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LM5008: Use the 3.3V GPIO to enableLM5008

Morris Chen82
Prodigy 50 points
Part Number: LM5008

Hi:

In the LM5008 datasheet, the pin6 RON/SD is with a pulled-high resistance to VIN and makes LM5008 output voltage.

Can I use a MCU GPIO 3.3V to pull high the pin6 RON/SD and enable the LM5008 and pull low to disable LM5008?

Thanks

  • 0
    TI__Mastermind 25590 points
    Morris,

    Ron sets the on time so it must be connected through a resistor to VIN for proper regulation.

    You can have Ron connected to VIN, then connect the GPIO directly to the pin. Set the GPIO to 0V to disable the device. Set the GPIO to high-Z to enable the device. The Ron/SD pin should be between 1.5V and 3V so your 3.3V MCU should not have an issue with overvoltage.

    You could also add a transistor controlled by the MCU to pull this pin down to disable the device.

    -Sam
  • 0 in reply to Samuel Jaffe
    Prodigy 50 points

    Sam:

    If I add a resistor(249K ohm) between the GPIO and Ron/SD pin instead of a resistor connection to VIN.

    Please see the below picture.

    When GPIO is pulled high to 3.3V, LM5008 is active. When GPIO is pulled low to 0V, LM5008 is disable.

    Does it work?

    Thanks

  • 0 in reply to Morris Chen82
    TI__Mastermind 25590 points
    Morris,

    I cannot see the picture but I don't believe this will work. The on-time is proportional to the Ron resistor and the input voltage. A constant 3.3V on the Ron/SD pin will create a constant Ton which will not behave as the part was designed.

    -Sam