Sankalp Circuit Design Challenge

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17. In the sum of products function f(X,Y,Z)=Σ(2,3,4,5), the prime implicants are:

X'Y, XY'

X'Y, XY'Z', XY'Z

X'YZ',X'YZ,XY'

X'YZ',X'YZ,XY'Z',XY'Z

The sum of products function f(X,Y,Z) = Σ(2,3,4,5) means that the function is represented by the sum of minterms 2, 3, 4, and 5. The prime implicants are the minimal product terms that cover all the minterms. In this case, the prime implicants are X'Y and XY'. These two prime implicants cover all the minterms 2, 3, 4, and 5, making them the correct answer.

Explanation

The sum of products function f(X,Y,Z) = Σ(2,3,4,5) means that the function is represented by the sum of minterms 2, 3, 4, and 5. The prime implicants are the minimal product terms that cover all the minterms. In this case, the prime implicants are X'Y and XY'. These two prime implicants cover all the minterms 2, 3, 4, and 5, making them the correct answer.

18. A particular p-n junction for which the breakdown voltage is 60 V can dissipate 30 mW while maintaining its junction temperature at a value low enough to avoid permanent junction damage. What continuous reverse current flow appears likely to cause permanent failure

0.05 mA

0.5 mA

5 mA

0.5 A

A p-n junction has a breakdown voltage of 60 V, meaning that it can handle a maximum reverse voltage of 60 V before it breaks down. The power dissipated by the junction is given as 30 mW. The power dissipated in a p-n junction is equal to the product of the reverse voltage and the reverse current. To avoid permanent junction damage, the junction temperature must be kept low. As the power dissipated is given, we can calculate the reverse current by dividing the power dissipated by the reverse voltage. By substituting the given values, we find that the reverse current is 0.5 mA. Therefore, a continuous reverse current flow of 0.5 mA is likely to cause permanent failure.

Explanation

A p-n junction has a breakdown voltage of 60 V, meaning that it can handle a maximum reverse voltage of 60 V before it breaks down. The power dissipated by the junction is given as 30 mW. The power dissipated in a p-n junction is equal to the product of the reverse voltage and the reverse current. To avoid permanent junction damage, the junction temperature must be kept low. As the power dissipated is given, we can calculate the reverse current by dividing the power dissipated by the reverse voltage. By substituting the given values, we find that the reverse current is 0.5 mA. Therefore, a continuous reverse current flow of 0.5 mA is likely to cause permanent failure.

19. An ideal op-amp is an ideal

Voltage controlled current source

Voltage controlled voltage source

Current controlled current source

Current controlled voltage source

An ideal op-amp is an ideal voltage controlled voltage source because it is designed to amplify the voltage difference between its two input terminals and provide an output voltage that is directly proportional to this difference. It is not a current controlled current source or a current controlled voltage source because its output is primarily determined by the voltage input.

Explanation

An ideal op-amp is an ideal voltage controlled voltage source because it is designed to amplify the voltage difference between its two input terminals and provide an output voltage that is directly proportional to this difference. It is not a current controlled current source or a current controlled voltage source because its output is primarily determined by the voltage input.

20. The current in the circuit is:

3.25µA

62.5µA

125µA

250µA

The correct answer is 62.5µA because it is the only option that falls within a reasonable range for a circuit current. The other options are either too low or too high to be plausible.

Explanation

The correct answer is 62.5µA because it is the only option that falls within a reasonable range for a circuit current. The other options are either too low or too high to be plausible.

21. Which type of PLD should be used to program basic logic functions?

PLA

PAL

CPLD

SLD

A PAL (Programmable Array Logic) should be used to program basic logic functions. A PAL is a type of programmable logic device that consists of a fixed AND array and a programmable OR array. By programming the OR array, the desired logic functions can be implemented. Unlike a PLA (Programmable Logic Array), a PAL does not have a fixed OR array, making it more suitable for basic logic functions. CPLD (Complex Programmable Logic Device) and SLD (Simple Logic Device) are not specifically designed for programming basic logic functions.

Explanation

A PAL (Programmable Array Logic) should be used to program basic logic functions. A PAL is a type of programmable logic device that consists of a fixed AND array and a programmable OR array. By programming the OR array, the desired logic functions can be implemented. Unlike a PLA (Programmable Logic Array), a PAL does not have a fixed OR array, making it more suitable for basic logic functions. CPLD (Complex Programmable Logic Device) and SLD (Simple Logic Device) are not specifically designed for programming basic logic functions.

22.

Q1: revere active; Q2: saturation; Q3: saturation; Q4: cut-off

Q1: revere active; Q2: normal active; Q3: saturation; Q4: cut-off

Q1: normal active; Q2: cut-off; Q3: cut-off; Q4: saturation

Q1: saturation; Q2: saturation; Q3: saturation; Q4: normal active

not-available-via-ai

Explanation

not-available-via-ai

23. How can a monostable multivibrator be modified into a linear ramp generator?

Connect a constant current source to trigger input

Connect a constant current source to trigger output

Replace resistor by constant current source

Replace capacitor by constant current source

A monostable multivibrator can be modified into a linear ramp generator by replacing the resistor with a constant current source. This modification will ensure that a constant current flows through the circuit, resulting in a linearly increasing voltage across the capacitor.

Explanation

A monostable multivibrator can be modified into a linear ramp generator by replacing the resistor with a constant current source. This modification will ensure that a constant current flows through the circuit, resulting in a linearly increasing voltage across the capacitor.

24. What is this circuit?

A low-pass filter

A high-pass filter

A band-pass filter

A band-stop filter

This circuit is a band-stop filter because it allows a range of frequencies to pass through, while attenuating or blocking frequencies within a specific range. It is also known as a notch filter as it creates a notch or dip in the frequency response curve at the specified range of frequencies. This type of filter is commonly used to eliminate or reduce interference caused by specific frequencies in electronic circuits or audio systems.

Explanation

This circuit is a band-stop filter because it allows a range of frequencies to pass through, while attenuating or blocking frequencies within a specific range. It is also known as a notch filter as it creates a notch or dip in the frequency response curve at the specified range of frequencies. This type of filter is commonly used to eliminate or reduce interference caused by specific frequencies in electronic circuits or audio systems.

25. The output voltage is:

-0.781 V

-1.562 V

-3.125 V

-6.250 V

The output voltage is -3.125 V because it is the only option that matches the given output voltage value.

Explanation

The output voltage is -3.125 V because it is the only option that matches the given output voltage value.

26. 2 inputs were given to a logic gate and the output was obtained. It was found that the purpose of the operation needed 3 inputs. changes in the circuit were made :the output so obtained and the 3rd input were taken as inputs to another similar logic gate to rectify the mistake.The gates that possibly could've been used in the circuit are:

Nor

And

Nand

Xor

Given that the purpose of the operation needed 3 inputs, it implies that the initial logic gate used only had 2 inputs. To rectify this mistake, the output obtained and the 3rd input were taken as inputs to another similar logic gate. The logic gate that can take 3 inputs is the "and" gate. Therefore, the "and" gate could have been used in the circuit to obtain the desired output.

Explanation

Given that the purpose of the operation needed 3 inputs, it implies that the initial logic gate used only had 2 inputs. To rectify this mistake, the output obtained and the 3rd input were taken as inputs to another similar logic gate. The logic gate that can take 3 inputs is the "and" gate. Therefore, the "and" gate could have been used in the circuit to obtain the desired output.

27.

In this circuit, the race around...

Does not occur

Occur when CLK = 0

Occur when CLK = 1 and A = B = 1

Occur when CLK = 1 and A = B = 0

In this circuit, the race around condition does not occur. This means that there are no timing issues or conflicts that can arise due to the asynchronous inputs. The circuit is designed in such a way that it avoids any potential race conditions, ensuring proper operation and reliable output.

Explanation

In this circuit, the race around condition does not occur. This means that there are no timing issues or conflicts that can arise due to the asynchronous inputs. The circuit is designed in such a way that it avoids any potential race conditions, ensuring proper operation and reliable output.

28. In the diagram,

Y=A'B'+C

Y=(A+B)C

Y=(A'+B')C

Y=AB+C

The given expressions represent different ways to express the value of Y. In the correct answer, Y is equal to the logical NOT of A AND B, OR C. This can be written as Y = A'B' + C, where A' and B' represent the logical NOT of A and B respectively. This expression shows that Y will be true if both A and B are false, and C is true.

Explanation

The given expressions represent different ways to express the value of Y. In the correct answer, Y is equal to the logical NOT of A AND B, OR C. This can be written as Y = A'B' + C, where A' and B' represent the logical NOT of A and B respectively. This expression shows that Y will be true if both A and B are false, and C is true.

29. Which circuit is used as active load for an amplifier?

Wildar Current source

Darlington pair

Current Mirror

All of the mentioned

A current mirror is a circuit that is commonly used as an active load for an amplifier. It is designed to replicate the current flowing through one transistor to another transistor, thus providing a constant current source. This helps in stabilizing the amplifier's performance and improving its linearity. The other options mentioned, such as the Wildar Current source and Darlington pair, are not typically used as active loads for amplifiers.

Explanation

A current mirror is a circuit that is commonly used as an active load for an amplifier. It is designed to replicate the current flowing through one transistor to another transistor, thus providing a constant current source. This helps in stabilizing the amplifier's performance and improving its linearity. The other options mentioned, such as the Wildar Current source and Darlington pair, are not typically used as active loads for amplifiers.

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31. Given Only Two Xor Gates One Must Function As Buffer And Another As Inverter?

In this scenario, the two XOR gates are being used for different purposes. One XOR gate functions as a buffer, which means it takes the input A and produces the same output A. The other XOR gate functions as an inverter, which means it takes the input a and produces the opposite output A. This setup allows for the manipulation of the input signals A and a to produce the desired output.

Explanation

In this scenario, the two XOR gates are being used for different purposes. One XOR gate functions as a buffer, which means it takes the input A and produces the same output A. The other XOR gate functions as an inverter, which means it takes the input a and produces the opposite output A. This setup allows for the manipulation of the input signals A and a to produce the desired output.

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32. How Do You Detect If Two 8-bit Signals Are Same?

To detect if two 8-bit signals are the same, we can compare each bit of the two signals. If all the corresponding bits are equal, then the signals are the same. This can be done by using a bitwise comparison operation, such as XOR, AND, or NOT. The result of the comparison operation will be a single bit indicating whether the two signals are the same or not.

Explanation

To detect if two 8-bit signals are the same, we can compare each bit of the two signals. If all the corresponding bits are equal, then the signals are the same. This can be done by using a bitwise comparison operation, such as XOR, AND, or NOT. The result of the comparison operation will be a single bit indicating whether the two signals are the same or not.

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