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Original question:

INA126: INA126

INA126: INA126

Murugavel S
Prodigy 115 points
Part Number: INA126
Other Parts Discussed in Thread: OPA2342, TLV3201,

Tool/software:

Hi TI team,

We are using INA126, TLV3201, and OPA2342 for thermocouple measurements. The thermocouple output is fed into the INA126 input (0.951mV). The gain is what we have set as 300. Hence the output of the OPA2342(U549 as a voltage buffer) is 605.2mV when we back-calculated (605.2mV/300) it came to 2.02mV.

Here, the concern is input :0.951mV(measured through a calibrated multimeter) but when back-calculated, the input is 2.02mV. Could you please provide a detailed answer as that where the circuit is taking 1.07mV input additionally?

Below is the circuit representation for your reference.

Regards,

Murugavel.S

  • 0
    TI__Mastermind 33593 points

    Hello Murugavel,

    Thank you for posting again on E2E! There are a couple culprits here.

    1. First, please replace R1390 with a 0 ohm resistor, or solder bridge. Resistance on this node will introduce gain error to the output
    2. Second, your offset voltage referred to output will be in the range of 30mV
      1. Datasheet offset voltage is 100uV typically, referred to input. This input offset voltage of 100uV times 300V/V will give a +/-30mV offset voltage.

    A side note, but one that is not causing your issue, if channel B of OPA2342 is unused, please populate R1515 and R1516. Not applying a voltage state could cause the input state to be undefined, and may cause extra unnecessary current draw from your op amp.

    Best,
    Gerasimos

  • 0 in reply to Gerasimos Madalvanos
    Prodigy 115 points

    Hi Gerasimos,

    Thanks for your reply. I found the statement below from the TI manual on thermocouple measurement.

    From my understanding, The PCB traces is working as a cold junction because of the "law of intermediate metals". If my interpretation is correct, can't we consider PCB traces working as a cold junction?. If so,  no need for RTD and other detection techniques. kindly correct me, if I am wrong.

    Regards,

    Murugavel.S

  • 0 in reply to Murugavel S
    TI__Mastermind 33593 points

    Hey Murugavel,

    The important part of this statement is later in the paragraph you have copied. Yes, this intermediate connection can be treated as your reference junction, however:

    "When the temperature of the reference junction is known, the absolute temperature at junction A can be calculated."

    This is the function the RTD serves. It provides an absolute temperature scale that establishes your reference temperature. In the previous thread, the time that cold junction compensation is not needed to be done with any external components is when a reference point is established.

    Imagine if I have 10 sensors, and I tell you that they all respond to temperature with a consistent 1mV/degK, but do not tell you what the starting voltage of each sensor is at 273K. Now imagine this starting voltage can range anywhere from +/-1V. If you set your reference point to be 0V = 273K (0C), one sensor reads out 500mV and the other reads out -500mV at 0C. Trying to backsolve the absolute temperature from the random offset of each sensor will yield a calculated temperature of 737K or -227K (below absolute 0!), when in actuality the room is 273K. These values are exaggerated, but are meant to illustrate the importance of relativistic vs absolute measurement.

    Best,
    Gerasimos

  • 0 in reply to Gerasimos Madalvanos
    Prodigy 115 points

    Thank you Gerasimos.

    As you know, In our present circuit the cold junction is not present and the circuit has already freezed for the PCB fabrication. FYI I am sharing the circuit again.

    Hence, I have made a calculation that how much error the circuit will produce without a cold junction. Kindly find the calculation below.

    For example: Room Ambient temperature 10degC, the actual cold junction temp 10 deg C, the correct temperature(original actual temperature) is 130degC.Due to the absence of the cold junction, the temperature will show as 145 degC at 25degC.

    25degC-10degC=15 degC 

    15degC+130degC=145degC

    Kindly find your insight regarding this calculation. If you have any other calculation methodology please let me know.

    Alos, please let me know if TI has any INA with cold compensation

    Regards,

    Murugavel.S

  • 0 in reply to Murugavel S
    TI__Mastermind 33593 points

    Hey Murugavel,

    I may have understood this incorrectly. The seebeck effect will generate a voltage based on the temperature difference between the hot side and reference side (cold junction) of your thermocouple.

    If the reference side temperature is unknown and varying you will have varying error since you do not know your reference point, and therefore will be unable to accurately gauge the expected voltage.

    https://www.allaboutcircuits.com/technical-articles/thermocouple-basics-using-the-seebeck-effect-for-temperature-measurement/

    Best,
    Gerasimos

  • 0 in reply to Gerasimos Madalvanos
    Prodigy 115 points

    Hi Gerasimos,

    Could you briefly explain the point number one? How the resistor R1390- contributing the gain error?

  • 0 in reply to Murugavel S
    TI__Genius 16410 points

    Hi Murugavel, 

    I will be covering this thread while Gerasimos is out. In the datasheet in the Typical Application section, if the reference pin will be grounded, this connection needs to be low impedance in order to have good common-mode rejection. If you apply a resistor to ground to this pin directly, this will introduce error to your output.  

    Please let me know if you have further questions.
    Thank you!

    Regards,
    Ashley

  • 0 in reply to Ashley Kang
    Prodigy 115 points

    Hi Ashley,

    Thanks for the sweep response. Could you please explain, how the resistor addition in Vref pin causes the Gain error?

  • 0 in reply to Murugavel S
    TI__Genius 16410 points

    Hi Murugavel, 

    The additional resistor will add to the internal resistance directly connected to inverting input of the op amp A2, see figure below. This mismatch of internal resistors changes the gain equation as well as degrade CMRR. 

    Please let me know if you have further questions.
    Thank you!

    Regards,
    Ashley