• State TI Thinks Resolved
  • Locked Locked
  • Replies 2 replies
  • Answers 2 answers
  • Subscribers 63 subscribers
  • Views 443 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

LM2841: Value of calculated inductor?

Michele Bertoni1
Intellectual 340 points
Part Number: LM2841

Hi to all, I was taking a look to LM2841 datasheet where is reported example schematic:

- Vin = 4,5 to 42 Volt

- Vout = 3,3 Volt

- Iout = 0,1 Amp

In this case 10uH inductor has been used.

About determining inductance: "the maximum input voltage is always used to determine the
inductance"
Assuming we are using Y option, nominal Fsw is 1,25MHz, while for X option is 550kHz: in first case nominal inductor value is 23 to 81uH (minimum voltage, maximum voltage), in last case is 53 to 184uH.

So, why so different from 10uH in example?

  • 0
    TI__Expert 6860 points
    In the low-power step-down converters, ripple current can be set to higher percent :

    1. Lower power means that the current-limit set-point difference is usually either 0.1233A (Ipeak at Vin = 12V typical with L=100uH ) or ~0.25A (Ipeak at Vin = 12V typical with L=15uH). There is not as significant a jump with the increase in current ripple.

    2. The smaller size of low-power converters necessitates a higher switching frequency. Combined with advances in ceramic-capacitor technology, there is rarely significant ESR in the output capacitance. Instead, a small amount of ceramic capacitance can reliably handle larger ripple currents while delivering very-low-output voltage ripple.

    ---Ambreesh
  • 0
    TI__Genius 11085 points
    I'd suggest to use the rated current of the IC to calculate the inductance, instead of the needed Iout. The IC is designed and tested with component values to meet the rated current requirement. It is ok to work at lower load current. This part is a current mode controlled part, it needs certain amount of ripple current for proper signal to noise ratio. With very large inductance, the ripple could be too small.