother with actives.

BTW, I recently came across two hefty copper heatsinks, they were so beautiful! Skived from a single block of copper, no soldering fin stacks here! These must have been high-end or something, but it looks like their approach was rough fins (possibly unwanted effect from skiving, but I'm thinking it was either intentional or beneficial side effect of skiving metal) and reducing the number of thermal interfaces (by 1, but still...). But I digress, as usual.

 

[Cyparagon]

That's why black heatsinks work better

Black in the visible spectrum, or black in the mid IR spectrum?
I put three heatsinks in a toaster oven set to warm for 10 minutes, took them out and quickly snapped a picture.

It appears to me that anodizing is what makes the heat sink "black" and not necessarily the dye they use.

 

[qumefox]

It appears to me that anodizing is what makes the heat sink "black" and not necessarily the dye they use.

Actually anodizing is just the process. The colors are purely due to dyes. You can anodize without using any dies and have the same protection as with dyes. Not using a dye just results in a light grey finish, but you get the exact same hardening and corrosion protection. The only difference between black, red, green, purple, etc.. is purely the color of the dye used.

And with your experiment. You really need identical heatsinks for such tests.. with the only difference being the coatings.. to get any meaningful results.. The whole 'eliminating variables' thing and all.

 

[Cyparagon]

Actually anodizing is just the process. The colors are purely due to dyes.

I am aware. That's why I put black in quotes. The middle heat sink is evidently just as black in the mid-IR range as the heat sink on the left.

And with your experiment. You really need identical heatsinks for such tests.. with the only difference being the coatings.

While that would be ideal, the difference in cooling for the ~5 seconds it took for me to snap a photo is negligible.

... to get any meaningful results..

You really think these results are meaningless? :thinking: That's a little harsh

 

[qumefox]

It actually is as far as the question posed.

When you took the pic immediately out of the oven, you only demonstrated which one *absorbed* heat faster. We were discussing the opposite. The test here should be to heat them all to the same temperature, then let them cool and see which one cools faster.. Which will tell which is radiating more heat. Though mass and surface area will play more part here than coatings. Which is why I said you really need identical heatsinks.

Come to think of it.. Mass and surface area would play a role in heating speed too, since the lower mass heatsink will heat faster.

So yes. I guess, even if it is harsh.. I have to stand by my previous statement that not much in the way of meaningful data was attained.

 

[Cyparagon]

The test here should be to heat them all to the same temperature

And putting them all in the same constant-temperature environment for 10 minutes doesn't accomplish this? :shhh:

then let them cool and see which one cools faster.. Which will tell which is radiating more heat.

That's one way to do it. Another way is to just measure the temperature with a thermal camera (or even IR-thermometer).
The surfaces that have lower emissivity will appear cooler, even though they are not.

 

[BShanahan14rulz]

The ones that radiate more efficiently ought to be the better radiative (passive) HS, and since they are all a similar mass, we can take somewhat meaningful results from this. I'm not saying that there is "data" but the conclusion can still be made

edit: if you do the anodization at a stable, cold temperature using the right solution and for a much longer time, you can achieve oxide layers that are much thicker than your regular anodize process. I believe this type of anodize is harder to dye and is often either dyed black or left as-is before sealing. I thought, though, that it was neat that they could make a thicker layer by just cooling it down.

Edit 2: as long as I'm just adding barely relevant bits, might as well just add them to my other post that has a random semi-relevant bit.
Concerning black body radiation, these heatsinks should all be radiating in the IR, otherwise they'd be changing colors. So really, its not important that the heatsink be colored black. Instead, it should be able to absorb IR. From what I understand, it's not hard to make things absorb IR?