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LMQ66430: about Inductor selection

Part Number: LMQ66430
Other Parts Discussed in Thread: LMR54406, LMQ66410

Hello Team

I'd like to concfirm validity of inductor selection design.

Our circuit requirement

 Vin : 19V

 Vout : 5.15V

Iout : 150mA (max)

I calculated inductor inductance constant with eq.(7) in the datasheet (Revision A  Dec, 2022)

Do the parameters using calculation follow design guidline?

VIN=19V, VOUT=5.15V, Fsw=1.055MHz, K by Iout_max = 0.2

Then, calculation result is 17.8uH 

Also, are there any design guidelines for the current flowing into the Swichng Coil  immediately after power on?

*VIN and EN are used short-circuited.

Best Regard

Miyano Shinichi

 

  • Hi Shinichi

    The equation(7) uses the device Iomax, which is 3A to calculate the inductance. 

    You will need to select a inductor with current rate larger than the Ipk_max, which is 4.4A for LMQ66430. Considering all operate conditions including output-short-circuit.

    What's the end equipment and Is there any consideration that you select LMQ66430 ? since the Io needed is <0.2A will you take a look at the LMQ66410/LMR54406 ?

    Thanks and best regards.

  • Hello Gui 

    Thanks for your reply.

    If we changes tevice to LMQ66430, what Ipeak value should I consider? 

    The datasheet only listed LMQ66430 specifications.

    And I recognize that we choose Iout_max 1A and K=0.3 with Eq. (7).

    I have a new question related to this.Is it okay for the total value of Iout(board requirement ) and Ilipple/2 to exceed Iout_max(=1A)?

    *This question is caused another application using LMQ66430.

    Best Regard

    Miyano Shinichi

  • Hi Shinichi

    Do you mean change to LMQ66410 ? You can see the LMQ66410 spec in the LMQ664X0 datasheet

    The 1A Ioutmax is the average value.  The peak limit is 1.8A.  So the total value of Iout(board requirement ) and Ilipple/2  will not exceed the current limit.

    Thanks and best regards.