AIN to 8-bit accuracy to reduce the leakage current

I refer to  Measuring Lithium battery voltage with nRF52 and the answer answer given by AI referring me to you.

I need the lithium battery only to about 8-bit accuracy to reduce the leakage current.  Please provide the recalculated values for R1, R2 and Cext, and the new reduced leakage current.

c-: 

To scale the resistor and capacitor values for 8-bit accuracy, we need to consider the trade-offs between resolution, power consumption, and sampling rate. However, the knowledge sources don't provide specific information on scaling these values for 8-bit accuracy with the nRF52.

The [Measuring Lithium battery voltage with nRF52](devzone.nordicsemi.com/.../measuring-lithium-battery-voltage-with-nrf52) blog post focuses on 12-bit accuracy and doesn't mention 8-bit configurations. Similarly, the [Measuring Lithium battery voltage with nRF51](devzone.nordicsemi.com/.../measuring-lithium-battery-voltage-with-nrf51) post provides values for nRF51, which may not be directly applicable to nRF52.

Without specific guidance from Nordic Semiconductor for 8-bit accuracy on nRF52, I cannot provide a definitive answer on how to scale these values. Scaling for 8-bit accuracy would likely involve adjusting the resistor values to provide a wider voltage range at the ADC input, but this would need to be carefully calculated to ensure it meets the specific requirements of the nRF52's ADC.

If you need precise values for 8-bit accuracy with nRF52, I recommend contacting Nordic Semiconductor's technical support or posting a question on their DevZone forum for expert advice tailored to your specific use case.

  • Hi,

    According to the blog, you should be able to use the same resistances of 4MΩ and 10MΩ respectively, or you could even use higher resistances like 8MΩ and 20MΩ. 

    From the blog, you can calculate Cext like:

    Cext = Csample * (2 ^ your bit resolution)

    So your Cext = Csample * 2^8 = 640pF. You can use any standard value, say 680pF.

    And your maximum leakage current should now be 4.2 / (8MΩ + 20MΩ) = 150nA

    Regards,
    Priyanka

     

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