I know that mathematically the k-V analog input on an A to D converter could be converted to the digital code of \$\dfrac{k(2^n-1)}{VDD}\$. But I really don't understand where it comes from and the logic behind it. Any help would be appreciated.
\$\begingroup\$
\$\endgroup\$
14
-
\$\begingroup\$ The values of a n-bit number range from 0 to 2**n-1. \$\endgroup\$– Ignacio Vazquez-AbramsCommented Sep 30, 2015 at 11:54
-
\$\begingroup\$ yes but how we get to that formula in the question? \$\endgroup\$– JackCommented Sep 30, 2015 at 11:56
-
\$\begingroup\$ The minimum value is represented by 0V and the maximum value is represented by Vdd. The rest is just linear interpolation. \$\endgroup\$– Ignacio Vazquez-AbramsCommented Sep 30, 2015 at 11:57
-
2\$\begingroup\$ I think my problem was that I didn't realize that the value of the 1024 level is 1023 because it is all ones. I think I get it now. Let me know if I am right. \$\endgroup\$– JackCommented Sep 30, 2015 at 12:55
-
2\$\begingroup\$ You are correct. \$\endgroup\$– AlexxxCommented Sep 30, 2015 at 13:40
|
Show 9 more comments
1 Answer
\$\begingroup\$
\$\endgroup\$
My problem was that I didn't realize that the value of the 1024 level is 1023 because it is all ones.