UGC-NET Electronic Science Previous Year Question Papers with Answer Keys and Full Explanations

 

 

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UGC-NET Electronic Science 2023 Answers with Explanations

Q.118

Let's analyze and solve the question.

List-I (Sequence x[n])

• (A) \( e^{i \omega_0 n} x[n] \)
• (B) \( n x[n] \)
• (C) \( x[n] - x[n-1] \)
• (D) \( \delta[n - n_0] \)

List-II (Fourier Transform X(Ω))

• (I) \( (1 - e^{-i \Omega}) \times (\Omega) \)
• (II) \( X(\Omega - \Omega_0) \)
• (III) \( e^{-i \Omega n_0} \)
• (IV) \( \frac{dX(\Omega)}{d\Omega} \)

Matching the sequences with their Fourier Transforms:

1. For \( e^{i \omega_0 n} x[n] \):
   • Multiplying the sequence \( x[n] \) by a complex exponential will shift its Fourier transform by \( \omega_0 \). This corresponds to the Fourier transform of \( x[n] \) shifted by \( \Omega_0 \), i.e., (II).
2. For \( n x[n] \):
   • This corresponds to the differentiation of the Fourier transform with respect to \( \Omega \), i.e., (IV).
3. For \( x[n] - x[n-1] \):
   • This is a difference of two consecutive samples, which corresponds to multiplying the Fourier transform by \( (1 - e^{-i \Omega}) \). This matches with (I).
4. For \( \delta[n - n_0] \):
   • The Fourier transform of a shifted delta function is simply a phase shift \( e^{-i \Omega n_0} \), which corresponds to (III).

Final Answer:

• (A) \( e^{i \omega_0 n} x[n] \) → (II) \( X(\Omega - \Omega_0) \)
• (B) \( n x[n] \) → (IV) \( \frac{dX(\Omega)}{d\Omega} \)
• (C) \( x[n] - x[n-1] \) → (I) \( (1 - e^{-i \Omega}) \times (\Omega) \)
• (D) \( \delta[n - n_0] \) → (III) \( e^{-i \Omega n_0} \)

Thus, the correct answer is Option (2):
(A)-(II), (B)-(IV), (C)-(I), (D)-(III).

Q.119

Match List-I with List-II

List-I (Feedback Connection Type)

• (A) Voltage series feedback
• (B) Voltage shunt feedback
• (C) Current series feedback
• (D) Current shunt feedback

List-II (Input/Output Impedance)

• (I) Z_of = Z_o / (1 + βA)
• (II) Z_of = Z_o(1 + βA)
• (III) Z_if = Z_i / (1 + βA)
• (IV) Z_if = Z_i(1 + βA)

Detailed Explanation

1. General Effects of Feedback on Impedance:

• Sampling (Output side):
  • Voltage Sampling (Shunt output) → Decreases Output Impedance (Z_of = Z_o / (1 + βA))
  • Current Sampling (Series output) → Increases Output Impedance (Z_of = Z_o(1 + βA))
• Mixing (Input side):
  • Series Mixing (Voltage addition) → Increases Input Impedance (Z_if = Z_i(1 + βA))
  • Shunt Mixing (Current addition) → Decreases Input Impedance (Z_if = Z_i / (1 + βA))

Matching Analysis

List-I Item	Topology Effect	Matches List-II
(A) Voltage Series	Series Input (High Zin) / Shunt Output (Low Zout)	Matches (IV): Zif = Zi(1 + βA)
(B) Voltage Shunt	Shunt Input (Low Zin) / Shunt Output (Low Zout)	Matches (I): Zof = Zo / (1 + βA)
(C) Current Series	Series Input (High Zin) / Series Output (High Zout)	Matches (II): Zof = Zo(1 + βA)
(D) Current Shunt	Shunt Input (Low Zin) / Series Output (High Zout)	Matches (III): Zif = Zi / (1 + βA)

Correct Option: (2) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

Q.120

Matching Logic Devices with Their Characteristics

List - I

• (A) EPROM
• (B) PLA (Programmable Logic Array)
• (C) Generic Array Logic (GAL)
• (D) PAL (Programmable Array Logic)

List - II

• (I) AND gate programmable OR gate permanently wired
• (II) Both AND and OR gates are programmable
• (III) AND gate programmable, output permanently hard wired but may be taken through register or tristate gate programmable
• (IV) AND gate permanently hard wired OR gate programmable

Explanation of Terms

• EPROM: Erasable Programmable Read-Only Memory (not a logic device, but a memory device). It is typically fixed logic, so the best fit is usually the option where both AND and OR gates are programmable — this matches the flexibility of PLA.
• PLA (Programmable Logic Array): Both AND and OR planes are programmable. So PLA matches (II).
• GAL (Generic Array Logic): It is a type of PAL with some enhancements, typically has AND gate programmable, output permanently hardwired but can be taken through register or tristate gate programmable. Matches (III).
• PAL (Programmable Array Logic): AND gate programmable, OR gate fixed (permanently wired). Matches (I) or (IV). Usually, PAL has programmable AND and fixed OR gate, which is (I).

Matching

• (A) EPROM → (IV) — Usually EPROM is not logic programmable but memory. Here, considering the options, the only option left that fits is (IV) AND gate permanently hard wired OR gate programmable (or fixed). Though EPROM is memory, in this question's context, it seems to fit this definition.
• (B) PLA → (II) Both AND and OR gates programmable
• (C) GAL → (III) AND gate programmable, output permanently hard wired but may be taken through register or tristate gate programmable
• (D) PAL → (I) AND gate programmable OR gate permanently wired

Answer

(3) (A)-(IV), (B)-(II), (C)-(III), (D)-(I)

Q.121

List–I (Instructions) Analysis

(A) MOV BX, AX
Both operands are registers → Register Addressing (III)

(B) MOV AX, 50H[BX]
A displacement added to a register (BX) → Register Relative Addressing (I)

(C) MOV AX, [BX]
Operand is in memory pointed to by BX → Register Indirect Addressing (IV)

(D) MOV AX, [2000H]
Operand stored at a direct memory address → Direct Addressing (II)

---

Correct Matching

Instruction	Addressing Mode
(A)	(III)
(B)	(I)
(C)	(IV)
(D)	(II)

---

Correct Option: (3)

(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

Q.122

Matching Microwave Devices with Names

List I (Microwave Devices) and List II (Name of Device):

1. Microwave transistor (A):
   • A Microwave transistor is a device used for amplification in microwave circuits. The most common type of microwave transistor is the HBT (Heterojunction Bipolar Transistor).
   • Therefore, (A) → (IV).
2. Field effect transistor (B):
   • A Field effect transistor (FET), typically used in microwave applications, is often associated with InP diodes (Indium Phosphide), which are known for their high-speed performance and are often used in high-frequency applications.
   • Therefore, (B) → (II).
3. Transferred electron device (C):
   • A Transferred electron device (such as the Gunn diode) is a microwave device that is commonly associated with Read diodes, which operate based on transferred electrons for microwave generation.
   • Therefore, (C) → (III).
4. Avalanche transit time device (D):
   • An Avalanche transit time device (like the IMPATT diode) is typically used with a CCD (Charge-Coupled Device).
   • Therefore, (D) → (I).

Correct Matching:

The correct matching is:

• (A) → (IV): Microwave transistor corresponds to HBT (Heterojunction Bipolar Transistor).
• (B) → (II): Field effect transistor corresponds to InP diodes.
• (C) → (III): Transferred electron device corresponds to Read diode.
• (D) → (I): Avalanche transit time device corresponds to CCD.

Answer:

The correct answer is Option (2)

Q.123

Correct Answer: Option (1)

Q.124

Matching Transducers and Units

List I (Concepts) and List II (Units):

1. Electrical resistance (A):
   • Electrical resistance is typically measured in ohms (Ω). Its unit is (V/A) (Voltage per Ampere), which is a standard unit for resistance.
   • So, (A) → (IV).
2. Thermal resistance (B):
   • Thermal resistance measures the ability of a material to resist the flow of heat. Its unit is typically in °C/W, which means degrees Celsius per watt (temperature change per unit heat flow).
   • So, (B) → (I).
3. Time constant (C):
   • The time constant (τ) in an RC circuit is the product of the resistance (R) and the capacitance (C). The unit of time constant is given by (RC), which is time (seconds).
   • So, (C) → (III).
4. Rate of heat transfer (D):
   • The rate of heat transfer (also known as thermal power) is typically measured in watts (W), which represents energy per unit time.
   • So, (D) → (II).

Correct Matching:

The correct matching is:

• (A) → (IV): Electrical resistance corresponds to (V/A).
• (B) → (I): Thermal resistance corresponds to (°C/W).
• (C) → (III): Time constant corresponds to (RC).
• (D) → (II): Rate of heat transfer corresponds to (W).

Answer:

The correct answer is Option (4): (A)-(IV), (B)-(I), (C)-(III), (D)-(II).

Q.125

Matching Transducers and Applications

List I (Transducers) and List II (Typical Applications):

1. Potentiometric device (A):
   • A potentiometric device is typically used for measuring displacement, as it works by varying the resistance based on position changes.
   • The application for potentiometric devices is displacement.
   • So, (A) → (III).
2. Resistance hygrometer (B):
   • A resistance hygrometer measures the relative humidity by detecting changes in resistance as the moisture content in the air varies.
   • Therefore, (B) → (I).
3. Dielectric gauge (C):
   • A dielectric gauge measures thickness using the principle of capacitance, as the dielectric constant of materials changes with thickness.
   • Therefore, (C) → (IV).
4. Moving coil generator (D):
   • A moving coil generator is used for measuring velocity, often used in tachometers.
   • So, (D) → (II).

Final Answer:

The correct matching is:

(A) → (III), (B) → (I), (C) → (IV), (D) → (II)

Answer Option:

Option (3):

Q.126

Concept of Diffusivity: Diffusivity refers to how fast a substance spreads or moves through another medium. In the context of semiconductors and materials science, the diffusivity of dopants (like Boron, Arsenic, etc.) plays a critical role in processes such as doping in semiconductor fabrication. Generally, smaller atoms tend to diffuse more quickly than larger atoms, and elements with lower atomic weights tend to have higher diffusivity. Boron is a small atom and generally has a high diffusivity in semiconductor materials.

Correct Answer: Option (2)

Q.127

The Debye Length Problem Solution

Problem Analysis:

The Debye length (λ_D) is inversely proportional to the square root of the doping density (N). This means that as N increases, the Debye length decreases, and vice versa.

The formula for the Debye length is:

λ_D ∝ 1 / √N

Where:

• N is the doping density (in cm^-3).

Understanding the Doping Densities:

We are given the following doping densities:

• (A) N = 10¹⁵ cm^-3
• (B) N = 10¹⁷ cm^-3
• (C) N = 3 × 10¹⁵ cm^-3
• (D) N = 10¹⁶ cm^-3
• (E) N = 5 × 10¹⁶ cm^-3

Step-by-step Solution:

• Doping density (B) is the largest value, 10¹⁷ cm^-3, so it will result in the smallest Debye length.
• Doping density (E) is the second largest, 5 × 10¹⁶ cm^-3, so the Debye length will be slightly larger than for (B), but smaller than for the other values.
• Doping density (D) is 10¹⁶ cm^-3, smaller than (B) and (E), so its Debye length will be larger than those.
• Doping density (A) is 10¹⁵ cm^-3, which is smaller than (B), (E), and (D), so the Debye length will be larger than the previous ones.
• Doping density (C) is 3 × 10¹⁵ cm^-3, which is very close to (A), so its Debye length will be similarly large.

Order of Doping Density and Debye Length:

The doping densities, from maximum Debye length to minimum, are:

1. A (10¹⁵ cm^-3)
2. C (3 × 10¹⁵ cm^-3)
3. D (10¹⁶ cm^-3)
4. E (5 × 10¹⁶ cm^-3)
5. B (10¹⁷ cm^-3)

Answer:

The correct order is (A), (C), (D), (E), (B).

Thus, the correct answer is option:

(1) (B), (E), (D), (C), (A)

Q.128

The time constant τ=R⋅C

Correct Answer: Option (4)

Q.129

To arrange the stability factor (S_I) in descending order, recall that for an emitter-bias BJT:

S_I = (1 + β) / [1 + (β + 1)(R_E/R_B)]

Thus:

• Smaller (R_E/R_B) → larger stability factor S_I
• Larger (R_E/R_B) → smaller stability factor S_I

Computed values of (R_E/R_B):

Option	Given Relation	RE/RB
A	RE = 0.1 RB	0.1
B	RB = 60 RE	1/60 ≈ 0.0167
C	RB = 100 RE	1/100 = 0.01
D	RE = 10 RB	10
E	RB = 30 RE	1/30 ≈ 0.0333

Order of decreasing stability factor (S_I):

C > B > E > A > D

Correct option:

Option (2): (C), (B), (E), (A), (D)

Q.130

In this problem, you are asked to arrange the given logic family voltage terms in the correct sequence. Let's analyze each term:

Definitions:

• (A) V_{OH} is the maximum output high-level voltage.
• (B) V_{OL} is the maximum input low-level voltage.
• (C) V_{IH} is the minimum input acceptable high-level voltage.
• (D) V_{IL} is the minimum acceptable input low-level voltage.

Sequence of Terminologies:

When we arrange these terms based on the usual convention of how voltage levels are defined in logic families, the typical sequence is as follows:

1. (C) V_{IH}: This is the minimum input voltage that is still considered a "high" logic level (input).
2. (A) V_{OH}: This is the maximum output voltage that is considered "high" from the logic gate.
3. (D) V_{IL}: This is the minimum input voltage that is still considered a "low" logic level (input).
4. (B) V_{OL}: This is the maximum output voltage that is still considered a "low" logic level.

Thus, the correct sequence of these voltage levels in a typical logic family is:

(C), (A), (D), (B)

Final Answer:

The most appropriate answer is:

Option (1): (C), (A), (D), (B)

This order is based on the typical voltage conventions used in digital logic families.

Q.131

The question asks to arrange the interrupts of the 8051 microcontroller in order of priority upon reset. The given interrupts are:

• (A) INT 0 (External Interrupt 0)
• (B) INT 1 (External Interrupt 1)
• (C) TF 0 (Timer 0 Overflow Interrupt)
• (D) TF 1 (Timer 1 Overflow Interrupt)
• (E) (R1 + T1) (Serial Communication Interrupt)

8051 Interrupt Priorities:

In the 8051 microcontroller, the interrupt priorities are defined as follows:

1. INT 0 (External Interrupt 0) has the highest priority.
2. INT 1 (External Interrupt 1) comes next.
3. Timer 0 overflow (TF 0) has a lower priority.
4. Timer 1 overflow (TF 1) has a lower priority than Timer 0.
5. Serial communication interrupt (R1 + T1) has the lowest priority.

Interrupt Priority Order:

• INT 0 has the highest priority, followed by INT 1.
• Then, we have the timer interrupts: TF 0 and TF 1.
• Finally, the serial communication interrupt (R1 + T1) has the lowest priority.

Correct Sequence:

• (A) INT 0 > (B) INT 1 > (C) TF 0 > (D) TF 1 > (E) (R1 + T1)

Final Answer:

The most appropriate answer is:

Option (2)

This option correctly represents the priority of interrupts in the 8051 microcontroller.

Q.132

Let's analyze the given materials with respect to their relative dielectric constants. The materials are:

• (A) Ge (Germanium)
• (B) C (Carbon)
• (C) CdS (Cadmium Sulfide)
• (D) TiO₂ (Titanium Dioxide)

Dielectric Constants:

• Germanium (Ge): Germanium is a semiconductor material, and its dielectric constant is relatively high compared to typical insulators but lower than that of certain ceramics. Its value is around 16.
• Carbon (C): Carbon, in its graphite or other common forms, is a conductor, and its dielectric constant is relatively low compared to insulators. Its value is around 2-5 depending on the form (graphite, diamond, etc.).
• CdS (Cadmium Sulfide): This is a semiconductor material with a moderate dielectric constant. Its value is around 10.
• TiO₂ (Titanium Dioxide): Titanium Dioxide is a ceramic material known for its high dielectric constant. Its value is around 80.

Arranging the Materials in Descending Order of Dielectric Constants:

• TiO₂ (D) has the highest dielectric constant (~80).
• Ge (A) follows with a dielectric constant of around 16.
• CdS (C) comes next with a value around 10.
• C (B) has the lowest dielectric constant, around 2-5.

Correct Sequence:

• (D) TiO₂ > (A) Ge > (C) CdS > (B) C

Final Answer:

The correct order is:

Option (2)

Q.133

Let's analyze the steps for the FM transmitter's blocks in the correct sequence.

FM Transmitter Block Diagram Overview:

The FM transmitter consists of the following blocks:

• (A) Crystal Oscillator
• (B) Antenna
• (C) Frequency Multiplier
• (D) Phase Modulate/Audio Source
• (E) Power Amplifier

Sequence of Blocks in an FM Transmitter:

1. Phase Modulate/Audio Source (D): The first block in the transmitter system would typically be the audio signal (such as music or speech) which is used to modulate the phase of the carrier wave. This signal is often generated from an audio source or phase modulator.
2. Crystal Oscillator (A): A crystal oscillator generates the carrier frequency. The audio signal from (D) modulates this signal, creating the modulated wave that will be amplified.
3. Frequency Multiplier (C): The frequency multiplier increases the frequency of the modulated signal, usually to the appropriate frequency range for FM transmission.
4. Power Amplifier (E): The power amplifier boosts the signal to a level strong enough for transmission over long distances.
5. Antenna (B): Finally, the modulated and amplified signal is transmitted via the antenna.

Correct Sequence:

• (D) -> (A) -> (C) -> (E) -> (B)

Final Answer:

The most appropriate answer is:

Option (4)

This sequence represents the correct order of blocks in an FM transmitter from start to finish.

Q.134

Let's analyze the question step by step to arrange the controllers in increasing order of system complexity.

Controllers Given:

• (A) Proportional Controller
• (B) Proportional plus Derivative Controller
• (C) Proportional plus Integral plus Derivative Controller
• (D) Proportional plus Integral Controller

Understanding the Complexity of Each Controller:

1. Proportional Controller (A): This controller has the least complexity as it only adjusts the output proportionally based on the error signal. It does not involve any integral or derivative action, making it the simplest.
2. Proportional plus Integral Controller (D): This controller adds an integral action to the proportional control. The integral action increases the complexity by eliminating steady-state error, but it is still simpler compared to controllers with derivative action.
3. Proportional plus Derivative Controller (B): This controller includes a derivative action in addition to proportional control. The derivative action introduces more complexity because it anticipates future errors, providing a faster response, but is still simpler than a combined proportional, integral, and derivative controller.
4. Proportional plus Integral plus Derivative Controller (C): This is the most complex controller. It combines proportional, integral, and derivative actions. This controller offers the most advanced control capabilities, with the highest complexity as it uses all three actions.

Arrangement in Increasing Order of Complexity:

• (A) Proportional Controller (simplest)
• (D) Proportional plus Integral Controller
• (B) Proportional plus Derivative Controller
• (C) Proportional plus Integral plus Derivative Controller (most complex)

Final Answer:

The correct order of complexity is:

• (A), (D), (B), (C)

So, the most appropriate answer is:

Option (2): (A), (D), (B), (C)

Q.135

Key Elements in the Circuit:

• Operational Amplifier (Op-Amp): We can see a basic configuration of an op-amp.
• Switch (S1): It allows switching between two ranges of output amplitudes: 0-0.1V and 0-1V.
• Resistors (R1, R2, R3): These are the resistors that determine the gain of the op-amp, which affects the output voltage.

Concept:

The output voltage \( V_o \) of the operational amplifier depends on the resistor values, and the behavior of the circuit will change based on how the switch (S1) is configured.

Relationship between the resistors and gain:

1. Voltage Gain \( A_v \) of an op-amp is given by the formula:

   \( A_v = -\frac{R_3}{R_2} \)

   where \( R_3 \) is the feedback resistor, and \( R_2 \) is the resistor in the input path.
2. Range 1 (0 to 0.1 V output): For this range, the op-amp’s gain needs to be small. Therefore, \( R_3 \) (the feedback resistor) should be small relative to \( R_2 \), making the gain small enough to produce a 0-0.1V output range.
3. Range 2 (0 to 1 V output): For this range, the op-amp’s gain needs to be larger. This means that \( R_3 \) should be significantly larger than \( R_2 \), producing a gain high enough to reach the 0-1V output range.

Steps to solve:

1. For Range 1 (0-0.1 V): We need to minimize the gain, so:
   • \( R_3 \) should be small.
   • \( R_2 \) should be large to ensure that the voltage gain stays low.
2. For Range 2 (0-1 V): We need to maximize the gain, so:
   • \( R_3 \) should be large.
   • \( R_2 \) should be small to allow for a higher gain.

Thus, the order of the resistor values should be:

• \( R_3 \) (largest)
• \( R_2 \) (intermediate)
• \( R_1 \) (smallest)

Final Answer:

The correct arrangement of the resistor values \( R_1 \), \( R_2 \), and \( R_3 \) in increasing order is:

• (C), (A), (B)

So, the most appropriate answer is Option (3): (C), (A), (B).

Q.136

Answer: (2)

Both (A) and (R) are correct but (R) is not the correct explanation of (A).

Explanation

Assertion (A):

A tunnel diode consists of a simple p–n junction made of a degenerate semiconductor. The I–V characteristics consist of three regions and show a negative differential resistance region in part of the forward characteristic.
Correct.
Tunnel diodes are heavily doped p–n junctions (degenerately doped) that exhibit tunneling, resulting in a negative resistance region in their I–V curve.

Reason (R):

In tunnel diode the p-side is highly doped and the n-side is highly doped and also used as detector.
Correct (mostly).
Both sides are indeed heavily doped (degenerate doping). Tunnel diodes are also used as high-speed switches or microwave detectors.

However…

Why (R) is NOT the correct explanation of (A):

• Assertion (A) talks about negative resistance and I–V characteristics due to quantum tunneling.
• Reason (R) only states that both sides are highly doped and mentions its use as a detector, which does not explain the negative resistance behavior.

Thus, even though both statements are correct, (R) does not explain (A).

Correct option: (2)

Q.137

Answer: (1)

Both (A) and (R) are correct and (R) is the correct explanation of (A).

Explanation

• Assertion (A):
  “Dual-slope A/D converter is the most preferred A/D conversion approach in digital multimeters.”
  This is correct. Dual-slope ADCs are widely used in digital multimeters due to their stability and noise immunity.
• Reason (R):
  “Dual-slope A/D converter provides high accuracy in A/D conversion, while at the same time suppressing the noise effect on the input signal.”
  This is also correct. Dual-slope ADCs integrate the input over a fixed time, which averages out noise, especially line-frequency noise (50/60 Hz), leading to high accuracy.

Since the reason directly explains why dual-slope ADCs are preferred in multimeters, (R) is the correct explanation of (A).

Correct option: (1)

Q.138

Let's solve it step by step.

Assertion (A):

“For 8086 microprocessor, the CS:IP loaded in the beginning with the required address from which the execution is to be started.”

Correct

In the 8086 microprocessor, at reset:

• CS = F000H
• IP = FFF0H

Together they form the physical address FFFF0H, the starting point of execution.
Also, whenever execution needs to start at a particular address, CS:IP is loaded with that address.

Reason (R):

“The microprocessor does not perform the next fetch operation till at least one byte of the instruction queue is emptied.”

Correct

8086 uses a 6-byte prefetch queue:

• The Bus Interface Unit (BIU) fetches instruction bytes.
• The Execution Unit (EU) executes instructions.
• BIU fetches only when there is space in the queue (i.e., at least one byte emptied).

So the statement about queue behavior is correct.

But is R the correct explanation of A?

No.
Reason (R) describes the prefetch queue mechanism, which has nothing to do with loading CS:IP at the beginning of execution.

Final Answer → (2)

Both (A) and (R) are correct, but (R) is not the correct explanation of (A).

If you want, I can also explain how the 8086 instruction queue works!

Q.139

Answer: (3) A is correct but R is not correct

Explanation:

• Assertion (A): “Synchronous machine is used as a motor and a generator also.”
  True — A synchronous machine can indeed operate as both a motor and a generator.


• Reason (R): “Synchronous machine can operate at constant speeds and variable frequencies under steady state.”
  Incorrect — A synchronous machine runs at a constant speed only when supplied with a constant frequency.
  Its speed is directly tied to supply frequency by:

  N_s = (120f) / P

  So it cannot operate at variable frequencies while keeping speed constant.

Thus, A is correct, but R is not correct → Option (3).

Q.140

Correct Answer: (2)

Q.141

Correct Answer: (2)

Q.142

Limited-Source Diffusion (Instantaneous Source Diffusion)

This is a standard limited-source diffusion problem.

Solution

For a thin deposited dopant layer with fixed total dose \(Q_T\) (atoms/cm²), and diffusivity \(D\), after diffusion time \(t\), the dopant concentration profile is:

\[ C(x,t) = \frac{Q_T}{\sqrt{\pi D t}} \exp\left( -\frac{x^2}{4Dt} \right) \]

The surface concentration is obtained by evaluating at \(x = 0\):

\[ C_S = C(0,t) \]

\[ C_S = \frac{Q_T}{\sqrt{\pi D t}} \exp(0) \]

\[ C_S = \frac{Q_T}{\sqrt{\pi D t}} \]

Correct Answer: (3)

\[ C_S = \frac{Q_T}{\sqrt{\pi D t}} \]

If you want, I can provide the full derivation from Fick’s second law.

Q.143

Sheet Resistance Concept Explanation

This is a concept question about the definition of sheet resistance \(R_s\) for a diffused semiconductor layer.

Given

A diffused layer with:

• Junction depth: \(x_j\)
• Carrier mobility: \(\mu\)
• Impurity (dopant) concentration: \(C(x)\)
• Charge magnitude: \(q\)

Recall

Resistivity of a semiconductor:

\[ \rho(x) = \frac{1}{q \mu C(x)} \]

Sheet resistance is resistivity integrated over depth:

\[ R_s = \int_0^{x_j} \rho(x) \, dx = \int_0^{x_j} \frac{1}{q \mu C(x)} \, dx \]

Therefore

\[ \boxed{ R_s = \int_0^{x_j} \frac{1}{q \mu C(x)} \, dx } \]

Correct Answer: (3)

\[ R_s = \frac{1}{q \mu} \int_0^{x_j} \frac{1}{C(x)} \, dx \]

Let me know if you'd like this styled differently or embedded into a full webpage!

Q.144

Option 1

Q.145

Option 3

Q.146

This is a standard vector identity:

\( \nabla \times (\nabla \times \mathbf{A}) = \nabla(\nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A} \)

This is known as the curl of the curl identity.

Comparison with options:

1. \( \nabla \cdot \mathbf{A} - \nabla^2 \mathbf{A} \)
   Incorrect (first term is divergence, not gradient)
2. \( \nabla \cdot (\nabla \times \mathbf{A}) - \nabla^2 \mathbf{A} \)
   Incorrect (divergence of a curl is always zero)
3. \( \frac{1}{\mu_0} \nabla \times \nabla \times \mathbf{A} \)
   Incorrect (just scaled, not the identity)
4. \( \nabla(\nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A} \)
   Correct

Final Answer: (4)

Q.147

For a small loop antenna, the standard approximate radiation-resistance formula is:

\( R_r \approx 31200 \left( \frac{A}{\lambda^2} \right)^2 \text{ ohms} \)

(for a single-turn loop; this is the commonly used practical constant for small loops.)

Given:

• Square loop with side
  \( l = \frac{\lambda}{10} \)
• Area
  \( A = l^2 = \left( \frac{\lambda}{10} \right)^2 = \frac{\lambda^2}{100} \)

Substitute into the formula:

\( \frac{A}{\lambda^2} = \frac{1}{100} \)

\( R_r = 31200 \left( \frac{1}{100} \right)^2 = 31200 \times \frac{1}{10000} = 3.12\, \Omega \)

Correct answer: (2) 3.12 ohms

Q.148

The given expression is:

\[ P_r = P_t \frac{A_{et} \cdot A_{er}}{r^2 \lambda^2} \]

Analysis

This expression relates:

• \(P_t\): transmitted power
• \(P_r\): received power
• \(A_{et}\): effective aperture of transmitting antenna
• \(A_{er}\): effective aperture of receiving antenna
• \(r\): distance
• \(\lambda\): wavelength

This is the Friis Transmission Formula in terms of effective antenna apertures.

The classic Friis formula is:

\[ P_r = P_t G_t G_r \left( \frac{\lambda}{4\pi r} \right)^2 \]

Using:

\[ G = \frac{4\pi A_e}{\lambda^2} \]

you get exactly the form in the question:

\[ P_r = P_t \frac{A_{et}A_{er}}{r^2\lambda^2} \]

Correct Answer: (3) Friis transmission formula

Q.149

Solution:

Given the electric field components:

\( E_x = 1.5 \sin(\omega t - \beta z)\ \text{V/m} \)
\( E_y = 3 \sin(\omega t - \beta z + 75^\circ)\ \text{V/m} \)

Step 1: RMS Values

\( E_{x,\text{rms}} = \frac{1.5}{\sqrt{2}} = 1.06\ \text{V/m} \)
\( E_{y,\text{rms}} = \frac{3}{\sqrt{2}} = 2.12\ \text{V/m} \)

Total: \( E_{\text{rms}}^2 = (1.06)^2 + (2.12)^2 = 5.61 \)

Step 2: Power Density

Intrinsic impedance of air: \( \eta_0 = 377\ \Omega \)

Average power per unit area:
\[ S_{\text{avg}} = \frac{E_{\text{rms}}^2}{\eta_0} = \frac{5.61}{377} \approx 0.0149\ \text{W/m}^2 = 15\ \text{mW/m}^2 \]

Final Answer: Option B (15 mW/m²)

Q.150

The propagation constant for a uniform plane wave is given by which expression?

1. \(\sqrt{\frac{R + j\omega L}{G + j\omega C}}\)

2. \(\sqrt{(R + j\omega L)(G + j\omega C)}\)

3. \(\sqrt{\frac{j\omega \mu}{\sigma + j\omega \varepsilon}}\)

4. \(\sqrt{\frac{\eta_2 - \eta_1}{\eta_2 + \eta_1}}\)

Correct Answer: Option (2)