NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits
NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics provides comprehensive coverage of energy band theory, intrinsic and extrinsic semiconductors, p-n junction formation, diode characteristics under biasing, and its practical application as rectifiers.
Overview of Topics & Subtopics (Click to Jump)
1. Classification of Solids & Energy Band Theory
Short Note:
• Energy Bands: When atoms come together to form a solid, discrete electronic energy levels split into continuous energy bands—Valence Band (VB) and Conduction Band (CB), separated by an Energy Band Gap ($E_g$).
• Classifications Based on $E_g$:
- Conductors (Metals): VB and CB overlap or CB is partially filled ($E_g \approx 0$). High density of free electrons.
- Insulators: Very large energy gap ($E_g > 3\text{ eV}$). VB is completely filled, CB is completely empty at room temperature.
- Semiconductors: Small energy gap ($E_g < 3\text{ eV}$; Silicon $E_g = 1.1\text{ eV}$, Germanium $E_g = 0.72\text{ eV}$). Thermal agitation can excite electrons across the gap.
2. Intrinsic & Extrinsic Semiconductors (n-type and p-type)
Short Note:
• Intrinsic Semiconductors: Pure semiconductors (Si, Ge) with equal numbers of free electrons ($n_e$) and holes ($n_h$). Total charge carrier concentration: $n_i = n_e = n_h$.
• Extrinsic Semiconductors: Doped semiconductors to increase conductivity drastically.
- n-type Semiconductor: Doped with pentavalent impurities (As, Sb, P). Majority carriers are electrons ($n_e \gg n_h$). Impurity atom is a donor.
- p-type Semiconductor: Doped with trivalent impurities (B, Al, In, Ga). Majority carriers are holes ($n_h \gg n_e$). Impurity atom is an acceptor.
• Mass Action Law: Under thermal equilibrium, $n_e \cdot n_h = n_i^2$.
3. p-n Junction Diode, Biasing & Application as Rectifier
Short Note:
• p-n Junction Formation: Diffusion of majority carriers creates a neutral boundary layer devoid of free carriers called the depletion region and establishes a barrier potential ($V_0$) (~0.7 V for Si, ~0.3 V for Ge).
• Biasing Modes:
- Forward Bias: p-side to positive terminal, n-side to negative terminal. Reduces barrier height; large current flows in $\text{mA}$.
- Reverse Bias: p-side to negative terminal, n-side to positive terminal. Increases barrier height; negligible reverse saturation current flows in $\mu\text{A}$.
• Diode as Rectifier: Converts Alternating Current (AC) to Direct Current (DC) due to unidirectional conduction property.
- Half-Wave Rectifier: Conducts during positive half cycles only. Output ripple frequency equals input frequency ($f_{out} = f_{in}$).
- Full-Wave Rectifier: Uses two center-tapped diodes or bridge circuit. Conducts during both half cycles. Output ripple frequency is double the input ($f_{out} = 2 f_{in}$).
Intext & Conceptual Questions
| Property | n-type Semiconductor | p-type Semiconductor |
|---|---|---|
| Dopant Material | Doped with Pentavalent impurities (e.g., Phosphorus, Arsenic, Antimony). | Doped with Trivalent impurities (e.g., Boron, Aluminum, Indium). |
| Charge Carriers | Majority carriers are Electrons ($n_e \gg n_h$), minority are Holes. | Majority carriers are Holes ($n_h \gg n_e$), minority are Electrons. |
| Energy Level | Donor energy level ($E_d$) lies just below the bottom of the conduction band. | Acceptor energy level ($E_a$) lies just above the top of the valence band. |
| Overall Electrical Neutrality | Electrically neutral as a whole. | Electrically neutral as a whole. |
• When light of energy $h\nu > E_g$ falls on a photodiode, electron-hole pairs are created in or near the depletion region.
• In forward bias, the initial majority current ($I$) is very high ($\text{mA}$). The fractional change in current due to light-generated minority carriers $(\Delta I / I)$ is extremely small and very difficult to detect.
• In reverse bias, the baseline current is only a minute reverse saturation current ($I_0$) in $\mu\text{A}$. Thus, the fractional change in current $(\Delta I / I_0)$ caused by incident photons is very large and easily measurable.
• Therefore, photodiodes are operated in reverse bias for high sensitivity optical detection.
Textbook Chapter End Exercise Questions
Given Input AC frequency $f_{in} = 50\text{ Hz}$.
(a) For Half-Wave Rectifier:
A half-wave rectifier allows only one half-cycle of the input AC to pass through per cycle.
Output Frequency $f_{out} = f_{in} = 50\text{ Hz}$
(b) For Full-Wave Rectifier:
A full-wave rectifier rectifies both positive and negative half-cycles of the input AC into positive pulses.
Output Frequency $f_{out} = 2 \times f_{in} = 2 \times 50 = 100\text{ Hz}$
• Half-wave output ripple frequency = 50 Hz
• Full-wave output ripple frequency = 100 Hz
Given:
• Band gap energy $E_g = 2.8\text{ eV} = 2.8 \times 1.6 \times 10^{-19}\text{ J} = 4.48 \times 10^{-19}\text{ J}$
• Wavelength of incident light $\lambda = 6000\text{ \AA} = 6000 \times 10^{-10}\text{ m} = 6 \times 10^{-7}\text{ m}$
Step 1: Calculate Energy of Incident Photon (E)
E = (h c) / \lambda
E = (6.63 \times 10^{-34} \times 3 \times 10^8) / (6 \times 10^{-7})\text{ Joules}
E = 3.315 \times 10^{-19}\text{ J}
Step 2: Convert Energy to eV
E = (3.315 \times 10^{-19}) / (1.6 \times 10^{-19})\text{ eV} \approx 2.07\text{ eV}
Conclusion:
To excite electrons across the energy gap and detect light, the photon energy $E$ must be greater than or equal to the band gap $E_g$ ($E \ge E_g$).
Here, $E = 2.07\text{ eV} < E_g = 2.8\text{ eV}$.
Hence, the semiconductor cannot detect the wavelength of 6000 Å.
Read More & Chapter Resources
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