NEET Physics: Semiconductor Electronics questions with solutions

Q1. Which of the following is unipolar transistor?

1. p-n-p transistor
2. n-p-n transistor
3. Field effect transistor
4. Point contact transistor

Answer: Field effect transistor

A field effect transistor (FET) is called unipolar because conduction occurs primarily through one type of majority carrier. In contrast, p-n-p and n-p-n transistors are bipolar devices, and point-contact transistor is not the standard unipolar choice here.

Q2. Band gap in a semiconductor is of the order

1. 6 eV
2. 0.60 eV
3. 60 eV
4. 0 eV

Answer: 0.60 eV

Semiconductors have a band gap that is small enough for some electrons to be thermally excited across it at ordinary temperatures. A value around 0.6 eV is typical, whereas 0 eV corresponds to metals and several eV is more like an insulator.

Q3. The transistor are usually made of

1. metal oxides with high temperature coefficient of resistivity.
2. metals with high temperature coefficient of resistivity.
3. metals with low temperature coefficient of resistivity.
4. semiconducting materials having low temperature coefficient of resistivity.

Answer: semiconducting materials having low temperature coefficient of resistivity.

Transistors are built from semiconductors because their conductivity can be precisely controlled by doping and electric fields. A low temperature coefficient of resistivity is desirable so the device remains stable as temperature changes.

Q4. Electric conduction in a semiconductor takes place due to

1. electrons only
2. holes only
3. both electrons and holes
4. neither electrons nor holes

Answer: both electrons and holes

Semiconductors conduct electricity through two kinds of charge carriers: electrons and holes. Both can move under an electric field and contribute to current flow.

Q5. The energy band diagrams for three semiconductor sample of silicon are shown. It follows that.

1. Sample X is undoped while samples Y and Z have been doped with a third group and a fifth group impurity respectively
2. Sample X is undoped while both samples Y and Z have been doped with a fifth group impurity
3. Sample X has been doped with equal amounts of third and fifth group impurities while samples \( Y \) and \( z \) are undoped
4. Sample X is undoped while samples Y and Z have been doped with a fifth group and a third group impurity respectively

Answer: Sample X is undoped while samples Y and Z have been doped with a fifth group and a third group impurity respectively

An undoped silicon sample shows only the valence and conduction bands with the Fermi level near midgap. A fifth-group impurity creates a donor level close to the conduction band, while a third-group impurity creates an acceptor level close to the valence band, matching Y and Z respectively.

Q6. A pure semiconductor at absolute zero has

1. Absence of electrons in the conduction band.
2. All the electrons occupying the valence band.
3. Large \( E_{g} \) value.
4. All of the above.

Answer: All of the above.

In a pure semiconductor at absolute zero, the valence band is completely filled and the conduction band is empty, so there are no free charge carriers. Also, semiconductors are characterized by a finite band gap, which is large enough to prevent electrons from reaching the conduction band at 0 K.

Q7. A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be

1. 8f
2. f / 2√2
3. f / 2
4. f / 4

Answer: f / 2√2

The frequency of oscillation for an LC circuit is given by f = 1 / (2π√(LC)). If L is doubled and C is changed to 4C, the new frequency becomes f' = 1 / (2π√(2L × 4C)) = 1 / (2π√(8LC)) = f / 2√2.

Q8. The increase in the width of the depletion region in a p-n junction diode is due to:

1. reverse bias only
2. both forward bias and reverse bias
3. increase in forward current
4. forward bias only

Answer: reverse bias only

The width of the depletion region in a p-n junction diode increases under reverse bias because the applied voltage widens the region by pulling majority carriers away from the junction.

Q9. For a p-type semiconductor, which of the following statements is true?

1. Electrons are the majority carriers and trivalent atoms are the dopants.
2. Holes are the majority carriers and pentavalent atoms are the dopants.
3. Electrons are the majority carriers and pentavalent atoms are the dopants.
4. Holes are the majority carriers and trivalent atoms are the dopants.

Answer: Holes are the majority carriers and trivalent atoms are the dopants.

In a p-type semiconductor, trivalent atoms are used as dopants, creating holes as the majority carriers due to the absence of one electron in the valence shell of the dopant atoms.

Q10. In a p-n junction diode, change in temperature due to heating:

1. Affects only reverse resistance
2. Affects only forward resistance
3. Affects the overall V-I characteristics of p-n junction
4. Does not affect resistance of p-n junction

Answer: Affects the overall V-I characteristics of p-n junction

Temperature changes affect the overall V-I characteristics of a p-n junction diode because both forward and reverse resistances are influenced by temperature, altering the current flow.

Q11. The barrier potential of a p-n junction depends on:

1. type of semiconductor material
2. amount of doping
3. temperature
4. (A) and (B) only

Answer: (A) and (B) only

The barrier potential of a p-n junction depends on the type of semiconductor material, the amount of doping, and the temperature. Since option D includes both (A) and (B), it is the correct choice.

Q12. In a n-type semiconductor, which of the following statement is true?

1. Electrons are minority carriers and pentavalent atoms are dopants.
2. Holes are minority carriers and pentavalent atoms are dopants.
3. Holes are majority carriers and trivalent atoms are dopants.
4. Electrons are majority carriers and trivalent atoms are dopants.

Answer: Holes are minority carriers and pentavalent atoms are dopants.

In an n-type semiconductor, pentavalent atoms (donors) are used as dopants, and electrons are the majority carriers, while holes are the minority carriers.

Q13. In an unbiased p-n junction, holes diffuse from the p-region to n-region because of:

1. the potential difference across the p-n junction
2. the attraction of free electrons of n-region
3. the higher hole concentration in p-region than that in n-region
4. the higher concentration of electrons in the n-region than that in the p-region

Answer: the higher hole concentration in p-region than that in n-region

Holes diffuse from the p-region to the n-region due to the concentration gradient, as the p-region has a higher hole concentration compared to the n-region. This diffusion is a natural process to equalize the concentration.

Q14. The solids which have the negative temperature coefficient of resistance are:

1. insulators only
2. semiconductors only
3. insulators and semiconductors
4. metals

Answer: insulators and semiconductors

Semiconductors and insulators both exhibit a negative temperature coefficient of resistance, meaning their resistance decreases as temperature increases, due to increased charge carrier availability.

Q15. Si and Cu are cooled to a temperature of 300 K. Then resistivity?

1. For Si increases and for Cu decreases
2. For Cu increases and for Si decreases
3. Decreases for both Si and Cu
4. Increases for both Si and Cu

Answer: For Si increases and for Cu decreases

For silicon (Si), a semiconductor, resistivity increases as temperature decreases because fewer charge carriers are thermally excited. For copper (Cu), a metal, resistivity decreases as temperature decreases due to reduced lattice vibrations.

Q16. Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is:

1. 0.75 A
2. zero
3. 0.25 A
4. 0.5 A

Answer: 0.5 A

The current depends on the orientation of the diodes. Assuming one diode is forward-biased and the other is reverse-biased, only the forward-biased diode conducts. The current can be calculated based on the circuit's resistance and voltage, leading to 0.5 A.

Q17. C and Si both have same lattice structure, having 4 bonding electrons in each. However, C is insulator whereas Si is intrinsic semiconductor. This is because:

1. In case of C the valence band is not completely filled at absolute zero temperature.
2. In case of C the conduction band is partly filled even at absolute zero temperature.
3. The four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in the third.
4. The four bonding electrons in the case of C lie in the third orbit, whereas for Si they lie in the fourth orbit.

Answer: The four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in the third.

Carbon's bonding electrons are in the second orbit, which has a higher energy gap between the valence and conduction bands, making it an insulator. Silicon's bonding electrons are in the third orbit, resulting in a smaller energy gap, allowing it to behave as an intrinsic semiconductor.

Q18. If a small amount of antimony is added to germanium crystal:

1. it becomes a p–type semiconductor
2. the antimony becomes an acceptor atom
3. there will be more free electrons than holes in the semiconductor
4. its resistance is increased

Answer: there will be more free electrons than holes in the semiconductor

Adding antimony (a pentavalent impurity) to germanium introduces extra electrons, making it an n-type semiconductor with more free electrons than holes.

Q19. Which one of the following statement is FALSE?

1. Pure Si doped with trivalent impurities gives a p-type semiconductor
2. Majority carriers in a n-type semiconductor are holes
3. Minority carriers in a p-type semiconductor are electrons
4. The resistance of intrinsic semiconductor decreases with increase of temperature

Answer: Majority carriers in a n-type semiconductor are holes

In an n-type semiconductor, the majority carriers are electrons, not holes. Holes are the minority carriers in n-type semiconductors.

Q20. Pure Si at 500K has equal number of electron (nₑ) and hole (nₕ) concentrations of 1.5 × 10¹⁶ m⁻³. Doping by indium increases nₕ to 4.5 × 10²² m⁻³. The doped semiconductor is of

1. n–type with electron concentration nₑ = 5 × 10²² m⁻³
2. p–type with electron concentration nₑ = 2.5 × 10¹⁰ m⁻³
3. n–type with electron concentration nₑ = 2.5 × 10¹⁰ m⁻³
4. p–type having electron concentration nₑ = 5 × 10⁻⁹ m⁻³

Answer: p–type with electron concentration nₑ = 2.5 × 10¹⁰ m⁻³

Indium is a trivalent impurity, so it introduces holes, making the semiconductor p-type. The product of electron and hole concentrations remains constant (nₑ × nₕ = constant). Using this, nₑ = (1.5 × 10¹⁶)² / 4.5 × 10²² = 2.5 × 10¹⁰ m⁻³.