Thermal considerations on a switch-mode regulator? - Page 1
Forum section: electronics / beginners

author: Kold
author_rank: Newbie
timestamp: January 30, 2015, 03:57:59 pm
content: Hi everyone,I'm working on my very first design which incorporates a switch-mode regulator and am curious to know what (if any) thermal considerations need to be taken.This is the regulator in consideration:http://www.digikey.ca/product-detail/en/TPS55340PWP/296-35493-5-ND/3768420.It comes in HTSSOP and QFN packages. The QFN appears to have the lower overall thermal resistance (including the junction-to-board resistance), but I would prefer to use the HTSSOP if possible.I also intend to use this regulator as a SEPIC.First, I notice this device has three thermal resistances noted:- Junction-to-ambient thermal resistance- Junction-to-case (top) thermal resistance- Junction-to-board thermal resistance.Before I can even get into the thermal calculation, though, I'm not even sure how to calculate the amount of Watts this thing will be consuming. I will make a back-of-the-envelope attempt though... and as a sanity check I could really use your guys' feedback.The maximum input voltage this regulator will see is 25V, and the minimum output will be 3V, so that's a 22V drop it needs to create. It might also draw 2A as the maximum spec.Now, from the graph on the first page, I note that, above 0.4A and 15V input, the regulator has about a 95% efficiency. So, the power it will be dissipating as heat is 22V * 2A * 0.05 = 2.2W. Is that fair?From there, I note that, with no heat sink, the junction temperature might rise to ambient temperature + (junction-to-ambient thermal resistance*2.2W) = 25 + 43.2*2.2 = 120 'C. And with a heat sink it would be ambient temperature + (heatsink-to-ambient thermal resistance*2.2W) + (heatsink-to-case thermal resistance*2.2W) + (junction-to-case thermal resistance*2.2W) = 25 + (heatsink-to-ambient thermal resistance*2.2W) + (heatsink-to-case thermal resistance*2.2W) + 33.3*2.2= 91.6 + maybe 10 'C?.And does the junction-to-board thermal resistance then mean that the BOARD temperature would be (junction-to-board thermal resistance*2.2W)higherthan the junction? I'm very unsure about this parameter...I'm sorry if that's a bit of an overload of info. I'm trying to not leave things out. Let me know if there's any additional info you need!Thanks, everyone! I really appreciate the help.Kold

author: raphaelcasimir
author_rank: Contributor
timestamp: July 07, 2017, 01:40:53 pm
content: Hey, did you have any answer elsewhere? (I'm having the same kind of issue)

author: t_ryner
author_rank: Regular Contributor
timestamp: July 07, 2017, 02:33:19 pm
content: For clarification, does the device draw 2A through the regulator? With current that high I would likely use something like this instead.-https://www.digikey.com/product-detail/en/rohm-semiconductor/BD00D0AWHFP-TR/BD00D0AWHFPDKR-ND/3663912This device fits the constraints and has a thermal protection shutoff feature, but may require additional circuitry to function as desired.If you're unsure about heat problems with high current applications, you should usually add a heatsink in anticipation, and see if it is needed.Hope I helped!