• State Resolved
  • Locked Locked
  • Replies 1 reply
  • Subscribers 47 subscribers
  • Views 360 views
  • Users 0 members are here
Support feedback
Options
Options
Related

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.

INA149: INA149 gain error

Victor Yu
Expert 5321 points
Part Number: INA149


Hi team，

1, according to the screenshot, if Vs=±15V, the output is ±4V

RL＝1K seems have smaller gain error than RL＝2K.

Can we say：the smaller the RL, the smaller the gain error?


2. Could you share typ  or max gain error for situation below?

Vs=±15V, RL=2Kohm and output is ±4V

we only see the data about output＝±10V.

  • +1
    TI__Mastermind 25648 points

    Victor,

    1. You are correct.  For higher output current (large impedance loads) the output impedance of the op amp will decrease compared to low output current.  Lower output impedance translates into better gain error because the feedback is less impacted by the output impedance when it is low.
    2. In general, for any specification the only place that the maximum limit is in the data sheet table.  The chrematistic curves provide an idea of how typical parts will operate over conidiations not specified in the table.   In your particular case, the data sheet table does cover the conditions that you will use (i.e. +/-15V supplies with 2k ohm load.  The gain error for this condition is specified as a maximum of ±0.02% of full-scale range.  Gain error is calculated as GE = (1 - ΔVout/ΔVin)*100.  Gain error is essentially the error in slope for the gain transfer function.  The gain is 1V/V for this amplifier, so gain error is the percentage error if the ideal versus the measured slope.  The specification indicates that the gain error is a "percentage of full-scale range".   However, since the slope is linear the gain error will be the same percentage for a subset of the output range (i.e. the gain error percentage is the same for ±4V as it is for ±10V because the slope is linear).  
    3. If you wanted to convert gain error as a percentage to a voltage, you could multiply by the percentage.  Thus, for a 10V range the error would be ±10V(0.02/100)= ±2mV, and for a ±4V range it would be ±4V(0.02/100)= ±0.8mV.  This is why the graphs versus load scale the axis in 2mV increments (this scale range makes sense according to the device expected gain error).

    I hope this helps.  Best regards,  Art