If the space between two concentric spheres, r1 > r2 is filled with a conductor with the resistivity rho, the resistance between the inner and outer surfaces is
R = rho(1/r2 - 1/r1)/4*pi
so as r2 approaches zero, R becomes infinite. That's electrical, but the thermal situation is the same.
(cribbed from
John
We ran it for days afair.
-- Tony Williams.
This neglects the thermal resistance of the surrounding air, mostly film resistance on the surface of your sphere. A reasonable first approximation of the air boundary thermal resistance for the laminar flow case typical for potted assemblies would be h = 1.32(deltaT/d)^.25 watt/(m^2 C) deltaT = surface temp - ambient air temp deg. C d = sphere diameter in meters
(This is actually an approximation for convection of a cylinder in free air from J.P. Holman, but it should be close enough for a sphere considering the very approximate nature of these simplified equations
- there is not much variation in h with changes in shape with laminar flow eg 1.32 changes to 1.42 for vertical cylinder.)
If you figure a spherical source with fixed heat dissipation and covered with potting compound of uniform thickness and include the thermal resistance of the potting compound and the air film, you can calculate the potting compound thickness which will result in minimum temperature of the source, and also the source size where potting compound will no longer be able to reduce the source temp.
Indeed, with zero radius it can neither conduct or radiate, it has no mass no energy and no temperature. But these bulk material calculations give way to quantum mechanics long before zero radius!
This still works out to the crude practical approximation of 1 degree C rise for every milliwatt dissipated from a cm^2 of surface area.
The problem is not establishing the surface temperature, but determining the the heat source temperature.
What is needed is a crude practical approximation of average 'l/w' for the path between a sphere's center and it's surface. This will be a fixed number related to the radius. This could then be combined with the thermal impedance of the potting compound to determine a rough deltaT from the surface to the center of a sphere for a known power transfer.
The same approximation for other simple shapes might also be useful.
It will be more accurate for any situation where the thermal impedance of the potting compound is much higher than the thermal impedance of the air - ie in most practical situations. The average surface temperature of the radiating shape is driven by this difference to be reasonably uniform, if the surface is plain.
RL
The conductive path can be divided up into a bunch of tapered sections, and they get very small close-in to the inner sphere. For a small inner sphere, most of the temperature gradient is very close to the inner sphere, so the "average" cross-sectional area of the thermal path is not meaningful.
I posted the exact equation, um, somewhere nearby. For smallish inner radius r, thermal resistance goes up as 1/r. That means the OP should try to spread the heat source out as much as possible, maybe with PCB foil and vias, if cooling is critical. Or maybe mount the TO-220 upside down?
John
On Tue, 29 Jul 2003 16:53:25 -0700, John Larkin Gave us:
If he operates it inside of its spec, it will never get that hot, or shouldn't as device specs have limits to their allowable characteristics in the industry as well.
We have products where the main transistor are only potted up to the hole, which means that their dies are coupled to the potting.
We have some, that we only dipped a couple of few times, and the potting still conducts the heat away fast enough.
A 2n4923 can only pass one amp. The thing is whether you are passing the amp, or dissipating some losses due to poor circuit efficiency.
The average cross-sectional area is the meaningful and effective dimension. The fact that the area and the impedance varies along the path is the irrelevent complication that this calculation is designed to avoid.
resistance increases proportionally to 1 / (r^2) - the increase is related to change of the surface area of the concentric sphere at the radius being examined. Surface are is 4 . pi . r^2.
If you mean that at similar r values, the squaring effect is minimized, I can agree with you, but then most other effects become irrelevent too, including thermal impedance of the intervening material. Potting isn't thin-all insulation.
If he finds out that the part IS overtemped, then of course some kind of action is required to reduce the delta through the potting material. This is the effect of making the radiator r being 1/2 the external surface r.
The two examples at 1/10th and 1/100th r were intended to show that point-source radiation was a mathematical abstraction, and that real effects were not going to outweigh published thermal impedance from junction to case at the source when this scale was present.
The other observation that I hope was noted, and may have been glossed over as being too obvious, was that the other limit on achievable temperatures was the mechanical limit of the external surface area.
Averages are your friend. Extra decimal places are hogwash. The sexier terms for calculating this effective average do exist; leave them those who want to impress their girlfriends and co-workers. We are attempting a practical understanding here.
RL
You were correct to question the 33% figure. I'm sort of dissapointed that you didn't take the time to do a bit of figuring and suggest a better one.
The problem with these numbers was that they looked at relative conductivity - not the reciprocal impedance.
Looking at impedance, internal source radii below 33% of the external radius will actually dominate the impedance. There is more than a 50% increase for every 10% (of the external radius) that is subtracted from the internal radius.
Looking at the increasing effect of the differing source radii, and using a 10% difference as the base-line, the incremental cross-sectional area effect on the ball source path is:
shell deltaZ accum unpotted potted effect .9 - 1 1 1 1 1 .8 - .9 1.23 2.23 1.10 1.12 x.98 .7 - .8 1.59 3.82 1.56 1.28 x.82 .6 - .7 2.14 5.96 2.04 1.50 x.75 .5 - .6 3.82 9.78 2.78 1.88 x.68 .4 - .5 4.64 14.42 4.0 2.34 x.58 .3 - .4 7.91 22.3 6.25 3.23 x.52 .2 - .3 16.45 38.8 11.1 4.78 x.43 .1 - .2 34.1 72.9 25.0 8.19 x.33
The relative impedance of the epoxy layers increase as they approach the center of the structure. (deltaZ)
The accumulated effect of the epoxy thickness can be summarized.(accum)
An unpotted single-temperature source of constant power experiences a surface temperature increase as it's radius and surface area are reduced. (unpotted)
The thermal conductivity of epoxy will be somewhere between 10x to 50x that of air (0.2 to 1.0 W/mK vs .025W/mK). This forces a lower relative increase in the temperature rise of the lowest layer, dependant on depth. (potted)
The relationship between free air source and potted source is the ratio between the potted and unpotted change. (effect)
These numbers suggest that with a commodity potting material (conductivity 0.2 W/mK), the source surface rise will be less than double the potted assy surface rise, providing that it is distributed evenly to a radius of at least 30% of the potted sphere's outer surface dimension.
I may be wrong. Perhaps you have a different perspective to offer, with some ball-park numbers. Please don't use plumbing analogies, these only go so far before you have to do things like relate delta T to water velocity - if you have to, use current flow in conductors.
RL
Because "average cross-sectional area" is useless here; it tells you nothing about the actual thermal resistance. Especially when the actual equation is right there.
My point was that averaging areas makes no sense, for any kind of flow. Why ballpark when you can have the real thing?
John
I guess it's not just angles that can be obtuse. Piece-wise linear approximation has been used for thousands of years - even your plumbing example depended upon it. It's often the easiest way to explain or to prove a mathematical expression. A rule of thumb is only useful if it's easy to remember and apply in practical situations.
Are you actually trying to say that a calculation makes no sense, that a measurement is better?
I think they both have their places, particularly where economy of time and materials is a concern. A calculation is intended to allow you to avoid the trouble of having to re-invent the wheel, every time you want to use one. A rule of thumb is intended to allow you to apply a formula's result without having to derive and prove it each time it's needed.
From all your many measurements, in permafrost or otherwise, have you developed no general understanding that could have been passed on succinctly to Ziguy? It appears that he 'left the biulding' about 4 days ago.
RL
While we are on this subject, I'm looking for a mould/box/whatever that I can put my small circuit board into, and fill it with epoxy, then remove the mould/box/whatever, and just leave the epoxy cube. What material can I use, so the epoxy won't stick to it?
Dave
Use a potting shell if you possibly can; just stick the board in the shell, fill, and you're done. Robison makes a jillion sizes and shapes of potting shells.
Molds, especially hard molds, are a nasty mess. If you do want to use a mold, use silicone rubber. Plastics supply houses (Tap Plastics, out here) have cool molding silicone. Make a prototype shapr out of anything, cast the silicone around it, and pull it out when it cures. Voila, a mold that *releases*. Doesn't even need draft angles, and you can include fancy things like logos or bosses or textures.
Vacuum degass the epoxy for the ultimate bubble-free fill. Messy, too.
Potting loses its charm fast.
John
A re-usable mold, made of any material, will benefit from a non-stick film of releasing agent. This can be as simple as oil or wax, but silicone is probably used most often nowadays. You would have to check for chemical compatability. A popular hobby mold material is thick-walled latex or silicone rubber, flexible enough to peel off the set material, but dimensional stability may be a problem, particularly with repeated use.
Alternatives are one-shot molds that you either destroy to recover the molded object, or that you leave permanently applied. The latter can give a more cosmetically attractive appearance. A flow-mark and void-free epoxy surface finish is not always easy to achieve, particularly if filled epoxy is necessary for thermal or other reasons.
Bulk cured epoxy is also prone to mechanical instability, particularly when foreign material or relatively large objects are imbedded, and the reagent is either too strong or poorly mixed. A permanent shell can protect it from shock that might cause fracture along flow lines, due to violent handling. Well-formulated epoxy, on the other hand, can be machined to a certain extent.
RL
On Sun, 03 Aug 2003 10:51:50 +0800, Dave Baker Gave us:
Dry mold release agent. Sold in spray cans. For plastics, and even for the food processing industry. Wet mold release will inhibit epoxy curing, or can.
On Sat, 02 Aug 2003 20:11:29 -0700, John Larkin Gave us:
Not if it is required. Doh!
On 3 Aug 2003 01:42:45 -0700, snipped-for-privacy@magma.ca (R.Legg) Gave us:
Very informative. The moral:
Always mix the proper amounts of the constituents, and always mix very thoroughly. Evacuate with vacuum where required or desired for maximum density, consistent grain structured potting matrices.
I'm still here... It's just that a have hard time to follow/understand the discussion... I think I will make tests/mesurements of my setup!
But, can I conclude that the case to ambient resistance will not increase as long as the additional layers that I put are more conductive than the air?
Thanks again guys!
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