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TIDA-010057: Calculation condition for inductor values

Satoshi
Guru 19645 points
Part Number: TIDA-010057

I am calculating the Inductance value with reference to page 14 of the tiduep0 document, but it does not match the listed value.

https://www.ti.com/jp/lit/pdf/tiduep0

Please let me know the various coefficients to make L1>70.1μF and L2>1020.8μH.

I am referring to the values listed in the document below, but are there any corrections that need to be made?

⇒Vin_min: 4.25V, Vout: 80V, Vd: 0.7V, Iout: 30mA, R: 0.4, fsw: 250kHz, Duty: 0.95, Pout: 2.4W

Best regards,

Satoshi

  • 0
    TI__Genius 12961 points

    I am attaching a calculator which can be used for the calculations, let me know if it is useful.

    Regards,

    Sanjay R. PithadiaSEPIC Power Supply Design Calculator.xlsx

  • 0 in reply to Sanjay Pithadia
    Guru 19645 points

    Pithadia

    Thank you for reply,

    I entered numerical values into the calculation tool assuming various conditions (Min/Max, etc.), but I could not arrive at the values of L1=70.1μF and　L2=1020.8μH.

    Would you please let me know what values are set for the following in the tiduep0 document to derive L1=70.1μF and L2=1020.8μH?

    ⇒Vin_min, Vout, Vd, Iout, R, fsw, (Duty, Pout)

    Best regards,

    Satoshi

  • 0 in reply to Satoshi
    TI__Genius 12961 points

    Hello, I am checking with the designer who designed this design.

  • 0
    TI__Expert 6980 points

    Hi Satoshi-san,

    I am sorry for so much delay in response. I looked at your question and here is the calculation and explanation for the chosen values of primary & secondary inductors.

    Vin  - 4.25 - 5.5 V 

    Vout max - 80V

    Iout  = 25 mA (50 mA for 2 rails)

    Fsw - 250 kHz

    Lpri (min) = Vin^2 x D / (Ripple factor x Fsw x Vout x Iout)  (Ripple factor is 0.4, D is 0.95)

    Substituting these values, we get:

    Lpri (Vin, min) = 43uH and Lpri (Vin, max) = 71.8uH

    Hence for the entire input range, we should use 71.8 uH. Also, note that this calculated value is the minimum inductance, hence in the design we have used commonly available 100 uH inductors.

    Similarly for the secondary inductors, we have

    Lsec = (1-D)xVout^2/(Ripple factor x Fsw x Vout x Iout) 

    Lsec (Dmax) = 797 uH and Lsec (Dmin) = 1017 uH. Hence selected value of Lsec is commonly available 1000 uH.

    I hope this answers your question.

    Thanks & regards,

    Abhishek