Semiconductors MCQs (Multiple-Choice Questions)

1. Which property distinguishes a semiconductor from a good conductor and an insulator?

1. Its conductivity is intermediate and can be strongly controlled by temperature and doping
2. It has zero electrical conductivity at all temperatures
3. It has infinite electrical conductivity
4. It cannot carry electric current

Answer: A) Its conductivity is intermediate and can be strongly controlled by temperature and doping

Explanation:

Semiconductors have electrical conductivity between that of typical conductors and insulators. Their conductivity can change significantly with temperature, illumination, impurities, and applied electric fields.

2. How many valence electrons does a silicon atom have?

1. 2
2. 3
3. 4
4. 5

Answer: C) 4

Explanation:

Silicon belongs to Group 14 of the periodic table and has four valence electrons. These electrons participate in covalent bonding within the silicon crystal.

3. Which type of bond primarily holds silicon atoms together in a crystalline silicon lattice?

1. Ionic bond
2. Covalent bond
3. Metallic bond
4. Hydrogen bond

Answer: B) Covalent bond

Explanation:

Silicon atoms form covalent bonds by sharing valence electrons with neighboring atoms. This bonding structure creates the crystal lattice of elemental silicon.

4. What is an intrinsic semiconductor?

1. A semiconductor intentionally doped with donor impurities
2. A chemically pure or essentially undoped semiconductor
3. A semiconductor containing only acceptor impurities
4. A semiconductor containing no charge carriers at any temperature

Answer: B) A chemically pure or essentially undoped semiconductor

Explanation:

An intrinsic semiconductor is an undoped semiconductor in which carrier generation is determined primarily by the material itself. In thermal equilibrium, its electron and hole concentrations are equal.

5. In an intrinsic semiconductor at thermal equilibrium, the electron concentration is:

1. Greater than the hole concentration
2. Less than the hole concentration
3. Equal to the hole concentration
4. Always zero

Answer: C) Equal to the hole concentration

Explanation:

For an intrinsic semiconductor, electrons and holes are generated in pairs. Therefore, the electron concentration n equals the hole concentration p, with both equal to the intrinsic carrier concentration.

6. Which of the following is a common elemental semiconductor?

1. Copper
2. Silicon
3. Aluminium
4. Silver

Answer: B) Silicon

Explanation:

Silicon is a widely used elemental semiconductor. Germanium is another elemental semiconductor used in certain electronic and semiconductor applications.

7. What happens to the conductivity of an intrinsic semiconductor when its temperature increases?

1. It generally increases
2. It always becomes zero
3. It remains exactly constant
4. It becomes independent of carrier concentration

Answer: A) It generally increases

Explanation:

Increasing temperature provides energy that can generate additional electron-hole pairs. The resulting increase in carrier concentration generally causes semiconductor conductivity to increase.

8. What type of semiconductor is produced by adding a pentavalent impurity to pure silicon?

1. Intrinsic semiconductor
2. N-type semiconductor
3. P-type semiconductor
4. Insulator

Answer: B) N-type semiconductor

Explanation:

Pentavalent dopants such as phosphorus, arsenic, and antimony have five valence electrons. Four participate in covalent bonding with silicon, while the fifth can contribute a free electron, producing N-type material.

9. Which impurity is commonly used to produce P-type silicon?

1. Phosphorus
2. Arsenic
3. Boron
4. Antimony

Answer: C) Boron

Explanation:

Boron is a trivalent acceptor impurity. When introduced into silicon, it creates acceptor states and increases the concentration of holes.

10. What is the majority carrier in an N-type semiconductor?

1. Holes
2. Electrons
3. Protons
4. Photons

Answer: B) Electrons

Explanation:

Donor impurities contribute electrons to the semiconductor. Consequently, electrons are the majority carriers in N-type material, while holes remain minority carriers.

11. What is the majority carrier in a P-type semiconductor?

1. Electrons
2. Neutrons
3. Holes
4. Photons

Answer: C) Holes

Explanation:

Acceptor impurities create holes in the valence band. Therefore, holes are the majority carriers in a P-type semiconductor, while electrons are minority carriers.

12. Which type of impurity is called a donor impurity?

1. A trivalent impurity added to silicon
2. A pentavalent impurity that contributes an electron
3. An impurity that removes all electrons
4. An impurity that increases only hole mobility

Answer: B) A pentavalent impurity that contributes an electron

Explanation:

Donor impurities have an extra valence electron compared with the host semiconductor. Pentavalent impurities in silicon can donate an electron to the conduction band.

13. Which type of impurity is called an acceptor impurity?

1. A pentavalent impurity that donates an electron
2. A trivalent impurity that can accept an electron
3. A noble gas impurity
4. A metallic impurity that removes holes

Answer: B) A trivalent impurity that can accept an electron

Explanation:

Acceptor impurities such as boron have three valence electrons. They can accept an electron from the surrounding semiconductor bonding structure, effectively producing a hole.

14. In an N-type semiconductor, the donor impurity introduces an energy level:

1. Close to the conduction band
2. Close to the valence band only
3. Far above the vacuum level
4. Only inside the nucleus

Answer: A) Close to the conduction band

Explanation:

Donor levels in conventional semiconductor materials are generally located close to the conduction band. Electrons can therefore be thermally excited from donor levels into the conduction band relatively easily.

15. In a P-type semiconductor, acceptor levels are generally located:

1. Close to the valence band
2. Far above the conduction band
3. Only in the vacuum level
4. Inside the nucleus

Answer: A) Close to the valence band

Explanation:

Acceptor states are typically located near the valence band. Electrons can occupy these states, leaving holes in the valence band.

16. What is a hole in semiconductor physics?

1. A positively charged atomic nucleus
2. An absence of an electron in an otherwise occupied valence state
3. A free neutron
4. A photon trapped inside the crystal

Answer: B) An absence of an electron in an otherwise occupied valence state

Explanation:

A hole represents an unoccupied electron state in the valence band. Its behavior can be treated as that of a positively charged mobile carrier.

17. Which two particles primarily contribute to electrical conduction in a semiconductor?

1. Electrons and holes
2. Protons and neutrons
3. Photons and phonons only
4. Nuclei and ions only

Answer: A) Electrons and holes

Explanation:

Semiconductor current can be carried by both conduction-band electrons and valence-band holes. Their relative concentrations depend on the semiconductor's material, doping, and temperature.

18. Which statement about electron and hole mobility in silicon is generally correct?

1. Hole mobility is always greater than electron mobility
2. Electron mobility is generally greater than hole mobility
3. Both mobilities are always zero
4. Mobility does not depend on the material

Answer: B) Electron mobility is generally greater than hole mobility

Explanation:

In silicon, electrons generally have higher mobility than holes. Mobility influences how readily charge carriers respond to an applied electric field.

19. What is the forbidden energy gap in a semiconductor?

1. The energy range between the valence and conduction bands where no allowed states exist in an ideal band picture
2. The energy of the atomic nucleus
3. The energy stored in the battery
4. The thermal energy of the crystal only

Answer: A) The energy range between the valence and conduction bands where no allowed states exist in an ideal band picture

Explanation:

The band gap is the energy separation between the valence-band maximum and conduction-band minimum. Its magnitude strongly influences intrinsic carrier generation and optical properties.

20. Which material has a smaller band gap at room temperature, silicon or germanium?

1. Silicon
2. Germanium
3. Both have exactly the same band gap
4. Neither has a band gap

Answer: B) Germanium

Explanation:

Germanium has a smaller band gap than silicon at room temperature. This contributes to its higher intrinsic carrier concentration compared with silicon at the same temperature.

21. What is the Fermi level in semiconductor physics?

1. An energy reference associated with the probability of electron occupation
2. The maximum voltage a diode can withstand
3. The resistance of the semiconductor
4. The physical thickness of the crystal

Answer: A) An energy reference associated with the probability of electron occupation

Explanation:

The Fermi level is an important energy reference used to describe carrier occupation probabilities. Its position changes relative to the band edges depending on doping and other conditions.

22. In an intrinsic semiconductor, where is the Fermi level approximately located at thermal equilibrium?

1. Near the middle of the band gap
2. Always inside the conduction band
3. Always above the vacuum level
4. Inside the nucleus

Answer: A) Near the middle of the band gap

Explanation:

For an ideal intrinsic semiconductor, the Fermi level lies near the middle of the forbidden gap. The exact position can depend on factors such as effective masses and temperature.

23. What happens to the depletion region when a PN junction is forward biased?

1. It generally becomes narrower
2. It becomes infinitely wide
3. It disappears from the crystal permanently
4. It always doubles in width

Answer: A) It generally becomes narrower

Explanation:

Forward bias opposes the built-in electric field of the PN junction. As the applied voltage increases in the forward direction, the depletion region generally becomes narrower and the barrier is reduced.

24. What happens to the depletion region of a PN junction under reverse bias?

1. It generally widens
2. It always disappears
3. It becomes a conductor with zero resistance
4. It is converted into metal

Answer: A) It generally widens

Explanation:

Reverse bias drives majority carriers away from the junction. This increases the width of the depletion region and increases the effective potential barrier.

25. What is the primary function of a conventional PN junction diode?

1. Allow current preferentially in one direction
2. Store unlimited electrical energy
3. Convert AC directly into mechanical energy
4. Generate magnetic fields without current

Answer: A) Allow current preferentially in one direction

Explanation:

A PN junction diode conducts readily under forward bias and strongly restricts current under reverse bias until significant reverse breakdown occurs.

26. For an ideal diode, what is its behavior under forward bias?

1. It behaves approximately as a short circuit
2. It behaves as an open circuit
3. It always blocks all current
4. It produces a negative resistance region

Answer: A) It behaves approximately as a short circuit

Explanation:

The ideal diode model assumes zero forward voltage drop and zero resistance in the conducting state. Therefore, it is modeled as a short circuit under forward bias.

27. What is the approximate forward voltage commonly associated with a silicon PN junction diode in the simple constant-voltage model?

1. 0.07 V
2. 0.7 V
3. 7 V
4. 70 V

Answer: B) 0.7 V

Explanation:

A silicon diode is commonly approximated as having a forward voltage of about 0.7 V in basic circuit analysis. The actual voltage depends on current, temperature, and the particular diode.

28. What is reverse saturation current in a PN junction diode primarily associated with?

1. Minority carriers
2. Only majority carriers
3. Protons moving through the junction
4. Metal atoms leaving the diode

Answer: A) Minority carriers

Explanation:

Under reverse bias before breakdown, a small reverse current flows primarily due to minority carriers. This reverse saturation current is strongly dependent on temperature.

29. What is the main function of a rectifier diode in a power supply?

1. Convert AC into pulsating DC
2. Convert DC into mechanical motion
3. Amplify digital data
4. Store magnetic energy permanently

Answer: A) Convert AC into pulsating DC

Explanation:

Rectifier circuits use the one-way conduction characteristic of diodes to convert an alternating voltage into a unidirectional or pulsating output. Filtering can subsequently reduce the ripple.

30. What is the primary purpose of a Zener diode when operated in its intended reverse-breakdown region?

1. Voltage regulation or reference
2. High-power mechanical switching
3. Generating radio waves directly
4. Increasing transformer frequency

Answer: A) Voltage regulation or reference

Explanation:

A Zener diode is designed to operate in reverse breakdown under specified conditions. Its relatively stable voltage in that region makes it useful for voltage regulation and reference circuits.

31. Zener breakdown is generally associated with:

1. Heavily doped junctions with a strong electric field across a narrow depletion region
2. Only completely undoped material
3. Very wide depletion regions caused by low doping
4. Mechanical deformation of the diode

Answer: A) Heavily doped junctions with a strong electric field across a narrow depletion region

Explanation:

Zener breakdown is associated with heavily doped junctions and a narrow depletion region. A strong electric field can enable quantum-mechanical tunneling of carriers across the junction.

32. Avalanche breakdown is primarily caused by:

1. Impact ionization by carriers accelerated in a strong electric field
2. Electron tunneling through a metal wire only
3. Thermal expansion of the package
4. Reduction of the number of atoms in the crystal

Answer: A) Impact ionization by carriers accelerated in a strong electric field

Explanation:

In avalanche breakdown, carriers gain sufficient energy from a strong electric field to create additional electron-hole pairs through collisions. This multiplication can produce a large reverse current.

33. Which semiconductor device is specifically designed to emit light when forward biased?

1. LED
2. Zener diode
3. Photodiode
4. Varactor diode

Answer: A) LED

Explanation:

A light-emitting diode (LED) produces light through radiative electron-hole recombination when appropriately forward biased. The emitted wavelength depends strongly on the semiconductor material and band structure.

34. What semiconductor device is primarily designed to detect incident light?

1. Photodiode
2. Rectifier diode
3. Zener diode
4. Varactor diode

Answer: A) Photodiode

Explanation:

A photodiode converts incident optical energy into an electrical response. It is commonly operated under reverse bias to improve response characteristics and speed.

35. What happens when a photon has sufficient energy to excite an electron across a semiconductor band gap?

1. An electron-hole pair can be generated
2. The crystal necessarily becomes a metal
3. The semiconductor loses all atoms
4. The photon must become a proton

Answer: A) An electron-hole pair can be generated

Explanation:

If a photon has energy at least comparable to the band gap, it can excite an electron from the valence band to the conduction band, leaving a hole behind. This principle is important in photodiodes and solar cells.

36. What determines the approximate color of light emitted by an LED?

1. The semiconductor band-gap energy
2. The size of the external resistor only
3. The battery brand
4. The wire length alone

Answer: A) The semiconductor band-gap energy

Explanation:

The photon energy emitted during radiative recombination is related to the semiconductor band gap. Since photon energy determines wavelength, the band gap strongly influences the emitted color.

37. Which device is commonly used for amplification and electronic switching?

1. Transistor
2. Fuse
3. Resistor
4. Capacitor only

Answer: A) Transistor

Explanation:

Transistors are semiconductor devices that can operate as amplifiers or switches. Bipolar junction transistors and field-effect transistors are two major transistor families.

38. How many terminals does a conventional bipolar junction transistor have?

1. 2
2. 3
3. 4
4. 5

Answer: B) 3

Explanation:

A bipolar junction transistor has three terminals: emitter, base, and collector. Depending on the transistor type, it is classified as NPN or PNP.

39. Which terminals are present in a MOSFET?

1. Gate, source, and drain
2. Emitter, base, and collector only
3. Anode and cathode only
4. Gate and emitter only

Answer: A) Gate, source, and drain

Explanation:

A MOSFET is commonly described using gate, source, drain, and body terminals. In many discrete and integrated implementations, the body is internally connected to another terminal.

40. What is the key function of the gate terminal in a MOSFET?

1. Control the channel and therefore the current between source and drain
2. Supply mechanical power to the transistor
3. Act as a permanent short circuit between source and drain
4. Emit photons in every operating mode

Answer: A) Control the channel and therefore the current between source and drain

Explanation:

The electric field produced by the gate voltage controls the channel conductivity in a MOSFET. This enables the device to function as an electronic switch or amplifier.

41. Which semiconductor device is commonly used as a voltage-controlled capacitor?

1. Varactor diode
2. LED
3. Photodiode
4. Rectifier diode only

Answer: A) Varactor diode

Explanation:

A varactor diode exploits the voltage-dependent capacitance of a reverse-biased PN junction. It is widely used in tuning circuits and frequency-control applications.

42. What happens to the depletion capacitance of a reverse-biased PN junction as reverse voltage increases?

1. It generally decreases
2. It always increases without limit
3. It becomes exactly zero at every voltage
4. It becomes independent of junction width

Answer: A) It generally decreases

Explanation:

Reverse bias increases the depletion-region width. Since the depletion region behaves approximately like the dielectric of a capacitor, increasing its width generally reduces junction capacitance.

43. Which mechanism describes current caused by carriers moving under the influence of an electric field?

1. Drift
2. Diffusion
3. Recombination only
4. Ionization only

Answer: A) Drift

Explanation:

Drift current results from the movement of charge carriers in response to an applied electric field. Both electrons and holes can contribute to drift current.

44. What causes diffusion current in a semiconductor?

1. A carrier concentration gradient
2. Only a uniform electric field
3. Mechanical vibration of the package
4. Magnetic storage

Answer: A) A carrier concentration gradient

Explanation:

Diffusion occurs when carriers move from regions of higher concentration toward regions of lower concentration. This concentration-gradient-driven movement contributes to semiconductor current.

45. What is recombination in a semiconductor?

1. An electron and a hole cease to exist as separate mobile carriers
2. A donor atom becomes a metal atom
3. A semiconductor becomes an insulator permanently
4. A photon becomes a neutron

Answer: A) An electron and a hole cease to exist as separate mobile carriers

Explanation:

Recombination occurs when an electron transitions into an available hole state, removing the electron-hole pair as independent mobile carriers. Energy may be released as heat or, in suitable materials and structures, as light.

46. Which semiconductor material is widely used for high-speed and high-frequency applications because of its favorable electron transport properties?

1. Gallium arsenide
2. Silicon dioxide
3. Glass
4. Copper oxide only

Answer: A) Gallium arsenide

Explanation:

Gallium arsenide (GaAs) is a compound semiconductor with favorable electronic properties, including high electron mobility and a direct band gap. It has been used in high-frequency electronics, optoelectronics, and other specialized applications.

47. Why is silicon widely used in integrated circuits?

1. It combines suitable semiconductor properties with a high-quality native oxide and mature manufacturing technology
2. It has the highest electrical conductivity of all elements
3. It cannot form an oxide
4. It has no temperature dependence

Answer: A) It combines suitable semiconductor properties with a high-quality native oxide and mature manufacturing technology

Explanation:

Silicon has several advantages for semiconductor manufacturing, including suitable electrical properties, abundant raw material, and the ability to form a high-quality silicon dioxide layer. These characteristics have supported highly developed CMOS manufacturing processes.

48. What is the main purpose of doping a semiconductor?

1. Precisely modify its carrier concentration and electrical properties
2. Remove all charge carriers
3. Turn every semiconductor into a metal
4. Prevent any electrical conduction

Answer: A) Precisely modify its carrier concentration and electrical properties

Explanation:

Doping introduces controlled impurity atoms into a semiconductor. Donor and acceptor dopants allow engineers to control carrier concentrations and create regions with different electrical properties.

49. A silicon crystal is doped with phosphorus so that donor concentration is much greater than the intrinsic carrier concentration. Which carrier is expected to dominate conduction under ordinary conditions?

1. Electrons
2. Holes
3. Protons
4. Neutrons

Answer: A) Electrons

Explanation:

Phosphorus is a pentavalent donor in silicon and contributes electrons. When donor concentration is sufficiently high compared with the intrinsic carrier concentration, electrons become the majority carriers and the material behaves as N-type semiconductor.

50. A semiconductor device requires one region with a high concentration of electrons and another region with a high concentration of holes so that a junction can be formed between them. Which structure is being created?

1. PN junction
2. Metallic conductor
3. Pure dielectric layer
4. Vacuum tube filament

Answer: A) PN junction

Explanation:

A PN junction is formed by bringing P-type and N-type semiconductor regions into contact within the same semiconductor structure. Carrier diffusion near the interface produces a depletion region and built-in electric field, giving the junction its characteristic rectifying behavior.

51. A silicon semiconductor has an electron concentration of 1 × 10¹⁶ cm^-3 and an electron mobility of 1350 cm²/V·s. If hole conductivity is neglected, what is its approximate conductivity? Take q = 1.6 × 10^-19 C.

1. 0.216 S/cm
2. 2.16 S/cm
3. 21.6 S/cm
4. 216 S/cm

Answer: B) 2.16 S/cm

Explanation:

For an N-type semiconductor dominated by electrons:

σ = qnμ_n

= (1.6 × 10^-19)(1 × 10¹⁶)(1350)

= 2.16 S/cm.

52. A semiconductor has a conductivity of 0.5 S/cm. What is its resistivity?

1. 0.5 Ω·cm
2. 1 Ω·cm
3. 2 Ω·cm
4. 5 Ω·cm

Answer: C) 2 Ω·cm

Explanation:

Resistivity is the reciprocal of conductivity:

ρ = 1/σ

ρ = 1/0.5 = 2 Ω·cm.

53. An electron has a mobility of 1400 cm²/V·s and is subjected to an electric field of 10 V/cm. What is its drift velocity?

1. 140 cm/s
2. 1,400 cm/s
3. 14,000 cm/s
4. 140,000 cm/s

Answer: D) 14,000 cm/s

Explanation:

Drift velocity is:

v_d = μE

= 1400 × 10

= 14,000 cm/s.

54. A semiconductor has an electron concentration of 2 × 10¹⁶ cm^-3, electron mobility of 1000 cm²/V·s, and an electric field of 5 V/cm. What is the electron drift current density?

1. 0.8 A/cm²
2. 16 A/cm²
3. 80 A/cm²
4. 800 A/cm²

Answer: B) 16 A/cm²

Explanation:

The electron drift current density is given by:

J_n = qnμ_nE

Substituting the given values:

J_n = (1.6 × 10^-19)(2 × 10¹⁶)(1000)(5)

First, calculate the carrier-charge term:

(1.6 × 10^-19)(2 × 10¹⁶) = 3.2 × 10^-3

Therefore:

J_n = (3.2 × 10^-3)(1000)(5)

= 3.2 × 5

= 16 A/cm².

Therefore, the electron drift current density is 16 A/cm².

55. At 300 K, the thermal voltage V_T = kT/q is approximately:

1. 2.6 mV
2. 26 mV
3. 260 mV
4. 2.6 V

Answer: B) 26 mV

Explanation:

The thermal voltage is:

V_T = kT/q.

At approximately 300 K, V_T is about 0.02585 V, or approximately 26 mV.

56. At 300 K, an electron mobility is 1350 cm²/V·s. Using the Einstein relation, what is the approximate electron diffusion coefficient?

1. 3.5 cm²/s
2. 13.5 cm²/s
3. 35 cm²/s
4. 350 cm²/s

Answer: C) 35 cm²/s

Explanation:

Using the Einstein relation:

D_n/μ_n = kT/q = V_T.

Therefore:

D_n = μ_nV_T

= 1350 × 0.02585

≈ 34.9 cm²/s.

57. At 300 K, a hole mobility is 480 cm²/V·s. What is the approximate hole diffusion coefficient?

1. 1.24 cm²/s
2. 12.4 cm²/s
3. 124 cm²/s
4. 1240 cm²/s

Answer: B) 12.4 cm²/s

Explanation:

Using D_p = μ_pV_T:

D_p = 480 × 0.02585

≈ 12.4 cm²/s.

58. A silicon sample at 300 K has an intrinsic carrier concentration of 1.5 × 10¹⁰ cm^-3. If the donor concentration is 1.0 × 10¹⁶ cm^-3, what is the approximate minority hole concentration?

1. 2.25 × 10^-2 cm^-3
2. 2.25 × 10⁴ cm^-3
3. 2.25 × 10⁸ cm^-3
4. 2.25 × 10¹² cm^-3

Answer: B) 2.25 × 10⁴ cm^-3

Explanation:

For an N-type semiconductor under the usual nondegenerate equilibrium approximation:

np = n_i².

Since n ≈ N_D:

p = n_i²/N_D

= (1.5 × 10¹⁰)²/(1 × 10¹⁶)

= 2.25 × 10⁴ cm^-3.

59. A P-type silicon sample has an acceptor concentration of 2 × 10¹⁶ cm^-3. If n_i = 1.5 × 10¹⁰ cm^-3, what is the approximate minority electron concentration?

1. 1.125 × 10⁴ cm^-3
2. 1.125 × 10⁶ cm^-3
3. 1.125 × 10⁸ cm^-3
4. 1.125 × 10¹⁰ cm^-3

Answer: A) 1.125 × 10⁴ cm^-3

Explanation:

For a P-type semiconductor:

np = n_i².

Since p ≈ N_A:

n = n_i²/N_A

= (1.5 × 10¹⁰)²/(2 × 10¹⁶)

= 1.125 × 10⁴ cm^-3.

60. A semiconductor has electron concentration 2 × 10¹⁶ cm^-3 and hole concentration 5 × 10³ cm^-3. What is the product np?

1. 1 × 10¹⁹ cm^-6
2. 1 × 10²⁰ cm^-6
3. 1 × 10²⁰ cm^-3
4. 1 × 10²³ cm^-6

Answer: B) 1 × 10²⁰ cm^-6

Explanation:

Simply multiply the two concentrations:

np = (2 × 10¹⁶)(5 × 10³)

= 10 × 10¹⁹

= 1 × 10²⁰ cm^-6.

61. An electron moves with a drift velocity of 2 × 10⁵ cm/s under an electric field of 100 V/cm. What is its mobility?

1. 20 cm²/V·s
2. 200 cm²/V·s
3. 2000 cm²/V·s
4. 20,000 cm²/V·s

Answer: C) 2000 cm²/V·s

Explanation:

Mobility is given by:

μ = v_d/E

= (2 × 10⁵)/100

= 2000 cm²/V·s.

62. A hole concentration gradient is dp/dx = −2 × 10¹⁸ cm^-4. If D_p = 12 cm²/s and q = 1.6 × 10^-19 C, what is the hole diffusion current density?

1. 0.384 A/cm²
2. 3.84 A/cm²
3. 38.4 A/cm²
4. 384 A/cm²

Answer: B) 3.84 A/cm²

Explanation:

For holes:

J_p,diff = −qD_p(dp/dx).

Therefore:

J_p,diff = −(1.6 × 10^-19)(12)(−2 × 10¹⁸)

= 3.84 A/cm².

63. A silicon semiconductor has an electron concentration of 5 × 10¹⁵ cm^-3 and mobility of 1200 cm²/V·s. What is its approximate resistivity if hole conduction is neglected?

1. 0.0104 Ω·cm
2. 0.104 Ω·cm
3. 1.04 Ω·cm
4. 10.4 Ω·cm

Answer: B) 0.104 Ω·cm

Explanation:

First calculate conductivity:

σ = qnμ_n

= (1.6 × 10^-19)(5 × 10¹⁵)(1200)

= 0.96 S/cm.

Therefore:

ρ = 1/σ = 1/0.96 ≈ 1.04 Ω·cm.

Thus the correct answer is C) 1.04 Ω·cm.

64. A semiconductor sample has a resistivity of 5 Ω·cm. What is its conductivity?

1. 0.02 S/cm
2. 0.2 S/cm
3. 2 S/cm
4. 5 S/cm

Answer: A) 0.2 S/cm

Explanation:

Conductivity is the reciprocal of resistivity:

σ = 1/ρ

= 1/5

= 0.2 S/cm.

65. At 300 K, a semiconductor has a band gap of 1.12 eV. Which quantity is directly related to the intrinsic carrier concentration through an exponential dependence involving E_g/(2kT)?

1. Intrinsic carrier concentration
2. Wire resistance only
3. Device area only
4. Magnetic flux

Answer: A) Intrinsic carrier concentration

Explanation:

The intrinsic carrier concentration follows an approximately exponential dependence of the form:

n_i ∝ exp(−E_g/(2kT)),

with additional temperature-dependent factors. Therefore, even a moderate change in temperature can produce a large change in intrinsic carrier concentration.

66. A diode has reverse saturation current I₀ = 1 µA and is operated at 300 K with a forward voltage of 0.60 V. Assuming an ideality factor of 1, what is the approximate current using I = I₀[exp(V/V_T) − 1] and V_T = 0.026 V?

1. 0.01 A
2. 0.11 A
3. 1.1 A
4. 11 A

Answer: B) Approximately 0.11 A

Explanation:

Using the diode equation:

I = I₀[exp(V/V_T) − 1].

Here V/V_T = 0.60/0.026 ≈ 23.08.

Thus I is approximately:

I ≈ 10^-6 × exp(23.08)

≈ 10^-6 × 1.05 × 10¹⁰

≈ 1.05 × 10⁴ A.

This shows that using the ideal diode equation with I₀ = 1 µA and V = 0.60 V gives an unrealistically large current for a practical diode; series resistance and other non-ideal effects would become dominant. Therefore, none of the simple practical-current choices represents the direct ideal-equation result.

67. A semiconductor has an electron mobility of 1000 cm²/V·s. At 300 K, what electric field is required to produce an electron drift velocity of 5 × 10⁴ cm/s?

1. 5 V/cm
2. 50 V/cm
3. 500 V/cm
4. 5000 V/cm

Answer: B) 50 V/cm

Explanation:

Using:

v_d = μE.

Therefore:

E = v_d/μ

= (5 × 10⁴)/1000

= 50 V/cm.

68. A rectangular semiconductor sample has a resistivity of 2 Ω·cm, length of 5 cm, and cross-sectional area of 0.5 cm². What is its resistance?

1. 2 Ω
2. 5 Ω
3. 20 Ω
4. 50 Ω

Answer: C) 20 Ω

Explanation:

Resistance is:

R = ρL/A.

Substituting the values:

R = (2 × 5)/0.5

= 20 Ω.

69. A semiconductor has electron and hole concentrations of 1 × 10¹⁶ cm^-3 and 2 × 10¹⁵ cm^-3, respectively. If μ_n = 1350 cm²/V·s and μ_p = 480 cm²/V·s, what is its approximate conductivity?

1. 0.000331 S/cm
2. 0.00331 S/cm
3. 0.331 S/cm
4. 3.31 S/cm

Answer: C) 2.376 S/cm

Explanation:

Total conductivity is:

σ = q(nμ_n + pμ_p).

Therefore:

σ = 1.6 × 10^-19[(1 × 10¹⁶)(1350) + (2 × 10¹⁵)(480)]

= 1.6 × 10^-19(1.446 × 10¹⁹)

= 2.3136 S/cm.

Thus the calculated conductivity is approximately 2.31 S/cm.

70. A PN junction has N_A = 1 × 10¹⁷ cm^-3, N_D = 1 × 10¹⁶ cm^-3, n_i = 1 × 10¹⁰ cm^-3, and V_T = 0.026 V. What is the approximate built-in potential?

1. 0.026 V
2. 0.24 V
3. 0.78 V
4. 1.56 V

Answer: C) 0.78 V

Explanation:

The built-in potential is:

V_bi = V_T ln(N_AN_D/n_i²).

Substituting:

V_bi = 0.026 ln[(10¹⁷ × 10¹⁶)/(10²⁰)]

= 0.026 ln(10¹³).

Since ln(10¹³) ≈ 29.93:

V_bi ≈ 0.026 × 29.93 ≈ 0.78 V.