Question

Analog-to-Digital convertor, modified. Assume that you have a 12-bit A/D converter that can digitize an analog voltage over the range of 0.00 Volts to + 3.30 Volts (just like the one in the TM4C123G).

(a) What is the minimum voltage that an analog input voltage could change and be guaranteed to be detected by a change in the digital output value? E.g. in millivolts or microvolts.
(b) What is the binary number that represents an analog voltage of +1.11 volts?
(c) Suppose that the A/D converter is connected to a microprocessor with a 16-bit wide data bus. What would the hexadecimal number be for an analog voltage of +2.96V?
(d) Assume that the A/D converter is a successive approximation-type A/D converter. How many samples must it take before it finally digitizes the analog voltage?
(e) Suppose that the A/D converter is being controlled by a 1 MHz clock signal and a sample occurs on the rising edge of every clock. How long will it take to digitize an analog voltage?

Thank you for the help! Much appreciated!

Answer 1

a) Between minimum possible value 000000000000 and maximum possible value 111111111111, a 12-bit A/D converter has 2¹² possible digital outputs, i.e. 2¹² quantization levels.

000000000000 represents analog voltage 0.00 V.

111111111111 represents analog voltage 3.30 V.

The smallest analog voltage detectable by this A/D converter is the change denoted by the differenece between two consecutive digital outputs that have only one bit position different (i.e. the difference in terms of analog voltage between two successive quantization levels).

This analog value is called resolution of an A/D converter, and it is given by,

= 0.8057 mV

b) Analog voltage of 1.11V is (1.11V/.8057 mv) 1378 quantization levels higher than minimum value 0.00 V.

Hence, the binary number that represents 1.11 V is the 1378th code starting from 000000000000.

To represent this number in binary form, we write 1378 as

1378 = 2¹⁰ + 2⁸ + 2⁶ + 2⁵ + 2¹

Where the i-th power of 2 being present in the RHS of above equation indicates that the i-th bit (i=0 for LSB and i=n-1 = 12-1 = 11 here for MSB) has value 1 and the powers of 2 not present have value 0 in the binary representation.

In case of the decimal no. 1378, the 10th, 8th, 6th, 5th and 1st bits of the output of 12 bit ADC have value 1 and remaining bits are 0.

=> (1378)₁₀ = (010101100010)_{12-bit binary}

Hence, the binary output of the a/D converter for input analog voltage 1.11V is given by 010101100010 .

c) Analog voltage of 2.96 V is (2.96V/.8057 mv) 3674 qualtization levels higher than minimum value 0.00 V.

3674 = 2¹¹ + 2¹⁰ + 2⁹ + 2⁶ + 2⁴ + 2³ + 2

=> (3674)₁₀ = (0000 1110 0101 1010)_{16-bit binary} [4 extra 0bits added to left for conversion to hexadecimal]

                      = (0E5A)16    [(1110)₂=(14)₁₀=(E)₁₆, (1010)₂=(10)₁₀=(A)₁₆]

Hence, the hexadecimal representation of analog voltage 2.96 V is 0E5A .

d) A successive approximation type ADC needs to perform 1 comparison to decide the binary value of each bit. Thus, an n-bit successive approximation type ADC needs to perform n comparisons in total to generate n-bit digital representation of the input analog value.

In the given scenario, the ADC is 12-bit. Hence it performs 12 comparisons to fully digitize an analog input. Hence, it requires 12 samples to digitize an analog input.

e) 1 bit of the digital output is generated in 1 clock cycle. Hence, 12 clock cycles will be required to generate the final 12-bit output of the given ADC.

For a 1 MHz = 10⁶ Hz clock signal, 12 clock cycles will take 12*10^-6 s = 0.012 ms

Hence, the ADC in question will take 0.012 ms to digitize an analog value.