Class 12 Physics Notes – Chapter 14: Semiconductor Electronics

1. Classification of Metals, Conductors and Semiconductors

Based on the relative values of electrical conductivity (σ) or resistivity (ρ), solids are classified as metals, semiconductors, and insulators [5].

• Metals: They possess very low resistivity (or high conductivity) [5].
• Insulators: They have high resistivity (or low conductivity) [5].
• Semiconductors: They have resistivity or conductivity intermediate to metals and insulators [5]. Examples include elemental semiconductors like Si and Ge, and compound semiconductors like GaAs and CdS [5].

Energy Bands

In a crystal, the closely spaced energy states form energy bands [6]. The lowest energy level in the conduction band is E_C and highest energy level in the valence band is E_V [7]. The gap between them is called the energy band gap (E_g) [7].

• Metals: The conduction band is partially filled, or the conduction and valence bands overlap (E_g ≈ 0) [7].
• Insulators: The energy gap is very large (E_g > 3 eV), and there are no electrons in the conduction band [7, 8].
• Semiconductors: The energy gap is small (E_g < 3 eV). At room temperature, some electrons can acquire enough energy to cross the gap [8].

2. Intrinsic and Extrinsic Semiconductors

Intrinsic Semiconductors: These are pure semiconductors free of impurities [8]. In these, the number of free electrons (n_e) is equal to the number of holes (n_h), i.e., n_e = n_h = n_i [8].

Extrinsic Semiconductors: When a small amount of suitable impurity is added to a pure semiconductor to increase its conductivity, it is called an extrinsic semiconductor [9].

• n-type Semiconductor: Formed by doping with a pentavalent impurity (like As, Sb, P) [9, 10]. Here, electrons are the majority carriers (n_e >> n_h) [11].
• p-type Semiconductor: Formed by doping with a trivalent impurity (like In, B, Al) [10, 11]. Here, holes are the majority carriers (n_h >> n_e) [11].

3. p-n Junction

A p-n junction is formed when a p-type semiconductor is joined to an n-type semiconductor [12]. Two important processes occur: diffusion and drift [12].

Depletion Region: Due to the diffusion of electrons from n-side to p-side and holes from p-side to n-side, a space-charge region is developed on either side of the junction which is depleted of mobile charges [12].

Barrier Potential: The accumulation of negative charges in the p-region and positive charges in the n-region sets up a potential difference across the junction that opposes further diffusion [12, 13].

4. Semiconductor Diode

A semiconductor diode is basically a p-n junction with metallic contacts at the ends for the application of an external voltage [13].

• Forward Bias: When the p-side is connected to the positive terminal and the n-side to the negative terminal of a battery [13]. The depletion layer width decreases, and the barrier height is reduced [14]. Current flows mainly due to majority carriers [14].
• Reverse Bias: When the n-side is connected to the positive terminal and the p-side to the negative terminal [14]. The depletion layer widens, and the barrier height increases [14]. A very small current (in μA) flows due to minority carriers [14, 15].

5. Application of Junction Diode as a Rectifier

A rectifier is a device that converts an alternating current (ac) into a direct current (dc) [16]. Diodes are used as rectifiers because they allow current to flow only when they are forward biased [16].

• Half-wave Rectifier: It rectifies only one half of the ac input wave [16]. It uses a single diode [16].
• Full-wave Rectifier: It gives output voltage corresponding to both the positive as well as negative half of the ac cycle [16]. It uses two diodes and a centre-tap transformer [16, 17].

Multiple Choice Questions (MCQs)

1. The energy band gap for an insulator is generally:
a) E_g = 0
b) E_g < 3 eV
c) E_g > 3 eV
d) E_g ≈ 1 eV
Answer: (c) E_g > 3 eV [7, 8]

2. In an intrinsic semiconductor, the number of free electrons (n_e) and holes (n_h) are related as:
a) n_e > n_h
b) n_e < n_h
c) n_e = n_h
d) n_e + n_h = 0
Answer: (c) n_e = n_h [8]

3. Which impurity is added to obtain an n-type semiconductor?
a) Trivalent
b) Pentavalent
c) Tetravalent
d) Monovalent
Answer: (b) Pentavalent [9, 10]

4. In a p-type semiconductor, the majority charge carriers are:
a) Electrons
b) Protons
c) Neutrons
d) Holes
Answer: (d) Holes [11]

5. The depletion region of a p-n junction contains:
a) Only electrons
b) Only holes
c) Both electrons and holes
d) Immobile ions (space-charge)
Answer: (d) Immobile ions (space-charge) [12]

6. During the forward bias of a p-n junction diode, the width of the depletion layer:
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero immediately
Answer: (b) Decreases [14]

7. A full-wave rectifier circuit uses:
a) One diode
b) Two diodes
c) Three diodes
d) No diodes
Answer: (b) Two diodes [16]

8. The reverse bias current in a p-n junction diode is primarily due to:
a) Majority carriers
b) Minority carriers
c) Immobile ions
d) Thermal expansion
Answer: (b) Minority carriers [14]

9. Which of the following is an elemental semiconductor?
a) GaAs
b) CdS
c) Silicon (Si)
d) Anthracene
Answer: (c) Silicon (Si) [5]

10. The fundamental process responsible for the formation of a p-n junction is:
a) Diffusion and Drift
b) Reflection and Refraction
c) Emission and Absorption
d) Polarisation
Answer: (a) Diffusion and Drift [12]

Textbook Exercise Solutions

Q 14.1: In an n-type silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
Solution:
Answer: (c) Holes are minority carriers and pentavalent atoms are the dopants.
Explanation: In an n-type semiconductor, electrons are the majority carriers and holes are the minority carriers. It is formed by doping with pentavalent impurities [2, 4, 11].

Q 14.2: Which of the statements given in Exercise 14.1 is true for p-type semiconductors.
Solution:
Answer: (d) Holes are majority carriers and trivalent atoms are the dopants.
Explanation: In a p-type semiconductor, holes are the majority carriers and electrons are the minority carriers. It is formed by doping with trivalent impurities [2, 4, 11].

Q 14.3: Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (E_g)_C, (E_g)_Si and (E_g)_Ge. Which of the following statements is true?
(a) (E_g)_Si < (E_g)_Ge < (E_g)_C
(b) (E_g)_C < (E_g)_Ge > (E_g)_Si
(c) (E_g)_C > (E_g)_Si > (E_g)_Ge
(d) (E_g)_C = (E_g)_Si = (E_g)_Ge
Solution:
Answer: (c) (E_g)_C > (E_g)_Si > (E_g)_Ge
Explanation: Carbon (diamond) is an insulator with a very high energy gap (about 5.4 eV). Silicon has an energy gap of about 1.1 eV, and Germanium has about 0.7 eV. Therefore, (E_g)_C > (E_g)_Si > (E_g)_Ge [2-4, 18].

Q 14.4: In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All the above.
Solution:
Answer: (c) hole concentration in p-region is more as compared to n-region.
Explanation: Diffusion is the process of movement of charge carriers from a region of higher concentration to a region of lower concentration [3, 4, 12].

Q 14.5: When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) None of the above.
Solution:
Answer: (c) lowers the potential barrier.
Explanation: Under forward bias, the applied voltage opposes the built-in potential barrier, effectively reducing the barrier height and narrowing the depletion region [3, 4, 14].

Q 14.6: In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency.
Solution:
Answer: 50 Hz for half-wave, 100 Hz for full-wave.
Explanation: In a half-wave rectifier, there is one output pulse for each cycle of the input ac, so the output frequency is the same as the input frequency (50 Hz). In a full-wave rectifier, there are two output pulses for each cycle of the input ac, so the output frequency is twice the input frequency (2 × 50 = 100 Hz) [3, 4, 16, 17].