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.

over 5 years ago

0 Garrett Roecker over 5 years ago

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 S.Kanda over 5 years ago 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!

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LM5118: Inductor selection in buck-boost mode