Semiconductor Electronics — Class 12 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Physics chapter "Semiconductor Electronics" — 5 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Semiconductor Electronics" — 5 important questions with detailed answers for CBSE boar…
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NCERT Solutions for Class 12 Physics — Semiconductor Electronics. Step-by-step answers to all 5 textbook questions from this chapter, written for CBSE board preparation and free to use.
Semiconductor Electronics — All 5 Questions Solved
Q1. Explain the band theory of solids. Distinguish between conductors, insulators, and semiconductors.
Band Theory of Solids:
In isolated atoms, electrons occupy discrete energy levels. In solids, atoms are closely packed, and these levels split into bands of closely-spaced levels.
Formation of Bands:
1. When N atoms combine, each level splits into N levels
2. Levels so close they form continuous band
3. Band width depends on atomic spacing
4. Bands separated by energy gaps (forbidden zones)
Key Bands:
Valence Band: Outermost electrons, partially filled
Conduction Band: Higher energy band, electrons can move freely
Forbidden Gap: Energy region with no allowed states
Classification Based on Band Structure:
1. CONDUCTORS (Metals):
- Valence and conduction bands overlap
- Electrons easily promoted to conduction band
- Eg = 0 (no band gap)
- High electrical conductivity (σ ≈ 10⁷ S/m)
- Examples: Cu, Al, Ag
2. INSULATORS:
- Large forbidden gap (Eg > 3 eV)
- Valence band full, conduction band empty
- Electrons cannot easily cross gap
- Very low conductivity (σ < 10⁻¹⁰ S/m)
- Examples: diamond (Eg = 5.5 eV), glass
3. SEMICONDUCTORS:
- Moderate forbidden gap (Eg = 0.1 - 3 eV)
- At T = 0: valence band full, conduction band empty
- At higher T: thermal energy excites electrons to conduction band
- Conductivity between conductors and insulators
- Conductivity increases with temperature
- Examples: Si (Eg = 1.1 eV), Ge (Eg = 0.66 eV)
Comparison Table:
Conductor Semiconductor Insulator
Band gap 0 0.1-3 eV > 3 eV
Valence band Partially filled/overlapping Full
Conduction band Empty/overlapping Empty Empty
Conductivity High Medium Very Low
Temp effect Decreases Increases Unchanged
Intrinic vs Extrinsic:
Intrinsic: Pure semiconductor, conductivity from thermal generation
Extrinsic: Doped semiconductor, conductivity from impurities
Q2. Explain the p-n junction and how it rectifies current (diode action).
p-n Junction: Contact between p-type and n-type semiconductor materials.
Formation of p-n Junction:
1. p-type: doped with acceptors (holes are majority carriers)
2. n-type: doped with donors (electrons are majority carriers)
3. At junction: electrons diffuse from n to p, holes diffuse from p to n
4. Diffusion leaves positive ions in n-type, negative ions in p-type
5. Creates electric field opposing further diffusion
6. Equilibrium reached (built-in potential V₀ ≈ 0.7 V for Si)
Depletion Region (Space Charge Region):
- Near junction, no mobile charge carriers
- Contains only fixed ions
- Width: ~0.1 to 1 μm for typical doping
- Electric field prevents carrier diffusion
- Built-in potential = V₀ = (kT/e) ln(N_a × N_d/n_i²)
where N_a, N_d = doping concentrations, n_i = intrinsic concentration
Forward Bias:
Applied voltage opposes built-in field
- Reduces depletion width
- Lowers barrier for carrier injection
- Current increases exponentially
- Ideal diode equation: I = I_s[e^(eV/kT) - 1]
Where I_s = saturation current
Reverse Bias:
Applied voltage aids built-in field
- Increases depletion width
- Increases barrier height
- Current very small (reverse saturation current I_s)
- Limited by generation of e-h pairs in depletion region
- Independent of voltage magnitude (up to breakdown)
Diode Characteristics:
Forward direction (conducting):
- Current increases with voltage
- Voltage drop ≈ 0.7 V (Si), 0.3 V (Ge)
- Acts as conductor
Reverse direction (blocking):
- Current ≈ I_s (very small)
- Acts as insulator
- Until breakdown voltage reached
Rectification:
- Forward current: large
- Reverse current: negligible
- Allows current in one direction only
- Used to convert AC to DC
- Pulsating DC output needs filtering
Applications:
1. Rectifier (AC to DC conversion)
2. Voltage regulator (Zener diode)
3. Photodiode (light detection)
4. LED (light emission)
5. Solar cell (energy conversion)
Q3. Explain the transistor (BJT). Describe the three operating regions and derive the current gain relationship.
Bipolar Junction Transistor (BJT): Three-terminal device with two back-to-back p-n junctions for current and voltage amplification.
Structure:
- Emitter (E): heavily doped, source of charge carriers
- Base (B): lightly doped, controls carrier injection
- Collector (C): lightly doped, collects carriers
- Two types: npn and pnp
Basic Operating Principle:
1. Small base current controls large collector current
2. Base-emitter junction: forward biased (conducts)
3. Base-collector junction: reverse biased (blocks)
4. Electrons injected into base recombine with holes
5. Most electrons reach collector before recombining
Current Relationships:
I_E = I_C + I_B (current continuity)
Current Gain (β):
β = I_C/I_B = DC current gain (typically 50-300)
Alpha (α):
α = I_C/I_E = collector efficiency (0.98-0.99)
Relation: β = α/(1 - α)
Three Operating Regions:
1. CUTOFF Region:
- Base current I_B = 0
- Both junctions reverse biased
- Collector current ≈ 0 (leakage current only)
- V_CE ≈ V_CC (maximum)
- Transistor OFF (switch open)
2. ACTIVE Region (Linear):
- Base-emitter: forward biased
- Base-collector: reverse biased
- I_C = β × I_B (linear relationship)
- V_BE ≈ 0.7 V (constant)
- 0 < V_CE < V_CC
- Used for amplification
- Stable operating point
3. SATURATION Region:
- Both junctions forward biased
- Base-collector junction forward biased (unusual)
- I_C reaches maximum (supply limited)
- Further increase in I_B doesn't increase I_C
- V_CE ≈ 0.2 V (minimum)
- Transistor ON (switch closed)
- Large I_B exists
Current Gain Derivation:
In active region, base acts as control:
- Emitter injects I_E = I_B + I_C
- For weak base: most carriers reach collector
- Collector current: I_C = α × I_E
Substituting I_E = I_B + I_C:
I_C = α(I_B + I_C)
I_C = αI_B + αI_C
I_C(1 - α) = αI_B
I_C = [α/(1 - α)] × I_B
Defining β = α/(1 - α):
I_C = β × I_B
Or from α: if α = 0.99, then β = 0.99/0.01 = 99
Applications:
1. Amplifier: AC signal to base → amplified AC at collector
2. Switch: ON/OFF control with small base current
3. Oscillator: generates AC signals
4. Power amplifier: drives loads like speakers
5. Logic circuits (digital)
Q4. An npn transistor has β = 100. If base current is 20 μA, find collector and emitter currents in active region.
Given:
Current gain: β = 100
Base current: I_B = 20 μA = 20 × 10⁻⁶ A
Transistor in active region
Collector Current:
In active region: I_C = β × I_B
I_C = 100 × 20 × 10⁻⁶
I_C = 2000 × 10⁻⁶ A = 2 mA = 0.002 A
Emitter Current:
From current continuity: I_E = I_B + I_C
I_E = 20 × 10⁻⁶ + 2000 × 10⁻⁶
I_E = 2020 × 10⁻⁶ A = 2.02 mA = 0.00202 A
Alpha (current transfer ratio):
α = I_C/I_E = 2000/2020 = 0.9901 ≈ 0.99
Verification:
β = α/(1 - α) = 0.99/0.01 = 99 ≈ 100 ✓
Percentage Distribution:
- Base current: I_B/I_E × 100 = 20/2020 × 100 = 0.99%
- Collector current: I_C/I_E × 100 = 2000/2020 × 100 = 99.01%
Power Considerations (if V_CE = 5 V):
Power dissipated: P = V_CE × I_C
P = 5 × 0.002 = 0.01 W = 10 mW
Amplification Factor:
Voltage gain = (R_C/R_E) × (V_T/I_E)
where V_T = thermal voltage ≈ 26 mV at 300 K
Answers:
- Collector current: I_C = 2 mA
- Emitter current: I_E = 2.02 mA
- Alpha: α = 0.99
- Base carries ~1% of total current
- Collector carries ~99% of total current
Q5. What is feedback in amplifiers? Explain negative feedback and its effects on gain and stability.
Feedback: Returning a portion of output signal to input, modifying overall behavior.
Types of Feedback:
1. Negative Feedback: Output opposes input (reduces gain but improves stability)
2. Positive Feedback: Output aids input (increases gain, can cause oscillation)
Negative Feedback System:
Input → [Gain Block] → Output
↑
[Feedback Network]
Feedback Factor β_f = V_f/V_o (fraction fed back)
Loop Gain: L = A × β_f (open-loop gain × feedback factor)
Closed-loop gain: A_f = A/(1 + Aβ_f) = A/(1 + L)
For high loop gain (L >> 1):
A_f ≈ 1/β_f (gain depends mainly on feedback network)
Effects of Negative Feedback:
1. REDUCES GAIN:
A_f = A₀/(1 + β_f A₀)
Closed-loop gain < open-loop gain
Trade-off: sacrifice gain for stability
2. INCREASES BANDWIDTH:
f_-3dB (closed loop) = f_-3dB (open loop) × (1 + β_f A₀)
Frequency response extended
Signal distortion decreased
3. REDUCES DISTORTION:
T.H.D. reduced by factor (1 + L)
Non-linearity cancelled
Better linearity over wide range
4. IMPROVES INPUT IMPEDANCE:
Series feedback (voltage feedback):
Z_in(f) = Z_in(0) × (1 + L)
Higher input impedance
5. REDUCES OUTPUT IMPEDANCE:
Shunt feedback (current feedback):
Z_out(f) = Z_out(0) / (1 + L)
Lower output impedance
6. IMPROVES STABILITY:
Reduces sensitivity to component variations
Gain becomes less dependent on transistor parameters
Temperature effects minimized
Oscillation prevented (if L positive and < 1)
7. REDUCES NOISE:
Input noise affected by noise figure
Output noise reduced by (1 + L)
Signal-to-noise ratio improved
Quantitative Example:
Open-loop gain: A₀ = 10,000
Feedback factor: β_f = 0.1
Open-loop gain: A₀ = 10,000
Without feedback: A = 10,000
With feedback: A_f = 10,000/(1 + 0.1 × 10,000) = 10,000/1001 ≈ 9.99
Gain reduction: ~1000×
Bandwidth increase: ~1000×
Loop gain: L = 1000
Applications of Negative Feedback:
1. Op-amp circuits (stable gain)
2. Audio amplifiers (low distortion)
3. Power amplifiers (improved efficiency)
4. Voltage regulators (stability)
5. Oscillators (frequency stability)
Stability Criterion:
- Phase margin > 45° for stable operation
- Gain margin > 6 dB for stable operation
- Negative feedback alone doesn't guarantee stability
- Proper design needed to avoid oscillation
Semiconductor Electronics — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which property distinguishes semiconductors from conductors and insulators?
- A. Very low electrical resistivity.
- B. Very high electrical resistivity.
- C. Electrical resistivity lies between conductors and insulators.
- D. Zero electrical resistivity at room temperature.
Answer: C. Electrical resistivity lies between conductors and insulators. — Semiconductors have electrical resistivity values that fall between those of good conductors (low resistivity) and good insulators (high resistivity), typically ranging from 10^-5 to 10^6 ohm-meter.
MCQ 2. What is the approximate energy gap for Silicon (Si) at room temperature?
- A. 0.7 eV
- B. 1.1 eV
- C. 5.5 eV
- D. 0.2 eV
Answer: B. 1.1 eV — The energy band gap for Silicon (Si) is approximately 1.1 eV at room temperature, which is significantly larger than Germanium (0.7 eV) but much smaller than insulators.
MCQ 3. In an intrinsic semiconductor, the number of free electrons is:
- A. Greater than the number of holes.
- B. Less than the number of holes.
- C. Equal to the number of holes.
- D. Dependent on the doping concentration.
Answer: C. Equal to the number of holes. — In an intrinsic (pure) semiconductor, electron-hole pairs are generated simultaneously by thermal excitation, meaning the concentration of free electrons is always equal to the concentration of holes.
MCQ 4. To obtain an N-type semiconductor, an intrinsic semiconductor like Silicon should be doped with a:
- A. Trivalent impurity.
- B. Pentavalent impurity.
- C. Divalent impurity.
- D. Monovalent impurity.
Answer: B. Pentavalent impurity. — Pentavalent impurities (e.g., Phosphorus, Arsenic) have 5 valence electrons. When doped into silicon, they contribute an extra free electron, creating an N-type semiconductor.
MCQ 5. Which of the following impurities would create a P-type semiconductor when added to pure Germanium?
- A. Phosphorus
- B. Arsenic
- C. Boron
- D. Antimony
Answer: C. Boron — Boron is a trivalent impurity (3 valence electrons). When doped into Germanium (tetravalent), it creates an electron deficiency or 'hole' to complete the covalent bond, forming a P-type semiconductor.
MCQ 6. The depletion region in a p-n junction is formed due to:
- A. Migration of majority carriers across the junction.
- B. Drift of minority carriers across the junction.
- C. Recombination of electrons and holes near the junction.
- D. Breakdown of covalent bonds at the junction.
Answer: C. Recombination of electrons and holes near the junction. — The depletion region forms when majority charge carriers (electrons from N-side, holes from P-side) diffuse across the junction and recombine, creating immobile positive and negative ions near the junction.
MCQ 7. In forward biasing of a p-n junction diode:
- A. The potential barrier increases.
- B. The depletion region width increases.
- C. The potential barrier decreases.
- D. The majority carriers are repelled from the junction.
Answer: C. The potential barrier decreases. — In forward bias, the external voltage opposes the built-in potential barrier, reducing its height and narrowing the depletion region, allowing majority carriers to flow easily.
MCQ 8. A p-n junction diode is reverse biased. What happens to the current?
- A. Large current flows due to majority carriers.
- B. A very small current flows due to minority carriers.
- C. No current flows at all.
- D. The current increases exponentially.
Answer: B. A very small current flows due to minority carriers. — In reverse bias, majority carriers are pulled away from the junction. A very small reverse saturation current flows due to the drift of thermally generated minority carriers across the junction.
MCQ 9. If the temperature of a p-n junction diode is increased, the cut-in voltage (knee voltage) generally:
- A. Increases.
- B. Decreases.
- C. Remains the same.
- D. Increases for Si, decreases for Ge.
Answer: B. Decreases. — With an increase in temperature, the intrinsic carrier concentration in the semiconductor increases. This means a smaller forward bias voltage is required to overcome the potential barrier, causing the cut-in voltage to decrease.
MCQ 10. Which special purpose diode is primarily used for voltage regulation?
- A. Light Emitting Diode (LED)
- B. Photodiode
- C. Solar Cell
- D. Zener Diode
Answer: D. Zener Diode — A Zener diode is specifically designed to operate in the reverse breakdown region, where it maintains a nearly constant voltage across its terminals despite significant changes in current, making it ideal for voltage regulation.
MCQ 11. In an LED, light is emitted when:
- A. Electrons jump from valence band to conduction band.
- B. Holes jump from conduction band to valence band.
- C. Electrons and holes recombine.
- D. Photons are absorbed by the semiconductor.
Answer: C. Electrons and holes recombine. — Light in an LED is produced through electroluminescence, a process where electrons from the conduction band recombine with holes in the valence band across the p-n junction, releasing energy as photons.
MCQ 12. A photodiode is used in reverse bias because:
- A. It increases the dark current.
- B. It decreases the response time.
- C. It increases the sensitivity to light.
- D. It allows for larger reverse breakdown voltage.
Answer: B. It decreases the response time. — Operating a photodiode in reverse bias increases the width of the depletion region and reduces the junction capacitance. This leads to a faster response time and improved sensitivity to incident light.
MCQ 13. In a Bipolar Junction Transistor (BJT), the base region is:
- A. Heavily doped and very wide.
- B. Lightly doped and very narrow.
- C. Moderately doped and wide.
- D. Heavily doped and narrow.
Answer: B. Lightly doped and very narrow. — The base region in a BJT is intentionally designed to be very thin and lightly doped to minimize recombination of injected charge carriers and ensure a high current gain.
MCQ 14. For an NPN transistor to operate in the active region, the emitter-base junction is _____ and the collector-base junction is _____.
- A. Forward biased, reverse biased.
- B. Reverse biased, forward biased.
- C. Forward biased, forward biased.
- D. Reverse biased, reverse biased.
Answer: A. Forward biased, reverse biased. — In the active region, the emitter-base junction of an NPN transistor is forward biased to inject electrons into the base, while the collector-base junction is reverse biased to collect these electrons.
MCQ 15. If the base current (IB) is 100 μA and the collector current (IC) is 5 mA, what is the current gain (β) for a common emitter configuration?
- A. 0.02
- B. 50
- C. 20
- D. 100
Answer: B. 50 — The current gain (β) in a common emitter configuration is calculated as β = IC / IB. Here, β = (5 mA) / (100 μA) = (5 × 10^-3 A) / (100 × 10^-6 A) = 50.
MCQ 16. In a common emitter amplifier, the output voltage is out of phase with the input voltage. What is this phase difference?
- A. 0°
- B. 90°
- C. 180°
- D. 270°
Answer: C. 180° — A distinguishing characteristic of a common emitter amplifier is that its output signal is 180° out of phase with the input signal, meaning it is inverted.
MCQ 17. Which logic gate produces a high output only when all of its inputs are high?
- A. OR gate
- B. NOT gate
- C. AND gate
- D. NOR gate
Answer: C. AND gate — An AND gate's output is HIGH (1) only if all of its inputs are HIGH (1); if any input is LOW (0), the output will be LOW (0).
MCQ 18. What is the output of an OR gate if its inputs are A=0 and B=1?
- A. 0
- B. 1
- C. Indeterminate
- D. Depends on the previous state
Answer: B. 1 — For an OR gate, the output is HIGH (1) if at least one of its inputs is HIGH (1). Therefore, 0 OR 1 = 1.
MCQ 19. The output of a NOR gate is HIGH if:
- A. All inputs are HIGH.
- B. All inputs are LOW.
- C. Any one input is HIGH.
- D. Exactly one input is HIGH.
Answer: B. All inputs are LOW. — A NOR gate is an inverted OR gate. Its output is HIGH (1) only when all of its inputs are LOW (0); otherwise, the output is LOW (0).
MCQ 20. Which of the following are known as universal gates?
- A. AND, OR, NOT
- B. NAND, NOR
- C. XOR, XNOR
- D. AND, NAND, XOR
Answer: B. NAND, NOR — NAND and NOR gates are called universal gates because any other logic gate (AND, OR, NOT, XOR, XNOR) can be constructed solely using either NAND gates or NOR gates.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.