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MSP430AFE253: SD24GAINx usage-time issue in SD24

Cherry Zhou
Mastermind 22235 points
Part Number: MSP430AFE253

Hi team,

Here's an issue from the customer may need your help:

When SD24GAINx is set to 0 time in the program, the exact voltage value can be calculated correctly using the formula (ad value / 7fff)*0.6V.

The customer now want to reduce the range -0.6V to 0.6V. Is it possible by modify SD24GAINx to achieve this? If so, set SD24GAIN x to 011 or x8. How much does 0.6 in the calculation formula (ad/7fff)*0.6 need to be modified?

They've tried 0.075 but there was a deviation. Around 0.0772 is probably correct. Why is that? Is the deviation from using the gain? Or is it not possible to implemented by this way?

Could you help check this case? Thanks.

Best Regards,

Cherry

  • 0
    TI__Mastermind 23520 points

    Hi Cherry,

    Since they have the gain set then if they want to convert to voltage they would want to take into account the 8x gain. So they can divide by the gain in their formula to get the result. The real formula would be Voltage = ( (AD Value) / (Value at FSR)) * (VREF/2) / Gain

    What are their other settings? Assuming they are using the Bipolar Two's compliment with a gain of x8 they would have the formula

    Voltage = AD / 7FFF * 0.6 / 8

    Regards,

    Luke

  • 0 in reply to Luke Ledbetter
    TI__Mastermind 22235 points

    Hi Luke,

    Thanks for your support.

    The customer is using a complement format and they tested two boards again with the same hardware, using the formula Voltage = ad / 7FFF * 0.6 / 8. Both boards have the same error as follows:

    • Changing gain to x1, use the formula Voltage = ad / 7FFF * 0.6. When actually enter 1mV, the result is 1.03mV. When actually enter 10mV, the result is 10.06mV
    • Changing gain to x2, use the formula Voltage = ad / 7FFF * 0.6 / 2. When actually enter 1mV, the result is 1.02mV. When actually enter 10mV, the result is only 9.9mV
    • Changing gain to x4, use the formula Voltage = ad / 7FFF * 0.6 / 4. When actually enter 1mV, the result is 1.02mV. When actually enter 10mV, the result is only 9.7mV
    • Changing gain to x8, use the formula Voltage = ad / 7FFF * 0.6 / 8. When actually enter 1mV, the result is 1.02mV. When actually enter 10mV, the result is only 9.5mV

    The greater the actual input voltage, the greater the error; the greater the gain, the greater the error.

    When the input signal drive is weak, use the formula (AD Value) / (Value at FSR)) * (VREF/2) / Gain without modifying gain to 1, calculations may have deviations (using a signal generator to output the test voltage).

    Thanks and regards,

    Cherry

  • +1 in reply to Cherry Zhou
    TI__Mastermind 23520 points

    Hi Cherry,

    There is a typical associated with the gain settings, so the gain isn't exactly 8. 

    I don't have their AD value but doing some calculations I have created a formula to get the actual gain. Their actual gain seems to be 7.6 (it could be 7.62 due to rounding) which is the typical in the datasheet.

    For reference to get the actual gain I did Actual Gain = Actual Read Voltage/ Expected Voltage * Expected Gain. Here the actual read voltage was 9.5mV the expected voltage was 10mV and the expected gain was 8.

    Please update the formula to use the typical gain instead of expected gain. (AD Value) / (Value at FSR)) * (VREF/2) / Typical Gain

    Regards,

    Luke

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