Step-Down AOZ1280CI - inductor selection - Page 1
Forum section: electronics / projects

author: up8051
author_rank: Frequent Contributor
timestamp: March 12, 2020, 10:03:43 am
content: I am designing a device where the AOZ1280CI step down converter will be used for the power supply.http://aosmd.com/res/data_sheets/AOZ1280CI.pdfVin max 16VVout 3.3VIout 50mAFrom calculations:deltaIL ~= 800mAIKpeak = Io + deltaIL/2 = 450mAI will want to mount the boards in JLCPCB and there are not many 2.2uH chokes to choose from.One of them is:Murata LQM21PN2R2NGCD 2.2uH 0.23Ohm 0.8A SMD0805https://psearch.en.murata.com/inductor/product/LQM21PN2R2NGC%23.pdfIs this type of choke (case 0805 & 0.8A ?) even suitable for this type of converter?Or rather choose:MWSA0402-2R2MThttps://datasheet.lcsc.com/szlcsc/Sunlord-MWSA0402-2R2MT_C169184.pdforMPH201210S2R2MT Isat=0.6A T=0.2Ohmhttps://datasheet.lcsc.com/szlcsc/Sunlord-MPH201210S2R2MTC55755.pdfMaybe I should choose a choke with higher inductance:For 4.7uHdeltaIL ~= 380mA = (VOut/(f*L)) * (1- Vout/Vin)IKpeak = Io + deltaIL/2 = 240mALQH32CN4R7M23L 4.7uH 450mAhttps://datasheet.lcsc.com/szlcsc/Murata-Electronics-LQH32CN4R7M23L_C86070.pdfPlease advise on this topic.Regards,up8051 aka JarekC.DIY

author: up8051
author_rank: Frequent Contributor
timestamp: March 12, 2020, 08:34:29 pm
content: Maybe another question:Which converter to choose for a big difference between VIn and Vout a small IOut?(Best available in JLCPCB)Vout 3.3V, Iout Max 50mA Iout Min 2mAVin Min 12V, Vin max 22VRegards,up8051

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 12, 2020, 08:46:06 pm
content: Why Ipk so many times higher than Io?Tim

author: Siwastaja
author_rank: Super Contributor
timestamp: March 12, 2020, 09:04:09 pm
content: For small I_out you'd generally scale dI_L down, as well, so that dI_L is some percentage of I_out, typically something around 30-50%. What this means is you need to scale the inductance up when you scale the output current down, which may seem counter-intuitive first. On the other hand, you can scale the inductor current rating down.But if you think about it closer, it makes sense. Current ripple is dI = V/L * dt, but energy stored in the inductor is E = 0.5*L*I^2, see the square of the current. This means, if you scale down the output current to 1/10th, and hence, scale dI_L down to 1/10th as well, and you want to scale energy stored in the inductor to 1/10th as well, L needs to go up by 10x!(For really robust design practice, youshouldselect the inductor saturation current based on thecurrent limitof the chip, not the calculated steady-state peak inductor current. This makes it rather difficult to obtain small output currents with the available ICs: you are limited in IC choice and may need to pick something with fixed, say, 2-3A worst-case current limit, when you'd actually want to design with a current limit somewhat above your actual steady-state peak inductor current. Luckily, the current limit circuits tend to be fast enough so that losing some inductance due to saturation doesn't automatically mean device destruction - as long as you still have some -, so you can escape this "robust" design requirement a bit. Additionally, "abusing" a larger current converter IC to do much smaller currents than it's designed for, now means you have excessively large MOSFETs compared to the increased parasitic resistance (mostly from the larger L), so it has better chances of surviving core saturation, resistively limiting current.)For really low-current converter, I would try to find the regulator IC with the lowest specified switch current limit! Some have adjustable current limits, but they tend to have much more pins anyway and may require other external components as well.

author: thm_w
author_rank: Super Contributor
timestamp: March 12, 2020, 10:10:29 pm
content: 2.2uH you've shown or 4.7uH should be fine as well.Quote from: up8051 on March 12, 2020, 08:34:29 pmThere is no best.AOZ1280 is rated to 26V so it should be OK unless you expect input spikes. In which case consider higher rated converter or overvoltage protection.

author: up8051
author_rank: Frequent Contributor
timestamp: March 12, 2020, 10:12:41 pm
content: I found something like this TPS62177 (Texas Instruments)http://www.ti.com/lit/ds/symlink/tps62177.pdfTheTPS6217x automatically enters power save mode at light loads ,to maintain high efficiency across whole load range.Output Current Limit 1A.and for a coupleMWSA0402S-100MT10uH, Isaturation =2Ahttps://datasheet.lcsc.com/szlcsc/1908071506_Sunlord-MWSA0402S-100MT_C408339.pdfWill it be a good choice?

author: thm_w
author_rank: Super Contributor
timestamp: March 12, 2020, 10:38:06 pm
content: Should be fine. Although the inductor datasheet does not provide us with too many details.It is your choice in terms of a cost/efficiency/features/etc trade-off. TPS62177 is about 4x the cost of AOZ1280, but much better efficiency at low load. The ripple may be higher if using the extreme power saving features.

author: Siwastaja
author_rank: Super Contributor
timestamp: March 13, 2020, 02:56:24 pm
content: Note that picking a regulator with "good" low-load efficiency and deliberately choosing the inductor based on the rated regulator current (i.e., for example, picking the value from the datasheet example circuit), results in a circuit optimized for that load current. At significantly lower actual output current, the circuit always operates in DCM, and is severely pulse-skipping all the time. So expect a lot of output ripple voltage.I would likely go for a large inductance (calculated for the actual load, not the regulator maximum) and hope (or calculate) that the MOSFETs are safe in corner cases even if the inductor is already in severe saturation at the specified switch current limit level. Remember, when you have more inductance to begin with, you can allow dropping it down to lower percentages and still have enough inductance left for the current limit to work. Say, for example, if the converter example circuit uses 1uH, it must mean they can react to output short with something like 0.7uH actual (because a saturation current of a 1uH inductor would be specified at 30% drop in inductance). Now if you use 10uH instead, you can still drop to that 0.7uH, or 93% drop in inductance, and the current limit should still be able to do the job! Higher DCR in larger inductance parts also helps limit dissipated energy in current limit event.