Switching Regulator Heat Question - Page 1
Forum section: electronics / beginners

author: bostonman
author_rank: Super Contributor
timestamp: December 29, 2020, 02:07:53 am
content: I've never dealt with switching regulators in the sense of designing.I'm looking at using the LM2678. My circuit will have between 9V and 12V on the input, 5V on the output, and about 1.3A - 2A load.This will actually be a power supply I plan to design for a Commodore 64 (I know I can buy them online, but looking to design/build one myself).If I understand the datasheet correctly, and using conservative numbers, the efficiency is 85%. I take the output power of (worst case) 5V * 2A = 10W. At 85% efficiency, I divide 10W by 0.85 to get an input wattage of 11.76W which means 1.76W gets dissipated from the regulator.The datasheet for a 220 package (the one I plan to use) is 2 degrees C / W, so the heat is 2 * 1.76 = 3.53 degrees C.I add this to my ambient temperature of (using a rough high number) 30 degrees C to make 33.5 degrees C.This number seems very low, is this correct? Also, the datasheet states it can work up to 125 degrees C, but it doesn't state whether I need a heatsink. Does 125 degrees C mean I'm 91.5 degrees C away from maximum temperature and how do I know whether it needs a heatsink?

author: Ian.M
author_rank: Super Contributor
timestamp: December 29, 2020, 02:47:11 am
content: Some of the losses will be in the external Schottky diode, so its probably not going to dissipate much over 1W in the regulator itself. However on its own in free air, 1W will heat the junction 65 deg C, and you don't want to get too close to the max operating temperature or you risk it going into thermal shutdown, so unless you can guarantee to keep the air inside its enclosure under say 45 deg C even on a hot summer day, (leaving a 15 degree safety margin), you'll need to heatsink it.It won't need much as a TO-220 style and size package is normally good for about a watt in free air so one of those clip-on twisted fin jobbies or a board-mount extruded heatsink little wider than the package and maybe twice as tall should be sufficient.

author: langwadt
author_rank: Super Contributor
timestamp: December 29, 2020, 02:48:20 am
content: Quote from: bostonman on December 29, 2020, 02:07:53 am2 C/W is junction-to-case, you need an infinitely good heatsink to use that number, you need to either look at the examples for junction-to-ambient, or calculate what the number is with the 2 C/W plus what ever heatsink you use

author: bostonman
author_rank: Super Contributor
timestamp: December 29, 2020, 03:03:24 am
content: QuoteHow did you come up with that?

author: Ian.M
author_rank: Super Contributor
timestamp: December 29, 2020, 03:07:03 am
content: https://www.ti.com/lit/ds/symlink/lm2678.pdfTable 6.4 Thermal Information, first data line

author: bostonman
author_rank: Super Contributor
timestamp: December 29, 2020, 03:09:56 am
content: I saw that, but how is that based on 1W?Also, were my calculations wrong?

author: langwadt
author_rank: Super Contributor
timestamp: December 29, 2020, 03:23:49 am
content: Quote from: bostonman on December 29, 2020, 03:09:56 amyou used junction to case, not junction to ambientit is 65C/W for: Junction to ambient thermal resistance (no external heat sink) for the 7-lead TO-220 package mounted vertically, with 1⁄2 inch leads in a socket, or on a PCB with minimum copper area.

author: Ian.M
author_rank: Super Contributor
timestamp: December 29, 2020, 03:28:21 am
content: Because RθJAis in units of °C/W, and my S.W.A.G is that the 1.76W of losses you calculated, split between the diode and the chip will have the chip dissipating a bit under 1W.==> up to 65°C junction temperature rise above local ambient without a heatsink.Edit: Langwadt just beat me to it with more detail.

author: bostonman
author_rank: Super Contributor
timestamp: December 29, 2020, 03:33:57 am
content: Then what is the purpose of junction-to-case?I thought we take the ambient temperature of 30 degrees C and add that to whatever the regulator temperature rise is - junction-to-case, thus giving me the temperature I had initially calculated in my first message.

author: langwadt
author_rank: Super Contributor
timestamp: December 29, 2020, 03:36:52 am
content: Quote from: bostonman on December 29, 2020, 03:33:57 amyou use that number when you add a heatsink,https://youtu.be/8ruFVmxf0zs

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: December 29, 2020, 03:49:17 am
content: Here's an old tool but still useful, if only for the simple thermal calculator.http://www.hp.woodshot.com/Turn off the sounds in the preferences cuz it's gonna annoy you.

author: bostonman
author_rank: Super Contributor
timestamp: December 29, 2020, 03:52:45 am
content: I'll watch the video.Just to clarify in simple terms, assuming all external circuitry (the diode in this case) has dropped whatever wattage and 1W is dissipated (based on 88% efficiency on the graph in figure 4), the heat will be 65 degrees C and the maximum it can run is 125 degrees C based on table 6.3?Again, just to understand this, if say the load was 5A at 5V, that would be 25W. According to the graph, the regulator would run at approximately 82% efficiency giving 5.5 C dissipated from the regulator. Multiply that by 65 per table 6.4, that would be a temperature of 357 C thus being over the maximum temperature by 232C?

author: Ian.M
author_rank: Super Contributor
timestamp: December 29, 2020, 04:31:54 am
content: Well, no, you wouldn't use that row in the table if you were expecting 25W dissipation. Its got over-temperature protection so without a heatsink at somewhere a little above 125°C TJit would shut down and stop delivering 5A (or any) load current until it had cooled down some. The datasheet doesn't specify whether it would then auto-restart or whether you'd need to toggle the power or the enable pin.You know that the junction will be running 50°C hotter than the mounting surface of its case (25W * 2°C/W). You've also got to allow for RθCS, the interface thermal resistance from the case mounting surface to the heatsink, which if its direct mounted with no thermal washer, using a minimum thickness layer of heatsink compound is *probably* under 1.5°C per watt. so you've got the problem of keeping the heatsink under 37.5°C with no safety margin. In 25°C ambient air, that's going to need a heatsink with RθSAless than 0.5 °C/W, which will be large and expensive. If you want to avoid active cooling while dissipating 25W, you really need to beat down RθCSand RθSAmaybe with liquid metal thermal compound between the chip and a large pure copper heatsink with an internal heatpipe to spread the heat out more effectively and a convection chimney to aid natural airflow over its fins.

author: bostonman
author_rank: Super Contributor
timestamp: December 29, 2020, 04:50:50 am
content: I was talking mathematically and not how to actually dissipate the 25W.I watched the video. It was quite informative and believe I now have a better understanding. At work we perform reliability and always use junction-to-case but most of the stuff is based on conservative numbers and general pre-entered formulas.Basically you use either junction-to-case if it's mounted to a heat sink, and junction-to-ambient if it's sitting in open air without a heat sink. But in both cases you always add in ambient air, correct?So generally speaking, if a datasheet stated 1C/W junction-to-case and 50C/W junction-to-ambient, my ambient is 30 C, and the regulator is dissipating 1W. Mounting to a heat sink would see 31C and ambient would see 80C?I didn't grasp something stated in the video. The square of the temp drops the reliability of the regulator? I"m uncertain if I have the statement correct, but he was trying to state how much the reliability of the MOSFET (in his case) would drop the closer to the maximum allowed junction temperature per the datasheet.

author: langwadt
author_rank: Super Contributor
timestamp: December 29, 2020, 05:24:05 am
content: Quote from: bostonman on December 29, 2020, 04:50:50 amonly 31'C if the heatsink is 0C/W iow infinitely good. As Dave explained you need to add the thermal resistance of the heatsinkso with 1 C/W and say a 5C/W heatsink the best you can get is 1+5 = 6 C/W -> 1W * 6C/W + 30'C = 36'C

author: Ian.M
author_rank: Super Contributor
timestamp: December 29, 2020, 05:28:53 am
content: Quote from: bostonman on December 29, 2020, 04:50:50 amPerhaps you are looking for the application of theArrhenius equationto semiconductor reliability.See:https://www.jedec.org/standards-documents/dictionary/terms/arrhenius-equation-reliabilityandhttps://www.electronics-cooling.com/2017/08/10c-increase-temperature-really-reduce-life-electronics-half/which demystifies it a bit.

author: bostonman
author_rank: Super Contributor
timestamp: December 29, 2020, 03:38:52 pm
content: QuoteCorrect. I actually meant in an ideal world with no other factors. Usually I like to start off with the basics, and, once I grasp the concept, I begin adding the other factors.I'll do more reading tonight. At my company, I don't really do an analysis of how long a component will last; although I'm sure others do. This regulator has a maximum of 125 degrees C. What is the standard practice for how much cooler it should be than 125?