Digital Electronics MCQ Quiz - Objective Question with Answer for Digital Electronics - Download Free PDF

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Explanation:

8-bit Digital-to-Analog Converter (DAC) using Two 4-bit DACs

Problem Understanding: In the given question, an 8-bit DAC is implemented using two identical 4-bit DACs with equal reference voltage. The binary inputs b<sub>7</sub> to b<sub>0</sub> are represented, where b<sub>7</sub> is the Most Significant Bit (MSB) and b<sub>0</sub> is the Least Significant Bit (LSB). To achieve the correct analog output voltage (V<sub>0</sub>) corresponding to an 8-bit DAC, we need to determine the value of the resistor R in the circuit. The operational amplifier (op-amp) is assumed to be ideal.

Working Principle:

To understand the operation of this circuit, note the following:

• The 8-bit input is divided into two 4-bit groups: the higher 4 bits (b<sub>7</sub> to b<sub>4</sub>) and the lower 4 bits (b<sub>3</sub> to b<sub>0</sub>).
• Each 4-bit group is fed into one of the two identical 4-bit DACs.
• The higher 4 bits produce an analog output proportional to their binary value, scaled to the range of the higher nibble.
• The lower 4 bits produce an analog output proportional to their binary value, scaled to the range of the lower nibble.
• The outputs of the two DACs are combined using a resistor network and an operational amplifier to produce a final output voltage V<sub>0</sub> that corresponds to the 8-bit binary input.

Analysis:

The higher nibble (b<sub>7</sub> to b<sub>4</sub>) represents the most significant part of the binary input. Its contribution to the output voltage should dominate over the lower nibble (b<sub>3</sub> to b<sub>0</sub>). To achieve this, the output of the DAC handling the higher nibble is scaled by a factor of 16 (since 2<sup>4</sup> = 16) relative to the output of the DAC handling the lower nibble.

The resistor R in the circuit is responsible for ensuring this scaling. Specifically, the resistor network and the ideal op-amp configure the circuit such that:

• The output voltage of the higher nibble DAC is multiplied by 16.
• The output voltage of the lower nibble DAC is added directly.

Mathematical Derivation:

Let the outputs of the two 4-bit DACs be V<sub>H</sub> (higher nibble) and V<sub>L</sub> (lower nibble), respectively. The final output voltage V<sub>0</sub> is given by:

V<sub>0</sub> = 16 × V<sub>H</sub> + V<sub>L</sub>

Both DACs have the same reference voltage and identical resistor networks. Therefore, the scaling factor of 16 is achieved by appropriately choosing the value of resistor R. For a standard 4-bit DAC, the output voltage is proportional to the binary input (b<sub>3</sub> to b<sub>0</sub> or b<sub>7</sub> to b<sub>4</sub>) scaled by the reference voltage and the DAC’s resolution.

The resistor R is chosen such that the higher nibble’s output voltage is effectively scaled by 16 relative to the lower nibble’s output. This is achieved by using a resistor value of 1 kΩ.

Correct Answer: The value of R should be 1 kΩ.

Additional Information

To further understand the reasoning, let’s analyze why the other options are incorrect:

Option 1: 8 kΩ

This value is too high. If R were 8 kΩ, the scaling factor would not properly match the requirement of multiplying the higher nibble’s output by 16. This would result in incorrect analog values at the output.

Option 2: 0.25 kΩ

This value is too low. A resistor value of 0.25 kΩ would result in an incorrect scaling factor, leading to a mismatch between the contributions of the higher and lower nibbles.

Option 4: 0.5 kΩ

Similar to Option 2, this value is also too low. The scaling factor would not correctly match the requirement, resulting in incorrect output voltages.

Option 5: Not provided in the question

As the correct scaling requires R = 1 kΩ, any other value not listed in the options would also fail to produce the correct output.

Conclusion:

The resistor R plays a crucial role in ensuring the correct scaling of the higher nibble’s output relative to the lower nibble’s output in the 8-bit DAC circuit. By choosing R = 1 kΩ, the circuit achieves the desired scaling, resulting in accurate analog values corresponding to the 8-bit binary inputs. This understanding is essential in designing such circuits to ensure proper functionality and accuracy.

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Concept:

The circuit contains a T Flip-Flop (TFF) whose T input is selected via a 2:1 MUX. The selection line of the MUX is S.

We know that the behavior of a T Flip-Flop is as follows:

• If T = 0 → No change in Q (hold state)
• If T = 1 → Toggle the state of Q

The MUX selects between two inputs: i0 and i1. Let’s assume:

• i0 = 0 (for S = 0) → T = 0 → Q holds
• i1 = 1 (for S = 1) → T = 1 → Q toggles

State Behavior:

• When S = 0 → T = 0 → Q holds its current state
• When S = 1 → T = 1 → Q toggles (0 ↔ 1)

Transition Summary:

• If Q = 0 and S = 0 → Q remains 0
• If Q = 0 and S = 1 → Q toggles to 1
• If Q = 1 and S = 0 → Q remains 1
• If Q = 1 and S = 1 → Q toggles to 0

Answer:

Option 4: State transition diagram matches the toggling/hold behavior correctly

Explanation:

Given Problem:

The problem requires finding the values of (X₀, X₁, X₂, X₃, X₄, X₅, X₆, X₇) to realize the function F = A + B̅. The correct option is provided as Option 2.

Step-by-Step Solution:

Understanding the Function:

The given function F = A + B̅ is a logical OR operation between A and the complement of B (denoted as B̅).

• In Boolean algebra, the OR operation (denoted by '+') results in a value of 1 if either of the inputs is 1.
• The complement B̅ represents the inverse of B; if B = 0, then B̅ = 1, and if B = 1, then B̅ = 0.

Truth Table:

To realize this function, we construct a truth table for the inputs A and B and the output F:

Mapping Output to (X₀, X₁, ..., X₇):

The values of (X₀, X₁, ..., X₇) correspond to the output F for all possible combinations of 3-bit binary inputs. For inputs A and B, the truth table above gives the first two bits. The third bit (C) is irrelevant for the function F = A + B̅, meaning the output F remains the same regardless of the value of C.

In a 3-bit binary system:

• X₀ corresponds to A = 0, B = 0, C = 0 → F = 1
• X₁ corresponds to A = 0, B = 0, C = 1 → F = 1
• X₂ corresponds to A = 0, B = 1, C = 0 → F = 0
• X₃ corresponds to A = 0, B = 1, C = 1 → F = 0
• X₄ corresponds to A = 1, B = 0, C = 0 → F = 1
• X₅ corresponds to A = 1, B = 0, C = 1 → F = 1
• X₆ corresponds to A = 1, B = 1, C = 0 → F = 1
• X₇ corresponds to A = 1, B = 1, C = 1 → F = 1

Thus, the values of (X₀, X₁, ..., X₇) are (1, 1, 0, 0, 1, 1, 1, 1), which matches Option 2.

Correct Option:

Option 2: (1, 1, 0, 0, 1, 1, 1, 1)

This option correctly represents the output values for the function F = A + B̅ based on the mapping of the truth table.

Important Information

To further understand why other options are incorrect, let’s analyze them:

Option 1: (1, 1, 0, 0, 1, 1, 1, 1)

This option represents the output values for the function F = A + B̅, but it is identical to Option 2. Therefore, it is redundant and not the correct answer.

Option 3: (1, 1, 0, 1, 0, 1, 1, 1)

This option misrepresents the output values of F. For example:

• X₃ (A = 0, B = 1, C = 1) is listed as 1, but based on F = A + B̅, the correct value should be 0.
• X₄ (A = 1, B = 0, C = 0) is listed as 0, but the correct value is 1.

Thus, Option 3 does not match the truth table of F = A + B̅ and is incorrect.

Option 4: (0, 0, 1, 1, 1, 1, 1, 1)

This option completely misrepresents the output values of F. For example:

• X₀ (A = 0, B = 0, C = 0) is listed as 0, but based on F = A + B̅, the correct value should be 1.
• X₂ (A = 0, B = 1, C = 0) is listed as 1, but the correct value is 0.

Thus, Option 4 does not match the truth table of F = A + B̅ and is incorrect.

Conclusion:

Understanding Boolean functions and their truth tables is critical for solving such problems. The correct values of (X₀, X₁, ..., X₇) for the function F = A + B̅ are (1, 1, 0, 0, 1, 1, 1, 1), which matches Option 2. Other options fail to accurately represent the truth table for the given function, leading to incorrect results.

Explanation:

Implementation of a One-Bit Full Adder Using a 3 ∶ 8 Active High Decoder

Definition: A one-bit full adder is a combinational circuit designed to perform the addition of three binary inputs: the two significant bits (A and B) and a carry-in (Cin). It produces two outputs: the sum (S) and carry-out (Cout).

Overview: To implement a one-bit full adder using a 3 ∶ 8 active high output decoder, the decoder is used to generate specific combinations of outputs corresponding to the truth table of the full adder. The outputs of the decoder are then combined using OR gates to derive the required sum and carry outputs.

Working of a 3 ∶ 8 Decoder:

A 3 ∶ 8 decoder takes 3 inputs (A, B, Cin) and generates 8 distinct outputs (Y0 to Y7). Each output corresponds to one unique combination of the inputs. The output is active high, meaning that for a given combination of inputs, only one output is high (logic 1) while the others remain low (logic 0).

For example:

• If A = 0, B = 0, Cin = 0, then output Y0 is high.
• If A = 0, B = 0, Cin = 1, then output Y1 is high.
• If A = 1, B = 1, Cin = 1, then output Y7 is high.

The truth table for a one-bit full adder is as follows:

Implementation:

To implement the full adder using the 3 ∶ 8 decoder, the outputs of the decoder are used to construct the Sum (S) and Carry (Cout) outputs according to the truth table. Each output is obtained by combining specific outputs of the decoder using OR gates.

Deriving Sum (S):

The Sum (S) is high (logic 1) for the following combinations of inputs:

• A = 0, B = 0, Cin = 1 → Decoder output Y1
• A = 0, B = 1, Cin = 0 → Decoder output Y2
• A = 1, B = 0, Cin = 0 → Decoder output Y4
• A = 1, B = 1, Cin = 1 → Decoder output Y7

Thus, Sum (S) = Y1 + Y2 + Y4 + Y7.

Deriving Carry (Cout):

The Carry (Cout) is high (logic 1) for the following combinations of inputs:

• A = 0, B = 1, Cin = 1 → Decoder output Y3
• A = 1, B = 0, Cin = 1 → Decoder output Y5
• A = 1, B = 1, Cin = 0 → Decoder output Y6
• A = 1, B = 1, Cin = 1 → Decoder output Y7

Thus, Carry (Cout) = Y3 + Y5 + Y6 + Y7.

Minimum Number of OR Gates:

To implement the Sum (S) and Carry (Cout) outputs:

• Sum (S) requires four OR gates to combine Y1, Y2, Y4, and Y7.
• Carry (Cout) requires four OR gates to combine Y3, Y5, Y6, and Y7.

However, one OR gate can be shared between Sum (S) and Carry (Cout) for Y7, reducing the total number of OR gates required.

Final Calculation:

Total OR gates = 4 (for Sum) + 4 (for Carry) - 1 (shared gate for Y7) = 7.

Since the question specifies the minimum number of 3-input OR gates, we can combine multiple outputs into groups of three, further reducing the number of gates.

Using this grouping approach:

• Sum (S): Combine Y1, Y2, and Y4 into one 3-input OR gate, and Y7 into another OR gate → 2 OR gates.
• Carry (Cout): Combine Y3, Y5, and Y6 into one 3-input OR gate, and Y7 into another OR gate → 2 OR gates.

Thus, the minimum number of 3-input OR gates required = 4.

Correct Option:

The correct answer is option 3) 4.

Important Information

To analyze the other options:

Option 1 (2 OR gates):

This option is incorrect because it underestimates the number of OR gates required. Each of the Sum (S) and Carry (Cout) outputs requires at least two OR gates to combine the necessary decoder outputs. Therefore, a total of two OR gates is insufficient.

Option 2 (5 OR gates):

This option is incorrect because it does not represent the minimum number of 3-input OR gates required. While 5 OR gates may work, combining outputs into groups of three allows the implementation to be optimized to 4 gates.

Option 4 (3 OR gates):

This option is incorrect because it underestimates the number of OR gates required. Even with optimal grouping, at least 4 OR gates are needed to derive the Sum (S) and Carry (Cout) outputs.

Conclusion:

The minimum number of 3-input OR gates required to implement a one-bit full adder using a 3 ∶ 8 active high decoder is 4, making option 3 the correct choice.

Explanation:

Analysis of the 4-bit Unipolar DAC and Comparator System:

The given problem involves a 4-bit unipolar DAC (Digital-to-Analog Converter) with a 10V reference voltage and a comparator whose threshold voltage is set at 7.5V. The DAC counts from 0 to 15 (since it is a 4-bit DAC, it can represent 2⁴ = 16 levels). The comparator monitors the output of the DAC and determines when it crosses the threshold voltage.

It is observed that when the DAC count reaches 8, the comparator erroneously indicates threshold crossing. Upon analysis, it is found that one of the input bits of the DAC is stuck at '1'. We need to determine which bit is stuck at '1'.

Step-by-Step Solution:

1. Understanding the DAC Operation:

A 4-bit unipolar DAC converts a digital value (binary) into an analog voltage. The output voltage is given by:

V_out = (Digital Value / Maximum Digital Value) × Reference Voltage

Since the DAC is 4-bit, the maximum digital value is 15 (binary 1111). The reference voltage is 10V. Therefore:

V_out = (Digital Value / 15) × 10

For each binary value, the corresponding analog voltage is calculated as follows:

2. Comparator Threshold Analysis:

The comparator is set to a threshold voltage of 7.5V. It should ideally indicate threshold crossing when the DAC output voltage exceeds 7.5V. From the table above, this happens when the binary value exceeds 1011 (decimal 11), as for binary 1100 (decimal 12), V_out becomes 8V.

However, the problem states that the comparator erroneously indicates threshold crossing at binary 1000 (decimal 8). This suggests that the DAC is producing an incorrect output voltage due to one of its input bits being stuck at '1'.

3. Identifying the Stuck-at-'1' Bit:

To determine which bit is stuck at '1', we analyze the effect of each bit being stuck at '1' on the output voltage:

• B₃ (MSB): If the most significant bit (B₃) is stuck at '1', the output voltage would be significantly higher than expected for lower binary values, as B₃ contributes the largest weight (8 × Reference Voltage / 15). For binary 1000 (decimal 8), the output voltage would be correct, so B₃ is not the stuck bit.
• B₂: If B₂ is stuck at '1', it adds an additional weight (4 × Reference Voltage / 15) to the output voltage. For binary 1000 (decimal 8), the output voltage would incorrectly include this additional weight, resulting in an erroneous threshold crossing at the comparator. This matches the problem description, so B₂ is the stuck bit.
• B₁: If B₁ is stuck at '1', it adds a smaller weight (2 × Reference Voltage / 15). For binary 1000 (decimal 8), the output voltage would not exceed the threshold, so B₁ is not the stuck bit.
• B₀ (LSB): If the least significant bit (B₀) is stuck at '1', it adds the smallest weight (Reference Voltage / 15). This would not cause the erroneous threshold crossing at binary 1000 (decimal 8), so B₀ is not the stuck bit.

Correct Answer: The stuck-at-'1' bit is B₂.

Important Information

To further understand the analysis, let’s evaluate the other options:

• B₀ (LSB): The least significant bit contributes the smallest weight to the output voltage. If it is stuck at '1', the error in output voltage would be minimal and insufficient to cause the threshold crossing at binary 1000 (decimal 8).
• B₁: The second least significant bit contributes a slightly larger weight than B₀. If it is stuck at '1', the output voltage would still not exceed the threshold at binary 1000 (decimal 8).
• B₃ (MSB): The most significant bit contributes the largest weight to the output voltage. If it is stuck at '1', the output voltage would be significantly higher than expected for lower binary values, but it would not match the observed erroneous behavior at binary 1000 (decimal 8).

Conclusion:

By analyzing the effect of each bit being stuck at '1', we identify that B₂ is the stuck bit, as its incorrect contribution to the output voltage causes the comparator to erroneously indicate threshold crossing at binary 1000 (decimal 8). This analysis highlights the importance of understanding DAC operation and bit weighting in digital systems.

1 0 0 1 0 0 0 0
-  1 1 1 1 0 0 1
0 0 0 1 0 1 1 1

Step 1: 0 – 1 = Borrow to make 10 – 1 = 1.
Step 2: 1 – 0 = 1.
Step 3: 1 – 0 = 1.
Step 4: 1 – 1 = 0.
Step 5: 0 – 1 = Borrow to make 10 – 1 = 1.
Step 6: 1 – 0 = 1.
Step 7: 1 – 0 = 1.

Remember: When zero takes 1 as carry from its left side number, '0' will become '10' which is equal to '2' (2-1=1) and if that '10' further gives carry then it will become '1' not '0'.

Concept:

XNOR Gate:

Symbol:

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Truth Table:

Output Equation: 

1) If B is always Low, the output is the inverted value of the other input A, i.e. A̅.

2) The output is low when both the inputs are different.

3) The output is high when both the inputs are the same.

4) XNOR gate produces an output only when the two inputs are same.

Analysis:

Solution:

The binary representation of 257 is 100000001

∴ The total number of the binary digit in 257 is 9

The correct answer is 'option 2'

Concept

Divide 127 by 2. Use the integer quotient obtained in this step as the dividend for the next step. Repeat the process until the quotient becomes 0.

Solution:

Write the remainder from bottom to top i.e. in the reverse chronological order.

This will give the binary equivalent of 127. 

Therefore, the binary equivalent of decimal number 127 is 1111111.

F(P, Q, R) = PQ + QR' + PR'

= PQ (R + R') + (P + P')QR' + P(Q + Q')R'

= PQR + PQR' + PQR' + P'QR' + PQR' + PQ'R'

= PQR + PQR' + P'QR' + PQ'R'

= m₇ + m₆ + m₂ + m₄

= m₂ + m₄ + m₆ + m₇

• AND, OR, NOT gates are the basic gates.
• The logic gates which are derived from the basic gates like AND, OR, NOT gates are known as derived gates. NAND, NOR, XOR, and XNOR are the derived gates.
• A universal gate is a gate which can implement any Boolean function without need to use any other gate type. The NAND and NOR gates are universal gates.

The correct answer is option 1): 45

Concept:

To convert the binary number 101101 to decimal, follow these two steps:

• Start from one's place in 101101: multiply one place with 2^0, tens place with 2^1, hundreds place with 2^2 and so on from right to left
• Add all the products we got from step 1 to get the decimal equivalent of 101101. Using the above steps, here is the work involved in the solution for converting 101101 to a decimal number (Don't forget that we start from one place to so on...)
  • Decimal equivalent of "1" = 1 × 2^0 = 1
  • Decimal equivalent of "0" = 0 × 2^1 = 0
  • Decimal equivalent of "1" = 1 × 2^2 = 4
  • Decimal equivalent of "1" = 1 × 2^3 = 8
  • Decimal equivalent of "0" = 0 × 2^4 = 0
  • Decimal equivalent of "1" = 1 × 2^5 = 32
  • The decimal equivalent of "101101" =   45
  • Here is the final answer, The binary number 101101₂ converted to decimal is therefore equal 45₁₀

(A.A̅) + A

= 0 + A = A

All Boolean algebra laws are shown below:

The Correct Answer is "Cache Memory".

Important Points

Cache Memory :

• Cache Memory is a special very high-speed memory.
• It is used to speed up and synchronizing with a high-speed CPU. Cache memory is costlier than main memory or disk memory but economical than CPU registers.
• Cache memory is an extremely fast memory type that acts as a buffer between RAM and the CPU.
• It holds frequently requested data and instructions so that they are immediately available to the CPU when needed.
• Cache memory is used to reduce the average time to access data from the Main memory.

Additional Information

Secondary Memory​ :

• It is non-volatile, i.e. it retains data when power is switched off.
• It is large capacities to the tune of terabytes.
• It is cheaper as compared to the primary memory.
• Depending on whether the Secondary memory device is part of the CPU or not, there are two types of secondary memory – fixed and removable.

 Auxiliary Memory :

• Auxiliary memory is the non-volatile memory lowest-cost, highest-capacity, and slowest-access storage in a computer system.
• It is where programs and data kept for long-term storage or when not in immediate use.
• Such memories tend to occur in two types-sequential access (data must access in a linear sequence) and direct access (data may access in any sequence).
• The most common sequential storage device is the hard disk drives, whereas direct-access devices include rotating drums, disks, CD-ROMs, and DVD-ROMs.
• It used as permanent storage of data in mainframes and supercomputers.

Virtual Memo :

• A computer can address more memory than the amount physically installed on the system.
• This extra memory is actually called virtual memory and it is a section of a hard disk that's set up to emulate the computer's RAM.
• The main visible advantage of this scheme is that programs can be larger than physical memory.
• Virtual memory serves two purposes.
  • First, it allows us to extend the use of physical memory by using the disk.
  • Second, it allows us to have memory protection because each virtual address is translated to a physical address.

The number of 2-input NAND gates required to implement a 2-input XOR gate is 4.

Similarly, the number of 2-input NOR gates required to implement a 2-input XNOR gate is 4.