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FOM of sigma-delta ADC (Read 2954 times)

ywguo

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Posts: 943
Shanghai, PRC

FOM of sigma-delta ADC
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...... Huh

Am I wrong?

Best regards,
Yawei

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Visjnoe

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Posts: 233

Re: FOM of sigma-delta ADC
Reply #1 - 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

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ywguo Community Fellow Offline Posts: 943 Shanghai, PRC	Re: FOM of sigma-delta ADC Reply #2 - Jan 22^nd, 2007, 6:11am Hi, Peter, I agree with you. Unfortunately, the authors didn't give their defintion of FOM. Thanks Yawei
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