ENOB and OSR ratio

[yefj]

Hello, In the lecture they say that if we double the OSR we get 0.5Bit increase.
I assumed its FOM of Walden as shown bellow but as you can see bellow We have 2^B goes down when Fs goes up so i cant see its happening.
Maybe they mean a different formula?
Where did i go wrong?
Thanks.

 

[danadakk]

Maybe this will help -

ADC performance evolution: Walden figure-of-merit (FOM)

POWER EFFICIENCY TRENDS: A series of blog posts on A/D-converter performance trends would not be complete without an analysis of figure-of-merit (FOM) trends, would it? We will therefore take a loo…

converterpassion.wordpress.com

Regards, Dana.

 

[sutapanaki]

If we don't oversample we have quantization noise power of LSB^2/12 which is distributed from 0 to Fs/2. If we keep the same FSR and physical bts out of the ADC but we sample twice as fast, that is the sampling frequency now is 2*Fs, then quantization noise power is distributed from 0 to Fs. If the signal BW is Fb, then now we will have 2x smaller Qnoise power in this BW compared to the case with no oversampling. So, Qnoise power drops by 3dB and since 1 bit is worth 6dB in Q noise power, then 3dB is worth half a bit.

 

[yefj]

Hello Sutapanaki, Yes i agree if we do the degital filtering on OSR=2 we qet SQNR ratio increase by 2 which is 3db so
ENOB=(SQNR_new-1.76)/6.02=(SQNR_old+3db-1.76)/6.02=enob_old+3/6.02

Is this the meaning?

Thanks.

 

[sutapanaki]

Yes

 

[KlausST]

Hi,

OSR=2 we qet SQNR ratio increase by 2 which is 3db so

factor of 2 (in voltage) is 6dB.
factor of 2 (in power ) is 3dB.

in my eyes:
* increase by two means " +2" (which can not expressed in dB)
* factor of 2 means " x 2 "

Klaus