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Original question:

Input EMI filter for step down converter

SNVA538: Question about result

Dominik Lorenz25
Prodigy 140 points
Other Parts Discussed in Thread: LM5005

Hello TI,

I have a short question about the SNVA538:

At page 10, the result of the mentioned optimization of the two stage filter is shown as:

L1 = L/2

L2 = 7*L1

Ld = L2/2

C2 = 4*C1

Rd = sqrt(L1/(4*C1))

Further below, at page 19, I found a example for this filter. But the formulas used for this are different:

L1 = L/4

C1 = C/4

L2 = 7*L1

C2 = 4*C1

Rd = sqrt(L1/C2)     --->   Rd = sqrt(L/(4*C1))

Ld = L1/8                   ---> Ld = (L2/7)/8 = L2 / 56

For me, there main difference is the big difference in the calculation of Ld by a factor of 28.

The difference in choosing of L1 is in my opinion not that serious, as it does not change the ratios of the components.

Could you please explain the differences i beetween these formulas? I would try to perform the mentioned optimization by myself, but the assumtions used for it are not stated in this paper (or at least I did not recognized them)

Thank you a lot,

Dominik Lorenz

  • 0
    TI__Genius 14455 points

    hi Dominik,

        We will get back to you on this? Can you please let us know the part number that you are trying to design with. For some of the products, WEBENCH will let you design the input EMI filter as well.

    regards,

    Gerold

  • 0 in reply to Gerold Dhanabalan - Online Design Tools
    Prodigy 140 points

    After the filter there are 3x lm5005, 1x tps65381a and a other load with approx 2A.

    So i guess the standard-tools do not work out for me.

    The paper is very highly ranked on google, so i thought maybe it is in your interest to correct a mistake (if there is one), too.

    regards,

    Dominik

  • 0 in reply to Dominik Lorenz25
    TI__Genius 14455 points

    Hi Dominik,

        We do not have WEBENCH input EMI filter for LM5005. 

    Regards,

    Gerold

  • 0
    TI__Guru 56825 points

    Dominick,

    After reviewing the paper, and the Mathcad results, I think the difference in the selection of the damping inductor between the optimal questions (Page 10) and the Example (page 19) may have been a mistake on the part of the author, rather than an intentional design choice.

    In figure 10, you can see that there is significant peaking, close to 6dB from the optimized 2-stage input filter with Ld = L2/2, but in the design example on page 20, you can see that with the much smaller Ld (L2/56) there is almost no peaking in the 2-stage input filter design.  The lower Ld allows the Rd to damp the filter more strongly because the Rd is a greater portion of the series resistance at the resonance frequency. 

    The example demonstrates that the circuits performance is not critically sensitive to the precise value of Ld, and lower values of Ld provide higher damping, though they also impose greater power loss.  Similarly with Cd in the parallel damping example, a smaller Cd capacitor will increase the damping factor generated by Rd, but will also increase it's power dissipation by increasing the ripple voltage on Cd and the ripple current through Rd.