This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

ADS1293: AC-Lead Off Electrode Disconnect Detection

Part Number: ADS1293

Hi TI Experts, 

I am using the ADS1293 dev board and its digital AC-Lead off detection feature to measure the skin-to-electrode impedance and to also determine when leads are connected/disconnected.   I was able to measure the skin-to-electrode impedance (tested using resistors and on body) and the results were accurate when looking at the FFT. 

The problem I am having is how do I determine to determine the electrode connectivity ( connected or disconnected) ?  When they are disconnected I would expect that signal amplitude saturates and I should get 0.4 V , but instead I get 0V.  Also an alarm goes off for registry 0x19 bit 4 : Channel 1 out-of-range error. Is this what should happen when their is an open circuit  I get an excitation signal of 0V?    

 I read through the application notes and in section 4.3 of this document:  http://www.ti.com/lit/an/sbaa196a/sbaa196a.pdf  It shows that when their is an open circuit that the excitation signal is expected to be high.  I also read through https://www.planetanalog.com/signal-chain-basics-149-how-to-accurately-measure-electrode-impedance-for-lead-off-detection-in-ecg-systems/# and didn't find anything that helped. 

Thanks in advance for the help :) 

  • Hello Gurjant, 

    Thank you for your post and please excuse the delay.

    Can you share a schematic of how the ADS1293 inputs are configured? I understand you are using our ADS1293EVM, but please show how you've connected and disconnected the external resistors from the inputs. Also, a complete register map table with the settings for the test would help as well.

    The lead-off detection circuitry in the ADS1293 operates differently than the other devices in the ADS129x family. All inputs are monitored using lead-off current that flows in the same direction. Therefore, if both input pins of a given channel become disconnected, they will both be pulled to the same supply voltage. Each input will indicate "out of range", while differentially the channel should read 0 V.

    Generally, the AC lead-off detection methods are to monitor the electrode contact quality. You may analyze the magnitude of the voltage produced by the excitation current by performing an FFT on the raw data.

    Best regards,