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[讨论] sigma delta ADC中 第一级采样电容大小确定 |
1000资产
最佳答案as far as i know, the input signal should be 2.5V and differential. So, the input power will be Vpower=(2.5^2)/2.
Then the KT/C noise power is: Npower=(1.38x10^-23*300)/(1024*12pF), if OSR=1024 & sampling cap is 12pF.
So, the SNR will be: 10*log(Vpower/Npower)=129.7dB, that will be what you want.
Overall, the input signal is too small in your assumption, that doesn't make sense.
I didn't desi ...
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