MP9488 Design review

DC-DC Power Converters
1

I could use some help with some of the component selection for an MP9488 DC-DC design. Schematic is attached.
Input is from Large solar modules anywhere from 30V to 250V DC. I would like this supply to output 24V @ 300mA (max of MP9488?).
Specifically, the Inductor choice of 470uH, the 0.47uF sampling cap, the RS1J diode, as well as my FB voltage divider total. Or if there’s other things that could be optimized or changed, please let me know.
PDF Schematic

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2

Hello tim1,

You seem to have followed the circuit shown in the datasheet for the part.

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Your resistor values seem to agree with the desired output voltage. The diode specs you chose seem to be close to those on the example circuit. I am not sure what you are looking for in the inductor.

I am not sure how the inductor fits into your application but changing the size seems to also change the switching frequency.

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For the sampling cap the equations in the datasheet also provide guidance for sizing.

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3

eduardo,
I still have a few questions;
let’s start with the easy one; the sampling cap equation, does my chosen cap value meet that equation? what are the consequences if i go higher or lower?

you say the inductor changes the switching frequency, i have a wide input range, what range of frequencies should i target? also could you show the other parameters for that equation? how do i calculate Ipeak?

what i am looking for in the inductor is the other parameters besides inductance to be able to choose a part to purchase. if the MP9488 can source 300mA then what peak current and/or saturation current should the inductor have? any other parameters to be aware of for selecting a part?

please advise.

4

Hello tim1,

I highly recommend visiting the page I have linked below. This explains the working principles of a buck converter. It explains where the values come from and how it is used to size the inductor.

How to Calculate a Buck Converter’s Inductance | Article | MPS

To provide a brief overview you calculate pk-pk current as percentage of output current as discussed on the page I linked which you decide. From this Ipeak = (Ipk-pk/2)+Iout .