selecting a physically smaller inductor for buck regulator - Page 1
Forum section: electronics / beginners

author: VEGETA
author_rank: Super Contributor
timestamp: July 22, 2025, 06:03:30 am
content: Hello,I have a regulator circuit working fine and nice results. However, i want to add more stuff on the board which makes space more expensive than gold.I use 2.5mOhm 13mmx12mm inductor (from LCSC, a lot of similar replacements there too). This is so huge and now pretty much no space left to keep it especially that I have 2 regulators on board.I found smaller size 7.1x6.6 (standard) sized inductors, 2.2uH (same value), but their resistance is 11-15mOhms instead of 2.5-3.5 mOhms of currently used ones.Buck regulators have 15v input, and outputs 5v and 3.3v at a total of 2-2.5 amps max. current choice the inductors run ice cold.my question is: will the new replacement work as cold too? will there be any drop or so in the output voltage due to resistance (i think not, but nice to hear your opinions)?thanks!!

author: PGPG
author_rank: Frequent Contributor
timestamp: July 22, 2025, 07:41:15 am
content: More important than resistance is current at which inductor magnetics saturate. When inductor saturates the current begins to rise so quickly that internal IC protections reaction can be too slow and ...Many DCDC during start temporarily reach higher current than later in operation under typical load.I always select inductor with saturation current higher than internal IC current limiter (its highest possible value according to datasheet). This guarantees that IC current limit will always be reached before inductor saturation occurs.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: July 22, 2025, 02:31:04 pm
content: I agree with PGPG. Saturation current is one of the most important parameters of a power inductor, and it greatly depends on the size.What sort of SMPS controller are you using? LM2596 is huge, SOIC8 is already smaller, but if your current is small enough, you can go to the SOT23-6 sized regulators. These also have a maximum current that is below 1A, and if that is sufficient, you can also greatly reduce the inductor size. If PCB size is really at a premium, you can use toroidal inductors, and then place them over other parts. This way, you only need PCB area for the two wires.Also, there are a bunch of things that are quite critical in SMPS design, but wire length to the inductor is not. (loop area is, so be aware). I'm almost certain that it is completely OK to use 100mm or so of twisted wire to connect to the inductor. This is something I'd have to test and verify. Could be a fun afternoon...

author: scottapotamas
author_rank: Regular Contributor
timestamp: July 22, 2025, 02:50:28 pm
content: This depends a bit on your requirements, but if you're making a custom board and need the space then modules with integrated inductors are pretty hard to beat.TI TPSM863257 is one I've used a fair bit that meets your stated requirements: 3-17VIN, 0.6-5.5VOUT, 3A. It's < $3 in small quantities from Digikey/Mouser, <$1.50 from LCSC.I typically fit the module and input/output caps inside a 8x8mm area on one side, but you can space things out to your comfort level.

author: MrAl
author_rank: Super Contributor
timestamp: July 22, 2025, 02:51:27 pm
content: Quote from: VEGETA on July 22, 2025, 06:03:30 amHi,As others have pointed out you need to check all of the specs of the new inductor for the comparison.As to the resistance, you can estimate the power heating due to the resistance alone by using the current and the resistance:P=i*i*Rso it looks like about 0.1 watt or less. That does not require that much surface area so unless the package is very very small it will probably be ok.

author: VEGETA
author_rank: Super Contributor
timestamp: July 22, 2025, 04:46:12 pm
content: well, the controller is tps564247 which uses small inductors and caps already.my previous big inductors were:TMPC1265HP-1R5MG-DTAI-TECH TMPC1265HP-2R2MG-Dnew ones (should be both 2.2u) have nice specs as well in terms of current handling, I suspect saturation current will reach that far.Here are some of my findings:https://lcsc.com/product-detail/Power-Inductors_TAI-TECH-TMPA0603S-2R2MN-D_C357100.htmlhttps://lcsc.com/product-detail/Power-Inductors_XR-XR0650-2R2M_C5307644.htmlhttps://lcsc.com/product-detail/Power-Inductors_PROD-Tech-PSPMAA0605H-2R2M-ANP_C2983088.htmlhttps://lcsc.com/product-detail/Power-Inductors_TAI-TECH-TMPC0605H-2R2MG-D_C357209.htmlmost of them have +10 amps of saturation current while maximum load current is 2 to 2.5 amps only, never to reach 3 amps.___I like the idea of power module but it is just that one you linked which has good pricing and might not be always the same. Also, what about its heat at full load?my current design runs ice cold and would like that to continue.___I already have excellent performance in current design which is < 5mv p-p total ripple and no noise at full load. I just want to add more stuff and require more space now___Well, checking this one:https://lcsc.com/product-detail/Power-Inductors_PROD-Tech-PSPMAA0605H-2R2M-ANP_C2983088.htmlit has about 10-13 mOhm DCR, and its data sheet says Irms = 8A which increases the temp by 40 degrees. Thus, I have 2.5a max -> It will heat by 12.5 degrees above ambient =~ roughly 25+12.5 = 37.5 ~= 40 degrees. Maybe this is the coil but the case can reduce that a bit too.doing the same for "TAI-TECH TMPC0605H-2R2MG-D" will give 10 degrees above ambient =~ 35 degrees.is this correct?

author: radiolistener
author_rank: Super Contributor
timestamp: July 22, 2025, 07:59:07 pm
content: Quote from: PGPG on July 22, 2025, 07:41:15 amYes, and the saturation current depends on the core size - a smaller core results in a lower saturation current. Additionally, using thinner wire increases resistance and leads to higher heat losses.Using silver wire can slightly reduce resistance, but the small core size remains the main limitation.

author: VEGETA
author_rank: Super Contributor
timestamp: July 22, 2025, 08:09:25 pm
content: Quote from: radiolistener on July 22, 2025, 07:59:07 pmso kindly check my last reply for the linked inductors + calculation based on my current solution. I guess it should be ok, but more than one pair of eyes is always better.