Question

1- A12 bits analog to digital converter (ADC) has a sampling rate of IMHz. If the analog signal has a range of 5V, a) How many different digital values can this ADC outputs? (20 POINTS) b) What are the smallest and largest possible outputs of this ADC? c) What is the voltage/bit resolution? d) If an application requires 1u V/bit resolution, how m ADC have? ny bits should the

Answer 1

a) Voltage Range = 5V

n is number of bits

Different digital values this ADC can output =2ⁿ = 2¹²  = 4096.

b) Smallest possible value of this ADC in binary is 000000000001$\frac{4096}{5}=\frac{ADC Reading}{Analog Voltage Measurement}$

$\frac{4096}{5}=\frac{1}{Analog Voltage Measurement}$

Analog voltage = 1.22 mV

Smallest Value is 1.22 mV

Largest Analog voltage = 5 - Resolution of ADC

= 5 - 1.22 mV = 4.99878 V

c)Voltage Resolution per bit = Voltage Range/ 2ⁿ

n is number of bits

Voltage Resolution = 5/2¹² = 5/4096 = 1.22mV/bit

d) For 1$\mu V$/bit resolution

Voltage Resolution = Voltage Range/2ⁿ

1 $\mu V$ = 5/2ⁿ

2ⁿ = 5/1$\mu V$

= 5 x 10⁶

n = 19

So, the ADC should have 19 bits.