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ADS1299 - electrode impedance measurements and electrical circuit for lead off configuration

leo casal
Prodigy 90 points
Other Parts Discussed in Thread: ADS1299

I'm implementing impedance measurement and have some questions. I have read many posts and documentation but still coudn't get around some issues. Specifically, at Figure 48 from ADS1299 datasheet, I need to know which switches are closed for different lead_off configurations.

1) The selected current magnitude, corresponds to mean value at DC excitation and peak-to-peak at AC? I can't find this info on the datasheet.

2) At the figure below, Config 1 and 2 correspond to which models?

3) The example at page 38 of SLAU443, corresponds to which configuration for LOFF_SENSP, LOFF_SENSP, and which model A,B,C or D?

4) The model E is supposed to be the circuit used in the mentioned example. The documantation says 120 mVpp = 2 x (6nApp x 5kohm + 6nApp x 5kohm). which is the return path for excitation current? why does it multiplies by 2?

Thanks in advance

Leo

  • 0
    TI__Mastermind 38745 points
    Hey Leo,

    Great questions! The current magnitudes listed in the datasheet refer to DC values. The "AC current sources" are produced by hard switching back and forth between them. This means that the AC current magnitudes are peak currents rather than peak-to-peak values. For simplicity, you may model them as a single alternating current source.

    Your CONFIG1 and CONFIG2 setups correspond to models A and B respectively.

    The documentation does not say specifically which configuration it corresponds to, but I think your model B would produce such results.

    I think model E might produce unexpected results since there is no sink for that current. Ideally if you only wanted to use a single current source for AC lead off detection, you would also have the output of the bias amplifier connected to the bias electrode. The amplifier will sink the current from the lead-off detection source in that case. The reason that the expression on page 38 is doubled is because I_LEADOFF refers to a peak current magnitude and the resulting voltage is a peak-to-peak voltage. Refer to my answer in your first question.

    Regards,
    Brian Pisani