How to choose heatsink for project. - Page 1
Forum section: electronics / beginners

author: fc3000
author_rank: Regular Contributor
timestamp: July 25, 2023, 02:55:03 pm
content: Hi all.I am using a Voltage regulator that has a thermal resistance of 3degreeC/W.The regulator dissipates a max power of 3Watts, and the max room temp would be 30degreeC.So the Max expected Junction temp=39degreeC.Is that correct?If so how do you select Heat-sink from supplier?They are rated at something like - 4.0K/W, and the like.What does it mean?Thank youRegards

author: Roehrenonkel
author_rank: Frequent Contributor
timestamp: July 25, 2023, 03:09:07 pm
content: Hi fc3000,most probably the 3C/W is Rjc (Junction to case).You'r right with the 9°C rise in temperature, but just if you can keep the case at a steady 30°C.Thermal Resistances have to be added: Rjc, Risolation (maybe) and the f.e. 4K/W of the heatsink.So without any isolation-washer your junction-temp. would be (4+3)x3+30=51°CGood luck

author: Ian.M
author_rank: Super Contributor
timestamp: July 25, 2023, 03:16:09 pm
content: The thermal resistances add, so in your case, with that heatsink, you've got 3+4=7 °C/W so Tj @3W dissipation will be at least 21 °C above local ambient (not room temperature).You are missing the temperature differential between the inside of the enclosure and room temperature, which can be quite high if there is restricted ventilation, and unless the regulator is soldered directly to a copper heatsink, you are also missing the thermal resistance of its mounting interface, which may be quite high if you are using a silpad or similar to electrically isolate the tab. Also, its unwise to design to a room temperature under 40° C unless you can guarantee the device will only be used in air-conditioned areas, with an over-temperature cutout for its supply.

author: mariush
author_rank: Super Contributor
timestamp: July 25, 2023, 04:03:43 pm
content: Tj = Pd(Θjc + Θcs + Θsa) + Tawhere:Tj = junction temperature, °CPd = power dissipation, WΘjc = junction thermal resistance, °C/WΘcs = insulator thermal resistance, °C/WΘsa = heat sink thermal resistance, °C/WTa = ambient temperature, °CYou can say Θja = Θjc + Θcs + Θsa, where Θja = junction to ambient, °C/W and then it gets simplified to :Tj = Pd (Θja) + TaThis Tj must be below the regulator's maximum temperature (ex 125c or 150c) otherwise the chip will break.Let's take an example,1. A LM1085 linear regulator (fixed 5v version) :https://www.digikey.com/en/products/detail/texas-instruments/LM1085IT-5-0-NOPB/3635642. some random thermal paste (let's say it has 0.1 C/w thermal resistance)3. a basic TO-220 heatsink :https://www.digikey.com/en/products/detail/assmann-wsw-components/V6560Y/3511467The LM1085 has these specs :https://www.ti.com/lit/ds/symlink/lm1085.pdf(page 4)RθJAJunction-to-ambient thermal resistance 40.6RθJC(top) Junction-to-case (top) thermal resistance 43.0RθJB Junction-to-board thermal resistance 23.1ψJT Junction-to-top characterization parameter 9.9ψJB Junction-to-board characterization parameter 22.1RθJC(bot)Junction-to-case (bottom) thermal resistance 0.7The heatsink has a thermal resistance of7C/wwith natural airflow, no extra fans.So let's say you need to power it with 12v and it produces 5v and you need up to 1A of current. This means the power dissipated by the regulator will be approximately P = ( 12v - 5v ) x 1 A = 7 watts .So without a heatsink, we can use the thermal resistance junction to ambient to estimate how hot it would get :Tj = 7w ( 40.6 + 0.7) + 30c (ambient) = 319c ... so obviously not gonna work.But with heatsink :Tj = 7w ( 7c/w for heatsink + 0.1c/w thermal paste + 0.7c/w junction-case) + 30c = 54.6 + 30 = 84.6 C ... which is below 125c so it's fine.There are fancier heatsinks, like for example this one is rated 2.6c/w at natural airflow:https://www.digikey.com/en/products/detail/aavid-thermal-division-of-boyd-corporation/530002B02500G/1216384The datasheet of LM1085 goes through explanations in section 10.3 (page 20) and even has the formula above.

author: tszaboo
author_rank: Super Contributor
timestamp: July 25, 2023, 04:28:05 pm
content: Funnily enough, if you know basic electronics, you know basic thermal design. Temperatures are voltages, heat transfer (W) is current, and thermal resistance is... resistance.Then you can solve most problems with Kirchhoffs voltage law.

author: fc3000
author_rank: Regular Contributor
timestamp: July 25, 2023, 05:14:47 pm
content: Thank you to all of you.This forum is excellent.I am now calculating all the aspect that I did not consider.Thank a lotRegards

author: Terry Bites
author_rank: Super Contributor
timestamp: July 25, 2023, 06:26:59 pm
content: This'll help.www.heatsinkcalculator.com/heat-sink-size-calculator.html