Hello,


We know that ADC 1LSB=Vref/(2^N) where N is the resolution of the ADC.

We know that ADC Effective number of bits (ENOB) is always less than the resolution "N".

When calculating 1LSB do I need to take 1LSB = Vref/(2^ENOB) or I need to take 1LSB=Vref/(2^N)

 

1LSB = Vref/(2^N)

 

Hello,


We know that ADC 1LSB=Vref/(2^N) where N is the resolution of the ADC.

We know that ADC Effective number of bits (ENOB) is always less than the resolution "N".

When calculating 1LSB do I need to take 1LSB = Vref/(2^ENOB) or I need to take 1LSB=Vref/(2^N)

The difference between ENOB and true bits, is expressed as "noise"
so for an 8 bit DAC you always have 8 bits, but the bottom two bits might be "noise"
but then if you average 16 samples of the same "voltage" , then you regain the two bits, ..
https://www.ti.com/lit/pdf/sprad55#...conversions,how much oversampling takes place.

 

The difference between ENOB and true bits, is expressed as "noise"
so for an 8 bit DAC you always have 8 bits, but the bottom two bits might be "noise"
but then if you average 16 samples of the same "voltage" , then you regain the two bits, ..
https://www.ti.com/lit/pdf/sprad55#:~:text=Software oversampling performs multiple conversions,how much oversampling takes place.

Thank you.
" bottom two bits might be "noise" ".
Bottom two bits means LSB.Please correct me if I am wrong

 

Thank you.
" bottom two bits might be "noise" ".
Bottom two bits means LSB.Please correct me if I am wrong

correct