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ADS8688A: Input leakage current

Renato Rocha
Prodigy 30 points
Part Number: ADS8688A

On the datasheet, there are some formulas to calculate the input leakage current (7.5 Electrical Characteristics, page 6). 

The datasheet claims that the unit for the result value on the leakage current formula is in µA.

Considering Rin = 1MΩ, input range, 2.5xVref, Vin = 0V, we would have: -2.5/1000000 = -0.0000025 µA =  -2.5 pA. 

Should the leakage current be on the level of some pA or is it too low?

  • +1
    TI__Mastermind 45715 points

    Hi Renato,

    The unit in the table should be "A" instead of "uA", the input current under the condition you specified is -2.5uA. We will fix it when we have a chance to update the datasheet. Thank you for pointing it out.

    Best regards,

    Dale

  • 0 in reply to Dale Li
    Prodigy 30 points

    Hi Dale Li,

    Thank you for your prompt answer! Just one more question: In case I have a circuit connected to the ADC input and it has some pulldown resistance (i.e. 10kΩ), should I still consider Rin = 1M as defined on the datasheet or I should consider the equivalent input resistance (1MΩ in parallel with 10kΩ)? Thank you again for your attention.

    Best Regards,

    Renato

  • 0 in reply to Renato Rocha
    TI__Mastermind 45715 points

    Hi Renato,

    When you calculating the input current by using the equation in the table, a pull-down resistor does not have an impact because the Vin is the signal voltage applied to the ADC's input pin (AIN_xP). However, a resistor (Rflt) in series with the AIN_xP should be considered if you are using an output voltage (Vout) from your front-end circuity to calculate the current instead of the Vin in the equation.

    Best regards,

    Dale