How to calculate a resolution of a delta-sigma ADC?

[CHL]

Hello

I'm confused how to calculate the resolution.

2^N clock cycles must be counted in order to achieve N-bit effective resolution.

For example, averaging 4 samples gives 2 bits of resolution, while averaging 8 samples yields 3 bits.

However, I cannot understand the required frequency.

If the input frequency is 10kHz and the ADC provides 16bit output, the sampling frequency should be 2*10k*2^16 = 1,310,720,000 Hz

This looks like reasonable, but weird.

What is wrong with my calculation?

 

[helpmejerry]

Hello CHL,

While delta-sigma or sigma-delta is in essence an averaging process the oversampling is not the only thing responsible for the ENOB.
For example only OSR improves the number ob bits from n to ENOB=n+0.5*log2(OSR). Now this means that for a 5-bit gaing OSR of 1024 is needed.

Now if you look at delta-sigma it also produces a shaping function of the quantization noise, which enhanses the OSR. SNR of a 1-bit modulator with OSR = 1024 becomes:

SNR=6.02*n+1.78-5.17+9.03*log2(1024) this equals to aproximately 92 dB which translates to aprox ENOB of 15 bits. Hopefully my math was correct .

Threre are number of resoureces regardin SDM. Something to get you started link.

 

[CHL]

Hello CHL,

While delta-sigma or sigma-delta is in essence an averaging process the oversampling is not the only thing responsible for the ENOB.
For example only OSR improves the number ob bits from n to ENOB=n+0.5*log2(OSR). Now this means that for a 5-bit gaing OSR of 1024 is needed.

Now if you look at delta-sigma it also produces a shaping function of the quantization noise, which enhanses the OSR. SNR of a 1-bit modulator with OSR = 1024 becomes:

SNR=6.02*n+1.78-5.17+9.03*log2(1024) this equals to aproximately 92 dB which translates to aprox ENOB of 15 bits. Hopefully my math was correct .

Threre are number of resoureces regardin SDM. Something to get you started link.

Hello

As mentioned above, if we want to get 3bit signals, we should count 8 PDM outputs.

If the output is 00110011...., the 3bit signal is (0+0+1+1+0+0+1+1)/(2^3) = 4/8 = 1/2, whose binary code is '100'.

What you explained is at the spectrum aspect, so I'm confused how the spectrum aspect and counter aspect are related to each other.

Thanks for your help

 

[FvM]

SNR=6.02*n+1.78-5.17+9.03*log2(1024) this equals to aproximately 92 dB which translates to aprox ENOB of 15 bits. Hopefully my math was correct

The calculation applies to a first order modulator. SD modulators are often implementing second or third order. The basic pulse density calculation in post #3 refers to a "zero" order modulator (no noise shaping at all). See the diagram in this recent thread: https://www.edaboard.com/threads/332686/