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I am looking at SLTA055 by Texas Instruments. It discusses, among other items, how to determine the bulk capacitance required for a buck regulator if a LC filter is used. Since the "L" will not allow the high frequency current draw to come from the PSU, it needs to come on the buck side of the L.

Two things on this appnote that are vague:

  1. There is zero mention about where this equation comes from, or how it is derived. I have searched everywhere online, and can't find this anywhere else. Any idea where they got it from? Itr is the max transient current draw, L is the series inductor value, and delta V is the max voltage sag. List item
  2. Immediately above equation 9, they also say

The next step is to decide if a series filter inductor is going to be used. If using an inductor, pick a value no greater than 560 nH.

That also confuses me, as using equation 9, you could in theory use as large of an "L" as you want. You will just need to use large capacitors that wouldn't be practical. Maybe this is just a rule of thumb, that they don't mention as such?

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    \$\begingroup\$ look here fig4 eetimes.com/… \$\endgroup\$
    – G36
    Commented May 10, 2021 at 12:52
  • \$\begingroup\$ @G36 Well that was easy! Thanks! \$\endgroup\$ Commented May 10, 2021 at 12:59

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Practically speaking, you could in theory rely on an equation giving the voltage ripple on the front-end capacitor but it does not work like that. What you need to care about truly for the capacitor selection is the rms current in worst-case conditions and the acceptable peak-to-peak ripple current drawn from the power supply unit (PSU). Yes, this input capacitor is to be combined with an inductor to form the front-end \$LC\$ filter.

The design equations are shown below and start with the capacitor rms current calculations. They are coming from the book I published on switching power converters:

enter image description here

These equations give the rms current in the capacitor for this particular application (a 5-V/5-A buck converter). Here, you see that a 2.5-A rms current will flow in the cap. If you are looking for a compact design, you may parallel several MLCC types like those from AVX for which they specify an ESR and a ripple current at 85 °C:

enter image description here

Here, the allowed current is 415 mA per capacitor so I will parallel 7 of them with a bit of margin to form a capacitance of 154 µF and a total ESR of 57 mohms. Once we have the ESR value on hand, we can determine the front-end inductance and find a value of 2.4 µH. In this example, the maximum dc input current will be 2.5 A so the selection of the inductor must account for this value to minimize insertion losses. The peak-to-peak voltage at the buck converter input will mainly depend on the ESR and weakly from the capacitance. That is why calculating the capacitance is a nice theoretical exercise but practically speaking has no value since rms capability at a given operating temperature is what matters in the end.

Finally, a quick simulation of this buck in current mode will tell us if this is correct:

enter image description here

When this is done, you need to damp the input filter to limit its interaction with the negative incremental resistance of the loading converter. You can find plenty of information regarding this extra mile to run in my APEC 2017 seminar. I proposed an alternate method to determine the \$LC\$ filter and it is based on the fundamental attenuation. Don't skip damping or you may run into stability problems later on. Beside the seminar which is kind of heavy lifting, you may want to look at the condensed explanation I gave here.

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  • \$\begingroup\$ Thanks @Verbal Kint! I have that book right next to me. What page is this on? I'll check it out. \$\endgroup\$ Commented May 11, 2021 at 10:38
  • \$\begingroup\$ Though, I do think you are misunderstanding the capacitor values I am asking about. Not the input caps right at the input of the buck. It's the bulk caps to recharge those caps. In this scenario here: i.imgur.com/XPfGAbj.png With a series inductor to the left of the bulk cap I highlighted. That inductor forms a LC with the bulk caps. The inductor forces the current to come not from the PSU, but the bulk caps on the bulk side of the L. \$\endgroup\$ Commented May 11, 2021 at 10:45
  • \$\begingroup\$ The calculations of the rms currents for the buck converter are detailed page 41 of the green second edition book. The procedure I described applies to the local front-end capacitors, assuming that all the high-frequency current circulates in these caps and less at steady-state goes in the bulkier one. The other thing is that a transient with a given slope undergone by one of the outputs will not translate with the same slope on the input as regulation and crossover in particular enter the picture. \$\endgroup\$ Commented May 11, 2021 at 12:04

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