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INA188: Seeking clarification on Reference pin termination for using as a load cell amplifier with single supply operation

Vignesh Vignesh
Prodigy 20 points
Part Number: INA188

Hai,

      I think to use INA188 amplifier for load cell(weigh scale) with single supply operation mode. Here, i just connect reference pin of an op-amp to ground, then its output is connect to 5V level operating microcontroller's 10-bit ADC pin. But datasheet told to connect mid-supply voltage value to reference pin for single supply operation. Load cell as 5Kg capacity, 12V exitation voltage, 2mV/V rating. We need 5g resolution. Is this below mentioned schematic is suitable for my requirement, else it can be go with any change? Please anyone give me your valuable suggestion.

    I am also attach my schematic, our datasheet and Load cell specification here.

Regards

Vignesh

  • 0
    TI__Genius 10850 points

    Hi Vignesh,

    Thank you for your post, I can help you.

    Remember that the output equation on an instrumentation amplifier is: Vout = Vdiff * Gain + Vref

    If you have a 12V excitation voltage with a 2mV/V sensitivity, then the full scale input differential voltage is 24mV. The input differential voltage range is anywhere from 0V to 24mV. The output stage of an amplifier cannot swing down to 0V. Actually the output swing of the INA188 can only swing 0.22mV from either rail. So on a 12V single supply, the output can only ever be from 0.22mV to 11.78V.

    If you expect to be able to detect small input differential voltages near ground, then we suggest you supply a negative supply or level shift the output (set the Vref to mid-supply). 

    Here is an example with numbers: Assume 12V single supply with Vref=0V.  If you have an input differential voltage from the load-cell if 1mV, then with a gain of 207V/V, you expect to see 0.207V at the output. However, because of the output stage voltage limitation on amplifiers, you will see 0.22V on the output of the INA188. That is as far as the amplifier can go in that direction. So you've lost accuracy there.

    Here is what I would recommend: set Vref to 6V (mid-supply). You can see from the image below, as I sweep Vdiff from 0V to 24mV, your output will be 6V to 10.98V:

    -Tamara

  • 0 in reply to Tamara M Alani
    Prodigy 20 points

    Hi Tamara,

    Thanks for your reply. I agree that we should set some higher reference voltage than 0.22V. In case we keep the reference to mid point i.e. at 6V, since the output will swing from 6-12V but as MCU is operating at 5V hence to read we would need to go for potential divider to limit the ADC input to below 5V going to MCU. In doing so, we will loose too much resolution(assume we use 2.5:1 ratio of potential divider). In this case we would loose resolution of ADC from 0-2.4V(for 6V output from AMP, ADC input 6V/2.5=2.4V)

    Do you suggest any better solution to this? Should be set reference to 0.22V?

    Regards

    Vignesh

  • 0 in reply to Vignesh Vignesh
    Guru 140200 points

    Hi Vignesh,

    I would do it this way:

    3240.vignesh_ina188.TSC

    Kai