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ADS1299 Impedance FFT Analysis

rafael cruz
Prodigy 150 points

Hello

I have a question regarding the interpretation of the FFT Analysis for impedance measurement, figure 45&46 of this data sheet:

http://www.ti.com/lit/ug/slau443/slau443.pdf

shows the analysis for impedance measurement at 31.25Hz, how can one tell by looking at this plot that a 5KOhm impedance is present? I understand the reason there's something present at 31.25Hz but not that magnitude.

figure 46 also shows the impedance component at 1KHz after changing the frequency and data rate to fdr/4 and 4Ksps but how does one extract the impedance value from this graph too?

Any help would be appreciate it!

thanks

thanks again

Rafael

  • 0
    TI__Guru 63196 points

    Hi Rafael,

    The peak-to-peak voltage produced by the lead-off current sources can be calculated from the FFT using the equations below:

    VIN (dB) = 20*log(VIN / VFS ), where VIN (dB) = -37 dB from FFT, VFS is the full-scale voltage = VREF = 4.5 V.

    VIN (V) = 4.5 V * 10(-37/20) = 63.6 mV

    This corresponds with the amplitude of the sine wave in Figure 44 of about 64 mV (128 mVPP). By knowing the magnitude of the current sources, you can then calculate the source impedance of each electrode:

    VPP = 2*(ILEADOFF*ZINP + ILEADOFF*ZINN) -> (ZINP + ZINN) = 128 mV / (2 * 6 uA) = approx. 10k

    Regards,