Calculating effective bit precision of DAC due to range restriction
I think your result is correct. A 16 bit DAC has 2^16 = 65536 steps, but over 5V range. You are only using 1/5 of the range, so are using 13107 steps. Since 2^14 = 16384 it is clearly a bit less than 14 bits. Log base 2 (n) = log (n)/ log (2) so log base 2(13107) = 13.68 bits.