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LM2670: LM2670 - Inductor value for Iload less than 1 A

Renato Senador
Prodigy 10 points
Part Number: LM2670

I am planning to use LM2670 as a basic switching controller for different loads (from 400 mA to 2.5 A) using the same PCB layout. Of course, I will choice a Catch diode and an output capacitor for the worst case (Vout = 15 V @ Iout = 2,5 A). My question is: how to determine te inductor value for loads less than 1 A as long as, there is no nomograph for loads less than 1 A. Is the LM2670S suitable to light loads? Tks

  • 0
    TI__Guru 72035 points

    The inductor value is calculated for the worst case as the other components.  I would use 30% of your max load as ripple current in the calculation.

    The saturation current rating of the inductor must be around 5A or so, since that is the current limit for this device.

    This would be over-kill for the smaller loads, but will still work correclty at all loads.

    It would be best if you could lay out the PCB for different inductors, then you could optimize your solution size.

    Also, I would use the Webench design tool to help optimize the design.