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INA186: Output Swing to power rail

Gabriel Xu
Intellectual 2045 points
Part Number: INA186

Hi team

Customer have questions about output swing to power rail spec of INA186.

From INA186 datasheet, it indicate that it could output up to VCC-20mV with Vs=1.8V@10Kohm load.

However, output voltage swing diagram shown below indicate another situation.

Only 2mA will lead to output drop 400mV. I am not fully understand this diagram.

  • 0
    Guru 140200 points

    Hi Gabriel,

    0.4V voltage drop at 1.8mA output current means an ON resistance of output stage of about 220R. 1.8V at a 10k load means an output current of 180µA. 180µA times 220R gives a voltage drop of 40mV.

    Keep in mind, that going all too close with the output voltage to the supply rails will result in a degradation of performance. See the "Gain error" and "Nonlinearity error" specifications in section 6.5 of datasheet.

    Kai

  • 0
    TI__Intellectual 2960 points

    Hi Gabriel,

    Thanks for the question and using the E2E forum.

    The easiest way to think about this is how heavy a load the INA186 can drive. If we look at the initial datasheet conditions (VCC-20mV with Vs=1.8V@10Kohm load), this equates to just about 1.78V / 10,000Ω = 178uA (like Kai pointed out). This is a very light load. As the required current drive increases, the output voltage capability of the device decreases. This is the purpose of Figure 6-4, to show that as more and more current is required from the INA186, the maximum output will continue decreasing.

    However, keep in mind that as the output is driven to its limits, the device will not be operating in its linear range and will no longer behave to the datasheet specifications.

    Please let us know if you have any further questions.

    Louis