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Help me in calculating FOM of ADC

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coolstuff07

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adc fom

Hi,

I have designed 12-bit ADC with 30mw power dissipation, input frequency=100KHz and SNDR=70dB. Please any one help me how to find FOM. Please put formula and solve it with values.What is its best value.

Bye.
 

fom adc

There are 2 FOMs in literature. The old one is FOM1 = P/ min(fs, 2fin)*2^ENOB
If your ADC is pipelined than you should use FOM2 = P/ min(fs, 2fin)*2^(2ENOB)
where fs= sampling frequency.Note the precision of 2^2ENOB in the denominator.

So, your ENOB = (SNDR-1.76)/6.02 = 11.33

Therefore your FOM2 = 22.6fJ/step^2
and FOM1 = 0.58 pJ/step

I used the 2*Fin, whereas you can use the fs instead.

Please click "helped me" button if it helps you. Thanks
/Noman Hai
 
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    lamoun

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fom for adc

Hi,

Why you are saying FOM1 and FOM2.
Please can you explain what is the
significance of both.

Bye.

Added after 1 minutes:

Why both are different in your calculation
 

8kt.snr.fs

FOM1 is the most quoted FOM, however, FOM2 is better suited for thermal noise dominated ADCs like pipelined ADC.
FOM is the energy efficiency. In thermal noise dominated ADC, the low power consumption limit is P = 8KT. Fs. SNR^2
So for FOM = P/(Fs . SNR^2) thats why we have FOM2.

Also using FOM1, low resolution looks better. Whereas, using FOM2 give credit to higher resolution.

They are different because of the precision of 2^2ENOB. and their units are also different.

In the papers you will see they use FOM1. So you can use that FOM1 without any concern. However, I am more inclined towards FOM2 because of the reasons I mentioned above.
 

adc fom calculation

Hi,

FOM2 is reducing with increasing the frequency. Is it correct or some thing iam making worng.

Please help me.

Bye
 

power and area fom adc

both FOMs decrease with increase in frequency. However FOM2 is decreases at a rate 2^ENOB more than than FOM1.
Please look at the formulas of both the FOMs carefully
 

thermal-noise power-consumption adc

Thanx , you have provide very useful information.

Bye.
 

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