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TXU0304: Reg :- For part TXU0304RUTR required confirmation for externa pull-down Resistor

Kalaiyarasi Mani
Prodigy 150 points
Part Number: TXU0304


Hi Team,

From the datasheet the part TXU0304 input leakage current is 2uA with 5Mohm internal weak pull down. Here did not understand input leakage current of 2uA with 5M ohm pull down. So the Voltage across the pull down will be quite high (2uA * 5Mohm = 10V) ! and beyond the logic low threshold. Can you confirm whether the external 10K pull down is enough?

Regards,

Kalaiyarasi M

 

  • 0
    TI__Mastermind 46305 points

    Hey Kalaiyarasi,

    Is the 10K pull-down enough for what exactly?

  • 0
    Prodigy 150 points

    Hi Dylan,

    For the input pins.

    Regards,

    Kalaiyarasi M

  • 0 in reply to Kalaiyarasi Mani
    TI__Mastermind 46305 points

    Hi Kalaiyarasi,

    For the purpose of elimination floating nodes at the inputs, the 5 Mohm resistor should be enough. There would be no need to add an external one. 

    Also, to clear up some confusion the Ii spec will account for the pull-down current.

  • 0 in reply to Dylan Hubbard
    Prodigy 150 points

    Hi Dylan,

    As per the datasheet specification for the part TXU0304 input leakage current is 2uA with 5Mohm internal weak pull down. So if we calculate (2uA X 5M ohm =10V) the voltage at voltage at the input is 10v, the10V is not expected the input with 5M pull down. The input value expected should be less than VIL max. So there is something wrong in the specification. Can you please check and update the what is the maximum leakage current into 5Mohm pull down and what is maximum VIL?

    Regards,

    Kalaiyarasi M

  • 0 in reply to Kalaiyarasi Mani
    TI__Mastermind 46305 points

    Hey Kalaiyarasi,

    There isn't anything wrong with the spec or how its represented, I believe the problem comes from some confusion on your end.

    Your understanding of the Ohms law equation is correct, but your application of it in this scenario is incorrect. Your equation assumes you have an ideal current source (driving 2 uA) in parallel with the pull-down,  however that won't be the case. If you want to find the maximum current through the resistor, you use Ohms law in this way: I = Vin(max)/ 5 Mohm which comes out to be ~ 1 uA. As I stated previously, this 1 uA is accounted for in the spec due to it being a current measurement when a voltage is applied to the input as listed in the conditions.