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LM25018: Questions of LM25018

HAN LEE
Expert 1515 points
Part Number: LM25018

We have some question about the LM25018 , could you please help it ? Thanks

1. IC is LM25018 , Question : According to the formula below ,where can I find the Lp (MAX.)   and  Lp(MIN.) value from datasheet?

2.Whether it need the minimum Ripple current ? Suggestion Ripple current value ?

3. Internal High side integrated 325mA. How many current is the Internal High side and Low side MOSFET can support (max value) ?

    Because of current will rise to 648mA when OCP.   We worry that MOSFET will damaged…

4.   △IL1 is the Buck side => What value is it ?  Vpeak-peak or Vrms or Vmax. ?

  • 0
    TI__Genius 11040 points
    Hello,

    Please see my responses below.

    1. I do not think that I fully understand your question. You can calculate the Lp (MAX) and Lp (MIN) based on your equations and your operating conditions - VIN, VOUT, IOUT, FSW. You can also see an example of how to choose a transformer on page 20 of the datasheet.

    2. As mentioned on page 20 of the datasheet, the maximum inductor primary ripple current can be calculated with equation 26 - which you call in your fourth question.

    3. The 325 mA is the continuous current which can be supplied and not enter current limit. The current limit will protect the device from seeing a current that it cannot support.

    4. You can calculate this value based on your design. This is buck inductor ripple current (A).

    Please let me know if you have any additional questions or if anything remains unclear.

    Best Regards,
    Katelyn Wiggenhorn
  • 0 in reply to Katelyn Wiggenhorn
    Expert 1515 points

    Hi Katelyn,

    For Q3 reply , how many current will trigger the CURRENT LIMIT ? How to calculate the OCP point ? Is trigger Low side MOSFET or High side MOSFET when OCP ?

    For Q4 reply , Calculate result value define => Vpeak-peak or Vrms or Vmax. ?

     

    Thanks.

  • 0 in reply to HAN LEE
    TI__Genius 11040 points
    Hi,

    For Q3, the minimum current limit is 390mA. This is a peak current limit, which means the peak inductor current will be limited. This will consequently lower the average current.

    For Q4, this is calculating the ripple current, so the result is in Amps peak-peak.

    Best Regards,
    Katelyn