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LM25010: LM25010

Part Number: LM25010

Hi, 

we are using LM25010SDX/NOPB switch mode regulator for 19 to 30 V input and output max of 300mA/5Volts. 

refer below our schematic. 

As per calculation we estimating the peak current in inductor  is 330mA. so we using 570mA inductor in our circuit.

we like to understand further, is this current rating of the inductor is sufficient ? and other devide around switching regulator is rated suffieicient  to handle our power requirement? 

looking forward you advice. thanks.

Best Regards, 

S.Suresh. 

  • Hi:

    1. LM25010 has 1.25A valley current limit , when 5V outptu short, the load current will higher than 1.25A, pls make sure that the inductor won't saturate at short conditon.

    2. LM25010 is COT control , it need at least 25mV ripple into the FB , so pls make sure that the output cap ESR generate enough ripple voltage by ripple current.

    3. L2 is bead, have to put a ceramic close to chip's Vin pin and GND

    Thanks

  • Hi Thanks for reply,

    in our circuit it will not expect any short circuit from the Switching regulator. this only for internal board supply. in this case How much factor i have consider on inductor current?

    output capacitor which we used have ESR of 1.21Ohms. is this enough ? or need additional resistor ?

    Best Regards,

    S.Suresh.

  • Hello Suresh,

    I recommend designing for worst-case (short circuit), but that is system level decision you can make.

    I good rule of thumb for saturation current of inductor would be 20% greater than peak inductor current in steady state, or equal to the typical peak current limit of the device.

    Quick start calculator is good reference for understanding componet selection and influence of output capacitor (below).

    In addition to reference on ripple injection (below).

    https://www.ti.com/tool/LM25010QUICK-CALC

    http://www.ti.com/lit/an/snva874/snva874.pdf