Hi,

I'm trying to clarify my understanding of the relationship between ADC resolution and ENOB (Effective Number of Bits), particularly when it comes to practical calculations. From what I understand, ENOB represents the actual performance of an ADC accounting for real-world imperfections like noise and distortion, while the resolution is the theoretical bit depth of the converter. Since ENOB is invariably less than the stated resolution, I'm confused about which value should be used in calculations.

Specifically, when calculating the LSB (Least Significant Bit) step size, should I be using: 1 LSB = V_ref / 2^Resolution (the theoretical value) or 1 LSB = V_ref / 2^ENOB (based on effective performance)?

If ENOB is the more meaningful metric for actual performance, why do datasheets prominently feature the resolution specification? Is the resolution purely a marketing/theoretical spec, while ENOB tells the real story? Or do these two parameters serve different purposes in system design?

I'd appreciate any insights into when each specification is relevant and how you approach LSB calculations in your designs.

 

hi hoy,
this may help, Google clip.

https://www.google.com/search?client=firefox-b-d&q=relationship+between+adc+resolution+and+ENOB

AI Overview ADC resolution (\(N\)) represents the maximum possible digital bits in an ADC's output, while Effective Number of Bits (ENOB) represents its real-world performance, accounting for noise and distortion. ENOB is typically lower than resolution (\(N>ENOB\)) and is calculated from the SINAD (Signal-to-Noise and Distortion ratio) using: \(ENOB=(\text{SINAD}-1.76)/6.02\). Key Aspects of the Relationship: Theoretical vs. Real Performance: While a 12-bit ADC has a nominal resolution of 12 bits, its ENOB might be 10.5 bits due to noise.SNR Link: An ideal ADC's Signal-to-Noise Ratio (SNR) is defined as \(6.02\times N+1.76\text{\ dB}\). ENOB replaces \(N\) in this formula to reflect the real SNR.Impact of Noise: As noise increases, the SNR decreases, reducing the ENOB.Effective Resolution vs. ENOB: ENOB is used for AC performance (using SINAD/SNR), while "effective resolution" is often used for DC, referring to the noise-free code resolution.Significance: Analog Devices describes how ENOB provides a more accurate, honest representation of an ADC's precision in practical, noisy applications.

 

Hi,

I'm trying to clarify my understanding of the relationship between ADC resolution and ENOB (Effective Number of Bits), particularly when it comes to practical calculations. From what I understand, ENOB represents the actual performance of an ADC accounting for real-world imperfections like noise and distortion, while the resolution is the theoretical bit depth of the converter. Since ENOB is invariably less than the stated resolution, I'm confused about which value should be used in calculations.

Specifically, when calculating the LSB (Least Significant Bit) step size, should I be using: 1 LSB = V_ref / 2^Resolution (the theoretical value) or 1 LSB = V_ref / 2^ENOB (based on effective performance)?

If ENOB is the more meaningful metric for actual performance, why do datasheets prominently feature the resolution specification? Is the resolution purely a marketing/theoretical spec, while ENOB tells the real story? Or do these two parameters serve different purposes in system design?

I'd appreciate any insights into when each specification is relevant and how you approach LSB calculations in your designs.

why do manufactureres specify number of buts,
because, given perfect conditions, thats what the chip can produce a rssolution of, i.e tbe smalest signal input change they can detect .
but thats going to be at a very specific, slow, 'zero' jitter sample rate, probably at a given level in, 'zero' noise power supplies, and at specific temprature.

ENOB is related to the sample clock frequency, the frequency of the input signal and the jitter of the sample clock,
thus as these change , the ENOB goes down,

So both, and other numberes are needed in system design.

better data sheets, show graphs of how the different paramiteres relate to each other.

 

My current understanding:

Resolution defines the theoretical quantization - a 16-bit ADC has 2^16 discrete output codes, so with a 1V reference:

1 LSB = 1V / 2^16 = 15.26 µV (the step size between codes)

ENOB represents the actual usable performance accounting for real-world noise and distortion. If the same 16-bit ADC has ENOB = 14 bits, then:

Effective noise floor ≈ 1V / 2^14 = 61 µV RMS. Signals smaller than ~61 µV will be buried in the noise (hard to distinguish from random fluctuations)

This means that while the ADC outputs 16-bit codes with 15.26 µV steps, the inherent noise causes the readings to fluctuate by approximately ±2 LSB, making the effective resolution equivalent to a cleaner 14-bit converter.

Is my understanding correct?

 

https://www.analog.com/en/resources...tal-converter-circuits--maxim-integrated.html

 

This means that while the ADC outputs 16-bit codes with 15.26 µV steps, the inherent noise causes the readings to fluctuate by approximately ±2 LSB, making the effective resolution equivalent to a cleaner 14-bit converter.

There is still information in the noise, though, and it can be recovered via oversampling and digital filtering, i.e. you can trade time for effective resolution.

 

My current understanding:

Resolution defines the theoretical quantization - a 16-bit ADC has 2^16 discrete output codes, so with a 1V reference:

1 LSB = 1V / 2^16 = 15.26 µV (the step size between codes)

ENOB represents the actual usable performance accounting for real-world noise and distortion. If the same 16-bit ADC has ENOB = 14 bits, then:

Effective noise floor ≈ 1V / 2^14 = 61 µV RMS. Signals smaller than ~61 µV will be buried in the noise (hard to distinguish from random fluctuations)

This means that while the ADC outputs 16-bit codes with 15.26 µV steps, the inherent noise causes the readings to fluctuate by approximately ±2 LSB, making the effective resolution equivalent to a cleaner 14-bit converter.

Is my understanding correct?

that works ,
but the "noise" , lost bits are not neciseraily fixed by a better adc,
The sample clock jitter and signal bandwidth also affect the ENOB.

 

There is still information in the noise, though, and it can be recovered via oversampling and digital filtering, i.e. you can trade time for effective resolution.

https://forum.allaboutcircuits.com/threads/50nv-resolution.130468/post-1073894


From this



To this

 

https://forum.allaboutcircuits.com/threads/50nv-resolution.130468/post-1073894
View attachment 363265

From this

View attachment 363266

To this

Yep, every 4 times oversampling gains one inherent bit

 

My current understanding:

Resolution defines the theoretical quantization - a 16-bit ADC has 2^16 discrete output codes, so with a 1V reference:

1 LSB = 1V / 2^16 = 15.26 µV (the step size between codes)

ENOB represents the actual usable performance accounting for real-world noise and distortion. If the same 16-bit ADC has ENOB = 14 bits, then:

Effective noise floor ≈ 1V / 2^14 = 61 µV RMS. Signals smaller than ~61 µV will be buried in the noise (hard to distinguish from random fluctuations)

This means that while the ADC outputs 16-bit codes with 15.26 µV steps, the inherent noise causes the readings to fluctuate by approximately ±2 LSB, making the effective resolution equivalent to a cleaner 14-bit converter.

Is my understanding correct?

To a first approximation, yes. But there are games you can play to clean up the noise, depending on the kind of noise you are dealing with and how well you understand it. Having the extra bits of ADC resolution may facilitate doing so, but even without them you can often get more effective bits than the ADC resolution by oversampling and filtering the data.