demistifying instrument specifications: some help on maths?

[Garcas]

Hi,

I am not sure if this is the place to post this but I am quite desperate. In application note 155 (demistifying instrument specifications) it's said that the effective resolution of a voltimeter that has a noise of N counts or N volts peak-to-peak decays to 0.52xN counts because of the gaussian distribution of noise.

Does anybody know whre that 0.52 factor comes from and how is it obtained? I am really interested in it because it will help me understand how noise affects system performance.

I can make to the point that adding noise to the system is adding a random variable with standard deviation of X/6. But how can I relate that to resolution ?

Thankls a lot

[Will Denman]

Good Afternoon,
I spoke with the author of this application note, and he had the following explanation of the .52 factor:

"ENOD is actually defined as log10(range/(rms noise * sqrt(12))). The sqrt(12) factor comes about from computing the rms value of uniformly-distributed quantization noise. (ENOD answers the question of "how many ideal digits do I have." This implies an "ideal" converter of some sort. Such a converter would have no noise/error other than that due to quantization. The rms value of that error is 1/sqrt(12) of a quantization level. I can show the derivation if necessary.)

Anyway, the app note specifies p-p noise instead of rms. If we assume the noise is Gaussian, then p-p noise is roughly 6.6 times rms noise. S
o the ENOD equation becomes: ENOD = log10(range/((p-p noise/6.6)*sqrt(12))). Simplifying gets us ENOD = log10(range/(0.52 * p-p noise)), as stated in the app note."

Please let me know if you need additional information. Have a great day.

Sincerely,
Will Denman
Application Engineering
National Instruments

[Garcas]

Thanks a lot Will,

I was working just in the opposite direction. Now that is clear enough.

By the way, there�s something else I�ve found in this application note that I can�t understand. Don�t worry; this is the last one 🙂

In page 6 under the �Resolution, precision and accuracy� discussion, it is said that an ideal 24 bits ADC would provide up to 7 digits resolution, an 18 bits 5 digits and an 18.6 bits 5 1/2 digits. How is this derived?

I�ve tried reasoning it, but I don�t get the same results: a 24 bits ADC can count 16 777 216 steps, which means from � 8 388 608 to + 8 388 608. This would be a 6 �and something� digits, almost 7 digits. With the same reasoning I get +/- 131 072 counts for 18 bits and +/- 198 668 counts for 18,6 bits. Why one i
s 5 digits and the other 5 ½ digits?

Then I try the other way:
- A 7 digits DMM can count +/- 9 999 999 = 19 999 998 steps.
This would require and ADC with 25 bits at least (2^24=16 777 216 steps, not enough)
- A 5 digits DMM can count +/- 99 999 = 199 998 steps.
This would require and ADC with 18 bits at least
- A 5 1/2 digits DMM can count +/- 199 999 = 399 998 steps.
This would require and ADC with 19 digits at least

Thanks a lot again.

PS: by the way, I think this is one of the best and most valuable application notes I�ve ever found about these topics. There�s not much around, and it�s a really important matter.

[Will Denman]

The digits of resolution of a DMM is derived using formula (1) in this document and a voltage range of 1V:

ENOD = log(total range/resoution at that range)

for 7 digits:

= log ( 1V / 1677216)

= 7.22

Please let me know if you need additional clarification. Have a great day.

Sincerely,
Will Denman
Application Engineering
National Instruments