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ADS1299: Calculation of Electrode Impedance using LOFF AC Current Source

A Paul
Prodigy 20 points
Part Number: ADS1299

I have a question about how electrode impedance should be calculated using the AC current sources configured by the LOFF register. 

(1) When using the 6nA at 31.2 Hz AC current configuration, is this signal expected to be centered at zero with peaks and troughs at +3nA and -3nA, respectively? Or, is the signal centered at +3nA, with peaks and troughs at +6nA and 0nA, respectively? 

(2) When calculating the impedance values, should I divide the output voltage by 6nA or 3nA? When I divide the output voltage (Vrms) by 6nA, I consistently get impedance values that are on half what is expected.

Thank you in advance for the clarification! 

  • 0
    TI__Guru 66466 points

    Hi A Paul,

    Thank you for your post and welcome to our forum!

    The lead-off current is generated from current sources on each input pin. For each pair of inputs per channel, one pin will source current while the other pin will sink current of the same magnitude. The AC lead-off feature uses the same current sources as the DC lead-off feature, but the direction of the current is swapped back and forth to produce a square wave. Therefore, the current flowing from INxP to INxN will either be +6 nA or -6 nA at any given moment.

    The impedance in the signal path between INxP and INxN will produce a voltage potential difference between the two pins in the presence of the lead-off current. The ADC will convert this voltage times the PGA gain. Simply divide the peak ADC output by the PGA gain and by the current magnitude to get the differential impedance for that channel.



    Best Regards,

  • 0 in reply to Ryan Andrews
    Prodigy 20 points

    Hi Andrew,

    Thank you very much for the reply. Just to make sure I understand, the chip produces a square-wave centered at 0 with peaks at +6nA and troughs at -6nA? When I divide by the current magnitude, is this Vpeak (6nA) or Vpeak-to-peak (12nA)?

    When I measure the impedance of known resistor, 10Mohm for example, the voltage output from the chip is a square wave from -30mV to 30mV. My impedance calculation, (30mV/6nA) gives me a 5Mohm impedance, half of what is expected. Can you think of a reason why the current magnitude would be 3nA, rather than 6nA? 

    Thank you! 

  • 0 in reply to A Paul
    TI__Guru 66466 points
    Hi A Paul,

    How are you converting the raw ADC output to volts? Can you share an example of the data you're capturing and perhaps a schematic as well?

    The current from INxP to INxN will be +6 nA in once half of the period, and -6 nA in the other half. Your calculation is correct. Let's check the code-to-volts conversion. Please review the Data Format section of the datasheet.

    Best Regards,