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INA2332: Gain of Amplfier B in INA2332 is not achieved.

Kumar Sunilkumar Pawar
Prodigy 10 points
Part Number: INA2332
Other Parts Discussed in Thread: INA232

I'm building an IoT circuit for sensing resistance change using Kelvin's 4 wire. I have implemented a constant current source that supplies my two sensors connected in series with 50 mA and the output of each sensor is approximately 15 mV(I introduce common-mode voltage of 1 V by adding a 20ohm resistor in series). Now I have connected the two voltage o/p wires of two sensors to the two differential inputs of INA2332 and set the gain 100 as you can see in my schematics.

When I measure the voltage output of amplifier A it gives 1.39~1.41V which isn't ideal but ok while the output of amplifier B is just 150 mV. I even set the reference to the ground as suggested in some previous posts. Can anyone tell me what I'm doing wrong?

  • 0
    TI__Guru* 93105 points

    Hello Kumar,

    Reviewing the INA2332 schematic I am not sure based on the resistor part numbers what their resistance values are. Is the resistor connected from pin 1 (RGA) to pin 12 (REFA) a 10 kilohm resistor (103)?  Then there are two resistors in series from pin 1 (RGA) and pin 13 (VOUTA) a 10 kilom + 184 kilohm (103 + 184). Are they intended to be 194 kilohm total? If those resistor value are correct the INA2332 output stage op amp would be -19.4 V/V.

    If the gain of the INA2332 two op amp input stage stage is 5 V/V, and the second stage gain is -19.4 V/V, the total gain would be 97 V/V. You mention a gain of 100 V/V so if those resistor values are correct the gain would be close. Please verify the resistor values.

    Regards, Thomas

    Precision Amplifiers Applications Engineering

  • 0 in reply to Thomas Kuehl
    Prodigy 10 points

    The value of gain resistors is 10kOhm and 180kOhm. I just want to know if my schematic is correct or am I missing something because I have tried 3 ICs and all of them had some issues.

  • 0 in reply to Kumar Sunilkumar Pawar
    TI__Guru* 93105 points

    Hi Kumar,

    Ok, yes I now understand the resistor marking and how those two resistors sum to form a 190 kilohm feedback resistance.

    I've been attempting to trace the schematic out. It looks like the gain and feedback resistor connections, power supply and shutdown pins are correctly connected. However, it isn't clear how the 4-wire Kelvin circuit connects to the INA2332 inputs. Would it be possible for you to provide an illustration of the 4-wire sensor connections to J1, J2, JP4, JP5, etc.?

    Thanks, Thomas

    Precision Amplifiers Applications Engineering

  • 0 in reply to Thomas Kuehl
    Prodigy 10 points

    Hello Thomas,
    The problem has been partially solved. Still, I'll provide you with my schematic of the analogue part.

  • +1 in reply to Kumar Sunilkumar Pawar
    TI__Guru* 93105 points

    Hi Kumar,

    I am glad to hear your resolved the issue with the INA2332 circuit. And thanks for the more detailed schematic of the application.

    Please go ahead and close out this e2e inquiry.

    Thanks, Thomas

    Precision Amplifiers Applications Engineering

  • 0 in reply to Thomas Kuehl
    Prodigy 10 points

    If I may ask one more question(the problem which persists). In this scheme when I measure the output voltage of amplifier A I get around 1320.3 mV more or less fluctuating (Vin diff = 15.64 mV) and the output voltage for amplifier B I get is around 1663.2 mV (Vin diff = 15.2 mV). Why there's a negative gain error in amplifier A and a positive one in amplifier B? Also the output is not very stable.

  • 0 in reply to Kumar Sunilkumar Pawar
    TI__Guru* 93105 points

    Hi Kumar,

    Each amplifier's gain resistors internal to the INA2332 are precisely trimmed to minimize the gain error, but the finished gain error may be positive or negative as the Electrical Characteristics table indicates. Do recognize that the resistive bridge sensor you show in the schematics may not have perfectly matched resistors as well and that would provide a gain error.

    If you have a way to switch the INA2332 input pairs from where each set connects to the resistive sensor that could be used to determine if they are contributing to the gain error polarities. First, you would measure the gain errors with amplifier A and B in the intended positions, then switch the A and B them and see if the errors reverse.

    If the output is not stable the input lines from the sensor to the INA inputs may be picking up noise. Induced 50/60 Hz line noise is a common problem in analog circuits.

    Regards, Thomas

    Precision Amplifiers Applications Engineering

  • 0 in reply to Thomas Kuehl
    Prodigy 10 points

    Hi Thomas!
    I tried switching the inputs and it didn't work. I'm using the two sensor setup where one is the actual sensor and the other is for temperature compensation and my main concern is that when the input voltage increases(because I increase the temperature) the increase in output of amplifier A is faster than of amplifier B which defeats my purpose of having a temperature compensation sensor. Any idea on why this can happen?

    Thank you,
    Kumar Pawar

  • 0 in reply to Kumar Sunilkumar Pawar
    TI__Guru* 77541 points

    Hi Kumar,

    my main concern is that when the input voltage increases(because I increase the temperature) the increase in output of amplifier A is faster than of amplifier B which defeats my purpose of having a temperature compensation sensor.

    Thomas will be busy in the remaining of the week, I will try to assist you. Please be patient with me as I am trying to understand your inquiry. 

     If I understood your question correctly, the differential voltage across Rx is the measurement interest and you'd like to compensate the temperature changes in Rx that resulted from increasing in changes of differential input voltage. 

    In addition, INA2332-A has a gain of 1320.3/15.64 = 84.41816 V/V and INA2332-B has a gain of 1663.2/15.2=109.421 V/V. And you want to compensate the measured INA2332-A's Vout due to temperature change in in Rx. 

    There are several ways to compensate INA2332's output measurement. 

    1. Calculate the temperature change at Rx and add/subtract the change due to temperature rise/fall from Vout-A

    2. If Rx has positive temperature coefficient, say α, then you want to Rcomp 's to have a negative temperature coefficient, say, β. you can derive the output relationships from Vout_A and Vout_B after ADC and MCU calculation. 

    3.  If Rx has the positive temperature coefficient, say α, and Rcomp = Rx with the same negative temperature coefficient, then the output difference in Vout_A and Vout_B can derive and correlate to changes in temperature at Rx or Rcomp. 

    Note: INA2332's Vout: Vout_A = GainA * (Vin+ - Vin-) + VrefA, where VrefA=GND=0 from the schematic. 

              Vin+ - Vin- = Iconst * (Rx + ΔR), where ΔR is resistance change over temperature. 

    The final transfer function in each of above case is different. If you are unable to figure it out, please let me know. 

    Best,

    Raymond

  • 0 in reply to Raymond Zhang1
    Prodigy 10 points

    Hello Raymond!

    To help you better understand I'll try to describe my setup here. I have a PCB Sensor with one face Rx as the actual sensor and on the other pace Rcomp for temperature compensation. This sensor is placed in a climatic chamber where I change the temperature and the 4 wires are connected to the device outside the chamber. I hope this helps.
    Now coming to the problem:

    Raymond Zhang1 said:
    INA2332-A has a gain of 1320.3/15.64 = 84.41816 V/V and INA2332-B has a gain of 1663.2/15.2=109.421 V/V

    This is true although I made a mistake with a value. The INA2332-A has a gain of 1320.3/15.4 = 85.7337 V/V and the gain of INA2332-B is the same as stated 1663.2/15.2 = 109.421 V/V, was measured at a temperature of 26°C. Now at 45°C, the measured values are: INA2332-A has a gain of 1400.6/16.5 = 84.884 V/V and INA2332-B has a gain of 1662.2/16.3 = 101.975 V/V.
    It is already demonstrated that the sensor successfully compensates for the temperature.
    My idea is to make a portable device that measures the ratio: Rx/Rcomp (which should remain the same irrespective of the temperature) and perform some other calculations corresponding to that ratio.
    I tried to understand your solution but it was a bit confusing to me. If you can please explain it in detail.

    Thank you,
    Kumar Pawar

  • +1 in reply to Kumar Sunilkumar Pawar
    TI__Guru* 77541 points

    Hi Kumar,

    The INA2332-A has a gain of 1320.3/15.4 = 85.7337 V/V at 26°C; the measured values are: INA2332-A has a gain of 1400.6/16.5 = 84.884 V/V at 45C. 

    I am not sure if you had a typo or something else. The gains of INA2332 should not change over temperature. R1 and R2 resistors should use low or no temperature coefficient resistors. The temperature coefficient of R1 and R2 should be the same with 0.1% to 0.01% tolerance for your application. In other words, the circuit gains should not change significantly over temperature (the data is shown that 1.178% change in gains has occurred from 26C to 45C). 

    With thermal coefficient matching in R1 and R2, the R2/R1 changes in ratio should remain the unchanged.  The Gains of 5 + 5(R2/R1) should remain to be a constant. In addition, you are only interested to compensate Rx value or input differential voltage changes over temperature. 

    Let us assume that INA2332-A's circuit gain is 85.7337. The input differential voltage is changed from 15.4mV at 26C to 16.5mV at 45C. If Rx is temperature independent, you should see 15.4mV generated at the input. 16.5mV-15.4mV or 1.1mV increasing is due to change in temperature from 26C to 45C or ΔT = 19C. Assume the constant current Icont = y mA, thus, the temperature coefficient of Rx is: (1.1mV/ymA) /19C will be your ΔR/C change (assumed Rx vs. temperature is in linear relationship).

    If temperature change from 26C is known, you are able to calculate changes in ΔR  --> you will know how much change is occurred at the output of INA2332. This approach is what I mentioned in option 1 in the previous reply. 

    Let me make a following assumption for INA2332-A, so that we can perform the remaining calculation. 

    1. INA2332-A's circuit gain is 85.7337 V/V and differential input voltage is 15.4mV at 26C, where VoutA_26C = 15.4mV*85.7337 V/V = 1320.299 mV.

    2. INA2332-A's circuit gain is 85.7337 V/V and differential input voltage is 16.5mV at 45C --> VoutA_45C = 16.5mV* 85.7337 V/V = 1414.6061 mV.

    INA2332-B is the same as stated 1663.2/15.2 = 109.421 V/V at 26C; and INA2332-B has a gain of 1662.2/16.3 = 101.975 V/V at 45C. 

    This data set may have typo. Per your prior description, Rcomp has negative temperature coefficient, which means that Rcomp should decrease in value from 26C to 45C. Your output is reduced from 1663.2mV to 1662.2mV from 26C to 45, which is ok, but your input differential voltage is increased from 15.2mV to 16.3mV. If Rcomp's resistance is decreasing, then the input differential voltage should decrease as well. INA2332 is operating in linear responses. 

    Again, the circuit gains should not change over temperature. Assumed INA2332-B's circuit gains are 109.421 V/V and input differential signal is changed from 15.2mV to 16.3mV from 26C to 45C. At 45C, INA2332-B's Vout should be 109.421 V/V * 16.3mV = 1783.56mV (this implies Rcomp has positive temperature coefficient, which is not true). 

    Let me make a following assumption for INA2332-B, so that we can perform the remaining calculation. 

    3. INA2332-B's circuit gain is 109.421 V/V and differential input voltage is 15.2mV at 26C, where VoutB_26C = 15.2mV * 109.421 V/V = 1663.1992 mV

    4. INA2332-B's circuit gain is 109.421 V/V and differential input voltage is 15.1909mV (1662.2mV/109.421) at 45C, where VoutB_45C = 15.1909mV*109.421 V/V = 1662.20347 mV.

    From the ΔVout_B figure, you can calculate the ΔC change in temperature at Rx or Rcomp. If you know the ΔC change at location of Rx resistor, you should know the compensated VoutA at 45C, where 26C is the temperature reference point. 

    The Rcomp's VoutB vs. Temperature curve may not be linear, thus it will be save to calculated ΔC change at Rcomp and compensate the VoutA figure per 26C reference figure.  

    If Rx vs. temperature response is not linear either, then you need to characterize Rx curve or obtain from a manufacture, perform a curve fit in Rx vs. Temperature and calculate the compensated VoutA at elevated temperature (you may also linearize the Rx vs. temperature curve, but you have to use log scale or other curve fit method to simplify the relationship.). 

    I hope that this helps. If you have additional questions, please let me know. 

    Best,

    Raymond

  • 0 in reply to Raymond Zhang1
    Prodigy 10 points

    Hello Raymond!

    Raymond Zhang1 said:
    Per your prior description, Rcomp has negative temperature coefficient, which means that Rcomp should decrease in value from 26C to 45C.

    I apologise for not being clear but the Rcomp doesn't have a negative temperature coefficient. It is the exact same sensor as Rx electrically.

    Raymond Zhang1 said:
    INA2332-B's circuit gain is 109.421 V/V and differential input voltage is 15.1909mV (1662.2mV/109.421) at 45C, where VoutB_45C = 15.1909mV*109.421 V/V = 1662.20347 mV.

    At 45°C the differential input to INA2332-B was 16.3 V which is an error. I'll run the measurements again and confirm if this was a mistake.
    I now understand what you were trying to explain.

    Thank you,
    Kumar Pawar

  • 0 in reply to Kumar Sunilkumar Pawar
    TI__Guru* 77541 points

    Hi Kumar,

    It is the exact same sensor as Rx electrically.

    If Rx and Rcomp have the exact temperature coefficient, then it is a lot easier to compensate Rx resistor at 26C. The ratios of VoutA/Vout_B should not change over temperature, which VoutA is the output of INA2332-A and VoutB is the output of INA2332-B. If Rcomp's output at 26C is the reference, you can calculate Rx's output with respect to 26C by multiplying by the  VoutA/Vout_B ratio or VoutB_@26C * (VoutA/Vout_B). 

    Best,

    Raymond

  • 0 in reply to Raymond Zhang1
    Prodigy 10 points

    Hello Raymond!

    I took measurements again and the gain is still changing and I'm not sure why. I'm attaching the results where I took measurements at 26°C, 30°C, 35°C, 40°C, 45°C and 50°C. I'm also attaching my board pic for you to take a look at it, just in case if I might have missed something. Result121121.xlsx

    Is it a problem that I put a 20Ω resistor in series to introduce 1 V common-mode voltage because according to the datasheet the Vcm = Vs/2 and my supply to INA2332 is 3.2V from Arduino? 

    Thank you,
    Kumar Sunilkumar Pawar

  • 0 in reply to Kumar Sunilkumar Pawar
    TI__Guru* 77541 points

    Hi Kumar,

    INA2332's input is only interested the differential input voltage, so 20Ω resistor in series to introduce 1 V common-mode voltage is no a factor. 

    BTW, what is your constant current? 1V/20Ω = 50mA. Do you have self heating issues in Rx and Rcomp? 

    From the plot, the temperatures of Rx and Rcomp (no voltage across the resistors) are not tracking. Both resistor should go up and down together with changing in temperature. And I do not see that.

    Please check out the following lists. 

    1. Rx and Rcomp have to be next to each other in order to improve the temperature tracking. 

    2. The temperature chamber is required to have air flow or circulating air. If this is a static oven, you will have localized heating/cooling and variation of temperature can be high from point to point. The temperature tracking between Rx and Rcomp will be poor. 

    3. Use 6.5 or 7.5 digits portable DMM and verify that your ADC from Arduino is stable and repeatable. Check your LSBs over temperature. 

    4. Make sure that INA2332's configuration is stable over temperature (I assumed that the whole fixture is placed inside the temperature chamber). 

    5. Make sure that the current source is stable and precise over temperature. 

    Best,

    Raymond

  • 0 in reply to Raymond Zhang1
    Prodigy 10 points

    Hello Raymond!
    1. Rx is in the front and Rcomp is in the back of the same PCB like this (This is the front face)
    2. Yes the climatic chamber has air circulation.
    3. I'll do that.
    4. Only the sensor is placed inside the chamber, the circuit is outside.
    5. I'll do that
    I'll get back to you if something goes wrong.


    Thank you,
    Kumar Pawar

  • 0 in reply to Kumar Sunilkumar Pawar
    Prodigy 10 points

    Hello Raymond!

    I think I might have found the problem. Please tell me if I'm on the right way.
    All this time I was ignoring the fact that CMRR can be an issue because I thought it could be compensated in the calculation but I forgot a small detail.

    The common-mode voltage across Rcomp and Rx can be given as:
    VRcompCM = (VRcomp+VR+VRcm)/2
    VRxCM = (VRx+2VR+VRcomp+VR+VRcm)/2
    where VR is the voltage across the wire and VRcm is the voltage across 20Ω resistor.
    So when the resistance changes     due to any reason the voltages should also change: 

    ΔRcomp→ΔVRcomp ΔRx→ΔVRx

     and so does the common-mode voltage changes: 

    ΔVRcompCM = (ΔVRcomp)/2

    ΔVRxCM = (ΔVRx+ΔVRcomp)/2
    My guess on why does the gain of Rcomp and Rx are different for different temperatures is because of this and also because I was only measuring the output voltage of INA from my ADC and just converting those values based on my first reading of resistance values.
    Now this can be an issue, right?

    Thank you,
    Kumar Pawar

  • 0 in reply to Kumar Sunilkumar Pawar
    TI__Guru* 77541 points

    Hi Kumar,

    The INA2332's Vcm vs. Vout relationship is shown below, also see the attached analog engineer's calculation, where you can download and check it out. 

    As long as the common mode Vcm and Vout are operated within the boundaries, Vout vs. differential input relationship will be linear.  

    https://www.ti.com/tool/ANALOG-ENGINEER-CALC

    Is your current source over temperature? 

    Based on this block diagram, your power supply voltage is operated approx. 3.2-3.3V. If you add all the voltage up from the components marked in red box, it will not exceed the supply voltage. Please verify if you have a true current source that is able to drive the load, which is changing with temperature. 

    Rx and Rcomp's value should have change significantly for a short duration (say during the sampling period), but I see the Vout varies significantly in some temperature (e.g. 26C and 30C). The data from 50C looks better, at least it is not changing significantly over sampling duration.

    If you still have issues, please provide me R, Rx and Rcomp and constant current figure and I will check it out in a simulation. 

    Best,

    Raymond


  • 0 in reply to Raymond Zhang1
    Prodigy 10 points

    Hello Raymond!

    I'm sorry if it shows 3.3V. The supply is 11.1V (for the testing purpose from a DC Power Supply) to the circuit as well as Arduino(Arduino will be battery operated and thus such representation). I have verified that the current source is working and quite stable (doesn't change over temperature as the circuit is outside the chamber) and also it doesn't exceed the supply voltage.
    I've already performed the simulation and I'm attaching the file please have a look.Ver3_Sim.TSC

    Thank you,
    Kumar Pawar

  • +1 in reply to Kumar Sunilkumar Pawar
    TI__Guru* 77541 points

    Hi Kumar,

    I think that your calculation issues are.

    There is Vos specified in INA232, which is up to ±8mV (that is Refer-to-Input or RTI). Since Gain=100V/V, the actual voltage offset set seen at the output is gained in 100. From the simulation, 646.692mV/100 = 6.46mV (RTI) with the input is shorted. 

    In your Arduino processor, the measured Vout needs to be subtracted from the offset, which Rx's actual Vout_Rx = Vout_measured_Rx - Vout_offset_Rx, then you can ratio the output voltage between Rx and Rcomp. In addition, Voffset's temperature coefficient also have ±5 uV/C. If you want to be very precise, you will need to take into account of temperature drift as well.  

    If you have additional questions, please let me know. 

    Best,

    Raymond 

  • 0 in reply to Raymond Zhang1
    Prodigy 10 points

    Hello Raymond!

    Is it abnormal that the offset voltage is almost 41% of the output signal? Also when I check this on hardware the offset voltages are: Vout_offset_Rx = 151.53 mV and Vout_offset_Rcomp is 11.042 mV.

    Thank you,
    Kumar Pawar

  • 0 in reply to Kumar Sunilkumar Pawar
    TI__Guru* 77541 points

    Hi Kumar,

    Is it abnormal that the offset voltage is almost 41% of the output signal?

    When you short the inputs of INA2332-A and INA2332-B port, you Vout_offset_Rx = 151.53 mV and Vout_offset_Rcomp is 11.042 mV. Since you gain is configured at 100,  Vos_A = 151.53 mV/100 = 1.51mV (RTI), which is a typical figure with the datasheet's specification.  Vos_B =11.042 mV/100 = 0.11mV, which is not typical. 

    Please check your gain of INA2332-B port and make sure the gain circuit is working properly, and you do not have bad solder joints or connections etc.. 

    If the condition persists, swap a new INA2332 part and see if the part has gone bad. Your circuit was operating ok previously. For instance, the collected data at 50C is near identical in both channels, which means that channel A & B is behaving in similar manner. 

    If you have additional questions, please let us know. 

    Best,

    Raymond

  • 0 in reply to Raymond Zhang1
    Prodigy 10 points

    Hello Raymond!

    My question "Is it abnormal that the offset voltage is almost 41% of the output signal?" was regarding the values obtained from simulations. If you can explain why simulation data is showing this.

    Thank you,
    Kumar Pawar

  • 0 in reply to Kumar Sunilkumar Pawar
    TI__Guru* 77541 points

    Hi Kumar

    Is it abnormal that the offset voltage is almost 41% of the output signal? was regarding the values obtained from simulations.

    I am confused about "from simulations" phrase. 

    These parts are laser trimmed and the offset voltage may have such large spread from +/-1.5mV to 0.11mV, but it is abnormal. It could be possible, since one of influence variable is the epoxy modeling stress on Si die during the encapsulation. I do not want to speculating it. Let us check on the gain circuit first. 

    Best,

    Raymond

  • 0 in reply to Raymond Zhang1
    Prodigy 10 points

    Hello Raymond!

    I'm sorry for not being clear but by "from simulation" I meant that the Voffset results obtained from the simulations mentioned in few comments back were this:

    Raymond Zhang1 said:
    There is Vos specified in INA232, which is up to ±8mV (that is Refer-to-Input or RTI). Since Gain=100V/V, the actual voltage offset set seen at the output is gained in 100. From the simulation, 646.692mV/100 = 6.46mV (RTI) with the input is shorted. 

    For which I was asking this:

    Kumar Sunilkumar Pawar said:
    Is it abnormal that the offset voltage is almost 41% of the output signal?

    Also, you were correct about the bad soldering joint. The corrected offset values are: Vout_offset_Rx = 91.99 mV and Vout_offset_Rcomp = 108.7 mV (These fluctuate a lot if I move the circuit by mistake while measuring). I understood how to compensate the offset values in my calculation my only doubt remains of that of the simulation.

    Thank you,
    Kumar Pawar

  • +1 in reply to Kumar Sunilkumar Pawar
    TI__Guru* 77541 points

    Hi Kumar,

    I meant that the Voffset results obtained from the simulations mentioned in few comments back were this: 646.692mV/100 = 6.46mV (RTI) 

    The Vos in INA2332 part is specified from ±2mV (typical) to ±8mV at 25C. For simulation, PSpice model engineer simulates a typical Vos of ±2mV. In this case, the model person is using 6mV as a typical Vos figure. I do not know the reasons behind it. Perhaps Vos in the part is biased high and the modeling engineer decided to use 6mV (RTI) figure as typical figure instead.  

    I understood how to compensate the offset values in my calculation my only doubt remains of that of the simulation.

    INA2332 is designed for low cost and with early 2000 fab IC processing, and hence Vos seems to be high. For instrumentation amplifier and certain application, Vos will be removed under the real world operation. Once the Vos is removed, the linear behavior is excellent for the applications specified in the datasheet. 

    You mentioned that "Is it abnormal that the offset voltage is almost 41% of the output signal?" in INA2332 part. If a input signal is very small, the relative error will be considerable large, if you are talking about Vout/Vos (without removing the offset). If the input signal is large, then the relative error is getting smaller. The Vos errors does not change significantly with given gain settings and operating conditions, so if the input signal is large, the error will be small to insignificant. All op amps behave in this manner. 

    I hope that I answered your questions. If you have additional inquiries, please let us know. 

    Best,

    Raymond