ADC and DAC

Question 1
Consider the circuit shown with an ideal OPAMP. The output voltage V_o is __________V (rounded off to two decimal places).

A
0.5
B
0.75
C
-0.5
D
-0.75
GATE EC 2022   Digital Circuits
Question 1 Explanation: 


V_A=\frac{V_R}{2} \; and R_{th}=R
V_B=\frac{V_B}{2}

V_C=\frac{(V_R+V_R/4)}{2}=\frac{5}{8}V_R
R_{th}=2R||2R=R
Similarly,
V_D=\frac{V_C}{2}=\frac{5}{16}V_R
So, \frac{S}{16}V_R=\frac{5}{16} \times 1.6V
V_o=\frac{-3R}{(2R+R)} \times \frac{5}{16} \times 1.6=-\frac{3}{2} \times \frac{5}{16} \times 1.6
V_o=-0.5V
Question 2
An 8-bit unipolar (all analog output values are positive) digital-to-analog converter (DAC) has a full-scale voltage range from 0\; V to 7.68\:V. If the digital input code is 10010110 (the leftmost bit is \text{MSB}), then the analog output voltage of the DAC (rounded off to one decimal place) is ___________ V.
A
2.4
B
4.5
C
8.4
D
5.4
GATE EC 2021   Digital Circuits
Question 2 Explanation: 
\begin{aligned} \text { Given: } \qquad \qquad V_{p s}&=7.68 \mathrm{~V} \\ n &=8 \mathrm{bit} \\ \text { Resolution }(k) &=\frac{V_{F S}}{2^{n}-1}=\frac{7.68}{2^{8}-1}=0.03\\ \text { Now, } \quad V_{\mathrm{DAC}}&=k \times\{\text { Decimal equivalent }\}\\ &=0.03 \times\{150\} \\ &=4.5 \mathrm{~V} \end{aligned}
Question 3
A 10-bit D/A converter is calibrated over the full range from 0 to 10 V. If the input to the D/A converter is 13A (in hex), the output (rounded off to three decimal places) is _________ V.
A
3.069
B
1.214
C
2.124
D
0.257
GATE EC 2020   Digital Circuits
Question 3 Explanation: 
Given, n=10
V_{FS}=10\, V
Input Voltage=(13A)_{16}=(314)_{10}
Output Voltage=Resolution X Decimal Equivalent of input
V_{o}=\frac{10}{2^{10}-1}\times 314=3.069\, V
Question 4
In an N bit flash ADC, the analog voltage is fed simultaneously to 2^{N}-1 comparators. The output of the comparators is then encoded to a binary format using digital circuits. Assume that the analog voltage source V_{in} (whose output is being converted to digital format) has a source resistance of 75 \Omega as shown in the circuit diagram below and the input capacitance of each comparator is 8 pF. The input must settle to an accuracy of 1/2 LSB even for a full scale input change for proper conversion. Assume that the time taken by the thermometer to binary encoder is negligible.

If the flash ADC has 8 bit resolution, which one of the following alternatives is closest to the maximum sampling rate ?
A
1 megasamples per second
B
6 megasamples per second
C
64 megasamples per second
D
256 megasamples per second
GATE EC 2016-SET-2   Digital Circuits
Question 5
Consider a four bit D to A converter. The analog value corresponding to digital signals of values 0000 and 0001 are 0 V and 0.0625 V respectively. The analog value (in Volts) corresponding to the digital signal 1111 is ________.
A
0.7
B
0.93
C
1.13
D
1.07
GATE EC 2015-SET-1   Digital Circuits
Question 5 Explanation: 
Step size =0.0625 \mathrm{V}
Decimal equivalent =15
Analog output =15 \times 0.0625=0.9375 Volts
Question 6
The output of a 3-stage Johnson (twisted ring) counter is fed to a digital-to analog (D/A) converter as shown in the figure below. Assume all the states of the counter to be unset initially. The waveform which represents the D/A converter output v_{o} is

A
A
B
B
C
C
D
D
GATE EC 2011   Digital Circuits
Question 6 Explanation: 
Sequence of Johnson counter is
\begin{array}{ccccccc} Q_{2} & Q_{1} & Q_{0} & D_{2} & D_{1} & D_{0} & V_{0} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 4 \\ 1 & 1 & 0 & 1 & 1 & 0 & 6 \\ 1 & 1 & 1 & 1 & 1 & 1 & 7 \\ 0 & 1 & 1 & 0 & 1 & 1 & 3 \\ 0 & 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array}
Question 7
In the following circuit, the comparator output is logic "1" if V_{1} \gt V_{2} and is logic "0" otherwise. The D/A conversion is done as per the relations
V_{DAC}=\sum_{n=0}^{3}2^{n-1}b_{n} Volts, where b_{3} (MSB), b_{2}, b_{1}, b_{0}(LSB) are the counter outputs.
The counter starts from the clear state.

The magnitude of the error between V_{DAC} and V_{in} at steady state in volts is
A
0.2
B
0.3
C
0.5
D
1
GATE EC 2008   Digital Circuits
Question 7 Explanation: 
Magnitude of the error between V_{\mathrm{DAC}} and V_{\text {in }} at steady state
=6.5-6.2=0.3 \mathrm{V}
Question 8
In the following circuit, the comparator output is logic "1" if V_{1} \gt V_{2} and is logic "0" otherwise. The D/A conversion is done as per the relations
V_{DAC}=\sum_{n=0}^{3}2^{n-1}b_{n} Volts, where b_{3} (MSB), b_{2}, b_{1}, b_{0}(LSB) are the counter outputs.
The counter starts from the clear state.

The stable reading of the LED displays is
A
6
B
7
C
12
D
13
GATE EC 2008   Digital Circuits
Question 8 Explanation: 
\begin{aligned} V_{\mathrm{DAC}} &=2^{-1} b_{0}+2^{\circ} b_{1}+2^{1} b_{2}+2^{2} b_{3} \\ &=0.5 b_{0}+b_{1}+2 b_{2}+4 b_{3} \end{aligned}
Counter output will start from 0000 and will increase by 1 at every clock pulse.
Table for V_{\mathrm{DAC}} is shown below
\begin{array}{cccc|c} b_{3} & b_{2} & b_{1} & b_{0} & V_{\text {DAC }} \\ \hline 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0.5 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 1.5 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 1 & 0 & 1 & 2.5 \\ 0 & 1 & 1 & 0 & 3 \\ 0 & 1 & 1 & 1 & 3.5 \\ 1 & 0 & 0 & 0 & 4 \\ 1 & 0 & 0 & 1 & 4.5 \\ 1 & 0 & 1 & 0 & 5 \\ 1 & 0 & 1 & 1 & 5.5 \\ 1 & 1 & 0 & 0 & 6 \\ 1 & 1 & 0 & 1 & 6.5 \\ 1 & 1 & 1 & 0 & 7 \\ 1 & 1 & 1 & 1 & 7.5 \end{array}
Counter will increase till V_{\text {in }} \gt V_{\text {DAC. }}. So, when V_{\text {DAC }} =6.5V, the comparator output will be zero and the counter will be stable at that reading. The corresponding reading of LED display is 13.
Question 9
In the Digital-to-Analog converter circuit shown in the figure below, V_{R}=10 V and R = 10K\Omega

The voltage V_{o} is
A
-0.781 V
B
-1.562 V
C
-3.125 V
D
-6.250 V
GATE EC 2007   Digital Circuits
Question 9 Explanation: 
Net current in inverting terminal of op-amp
\begin{aligned} &=\frac{I}{4}+\frac{I}{16}=\frac{5 I}{16} \\ V_{0} &=-R \times \frac{5 I}{16} \\ =&-\frac{10 \times 10^{3} \times 5 \times 1 \times 10^{-3}}{16}=-3.125 \mathrm{V} \end{aligned}
Question 10
In the Digital-to-Analog converter circuit shown in the figure below, V_{R}=10 V and R = 10K\Omega

The current i is
A
31.25 \muA
B
62.5 \muA
C
125 \muA
D
250\muA
GATE EC 2007   Digital Circuits
Question 10 Explanation: 


Last both 2R resister are in parallel and series with R then after


then again similar condition last both 2R are in parallel and series with R similarly after solving equivalent circuit is


I=\frac{V_{R}}{R}=\frac{10}{10 \mathrm{k} \Omega}=1 \mathrm{mA}


then i=\frac{I}{16}=\frac{1 \times 10^{-3}}{16}=62.5 \mu \mathrm{A}
There are 10 questions to complete.