The radiation emitted from a 'black' fin will be largely absorbed by the adjacent 'black' fin. (I mean 'black' in the infra-red sense - a good absorber and emitter).AndrewT said:
Radiation from a shiny fin will be largely reflected by the adjacent fin - this is what I called bouncing. However, some will be absorbed and later re-radiated. If 10% gets absorbed on each reflection then after 10 reflections only 35% is left. This is equivalent to a 65% absorber. So bouncing radiation around between good (but imperfect) reflectors achieves the same result as a good absorber/emitter. This is why a shiny finned heatsink radiates slightly better than the equivalent enclosing rectangular box. The same effect occurs for a black heatsink but there it makes less difference. Ridges on the fins may slightly amplifiy the effect.
My reference to anechoic chambers was because the wedges on the walls use the same effect. They are quite good absorbers anyway, but this is boosted by ensuring that sound has to bounce between their surfaces and gets further absorbed on each bounce.
In physics you can make a perfect black body by making a small hole in an infinite cavity. It doesn't matter how shiny the interior walls are, as any radiation escaping though the hole has had to bounce around an infinite number of times first. Similarly, any incoming radiation will bounce around infinitely so no matter how little gets absorbed on each bounce it all gets absorbed eventually.
No, because a lot of radiation escapes too. It is not confined to the fins.tinitus said:
If the heatsink is inside a box then you have to consider how the heat is eventually going to get outside the box. It can only be some combination of radiation, conduction and convection. Some black paint on the inside of the box will absorb internal radiation and transfer heat to the box. It doesn't all have to be black, but it should not all be shiny as otherwise the only place the radiation can go is back to the heatsink or into other components.
Except that the TiO paint was essentially identical to the black paint in the values given on the linked list of emmissivity...
If present, it helps if fins are black. I'm not sure what you mean by inner mounting surface. Are you talking about mounting the heatsink on the case, or mounting the device on the heatsink? In either case the dominant mechanism is conduction so paint should be avoided.tinitus said:
I assume this was a joke?tinitus said:
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I'm only interested in big heatsinks
and inner mounting surface is where the devices usually are mounted
don't know about any other way
only where the devices are mounted
should be milled, or left unpainted
but on a big heatsink maybe 90% of inner mounting surface is just empty
and from what you say, a very big area for efficient heat radiation, if it's painted
sounds to me like its where the black matter most, and easy to do with paint
and since it's been proven that paint is just as efficient ...
This will increase the temperature of nearby components.
A layer or two of aluminium foil (tin foil in the UK) placed between the heatsink backplate and the affected components will help keep the components cooler.
capacitors, particularly, are badly affected by radiated heat from the heatsink.
Self describes the problem of radiated heat affecting small signal transistors and suggests inserting a shield to attenuate temperature changes in the devices that can push the operating points away from what is intended.
Yes it does matter !
Self describes the problem of radiated heat affecting small signal transistors and suggests inserting a shield to attenuate temperature changes in the devices that can push the operating points away from what is intended.
Yes it does matter !
An attempt at poetry?tinitus said:
Yes, it does matter. If a heatsink is radiating heat then anything nearby which is a good absorber will collect some of that heat - especially other items which have been designed to be good radiators too! That is why it is helpful if the inside of the case has a black patch somewhere so that can absorb much of the heat - this assumes that the case will be cooler than the heatsinks etc. If a piece of electronics uses serious amounts of power then it will need serious thermal design.
agree with this
I use white plastic foil though. I also like it because it is an isolator.
Obviously you cannot see infrared light = heat.
referring to color-effected radiation, ofcourse
(ofcourse there is heat radiation at all times)
someone stated earlier that there is no color-effected radiation inside the box because of no light(dark)
but may not be true at all, and light have no influence
about white ...
I have a white bass guitar, and it clearly feels much colder than my black basses ... and it's just wood
I will try and 'test' it when it gets dark
(ofcourse there is heat radiation at all times)
someone stated earlier that there is no color-effected radiation inside the box because of no light(dark)
but may not be true at all, and light have no influence
about white ...
I have a white bass guitar, and it clearly feels much colder than my black basses ... and it's just wood
I will try and 'test' it when it gets dark
Now I am more confused!
Visible light is only a small part of the spectrum. For bodies which are not too hot the issue is how well they absorb/emit infra-red. This may not always be well correlated with how well they absorb/emit visible light. Visible light can have a heating effect when absorbed, but a heatsink is unlikely to emit much visible light!!
Visible light is only a small part of the spectrum. For bodies which are not too hot the issue is how well they absorb/emit infra-red. This may not always be well correlated with how well they absorb/emit visible light. Visible light can have a heating effect when absorbed, but a heatsink is unlikely to emit much visible light!!
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