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LM5118: Inductor selection in buck-boost mode

S.Kanda
Prodigy 20 points
Part Number: LM5118

Hi.  I have planed to use  LM5118-Q1 as buck-boost converter for  automotive application .

The requirement is Input  16 to 36[V],Output   24[V]/2.5[A] .

Now, I have a question about Inductor selection.

There is a formula (16)  In LM5118's datasheet  page 27.

I want to know why the 1st term of right side is multiplied by factor "(Vout + Vin(min))/Vin(min) ".

This factor makes an increase of current capacity request much more.

  • 0
    TI__Mastermind 21720 points

    Hello user1616626,

    Thanks for reaching out with your question and for considering the LM5118.

    In buck-boost mode when (both the switches are on) the duty cycle is (VOUT)/(VIN+VOUT). The average input current is (VOUT*IOUT)/VIN. To get the average inductor current the average input current is divided by the duty cycle. This how the term is derived.

    Please let me know if you have any questions

    Thanks,

    Garrett

  • 0 in reply to Garrett Roecker
    Prodigy 20 points

    Hi. Garrett,

    Thank you for  your answer. I understand the derivation.

    I had a misunderstanding that the relation of Vin and Vout is "Vout= 1/(1-D)*Vin"(i.e. same as boost converter). 

    Thank you very much!