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What is relation between Resolution and ADC Noise level

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Bjtpower_magic

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Hi Everyone,

I wanted to understand that ADC Noise level.
I am using ADC of 24bit ADS131M06

How to measure the ADC Noise level?
Are there anyways to relation in resolution vs Noise level

Marx
 

Hi,

Are there anyways to relation in resolution vs Noise level

the effective number of bits tells you how many bits of your ADC are actually "useable", depending on the prevailing noise and distortion, see [1].

How to measure the ADC Noise level?

A possebility for a DC coupled ADC input is the observation of a battery connected to the ADC input filter. As the voltage of the battery can be considered stable (clean), the standard deviation of a large number of measured samples displays the noise. Keep in mind, the noise will be also influenced by the stability of your (external) reference voltage and power supply.

[1] https://www.analog.com/media/en/training-seminars/tutorials/MT-003.pdf

BR
 


Hi,

ADC resolution is the smallest step size regarding digital code.
Noise is some kind of uncertainty in conversion. It may rsult from analog noise, from bad timing, from power supply from EMi/EMC...

Resolution has nothing to do with noise.

ENOB is really telling "how many bits" of resolution give useful information.

From my experience:
There are many 24 bit ADCs. I´ve not yet seen an ADC with 24 bit ENOB (but I did not search for one, either).
There are many 24 bit ADCs with ENOB < 14 bits.

(jokingly) I call them "true 14 bit ADCs with a 10 bit random generator"

Klaus
 

I will give one use case example
I wanted to measure Max AC Voltage with resolution of 1V.

How do i make sure that i will get 1V Resolution with ADC?
What parameters i need to look into it to make sure my ADC Resolution is correct?
Here i have used ADC AD131M06 Chip which is 24bit ADC.

Marks.
 

Hi,

as KlausST stated in reply #4
ADC resolution is the smallest step size regarding digital code.

This means with respect to your used reference voltage you need the correct (usable) number of bits N, where your step size is also known as LSB (least significant bit), LSB = Vref / (2^N).

BR
 

Hi,

Resolution is the step size.
So if you want a resolution of 1V then step size is 1V.
Now a 24Bit ADC has 2^24 = about 16,000,000 steps.

I doubt you want to build a measurement device designed for a neasurement range of 16 million volts.

--> your requirement does not match the chosen device.

******
Let's start from the beginning.
Define what your goal is.
You say you want to measure AC voltage. AC voltage is changing with time according it's wavefrom and frequency.
--> what waveform do you expect?
--> what fundamental frequency do you expect?
--> do you expect DC (DC superimposed by AC)? Are you interested in this DC value?
Now that the ADC only is able to measure an instantaneous value ... and it's the nature of AC that it's instantaneous value continously changes...the ADC output also changes with every measurement.
Thus you need to take many ADC samples and use them all to calculate an AC value.
Next problem: There are many "AC values". Thus you need to define which one you want to know.
--> select one: True RMS including DC, true RMS excluding DC, peak-to-peak, amplitude (of sine), average of rectified...or any other

An example:
You want to measure true "RMS excluding DC" of mains voltage with nominal value of 230VRMS, 50Hz sine with a resolution of 1V.
* excluding DC says you need to install a high pass filter (analog or digital) with a cutoff frequency way below 50Hz. Let's say 1Hz
* sampling frequency: since it's expected to be sine shape... I'd say 16 samples per full wave gives a good result: 16 × 50Hz = 800Hz sampling frequency
* voltage range: 230V RMS sine has 230V x sqrt(2) peak voltage, thus 2 x 230V x sqrt(2) peak-to-peak = about 650Vpp. But this is nominal only.
Maybe you expect the mains voltage to fluctuate +/-30% worst case.... you need a 30% bigger voltage range: 650Vpp x 1.3 = about 800Vpp.
* resolution: to decode an 800Vpp input signal with 1V resolution you need an ADC with 800V/ 1V = 800steps (or more). A 10 bit ADC has a resolution of 2^10 = 1024 steps.
* Now that the RMS algorithm takes at least 16 (16 or integer multiples of 16) values into account you win additional 3-4 bits of resolution. So indeed a simple 8 bit ADC is sufficient.

I've designed an RMS meter with a so called 24 bit ADC. Calculating an RMS value every 20ms shows a useful resolution down to almost 1mV at an 800Vpp range. This is about 20 bits overall ENOB or 20 bits of "useful resolution".

Klaus
 
Do a complete error analysis thru the signal path and the ADC to get at absolute
accuracy. Temp, PSRR effects, non linearity, component tolerances and drift....

Here is a good starting point, many more ap notes on various sites, especially Analog
Devices. Attached.




Regards, Dana.
 

Attachments

  • AN280.pdf
    2.1 MB · Views: 129

With reference to ADS131M06, i have seen in the datasheet that SNR=101dB
So, ENOB=(SNR-1.76dB)/6.02=(101-1.76)/6.02=~16Bits

So it means that i can only use 16Bits for resolution.
 
Last edited:

Hi,

ENOB means Effective Number Of Bits. The unit is "bits", not "dB".

so it´s about 16 bits of resolution. 16 bits = 2^16 steps = 65,536 steps.

Klaus
 
Hi,

ENOB means Effective Number Of Bits. The unit is "bits", not "dB".

so it´s about 16 bits of resolution. 16 bits = 2^16 steps = 65,536 steps.

Klaus
Got it.. Thanks :)
Why IC manufacturers not only mention ENOB rather than 8,16,24 Bits.

Resolution=Full scale voltage/(2^ENOB)..??
 

Hi,

many people think a 24 bit ADC is always better than a 16 bit ADC.
so "bit sells".

The bit value sometimes is useful, but sometimes it´s just a marketing value.

Klaus
 
Resolution=Full scale voltage/(2^ENOB)..??
No. As explained above, ENOB defines useful or effective resolution in contrast to numerical resolution. If you say just resolution, it's numerical resolution.
 
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