To whom it may concern:
So I have scouted through the internet for a while to figure out different ways to calculate the effective resolution.
Finally, I found Application Note AN-615, which helped a lot.
But I still do not understand 100% based on the AD7746 datasheet.
So this is more of a two-part conceptual question.
So I think I understand how to derive Tables 6 and 7. I just wanted to verify if I am on the right track.
To derive the general equation from AN-615,
SNR = 20 log ( RMS Noise (uV) / (full range of voltage reference) )
effective resolution (bits) = (SNR -- 1.76) / 6.02
So for Table 7, I tried to derive using one of the case examples:
RMS Noise = 2.1e-6 V and Full Range for External Voltage Reference = 5 V (2.5 *2 since range is +2.5 to -2.5)
Plugging it into the equation, I get 20.89 bits, which is close to the 21.1 bits on the datasheet.
And it seems continuing with this general equation yields values very close to the ones for TABLE 7.
As for Table 6, looking at the datasheet, it seems the internal voltage reference is between 1.169 and 1.171 V. Is this correct??
Using one of the provided values:
RMS Noise = 11.4e-6 V and External Reference as 2.34 V (1.171 * 2 since range is +1.17 to -1.17)
Plugging it into the equation, I get 17.36 bits, which is close to the 17.6 bits on the datasheet.
And it seems continuing with this general equation yields values very close to the ones for TABLE 6.
Naturally, it would seem logical to use the same equations to analyze TABLE 5; however, because it deals with Capacitive Input Noise, I am at a loss as to how to calculate. If I tried using the same equations, what would I divide the RMS Noise (aF) by? The accuracy value of 4fF? The resolution for ENOB of 21 derived as 4aF?
In other words, I do not know how to calculate the effective resolution that is shown in the AD7746 datasheet.
Thank you very much.
Message was edited by: Allan Morales