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ADXL1002 Accuracy Limitation

Hi engineer team,

 

I was looking on the ADXL1002 datasheet and I was wondering about something : is that significant and accurate to measure a voltage difference which is less than 2mV or 1mV with the ADXL1002 ? We know that the sensitivity is 40mV/g, and the ADXL1002 is made for a huge range +/-50g, so basically the main purpose is not to measure low variation.

I saw as well that the sensitivity due to the temperature is 5% of 40mV/g. So I assume than measuring an acceleration which is lower than 5% of 1g (which means 2mV) is not really accurate.

I am currently asking this cause we are reflecting about which ADC we are gonna use, and I am wondering if a 12bit ADC with a 4V VREF is enough (which allow us to have a resolution of 0.97mV).


Thanks for your help,

Florent

Parents
  • Thank you for your post.  While you are correctly citing the sensitivity error and measurement range from the ADXL1002 datasheet, I do not understand how that limits the capability of the ADXL1002 as a vibration sensor, nor do I understand why the sensitivity error would drive the resolution requirement for a complementary ADC.  Page 11 of the ADXL1002 datasheet actually recommends the use of a 16-bit ADC (AD4000). 

    I didn't write the ADXL1002 datasheet, nor did I select the AD4000 for this purpose, but I can guess at what the author's motivation was.  I suspect that the resolution of this ADC was driven by making sure that its (ADC) quantization noise was less than the total noise contribution from the ADXL1002, for the stated bandwidth of the application (5kHz  in this context).  This is often an important criteria for those who do not want the ADC to limit the resolution of a particular signal. So let's do some math and see how this works out:

    • Sensor noise: assume single pole filter, which has a -3dB frequency of 5kHz and a nominal noise density (25ug/sqrt(Hz)):
      • Total Noise = Noise Density x sqrt(Noise Bandwidth)
      • Total Noise = 25 x 10^-6 g/sqrt(Hz) x sqrt[1.57 x 5000Hz] = 2215ug
      • Total Noise = 2215 ug x 40mV/g x 1g/1000000ug = 0.0886mV
    • ADC Quantization noise, assume Vref = 5V, 16-bits and full use of the ADC input range
      • q = 5V / (2^16-1) = 0.0763mV
      • Quantization noise = q/sqrt(12) = 0.022mV
    • So, it would appear like the (5kHz) total noise of the ADXL1002 (0.0886mV) is greater than the quantization noise of the ADC (0.022mV). 
    • If we make the same assumptions about a 12-bit ADC, its quantization noise (0.352mV) is actually larger than the total noise of the ADXL1002 (0.022mV), which means that the ADC will limit the resolution of the vibration observation. 
      • q = 5V / (2^12-1) = 1.22mV
      • Quantization noise = q/sqrt(12) = 0.352mV

    : Am I thinking about this correctly? 

Reply
  • Thank you for your post.  While you are correctly citing the sensitivity error and measurement range from the ADXL1002 datasheet, I do not understand how that limits the capability of the ADXL1002 as a vibration sensor, nor do I understand why the sensitivity error would drive the resolution requirement for a complementary ADC.  Page 11 of the ADXL1002 datasheet actually recommends the use of a 16-bit ADC (AD4000). 

    I didn't write the ADXL1002 datasheet, nor did I select the AD4000 for this purpose, but I can guess at what the author's motivation was.  I suspect that the resolution of this ADC was driven by making sure that its (ADC) quantization noise was less than the total noise contribution from the ADXL1002, for the stated bandwidth of the application (5kHz  in this context).  This is often an important criteria for those who do not want the ADC to limit the resolution of a particular signal. So let's do some math and see how this works out:

    • Sensor noise: assume single pole filter, which has a -3dB frequency of 5kHz and a nominal noise density (25ug/sqrt(Hz)):
      • Total Noise = Noise Density x sqrt(Noise Bandwidth)
      • Total Noise = 25 x 10^-6 g/sqrt(Hz) x sqrt[1.57 x 5000Hz] = 2215ug
      • Total Noise = 2215 ug x 40mV/g x 1g/1000000ug = 0.0886mV
    • ADC Quantization noise, assume Vref = 5V, 16-bits and full use of the ADC input range
      • q = 5V / (2^16-1) = 0.0763mV
      • Quantization noise = q/sqrt(12) = 0.022mV
    • So, it would appear like the (5kHz) total noise of the ADXL1002 (0.0886mV) is greater than the quantization noise of the ADC (0.022mV). 
    • If we make the same assumptions about a 12-bit ADC, its quantization noise (0.352mV) is actually larger than the total noise of the ADXL1002 (0.022mV), which means that the ADC will limit the resolution of the vibration observation. 
      • q = 5V / (2^12-1) = 1.22mV
      • Quantization noise = q/sqrt(12) = 0.352mV

    : Am I thinking about this correctly? 

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