ADS1298: Electrode Impedance Measurement

Hi Tommy,

Thanks for your post and welcome to our forum.

Could you please share the register settings for the device and a schematic showing the circuitry at the channel inputs?

At first thought, if the DC is correct but the AC is not, it may be that you have some input RC filters which are attenuating the voltage formed by the AC lead-off current and any resistance in the signal path. This current is formed by the same resistor string as the DC current - only difference is that the current is swapped from one input to the other, producing equal current magnitudes in either direction.

Best Regards,

Here are the register settings.

If it's not listed, it is not being updated from default. Assumes AC Leadoff mode with "All P" config, RLD enabled.

#define AFE_SETT_CFG1_1298    0x06    // value for CFG1 on ADS1298

#define AFE_SETT_CFG2_1298    0x03    // value for CFG2 on ADS1298

#define AFE_SETT_CFG3            0xEE    // value for CFG3

#define AFE_SETT_LOFF            0x11    // value for lead off settings (enable AC lead-off)

#define AFE_SETT_LOFFSENSP    0xFF    // value for lead off sense positive select

#define AFE_SETT_LOFFSENSN    0x00    // value for lead off sense negative select

#define AFE_SETT_RLDSENSP    0xFF    // value for RLD sense positive select

#define AFE_SETT_RLDSENSN    0xFF    // value for RLD sense negative select

#define AFE_SETT_CFG4            0x02    // value for CFG4

#define AFE_SETT_CH_X        0x00    // value for active electrode channel

I set the AFE to enable AC Lead-off for only the 8 positive amplifier inputs.  To measure the electrode impedance for the negative inputs, I enabled only channel 8 since all of the negative inputs were bussed together.

Here is the schematic with the components around the input to the AFE.  I have changed the 22.1k and 10k resistors in series with each input to 3.01k resistors.   All of the EMI filter caps are 47pF.  The negative inputs are bussed together and connected to EAFE_REF which is wired to pin 1.  I did consider the possibility of interference from the 47pF capacitors on the inputs.  I modeled the input and found that once the connection was made to RLD the low pass filter formed by the 10M and 47pF was eliminated.  

ADS1298_Input Schematic 1.pdf

Hi Tommy,

Thank you for providing more details.

Another consideration to consider is the attenuation from the third-order digital sinc filter. At fDR / 4, the sinc filter response will reach approximately -2.75 dB, which equates to a gain of 0.7286. This certainly doesn't explain all of the attenuation you're seeing, but it's worth considering as well.

You said that all electrodes are tied together, including RLD. Therefore, if the DC common-mode of RLDOUT is set for mid-supply (i.e. 0 V), then I believe the expected current in each electrode will be approximately ((2.5 V - 0 V) / 10 M), ignoring the patient protection for now.

Have you tried enabling only one positive and one negative channel input for AC lead-off detection? I'm wondering how the imbalance might affect the results of each channel. Ideally, the current flowing from one INxP pin returns to the negative supply through the corresponding INxN pin. However, with 8 positive pins enabled and only one negative pin enabled, the extra lead-off current will have to return through the RLD amplifier. If it's unable to sink that much current, you'll notice the voltage at the RLDOUT pin deviate from the configured mid-supply setting.

Best Regards,

Ryan,

Thanks for getting back to me.  

I had not figured in the sinc filter attenuation.  

When I configure the system, I exclusively activate the lead-ff for all 8 INP then for the single IN8N.   So, in one case the excitation current flows from all 8 INxP channesl through the 6.02k patient protection resistors to the RLD input to common.  When I perform this with DC lead-off, I measure ~250 nA for each INxP channel and ~2 uA for the RLD input.  When I enable only the IN8N DC lead-off, I get ~250 nA for the INN Ch8 and RLD.   It appears that the RLD amplifier can handle the DC current.  We are sampling at 250 Hz so the excitation current squarewave frequency is 62.5Hz.  With such a small current at a low frequency, the RLD amplifier should be okay.  

I will repeat the INN Ch8 current measurement with AC lead-off to verify my results.  I will also repeat the AC lead-off with all 8 INxP channels enabled and will check the RLD input to see if it remains near zero volts.

I still suspect that I have a settings issue or some other basic misunderstanding about the lead-off.

Thanks for all of the helpful suggestions.

Best,

Tommy

Ryan,

I am still attempting to use the ADS1298 to measure AC impedance.  I removed all of the capacitors on the INP and INN inputs.  The patient protection resistor is 6.02k ohms.  I have a test fixture which can switch in a resistor in series with each of the INPs to common in the test fixture.  All of the INNs are shorted together internally on my pcb.  A single INN connection is taken to the test fixture where a resistor can be switched in series with it to common.  The RLD is connected to the test fixture common.  I set the resistor selector switches to 0 ohms (short circuit, so all the INPs and INNs are shorted to test fixture common.   Per the settings, the RLD input is connected to (AVDD-AVSS)/2. 

For case 1, INP8 and INN8 were enabled for AC impedance.  I evaluated two systems and was not able to measure the AC current at expected levels (250 nArms) on the enabled channel.  I measured the DC voltage between pcb common and the RLD connection to be 12 mV and -1.4 mV, respectively, on the two units.

For case 2, all 8 INPs and INNs were enabled for AC impedance.  I got similar results as for case 1 and was not able to measure excitation current at expected levels.  The dc voltage between pcb common and RLD was very similar to the case 1 measurements.

So, i have not been able to set up the ADS1298 to allow for measuring AC impedance.  To your knowledge, has anyone been able to successfully use the AC impedance measurement feature on the ADS1298?

Best regards,

Tommy Cooper

Hi Tommy,

Sorry for my delay. I was putting together the attached TINA-TI simulation for you and had some troubling with it converging. It's using ideal amplifiers now, which seems more reliable. Please give it a try and let me know if it helps you better visualize where the current is flowing in various configurations.

I've embedded some instructions in there as far as how to configure the switches for different setups. If you have any doubts, just let me know.

The results you see with DC lead-off should match the same amplitudes you see with AC lead-off, with the exception of some RC filter or digital filter attenuation, as we've discussed. You can think of the difference as if the device were toggling the "LOFF_FLIP" bit for you. In one phase, LOFF_FLIP = 0, and in the second phase, LOFF_FLIP = 1. You can set the bits this way yourself manually and check the DC current magnitude in each case.

3733.DC and AC LOFF - 8CH_Cooper.TSC

Best Regards,