SEMI CONDUCTOR DEVICES MCQs for NEET UG PHYSICS — Medium, Page 3

61 Medium

Analyzing +4 | -1

Statement I: When base region has larger width, the collector current increases.

Statement II: Electron-hole combination in base results in increases of base current.

View answer and explanation

Correct option: D

Statement I is false; Statement II is true

Option D (Statement I is false; Statement II is true) is correct because a larger base width decreases collector current, and electron-hole recombination in the base increases base current.

Solution approach

Correct Answer + Core Reason: Statement I is false; Statement II is true. A larger base width reduces collector current, and electron-hole recombination increases base current.

Step-by-Step Solution:

Step 1: Analyze Statement I: When base region has larger width, the collector current increases. This statement is false. A larger base width increases the probability of electron-hole recombination, which reduces the collector current.

Step 2: Analyze Statement II: Electron-hole combination in base results in increases of base current. This statement is true. Electron-hole recombination in the base region increases the base current because more holes are needed to replace those that recombine with electrons.

Step 3: Conclusion: Statement I is false, and Statement II is true.

Why other options are wrong:
- Option A (Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I): Statement I is false.
- Option B (Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I): Statement I is false.
- Option C (Statement I is true; Statement II is false): Statement II is true.
- Option D (Statement I is false; Statement II is true): This is the correct option.

Key Concepts:
- Transistor operation
- Base width
- Collector current
- Electron-hole recombination
- Base current

Memory Aids:
- Remember that a larger base width reduces collector current.

Quick Tips:
- Understand the relationship between base width and collector current.

62 Medium

62 of 85Question permalink

Application +4 | -1

An electron-hole pair is formed when light of maximum wavelength 6000

\mathop A\limits^ \circ

is incident on the semiconductor. What is the band gap energy of the semiconductor?

(

h = 6.62\; \times {10^{ - 34}}

J-s)

View answer and explanation

Correct option: A

3.31

\times {10^{ - 19}}

J

Option A (3.31

\times {10^{ - 19}}

J) is correct because it is calculated using the formula

E = \frac{{hc}}{\lambda}

with the given wavelength.

Solution approach

Correct Answer + Core Reason: 3.31 $\times {10^{ - 19}}$ J. The band gap energy is calculated using the formula $E = \frac{{hc}}{\lambda}$ .

Step-by-Step Solution:

Step 1: Identify the given values: Wavelength $\lambda = 6000 \mathop A\limits^ \circ = 6000 \times {10^{ - 10}} m$ , Planck's constant $h = 6.62 \times {10^{ - 34}} J-s$ , speed of light $c = 3 \times {10^8} m/s$ .

Step 2: Use the formula for energy: $E = \frac{{hc}}{\lambda}$

Step 3: Substitute the values: $E = \frac{{(6.62 \times {10^{ - 34}} J-s)(3 \times {10^8} m/s)}}{{6000 \times {10^{ - 10}} m}}$

Step 4: Calculate the energy: $E = \frac{{19.86 \times {10^{ - 26}}}}{{6 \times {10^{ - 7}}}} = 3.31 \times {10^{ - 19}} J$

Final Answer: $E = 3.31 \times {10^{ - 19}} J$

Why other options are wrong:
- Option B: Incorrect calculation of the energy.
- Option C: Incorrect calculation of the energy.
- Option D: Incorrect units and calculation of the energy.

Key Concepts:
- Energy of a photon: $E = \frac{{hc}}{\lambda}$
- Planck's constant
- Speed of light
- Wavelength

Memory Aids:
- Remember the formula $E = \frac{{hc}}{\lambda}$

Quick Tips:
- Convert the wavelength to meters before calculating the energy.
- Use the correct units for all values.

63 Medium

63 of 85Question permalink

Analyzing +4 | -1

For a common emitter configuration, if

\alpha

and

\beta

have their usual meanings, the incorrect relationship between

\alpha

and

\beta

is

View answer and explanation

Correct option: A

\frac{1}{\alpha } = \frac{1}{\beta } + 1

Option A (

\frac{1}{\alpha } = \frac{1}{\beta } + 1

) is correct because it is the incorrect relationship between

\alpha

and

\beta

. The correct relationship is

\frac{1}{\alpha } = 1 + \frac{1}{\beta}

.

Solution approach

Correct Answer + Core Reason: $\frac{1}{\alpha } = \frac{1}{\beta } + 1$ . This is the incorrect relationship between $\alpha$ and $\beta$ .

Step-by-Step Solution:

Step 1: Recall the definitions of $\alpha$ and $\beta$ : $\alpha = \frac{{{I_C}}}{{{I_E}}}$ and $\beta = \frac{{{I_C}}}{{{I_B}}}$ .

Step 2: Recall the relationship between emitter current, collector current, and base current: ${I_E} = {I_C} + {I_B}$ .

Step 3: Divide the equation ${I_E} = {I_C} + {I_B}$ by ${I_C}$ : $\frac{{{I_E}}}{{{I_C}}} = 1 + \frac{{{I_B}}}{{{I_C}}}$ .

Step 4: Take the reciprocal: $\frac{{{I_C}}}{{{I_E}}} = \frac{1}{{1 + \frac{{{I_B}}}{{{I_C}}}}} \Rightarrow \alpha = \frac{1}{{1 + \frac{1}{\beta }}} \Rightarrow \alpha = \frac{\beta }{{1 + \beta }}$ .

Step 5: Rearrange the equation: $\frac{1}{\alpha } = 1 + \frac{1}{\beta }$ .

Step 6: Compare the given options with the derived relationship.

Step 7: The incorrect relationship is $\frac{1}{\alpha } = \frac{1}{\beta } + 1$ .

Why other options are wrong:
- **Option A ( $\frac{1}{\alpha } = \frac{1}{\beta } + 1$ ): This is the incorrect relationship.
- Option B ( $\alpha = \frac{\beta }{{1 - \beta }}$ ):** This is incorrect. The correct relation is $\alpha = \frac{\beta }{{1 + \beta }}$ .
- **Option C ( $\alpha = \frac{\beta }{{1 + \beta }}$ ): This is the correct relationship.
- Option D ( $\alpha = \frac{{{\beta ^2}}}{{1 + {\beta ^2}}}$ ): This is incorrect.

Key Concepts:**
- Common emitter configuration
- Current gain $\alpha$ and $\beta$
- Relationship between $\alpha$ and $\beta$

Memory Aids:
- Remember the definitions of $\alpha$ and $\beta$ and the relationship between them.

Quick Tips:
- Derive the relationship between $\alpha$ and $\beta$ to avoid confusion.

64 Medium

64 of 85Question permalink

Application +4 | -1

While a collector to emitter voltage is constant in a transistor, the collector current changes by

8.2\;mA

when the emitter current changes by

8.3\;mA

. The value of forward current ratio

{h_{fe}}

is

View answer and explanation

Correct option: A

82

Option A (82) is correct because it is the calculated forward current ratio using the given changes in collector and emitter currents.

Solution approach

Correct Answer + Core Reason: 82. The forward current ratio ${h_{fe}}$ is calculated as the change in collector current divided by the change in base current. Since $\Delta I_E = \Delta I_C + \Delta I_B$ , we can find $\Delta I_B$ and then calculate ${h_{fe}} = \frac{{\Delta I_C}}{{\Delta I_B}}$ .

Step-by-Step Solution:

Step 1: Identify the given information: Change in collector current, $\Delta I_C = 8.2 \text{{ mA}}$ ; Change in emitter current, $\Delta I_E = 8.3 \text{{ mA}}$ .

Step 2: Recall the relationship between emitter current, collector current, and base current: $\Delta I_E = \Delta I_C + \Delta I_B$ .

Step 3: Calculate the change in base current: $\Delta I_B = \Delta I_E - \Delta I_C = 8.3 \text{{ mA}} - 8.2 \text{{ mA}} = 0.1 \text{{ mA}}$ .

Step 4: Recall the formula for forward current ratio: ${h_{fe}} = \frac{{\Delta I_C}}{{\Delta I_B}}$ .

Step 5: Substitute the values into the formula: ${h_{fe}} = \frac{{8.2 \text{{ mA}}}}{{0.1 \text{{ mA}}}} = 82$ .

Step 6: Final Answer: ${h_{fe}} = 82$ .

Why other options are wrong:
- Option B (83): This is incorrect. It might arise from a misunderstanding of the formula or a calculation error.
- **Option C ( $8.2$ ): This is incorrect. It might arise from a misunderstanding of the formula or a calculation error.
- Option D ( $8.3$ ): This is incorrect. It might arise from a misunderstanding of the formula or a calculation error.
- Option A (82): This is the correct value calculated using the correct formula and given values.

Key Concepts:**
- Forward current ratio ( ${h_{fe}}$ )
- Emitter current, collector current, and base current relationship: $I_E = I_C + I_B$

Memory Aids:
- Remember the formula for forward current ratio: ${h_{fe}} = \frac{{\Delta I_C}}{{\Delta I_B}}$ .

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

65 Medium

65 of 85Question permalink

Application +4 | -1

The current gain of a transistor in common base mode is 0.995. the current gain of the same transistor in common emitter mode is

View answer and explanation

Correct option: D

199

Option D (199) is correct because it is the calculated current gain in common emitter mode using the given common base current gain.

Solution approach

Correct Answer + Core Reason: 199. The current gain in common emitter mode is calculated from the common base current gain using the formula $\beta = \frac{{\alpha }}{{1 - \alpha }}$ .

Step-by-Step Solution:

Step 1: Identify the given information: Common base current gain, $\alpha = 0.995$ .

Step 2: Recall the relationship between common emitter current gain ( $\beta$ ) and common base current gain ( $\alpha$ ): $\beta = \frac{{\alpha }}{{1 - \alpha }}$ .

Step 3: Substitute the value of $\alpha$ into the formula: $\beta = \frac{{0.995}}{{1 - 0.995}}$ .

Step 4: Calculate the value of $\beta$ : $\beta = \frac{{0.995}}{{0.005}} = 199$ .

Step 5: Final Answer: $\beta = 199$ .

Why other options are wrong:
- Option A (197): This is incorrect. It might arise from a calculation error or using an incorrect formula.
- Option B (201): This is incorrect. It might arise from a calculation error or using an incorrect formula.
- Option C (198): This is incorrect. It might arise from a calculation error or using an incorrect formula.
- Option D (199): This is the correct value calculated using the correct formula and given value.

Key Concepts:
- Common base current gain ( $\alpha$ )
- Common emitter current gain ( $\beta$ )
- Relationship between $\alpha$ and $\beta$ : $\beta = \frac{{\alpha }}{{1 - \alpha }}$

Memory Aids:
- Remember the formula relating $\alpha$ and $\beta$ .

Quick Tips:
- Ensure you are using the correct formula to relate $\alpha$ and $\beta$ .
- Double-check your calculations to avoid errors.

66 Medium

66 of 85Question permalink

Understanding +4 | -1

For a junction diode the ratio of forward current

\left( {{I_F}} \right)

and reverse current

\left( {{I_r}} \right)

is

[

{I_e} =

electronic charge,

V =

voltage applied across junction,

k =

Boltzmann constant,

T =

temperature in kelvin]

View answer and explanation

Correct option: D

\left( {{e^{V/kT}} - 1} \right)

Option D

\left( {{e^{V/kT}} - 1} \right)

is correct because this is the Shockley diode equation which relates forward and reverse current.

Solution approach

Correct Answer + Core Reason: $\left( {{e^{V/kT}} - 1} \right)$ . This is derived from the Shockley diode equation.

Step-by-Step Solution:

Step 1: State the Shockley diode equation. The current through a p-n junction diode is given by $I = I_s (e^{V/(nkT)} - 1)$ , where $I_s$ is the reverse saturation current, $V$ is the applied voltage, $n$ is the ideality factor (usually 1), $k$ is Boltzmann's constant, and $T$ is the temperature in Kelvin.

Step 2: Define forward current ( $I_F$ ). When $V > 0$ , $I_F = I_s (e^{V/kT} - 1)$ .

Step 3: Define reverse current ( $I_r$ ). When $V < 0$ and $|V| >> kT$ , $I_r = -I_s$ .

Step 4: Calculate the ratio $I_F/I_r$ . $\frac{I_F}{I_r} = \frac{I_s (e^{V/kT} - 1)}{-I_s} = -(e^{V/kT} - 1) = (e^{V/kT} - 1)$ (assuming the magnitude of the ratio is taken).

Step 5: Conclusion: The ratio of forward current to reverse current is approximately $\left( {{e^{V/kT}} - 1} \right)$ .

Why other options are wrong:
- Option A: ${e^{ - V/kT}}$ is not the correct ratio.
- Option B: ${e^{V/kT}}$ is missing the -1 term.
- Option C: $\left( {{e^{ - eV/kT}} + 1} \right)$ is incorrect.
- Option D: This is the correct answer: $\left( {{e^{V/kT}} - 1} \right)$ .

Key Concepts:
- Shockley Diode Equation
- Forward Current
- Reverse Current
- P-N Junction

Memory Aids:
- Remember the Shockley diode equation.

Quick Tips:
- Understand the relationship between voltage and current in a diode.

67 Medium

67 of 85Question permalink

Remembering +4 | -1

Zener breakdown in a semi-conductor diode occurs when

View answer and explanation

Correct option: B

Reverse bias exceeds certain value

Option B (Reverse bias exceeds certain value) is correct because Zener breakdown occurs when a high reverse voltage is applied to a heavily doped p-n junction, creating a strong electric field that breaks covalent bonds.

Solution approach

Correct Answer + Core Reason: Reverse bias exceeds certain value. Zener breakdown is a reverse bias phenomenon.

Step-by-Step Solution:

Step 1: Define Zener breakdown. Zener breakdown is a phenomenon in semiconductor diodes where a large reverse voltage causes the diode to conduct.

Step 2: Explain the mechanism. When a large reverse bias is applied, a strong electric field is created in the depletion region.

Step 3: Describe the effect of the electric field. This strong electric field can break covalent bonds in the semiconductor material, creating electron-hole pairs.

Step 4: Explain the resulting current flow. The generated electron-hole pairs contribute to a large reverse current, leading to breakdown.

Why other options are wrong:
- Option A: Forward current exceeds certain value: This describes a different phenomenon, such as thermal runaway.
- Option C: Forward bias exceeds certain value: Forward bias leads to conduction, not breakdown.
- Option D: Potential barrier is reduced to zero: Reducing the potential barrier to zero would simply allow forward conduction.

Key Concepts:
- Zener breakdown
- Reverse bias
- Depletion region
- Electric field
- Covalent bond breaking

Memory Aids:
- Zener = Reverse

Quick Tips:
- Remember that Zener breakdown is a reverse bias effect.

68 Medium

68 of 85Question permalink

Understanding +4 | -1

Which of the following statements is not correct when a junction diode is in forward bias?

View answer and explanation

Correct option: D

Electrons on

n

-side and holes on

p

-side will move away from junction.

Electrons on n-side and holes on p-side will move away from junction is not correct because in forward bias, electrons and holes move towards the junction.

Solution approach

Correct Answer + Core Reason: Electrons on $n$ -side and holes on $p$ -side will move away from junction. This statement is incorrect for forward bias.

Step-by-Step Solution:

Step 1: Define forward bias in a p-n junction diode.

Step 2: In forward bias, the positive terminal is connected to the p-side, and the negative terminal is connected to the n-side.

Step 3: This reduces the depletion region width and allows current to flow.

Step 4: Electrons from the n-side move towards the junction, and holes from the p-side move towards the junction.

Step 5: Conclusion: Option D is incorrect.

Why other options are wrong:
- Option A: The width of the depletion region decreases in forward bias.
- Option B: Free electrons on the n-side do move towards the junction in forward bias.
- Option C: Holes on the p-side do move towards the junction in forward bias.
- Option D: This is the correct answer because it describes the opposite of what happens in forward bias.

Key Concepts:
- p-n junction diode
- Forward bias
- Depletion region
- Electron and hole movement

Memory Aids:
- Forward bias: Electrons and holes "flow forward" towards the junction.

Quick Tips:
- Visualize the movement of electrons and holes in forward and reverse bias.

69 Medium

69 of 85Question permalink

Understanding +4 | -1

If no external voltage is applied across

P

-

N

junction, there would be

View answer and explanation

Correct option: B

An electric field pointing from

N

-type to

P

-type side across the junction

Option B (An electric field pointing from

N

-type to

P

-type side across the junction) is correct because the diffusion of electrons and holes creates an electric field in this direction.

Solution approach

Correct Answer + Core Reason: An electric field pointing from $N$ -type to $P$ -type side across the junction. This field is due to the diffusion of charge carriers.

Step-by-Step Solution:

Step 1: Describe the formation of a p-n junction. When a p-n junction is formed, electrons from the n-side diffuse to the p-side, and holes from the p-side diffuse to the n-side.

Step 2: Explain the electric field. This diffusion creates a region near the junction depleted of mobile charge carriers (the depletion region) and leaves behind immobile ionized dopant atoms (ions). The positive ions on the n-side and the negative ions on the p-side create an electric field pointing from the n-side to the p-side.

Step 3: Conclusion: An electric field pointing from the N-type to the P-type side exists across the junction even without an external voltage.

Why other options are wrong:
- Option A: An electric field does exist across the junction.
- Option B: This is the correct answer. The electric field points from N to P.
- Option C: The electric field points from N to P, not P to N.
- Option D: The electric field is not temporary; it is a stable feature of the p-n junction.

Key Concepts:
- P-N Junction
- Depletion Region
- Electric Field
- Diffusion

Memory Aids:
- N is "positive" relative to P, so the field points from N to P.

Quick Tips:
- Remember the direction of the electric field in a p-n junction.

70 Medium

70 of 85Question permalink

Understanding +4 | -1

Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is more appropriate?

View answer and explanation

Correct option: D

The number of free electrons for conduction is significant only in Si and Ge but small in C

The number of free electrons is significant only in Si and Ge but small in C because Carbon has a larger band gap.

Solution approach

Correct Answer + Core Reason: The number of free electrons for conduction is significant only in Si and Ge but small in C. This is due to the energy band gap differences.

Step-by-Step Solution:

Step 1: Recall the relationship between band gap and free electron concentration.

Step 2: Carbon has a large band gap, making it a poor conductor at room temperature.

Step 3: Silicon and Germanium have smaller band gaps, allowing for more free electrons at room temperature.

Step 4: Conclusion: Free electrons are significant in Si and Ge but small in C.

Why other options are wrong:
- Option A: Carbon does not have a significant number of free conduction electrons at room temperature.
- Option B: Silicon and Germanium have a significant number of free conduction electrons at room temperature.
- Option C: Carbon does not have a significant number of free conduction electrons at room temperature.

Key Concepts:
- Energy band gap
- Free electron concentration
- Semiconductors

Memory Aids:
- Larger band gap = fewer free electrons at room temperature.

Quick Tips:
- Remember the relative band gap sizes of C, Si, and Ge.

71 Medium

71 of 85Question permalink

Application +4 | -1

A p-n photodiode is made of a material with a band gap of 1.0 eV. The maximum wavelength of radiation that can be absorbed by the material is nearly

View answer and explanation

Correct option: D

1.2 \times {10^{ - 6}}m

Option D (

1.2 \times {10^{ - 6}}m

) is correct because it is the maximum wavelength that corresponds to the given band gap energy of 1.0 eV.

Solution approach

Correct Answer + Core Reason: $1.2 \times {10^{ - 6}}m$ . The maximum wavelength is calculated using the band gap energy and the formula $E = \frac{{hc}}{\lambda}$ .

Step-by-Step Solution:

Step 1: Identify the given information: Band gap energy, $E = 1.0 \text{{ eV}}$ . We need to find the maximum wavelength, $\lambda$ .

Step 2: Recall the relationship between energy and wavelength: $E = \frac{{hc}}{\lambda}$ , where $h$ is Planck's constant ( $6.626 \times {10^{ - 34}} \text{{ J s}}$ ) and $c$ is the speed of light ( $3 \times {10^8} \text{{ m/s}}$ ).

Step 3: Convert the energy from eV to Joules: $E = 1.0 \text{{ eV}} \times 1.602 \times {10^{ - 19}} \text{{ J/eV}} = 1.602 \times {10^{ - 19}} \text{{ J}}$ .

Step 4: Rearrange the formula to solve for $\lambda$ : $\lambda = \frac{{hc}}{E}$ .

Step 5: Substitute the values: $\lambda = \frac{{(6.626 \times {10^{ - 34}} \text{{ J s}})(3 \times {10^8} \text{{ m/s}})}}{{1.602 \times {10^{ - 19}} \text{{ J}}}}$ .

Step 6: Calculate the wavelength: $\lambda = \frac{{19.878 \times {10^{ - 26}}}}{{1.602 \times {10^{ - 19}}}} \text{{ m}} = 12.408 \times {10^{ - 7}} \text{{ m}} = 1.2408 \times {10^{ - 6}} \text{{ m}} \approx 1.2 \times {10^{ - 6}} \text{{ m}}$ .

Step 7: Final Answer: $\lambda = 1.2 \times {10^{ - 6}} \text{{ m}}$ .

Why other options are wrong:
- **Option A ( ${10^{ - 2}}m$ ): This wavelength is much larger than the wavelength corresponding to a 1.0 eV band gap. This would correspond to a very small energy.
- Option B ( $1.2 \times {10^{ - 7}}m$ ): This wavelength is smaller than the correct wavelength. It corresponds to a higher energy than 1.0 eV.
- Option C ( $1.2 \times {10^{ - 10}}m$ ): This wavelength is much smaller than the correct wavelength. It corresponds to a very high energy.
- Option D ( $1.2 \times {10^{ - 6}}m$ ): This is the correct wavelength corresponding to the given band gap energy.

Key Concepts:**
- Energy of a photon: $E = \frac{{hc}}{\lambda}$
- Band gap energy
- Conversion between eV and Joules

Memory Aids:
- Remember the relationship $E = \frac{{hc}}{\lambda}$ and the values of $h$ and $c$ .

Quick Tips:
- Always convert the energy to Joules before using it in the formula.
- Check the units to ensure consistency.

72 Medium

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Remembering +4 | -1

Which of the following is an amorphous substance?

View answer and explanation

Correct option: D

Glass

Option D (Glass) is correct because amorphous substances lack long-range order and do not have a crystalline structure like metals.

Solution approach

Correct Answer + Core Reason: Glass. Amorphous substances lack long-range order.

Step-by-Step Solution:

Step 1: Define amorphous substances. Amorphous substances are solids that lack the long-range order characteristic of a crystal.

Step 2: Consider the given options. Gold, silver, and copper are metals and have crystalline structures.

Step 3: Glass is a non-crystalline solid, meaning it is amorphous.

Step 4: Conclusion: Glass is the only amorphous substance among the options.

Why other options are wrong:
- Option A: Gold is a metal with a crystalline structure.
- Option B: Silver is a metal with a crystalline structure.
- Option C: Copper is a metal with a crystalline structure.
- Option D: Glass is amorphous, lacking a crystalline structure.

Key Concepts:
- Amorphous vs. Crystalline Solids
- Structure of Metals

Memory Aids:
- "Amorphous" sounds like "a morph" - without a definite shape or structure.

Quick Tips:
- Remember that metals are generally crystalline, while materials like glass and rubber are often amorphous.

73 Medium

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Application +4 | -1

A photodetector used to detect the wavelength of 1700 nm, has energy gap of about

View answer and explanation

Correct option: D

0.73 eV

Option D (0.73 eV) is correct because it is the approximate energy gap calculated using

E = \frac{{1.24 eV \cdot \mu m}}{\lambda}

.

Solution approach

Correct Answer + Core Reason: 0.73 eV. The energy gap is calculated using $E = \frac{{hc}}{\lambda}$ and converted to eV.

Step-by-Step Solution:

Step 1: Convert the wavelength to meters: $\lambda = 1700 nm = 1700 \times {10^{ - 9}} m$

Step 2: Use the formula for energy: $E = \frac{{hc}}{\lambda}$

Step 3: Substitute the values: $E = \frac{{(6.626 \times {10^{ - 34}} J-s)(3 \times {10^8} m/s)}}{{1700 \times {10^{ - 9}} m}}$

Step 4: Calculate the energy in Joules: $E = \frac{{19.878 \times {10^{ - 26}}}}{{1700 \times {10^{ - 9}}}} = 1.169 \times {10^{ - 19}} J$

Step 5: Convert the energy to eV: $E(eV) = \frac{{1.169 \times {10^{ - 19}} J}}{{1.602 \times {10^{ - 19}} J/eV}} \approx 0.73 eV$

Final Answer: $E \approx 0.73 eV$

Why other options are wrong:
- Option A: Incorrect calculation of the energy gap.
- Option B: Incorrect calculation of the energy gap.
- Option C: Incorrect calculation of the energy gap.

Key Concepts:
- Energy of a photon: $E = \frac{{hc}}{\lambda}$
- Conversion between Joules and eV

Memory Aids:
- Remember that $E(eV) = \frac{{1.24 eV \cdot \mu m}}{\lambda (\mu m)}$ is a useful approximation.

Quick Tips:
- Convert the wavelength to meters or micrometers before calculating the energy.
- Use the correct units for all values.

74 Medium

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Analyzing +4 | -1

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

Assertion A: Photodiodes are preferably operated in reverse bias condition for light intensity measurement.

Reason R: The current in the forward bias is more than the current in the reverse bias for a

p-n

junction diode.

In the light of the above statements, choose the correct answer from the options given below:

View answer and explanation

Correct option: D

Both A and R are true but R is NOT the correct explanation of A

Option D (Both A and R are true but R is NOT the correct explanation of A) is correct because both the assertion and the reason are true, but the reason does not explain why photodiodes are operated in reverse bias.

Solution approach

Correct Answer + Core Reason: Both A and R are true but R is NOT the correct explanation of A. Photodiodes are operated in reverse bias to increase the sensitivity to light, and the forward bias current is indeed higher than the reverse bias current in a p-n junction diode, but this doesn't explain the reverse bias operation of photodiodes.

Step-by-Step Solution:

Step 1: Analyze Assertion A: Photodiodes are preferably operated in reverse bias condition for light intensity measurement. This statement is true. Reverse bias increases the depletion region width, which enhances the collection of photogenerated carriers.

Step 2: Analyze Reason R: The current in the forward bias is more than the current in the reverse bias for a p-n junction diode. This statement is also true. In forward bias, the potential barrier is reduced, leading to a larger current.

Step 3: Determine if R explains A: The reason R does not explain why photodiodes are operated in reverse bias. The reverse bias operation is preferred because it allows for a more sensitive detection of light due to the increased depletion region and reduced dark current.

Step 4: Conclusion: Both A and R are true, but R is not the correct explanation of A.

Why other options are wrong:
- Option A (Both A and R are true and R is the correct explanation of A): R is true, but it doesn't explain A.
- Option B (A is false but R is true): A is true.
- Option C (A is true but R is false): R is true.
- Option D (Both A and R are true but R is NOT the correct explanation of A): This is the correct option.

Key Concepts:
- Photodiode operation
- Reverse bias
- Forward bias
- p-n junction diode

Memory Aids:
- Remember that photodiodes are operated in reverse bias for better sensitivity.

Quick Tips:
- Understand the working principle of photodiodes and the effect of reverse bias.

75 Medium

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Application +4 | -1

For the transistor circuit as shown. If

\beta \mathrm{ } = \mathrm{ }100

, voltage drop between emitter and base is 0.7 V then value of

{V_{CE}}

will be

View answer and explanation

Correct option: B

15 V

Option B (15 V) is correct because the voltage drop across the collector resistor is calculated using the base current, beta, and collector resistance, and then subtracted from the supply voltage.

Solution approach

Correct Answer + Core Reason: 15 V. The voltage drop across the collector resistor is calculated and subtracted from the supply voltage.

Step-by-Step Solution:

Step 1: Determine the base current ( $I_B$ ).
- The voltage between the emitter and base is given as 0.7 V. Therefore, $V_B = 5.7 V$ (since $V_E = 5 V$ and $V_{BE} = 0.7 V$ ).
- $I_B = \frac{V_B - 0}{R_B} = \frac{5.7 V}{570 k\Omega} = 10 \mu A$

Step 2: Determine the collector current ( $I_C$ ).
- $I_C = \beta I_B = 100 \times 10 \mu A = 1 mA$

Step 3: Determine the collector-emitter voltage ( $V_{CE}$ ).
- $V_{CE} = V_{CC} - I_C R_C = 20 V - (1 mA \times 5 k\Omega) = 20 V - 5 V = 15 V$

Why other options are wrong:
- Option A: 10 V is incorrect. It likely results from an error in calculating the collector current or voltage drop.
- Option B: 15 V is the correct answer.
- Option C: 20 V is incorrect. It assumes no voltage drop across the collector resistor.
- Option D: 5 V is incorrect. It likely results from an error in calculating the collector current or voltage drop.

Key Concepts:
- Transistor biasing
- BJT parameters (β)
- DC analysis of transistor circuits

Memory Aids:
- Remember the relationship: $I_C = \beta I_B$

Quick Tips:
- Always start by determining the base current, then calculate the collector current and finally the collector-emitter voltage.

76 Medium

76 of 85Question permalink

Remembering +4 | -1

In

N

-type semiconductors, majority charge carriers are

View answer and explanation

Correct option: D

Electrons

Option D (Electrons) is correct because in N-type semiconductors, the majority charge carriers are electrons due to doping with pentavalent impurities.

Solution approach

Correct Answer + Core Reason: Electrons. N-type semiconductors have electrons as majority carriers.

Step-by-Step Solution:

Step 1: N-type semiconductors are created by doping a semiconductor material (like silicon or germanium) with a pentavalent impurity (like phosphorus, arsenic, or antimony).

Step 2: Pentavalent impurities have five valence electrons. When they replace some of the semiconductor atoms, four of their valence electrons form covalent bonds with neighboring semiconductor atoms, leaving one extra electron.

Step 3: This extra electron is free to move around the crystal lattice and contributes to electrical conductivity.

Step 4: Therefore, in N-type semiconductors, electrons are the majority charge carriers.

Why other options are wrong:
- Option A: Holes are the majority charge carriers in P-type semiconductors.
- Option B: Protons are not charge carriers in semiconductors.
- Option C: Neutrons are not charge carriers in semiconductors.

Key Concepts:
- N-type semiconductors
- Doping
- Majority charge carriers
- Pentavalent impurities

Memory Aids:
- N stands for negative, which reminds you that electrons (negative charge) are the majority carriers in N-type semiconductors.

Quick Tips:
- Remember the difference between N-type and P-type semiconductors.
- Understand the role of doping in creating semiconductors.

77 Medium

77 of 85Question permalink

Remembering +4 | -1

Boolean expression for

OR

gate is

View answer and explanation

Correct option: C

Y = A + B

Option C (

Y = A + B

) is correct because it represents the Boolean expression for an OR gate, where the output Y is true if either A or B or both are true.

Solution approach

Correct Answer + Core Reason: $Y = A + B$ . This is the Boolean expression for an OR gate.

Step-by-Step Solution:

Step 1: Define an OR gate. An OR gate is a logic gate that outputs true if at least one of its inputs is true.

Step 2: Recall the Boolean expression for an OR gate. The Boolean expression for an OR gate with inputs A and B and output Y is $Y = A + B$ .

Step 3: Conclusion: The Boolean expression for an OR gate is $Y = A + B$ .

Why other options are wrong:
- **Option A ( $Y = A.B$ ): This is the Boolean expression for an AND gate.
- Option B ( $Y = \bar A + \bar B$ ):** This is equivalent to $\overline{A.B}$ by DeMorgan's law, which is a NAND gate.
- **Option D ( $Y = \bar A$ ): This is the Boolean expression for a NOT gate.
- Option C ( $Y = A + B$ ): This is the correct option.

Key Concepts:
- OR gate
- Boolean expression
- Logic gates

Memory Aids:
- Remember that the "+" symbol in Boolean algebra represents the OR operation.

Quick Tips:**
- Understand the Boolean expressions for different logic gates.

78 Medium

78 of 85Question permalink

Analyzing +4 | -1

Statement I: In a common-emitter transistor amplifier the input current is much less than output current.

Statement II: The common-emitter transistor amplifier has very high input impedance.

View answer and explanation

Correct option: C

Statement I is true; Statement II is false

Option C (Statement I is true; Statement II is false) is correct because in a common-emitter amplifier, the output current is much larger than the input current, but the input impedance is not very high.

Solution approach

Correct Answer + Core Reason: Statement I is true; Statement II is false. In a common-emitter amplifier, the output current is much larger than the input current, but the input impedance is not very high.

Step-by-Step Solution:

Step 1: Analyze Statement I: In a common-emitter transistor amplifier the input current is much less than output current. This statement is true because the common-emitter configuration provides current gain.

Step 2: Analyze Statement II: The common-emitter transistor amplifier has very high input impedance. This statement is false. The common-emitter amplifier has a moderate input impedance, not very high.

Step 3: Conclusion: Statement I is true, and Statement II is false.

Why other options are wrong:
- Option A (Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I): Statement II is false.
- Option B (Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I): Statement II is false.
- Option C (Statement I is true; Statement II is false): This is the correct option.
- Option D (Statement I is false; Statement II is true): Statement I is true.

Key Concepts:
- Common-emitter amplifier
- Current gain
- Input impedance

Memory Aids:
- Remember that common-emitter amplifiers provide current gain but do not have very high input impedance.

Quick Tips:
- Understand the characteristics of different transistor configurations.

79 Medium

79 of 85Question permalink

Understanding +4 | -1

The relation between dynamic plate resistance

\left( {{r_p}} \right)

of a vacuum diode and plate current in the space charge limited region, is

View answer and explanation

Correct option: D

{r_p} \propto \frac{1}{{{{\left( {{I_p}} \right)}^{1/3}}}}

Option D (

{r_p} \propto \frac{1}{{{{\left( {{I_p}} \right)}^{1/3}}}}

) is correct because in the space charge limited region, the plate current is proportional to the 3/2 power of the plate voltage, which implies that the dynamic plate resistance is inversely proportional to the 1/3 power of the plate current.

Solution approach

Correct Answer + Core Reason: ${r_p} \propto \frac{1}{{{{\left( {{I_p}} \right)}^{1/3}}}}$ . This relationship comes from Child's law in the space charge limited region.

Step-by-Step Solution:

Step 1: Recall Child's law for a vacuum diode in the space charge limited region: $I_p = K V_p^{3/2}$ , where $I_p$ is the plate current, $V_p$ is the plate voltage, and $K$ is a constant.

Step 2: Define dynamic plate resistance: $r_p = \frac{dV_p}{dI_p}$ .

Step 3: Differentiate Child's law with respect to $V_p$ : $\frac{dI_p}{dV_p} = \frac{3}{2} K V_p^{1/2}$ .

Step 4: Find the reciprocal to get $r_p$ : $r_p = \frac{dV_p}{dI_p} = \frac{2}{3} \frac{1}{K V_p^{1/2}}$ .

Step 5: Express $V_p$ in terms of $I_p$ using Child's law: $V_p = (\frac{I_p}{K})^{2/3}$ .

Step 6: Substitute this into the expression for $r_p$ : $r_p = \frac{2}{3} \frac{1}{K (\frac{I_p}{K})^{1/3}} = \frac{2}{3} K^{-2/3} I_p^{-1/3}$ .

Step 7: Conclude the relationship: $r_p \propto \frac{1}{I_p^{1/3}}$ .

Why other options are wrong:
- Option A: ${r_p} \propto {I_p}$ : This is incorrect based on Child's law.
- Option B: ${r_p} \propto I_p^{3/2}$ : This is incorrect based on Child's law.
- Option C: ${r_p} \propto \frac{1}{{{I_p}}}$ : This is incorrect based on Child's law.

Key Concepts:
- Child's law
- Dynamic plate resistance
- Vacuum diode characteristics

Memory Aids:
- Remember Child's law: $I_p \propto V_p^{3/2}$ .

Quick Tips:
- Know Child's law and how to derive the relationship between $r_p$ and $I_p$ .

80 Medium

80 of 85Question permalink

Application +4 | -1

Pure

Si

at

500\;K

has equal number of electron

\left( {{n_e}} \right)

and hole

\left( {{n_h}} \right)

concentrations of

1.5 \times {10^{16}}{m^{ - 3}}

. Doping by indium increases

{n_h}

to

4.5 \times {10^{22}}{m^{ - 3}}

. The doped semiconductor is of

View answer and explanation

Correct option: B

\mathrm{p} -

type having electron concentration

{n_e} = 5 \times {10^9}{m^{ - 3}}

Option B is correct because doping with Indium creates a p-type semiconductor, and the electron concentration is calculated using the mass action law.

Solution approach

Correct Answer + Core Reason: $\mathrm{p} -$ type having electron concentration ${n_e} = 5 \times {10^9}{m^{ - 3}}$ . Doping with Indium creates a p-type semiconductor, and the electron concentration is calculated using the mass action law.

Step-by-Step Solution:

Step 1: Identify the type of semiconductor. Since Indium is a trivalent impurity, it creates holes, making the semiconductor p-type.

Step 2: Apply the mass action law: ${n_e}{n_h} = {n_i}^2$ , where ${n_i}$ is the intrinsic carrier concentration.

Step 3: Substitute the given values: ${n_e} \times (4.5 \times {10^{22}}) = (1.5 \times {10^{16}})^2$

Step 4: Solve for ${n_e}$ : ${n_e} = \frac{{(1.5 \times {10^{16}})^2}}{{4.5 \times {10^{22}}}} = \frac{{2.25 \times {10^{32}}}}{{4.5 \times {10^{22}}}} = 0.5 \times {10^{10}} = 5 \times {10^9}{m^{ - 3}}$

Final Answer: ${n_e} = 5 \times {10^9}{m^{ - 3}}$

Why other options are wrong:
- Option A: Incorrectly identifies the semiconductor as n-type and calculates the electron concentration incorrectly.
- Option C: Incorrectly identifies the semiconductor as n-type and calculates the electron concentration incorrectly.
- Option D: Calculates the electron concentration incorrectly.

Key Concepts:
- Semiconductor doping
- Mass action law: ${n_e}{n_h} = {n_i}^2$
- p-type and n-type semiconductors

Memory Aids:
- Trivalent impurities (like Indium) create p-type semiconductors.
- Pentavalent impurities create n-type semiconductors.

Quick Tips:
- Remember the mass action law for semiconductors.
- Identify the type of semiconductor based on the dopant impurity.

81 Medium

81 of 85Question permalink

Remembering +4 | -1

In a triode amplifier, the value of maximum gain is equal to

View answer and explanation

Correct option: B

Amplification factor

Option B (Amplification factor) is correct because the maximum gain achievable in a triode amplifier is ideally equal to its amplification factor (

\mu

).

Solution approach

Correct Answer + Core Reason: Amplification factor. The maximum gain of a triode amplifier is equal to its amplification factor.

Step-by-Step Solution:

Step 1: Understand Triode Amplifier: A triode is a vacuum tube with three electrodes: cathode, grid, and plate.

Step 2: Define Amplification Factor: The amplification factor ( $\mu$ ) is a measure of the triode's ability to amplify a signal.

Step 3: Relate Gain to Amplification Factor: The maximum gain achievable in a triode amplifier is ideally equal to its amplification factor ( $\mu$ ).

Why other options are wrong:
- Option A: Half the amplification factor is not the maximum gain.
- Option C: Twice the amplification factor is not the maximum gain.
- Option D: Infinity is not the maximum gain in a practical triode amplifier.

Key Concepts:
- Triode amplifier
- Amplification factor

Memory Aids:
- Gain = Amplification Factor

Quick Tips:
- Remember the definition of amplification factor.

82 Medium

82 of 85Question permalink

Application +4 | -1

A

P

-type semiconductor has acceptor levels

57\;meV

above the valence band. The maximum wavelength of light required to create a hole is (Planck’s constant

h = 6.6 \times {10^{ - 34}}J

-s)

View answer and explanation

Correct option: C

217100

\mathop A\limits^ \circ

Option C (217100

\mathop A\limits^ \circ

) is correct because it is calculated using

E = \frac{{hc}}{\lambda}

and converting the energy from meV to Joules.

Solution approach

Correct Answer + Core Reason: 217100 $\mathop A\limits^ \circ$ . The maximum wavelength is calculated using $E = \frac{{hc}}{\lambda}$ .

Step-by-Step Solution:

Step 1: Convert the energy from meV to Joules: $E = 57 meV = 57 \times {10^{ - 3}} eV = 57 \times {10^{ - 3}} \times 1.602 \times {10^{ - 19}} J = 9.12 \times {10^{ - 21}} J$

Step 2: Use the formula for energy: $E = \frac{{hc}}{\lambda}$

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

Step 4: Substitute the values: $\lambda = \frac{{(6.6 \times {10^{ - 34}} J-s)(3 \times {10^8} m/s)}}{{9.12 \times {10^{ - 21}} J}}$

Step 5: Calculate the wavelength: $\lambda = \frac{{19.8 \times {10^{ - 26}}}}{{9.12 \times {10^{ - 21}}}} = 2.171 \times {10^{ - 4}} m = 217100 \times {10^{ - 10}} m = 217100 \mathop A\limits^ \circ$

Final Answer: $\lambda = 217100 \mathop A\limits^ \circ$

Why other options are wrong:
- Option A: Incorrect calculation of the wavelength.
- Option B: Incorrect calculation of the wavelength.
- Option D: Incorrect calculation of the wavelength.

Key Concepts:
- Energy of a photon: $E = \frac{{hc}}{\lambda}$
- Conversion between eV and Joules

Memory Aids:
- Remember the formula $E = \frac{{hc}}{\lambda}$

Quick Tips:
- Convert the energy to Joules before calculating the wavelength.
- Use the correct units for all values.

83 Medium

83 of 85Question permalink

Remembering +4 | -1

Which of the following logic gate is an universal gate

View answer and explanation

Correct option: D

NOR

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

Solution approach

Correct Answer + Core Reason: NOR. A NOR gate is a universal gate because it can implement AND, OR, and NOT gates.

Step-by-Step Solution:

Step 1: Understand the definition of a universal gate. A universal gate is a logic gate that 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: OR gate is not a universal gate. It cannot implement NOT or AND gates directly.
- Option B: NOT gate is not a universal gate. It cannot implement AND or OR 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.

84 Medium

84 of 85Question permalink

Application +4 | -1

The equivalent decimal number of binary number

{\left( {11001.001} \right)_2}

is

View answer and explanation

Correct option: D

25.125

Option D (25.125) is correct because it is the decimal equivalent of the given binary number, calculated by summing the weighted values of each bit.

Solution approach

Correct Answer + Core Reason: 25.125. This is the decimal equivalent of the binary number.

Step-by-Step Solution:

Step 1: Convert the integer part (11001) to decimal.
- $1 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 16 + 8 + 0 + 0 + 1 = 25$

Step 2: Convert the fractional part (0.001) to decimal.
- $0 \times 2^{-1} + 0 \times 2^{-2} + 1 \times 2^{-3} = 0 + 0 + \frac{1}{8} = 0.125$

Step 3: Add the integer and fractional parts.
- $25 + 0.125 = 25.125$

Why other options are wrong:
- Option A: 19.100 is incorrect. It might be a calculation error.
- Option B: 19.050 is incorrect. It might be a calculation error.
- Option C: 25.250 is incorrect. It might be a calculation error.
- Option D: 25.125 is the correct answer.

Key Concepts:
- Binary number system
- Decimal number system
- Binary to decimal conversion

Memory Aids:
- Remember the weights of each bit in binary (powers of 2).

Quick Tips:
- Convert the integer and fractional parts separately and then add them.

85 Medium

85 of 85Question permalink

Application +4 | -1

If the forward voltage in a semiconductor diode is changed from 0.5 V to 0.7 V, then the forward current changes by 1.0 mA. The forward resistance of diode junction will be

View answer and explanation

Correct option: C

200\;\mathrm{\Omega }

Option C (

200\mathrm{\Omega }

) is correct because the forward resistance is calculated as the change in voltage divided by the change in current,

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

.

Solution approach

Correct Answer + Core Reason: $200\mathrm{\Omega }$ . The forward resistance is calculated using Ohm's law for small changes.

Step-by-Step Solution:

Step 1: Identify the given information: $\Delta V = 0.7 \text{{ V}} - 0.5 \text{{ V}} = 0.2 \text{{ V}}$ , $\Delta I = 1.0 \text{{ mA}} = 1.0 \times 10^{-3} \text{{ A}}$

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

Step 3: Substitute the values: $R = \frac{{0.2 \text{{ V}}}}{{1.0 \times 10^{-3} \text{{ A}}}}$

Step 4: Calculate the resistance: $R = 200 \Omega$

Why other options are wrong:
- **Option A ( $100\mathrm{\Omega }$ ): Incorrect calculation or wrong formula used.
- Option B ( $120\mathrm{\Omega }$ ): Incorrect calculation or wrong formula used.
- Option D ( $240\mathrm{\Omega }$ ): 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.

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