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WEBENCH® Tools/LM317A: LM317A - 24Vin - 17Vout , Load = 0.7A will it work?

Part Number: LM317A
Other Parts Discussed in Thread: LM317

Tool/software: WEBENCH® Design Tools

Hello,

I am trying to simulate the design on webench, for LM317A - 24Vin - 17Vout , Load = 0.7A. but I am not able to simulate it.

will the below design work properly?

please let me know the feedback as soon as possible.

  • Hi Swapnil,

    datasheet contains a formula for the calculation of programming resistors. From this formula you should choose R1=3k0 if using R3=240R.

    Have you read the datasheet regarding the input and output caps and the eventually needed protection diodes?

    (24V - 17V) x 700mA = 4.9W means that the regulator has lots of heat to dissipate. Which LM317 version do you intend to use? Which package? Do you intend to use a heat sink?

    Kai
  • I am using LM317AT/NOPB part.
    I can use Heat sink in my design
  • Hi Swapnil,

    TO-220 package is nice. A 10...15K/W heat sink should be ok. What is your ambient temperature?

    Kai
  • Hi Kai,

    Thanks for reply.

    Ambient Temp. will be near to 20C. it will be used in European countries.

    Regards,

    Swapnil.

  • Hi Swapnil,

    Like Kai suggested, you should follow Section 8.1's equation to choose the resistor values for the resistor divider. With the power requirement you provided, the LDO will burn 4.9W. With TO-220 package, the Junction-to-ambient thermal resistance of this package is 23.3C/W, the temperature increased from the power will be 114C. With the ambient temperature of 20C, the junction temperature of the device may exceed the max recommended operating temperature.

    You may need to consider lower your vin if application allows.

    Regards,
    Jason Song
  • Hi Jason,

    thermal resistance from junction to case is only 1.1°C/W for the LM317A in TO-220 package. So using a 10...15°C/W heat sink should allow him to keep the junction temperature well under 100°C.

    Kai
  • Hi Jason/Kai.

    Is it good practice to use 2 LDO in series.

    one will convert it from 24V to 20V and other will convert from 20V to 15V?

    output current will be max 0.5A

    can it be a better solution?

    Regards,

    Swapnil.

  • Hi Swapnil,

    with a load current of 0.5A a one-regulator-solution would only have to dissipate (24V-17V) x 0.5A = 3.5W. If you take a 10...15°C/W heat sink this is absolutely no a problem! You don't need to take two LDOs in series.

    Kai