• State Resolved
  • Locked Locked
  • Replies 6 replies
  • Answers 2 answers
  • Subscribers 48 subscribers
  • Views 269 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

INA129-EP: Understanding gain error and thermal drift

ArTzy
Genius 13999 points
Part Number: INA129-EP

Hi Experts,

Seeking your assistance on the following queries:

[1] In the INA129-EP DS (see below), 2 column in electrical characteristics are for TA=25°C, is it right that the second is for T=125°C?

[2] Gain error is given for different gain value, on my understanding, the error for G=1 is due to a different part than the one provided for G=10. So,  For a gain >1 (for example, 2), should I calculate the maximum gain by : 1*(1+0.001)+ 1*(1+0.005)=2.006 -> gain error(for G=2) = 0.3%

[3] Same question with the thermal drift. For example G=2, T=125°C and an ideal external resistor, gain (for G=2 @125°C) =1*(1+0.001+10^-5*dT)+1*(1+0.005+10^-4*dt)= 2.017 -> gain error (for G=2 @125°C)= 0.85% 

[4] Compared to [3], what is the meaning of max gain error provided for G=10 overtemperature? Should I use the min value of both? 

Thanks ahead for your support.

Regards,
Archie A.

  • +1
    TI__Mastermind 18535 points

    Hello, 

    The gain error in the right column states that this is over temperature. This is a measured parameter value with a statistically calculated max. This is a max for the specified temperature range of -55 to 125 degrees C as specified below. 

    We could use the temperature coefficient of the 49.4-kΩ term in the gain equation that is specified by wafer test to calculate the worst case. 

    An example of this calculation is as shown below. 

    Gain of 2 example

    Gain of 10 example

    Gain of 100 example

    Best Regards, 

    Chris Featherstone

  • 0 in reply to Chris Featherstone
    TI__Genius 13999 points

    Thank you, Chris.

    But we have follow up:

    Not sure to understand, when you say “a max for the specified temperature range”, does this take into account the drift of the 49k4 term and for example on a 10 gain, we should take the minimum between 0.9% (from your calculation) and the specified max 0.65% ?

    Or it doesn’t take into account the 49k4 term and it should be added -> 0.9%+0.65% => a gain error of 1.55%?

    Regards,
    Archie A.

  • 0 in reply to ArTzy
    TI__Mastermind 18535 points

    Hello, 

    Let me answer part of this question and I will do a follow up tomorrow. 

    There is Gain Error and Gain Error Drift. 

    ArTzy said:
    Not sure to understand, when you say “a max for the specified temperature range”,

    Below you can see the Max gain error specification for over temperature for gain of 1, 10 and 100. 

    The section below circled in blue controls the gain drift for gains greater than 1 and the section circled in red controls the gain drift for gain of 1. The initial gain error is not correlated with the gain drift. In addition the two stages shown below are not correlated so we must add the total gain drift error for each using the square root sum of the squares. 

    I will follow up tomorrow with the rest of the answer. 

    Best Regards, 

    Chris Featherstone

  • 0 in reply to Chris Featherstone
    TI__Genius 13999 points

    Thank you Chris, for this interpretation.

    Looking forward to the full response.

    Regards,
    Archie A.

  • 0 in reply to ArTzy
    TI__Mastermind 18535 points

    Hello, 

    Below you can see that I took the square root sum of the squares using the error that was calculated for the Gain of 10 and the Max over temperature specification for the gain of 1 of 0.15%. In other words I took the overtemperature gain error calculation and added it to the measured value in the table of the product datasheet for a gain of 1 gain error over temperature. This assumes a perfect gain resistor that is exactly 5.489 kohm. This calculation would be performed again subtracting the errors for a minimum and using a -100ppm/degC in the calculation below. Please note that this is a theoretical calculation where the values I circled in red in the prior post are measured values. 

    I always forget how to convert ppm to percentage if I don't use it for a while. I put together a quick 1 pager to remind myself. Just thought I would share if it is helpful to others reading this post in the future. 

    I hope this helps clarify. 

    Best Regards, 

    Chris Featherstone 

  • 0 in reply to ArTzy
    TI__Mastermind 18535 points

    Hello, 

    To follow up I hope this picture will help as well. 

    Best Regards, 

    Chris Featherstone