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LMR36506-Q1: Inductor selection for very high input voltage use case

Atsushi Hirai
Expert 3870 points
Part Number: LMR36506-Q1
Other Parts Discussed in Thread: LMR36506

Hi team,

Now my customer is going to use LM36506-Q1 with 50-60V input.
In the case of this level of input voltage, the switching frequency is reduced to around 1.3MHz by frequency foldback function.
In this case, should the customer select the inductor for 1.3MHz switching? It means the inductor should be 22uH rather than 6.8uH which is suggested for 2.2MHz.

Or, 6.8uH or 10uH is okay?

Regards,

  • 0
    TI__Genius 13735 points

    you could use TI Webench tool to calculate the inductor peak-peak current, example as the 10uh may have 45%, 6.8uh will lead to 70% , you can choose which meet your requirement. 

    Elena

  • 0 in reply to Elena Gao
    TI__Expert 3870 points

    Elena-san,

    Thank you for your advice.

    Is it okay to interpret LMR36506 can be used with such high input voltage and 6.8uH if the inductor current peak is lower than current limit threshold?

    Thanks,

  • 0 in reply to Atsushi Hirai
    TI__Genius 13735 points

    hi, Hirai,

    may I know under what condition this frequency foldback happen, is that Vout is very low? please input your Vin=? Vout=? Iout=?.

    Elena

  • 0 in reply to Elena Gao
    TI__Expert 3870 points

    Hi Elena-san,

    You can see the foldback characteristics on the datasheet P40 Figure 52.

    Customer's condition is Vin=50~60V, Vout=5V, Iout=100mA~300mA(roughly).

    Regards,

  • +1 in reply to Atsushi Hirai
    TI__Genius 13735 points

    hi, Hirai-san,

    60V to 5V is designed to enter frequency foldback, because it will first trigger Ton_min, frequency will drop to 1.6Mhz,  then the light load 100mA will further decrease the frequency ,to what you see for 1.3M, this is right performance. If your application always work under this condition, 22uh is better. 

    to answer another question about current limit, if we choose min inductor value, like 6.8uH , that means 70% p-p ripple for 50~60Vin ,5Vout application, it doesn't touch high-side current limit, (0.87A as min value in datasheet,)   it will guarantee inductor current peak is lower than current limit. this is right.

    Elena.