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| Step-Down AOZ1280CI - inductor selection |
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| up8051:
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.pdf Vin max 16V Vout 3.3V Iout 50mA From calculations: deltaIL ~= 800mA IKpeak = Io + deltaIL/2 = 450mA I 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 SMD0805 https://psearch.en.murata.com/inductor/product/LQM21PN2R2NGC%23.pdf Is this type of choke (case 0805 & 0.8A ?) even suitable for this type of converter? Or rather choose: MWSA0402-2R2MT https://datasheet.lcsc.com/szlcsc/Sunlord-MWSA0402-2R2MT_C169184.pdf or MPH201210S2R2MT Isat=0.6A T=0.2Ohm https://datasheet.lcsc.com/szlcsc/Sunlord-MPH201210S2R2MT C55755.pdf Maybe I should choose a choke with higher inductance: For 4.7uH deltaIL ~= 380mA = (VOut/(f*L)) * (1- Vout/Vin) IKpeak = Io + deltaIL/2 = 240mA LQH32CN4R7M23L 4.7uH 450mA https://datasheet.lcsc.com/szlcsc/Murata-Electronics-LQH32CN4R7M23L_C86070.pdf Please advise on this topic. Regards, up8051 aka JarekC.DIY |
| up8051:
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 2mA Vin Min 12V, Vin max 22V Regards, up8051 |
| T3sl4co1l:
Why Ipk so many times higher than Io? Tim |
| Siwastaja:
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, you should select the inductor saturation current based on the current limit of 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. |
| thm_w:
2.2uH you've shown or 4.7uH should be fine as well. --- Quote from: up8051 on March 12, 2020, 08:34:29 pm ---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 2mA Vin Min 12V, Vin max 22V --- End quote --- There 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. |
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