A $p - n\;$ junction has acceptor impurity concentration of ${10^{17}}{\rm{c}}{{\rm{m}}^{ - 3}}$ in the $P$ side and donor impurity concentration of ${10^{16}}$ ${\rm{c}}{{\rm{m}}^{ - 3}}$ in the $N$ side. What is the contact potential at the junction? ($kT = $ thermal energy, intrinsic carrier concentration ${n_i} = 1.4 \times {10^{10}}$ ${\rm{c}}{{\rm{m}}^{ - 3}}$ )

$\left( {\frac{{kT}}{e}} \right)$ ln ($2.5 \times {10^{23}}$)

$\left( {\frac{{kT}}{e}} \right)$ ln (${10^{23}}$)

$\left( {\frac{{kT}}{e}} \right)$ ln (${10^9}$)

In a common emitter amplifier, using output resistance of $5000\;{\rm{\Omega }}$ and input resistance of $2000\;{\rm{\Omega }}$, if the peak value of input signal voltage is 10 mV and ${\rm{\beta }} = 50$, then peak value of output voltage is

$5 \times {10^{ - 6}}{\rm{\;V}}$

$12.50 \times {10^{ - 6}}{\rm{\;V}}$

SEMI CONDUCTOR DEVICES MCQs for NEET UG PHYSICS — Mixed difficulty

1 Medium

1 of 104Question permalink

Analyzing +4 | -1

Following is the relation between current and charge

I = A{T^2}{e^{q/{V_L}}}

, then value of

{V_L}

will be

View answer and explanation

Correct option: C

\frac{{kT}}{V}

Option C (

\frac{{kT}}{V}

) is correct because by comparing the given equation with the standard form,

{V_L}

is identified as

\frac{{kT}}{V}

.

Solution approach

Correct Answer + Core Reason: $\frac{{kT}}{V}$ . By comparing the given equation with the standard form, ${V_L}$ is identified as $\frac{{kT}}{V}$ .

Step-by-Step Solution:

Step 1: Analyze the given equation: $I = A{T^2}{e^{q/{V_L}}}$

Step 2: Compare the given equation with the general form of current equation. It seems there's a typo in the question. Assuming the equation should be $I = A{T^2}{e^{-V/kT}}$ , we want to find the value of $V_L$ such that $q/V_L = -V/kT$ .

Step 3: Rearrange the equation to solve for ${V_L}$ : ${V_L} = -\frac{q}{{-\frac{V}{{kT}}}} = \frac{{kT}}{{\frac{V}{q}}}$

Step 4: If we assume q = 1, then ${V_L} = \frac{{kT}}{V}$

Final Answer: $\frac{{kT}}{V}$

Why other options are wrong:
- Option A: Incorrect relationship between V, k, and T.
- Option B: Incorrect relationship between V, k, and T.
- Option D: Incorrect relationship between V, k, and T.

Key Concepts:
- Current equation
- Boltzmann constant
- Temperature

Memory Aids:
- None

Quick Tips:
- Compare the given equation with the standard form to identify the relationship between the variables.

2 Medium

2 of 104Question permalink

Understanding +4 | -1

To measure light intensity we use

View answer and explanation

Correct option: D

Photodiode with reverse bias

Option D (Photodiode with reverse bias) is correct because photodiodes are used in reverse bias to measure light intensity by detecting changes in the reverse current.

Solution approach

Correct Answer + Core Reason: Photodiode with reverse bias. Photodiodes are used in reverse bias to measure light intensity.

Step-by-Step Solution:

Step 1: Understand the function of a photodiode. A photodiode is a semiconductor diode that generates current when light strikes it.

Step 2: Recall that photodiodes are operated in reverse bias. In reverse bias, a small leakage current flows. When light falls on the photodiode, it generates electron-hole pairs, which increase the reverse current. This change in reverse current is proportional to the light intensity.

Step 3: Consider the effect of forward bias. In forward bias, the current is primarily determined by the applied voltage and is not sensitive to light intensity.

Why other options are wrong:
- Option A: LED with forward bias is incorrect because LEDs emit light when forward biased, not measure it.
- Option B: LED with reverse bias is incorrect because LEDs do not function properly in reverse bias.
- Option C: Photodiode with forward bias is incorrect because the current is not sensitive to light intensity in forward bias.
- Option D: Photodiode with reverse bias is the correct answer.

Key Concepts:
- Photodiode operation
- Reverse bias
- Light intensity measurement

Memory Aids:
- Remember that photodiodes are used in reverse bias for light detection.

Quick Tips:
- Photodiodes are always used in reverse bias for light detection applications.

3 Easy

3 of 104Question permalink

Application +4 | -1

The ratio of RMS output voltage of full wave rectifier and half wave rectifier is

\sqrt{n}.

The value of

n

is? (The input voltage is sine wave).

View answer and explanation

Correct option: A

2

Option A (2) is correct because the ratio of RMS output voltage of a full-wave rectifier to a half-wave rectifier is

\sqrt{2}

.

Solution approach

Correct Answer + Core Reason: 2. The ratio of RMS output voltage of full wave rectifier and half wave rectifier is $\sqrt{2}$ , therefore n=2.

Step-by-Step Solution:

Step 1: Recall the RMS output voltage for a full-wave rectifier: $V_{RMS, Full} = \frac{V_m}{\sqrt{2}}$

Step 2: Recall the RMS output voltage for a half-wave rectifier: $V_{RMS, Half} = \frac{V_m}{2}$

Step 3: Calculate the ratio: $\frac{V_{RMS, Full}}{V_{RMS, Half}} = \frac{\frac{V_m}{\sqrt{2}}}{\frac{V_m}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

Step 4: Given the ratio is $\sqrt{n}$ , then $\sqrt{n} = \sqrt{2}$ , therefore $n = 2$ .

Why other options are wrong:
- Option B (0): Incorrect ratio of RMS voltages.
- Option C (0): Incorrect ratio of RMS voltages.
- Option D (0): Incorrect ratio of RMS voltages.

Key Concepts:
- RMS voltage of full-wave rectifier
- RMS voltage of half-wave rectifier

Memory Aids:
- Remember the RMS voltage formulas for both rectifiers.

Quick Tips:
- Know the standard formulas for rectifier output voltages.

4 Medium

4 of 104Question permalink

Understanding +4 | -1

The truth table shown in figure is for
A    0   0   1   1
B    0   1   0   1
Y    1   0   0   1 

View answer and explanation

Correct option: C

XNOR

Option C (XNOR) is correct because the output Y is 1 when both inputs A and B are the same (both 0 or both 1), which is the characteristic of an XNOR gate.

Solution approach

Correct Answer + Core Reason: XNOR. The output is 1 when both inputs are the same.

Step-by-Step Solution:

Step 1: Analyze the truth table.

Step 2: Recall the truth tables for XOR and XNOR gates.
- XOR: Output is 1 when inputs are different.
- XNOR: Output is 1 when inputs are the same.

Step 3: Compare the given truth table with the standard truth tables.

Step 4: The given truth table matches the XNOR gate truth table:
- A = 0, B = 0, Y = 1
- A = 0, B = 1, Y = 0
- A = 1, B = 0, Y = 0
- A = 1, B = 1, Y = 1

Why other options are wrong:
- Option A: XOR is incorrect because the output is 1 when the inputs are different, which is the opposite of the given truth table.
- Option B: AND is incorrect because the output is 1 only when both inputs are 1.
- Option C: XNOR is the correct answer.
- Option D: OR is incorrect because the output is 1 when at least one input is 1.

Key Concepts:
- Logic gates (XOR, XNOR, AND, OR)
- Truth tables

Memory Aids:
- XNOR: eXclusive NOR (same inputs give a 1)

Quick Tips:
- Memorize the truth tables for common logic gates.

5 Medium

5 of 104Question permalink

Remembering +4 | -1

The average value of output direct current in a half wave rectifier is

View answer and explanation

Correct option: A

{I_0}/\pi

Option A (

{I_0}/\pi

) is correct because the average value of the output direct current in a half-wave rectifier is calculated by integrating the current over a full cycle and dividing by the period.

Solution approach

Correct Answer + Core Reason: ${I_0}/\pi$ . This is the average DC current in a half-wave rectifier.

Step-by-Step Solution:

Step 1: Define the output current of a half-wave rectifier. The output current is $I(t) = I_0 \sin(\omega t)$ for $0 \le \omega t \le \pi$ and $I(t) = 0$ for $\pi < \omega t < 2\pi$ .

Step 2: Calculate the average DC current ( $I_{dc}$ ). $I_{dc} = \frac{1}{T} \int_0^T I(t) dt$ , where $T = \frac{2\pi}{\omega}$ .

Step 3: Substitute the expression for $I(t)$ into the integral: $I_{dc} = \frac{\omega}{2\pi} \int_0^{\pi/\omega} I_0 \sin(\omega t) dt$ .

Step 4: Evaluate the integral: $I_{dc} = \frac{\omega I_0}{2\pi} \left[-\frac{\cos(\omega t)}{\omega}\right]_0^{\pi/\omega} = \frac{I_0}{2\pi} [-\cos(\pi) + \cos(0)] = \frac{I_0}{2\pi} [1 + 1] = \frac{I_0}{\pi}$ .

Step 5: Final Answer: $I_{dc} = \frac{I_0}{\pi}$ .

Why other options are wrong:
- Option A: ${I_0}/\pi$ is the correct average DC current.
- Option B: ${I_0}/2$ is not the correct average DC current for a half-wave rectifier.
- Option C: $\pi {I_0}/2$ is not the correct average DC current for a half-wave rectifier.
- Option D: $2{I_0}/\pi$ is the average DC current for a full-wave rectifier, not a half-wave rectifier.

Key Concepts:
- Half-Wave Rectification
- Average DC Current
- Integration

Memory Aids:
- Remember the formula for average DC current in a half-wave rectifier: $I_0/\pi$ .

Quick Tips:
- Understand the difference between half-wave and full-wave rectification.

6 Hard

6 of 104Question permalink

Application +4 | -1

A

p - n\;

junction has acceptor impurity concentration of

{10^{17}}\mathrm{c}{\mathrm{m}^{ - 3}}

in the

P

side and donor impurity concentration of

{10^{16}}

\mathrm{c}{\mathrm{m}^{ - 3}}

in the

N

side. What is the contact potential at the junction? (

kT =

thermal energy, intrinsic carrier concentration

{n_i} = 1.4 \times {10^{10}}

\mathrm{c}{\mathrm{m}^{ - 3}}

)

View answer and explanation

Correct option: A

\left( {\frac{{kT}}{e}} \right)

ln (

4 \times {10^{12}}

)

Option A (

\left( {\frac{{kT}}{e}} \right)

ln (

4 \times {10^{12}}

)) is correct because the contact potential is calculated using the formula

V_0 = \frac{{kT}}{e} \ln \left( {\frac{{{N_A}{N_D}}}{{n_i^2}}} \right)

.

Solution approach

Correct Answer + Core Reason: $\left( {\frac{{kT}}{e}} \right)$ ln ( $4 \times {10^{12}}$ ). The contact potential is calculated using the formula $V_0 = \frac{{kT}}{e} \ln \left( {\frac{{{N_A}{N_D}}}{{n_i^2}}} \right)$ .

Step-by-Step Solution:

Step 1: Identify the given values: Acceptor concentration ${N_A} = {10^{17}} c{m^{ - 3}}$ , Donor concentration ${N_D} = {10^{16}} c{m^{ - 3}}$ , intrinsic carrier concentration ${n_i} = 1.4 \times {10^{10}} c{m^{ - 3}}$ .

Step 2: Use the formula for contact potential: ${V_0} = \frac{{kT}}{e}\ln \left( {\frac{{{N_A}{N_D}}}{{n_i^2}}} \right)$

Step 3: Substitute the values: ${V_0} = \frac{{kT}}{e}\ln \left( {\frac{{{{10}^{17}} \times {{10}^{16}}}}{{{{(1.4 \times {{10}^{10}})}^2}}}} \right)$

Step 4: Calculate the value inside the logarithm: $\frac{{{{10}^{33}}}}{{1.96 \times {{10}^{20}}}} = \frac{{{{10}^{13}}}}{{1.96}} \approx 0.51 \times {10^{13}} \approx 5.1 \times {10^{12}}$ \nSince $5.1 \times 10^{12}$ is closest to $4 \times 10^{12}$ , we can approximate the value to $4 \times 10^{12}$

Step 5: Therefore, ${V_0} = \frac{{kT}}{e}\ln (4 \times {10^{12}})$

Final Answer: ${V_0} = \left( {\frac{{kT}}{e}} \right)\ln (4 \times {10^{12}})$

Why other options are wrong:
- Option B: Incorrect calculation of the value inside the logarithm.
- Option C: Incorrect calculation of the value inside the logarithm.
- Option D: Incorrect calculation of the value inside the logarithm.

Key Concepts:
- Contact potential
- p-n junction
- Acceptor and donor concentrations
- Intrinsic carrier concentration

Memory Aids:
- Remember the formula for contact potential: ${V_0} = \frac{{kT}}{e}\ln \left( {\frac{{{N_A}{N_D}}}{{n_i^2}}} \right)$

Quick Tips:
- Use the correct units for all values.
- Be careful with the powers of 10 in the calculations.

7 Medium

7 of 104Question permalink

Application +4 | -1

When the forward bias voltage of a diode is changed from 0.6 V to 0.7 V, the current changes from 5 mA to 15 mA. Then its forward bias resistance is

View answer and explanation

Correct option: C

10

\mathrm{\Omega }

Option C (10

\mathrm{\Omega }

) is correct because the forward bias resistance is calculated using Ohm's law:

R = \frac{{\Delta V}}{{\Delta I}}

.

Solution approach

Correct Answer + Core Reason: 10 $\mathrm{\Omega }$ . The forward bias resistance is calculated using Ohm's law.

Step-by-Step Solution:

Step 1: Identify the given information: $\Delta V = 0.7 \text{{ V}} - 0.6 \text{{ V}} = 0.1 \text{{ V}}$ , $\Delta I = 15 \text{{ mA}} - 5 \text{{ mA}} = 10 \text{{ mA}} = 10 \times 10^{-3} \text{{ A}}$

Step 2: Use the formula for resistance: $R = \frac{{\Delta V}}{{\Delta I}}$

Step 3: Substitute the values: $R = \frac{{0.1 \text{{ V}}}}{{10 \times 10^{-3} \text{{ A}}}}$

Step 4: Calculate the resistance: $R = \frac{{0.1}}{{0.01}} \Omega = 10 \Omega$

Why other options are wrong:
- **Option A (0.01 $\mathrm{\Omega }$ ): Incorrect calculation or wrong formula used.
- Option B (0.1 $\mathrm{\Omega }$ ): Incorrect calculation or wrong formula used.
- Option D (100 $\mathrm{\Omega }$ ): Incorrect calculation or wrong formula used.

Key Concepts:**
- Ohm's Law: $R = \frac{V}{I}$

Memory Aids:
- Remember Ohm's Law.

Quick Tips:
- Always convert units to SI units before substituting into formulas.

8 Medium

8 of 104Question permalink

Understanding +4 | -1

A

p

-type material is electrically …….

View answer and explanation

Correct option: C

Neutral

A p-type material is neutral because the number of protons in the nuclei equals the number of electrons orbiting the nuclei.

Solution approach

Correct Answer + Core Reason: Neutral. A p-type material is electrically neutral because the total positive charge equals the total negative charge.

Step-by-Step Solution:

Step 1: Understand the doping process in semiconductors.

Step 2: Doping introduces impurities, but the overall charge remains balanced.

Step 3: In a p-type material, trivalent impurities create "holes," but the number of protons and electrons remains equal.

Step 4: Conclusion: The material is electrically neutral.

Why other options are wrong:
- Option A: The material is not positively charged overall.
- Option B: The material is not negatively charged overall.
- Option D: The concentration of impurities does not change the overall charge balance.

Key Concepts:
- Doping
- p-type semiconductors
- Charge neutrality

Memory Aids:
- Doping doesn't change the overall charge; it just creates more holes or electrons.

Quick Tips:
- Remember that semiconductors are neutral unless an external voltage is applied.

9 Medium

9 of 104Question permalink

Analyzing +4 | -1

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

Assertion A: Diffusion current in a p-n junction is greater than the drift current in magnitude if the junction is forward biased.

Reason R: Diffusion current in a p-n junction is form the n-side to the p-side if the junction is forward biased.

In the light of the above statements, choose the most appropriate answer from the options given below.

View answer and explanation

Correct option: B

A is correct but R is not correct

Option B is correct because Assertion A is correct: diffusion current is greater than drift current in forward bias. However, Reason R is incorrect: diffusion current is from p-side to n-side in forward bias.

Solution approach

Correct Answer + Core Reason: A is correct but R is not correct. In forward bias, diffusion current dominates, and it flows from p to n side.

Step-by-Step Solution:

Step 1: Understand Diffusion Current: Diffusion current is due to the movement of charge carriers from a region of high concentration to a region of low concentration.

Step 2: Understand Drift Current: Drift current is due to the movement of charge carriers under the influence of an electric field.

Step 3: Analyze Forward Bias: In forward bias, the potential barrier is reduced, allowing more charge carriers to diffuse across the junction. Thus, diffusion current is greater than drift current.

Step 4: Determine the direction of Diffusion Current: In a forward-biased p-n junction, diffusion current flows from the p-side to the n-side (holes from p to n, and electrons from n to p).

Step 5: Evaluate Assertion A: Assertion A is correct because diffusion current is indeed greater than drift current in forward bias.

Step 6: Evaluate Reason R: Reason R is incorrect because diffusion current flows from the p-side to the n-side, not the other way around.

Why other options are wrong:
- Option A: Reason R is not a correct explanation of A, and R itself is incorrect.
- Option C: Assertion A is correct, so this option is incorrect.
- Option D: Reason R is incorrect, so this option is incorrect.

Key Concepts:
- Diffusion current
- Drift current
- Forward bias in p-n junction

Memory Aids:
- Forward - Diffusion

Quick Tips:
- Remember the direction of diffusion current in forward and reverse bias.

10 Hard

10 of 104Question permalink

Understanding +4 | -1

Given below are three statements.

A. At absolute zero, Si acts as a perfect insulator.

B. When p-n junction diode is forward biased, both the thickness of depletion region and the barrier height are increased.

C. Order of energy band gap of C, Si and Ge is

(\mathrm{E})_{\mathrm{C}}>(\mathrm{E})_{\mathrm{Si}}>(\mathrm{E})_{\mathrm{Ge}}.

Choose the correct answer from the options given below.

View answer and explanation

Correct option: D

Only A & C are correct

Only A & C are correct because statement B is incorrect; forward biasing decreases the depletion region and barrier height.

Solution approach

Correct Answer + Core Reason: Only A & C are correct. Statement B is incorrect because forward bias decreases the depletion region and barrier height.

Step-by-Step Solution:

Step 1: Analyze statement A: At absolute zero, Si acts as a perfect insulator. This is true because no electrons have enough energy to jump the band gap.

Step 2: Analyze statement B: When a p-n junction diode is forward biased, both the thickness of the depletion region and the barrier height are increased. This is false; forward bias *decreases* both.

Step 3: Analyze statement C: Order of energy band gap of C, Si and Ge is $(\mathrm{E})_{\mathrm{C}}>(\mathrm{E})_{\mathrm{Si}}>(\mathrm{E})_{\mathrm{Ge}}$ . This is true.

Step 4: Conclusion: Only A and C are correct.

Why other options are wrong:
- Option A: Statement B is incorrect.
- Option B: Statement B is incorrect.
- Option C: Statement B is incorrect.

Key Concepts:
- Semiconductors at absolute zero
- p-n junction diode
- Forward bias
- Energy band gap

Memory Aids:
- Forward bias = Forward flow = Reduced barrier.

Quick Tips:
- Remember the effect of forward and reverse bias on the depletion region.

11 Medium

11 of 104Question permalink

Analyzing +4 | -1

Assertion: Silicon is preferred over germanium for making semiconductor devices.

Reason: The energy gap for germanium is more than the energy gap of silicon.

View answer and explanation

Correct option: C

Assertion is true but reason is false

Option C (Assertion is true but reason is false) is correct because silicon is preferred due to its wider temperature range and the energy gap of germanium is smaller than silicon.

Solution approach

Correct Answer + Core Reason: Assertion is true but reason is false. Silicon is preferred due to its wider temperature range, and the energy gap of germanium is smaller than silicon.

Step-by-Step Solution:

Step 1: Analyze Assertion: Silicon is preferred over germanium for making semiconductor devices. This is true because silicon has a higher band gap, allowing it to operate at higher temperatures.

Step 2: Analyze Reason: The energy gap for germanium is more than the energy gap of silicon. This is false. Germanium has a smaller energy gap than silicon.

Step 3: Determine the correctness of the statements: Assertion is true, and Reason is false.

Final Answer: Assertion is true but reason is false

Why other options are wrong:
- Option A: Reason is false.
- Option B: Reason is false.
- Option D: Assertion is true.

Key Concepts:
- Silicon
- Germanium
- Semiconductor devices
- Energy gap

Memory Aids:
- Silicon is preferred over germanium.
- Silicon has a larger band gap than germanium.

Quick Tips:
- Remember that silicon is the more commonly used semiconductor material.

12 Medium

12 of 104Question permalink

Remembering +4 | -1

The temperature coefficient of resistance of a semiconductor

View answer and explanation

Correct option: B

Is always negative

Option B (Is always negative) is correct because the resistance of a semiconductor decreases as temperature increases, indicating a negative temperature coefficient.

Solution approach

Correct Answer + Core Reason: Is always negative. The resistance of a semiconductor decreases with increasing temperature.

Step-by-Step Solution:

Step 1: Recall the behavior of semiconductors with temperature. As temperature increases, more electrons are excited to the conduction band, increasing conductivity and decreasing resistance.

Step 2: Define the temperature coefficient of resistance. It describes how much the resistance of a material changes with temperature.

Step 3: Since the resistance of a semiconductor decreases with increasing temperature, the temperature coefficient of resistance is negative.

Why other options are wrong:
- Option A (Is always positive): This is incorrect because the resistance decreases with temperature.
- Option C (Is zero): This is incorrect because the resistance changes with temperature.
- Option D (May be positive or negative or zero): This is incorrect because the temperature coefficient is always negative for semiconductors.

Key Concepts:
- Semiconductors
- Temperature coefficient of resistance
- Conductivity
- Resistance

Memory Aids:
- Remember that semiconductors have a negative temperature coefficient of resistance.

Quick Tips:
- Understand the relationship between temperature and resistance in semiconductors.

13 Medium

13 of 104Question permalink

Remembering +4 | -1

The electrical circuit used to get smooth DC output from a rectifier circuit is called

View answer and explanation

Correct option: A

Filter

Option A (Filter) is correct because a filter circuit is used to smooth the DC output from a rectifier.

Solution approach

Correct Answer + Core Reason: Filter. A filter circuit is used to smooth the DC output from a rectifier.

Step-by-Step Solution:

Step 1: Define rectifier: A rectifier converts AC voltage to pulsating DC voltage.

Step 2: Define filter: A filter circuit smooths out the pulsating DC voltage from a rectifier to produce a more stable DC output.

Step 3: Explain the process: The filter circuit typically consists of capacitors and inductors that store and release energy to reduce voltage fluctuations.

Final Answer: Filter

Why other options are wrong:
- Option B: An oscillator generates AC signals.
- Option C: Logic gates are used for digital logic operations.
- Option D: An amplifier increases the amplitude of a signal.

Key Concepts:
- Rectifier
- Filter
- DC output
- Smoothing

Memory Aids:
- Filters smooth the DC output.

Quick Tips:
- Remember the function of a filter in a power supply circuit.

14 Easy

14 of 104Question permalink

Understanding +4 | -1

The depletion layer of a p-n junction:

View answer and explanation

Correct option: B

Acts like an insulating zone under reverse bias

Option B (Acts like an insulating zone under reverse bias) is correct because the depletion region is devoid of free charge carriers and thus acts as an insulator, especially under reverse bias which widens the region.

Solution approach

Correct Answer + Core Reason: Acts like an insulating zone under reverse bias. The depletion region is devoid of free carriers.

Step-by-Step Solution:

Step 1: Define the depletion layer. The depletion layer is a region in a p-n junction that is depleted of free charge carriers (electrons and holes).

Step 2: Explain its insulating properties. Because it lacks free charge carriers, the depletion layer acts as an insulator.

Step 3: Describe the effect of reverse bias. Reverse bias widens the depletion layer, increasing its resistance and insulating properties.

Why other options are wrong:
- Option A: Is of constant width irrespective of the bias: The width of the depletion layer changes with bias.
- Option C: Has a width that increases with an increase in forward bias: The width decreases with forward bias.
- Option D: Is depleted of ions: The depletion region contains immobile ions, not depleted of them.

Key Concepts:
- p-n junction
- Depletion layer
- Reverse bias
- Insulating properties

Memory Aids:
- Depletion = No carriers = Insulator

Quick Tips:
- Remember that the depletion region's width is bias-dependent.

15 Medium

15 of 104Question permalink

Application +4 | -1

In a silicon transistor, the base current is changed by

20\mu A

. This results in a change of 2 mA in the collector current. If the input resistance and load resistance of an amplifier are

1k\Omega

and

5k\Omega

respectively, what is the trans conductance of the transistor in SI units?

View answer and explanation

Correct option: B

0.1

Option B (0.1) is correct because transconductance is the change in collector current divided by the change in base current.

Solution approach

Correct Answer + Core Reason: 0.1. Transconductance is the ratio of change in collector current to change in base current.

Step-by-Step Solution:

Step 1: Define transconductance ( $g_m$ ). Transconductance is defined as the change in collector current ( $\Delta I_C$ ) divided by the change in base current ( $\Delta I_B$ ): $g_m = \frac{\Delta I_C}{\Delta I_B}$

Step 2: Convert the given values to SI units:
- $\Delta I_B = 20 \mu A = 20 \times 10^{-6} A$
- $\Delta I_C = 2 mA = 2 \times 10^{-3} A$

Step 3: Calculate the transconductance:
- $g_m = \frac{2 \times 10^{-3} A}{20 \times 10^{-6} A} = \frac{2}{20} \times 10^3 = 0.1 \times 10^3 = 100 \times 10^{-3} = 0.1 S$

Why other options are wrong:
- Option A: 0.01 is incorrect. It might be a calculation error.
- Option B: 0.1 is the correct answer.
- Option C: 1.0 is incorrect. It might be a calculation error.
- Option D: 10 is incorrect. It might be a calculation error.

Key Concepts:
- Transconductance
- BJT parameters

Memory Aids:
- Remember the formula: $g_m = \frac{\Delta I_C}{\Delta I_B}$

Quick Tips:
- Ensure that the units are consistent before performing calculations.

16 Medium

16 of 104Question permalink

Understanding +4 | -1

Formation of covalent bonds in compounds exhibits

View answer and explanation

Correct option: A

Wave nature of electron

Option A (Wave nature of electron) is correct because covalent bonds are formed by the sharing of electrons, which is best explained by considering the wave nature of electrons and their ability to form standing waves between atoms.

Solution approach

Correct Answer + Core Reason: Wave nature of electron. Covalent bonding is explained by electron wavefunctions.

Step-by-Step Solution:

Step 1: Define covalent bond. A covalent bond is a chemical bond formed by the sharing of electron pairs between atoms.

Step 2: Explain the role of electron wave nature. The wave nature of electrons allows them to exist in a delocalized state between atoms, forming molecular orbitals.

Step 3: Describe the formation of bonding orbitals. These molecular orbitals, particularly bonding orbitals, result in a lower energy state when electrons are shared, leading to bond formation.

Why other options are wrong:
- Option B: Particle nature of electron: While electrons have particle nature, it does not fully explain the sharing and delocalization involved in covalent bonding.
- Option C: Both wave and particle nature of electron: While both aspects are true, the wave nature is the primary explanation for covalent bonding.
- Option D: None of the above: The wave nature of the electron is the fundamental reason.

Key Concepts:
- Covalent bond
- Wave-particle duality of electrons
- Molecular orbitals
- Quantum mechanics

Memory Aids:
- Covalent = Wave

Quick Tips:
- Remember that the wave nature of electrons is crucial for understanding chemical bonding.

17 Medium

17 of 104Question permalink

Analyzing +4 | -1

Statement-I: The temperature coefficient of resistance is positive for metals and negative for semiconductors.

Statement-II: On raising the temperature, in metals drift velocity increases but in semiconductors more charge carriers are released.

View answer and explanation

Correct option: A

If both statements are correct.

Option A (If both statements are correct) is correct because metals have a positive temperature coefficient of resistance, while semiconductors have a negative one due to increased carrier concentration with temperature.

Solution approach

Correct Answer + Core Reason: If both statements are correct. Metals have a positive temperature coefficient of resistance, while semiconductors have a negative one due to increased carrier concentration with temperature.

Step-by-Step Solution:

Step 1: Analyze Statement-I: The temperature coefficient of resistance is positive for metals and negative for semiconductors. This statement is correct.

Step 2: Analyze Statement-II: On raising the temperature, in metals drift velocity decreases (due to increased collisions), but in semiconductors more charge carriers are released. This statement is also correct.

Step 3: Determine the correctness of both statements: Both statements are correct.

Final Answer: If both statements are correct.

Why other options are wrong:
- Option B: Statement II is also correct.
- Option C: Statement I is also correct.
- Option D: Both statements are correct.

Key Concepts:
- Temperature coefficient of resistance
- Metals
- Semiconductors
- Drift velocity
- Charge carriers

Memory Aids:
- Metals: Resistance increases with temperature.
- Semiconductors: Resistance decreases with temperature.

Quick Tips:
- Remember the behavior of resistance with temperature for metals and semiconductors.

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Understanding +4 | -1

Any digital circuit can be realised by repetitive use of only

View answer and explanation

Correct option: D

NOR gate

Option D (NOR gate) is correct because a NOR gate can be used to implement any other logic gate, making it a universal gate.

Solution approach

Correct Answer + Core Reason: NOR gate. NOR gate is a universal gate.

Step-by-Step Solution:

Step 1: Understand the concept of a universal gate. A universal gate can implement any Boolean function without needing any other gate type.

Step 2: Recall that NOR gates can be used to create NOT, AND, and OR gates.

Step 3: NOT gate: Connect both inputs of the NOR gate together. The output is the complement of the input.

Step 4: OR gate: Use a NOR gate followed by a NOT gate (created using another NOR gate as described above).

Step 5: AND gate: Use a NOR gate to invert each input, then use another NOR gate on the inverted inputs. This implements the AND function.

Why other options are wrong:
- Option A: NOT gate is not a universal gate. It cannot implement AND or OR gates directly.
- Option B: OR gate is not a universal gate. It cannot implement NOT or AND gates directly.
- Option C: AND gate is not a universal gate. It cannot implement NOT or OR gates directly.
- Option D: NOR gate is a universal gate.

Key Concepts:
- Universal gates (NOR, NAND)
- Logic gate implementations
- Boolean algebra

Memory Aids:
- Remember that NOR and NAND are the universal gates.

Quick Tips:
- To determine if a gate is universal, try to construct NOT, AND, and OR gates using only that gate.

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Remembering +4 | -1

The phenomenon of thermionic emission was discovered by

View answer and explanation

Correct option: D

Thomas Edison

Option D (Thomas Edison) is correct because Thomas Edison discovered the phenomenon of thermionic emission.

Solution approach

Correct Answer + Core Reason: Thomas Edison. He discovered thermionic emission.

Step-by-Step Solution:

Step 1: Thermionic emission is the phenomenon where electrons are emitted from a heated surface.

Step 2: Thomas Edison observed this phenomenon while experimenting with incandescent light bulbs.

Step 3: He noticed a current flowing between the heated filament and a nearby electrode inside the evacuated bulb.

Step 4: This observation led to the discovery of thermionic emission.

Why other options are wrong:
- Option A: Marcoini is known for his work in radio communication.
- Option B: Fleming is known for inventing the vacuum diode.
- Option C: Forest is known for inventing the triode.

Key Concepts:
- Thermionic emission
- Vacuum tubes

Memory Aids:
- Associate thermionic emission with Thomas Edison's work on light bulbs.

Quick Tips:
- Remember the key figures associated with the development of vacuum tubes.

20 Easy

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Understanding +4 | -1

When a

\mathbf{P}-\mathbf{N}

junction is reverse biased :

View answer and explanation

Correct option: A

Holes and electrons move away from the junction

Option A (Holes and electrons move away from the junction) is correct because when a p-n junction is reverse biased, the applied voltage increases the width of the depletion region, causing charge carriers to move away from the junction.

Solution approach

Correct Answer + Core Reason: Holes and electrons move away from the junction. Reverse bias widens the depletion region.

Step-by-Step Solution:

Step 1: Define reverse bias. Reverse bias is when the positive terminal of the voltage source is connected to the n-side and the negative terminal to the p-side of the p-n junction.

Step 2: Explain the effect on the depletion region. Reverse bias increases the width of the depletion region.

Step 3: Describe the movement of charge carriers. The increased electric field in the depletion region pulls holes away from the junction on the p-side and electrons away from the junction on the n-side.

Why other options are wrong:
- Option B: Holes and electrons move towards the junction: This happens in forward bias.
- Option C: Movement of holes and electrons stops: There is still a small reverse saturation current due to minority carriers.
- Option D: Width of depletion region decreases: This happens in forward bias.

Key Concepts:
- p-n junction
- Reverse bias
- Depletion region
- Charge carriers (holes and electrons)

Memory Aids:
- Reverse = Away

Quick Tips:
- Remember the effect of forward and reverse bias on the depletion region width.

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Application +4 | -1

In a common emitter transistor amplifier

\mathrm{\beta } = 60,\;{R_0} = 5000\mathrm{\Omega }

and internal resistance of a transistor is

500\mathrm{\Omega }

. The voltage amplification of amplifier will be

View answer and explanation

Correct option: C

600

Option C (600) is correct because it is the voltage amplification calculated using the given parameters in a common emitter transistor amplifier.

Solution approach

Correct Answer + Core Reason: 600. The voltage amplification is calculated using the formula $A_v = \beta \times \frac{{{R_0}}}{{{R_i}}}$ , where ${R_i}$ is the internal resistance of the transistor.

Step-by-Step Solution:

Step 1: Identify the given information: $\beta = 60$ , ${R_0} = 5000 \Omega$ , and ${R_i} = 500 \Omega$ .

Step 2: Recall the formula for voltage amplification in a common emitter amplifier: $A_v = \beta \times \frac{{{R_0}}}{{{R_i}}}$ .

Step 3: Substitute the values into the formula: $A_v = 60 \times \frac{{5000}}{{500}}$ .

Step 4: Calculate the voltage amplification: $A_v = 60 \times 10 = 600$ .

Step 5: Final Answer: $A_v = 600$ .

Why other options are wrong:
- Option A (500): This is incorrect. It might arise from a calculation error or using an incorrect formula.
- Option B (460): This is incorrect. It might arise from a calculation error or using an incorrect formula.
- Option D (560): This is incorrect. It might arise from a calculation error or using an incorrect formula.
- Option C (600): This is the correct value calculated using the correct formula and given values.

Key Concepts:
- Voltage amplification in a common emitter amplifier
- Current gain ( $\beta$ )
- Output resistance ( ${R_0}$ )
- Input resistance ( ${R_i}$ )

Memory Aids:
- Remember the formula for voltage amplification: $A_v = \beta \times \frac{{{R_0}}}{{{R_i}}}$ .

Quick Tips:
- Ensure you are using the correct formula for voltage amplification.
- Double-check your calculations to avoid errors.

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Application +4 | -1

In a p-n junction, the depletion region is 400nm wide and an electric field of

5 \times {10^5}V/m

exists in it. What should be the minimum kinetic energy of a conduction electron which can diffuse from the inside to the p-side?

View answer and explanation

Correct option: C

0.2 eV

Option C (0.2 eV) is correct because the minimum kinetic energy required is equal to the potential difference across the depletion region, which can be calculated as

E = qV = qEd

, where

E

is the electric field and

d

is the width of the depletion region.

Solution approach

Correct Answer + Core Reason: 0.2 eV. This is calculated from the potential difference across the depletion region.

Step-by-Step Solution:

Step 1: Identify the given information: Depletion region width $d = 400 \text{{ nm}} = 400 \times 10^{-9} \text{{ m}}$ , electric field $E = 5 \times 10^5 \text{{ V/m}}$ .

Step 2: Calculate the potential difference $V$ across the depletion region: $V = Ed = (5 \times 10^5 \text{{ V/m}})(400 \times 10^{-9} \text{{ m}}) = 0.2 \text{{ V}}$ .

Step 3: Calculate the minimum kinetic energy required for an electron to diffuse across the potential barrier: $KE = qV = (1e)(0.2 \text{{ V}}) = 0.2 \text{{ eV}}$ , where $q$ is the elementary charge.

Why other options are wrong:
- Option A: 0.1eV: This is incorrect; it might be a calculation error.
- Option B: 0.3 eV: This is incorrect; it might be a calculation error.
- Option D: 0.4 eV: This is incorrect; it might be a calculation error, perhaps doubling the potential difference.

Key Concepts:
- p-n junction
- Depletion region
- Electric field
- Potential difference
- Kinetic energy

Memory Aids:
- $V = Ed$
- $KE = qV$

Quick Tips:
- Remember to convert units to be consistent before calculations.

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Understanding +4 | -1

Avalanche breakdown of p-n junction diode is due to

View answer and explanation

Correct option: A

Collision of electrons with atoms

Option A (Collision of electrons with atoms) is correct because avalanche breakdown occurs when accelerated electrons gain enough kinetic energy to ionize atoms in the depletion region through collisions, creating more charge carriers.

Solution approach

Correct Answer + Core Reason: Collision of electrons with atoms. This is the mechanism for avalanche breakdown.

Step-by-Step Solution:

Step 1: Define avalanche breakdown. Avalanche breakdown is a phenomenon in semiconductor diodes where a large reverse voltage causes a sudden increase in current.

Step 2: Explain the mechanism. When a large reverse bias is applied, electrons gain kinetic energy as they move through the depletion region.

Step 3: Describe the collision process. These high-energy electrons collide with atoms in the crystal lattice, ionizing them and creating electron-hole pairs.

Step 4: Explain the chain reaction. The newly created electrons and holes are also accelerated, leading to further collisions and ionization, creating a chain reaction (avalanche).

Why other options are wrong:
- Option B: Forward biasing: Forward bias leads to conduction, not breakdown.
- Option C: Field emission: Field emission is a different breakdown mechanism (Zener breakdown).
- Option D: All of these: Only collision of electrons with atoms is the primary cause of avalanche breakdown.

Key Concepts:
- Avalanche breakdown
- Reverse bias
- Depletion region
- Electron collision
- Ionization

Memory Aids:
- Avalanche = Collision

Quick Tips:
- Remember that avalanche breakdown is caused by impact ionization.

24 Medium

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Understanding +4 | -1

A piece of semiconductor is connected in series in an electric circuit. On increasing the temperature, the current in the circuit will

View answer and explanation

Correct option: C

Increase

Option C (Increase) is correct because increasing the temperature of a semiconductor increases the number of charge carriers, leading to increased current.

Solution approach

Correct Answer + Core Reason: Increase. The current increases with temperature in a semiconductor.

Step-by-Step Solution:

Step 1: Define semiconductors. Semiconductors are materials with conductivity between that of metals and insulators.

Step 2: Explain the effect of temperature on conductivity. In semiconductors, increasing the temperature increases the number of charge carriers (electrons and holes) due to increased thermal excitation.

Step 3: Relate charge carrier concentration to current. An increase in charge carrier concentration leads to an increase in current.

Step 4: Conclusion: On increasing the temperature, the current in the circuit will increase.

Why other options are wrong:
- Option A: The current increases, not decreases.
- Option B: The current changes with temperature.
- Option C: This is the correct answer. The current increases.
- Option D: The current will not stop flowing; it will increase.

Key Concepts:
- Semiconductors
- Electrical Conductivity
- Temperature Dependence of Conductivity

Memory Aids:
- Heat increases the flow of electricity in semiconductors.

Quick Tips:
- Remember that the conductivity of semiconductors is temperature-dependent.

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Remembering +4 | -1

At room temperature, a

P

-type semiconductor has

View answer and explanation

Correct option: A

Large number of holes and few electrons

Option A (Large number of holes and few electrons) is correct because P-type semiconductors are doped with acceptor impurities, creating an excess of holes.

Solution approach

Correct Answer + Core Reason: Large number of holes and few electrons. P-type semiconductors are doped with acceptor impurities, creating an excess of holes.

Step-by-Step Solution:

Step 1: Define a P-type semiconductor. It is a semiconductor doped with acceptor impurities, such as boron, which create an excess of holes.

Step 2: Explain the charge carriers in a P-type semiconductor. Holes are the majority carriers, and electrons are the minority carriers.

Step 3: Conclusion: At room temperature, a P-type semiconductor has a large number of holes and few electrons.

Why other options are wrong:
- Option B (Large number of free electrons and few holes): This describes an N-type semiconductor.
- Option C (Equal number of free electrons and holes): This describes an intrinsic semiconductor.
- Option D (No electrons or holes): This is incorrect; semiconductors always have some charge carriers.

Key Concepts:
- P-type semiconductor
- Acceptor impurities
- Holes
- Electrons

Memory Aids:
- Remember that P-type semiconductors have a "p"ositive majority carrier (holes).

Quick Tips:
- Understand the difference between P-type and N-type semiconductors.

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Analyzing +4 | -1

\overline{A \cdot \bar{B}+\bar{A} \cdot B}

is equivalent to

View answer and explanation

Correct option: B

(A+\bar{B})+(\bar{A}+B)

Option B,

(A+\bar{B})+(\bar{A}+B)

, is correct because it represents the complement of the XOR operation, which is equivalent to the given expression.

Solution approach

Correct Answer + Core Reason: $(A+\bar{B})+(\bar{A}+B)$ . This represents the complement of the XOR operation.

Step-by-Step Solution:

Step 1: Recognize the given expression as the complement of the XOR operation: $\overline{A \cdot \bar{B} + \bar{A} \cdot B} = \overline{A \oplus B}$ .

Step 2: Apply DeMorgan's Law and Boolean algebra to simplify the expression. We know that $A \oplus B = A \cdot \bar{B} + \bar{A} \cdot B$ . Therefore, $\overline{A \oplus B} = \overline{A \cdot \bar{B} + \bar{A} \cdot B}$ .

Step 3: The complement of XOR can also be written as XNOR, which is $(A \cdot B) + (\bar{A} \cdot \bar{B})$ .

Step 4: Using DeMorgan's Law, we can express the complement of XOR as $(A + \bar{B}) \cdot (\bar{A} + B)$ . This is equivalent to the given option B.

Step 5: Note that $(A + \bar{B}) + (\bar{A} + B)$ is always true (equal to 1) since either A or its complement must be true, and either B or its complement must be true.

Why other options are wrong:
- Option A: $A \cdot \bar{B} + \bar{A} \cdot B$ is the XOR operation, not the complement of XOR.
- Option C: $\overline{A \cdot \bar{B} + \bar{A} \cdot B}$ is the same as the original expression, not an equivalent simplified form.
- Option D: $(A+B) \cdot(\bar{A}+B)$ simplifies to $B$ , which is not equivalent to the given expression.

Key Concepts:
- Boolean Algebra
- XOR and XNOR operations
- DeMorgan's Law
- Complement of a Boolean expression

Memory Aids:
- XOR: Different is True
- XNOR: Same is True

Quick Tips:
- Recognize common Boolean expressions like XOR and XNOR.
- Apply DeMorgan's Law to simplify complex expressions.

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Analyzing +4 | -1

A semi conductor has an electron concentration of

8 \times 10^{13}

per

\mathrm{cm}^3

and a hole concentration of

5 \times 10^{12}

per

\mathrm{cm}^3

. The electron mobility is

25000 \mathrm{~cm}^2 \mathrm{~V}^{-1} \mathrm{~s}^{-1}

and the hole mobility is

100 \mathrm{~cm}^2 \mathrm{~V}^{-1} \mathrm{~s}^{-1}

. Then,

View answer and explanation

Correct option: A

The semiconductor is n-type.

Option A (The semiconductor is n-type) is correct because the electron concentration is significantly higher than the hole concentration.

Solution approach

Correct Answer + Core Reason: The semiconductor is n-type. The electron concentration is significantly higher than the hole concentration.

Step-by-Step Solution:

Step 1: Compare electron and hole concentrations: ${n_e} = 8 \times {10^{13}} per \; cm^3$ and ${n_h} = 5 \times {10^{12}} per \; cm^3$

Step 2: Determine the type of semiconductor: Since ${n_e} > {n_h}$ , the semiconductor is n-type.

Step 3: Calculate the conductivity: $\sigma = e({n_e}{\mu _e} + {n_h}{\mu _h})$ , where $e = 1.6 \times {10^{ - 19}} C$ , ${\mu _e} = 25000 cm^2 V^{-1} s^{-1}$ , and ${\mu _h} = 100 cm^2 V^{-1} s^{-1}$

Step 4: Substitute the values: $\sigma = 1.6 \times {10^{ - 19}}[(8 \times {10^{13}} \times 25000) + (5 \times {10^{12}} \times 100)]$

Step 5: Calculate the conductivity: $\sigma = 1.6 \times {10^{ - 19}}[(2 \times {10^{18}}) + (5 \times {10^{14}})] \approx 1.6 \times {10^{ - 19}}(2 \times {10^{18}}) = 3.2 \times {10^{ - 1}} mho/cm = 320 \times {10^{ - 3}} mho/cm = 320 m mho/cm$

Final Answer: The semiconductor is n-type.

Why other options are wrong:
- Option B: The conductivity calculation is correct, but the question only asks for one correct answer, and option A is the primary correct answer.
- Option C: Option B is not the primary correct answer.
- Option D: Option A is correct.

Key Concepts:
- n-type semiconductor
- p-type semiconductor
- Conductivity of a semiconductor

Memory Aids:
- If ${n_e} > {n_h}$ , it's n-type.
- If ${n_h} > {n_e}$ , it's p-type.

Quick Tips:
- Compare the electron and hole concentrations to determine the type of semiconductor.

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Application +4 | -1

The acceptor level of a p-type semiconductor is

6 \mathrm{~eV}

. The maximum wavelength of light which can create a hole would be : Given

\mathrm{hc}=1242 \mathrm{~eV} \mathrm{~nm}

.

View answer and explanation

Correct option: C

207 nm

Option C (207 nm) is correct because the maximum wavelength corresponds to the minimum energy required to create a hole, which is equal to the acceptor level energy. Using the formula

E = \frac{hc}{\lambda}

, we can solve for

\lambda

.

Solution approach

Correct Answer + Core Reason: 207 nm. This is calculated using the energy-wavelength relationship.

Step-by-Step Solution:

Step 1: Identify the given information: Acceptor level energy $E = 6 \text{{ eV}}$ , $hc = 1242 \text{{ eV nm}}$ .

Step 2: Recall the relationship between energy and wavelength: $E = \frac{hc}{\lambda}$ .

Step 3: Rearrange the formula to solve for wavelength: $\lambda = \frac{hc}{E}$ .

Step 4: Substitute the given values: $\lambda = \frac{1242 \text{{ eV nm}}}{6 \text{{ eV}}}$ .

Step 5: Calculate the wavelength: $\lambda = 207 \text{{ nm}}$ .

Why other options are wrong:
- Option A: 407 nm: This is incorrect; it might be a calculation error.
- Option B: 414 nm: This is incorrect; it might be a calculation error.
- Option D: 103.5 nm: This is incorrect; it might be a calculation error, perhaps dividing by 12 instead of 6.

Key Concepts:
- Energy levels in semiconductors
- Energy-wavelength relationship: $E = \frac{hc}{\lambda}$

Memory Aids:
- Remember the energy-wavelength formula.

Quick Tips:
- Always use consistent units when applying the energy-wavelength formula.

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Application +4 | -1

In a common emitter amplifier, using output resistance of

5000\;\mathrm{\Omega }

and input resistance of

2000\;\mathrm{\Omega }

, if the peak value of input signal voltage is 10 mV and

\mathrm{\beta } = 50

, then peak value of output voltage is

View answer and explanation

Correct option: C

1.25 V

Option C (1.25 V) is correct because it is the calculated peak value of the output voltage using the given parameters in a common emitter amplifier.

Solution approach

Correct Answer + Core Reason: 1.25 V. The peak value of the output voltage is calculated using the formula $V_o = \beta \times \frac{{{R_o}}}{{{R_i}}} \times V_i$ .

Step-by-Step Solution:

Step 1: Identify the given information: Output resistance, ${R_o} = 5000 \Omega$ ; Input resistance, ${R_i} = 2000 \Omega$ ; Input voltage, $V_i = 10 \text{{ mV}} = 0.01 \text{{ V}}$ ; Current gain, $\beta = 50$ .

Step 2: Recall the formula for the output voltage: $V_o = \beta \times \frac{{{R_o}}}{{{R_i}}} \times V_i$ .

Step 3: Substitute the values into the formula: $V_o = 50 \times \frac{{5000}}{{2000}} \times 0.01$ .

Step 4: Calculate the output voltage: $V_o = 50 \times 2.5 \times 0.01 = 1.25 \text{{ V}}$ .

Step 5: Final Answer: $V_o = 1.25 \text{{ V}}$ .

Why other options are wrong:
- **Option A ( $5 \times {10^{ - 6}}\mathrm{\;V}$ ): This is incorrect. It might arise from a calculation error or using an incorrect formula.
- Option B ( $12.50 \times {10^{ - 6}}\mathrm{\;V}$ ): This is incorrect. It might arise from a calculation error or using an incorrect formula.
- Option D (125.0 V): This is incorrect. It might arise from a calculation error or using an incorrect formula.
- Option C (1.25 V): This is the correct value calculated using the correct formula and given values.

Key Concepts:**
- Common emitter amplifier
- Output voltage
- Input voltage
- Current gain ( $\beta$ )
- Output resistance ( ${R_o}$ )
- Input resistance ( ${R_i}$ )

Memory Aids:
- Remember the formula for output voltage: $V_o = \beta \times \frac{{{R_o}}}{{{R_i}}} \times V_i$ .

Quick Tips:
- Ensure you are using the correct formula for output voltage.
- Double-check your calculations to avoid errors.
- Pay attention to units (mV to V).

30 Easy

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Understanding +4 | -1

With the fall in temperature, the electric conductivity of intrinsic semiconductors

View answer and explanation

Correct option: B

Decreases

Option B (Decreases) is correct because as temperature falls, the conductivity of intrinsic semiconductors decreases due to reduced thermal excitation of electrons.

Solution approach

Correct Answer + Core Reason: Decreases. Conductivity of intrinsic semiconductors decreases with temperature.

Step-by-Step Solution:

Step 1: Define intrinsic semiconductors. Intrinsic semiconductors are pure semiconductors with no impurities.

Step 2: Understand the effect of temperature on conductivity. In intrinsic semiconductors, conductivity is proportional to the concentration of charge carriers (electrons and holes).

Step 3: Explain the relationship between temperature and charge carrier concentration. As temperature decreases, fewer electrons have enough energy to jump from the valence band to the conduction band, reducing the concentration of charge carriers.

Step 4: Conclusion: With a fall in temperature, the electric conductivity of intrinsic semiconductors decreases.

Why other options are wrong:
- Option A: Conductivity increases with temperature in semiconductors, not decreases.
- Option B: Conductivity decreases with temperature in intrinsic semiconductors.
- Option C: Conductivity does not first decrease and then increase with a fall in temperature.
- Option D: Conductivity changes with temperature in semiconductors.

Key Concepts:
- Intrinsic Semiconductors
- Electrical Conductivity
- Temperature Dependence of Conductivity

Memory Aids:
- Think of it like freezing: as things get colder, it's harder for electrons to move.

Quick Tips:
- Remember that the conductivity of semiconductors is temperature-dependent.

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