Equation for converting thermal conductivity to thermal resistance? - Page 1
Forum section: electronics / projects

author: beaker353
author_rank: Unknown
timestamp: April 27, 2015, 05:12:08 am
content: I searched the forums to see if this has been answered already, but it looks like no. If I missed it somehow, my apologies.I'm comparing a number of thermal interface options for an upcoming project. Needs to be electrically non-conductive, for a TO-220 package, can be relatively expensive as this is a custom few-off project that only a few will be built, and decent availability through the likes of Mouser or DigiKey.I'm finding some option are spec'ed in thermal resistance (C/W) and some are in thermal conductivity (W/mK). From what I've been able to gather, thermal resistance is better when lower, thermal conductivity is better when higher. So within one type I can determine a winner and looser, but I can't figure out the math to convert between the two. I've tried converting with some equations I found online, but I have a feeling my units are screwy as I get numbers way outside anything related to reality.http://multimedia.3m.com/mws/media/122268O/characteristics-of-thermal-interface-materials.pdfI do understand that thermal conductivity is a physical property of the material so physical spec's must be taken into account to do the conversion properly.Anyone able to offer some guidance? Thanks!- EM

author: T3sl4co1l
author_rank: Super Contributor
timestamp: April 27, 2015, 05:47:47 am
content: In most real cases, you won't get more than 50W out of a TO-220, or 100W out of a TO-247. Good rubber pads can do that, but for most types, you'll want to cut those figures in half. If that won't afford you enough power dissipation for your expected use, you'll have to use multiple in parallel (which is usually pretty cheap, but mind the knock-on cost of hardware), or rethink your circuit smarter to dissipate less power.Typical thermal resistances are in the 1-3 C/W range, or conductivity 1-5 W/(m*K). More is better, of course, but you also want it as thin as possible.Do you know dimensional analysis? Just take the unit and, everywhere you see a unit you don't want, multiply or divide by some quantity you have to cancel them out until you get the right result. At worst, your result will be off by a fixed constant (like 2 or pi). In this case, as long as the pad is thin relative to its area, we can assume heat flow is normal and the constant happens to be 1, no worries.So...You have to take 1 / (1 W/(m*K) ) to get K/W. But you have an extra m.Divide by the length factor of the pad.Ah, but the pad has area (cross sectional) and thickness (length in the direction of heat transfer).So we actually need to multiply by length, then divide by cross sectional area.A lot of physical units do this, where there's a bulk property, like resistivity, with an associated length scale. And the resulting scaling is length over area, or whatever. Units that don't, usually have a different exponent (like work is J == N*m, or pressure is Pa == N/m^2).If you prefer customary units (or you only have them on hand, e.g. rubber sheets rated in mils or inches thickness), just throw it all at Google:https://www.google.com/search?q=%2820+mil+%2F+%280.6+in+*+0.4+in%29%29+%2F+%282+W%2F%28m*Kelvin%29%29+in+Kelvin%2FW(Note that Kelvin has to be spelled out since case-insensitive "K" is ambiguous, and it gives you the gross SI unit format if you don't tell it to format as a derived unit i.e. K/W. Also, since this is a thermal difference unit, K and C make no difference.)Tim

author: IanB
author_rank: Super Contributor
timestamp: April 27, 2015, 05:57:50 am
content: Quote from: beaker353 on April 27, 2015, 05:12:08 amYou can't directly convert one into the other. Thermal conductivity is a property of a material, like electrical conductivity."Thermal resistance", on the other hand, is not a physical quantity as such. It is an aggregate measure of the total performance of a system, taking into account different materials, interfaces between them, shape and geometry, surface area, and external air currents. There is no simple formula for thermal resistance. It has to be computed using complex models, or measured experimentally.Considering thermal conductivity: this measure how fast heat can flow through a uniform substance given a temperature gradient. Heat flow is measured in watts (W). The heat flow through a cube of material is proportional to the temperature difference between opposite faces, measured in kelvin (K). It is proportional to the cross-sectional area, measured in square meters (m2). And it is inversely proportional to the thickness (m). Putting it all together, we find that:(heat flow) = (thermal conductivity) x (temperature difference) x (area) / (thickness)orW = (thermal conductivity) x K x m2/ mhence(thermal conductivity) = W/m-KThermal conductivity is often given the symbolk.

author: johnwa
author_rank: Frequent Contributor
timestamp: April 27, 2015, 07:30:28 am
content: Quote from: T3sl4co1l on April 27, 2015, 05:47:47 amYes, this is a good way of approaching things when you can't remember the formula. In cases where there are multiple variables with the same unit, it may help to ask, "Will increasing this variable make the result bigger or smaller?" E.g, it makes sense that a larger contact area will reduce the thermal resistance, therefore you must divide by the area.

author: beaker353
author_rank: Unknown
timestamp: May 02, 2015, 06:13:05 am
content: Thank you T3sl4co1, IanB, & johnwa! I have a little Google doc spreadsheet setup with what I think might be the correct formulas given your advice. You can change the tan cells all you want and it should update the calculated numbers in real-time. Taking an actual part, Aavid 4169G, with 15.06W/mK thermal conductivity, 1.57mm thickness, and 259mm(sq) surface area. I get a calculated thermal conductivity of 2.484W/K and a calculated thermal resistance of 0.4025C/W.It looks plausible'ish to me, but I would like to get confirmation to the wiser collective here. Thanks!https://docs.google.com/spreadsheets/d/1aVSkXDToJmjPRyPu5n9L8gET1Sm1va2AXKfam315iFg/edit?usp=sharing-EM

author: Sniper1
author_rank: Regular Contributor
timestamp: October 01, 2024, 07:20:53 pm
content: So to make sure i understood:lets assume a to220 has a thermal contact area of 11x12mmhttps://eu.mouser.com/ProductDetail/Bergquist-Company/SPK4-0.006-00-104?qs=unwgFEO1A6t61wtBZCnD3w%3D%3Dthis is my thermal material with Thermal Conductivity 0.9 W/m-K 0.1524mm thickTHEN value of C/W = 1 /( Thermal_Conductivity_VALUE * MATERIAL_THICKNESS / CONTACT_AREA )= 1/ ( 0.9 * 0.0001524 / ( 0.011* 0.012))= 0.96237 C/WIn this case it is super close the conductivity but is this at least a plausible value.same product on herehttps://www.digikey.in/en/products/detail/bergquist/SPK4-0-006-00-104/529936?s=N4IgTCBcDaIMoAUDSAWAtABgHQYwNkwzQEYMUQBdAXyABUT Thermal Resistivity0.48°C/WIF we assume it is TO-247 then that is 17x14mm then =1/ ( 0.9 * 0.0001524 / ( 0.017* 0.014))= 1.735 that is not even closeI think i am doing something wrongNOTE: I wrote the formula the way that i wrote it since i think it is easier to follow for other bothered by this topic ( never found a proper formula that i could understand so i made on).

author: IanB
author_rank: Super Contributor
timestamp: October 01, 2024, 07:53:20 pm
content: Quote from: beaker353 on April 27, 2015, 05:12:08 amLooking back at my first reply to this thread, I realize my answer was not quite correct.If you have a value quoted with units of C/W, this is degrees Celsius per watt.A value quoted as W/mK is watts per milli-Kelvin, which is the same as kW/C or kilowatts per degree Celsius.In that case, the interconversion is simple: [C/W] = 10-3/ [W/mK]Edit: as pointed out below, W/mK probably really ought to have been read as "watts per meter per Kelvin", which is the traditional definition of thermal conductivity. So perhaps, ignore what I wrote above. On the other hand, if an electrical component on a datasheet is specified as having a "thermal conductance" of W/mK rather than a "thermal resistance" of K/W, then it would be as well to disambiguate exactly what is meant. By electrical analogy, conductance is the reciprocal of resistance.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: October 01, 2024, 08:42:55 pm
content: As a practical matter, anything that isn't metal is bad. That includes the most expensive silver loaded thermal paste, rubber pads or whatever. Thus, they have to be as thin as possible. Your surfaces need to be as flat as possible. Machine the interface surfaces and maybe even lap them. What voltage is involved? I've built HV amplifiers that needed fairly thick alumina ceramic insulators. Great isolation but only so-so thermal conductivity. If the voltage isn't too high, there are heavy anodized aluminum insulators. With a thin layer of grease, they're very good because the anodize layer is thinner than most other things. Speaking of grease, you might try a phase change material like Aavid Ultrastick. It's expensive, but it becomes liquid at 60C and gives you the thinnest possible interface. IMO, manufacturers of thermal interface products try to put things in different units, making it harder to determine what's actually the best. BTW, if you use rubber pads, pay a bit extra for the better grades as they're much better.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 01, 2024, 08:50:36 pm
content: Quote from: IanB on October 01, 2024, 07:53:20 pmNo, this is meters. It's a volumetric parameter, much like resistivity: R = ρL/A, where ρ ≡ Ωm, not milliohms.Tim

author: IanB
author_rank: Super Contributor
timestamp: October 01, 2024, 09:08:51 pm
content: Quote from: T3sl4co1l on October 01, 2024, 08:50:36 pmOh, you mean I was right the first time? As a matter of notation though, one ought to write W/m-K or W/m/K rather than W/mK, since the unit mK really is a milli-Kelvin.

author: TimFox
author_rank: Super Contributor
timestamp: October 01, 2024, 09:09:18 pm
content: Correction:R= ρL/A

author: TimFox
author_rank: Super Contributor
timestamp: October 01, 2024, 09:14:40 pm
content: Yes, in careful writing of units, a product of units should be denoted with a hyphen, such as N-m for Newton-meters for torque.Also, avoid using "C" or "F" in this context, "K" for kelvins (degrees not required) is much more appropriate.I carefully distinguish Cofor changes in temperature fromoC for temperature, but K works for both.A newspaper article a few months ago was discussing what he called a 2oC temperature increase in an article about global warming and inserted that into a units conversion calculator for Fahrenheit.I think we all agree that a temperature increase of 35 Fowould be unpleasant

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 01, 2024, 09:48:48 pm
content: Quote from: IanB on October 01, 2024, 09:08:51 pmYeah, it's not always well rendered, but the unit is well understood in context, so it gets a pass. SI malding...Tim

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: October 02, 2024, 12:07:18 am
content: Units drive me crazy! I usually use a dot for products. You also have to be careful with Newton-meters for work (1 Joule) vs torque, which is completely different. And then I'm never sure what to capitalize. Fortunately there's this-https://physics.nist.gov/cuu/Units/checklist.html

author: TimFox
author_rank: Super Contributor
timestamp: October 02, 2024, 02:06:57 am
content: Yes, both torque and energy units are force x distance, but we don’t measure torque in Joules.A bit confusing, but the distinction is important.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 02, 2024, 02:43:27 am
content: Incidentally, work and torque have the same relation as real and reactive power; you can express reactive power as the peak energy stored, uhhh... Q / (2 pi F) I think it is?Tim

author: ArdWar
author_rank: Frequent Contributor
timestamp: October 02, 2024, 03:01:29 am
content: Quote from: TimFox on October 02, 2024, 02:06:57 amIt's often easier (and safer) to initially assume dimensional analysis does not work across angular-linear boundary."distance" for example, are usually understood as (relative) translation in linear system, but (absolute) position in angular system. Slight difference with huge implication.

author: IanB
author_rank: Super Contributor
timestamp: October 02, 2024, 03:04:00 am
content: Quote from: T3sl4co1l on October 02, 2024, 02:43:27 amI'd have to think about that a bit. Torque can be used to transmit real power from A to B (think drive shaft in motors), so in a sense torque is analogous to voltage. Power = voltage x current = force x linear velocity = torque x angular velocity. So one joule (one newton-meter) is a unit of energy. But one newton-meter of torque is a unit of angular force, which only becomes energy when multiplied by a unit of angular distance (which is dimensionless).I wish you hadn't mentioned this, because now it's going to sit in the back of my mind until I untangle it.Quote from: ArdWar on October 02, 2024, 03:01:29 amI like this approach. It might stop my brain from melting.

author: Sniper1
author_rank: Regular Contributor
timestamp: October 02, 2024, 04:39:32 pm
content: [W/mK] was given since the product is itself a sheet from who you cut pieces , the fact that on mouser there are to220 shaped pieces is another problem i was doing the math for a to220Sadly now i am more confused as to how to make the conversion or if i am correct in my math or not.

author: TimFox
author_rank: Super Contributor
timestamp: October 02, 2024, 04:42:48 pm
content: When fabricating a heat conductor from a sheet, you need to specify the effective area (dominated by the metal face of the transistor) to get a thermal resistance.

author: Sniper1
author_rank: Regular Contributor
timestamp: October 02, 2024, 04:44:10 pm
content: I need to dissipate 5-7W max it is a APFC mosfet .I NEED to isolate the 400Vdc it has from the heat sink since this is a real safety problem ( i am afraid of making a mess with mica+paste but that is option 2 since i have a mica kit already .I will check that product on mouser and depending on stock and price and what it is i may try it.

author: Sniper1
author_rank: Regular Contributor
timestamp: October 02, 2024, 04:45:39 pm
content: Quote from: TimFox on October 02, 2024, 04:42:48 pmIn my post i did that 11x12 mm metal contact area.NOTE: i was trying to quote the upper messages , that is why they may not make sense out of context.

author: IanB
author_rank: Super Contributor
timestamp: October 02, 2024, 06:04:54 pm
content: Quote from: Sniper1 on October 01, 2024, 07:20:53 pmQuote from: Sniper1 on October 02, 2024, 04:39:32 pmIf you have a flat, thin sheet, and you wish to transfer heat from one side to the other, then what you get is a heat transfer coefficient, often denoted as h, with units of W/m²-K (watts per square meter per Kelvin).Now, heat transfer increases in proportion to area, and decreases in proportion to thickness.So if you have a thermal conductivity (κ) of 0.9 W/m-K, and the sheet is 0.1524 mm thick (= 1.524x10-4m), then you divide the the thermal conductivity by thickness (thicker is worse), to get the heat transfer coefficient. In this case:h = κ / t = 0.9 [W/m-K] / 1.524x10-4 = 5906 W/m²-KTo get an overall thermal conductance, you multiply h by the area. Suppose the area is 11x12 mm, then:A = 0.011  x 0.012  = 0.000132 m² = 1.32x10-4m²This gives you a thermal conductance (let's call it J) of:J = hA = 5906 [W/m²-K] x 1.32x10-4 = 0.780 W/KThe thermal resistance is the reciprocal of this, so:θ = 1 / J = 1 / 0.780 [W/K] = 1.28 K/WI hope this all makes sense.

author: Sniper1
author_rank: Regular Contributor
timestamp: October 02, 2024, 08:53:25 pm
content: Quote from: IanB on October 02, 2024, 06:04:54 pmso basically this:Rth​=thickness / (thermal_conductivity * are_contact​ )

author: Phil1977
author_rank: Frequent Contributor
timestamp: October 02, 2024, 09:12:53 pm
content: Quote from: Sniper1 on October 02, 2024, 04:44:10 pm7W from a TO220 are easy going if you just have a heat sink that´s large enough. For this dissipation you can also use a fully isolated package FET.

author: IanB
author_rank: Super Contributor
timestamp: October 02, 2024, 09:16:03 pm
content: Quote from: Sniper1 on October 02, 2024, 08:53:25 pmThat seems correct, yes. But make sure you keep all your units of measure consistent.

author: TimFox
author_rank: Super Contributor
timestamp: October 02, 2024, 09:22:59 pm
content: In Imperial units, thermal conductivity is measured in 1 Btu/(h⋅ft⋅°F) = 1.730735 W/(m⋅K), where "h" = hours.For engineering purposes, I have seen this converted to use thickness in inches and area in ft2, also time in seconds.(I usually distinguish between Imperial and US Customary units, but British Thermal Units sound Imperial to me.)

author: IanB
author_rank: Super Contributor
timestamp: October 02, 2024, 09:40:29 pm
content: Quote from: TimFox on October 02, 2024, 09:22:59 pmYup. This is why it is not fun to study engineering in the USA. There are so many conversion factors in the standard equations. You invariably need a standard table of formulas to avoid having to remember all the magic numbers that crop up everywhere.As for the Btu, US Customary units were derived from Imperial units, and the Btu was inherited from the Imperial system. Nobody has quite managed to agree on a universal definition of its value, but 1.055 kJ is close enough for government work.

author: TimFox
author_rank: Super Contributor
timestamp: October 02, 2024, 10:02:54 pm
content: I believe the original definition of a BTU is the heat required to raise the temperature of 1 lb av of water by 1 Fo, analogous to the original definitions of the metric calorie or kilocalorie.

author: IanB
author_rank: Super Contributor
timestamp: October 02, 2024, 10:15:19 pm
content: Quote from: TimFox on October 02, 2024, 10:02:54 pmYes, but there are different ways to do this, resulting in various different values. See Wikipedia for details:https://en.wikipedia.org/wiki/British_thermal_unit

author: TimFox
author_rank: Super Contributor
timestamp: October 02, 2024, 10:20:16 pm
content: "original definition"

author: Jeroen3
author_rank: Super Contributor
timestamp: October 03, 2024, 08:32:35 am
content: Quote from: TimFox on October 02, 2024, 09:22:59 pmToday in cursed units!British Thermal Unit per hour feet squared Fahrenheit

author: Sniper1
author_rank: Regular Contributor
timestamp: October 03, 2024, 07:34:06 pm
content: Quote from: Phil1977 on October 02, 2024, 09:12:53 pmI had a had time picking the proper fet , i went through 15-20 in order to get 4 usable ones. So finding a isolated one with proper characteristics is probably not wort the time since those sil pads are not that expensive.

author: SteveThackery
author_rank: Super Contributor
timestamp: October 04, 2024, 10:18:33 am
content: Quote from: Conrad Hoffman on October 02, 2024, 12:07:18 amI think it's a brilliant system, carefully designed to be clear and unambiguous, with simple numeric relationships between them.Yes, a force of 1 N applied over a distance of 1 m converts 1 Joule of energy, but youneverrefer to that work as Newton-metres.

author: Phil1977
author_rank: Frequent Contributor
timestamp: October 04, 2024, 10:34:47 am
content: Quote from: Sniper1 on October 03, 2024, 07:34:06 pmThe isolated FET´s for sure are more expensive than standard packages + glimmer sheets. It´s just for the convenience that it´s easier to assemble and potentially safer to handle.

author: TimFox
author_rank: Super Contributor
timestamp: October 04, 2024, 02:18:32 pm
content: Quote from: SteveThackery on October 04, 2024, 10:18:33 amThe N-m in torque refers to a force that is perpendicular to the radius.For N-m in energy (Joules), the "scalar product" of the force times distance traveled gives you the work done: maximum for force and distance vectors collinear.The normal unit for energy or work in English units is lb-ft.

author: TimFox
author_rank: Super Contributor
timestamp: October 04, 2024, 02:41:04 pm
content: A side comment about expressing equations and units:In careful usage, a letter in an equation for a variable or parameter should beitalicized.The same letter, used as a unit, should be in plain Roman.This is covered in books such as the mathematical typesetting chapter of University of Chicago PressChicago Manual of Stylehttps://www.chicagomanualofstyle.org/home.html(Some modern publishers, such as Springer, ignore this and potentially cause confusion.)For exampleVin an equation is a voltage or potential, while V represents the unit "Volts".