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TIDA-01512: How to calculate the LSET resistor value

Sai Mukhesh
Intellectual 390 points
Part Number: TIDA-01512
Other Parts Discussed in Thread: CSD95490Q5MC

Hi TI team,

I am rightly working with LS2088A and I am taking the reference design from TI to generate the Core and DDR rails. I have a doubt regarding the LSET value resistor. So can you please share me the formula for getting the LSET resistor value. 

In reference design the value of the inductor is 217nH and the LSET was 162K, but in BOM 300nH is been mentioned. So, I request you to suggest me whether it is fine to proceed with the 300nH inductor and if it is 300nH what is the LSET value that has to be connected.

  • 0
    TI__Mastermind 24075 points
    HI Sai,

    We will have our FET team to reply to you.

    Thanks
    Qian
  • 0 in reply to Qian Chen
    TI__Prodigy 510 points

    Hi Sai,

    The equation for calculating LSET is R[kOhm] = 2.04e-9*L^4 - 2.48e-6*L^3 + 1.79e-3*L^2 +3.99e-1*L + 1.61e1 where L is specified in nH.  For 300nH this equates to a resistor value of 249kOhm (assuming you are using 1% tolerance resistors).    

    Please consider requesting the full datasheet for the CSD95490Q5MC on the TI.com at the link below where this information and more can be found about this product.

    Regards,

    Evan

  • 0 in reply to Evan Reutzel
    Intellectual 390 points

    Hi TI team,

    I was not able to get the value of 249Kohm with the formula shared. Can you please provide in a little bit clear way, Because I might have done some calculation error(BODMAS). I was implementing the circuit with an LSET value of 243K. So kindly let me know whether there will be any big impact between these 2 resistor value, because we already have this resistor.

  • 0 in reply to Sai Mukhesh
    TI__Prodigy 510 points

    Hi Sai,

    I believe the formula gives you a value of 246K.  I was just converting that to the closest standard resistor value when I suggested 249K.  243K should work fine as well.  The formula gives you the ideal value, but the circuit has enough range to cover resistor tolerance.

    Regards,

    Evan