Question 1 |
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.
3.069 | |
1.214 | |
2.124 | |
0.257 |
Question 1 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
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 2 |
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 ?
If the flash ADC has 8 bit resolution, which one of the following alternatives is closest to the maximum sampling rate ?
1 megasamples per second | |
6 megasamples per second | |
64 megasamples per second | |
256 megasamples per second |
Question 3 |
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 ________.
0.7 | |
0.93 | |
1.13 | |
1.07 |
Question 3 Explanation:
Step size =0.0625 \mathrm{V}
Decimal equivalent =15
Analog output =15 \times 0.0625=0.9375 Volts
Decimal equivalent =15
Analog output =15 \times 0.0625=0.9375 Volts
Question 4 |
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 | |
B | |
C | |
D |
Question 4 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}
\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 5 |
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
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
0.2 | |
0.3 | |
0.5 | |
1 |
Question 5 Explanation:
Magnitude of the error between V_{\mathrm{DAC}} and V_{\text {in }} at steady state
=6.5-6.2=0.3 \mathrm{V}
=6.5-6.2=0.3 \mathrm{V}
Question 6 |
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
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
6 | |
7 | |
12 | |
13 |
Question 6 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.
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 7 |
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
The voltage V_{o} is
-0.781 V | |
-1.562 V | |
-3.125 V | |
-6.250 V |
Question 7 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}
\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 8 |
In the Digital-to-Analog converter circuit shown in the figure below, V_{R}=10 V and R = 10K\Omega
The current i is
The current i is
31.25 \muA | |
62.5 \muA | |
125 \muA | |
250\muA |
Question 8 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}
Question 9 |
A 4-bit D/A converter is connected to a free -running 3-bit UP counter,
as shown in the following figure. Which of the following waveforms will be
observed at v_{0} ?
A | |
B | |
C | |
D |
Question 9 Explanation:
\begin{array}{ccccccccc} \text { CP } & \rightarrow & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ & &\downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \downarrow \\ \text { Counter } &\rightarrow & 001 & 010 & 011 & 100 & 101 & 110 & 1000 \\ & & \downarrow & \dagger & \dagger & \downarrow & \downarrow & \downarrow & \downarrow \\ \text { Decoder } & \rightarrow & 0001 & 0010 & 0011 & 1000 & 1001 & 1010 & 0000 \\ & & 1 & 2 & 3 & 8 & 9 & 10 & 0 \end{array}
D_{2} is connected to ground and Q_{2} to D_{3}
D_{2} is connected to ground and Q_{2} to D_{3}
Question 10 |
A digital system is required to amplify a binary-encoded audio signal. The user should be able to control the gain of the amplifier from minimum to a maximum in 100 increments. The minimum number of bits required to encode, in straight binary, is
8 | |
6 | |
5 | |
7 |
Question 10 Explanation:
\begin{aligned} 2^{n}&\geq100\\ \therefore \quad n&\geq7 \end{aligned}
There are 10 questions to complete.