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DAC80004: Some confusion about its noise

Part Number: DAC80004

Hello, everyone. The confusion is as follows. If we use a 3.0V reference as the full-scale voltage, then 1LSB is about 3/(2^16-1)=46uV. If we want to guarantee the 16bits resolution we should make the total output peak-peak noise is much smaller than 1/2*LSB which is 23uVpp. So it means the total rms noise should be smaller than 1/3*1/6*1/2*LSB=1.3uVrms. The datasheet of dac says its broadband noise is 55nV/rtHz at 10KHz. Let's assume it to be 50nV/rtHz and ignore the 1/f noise. Now we have BW=(1.3uV/(50nV/rtHz))^2=676Hz as the noise bandwidth, which means the signal bandwidth should be below 676Hz to  guarantee the resolution. The conclusion seems confusing to me and i'm not quite sure the above analysis is correct or not. Can anyone help to tell how to understand it?

  • Hi Xueliang,

    Thank you for your query. I am not able to understand why you have done 1/3*1/6*1/2. The peak-2-peak to rms conversion is 1/6. That means, to limit the noise below 1/2LSB i.e. 23uVpp or (23uVpp/6 = 3.8uVrms), you need a BW of (3.8uVrms/55nV)^2 = 4.7kHz.

    However, this is the worst case calculation. It will purely depend on the application about how the noise impacts the final output.

    Hope that answers your question.

    You can go through some TI precision lab training on noise calculations:

    training.ti.com/.../1313 - Noise 3 - slides.pdf

    Regards,
    Uttam Sahu
    Applications Engineer, Precision DACs
  • Hi Uttam,

    Thanks for your reply. The factor 1/3 comes from my misunderstanding of this: when rms-adding two unrelated noise sources if noise-a is less than 1/3 of noise-b then we can ignore the effect of noise-a. So I made it into the calculation. It's a mistake. : )

    But, I'm not quite clear about your words "the worst case". Do you mean that in actual application the result of above estimation will be higher then 4.7KHz, e.g. 5.9KHz or 7.2KHz?

    Can you say something more about this? How will this happen and if it does happen how much increase will the result be? Does an order of magnitude, i.e. 47KHz, seem reasonable?

  • Hi Xueliang,

    By worst case, I meant that we are calculating the noise figure considering it never goes beyond 1/2 LSB. Many applications may not be so sensitive to the peak-to-peak thermal noise. Hence, you may get an advantage in the bandwidth. May I ask what is your application?

    Regards,
    Uttam
  • Hi Uttam,
    The application is to drive a piezo stack to generate displacement, and I think the displacement resolution is determined by the peak-peak noise. Is that right?
  • Hi Xueliang,

    Is the displacement control in closed-loop or open-loop? In open-loop, the peak-to-peak noise will matter. But in closed-loop, it won't.

    Regards,
    Uttam
  • Hi Uttam,

    In my opinion the peak-to-peak noise will also matter in a closed-loop control system. Assume that the displacement resolution which is determined by peak-to-peak noise is of 10 units, the wanted displacement accuracy is of 5 units, and there is a sensor with accuracy&resolution of 1 unit. In closed-loop control, the sensor can always find out the error between the objective position and the current position, but the actuator with inadequate resolution will step over the target and maybe the system will never settle.

    Regards,

    Xueliang

  • You can consider peak-to-peak, in that case.

    Hope that resolves your issue.

    Regards,
    Uttam
  • Hi Uttam,

    Yes, you had helped a lot.

    So, in that case, I dave to limit the bandwidth of DAC output signal within about 4.7kHz to guarantee the its resolution.

    Regards,

    Xueliang