Answer:
minimum of 12 bits is required
Explanation:
Full scale range Ef = 5v = L
Resolution [tex]R_{ADC} = \frac{L}{2^n -1}[/tex]
n = 4 bit
[tex]R_{ADC} = \frac{5}{2^4 -1} = 0.333 V
Quartization error is
[tex] \epsilon = \frac{1}{2} R_{ADC}[/tex]
[tex]= \frac{1}{2} 0.333 = 0.166 V[/tex]
we know if error is less than 0.001 V then we have
[tex]1/2 R_{ADC} \leq 0.001[/tex]
[tex]R-{ADC} \leq 0.002[/tex]
[tex]0.002\geq \frac{L}{2^n -1}[/tex]
[tex] \frac{L}{2^n -1} \leq 0.002[/tex]
[tex]\frac{L}{0.002} \leq 2^n -1[/tex]
Solving for n we get
[tex]n\geq 11.288[/tex]
n = 12
Therefore minimum of 12 bits is required
