# Ee 2401: Semiconductor Devices - Practice Quiz I

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Quizzes Created: 2 | Total Attempts: 330
Questions: 30 | Attempts: 203  Settings  This practice quiz has been created by Dr. Sandip Das for the students taking Semiconductor Devices (EE 2401) course in the Department of Electrical Engineering at Kennesaw State University. The quiz aims to help self evaluate student's preparation on the understanding of the subject matter in an efficient way and does NOT contribute to the final grade.

• 1.

### Which of the following crystal structure has highest atomic packing fraction (APF)?

• A.

Simple Cubic

• B.

Face Centered Cubic (FCC)

• C.

Body Centered Cubic (BCC)

• D.

All of the above have same APF

B. Face Centered Cubic (FCC)
Explanation
The face-centered cubic (FCC) crystal structure has the highest atomic packing fraction (APF) among the given options. In FCC, atoms are arranged in a close-packed manner, with atoms located at each corner and at the center of each face of the unit cell. This arrangement allows for efficient packing of atoms, resulting in a higher APF compared to the other structures. Body-centered cubic (BCC) and simple cubic structures have lower APFs as they do not have atoms located at the faces of the unit cell. Therefore, FCC has the highest APF among the options provided.

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• 2.

### In a Simple Cubic structure, the lattice parameters are:

• A.
• B.
• C.

Option 3

• D.
D.
Explanation
In a Simple Cubic structure, the lattice parameters are equal in all three dimensions. This means that the length of each side of the cube is the same. This is different from other types of crystal structures, such as Body-Centered Cubic or Face-Centered Cubic, where the lattice parameters may vary. In a Simple Cubic structure, the atoms are arranged at the corners of the cube, with no additional atoms in the center or on the faces of the cube. This arrangement results in a simple and symmetrical structure.

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• 3.

### Si has the following crystal structure:

• A.

Simple Cubic

• B.

Face Centered Cubic

• C.

Body Centered Cubic

• D.

Diamond

D. Diamond
Explanation
Diamond has a unique crystal structure in which each carbon atom is bonded to four neighboring carbon atoms in a tetrahedral arrangement. This results in a strong and rigid lattice structure, making diamond the hardest known natural substance. The face-centered cubic, simple cubic, and body-centered cubic structures do not possess the same arrangement of atoms as diamond, making them incorrect answers.

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• 4.

### An electric field is applied across a semiconductor sample. Which of the following is true?

• A.

Electrons will move along the direction of the electric field

• B.

Holes will move against the direction of the electric field

• C.

Electrons will move perpendicular to the electric field

• D.

Electrons will move against the direction of the electric field

D. Electrons will move against the direction of the electric field
Explanation
When an electric field is applied across a semiconductor sample, the movement of charge carriers depends on their nature. In the case of electrons, they are negatively charged particles and will experience a force in the opposite direction to the electric field. Therefore, electrons will move against the direction of the electric field. On the other hand, holes are considered as positive charge carriers, and they will move along the direction of the electric field.

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• 5.

### In the Ternary Compound Semiconductor AlxGa1-xAs, the value of  lies in the range:

C.
Explanation
The value of x in the Ternary Compound Semiconductor AlxGa1-xAs lies in the range of 0 to 1. This is because x represents the mole fraction of aluminum (Al) in the compound, and mole fractions range from 0 to 1. A value of x = 0 means there is no aluminum present, while x = 1 means the compound is pure aluminum arsenide (AlAs). Intermediate values of x represent varying compositions of aluminum and gallium arsenide (GaAs) in the compound.

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• 6.

### Consider the free electron concentration in an intrinsic semiconductor sample at temperature T1 is n1. When the temperature is increased to T2, the free electron concentration is found to be n2. Which one of the following is true?

B.
Explanation
The free electron concentration in an intrinsic semiconductor sample increases with increasing temperature.

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• 7.

### Identify the correct order of the each type of solid shown below:

• A.

(A) Polycrystalline, (B) Amorphous, (C) Single Crystal

• B.

(A) Single Crystal, (B) Amorphous, (C) Polycrystalline

• C.

(A) Amorphous, (B) Polycrystalline, (C) Single Crystal

• D.

(A) Single Crystal, (B) Polycrystalline, (C) Amorphous

C. (A) Amorphous, (B) Polycrystalline, (C) Single Crystal
Explanation
The correct order of the types of solid shown below is (A) Amorphous, (B) Polycrystalline, (C) Single Crystal. Amorphous solids lack a regular arrangement of atoms and have a disordered structure. Polycrystalline solids consist of many small crystals with different orientations. Single crystals have a regular and continuous arrangement of atoms, with a well-defined crystal structure. Therefore, the correct order reflects the increasing level of order and regularity in the arrangement of atoms.

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• 8.

### In an intrinsic semiconductor, the hole concentration is:

• A.

Equal to the atomic concentration

• B.

Equal to the free electron concentration

• C.

Higher than the free electron concentration

• D.

Lower than the free electron concentration

B. Equal to the free electron concentration
Explanation
In an intrinsic semiconductor, the hole concentration is equal to the free electron concentration. This is because an intrinsic semiconductor is a pure semiconductor with no impurities added. In this case, the number of free electrons and holes is equal due to thermal excitation. When an electron moves from the valence band to the conduction band, it leaves behind a hole in the valence band. The number of free electrons and holes created is balanced, resulting in an equal concentration of both.

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• 9.

### The Fermi level in a Boron-doped Si sample would be –

• A.

Closer to the conduction band edge

• B.

Above the conduction band edge

• C.

Closer to the valence band edge

• D.

Below the valence band edge

C. Closer to the valence band edge
Explanation
In a Boron-doped Si sample, Boron acts as an acceptor impurity, meaning it introduces holes into the valence band. The Fermi level is the energy level at which there is a 50% probability of finding an electron. In this case, the presence of Boron introduces more holes, shifting the Fermi level closer to the valence band edge where the majority of holes are located. Therefore, the Fermi level in a Boron-doped Si sample would be closer to the valence band edge.

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• 10.

### The free electron concentration in a semiconductor material depends on –

• A.

The effective density of states in the conduction band

• B.

Temperature

• C.

The Fermi level

• D.

All of the above

D. All of the above
Explanation
The free electron concentration in a semiconductor material depends on the effective density of states in the conduction band, temperature, and the Fermi level. The effective density of states in the conduction band determines the number of available energy states for electrons to occupy. Temperature affects the thermal energy of the electrons, which can increase or decrease their concentration. The Fermi level represents the energy level at which the probability of finding an electron is 50%, and it influences the distribution of electrons in the material. Therefore, all of these factors play a role in determining the free electron concentration in a semiconductor material.

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• 11.

### With increasing reverse bias, width of the depletion region in a p-n junction diode will –

• A.

Decrease

• B.

Increase

• C.

Remain same

• D.

Becomes zero

B. Increase
Explanation
As the reverse bias is increased in a p-n junction diode, the width of the depletion region increases. This is because the reverse bias causes the majority carriers to move away from the junction, resulting in an increase in the width of the region depleted of charge carriers. Therefore, the correct answer is "Increase."

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• 12.

### Which of the following is an indirect bandgap semiconductor?

• A.

CdTe

• B.

GaAs

• C.

Si

• D.

InP

C. Si
Explanation
Si is an indirect bandgap semiconductor because it requires an additional step for an electron to reach the conduction band from the valence band. In an indirect bandgap semiconductor, the minimum energy required for this transition is not satisfied by a single photon. Instead, a combination of photons or phonons is needed to facilitate the transition. CdTe, GaAs, and InP are all direct bandgap semiconductors, meaning that the minimum energy required for an electron to transition from the valence band to the conduction band can be satisfied by a single photon.

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• 13.

### The depletion region of a p-n junction is depleted of –

• A.

Atoms

• B.

Immobile charge carriers

• C.

Phonons

• D.

Mobile charge carriers

D. Mobile charge carriers
Explanation
The depletion region of a p-n junction is depleted of mobile charge carriers. In a p-n junction, when a p-type semiconductor and an n-type semiconductor are brought together, the free electrons from the n-type region diffuse into the p-type region, leaving behind positively charged ions. Similarly, the holes from the p-type region diffuse into the n-type region, leaving behind negatively charged ions. This creates a region near the junction where there are no mobile charge carriers, known as the depletion region. Therefore, the correct answer is mobile charge carriers.

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• 14.

### What is the miller indices of the following crystal plane (shaded)?

• A.

(111)

• B.

(001)

• C.

(100)

• D.

(010)

D. (010)
Explanation
The Miller indices represent the orientation of crystal planes in a crystal lattice. In this case, the shaded crystal plane is parallel to the y-axis and intersects the x-axis at a point, indicating that the plane is perpendicular to the x-axis. Therefore, the Miller indices of the shaded crystal plane are (010).

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• 15.

### Measured bandgap of an unknown material is found to be 8.2 eV. At room temperature, the material is expected to be –

• A.

A semiconductor

• B.

A metal

• C.

An insulator

• D.

A superconductor

C. An insulator
Explanation
The measured bandgap of 8.2 eV indicates that the energy required for an electron to move from the valence band to the conduction band is very high. In insulators, the bandgap is typically greater than 5 eV, which means that very few electrons can be excited to the conduction band at room temperature. Therefore, the material is expected to be an insulator.

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• 16.

### Current due to the concentration gradient of charge carriers is called –

• A.

Concentrated current

• B.

Drift current

• C.

Diffusion current

• D.

Eddy current

C. Diffusion current
Explanation
Diffusion current refers to the flow of charge carriers, such as electrons or ions, due to a concentration gradient. When there is a difference in the concentration of charge carriers between two regions, they tend to move from the region of higher concentration to the region of lower concentration. This movement creates a diffusion current. It is important to note that diffusion current is different from drift current, which is caused by an electric field.

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• 17.

### In a p-n junction diode, doping concentration in the n-side (ND) is higher than the doping concentration in the p-side (NA). xn and xp denote the depletion width in the n- and p-sides, respectively. Which one of the following is correct?

C.
Explanation
The correct answer is "xn is greater than xp". In a p-n junction diode, the depletion width is determined by the difference in doping concentrations between the n-side and the p-side. Since the doping concentration in the n-side (ND) is higher than the doping concentration in the p-side (NA), the depletion width in the n-side (xn) will be greater than the depletion width in the p-side (xp).

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• 18.

### At T>0K, the probability of finding an electron at the Fermi energy level is –

• A.

1

• B.

1/3

• C.

1/4

• D.

1/2

D. 1/2
Explanation
At T>0K, the probability of finding an electron at the Fermi energy level is 1/2. This is because at absolute zero temperature (T=0K), all energy levels below the Fermi energy level are filled, while all energy levels above it are empty. As the temperature increases, some electrons gain enough energy to move from the filled energy levels below the Fermi energy to the empty energy levels above it. At T>0K, there is an equal probability for an electron to be either below or above the Fermi energy level, resulting in a probability of 1/2.

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• 19.

### If the segregation coefficient of an impurity, k0 = 1; then the Czochralski grown crystal would have –

• A.

Higher impurity concentration at the middle of the ingot

• B.

Uniform impurity concentration throughout the ingot

• C.

Higher impurity concentration at the top of the ingot

• D.

Higher impurity concentration at the bottom of the ingot

B. Uniform impurity concentration throughout the ingot
Explanation
If the segregation coefficient of an impurity, k0 = 1, it means that the impurity has an equal tendency to distribute itself between the solid and liquid phases during crystal growth. This indicates that the impurity will be evenly distributed throughout the ingot, resulting in a uniform impurity concentration. Therefore, the correct answer is "Uniform impurity concentration throughout the ingot."

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• 20.

### With increase in reverse bias, the depletion capacitance –

• A.

Increases

• B.

Decreases

• C.

Remains unchanged

• D.

Becomes zero

B. Decreases
Explanation
As the reverse bias increases, the width of the depletion region in a pn junction diode also increases. This results in a decrease in the depletion capacitance. The depletion capacitance is directly proportional to the width of the depletion region, so as the width decreases, the capacitance decreases. Therefore, the correct answer is "Decreases".

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• 21.

### In a p-n junction diode, which of the following phenomena is responsible for forward current flow (under forward bias)?

• A.

Thermionic emission

• B.

Tunneling

• C.

Minority carrier injection

• D.

All of the above

C. Minority carrier injection
Explanation
Forward current flow in a p-n junction diode under forward bias is primarily due to minority carrier injection. When the diode is forward biased, the p-n junction is in a condition where the p-side is at a higher potential than the n-side. This causes the majority carriers (electrons in the n-side and holes in the p-side) to move towards the junction. As they cross the junction, some of these majority carriers recombine with the opposite majority carriers, creating a region near the junction with a lower concentration of majority carriers. This region is known as the depletion region. The presence of the depletion region allows minority carriers (holes in the n-side and electrons in the p-side) to cross the junction and contribute to the forward current flow.

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• 22.

### Which of the following metal will result in a Schottky contact with a p-type semiconductor?

• A.
• B.
• C.
• D.

All of the above

B.
• 23.

### Under reverse bias across a Schottky junction with ultra-thin barrier-width, which of the following quantum-mechanical phenomena is responsible for current conduction?

• A.

Avalanche multiplication

• B.

Tunneling

• C.

Thermionic emission

• D.

Recombination

B. Tunneling
Explanation
In a Schottky junction with an ultra-thin barrier-width under reverse bias, the quantum-mechanical phenomenon responsible for current conduction is tunneling. Tunneling occurs when particles pass through a barrier that they do not have enough energy to overcome according to classical physics. In this case, electrons can tunnel through the thin barrier and flow across the junction, resulting in current conduction.

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• 24.

### How many Bravais lattices are possible?

• A.

20

• B.

12

• C.

10

• D.

14

D. 14
Explanation
There are 14 possible Bravais lattices. Bravais lattices are the 14 unique three-dimensional lattice types that describe the arrangement of atoms or molecules in a crystal lattice. Each Bravais lattice is characterized by its lattice parameters, which include the lengths of the unit cell edges and the angles between them. These lattice types are based on different combinations of translation symmetry and rotational symmetry, resulting in 14 distinct possibilities.

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• 25.

### Which crystal growth technique is most widely employed for commercial Si wafer production?

• A.

Czochralski

• B.

Float Zone

• C.

Bridgman

• D.

All of the above

A. Czochralski
Explanation
The Czochralski technique is the most widely employed crystal growth technique for commercial Si wafer production. This method involves melting a polycrystalline silicon seed crystal and slowly pulling it out of a molten silicon melt, allowing a single crystal to grow. The Czochralski technique is preferred because it produces large, high-quality single crystal wafers that are suitable for semiconductor device fabrication. The Float Zone and Bridgman techniques are also used for crystal growth, but they are not as commonly employed for Si wafer production as the Czochralski technique.

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• 26.

### Drift current flowing through a sample of n-type semiconductor will depend on –

• A.

Doping concentration

• B.

Electron mobility

• C.

Applied electric field

• D.

All of the above

D. All of the above
Explanation
The drift current flowing through a sample of n-type semiconductor depends on the doping concentration, electron mobility, and the applied electric field. The doping concentration determines the number of charge carriers available in the semiconductor, which affects the current. The electron mobility determines how easily the charge carriers can move through the semiconductor, influencing the current flow. The applied electric field provides the driving force for the charge carriers, affecting the current. Therefore, all of these factors contribute to the drift current in an n-type semiconductor.

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• 27.

### If the segregation coefficient of an impurity, k0 < 1; then the Czochralski grown crystal would have –

• A.

Higher impurity concentration at the middle of the ingot

• B.

Uniform impurity concentration throughout the ingot

• C.

Higher impurity concentration at the top of the ingot

• D.

Higher impurity concentration at the bottom of the ingot

D. Higher impurity concentration at the bottom of the ingot
Explanation
If the segregation coefficient of an impurity, k0 < 1, it means that the impurity has a tendency to concentrate in the liquid phase rather than the solid phase during crystal growth. As a result, during the Czochralski growth process, the impurity will be more likely to be concentrated at the bottom of the ingot where the liquid phase is located. Therefore, the correct answer is that the crystal would have a higher impurity concentration at the bottom of the ingot.

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• 28.

### Reverse saturation current of a Schottky diode depends on –

• A.

Schottky barrier height

• B.

Temperature

• C.

Effective Richardson constant

• D.

All of the above

D. All of the above
Explanation
The reverse saturation current of a Schottky diode depends on all of the given factors. The Schottky barrier height affects the flow of current across the junction, with a higher barrier height resulting in a lower reverse saturation current. Temperature also plays a role, as an increase in temperature can increase the reverse saturation current. The effective Richardson constant is a parameter that relates to the thermionic emission current, which is a component of the reverse saturation current. Therefore, all of these factors contribute to the reverse saturation current of a Schottky diode.

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• 29.

### To obtain high Schottky barrier height, which of the following options would you chose?

• A.

A semiconductor with high work function

• B.

A semiconductor with high electron affinity

• C.

A Metal with low work function

• D.

A Metal with high work function

D. A Metal with high work function
Explanation
A metal with a high work function would be chosen to obtain a high Schottky barrier height. The Schottky barrier height is the energy difference between the Fermi level of the metal and the conduction band edge of the semiconductor. A high work function indicates that it is more difficult for electrons to escape from the metal, resulting in a larger energy difference and a higher Schottky barrier height.

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• 30.

### 1 eV =

• A.

1E-18 Joules

• B.

1.602E-19 Joules

• C.

1.602E+19 Joules

• D.

8.854E-14 Joules

B. 1.602E-19 Joules
Explanation
1 eV is a unit of energy equal to the amount of energy gained or lost by an electron when it moves across an electric potential difference of 1 volt. The conversion factor between eV and Joules is 1 eV = 1.602E-19 Joules.

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