Post Go back to editing

Looking to measure current noise in EIS - AD5941

Category: Software
Product Number: AD5941

Hi, I'm looking to measure open-circuit current noise for EIS using the AD5941. Can you please confirm if the following methods are correct?

My protocol is to run a two-wire setup for a known resistor across SE, RE, and CE. 

I run an EIS measurement at a set frequency and set RTIA, and then obtain the DFT values of that known resistor. From there, I use the following equation: 

I_{RMS} = (DFT_{MAGNITUDE} * V_{DD}) / ((2^{16} - 1) * R_{TIA})

Is that equation correct for calculating the I_RMS using the DFT? I know that I don't need to measure RCAL since I'm only measuring current, but I wanted to ask to make sure.

Lastly, is the theoretical value for the I_RMS equivalent to the AC Voltage RMS divided by the sum of the known resistor and the 100 Ohms in series with SE? 

Thank you!



Added AD5941 to title.
[edited by: kbaudue at 3:31 PM (GMT -5) on 6 Feb 2024]
Parents
  • Hi,

    Since you are measuring only current,

    if using EIS example code in GitHub,

    1) you may remove below sections:

    - /* RLOAD Measurement */

    -   /* RCAL Measurement */ 

    inside measure sequence (AppIMPSeqMeasureGen()).

    2) To reduce complexity, you may not enable DFT block in measure sequence:

    AD5940_AFECtrlS(AFECTRL_ADCCNV|AFECTRL_DFT, bTRUE);  /* Start ADC convert and DFT */

    3) You may just do data process as below. It returns current input from SE. (load includes the 100ohm resistor at SE)

    static AD5940Err AppIMPDataProcess(int32_t * const pData, uint32_t *pDataCount)
    {
    uint32_t i, datacount;
    datacount = *pDataCount;
    float *pOut = (float *)pData;
    for(i=0;i<datacount;i++)
    {
    pData[i] &= 0xffff;
    pOut[i] = AppAMPCalcCurrent(pData[i]);
    }
    return AD5940ERR_OK;
    }

  • Hi Akila, 

    Is this still accounting for the AC amplitude? If I understand correctly, I use the following method: 

    1. Apply AC signal at set frequency to two-electrode configuration

    2. Poll the ADC register and NOT the DFT one

    3. Use the Current and Voltage code from the Amperometry example 

    I also have a few questions about the amperometry example: ad5940-examples/examples/AD5940_Amperometric/Amperometric.c at master · analogdevicesinc/ad5940-examples (github.com)

    AppAMPCalcVoltage uses a k-factor: 1.835/1.82 - what is the origin of these values?

    It also uses a 1.8V Ref Voltage, what pin do I measure that from? 

    The RTIA magnitude it uses, in the EIS case, is this just the RTIA I set for that measurement?

    From the current measurements, how would I obtain RMS from these values since I'm using an AC signal instead of the DC one used in Amperometry? 

    Thank you!

  • Hi Kevin,

    Yes. It is a valid method.

    If the aim is to measure only open circuit current,

    I was just wondering why VRE0 is required to be measured here as open circuit current input can be measured at SE0 alone and processed.

    Since this is open circuit, CE0 is irrelevant as there is no excitation.

    RE0 is irrelevant as this feedback is not going to be used by excitation amplifier for providing any excitation.

    Of course, for measuring Open circuit voltage across RE0/CE0 and SE0, your method is perfect.

  • Oh that's interesting. So if I just wanted open-circuit current input noise, I can just measure at SE0 using: 

     sw_cfg.Dswitch = SWD_CE0;
      sw_cfg.Pswitch = SWP_RE0;
      sw_cfg.Nswitch = SWN_SE0;
      sw_cfg.Tswitch = SWT_SE0LOAD|SWT_TRTIA;
      AD5940_SWMatrixCfgS(&sw_cfg);
      adc_base.ADCMuxP = ADCMUXP_HSTIA_P;
      adc_base.ADCMuxN = ADCMUXN_HSTIA_N;
      adc_base.ADCPga = ADCPGA_1;
    And I would be able to get the input current noise this way?
  • Hi,

    Yes. To ensure that no excitation is given to the load, sw_cfg.Dswitch can be set to SWD_OPEN.

  • Oh okay, I see. And this is valid even with 1.11V at the HSTIA_N because the SE0_Pin also has the same zero voltage? One last question though, can you explain further why we aren't accounting for RE0 in measuring current noise here since it is the reference when we do electrochemical measurements, but we account for it when we calculate OCP? 

    Thanks Akila!

  • Hi,

    RE0 is an input to AD5940. It is not used anywhere in current measurement.

    (I assumed that no bias potential is required for the electrochemical to generate this open circuit current.

    If my assumption is wrong. then CE/RE play role is providing the bias potential to the electrochemical)

  • Hi, 

    When using the HSTIA, isn't the cell always biased at 1.11V when there is no external bias is applied? 

    Given this, and that CE/RE do play a role, shouldn't the method be measuring Vocp and then dividing by the HSTIA as I've addressed above? 

  • Hi,

      adc_base.ADCMuxP = ADCMUXP_HSTIA_P;
      adc_base.ADCMuxN = ADCMUXN_HSTIA_N

    Differential voltage measured by ADC = VHSTIA_P  - VHSTIA_N 

                                                                 = VHSTIA_P  - VSE0

                                                                 = Open circuit Input current X RTIA

    Therefore, Open circuit input current = ADC Measured value / RTIA

    CE/RE doesn't come into picture here if no bias or excitation potential is required to control the current.

  • That makes sense, but what do you mean by bias? Measuring the SE0 pin w.r.t. yields a voltage equal to 1.1V, which means the pins prior to any electrochemical cell being attached already has a potential applied ot them.

    by bias do you mean the use of VBIAS0 and VZERO0 across RE/CE? I think this is where my confusion is coming through. Because without VBIAS and VZER0, assuming normal conditions, RE/CE should have the same voltage as SE0 (1.1V)

  • Hi,

    Yes. The pins prior to any electrochemical cell being attached already has a potential applied to them w.r.t. gnd. But the electrochemical is not necessarily connected to a gnd reference.

    If biasing is required, (VBias0 - Vzero) voltage is applied to sensor or electrochemical across CE and SE such that

    - SE is at Vzero w.r.t. gnd and

    - CE is at Vbias w.r.t. gnd.

  • Oh okay. So to clarify: 

    If I apply a bias: use the method I've described above to account for VRE0 / VCE0.

    Without a bias (i.e. just using 1.1V), then simply just measure SE0 through the HSTIA. 

    Is this correct?

Reply Children