The data sheet of the Buck Converter I plan to use gives a formula for calculating the inductor based on a ripple current that is 30% of the DC load current. I have 13V to 5V with 1A and plan to use the 420kHz MPQ4420H. Using 0.3A as ripple target, I will need a 25uH inductor. Is that really needed? Does the inductor need to be a 2A part?
Hi Jens.
A 25uH inductor will give you just shy of 30% ripple–a fine place to be. The rule of thumb of maintaining ~30-40% ripple current gives us a good balance between DC ripple at steady state and AC performance during load transients.
You should make sure the inductor you select has a saturation current greater than your expected max inductor current with some margin. With your setup and a 25uH inductor you should expect a max inductor current of 1.15A, so a 2A inductor should be sufficient.
For low currents you can easily go higher with the ripple current even into discontinuous mode.
The ripple current is an arbitrary point you optimize for your component cost, size and performance.
On low currents output capacitor ripple is still low for the MLCC used, so it is no issue.
BR
Christian
The 30% to 40% inductor ripple current (of average current) formula used in most data sheets, is a good start-point, if the converter IC is used in the higher current area of its capability. If a 2A or 3A converter IC, like MPQ4420H or MPQ4423H is used for a 0.5A or 1A application, the 30% or 40% rule based on this 2A or 3A can be used. The selected inductor has to be able to handle the converters peak current in case of a fault and maintain a certain rest inductivity at this peak current level. Here the output short circuit strategy of the converter is very important and has to be considered. (Hiccup, latch-OFF or full current) Typically, there are two limits that define the minimum or maximum inductor value: 1) a minimum di/dt ramp for current mode control, to provide enough signa; 2) a maximum di/dt slope, that is covered by the IC internal slope compansation.