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Original question:

[FAQ] Getting Started on TI E2E™︎ design support forums

LM3488: How to design 5V, 2A output on Rf1, Rf2

Tseng Ruihong
Prodigy 100 points
Part Number: LM3488

Hi Supporter

I am confuse on LM3488 Rf1 Rf2

On data sheet, your suggestion on Rf2, Rf1 is 60K and 20K, the ratio is 3. However, in webench, the Rf2, Rf1 is 2.94K and 1K, the ratio is 2.94.

my circuit parameter is 3.7V input. 

Please clarify tell me, ratio is 3 or 2.94 and why?

Thanks,

Tseng

  • 0
    TI__Guru 61135 points

    Hello Tseng,

    The feedback voltage Vfb of LM3488 is 1.26V.

    Vout / Vfb = (Rfb1+Rfb2) / Rfb2

    In the datasheet, Vout / Vfb = (60 kohm + 20 kohm) / 20 kohm = 4. Therefore the output voltage is slightly higher than 5V (0.8%): 1.26V * 4 = 5.04V.

    In the Webench design, Vout / Vfb = (2.94 kohm + 1 kohm) / 1 kohm = 3.94. Therefore the output voltage is slightly lower than 5V (0.712%): 4.944V.

    Whichever you prefer is up to your system needs.
    Webench calculates a lot different versions and takes the one that is closest to the target voltage even if it is just slightly closer as in the above example.