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LM2674 inductor selection

Juan Cerrudo
Prodigy 180 points
Other Parts Discussed in Thread: LM2674

I'm designing a power supply with the LM2674;

Vin=12 V

Vout=5 V

Iout=100 mA

I'll be using the 5 V fixed output version. According to figure 24 in the datasheet I have to use a 150 uH inductor.  Then I used the WEBENCH design tool and the inductor value is 680uH (see attached file).

Wich one should I use? Am I doing something wrong with the WEBENCH design tool  ? 

Thanks!

1258.webench_design_1111956_10_335734933.pdf

  • 0
    TI__Genius 11085 points

    We've forwarded this question to the webench team to look into.

    You are correct that based on datasheet, 150uH inductor should be used.

    Yang

  • 0 in reply to YangZhang
    TI__Genius 11085 points

    Juan,

    I talked to Webench team and understand why there is a difference between the ds and calcuation.

    In Webench calculation, the target inductance is picked such that the peak-2-peak inductor current ripple is 30% of the maximum DC load current. Based on Vin = 12V, Vout = 5V, fs=260kHz, the calculated target inductance is 370uH. Since it is not a standard number, webench picked an standard number together with considerations of size and cost. The calculated peak2peak inductor current ripple is 75mA (75% with 100mA DC load) with 150uH inductor, and 16% with 680uH.

    The LM2674 is a voltage mode regulator. It is not sensitive to inductor ripple magnitude. It is ok to follow Webench design, since the entire solution is ready. It should also be stable with 150uH inductor. Output capacitance should be higher to reduce vout ripple is that case.

    regards,

    Yang 

  • 0 in reply to YangZhang
    Prodigy 180 points

    Great! thanks Yang for the detailed answer, you have been really helpful.

    Juan