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TL2575-05: TL2575 Inductor Dimensioning

Part Number: TL2575-05
Other Parts Discussed in Thread: LM2574, LM2674, LM2675, LM22675

Hello,

I'm considering to use a TL2575 for a application with the following values:

Vin = 24V

Vout = 5V

Iout = 375mA

From the Graph on the datasheet (Figure 14) it the inductor should be 680uH.

Can you please clarify if i only have to take into account the maximum current?

I'm asking because the 375mA is the peak current but it normal functioning it can be lower and as such i wanted to make sure that it would work properly( lets say if nominal Iout = 200mA or less ).

Thank you.

  • In general you will always want to select the inductor for you maximum output current.

    In fact, it is best to select for the current rating of the device; in this case 1A.

    It may be difficult to find such a large inductor in that current rating.

    You may want to look at the LM2574.  The current rating is lower and you may be 

    able to use a smaller inductor.

    Also, we have new devices that use faster switching frequency and they allow 

    a much smaller inductor.  This will save space and cost.

    You could look at the LM2675 or LM2674 family.  Or the LM22675 family.

  • Hello,

    So In this case, if i don't expect the current to be higher then 0.4A, the 680uH inductor will be enough ? and it is not expected to have any problem while working with smaller currents? 

    The other devices seem to be twice the price of the TL.

    In terms of thermal management do i need to take any special care ? I'm thinking on placing thermal vias underneath the GND tab of the package.

    Thank you.

  • The 680uH is OK for your currents.

    Using vias under the device to connect top and bottom grounds is a good idea.

    By sure to follow the PCB layout guidelines in the data sheet.

  • Closing this post