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SLVA882

Hi Carmen,

I would like to calculate the ripple current for the output capacitors by using Equation 2.

But I can’t find more detail from section 5 (Multiphase Design Example - Component Selection)

Could you teach me how to use this formula at the two-phase buck converter?

Thank you.

Vin=12.5V, Vout =0.82V, Iout = 45A

Fsw = 600k Hz

N= 2 (two-phase)

D= 0.82V/12.5V = 0.0656

m = floor (2x 0.656) =0

The output ripple current = ICOUTnom ?! (r.m.s)

  • HI Nestor,

    We will have someone to support you.

    Thanks
    Qian
  • Hi Nestor,

    You seem to have everything you need to calculate the output ripple current. Just use n, m, and D to get the normalized value and then multiply by your max output current for the worst case current. Using your numbers I'm getting the normalized ripple current to be 0.927 which makes sense because D is so low (Figure 6) there isn't a lot of cancellation between phases. With a max current of 45A this gives an current through the caps of 41.7Arms. 

    Please note, if you're going through these calculations to determine the output capacitance you'll need picking a number to handle just the RMS current alone isn't enough. You'll likely underestimate how much is actually needed. Typically transient response dominates your performance and so going through the steps in Section 5.5 will give you a much more realistic picture of how much COUT you'll actually need.

    Cheers,

    Carmen

  • Hi Carmen,

    Thanks for your reminder and reply.

    According to equation 2, we can estimate the current through the output capacitors is 41.7Arms.

    But I was confused, why the calculation result is some big.
    As I know, the ripple current of the output capacitor should be similar to the inductor current.

    https://www.richtek.com/Design%20Support/Technical%20Document/~/media/AN%20PDF/AN041_EN.ashx(Figure 3)

    Or I misunderstand the meaning of this formula.

    I try to use the other formula to calculate the output ripple current of the capacitor and it looks like only 7.29Arms. (mic2155@page21)

    http://ww1.microchip.com/downloads/en/DeviceDoc/mic2155.pdf

    Btw, I also use webench to do the rough simulation.

    The “Cout IRMS” isn’t big even the inductor or switch frequency is a little different.

    Could you explain more about the calculate result and how to use it to select the output capacitor? (permissible ripple current?!)

    Thank you.

  • Hey Nestor,

    You're entirely right about the current through the caps. I misunderstood what you were asking but also apparently mislabeled the equation in my app note. It should should read ICOUT,ripple,norm for Equation 2 to make that point more clear. I did get it right in Figure 6 though at least :).

    For your application, since the duty cycle is pretty small there isn't a lot of ripple cancellation with only two phase so that 0.927 gets multiplied by the inductor ripple current for a single phase, not the output current like I said originally. I'm getting ~4.7App for one phase so you'd see ~4.4App through the output caps. I was also incorrect about this being an RMS value too which is why it doesn't match the sims.

    My one point does still stand though, be sure to check transient response too to get a clear picture of COUT. 4.4A of ripple current is pretty easy to handle so dividing the rating for your chosen cap by 4.4 will give a very low estimate.

    This seems to put us in pretty good agreement with the other app notes now. I've got it on my plate to update this app note anyways, ideally this month, to fix some other typos that have been caught so I'll add this to the list.

    Sorry about that confusion. I appreciate you pointing out the error though!

    Cheers,

    Carmen
  • You can also check my work against the source app note I used here.

    www.ti.com/.../slyt449.pdf

    Carmen
  • Hi Carmen, 

    Thank very much for your clarification and sharing more information with me. ^^