getting "clear" 8 bit

[yxo]

Let's take a non-ideal ADC, for example 10 bit with ENOB 9.5 But will use only 8 upper bits. Will I get ENOB = 8bit? I think it won't because of I have non-linear distortion, kT/C noise. As a result, I would have ENOB = 7.5 bits. What do you thnik? thanks

 

[gevy]

ENOB is defined by Signal/Noise Ratio (SNR).
As SNR is large enough as you will get 8 bits.

 

[yxo]

It is not so. ENOB = (SNDR - 1.76)/6.02. (1.76 and 6.02 is appeared after integration of quantization error )
At the same time SNDR is a _SUM_ of four-factors SNR quantization error + input reffered noise +Phase noise +harmonic distortion. Anyway, the three last factors will decrease SNDR.. and decrease ENOB.

 

[electronspin]

Assuming these noise powers added. Then we see that the original limit of 10 bits (quantization noise) was impacted negatively by 3 dB reducing the total performance by half a bit down to 9.5 bits. The extra noise then, injected by the circuit somehow (again, assuming it is additive at this particular voltage magnitude/signal and frequency) has the same magnitude as the quantization noise, i.e. 10 bit noise, so that the two together equal 9.5 bits.

If the above is true then if the lower two bits are truncated the quantization noise alone would now give 8 bits plus our 10 bit inherent circuit noise with gives us the total output noise of: 7.96.

Again, this assumes much, but this whole conversation needs some boundaries.