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ADS131M08EVM: ADC

HH1992
Prodigy 120 points
Part Number: ADS131M08EVM

Hi,

I'm using a load cell with these specifications: 

Full-scale load: 4400N

Rated Output; 4.5mv/v

Excitation: max 15.

ADC res, 24 Bit

I want to have a minimum 10N accuracy. I connected to excitation to a 2.5V accurate reference voltage. The ADC reference voltage(max 1.3v) is supplied from another accurate 1.2V. (LT6658).

I know that this is not a radiometric reading. But I'm confused about how can I convert the ADC value to the analog voltage.

For example, for a 22N weight, the ADC value is 162. Is the below formula is correct?

(162/2^24)*1.2=11.58uV 

should I use 1.2 v for this formula or 2.5V?

Thanks

  • 0
    TI__Guru 66446 points

    Hello Hamed,

    Welcome to our forum and thank you for your post!

    The ADC full-scale range is based on the reference voltage. Equation 3 in the data sheet calculates this as FSR = +/-1.2 V / Gain when the internal reference is used.

    When using an external reference, there is some additional scaling to that equation as written in the description in section 8.3.4. The full-scale range becomes (1.2 V / 1.25 V) x Vref_external. If Vref_external = 1.2 V, then FSR = +/- 1.152 V / Gain. Use these equations to translate the binary two's complement output code back to input-referred voltage. This voltage is the differential voltage between the input pins of the channel.

    Once you have the input-referred voltage calculated, you'll need to translate that to a load force based on your specific load cell's transfer function and the provided excitation voltage.

    Regards,

    Ryan

  • 0
    Prodigy 120 points

    Hi Rayan,

    thanks for your information.

    So considering that gain is one, I should use 1.2V for the formula.

    (162/2^24)*1.2=11.58uV 

    Is it correct?

    Thanks

  • 0 in reply to HH1992
    TI__Guru 66446 points

    Hello Hamed,

    You mentioned that you are using an external 1.2-V reference in your original post. Therefore, you must scale FSR by the ratio I explained above. Therefore,

    FSR = +/-(1.2 / 1.25) x Vref_external / Gain

    The LSB formula becomes 1 LSB = (2 x (1.2 / 1.25) x 1.2) / (2^24 - 1)

    162 codes is equal to approx. 22.25 uV.

    Best regards,

    Ryan

  • 0 in reply to Ryan Andrews
    Prodigy 120 points

    Hello Ratan, 
    Just another question.

    As you mentioned the formula is "2 x (1.2 / 1.25) x 1.2) / (2^24 - 1)",

    The last  one is (2^24-1), is it 2^23 or (2^24)-1

    Because with 2^24, 162 equals 11.12uV.
    Thanks

  • 0 in reply to HH1992
    TI__Guru 66446 points

    Hello Ratan,

    Exponents are computed before addition. (2^24 - 1) = (2^24) - 1. 

    We have (2^23 - 1) positive codes, ranging from 0 to 8,388,607, and (2^23) negative codes ranging from -8,388,608 to 0. The "-1" is to avoid counting the step change to code 0 twice for a total of 16,777,215 steps in the transfer function. Slight smile

    Regards,

    Ryan