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 $\mu$A B 62.5 $\mu$A C 125 $\mu$A D 250$\mu$A
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.