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Ad5933 analog front end - 4 electrode measuring

Hey everyone,

I need to implement a four electrode measuring circuit to interface with the Ad5933.

Ive used the suggested analog front end in the attached link, which is very popular throughout the references:

www.instructables.com/.../

My problem:

Im able to calibrate a 1k resistor but whenever I change the load after calibration it shows very little difference, for instance an avg of 1040 ohm instead of 1000 ohm when I plug a 2k resistor.

Sometimes it shows even a lower avg value.

My setup:

For the opamps I ised TL072 and for the inamp I used INA118P. both are have suitable specs for the job.

I used 1M resistor for R current and R protect and used Rin of 2k and Rfb of 1k.

Ive fed the inamp and opamps with v+ of 10v and v- of negative 10v.

Ive attached all grounds in the system.

I short circuited all the ad5933 vdd pins and have them 5v vdd from an arduino, and did the same for the ground pins and attached to the common ground.

The ad5934 is controlled via the arduino by attaching a4 and a5 pins to scl and sda pins.

Finally ive attached a 1k resistor as the load between the four electrodes. Each pole of the resistor gets two parallel electrodes.

Ive checked my wiring a hundred times and Ive tried to add Rg to the inamp.

Ive tried to play with rfb and r in.

Ive chexked with a scope in every point to see that the signal is what its supposed to be. Im sure its good until the inamps input, where im not quite sure because I get very low voltage different across the input pins (20mv).

I would cery much appreciate guidance

Ben

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  • It is had to suggest much without seeing the exact schematic with all connections and component values. Referring to a general sketch in your link, first suspect to check would be the DC working point for the entire schematic as the AD5933, generally speaking, requires the AC signals to have zero line at Vdd/2 = 2.5V in your case. Can you check if the DC voltage is 2.5V at:
    1. The left pin of Rcurrent resistor
    2. The output of the Vref Generator
    3. The right pin of Rprotect
    4. The left pin of Rin
    5. The RF pin of the AD5933
    ?
    If all is well, can you check that the AC voltage at the RFB pin of the AD5933 is a well-formed sinewave without any distortion or clipping at 0V or Vdd or both?
    Also, what settings are you programming into the AD5933? Excitation voltage, PGA gain, frequency range, number of settling cycles, etc.? 

  • Hey  Snorlax, Thanks for your reply!

    Im attaching an image of the schematics to make it more clear.

    regarding your questions, I have checked for vdc of 2.5v at all the points you suggested and it is indeed 2.5v.

    I do see some kind of clipping at the rfb pib and I cant seem to understand why it happens. it clips the lower part of the wave.

     my settings:

    supply voltage 5v

    Excitation voltage: 2.24v dc offset with vpp of 3v

    PGA gain=1

    frequency range 50Khz-52Khz

    default number of settling cycles 

  • Thanks

    I'll give a try and I'll update

  • I forgot to mention that on your schematic diagram the inputs of the  U8.1 seem to be flipped: the + input should be connected to Vref, not the "-". Assumed it was a graphics "typo," but could you please double check the actual circuit?

  • yes you're correct , it's a typo, the actual wiring is correct

    I've done some tests yesterday and I discovered that the TL072's slew rate wasn't good enough for my freq range, so now I lowered the freq range and that part worked out

    additionally I've used 1k resistor for the first voltage divider and Ive discovered I have to use larger resistors, otherwise it attenuates the voltage in the first opamp input.

    Finally I've added a 2k resistor as Rg to the inamp to get a stronger signal.

    I still dont get the results I expected, Im going to go over the circuit again now and see where the signal goes wrong

  •  OK  so I checked the circuit and I get a nice sine wave in the amplitude I expect, see figure attached.

    Could you help me out figuring what my total gain should be? I still get wrong readings

    the resistor values I used:

    R_V11=R_V22= 1M

    c= 100nF

    R_current = 10k

    R_protect =1M

    R_V1=R_V2= 1k

    R_IN=2k

    R_FB=1k

  • TL072's slew rate wasn't good enough for my freq range

    Your OPAMPs are operating at/about unity gain and the output is AC 2V p-p, so the estimate for the min slew rate required is 2*Pi*50Khz*1V ≅ 0.3 V/μs, so TL072 should operate OK.

    I've used 1k resistor for the first voltage divider

    Assuming you mean R_V11 and R_22, with the C1 they do produce a high-pass filter with a cut-off frequency of 1 / (2*Pi*(R_V11 || R_22)*C1) . With 10k resistors, as indicated on the diagram, the cut-off would be at about 320 Hz - significantly lower than your operating frequency of 50 KHz. Even with 1k instead of 10k the cut off would be 10 times higher, 3.2 KHz, still considerably lower than 50 KHz to affect your signal much (unless the C1 is also not the 100 nF indicated on the schematic).

    2k resistor as Rg to the inamp

    I would not do that until you get your cricut to work at unity gain with a known resistor load between E1 and E2. It would make sense to lower R_CURRENT to, say, 10k, attach 10k between E1 and E2, make R_IN and RFB resistors equal and see if the same 2V p-p excitation voltage can be found at all OPAMP and INA outputs as well as at the RFB pin of the AD5933. With these values you should not need INA gain. Once everything works as expected it would be time to attach various loads simulating expected practical values between E1 and E2 and change R_CURRENT and INA gain to get reasonable signal into the AD5933 for measuring.

  • R_V11=R_V22= 1M

    c= 100nF

    No point having these two resistors at 1M, much lower should work better. When you send the command to the AD5933 to start outputting the excitation voltage, the high pass filter we discussed earlier takes some time getting to VCC/2: in this case the time constant is (1MΩ ||1MΩ) * 100nF = 50 msec and your code should wait 7-10 times that for the transient to reasonably settle. If your code start measuring the signal before the transient settles, the the AD5933 will be processing "tilted" sinewave with unsettled baseline, so the output will be erroneous.

    what my total gain should be?

    The total gain should be set so that at the highest value of your load impedance, the voltage on the AD5933 RFB pin is about 2V p-p. This way the internal ADC available dynamic range will be mostly utilized and as lower load impedance is measured, the ADC will not be saturating/clipping. As lower and lower impedance is measured the discretization errors from the ADC will start showing in the data, which your circuit can be compensated for by reducing the value of R_CURRENT, adding gain resistor to INA or programming 5x gain into the AD5933 internal PGA or doing all of the above (bringing the ADC input signal back to similar 1-2V p-p, where the discretization error is low). 

  • Thanks ! that helps alot, Ill try all that out

  • Speaking of gain again, in addition to changing R_CURRENT, adding gain resistor to INA and programming 5x gain into the AD5933 there are two more "knobs" to turn: programming different excitation voltages into the chip and changing RFB / R_IN ratio. The latter two resistors turn the AD5933 internal OPAMP into an inverting amplifier with gain equal RFB / R_IN (with the values on your schematics this gain is 0.5, gain of 1 could be more convenient for now).
    To summarize: there is plenty of available gain throughout the circuit from VOUT to RFB along with the ways to change it. The goal is to keep the AC voltage at the ADC input high enough to capture a substantial part of the available dynamic range, but not too high to avoid clipping at the power supply rails.
    Once you get comfortable with gain setting, calibration and measuring various load impedances with the hardware we can get into the details of extracting complex impedance from the chip with the software.

  • Hey would be too much to ask for a quick zoom call ?

  • The idea discussing issues in this forum is for everybody interested to see the matters  discussed and hopefully learn and find solutions to similar problems. So, no zoom calls, but please do not hesitate to ask anything here.

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  • The idea discussing issues in this forum is for everybody interested to see the matters  discussed and hopefully learn and find solutions to similar problems. So, no zoom calls, but please do not hesitate to ask anything here.

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  • OK thank you, It just seems that I keep going in circles.
    When I change my voltage divider resistors (R_v1 &Rv_2) to 1k as you said , and activate the opamp I get a potential of 1.67V (or even 3.33V sometimes). Do you know what can be the cause for that? (If the OPamp doesn't get voltage supply the potential is 2.5V as it should)

  • (If the OPamp doesn't get voltage supply the potential is 2.5V as it should

    On your schematic it shows that the inputs of this OPAMP are swapped too, "-" should be connected to the output and "+" should be connected to the center point of the resistor divider, could you please check? If the physical circuit wiring is correct, the OPAMP might be fried.
    Something also could be wrong with the resistors or the capacitor. I would try disconnecting the capacitor and the OPAMP and checking whether the voltage between the two resistors is 2.5V, if not - something is wrong with the resistors or the power supply. If yes, add the capacitor and, if it is still 2.5V, add the opamp, it should still be 2.5V at both the OPAMP input and output.
    Using 1k resistors might be suboptimal because the AD5933 is not a very good voltage source, it has internal resistance that creates a voltage divider with R_v1 in parallel with R_v2, please see datasheet Table 17 on p. 30. Moreover, this internal resistance is dependent on the excitation voltage programmed into the chip. When you program in 2V p-p this internal resistance is 200Ω, which divides output voltage by a factor of (R_v1 || R_v2) / (200Ω + R_v1 || R_v2 ) ≅ 0.71, so you will see only 70% of the signal. If 1V p-p excitation voltage is programmed in, the internal resistance goes to 2.4k and the dividing factor becomes  ≅0.17, so only 17% of the excitation voltage is passed through the OPAMP. There is nothing wrong using higher R_v1 and R_v2, say 100k or even 1M to make the effect of this internal resistor negligible, only you need to reduce the C capacitance proportionately to keep the time constant the same: 1nF for 100k and/or 0.1nF for 1M instead of 100nF.