# 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.