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INA330 / Impact for output error by tolerance of Rset and Rg

Other Parts Discussed in Thread: INA330, ADS7953, OPA320, OPA365

Hi All,

Our customer hope to know the impact for output error by tolerance of Rset and Rg. I understand that the electrical characteristics and typical characteristics in datasheet is specified with ideal resistor. The variation of resistance of Rset and Rg will affect to performance directly, right?

And the performance in datasheet is specified with ideal Vexcite also, right?


Best Regards,

Sonoki / Japan Disty

  • Hi Sonoki,

    Please know that the INA330 is intended to be calibrated so that initial errors such as those caused from Rset, Rg and Vexcited will be reduced. By calibrating, the main source of error is caused from the drift of Rset. Please see sections "SOURCES OF ERRORS" and "SELECTING COMPONENTS" in the datasheet where this is discussed.

    To calculate the output errors due to only Rg and Rset please see the equation for Vo in the section "SELECTING COMPONENTS" on page 7 of the datasheet. Vo = Vadj + Vexcite*Rg*(1/Rtherm - 1/Rset). If you plan on using 1% resistors for Rset and Rg, the equation then becomes Vo = Vadj + Vexcite*(Rg +/-1%)*(1/Rtherm - 1/(Rset +/-1%)).

    I will now show an example of output error calculations due to ONLY Rset and Rg using the values shown in Figure 3 of the datasheet. Where Rtherm=Rset=10k ohm, Rg=200k ohm, Vadj=2.5V, and Vexcited=1V

    Vo = Vadj + Vexcite*(Rg +/-1%)*(1/Rtherm - 1/(Rset +/-1%))

    Vo = 2.5V + 1V*(200k ohms +/-2k ohms)*(1/10k ohms - 1/(10k ohms +/-100 ohms))

    If RG has a positive tolerance (+1%) and Rset has a negative tolerance (-1%) the total output error due to ONLY Rset and Rg is,

    Vo = 2.5V + 1V*202k ohms*(1/10k ohms - 1/9.5k ohms)

    Vo = 2.296V

    The ideal Vo is 2.5V, therefore you will have approximately 204mV of error due to using 1% resistors for Rset and Rg.

    This example was only done for a +1% tolerance for Rg and -1% for Rset, please know that resistor tolerances can go both in the positive and negative direction. They can also both have a positive tolerance or both have a negative tolerance.

    -Tim Claycomb 

  • Sonoki-san,

    Is the customer using the INA330 as a temperature set-point controller similar to the circuit in Figure 1? If so, I believe the customer will be confused by Tim's answer.

    The customer is probably most interested in knowing the variation in set-point temperature that would be created by errors in Rset and Rg. Is that correct? Because the output voltage of the INA330 is applied to an integrator (PID) feedback loop, the output voltage error calculated by Tim is not really relevant. Please verify that the customer is using a circuit similar to one shown in the INA330 data sheet. If not, please provide details on the customers circuit.

    Regards—  Bruce

  • Hi Tim-san, Bruce-san,

    Thank you for your response. I'm asking our customer which circuit they'll plan to use. I'll get back to you soon.

    Best Regards,


  • Hi Bruce-san,

    I confirmed the circuit and value of external components that our customer has planed to use. Please see the following setting with inserted schematic.

    Rset = 7.5kohm +/- 0.1%

    RG = 4.7kohm +/- 0.1%

    CFILTER = 330pF +/- 5%, +/-60ppm/degC

    Vexcite = Vadjust = +2.5V

    Thermistor : PT3-51F (

    Can you let me know hot to calculate the variation in set-point temperature that would be created by errors in these parameters?


    Your advice will be appreciated.


    Best Regards,

    Sonoki / Japan Disty

  • Sonoki-san,

    Thank you for this additional information. Still, however, there are important details about the application that are not yet provided. How is the output of the INA330 used? Is it used in a set-point controller as shown in figure 1 of the data sheet?

    The INA330 is intended to be used as a temperature set-point controller. Referring to figure 1 of the data sheet, Vo drives an op amp connected as an integrator. When connected in a temperature controller loop, as shown, the control loop will drive the heater (or cooler) to force Vo of the INA330 to 2.5V. This Vo control-point voltage is set by the voltage applied to the non-inverting input of the integrator op amp.

    The control loop will come into balance when Rtherm is equal to Rset. This set-point will be approximately 73'C for the specified thermistor—the temperature at which the thermistor resistance is equal to 7.5k-ohms. A different set-point temperature should be made by making Rset equal to the resistance of the thermistor at the desired set-point temperature.

    This assumes that Vadjust is also equal to 2.5V, the same voltage source applied to the non-inverting input of the integrator. When operated in this way, the resistance of Rg is not particularly critical and will not affect the set-point temperature. Likewise, the voltage applied to V1 and V2 is not particularly critical and will not affect the set-point temperature. When operated this way, the temperature control point is determined only by the thermistor and Rset.

    The thermistor resistance changes approximately 4% for a 1'C change in temperature. So a 1% change in Rset would change the set-point temperature by approximately 0.25'C.

    If the customer is not making a set-point temperature controller, I would recommend using a more conventional type of instrumentation amplifier.

    Regards—  Bruce

  • Hi Bruce-san,

    I confirmed that our customer use INA330 as temperature set-point controller as you commented and use ADS7953 at the output of INA330. They already decided to use ADS7953 at other circuit and use it one channel for this usage.

    I'll recommend to add buffer amp at input stage of ADS7953, the do you have recommended OP amp to use this usage?

    Best Regards,


  • Hi Sonoki-san,

    I recommend using the OPA320 or the OPA365.

    Thank you,

    Tim Claycomb