Help with thermal calculations

I was trying to figure out (measure) how much of the heat in a

George H.

I got this some place off the Web in 2010... probably here at SED.

========== I once worked at a company where we wanted to know what the average temperature inside a box would be (for MTBF calculations). An older enginee r pulled a diagram out of a folder - a photocopy of a photocopy of an old book - and said "well... we're dissipating an average of 100 watts in that box. Its surface area is 1.3 square metres. So it'll be... um... about 20 degrees hotter than ambient in there."

This took all of 15 minutes to gather the data and look at the graph.

This was Too Easy and therefore couldn't be dependable, so the managers bou ght a thermal CAD package and detailed an engineer to learn how to use it, model the cards in the 3D space, and give them an answer they could believe in.

This is suspiciously close to the non-credible first result, so the manager s now order a full scale mock up be built (we don't have the cards yet, we had to make dummy ones with resistor loads).

This gives an answer of 20 degrees.

Average surface rise above ambient in free air

delta T = 20000 / 337.5 = 69degC

So if it's running at 40C ambient, expect 110C case temperature.

Increase the surface area or reduce loss to bring the surface temperature down.

RL

transcription error from paper to keyboard...

RL

Can you say more about dividing power by surface area and getting degrees C. You've got some "secret" factor... 1mW of power, into 1 cm^2 gives a 1 C rise? I'm not disputing your answer, just wondering where you got it from.

George H.

In one application, my customer stacks 6 boxes in a 2x3 array such that the middle two get no cooling. Dissipation is about 7 watts per box. The middle boxes run around 60C surface temp, and the internal parts are way past their max temp specs. After great management deliberation, this is the required fix:

We had to order those specific labels from europe. They are not visible on installed units.

I have a name for that company's management structure: autoparalytic.

(...)

Foundit in Oct 2010:

Jan Panteltje posted the graph to his server at:

I stole it, cleaned it up a bit, converted to PDF, and posted it to my web pile:

Y'er right, I'm wrong. I forgot about convection. Thanks for the correction.

Gone sulking...

Yes. In addition to the bulk theta, you may have hot spots where the dissipative parts are, and lateral heat spreading resistance.

This is roughly suggestive:

The thicker the metal or the bigger the component, the less hot spot problem. Bolting a finned heat sink to the panel improves average theta and also helps the spreading thing.

Just a couple of square inches of added aluminum or copper inside would help spread out a hot spot.

Nice, Thanks. The slope is less than 1... ~3/4 which makes sense As it gets hotter there is more convection and that cools it more.

George H.

Yeah something like that. Looking at the graph Jeff linked to up stream it seems like my numbers are at least reasonable.

George H.

I developed this rule of thumb in my own work because I had to.

It is based on studies of plastic and bare/painted aluminum cases with variously distributed controlled loads, with differing relationships between load sources and the container - measured with point-contact thermocouples and IR imaging. It has since been applied to any sealed container regardless of material, in free air, with no appreciable errors resulting from the one simple calculation.

Furthermore, on examination of established mfr's data sheets for common magnetic and capacitive components, I've found corresponding trends tha seem to confirm it.

This thermal impedance is based on the shape of the part. However, if the surface area of the part is factored into the listed values, a fairly stable constant appears for all parts listed.

This varies around 163degC x in2 / W by about 3%.

This rise of surface area temperature for power density in a surface area, can be used on all other capacitors to give a consistent temperature rise for Iripplesquared x Resr calculations for all parts in any series of capacitor, by most manufacturers, when surface area is considered.

For every iteration the relationship between power loss and surface temperature rise is 1.05 degC x cm2 / mW.

The same power rating versus surface area for Ecore transformers can be applied for consistent temperature rise limits in other suppliers catalogs.

These two constants are equivalent, when converted for units, in spite of the wide variation in component shapes and sizes (below delta40degC).

This can be explained by the relatively small variations in this characteristic that apply to simpler solid shapes, that have no wide ratio between the dimensions of Length,Width and Height (between .48 and .63). A good reference is chapter5 of "Cooling Techniques for Electronic Equipment" by D.S.Steinberg, although imperial units are used, in all the local reference libraries.

(Emissivity, of course is fairly constant over a narrow temperature range for simple unpolished surfaces.)

Getting a ballpark figure, even with tolerances exceeding 10%, is usually an eye-opener for designers attacking their first physically constrained dissipator.

Everything inside the box will have rises elevated above the average surface temperature of the external skin, based on their thermal impedance to the skin. So an open air prototype, relocated to within the container, can simply add previous spot temperatures to the skin temperature, if it does not directly contact the skin.

With ideal thermal conductivity, no part will have a temperature lower than the envelope.

RL

I recently did what boils down to two SMPS in series, 110W, 3x5" pcb in a sealed case. It has to work in direct sunlight.

The first task was to reduce heat generated as much as possible, the second to get it off the board to the case, and third from case to ambient.

o A several-pass effort got dissipation down to ~5W. o That's carried from topside to the board's bottom with thermal vias into bottomside pours. o The bottom, electrically-hot pours are bare metal. That surface bolts directly to a raised island on the milled case, hard-anodized for insulation. o The external surface of the casing has milled fins.

In a thread here I posted measurements of an old ROT of 60K rise, for 1W into a 1"^2 heatsink. Empirical results were close, and I quoted some figures backed-out of some heatsink (Aavid?) specs.

R. Legg's ROT works out to 52K/W/in^2.

All of these are in the same ballpark. Vertical surfaces cool MUCH better than horizontal, due to natural convection.

Hope that sets the table somewhat...

I'd echo Mike's advice with a twist--mockup a metal case, and put a 20W light bulb in it. Get out the IR thermometer. And don't leave the box on anything flammable!

Cheers, James Arthur

Wooah! Thanks. I took your number 1 deg C = 1mW/cm^2 and stuck it on the graph posted by Jeff, via Jan up-thread. It's about 155 C = 1W/in^2. Which is just about smack dab in the middle of the graph... 15.5 C at 0.1 W/ in^2. Your number is a lot easier to remember. At high temperatures radiation has to take over and dominate convection...At which point you want to paint it black.. I wonder about what temperature that is? OK your number is 10W / (C*m^2) That seems small compared to radiation. (at 300 K into 0 K with emissivity equal to 1 is about 500W/m^2... I got only a few degree's for 10W.. But I might have made a maht mistake..)

George H.

If it sits in the sun you might not want to paint it

Hi James, I'm not sure what ROT is.. but I got 155 K/W/in^2. From R. Legg's 1K/mW/cm^2 number... I might have made a math mistake.

George H. (

If it sits in the sun you might not want to paint it

Hi James, I'm not sure what ROT is.. but I got 155 K/W/in^2. From R. Legg's 1K/mW/cm^2 number... I might have made a math mistake.

George H. (

For something out in the sun, can I ask what you did on the outside surface?

George H.

George, converting Mr. Legg's figure below, I get 52K/W/in^2. Oh, wait, no, it's 154K/W/in^2.

I have to wonder how the various rules of thumb incorporate geometry, which matters quite a lot.

The 60K/W/in^2 ROT is for a vertical surface in free air. A horizontal surface would heat up a great deal more.

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Cheers, James Arthur

I don't want to paint it either :-). The sun part was significant because the case itself may have external heat input (from the sun), making dissipation & thermal design all the more critical.

Rule Of Thumb.

Yep, I goofed. Operator error. 153.3 K/W/in^2, nearly triple the ROT I got from an olde Tymer, and recently verified empirically, and verified against a published heatsink's datasheet.

I can't explain the variation with Legg. I can only imagine it's due to vertical vs. horizontal mounting.

Cheers, James Arthur