Small Buck Converter / Regulator Design Advice? - Page 1
Forum section: electronics / projects

author: fourfathom
author_rank: Super Contributor
timestamp: September 23, 2023, 05:47:23 am
content: I want to add a small switching regulator to a design, input between 9V and 16V (or greater range), and output at 5V, 4A. I don't need isolation or anything fancy, but it should be cheap and reliable, and be manufacturable by JLCPCB. I've built a prototype using the LMR14050 (from TI), using the datasheet recommended component values and it works but is getting too hot above 3A. No doubt I could improve the thermal layout, but I would rather have a design with more margin. I would like to have a clean output, but I will have distributed power filtering so the supply ripple requirements are only modest.Any recommendations?

author: PCB.Wiz
author_rank: Super Contributor
timestamp: September 23, 2023, 08:05:27 am
content: Quote from: fourfathom on September 23, 2023, 05:47:23 amThat has an external flyback diode, you should target a sync chip with two FETs these days.Look for the RDS of the two FETS and go for a lot lower than the 90mOhms of that part.Do not focus so much 'on paper' current rating, as you have already found, thermal issues often matter moreeg a AOZ1267 hasLow RDS(ON) internal NFETs– 35mOhm high-side– 12mOhm low-side

author: fourfathom
author_rank: Super Contributor
timestamp: September 23, 2023, 11:26:13 pm
content: Thanks for the info! I actually did look at the AOZ1267, but just the headline specs. I decided to try the LMR14050 instead, mainly because of the simpler PCB footprint (not the best reason, I will admit). So now I have built a KiCad footprint for the AOZ1267, found a low-R high-current inductor, and will make another test board.Again, I've not dug deeply into the IC specs, but while I can see how using a flyback diode rather than a synchronous switch will reduce overall efficiency and increase total heat, does this really affect the switcher IC power dissipation? I am less worried about overall efficiency than I am about the IC die temperature.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: September 24, 2023, 05:14:43 am
content: Quote from: fourfathom on September 23, 2023, 11:26:13 pmA high loss external part does not affect IC power loss, but it does affect IC die temperature, which you are most worried about.It does that, by raising the board temperature near the SMPS chip.You are correct to find a low-R inductor, for the same reason. Lowest overall PCB heat, helps get to lowest die temperature.

author: jwet
author_rank: Frequent Contributor
timestamp: September 25, 2023, 02:30:23 am
content: 5V at 4 amps (20W) is pushing that LMR14050 but it should be able to handle. The freewheel diode is costing you about 5-7% efficiency (33% duty cycle means diode is on 66%). This is 5-7 is the hit over a sync rectifier. In their graphs, their showing overall efficiencies in this circuit of over 90%, so the chip with .09 ohm FET is doing pretty good. Get a big fast diode that can take lots of current without too much drop and has a low thermal theta. Get a very low R inductor and get as much heat as you can with the exposed thermal pad on the LMR. If you can get to numbers like 90%, this only 2W and if you sink the part well, it will rise some but not over 55C rise. Die max is 170C. At 5v out, you're on the border of whether a sync rectifier is really needed or just a good diode. At 3.3v and below, you need to go synchronous at these currents.

author: fourfathom
author_rank: Super Contributor
timestamp: October 12, 2023, 01:35:46 am
content: I designed a test board for the AOZ1267 and I just can't get it to turn on. I am particularly puzzled by the "AIN" pin (pin 7) , it appears from the "Typical Application" diagram in the spec-sheet that I should leave it disconnected, as I have no external 5V bias. But the internal diagram and the example layout show it connected to the "IN" pins (8 and 9). Regardless, connected or not, the regulator still won't start up and I get zero volts on the output. Would someone care to look at my schematic and tell me what I'm doing wrong?The input is a nominal 12V, the output is supposed to be 5V. That FET at the input is for reverse-voltage protection (and that at least is working.)

author: PCB.Wiz
author_rank: Super Contributor
timestamp: October 12, 2023, 02:41:58 am
content: Quote from: fourfathom on October 12, 2023, 01:35:46 amI think it needs that 5V, and how it gets that varies with part - it looks like AOZ1267-02 has no LDO, but AOZ1267QI-01 has a LDO connected to AIN(p7)