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Design >> Mixed-Signal Design >> FOM of sigma-delta ADC
https://designers-guide.org/forum/YaBB.pl?num=1169457500

Message started by ywguo on Jan 22^nd, 2007, 1:18am

Title: FOM of sigma-delta ADC
Post by ywguo on Jan 22^nd, 2007, 1:18am

Hi, Guys,

I read a paper in ISSCC 2006, "An 80/100MS/s 76.3/70.1dB SNDR ΔΣ ADC for Digital TV Receivers" .

It reads that
Quote:

The ΔΣ ADC achieves a 76.3/70.1dB peak SNDR over a 3.2/4MHz bandwidth while consuming 23.8/34.4mW from a 1.8V supply, which leads to a 0.7/1.64pJ/conversion FOM

I calculated and found that the FOM should be 1.4/3.29pJ/conversion, which are twice of the FOM in that paper. My formula is
Power
FOM=---------------------
2^ENOB * bandwidth
where bandwidth is 3.2/4MHz respectively.

I am sure this formula is correct when I calculate FOM of nyquist ADCs, where bandwidth is ERBW or fs. But for an oversampling ADC, Mm...... :-? Am I wrong?

Best regards,
Yawei

Title: Re: FOM of sigma-delta ADC
Post by Visjnoe on Jan 22^nd, 2007, 3:38am

Hi Ygwuo,

first of all, there are a lot of FOM definitions, but the one you use is a very common one.

The difference between your calculated value and the one in the paper lies in the 'bandwidth' term in your equation: most likely, that should be the sampling rate Fs and according to the Nyquist theorem, this should be at least twice the bandwidth, hence the factor 2.

Kind Regards

Peter

Title: Re: FOM of sigma-delta ADC
Post by ywguo on Jan 22^nd, 2007, 6:11am

Hi, Peter,

I agree with you. Unfortunately, the authors didn't give their defintion of FOM. :)

Thanks
Yawei