Materials, Devices & Simple Circuits — From Energy Bands to p-n Junctions & Rectifiers for AAI ATC Written Exam
📚 NCERT Class 12 Physics
✈️ AAI ATC Relevant
🎯 18 Numerical MCQs
⏱️ 6 Subtopics Covered
📌 Chapter Overview
⚡ Classification of Solids & Energy Bands
🔬 Intrinsic Semiconductors
🧪 Extrinsic Semiconductors (n & p type)
🔗 p-n Junction Formation
📊 Semiconductor Diode & V-I Characteristics
🔄 Rectifiers (Half-wave & Full-wave)
1
Classification of Solids & Energy Bands
Solids are classified based on their electrical conductivity (σ) or resistivity (ρ = 1/σ):
Type
Resistivity (ρ)
Conductivity (σ)
Energy Gap
Metals
10⁻² – 10⁻⁸ Ω·m
10² – 10⁸ S/m
Eᵍ ≈ 0
Semiconductors
10⁻⁵ – 10⁶ Ω·m
10⁵ – 10⁻⁶ S/m
Eᵍ < 3 eV
Insulators
10¹¹ – 10¹⁹ Ω·m
10⁻¹¹ – 10⁻¹⁹ S/m
Eᵍ > 3 eV
Energy Band Theory
In a solid, closely spaced energy levels form energy bands. The two key bands are the Valence Band (filled, lower energy) and the Conduction Band (empty or partially filled, higher energy). The gap between them is the Energy Band Gap (Eᵍ).
🔑 Si energy gap: 1.1 eV | Ge: 0.7 eV | C (diamond): 5.4 eV | Sn (metal): 0 eV
For Si or Ge crystal with N atoms: 4N valence electrons occupy the lower band (valence band) at 0 K. The conduction band is empty at 0 K. The two bands are separated by Eᵍ.
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ADD IMAGE HERE Energy band diagrams of metals, insulators, and semiconductors (Fig. 14.2)
🎯 Practice MCQs — Energy Bands & Classification
Q1 The energy gap of Silicon is 1.1 eV. What is the minimum frequency of photon required to excite an electron from the valence band to the conduction band? (h = 6.63 × 10⁻³⁴ J·s)
Q3 Which of the following correctly orders the energy gaps of C (diamond), Si, and Ge?
A Ge > Si > C
B Si > Ge > C
C C > Si > Ge (5.4 > 1.1 > 0.7 eV)
D C > Ge > Si
✅ Valence electrons of C are in 2nd orbit (tightly bound), Si in 3rd, Ge in 4th orbit (loosely bound).
Hence ionisation energy: C (5.4 eV) > Si (1.1 eV) > Ge (0.7 eV)
2
Intrinsic Semiconductors
A pure semiconductor (without any dopant) is called an intrinsic semiconductor. Si and Ge form diamond-like structures with each atom bonded to 4 neighbours via covalent bonds.
Thermal Generation of Carriers
At T > 0 K, thermal energy breaks some covalent bonds. A free electron (charge −q) is released and leaves behind a hole (effective charge +q). Hole = electron vacancy that behaves like a positive charge carrier.
nₑ = nₕ = nᵢ (intrinsic carrier concentration)
Current in Intrinsic Semiconductor
I = Iₑ + Iₕ (electron current + hole current)
📌 At T = 0 K → intrinsic semiconductor behaves like an insulator (no free carriers).
At T > 0 K → thermal energy generates electron-hole pairs. Rate of generation = rate of recombination at equilibrium.
🎯Holes move towards negative potential; electrons move towards positive potential under an applied electric field.
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ADD IMAGE HERE 2D covalent bond structure of Si/Ge & hole generation (Fig. 14.4, 14.5)
🎯 Practice MCQs — Intrinsic Semiconductor
Q4 In an intrinsic semiconductor at 300 K, the intrinsic carrier concentration is 1.5 × 10¹⁶ m⁻³. The number of holes per m³ is:
A 0
B 3.0 × 10¹⁶ m⁻³
C 1.5 × 10¹⁶ m⁻³
D 2.25 × 10³² m⁻³
✅ In intrinsic semiconductor: nₑ = nₕ = nᵢ
∴ Number of holes = 1.5 × 10¹⁶ m⁻³
Q5 For an intrinsic semiconductor, nenh = nᵢ². If nᵢ = 2 × 10¹⁶ m⁻³, the product nenh equals:
A 2 × 10¹⁶ m⁻⁶
B 4 × 10³² m⁻⁶
C 4 × 10¹⁶ m⁻⁶
D 8 × 10⁴⁸ m⁻⁶
✅ nₑnₕ = nᵢ² = (2×10¹⁶)² = 4 × 10³² m⁻⁶
Q6 Why does an intrinsic semiconductor behave like an insulator at 0 K?
A All electrons are in the conduction band
B Valence band is completely filled and conduction band is empty — no free carriers
C Holes and electrons recombine instantly
D Resistivity becomes zero
✅ At 0 K, all covalent bonds are intact. Valence band is completely filled, conduction band is completely empty. No thermal energy → no free carriers → behaves like insulator.
3
Extrinsic Semiconductors — n-type & p-type
Adding a small impurity (dopant) to a pure semiconductor increases conductivity manifold. This is called doping. The result is an extrinsic semiconductor.
n-type Semiconductor (Pentavalent dopant)
Dopants: As, Sb, P (valency 5). The 5th electron is very weakly bound (~0.05 eV for Si, ~0.01 eV for Ge) and becomes a free electron at room temperature. These dopants are called donor impurities.
A p-n junction is the basic building block of all semiconductor devices (diodes, transistors, etc.).
Formation Process — Diffusion & Drift
When p-type and n-type materials are joined, two processes occur simultaneously: Diffusion (due to concentration gradient) and Drift (due to electric field).
Holes diffuse from p→n; electrons diffuse from n→p. This leaves behind immobile ions (positive on n-side, negative on p-side), forming the depletion region.
Depletion Region
🔑 The depletion region is devoid of free charge carriers. Thickness ≈ one-tenth of a micrometre (0.1 μm). It creates a built-in electric field that opposes further diffusion, leading to barrier potential V₀.
Equilibrium
At equilibrium: Diffusion current = Drift current → Net current = 0. The n-side is positive relative to p-side (barrier potential).
⚠️You CANNOT make a p-n junction by physically joining two slabs — surface roughness is much larger than inter-atomic spacing (~2–3 Å). Junction must be formed within a single crystal wafer.
Q10 The depletion region in a p-n junction is formed due to:
A Drift of majority carriers only
B Diffusion of majority carriers leaving behind immobile ions
C Recombination of all electrons with all holes
D External applied voltage
✅ Diffusion of holes (p→n) and electrons (n→p) leaves behind immobile ionised cores on both sides. This space-charge region (devoid of free carriers) is the depletion region.
Q11 In a p-n junction at equilibrium, the barrier potential V₀ is 0.7 V for silicon. If the junction is forward biased with V = 0.5 V, the effective barrier height is:
A 1.2 V
B 0.7 V
C 0.2 V
D 0 V
✅ Forward bias reduces barrier: Effective barrier = V₀ − V = 0.7 − 0.5 = 0.2 V
Q12 In a reverse-biased p-n junction with barrier potential V₀ = 0.7 V and applied reverse voltage V = 5 V, the effective barrier height is:
A 4.3 V
B 5.7 V
C 0.7 V
D 5.0 V
✅ Reverse bias increases barrier: Effective barrier = V₀ + V = 0.7 + 5.0 = 5.7 V
5
Semiconductor Diode & V-I Characteristics
A semiconductor diode is a p-n junction with metallic contacts at both ends. It is a two-terminal device. Current flows conventionally from p to n (arrow direction in symbol).
Forward Bias (p → +ve terminal, n → −ve terminal)
Barrier height decreases (V₀ − V). Depletion region narrows. Majority carriers are injected across junction → minority carrier injection. Current: milliamperes (mA).
Threshold voltage: ~0.2 V (Ge) | ~0.7 V (Si)
Reverse Bias (n → +ve terminal, p → −ve terminal)
Barrier height increases (V₀ + V). Depletion region widens. Only minority carriers drift → very small reverse saturation current (~μA). This current is almost independent of applied voltage (up to breakdown).
Breakdown Voltage (Vbr)
At very high reverse bias = Vbr, reverse current suddenly increases sharply. Beyond this, the diode may be destroyed if not current-limited.
Dynamic Resistance
rᵈ = ΔV / ΔI
Forward bias resistance (rᶠᵇ) is very low; Reverse bias resistance (rʳᵇ) is very high (~MΩ).
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ADD IMAGE HERE V-I characteristics of silicon diode showing forward & reverse bias regions (Fig. 14.16c, 14.17)
🎯 Practice MCQs — Diode V-I Characteristics
Q13 A silicon diode's V-I curve shows: at I = 20 mA, V = 0.8 V; at I = 10 mA, V = 0.7 V. The dynamic (AC) resistance of the diode is:
Q14 A diode under reverse bias shows I = −1 μA at V = −10 V. Its reverse bias resistance is:
A 10 kΩ
B 1 MΩ
C 10 MΩ
D 100 MΩ
✅ rʳᵇ = V/I = 10 / (1×10⁻⁶) = 10⁷ Ω = 10 MΩ
Q15 In a forward-biased p-n junction diode, the majority carriers crossing the junction are:
A Drift carriers
B Minority carrier injection — electrons to p-side, holes to n-side
C Immobile ion cores
D Only holes cross the junction
✅ Under forward bias, electrons from n-side and holes from p-side cross the junction. On reaching the other side they become minority carriers — this is minority carrier injection. Both contribute to forward current (mA range).
6
Application of Diode — Rectifiers
A diode allows current only in one direction (forward bias). This property converts AC to DC — called rectification.
Half-Wave Rectifier
Single diode in series with load Rₗ. Conducts only during the positive half-cycle of AC. Output is a pulsating DC with only positive half-cycles. Output frequency = Input frequency.
Output frequency (half-wave) = Input frequency (e.g., 50 Hz input → 50 Hz output)
Full-Wave Rectifier
Two diodes + centre-tap transformer. D₁ conducts in positive half, D₂ conducts in negative half. Both half-cycles appear at output → more efficient. Output frequency = 2 × input frequency.
Output frequency (full-wave) = 2 × Input frequency (e.g., 50 Hz → 100 Hz output)
Capacitor Filter
A capacitor connected across Rₗ (in parallel) charges during peak and discharges slowly through load. This smooths pulsating DC into near-steady DC. Time constant τ = C × Rₗ should be large for better filtering.
📌 Key comparison:
Half-wave: 1 diode, output for 50% of cycle, ripple frequency = fin
Full-wave: 2 diodes + centre-tap, output for 100% of cycle, ripple frequency = 2×fin
Bridge rectifier: 4 diodes, no centre-tap needed, full-wave output
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ADD IMAGE HERE Half-wave rectifier circuit & waveforms (Fig. 14.18)
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ADD IMAGE HERE Full-wave rectifier circuit, input/output waveforms, and capacitor filter (Fig. 14.19, 14.20)
🎯 Practice MCQs — Rectifiers
Q16 In a half-wave rectifier, the input AC frequency is 50 Hz. What is the output ripple frequency?
A 50 Hz
B 100 Hz
C 25 Hz
D 0 Hz (pure DC)
✅ In half-wave rectifier, only one pulse per input cycle → Output frequency = Input frequency = 50 Hz
Q17 In a full-wave rectifier with 50 Hz input and load resistance Rₗ = 1 kΩ, a capacitor C = 100 μF is connected across Rₗ. The approximate time constant (τ) of the filter is:
A 10 ms
B 1 ms
C 100 ms
D 0.1 ms
✅ τ = C × Rₗ = 100×10⁻⁶ × 1×10³ = 100×10⁻³ = 100 ms = 0.1 s
Since τ >> T (period = 10 ms for 100 Hz full-wave ripple), filtering is effective.
Q18 A full-wave rectifier uses a centre-tap transformer with peak secondary voltage 20 V (total). The peak voltage available across the load is:
A 20 V
B 10 V
C 14.14 V
D 40 V
✅ In a centre-tap full-wave rectifier, each diode only uses half the secondary voltage.
Peak voltage across load = 20/2 = 10 V (minus diode drop of ~0.7 V in practice)
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