A-D and D-A Converters


Question 1
An 8-bit ADC converts analog voltage in the range of 0 to +5 \mathrm{~V} to the corresponding digital code as per the conversion characteristics shown in figure. For \mathrm{V}_{\text {in }}=1,9922 \mathrm{~V}, which of the following digital output, given in hex is true?

A
64 \mathrm{H}
B
65 \mathrm{H}
C
66 \mathrm{H}
D
67 \mathrm{H}
GATE EE 2023   Digital Electronics
Question 1 Explanation: 
Resolution =\frac{V_{F S}}{2^{n}-1}=\frac{5}{2^{8}-1}=0.0196 \mathrm{~V}
Now, No. of step =\frac{\mathrm{V}_{\text {in }}}{\text { Resolution }} =\frac{1.9922}{0.0196}=(101.64)_{10}

For (101)_{10} the analog input is less than 1.9922 \mathrm{~V}.
So, here we take (102)_{10}.
\therefore \quad(102)_{10}=66 \mathrm{H}
Question 2
A 2-bit flash Analog to Digital Converter (ADC) is given below. The input is 0 \leq V_{IN} \leq 3 Volts. The expression for the LSB of the output B_0 as a Boolean function of X_{2},X_{1} \; and \; X_{0} is
A
X_{0}[\overline{X_{2}\oplus X_{1}}]
B
\bar{X_{0}}[\overline{X_{2}\oplus X_{1}}]
C
X_{0}[{X_{2}\oplus X_{1}}]
D
\bar{X_{0}}[X_{2}\oplus X_{1}]
GATE EE 2016-SET-1   Digital Electronics
Question 2 Explanation: 
The input to digital circuuit is X_2,X_1,X_0 and output is B_1,B_0

B_0=\bar{X_2}\bar{X_1}X_0+X_2X_1X_0
\;\;=X_0(\bar{X_2}\bar{X_1}+X_2X_1)
\;\;=X_0(\overline{X_2\oplus X_1})


Question 3
A temperature in the range of -40^{\circ}C to 55^{\circ}C is to be measured with a resolution of 0.1^{\circ}C. The minimum number of ADC bits required to get a matching dynamic range of the temperature sensor is
A
8
B
10
C
12
D
14
GATE EE 2016-SET-1   Digital Electronics
Question 3 Explanation: 
Temperature range of -40^{\circ}C \; to \; 55^{\circ}C
So. Total range in 95^{\circ}C
Since, resolution is 0.1^{\circ}C
So, number of steps will be 950
To have 950 steps, we need at least 10 bits.
Question 4
An 8-bit, unipolar Successive Approximation Register type ADC is used to convert 3.5 V to digital equivalent output. The reference voltage is +5 V. The output of the ADC, at the end of 3rd clock pulse after the start of conversion, is
A
1010 0000
B
1000 0000
C
0000 0001
D
1000 0000
GATE EE 2015-SET-1   Digital Electronics
Question 4 Explanation: 
The reference voltage is 5 V.
The number of bits in ADC are 8.
So, the resolution will be =\frac{5}{2^8-1}=\frac{5}{255}
The applied input is 3.5 V.
The succesive approximation ADC start working from the MSB so,

After one clock:
SAR will toggle it's MSB from 0\rightarrow 1. So, output of SAR will be 1000 0000.

After second clock:
SAR will toggle its 7th bit from 0\rightarrow 1 but 1100 0000 will result in value greater than 3.5. So, output of SAR after 2nd clock will be 1000 0000.

After third clock:
SAR will toggle it's 6th bit from 0\rightarrow 1 and output will be 1010 0000.
Question 5
The Octal equivalent of HEX and number AB.CD is
A
253.314
B
253.632
C
526.314
D
526.632
GATE EE 2007   Digital Electronics
Question 5 Explanation: 
Hex Number (AB.CD)
1010\;\;1011\;.\; 1100\;\;1101
For finding its octal number, we can add one zero in both extreme and grouping.
010\;101\;011\;.\;110\;011\;010
Its equavalent octal number is (253.632)_8


There are 5 questions to complete.