CBSE Important Questions Class 12 Physics Chapter 14: Semiconductor Electronics With Solutions

Semiconductor electronics studies materials whose conductivity lies between metals and insulators. A small amount of heat, light, impurity, or applied voltage can change charge-carrier flow in semiconductors.

Semiconductors power modern electronic circuits because they control charge flow inside solids at low voltage. CBSE 2026 tests CBSE Important Questions Class 12 Physics Chapter 14 through conductivity, energy bands, intrinsic and extrinsic semiconductors, p-n junction formation, diode biasing, V-I characteristics, and rectifier circuits. The NCERT Reprint 2025-26 scope includes semiconductor materials, junction diodes, and simple rectifier applications. Students must know carrier movement, majority carriers, depletion region, barrier potential, and output frequency in rectifiers.

Key Takeaways

• Energy Gap Rule: Semiconductors have a small forbidden energy gap below 3 eV.
• Carrier Balance: Intrinsic semiconductors have equal electron and hole concentrations.
• Biasing Trap: Forward bias lowers the barrier, while reverse bias increases it.
• Rectifier Output: Full-wave rectifier output frequency is twice the input frequency.

CBSE Important Questions Class 12 Physics Chapter 14 Structure 2026

CBSE Important Questions Class 12 Physics Chapter 14: Key Concepts

These class 12 physics chapter 14 important questions depend on definitions, carrier movement, and energy band logic. Learn each rule before solving diode and rectifier questions.

Q1. What Is Semiconductor Electronics?

Semiconductor electronics studies materials and devices that control electron and hole flow inside solids. It includes semiconductors, p-n junctions, diodes, and rectifiers.

1. Key Point:
   Semiconductors conduct better than insulators but worse than metals.
2. Main Devices:
   p-n junction diode, transistor, and rectifier circuits.
3. Final Result:
   Semiconductor electronics controls charge flow in solid-state devices

Q2. Why Did Semiconductor Devices Replace Vacuum Tubes?

Semiconductor devices replaced vacuum tubes because they are smaller, consume less power, work at low voltages, and have higher reliability.

1. Vacuum Tube Issues:
   Large size, high voltage, high power consumption, limited life.
2. Semiconductor Advantage:
   No heated cathode or evacuated space is required.
3. Final Result:
   Semiconductors made electronic circuits compact and reliable

Q3. What Are The Main Topics In CBSE Important Questions Class 12 Physics Chapter 14?

Chapter 14 focuses on materials, charge carriers, p-n junctions, diodes, and rectifiers. These topics follow NCERT Reprint 2025-26 for CBSE 2026.

1. Core Topics:
   Conductivity, energy bands, intrinsic semiconductors, extrinsic semiconductors.
2. Device Topics:
   p-n junction diode, forward bias, reverse bias, rectification.
3. Final Result:
   Chapter 14 covers semiconductor materials, devices, and simple circuits

Semiconductor Electronics Class 12 Important Questions MCQ With Answers

These semiconductor electronics class 12 important questions test direct NCERT facts. MCQs usually focus on carriers, dopants, energy gaps, biasing, and rectifiers.

Q1. In n-Type Silicon, Which Statement Is Correct?

Holes are minority carriers, and pentavalent atoms are dopants. Pentavalent dopants donate extra electrons.

1. Options:
   (A) Electrons are majority carriers and trivalent atoms are dopants.
   (B) Electrons are minority carriers and pentavalent atoms are dopants.
   (C) Holes are minority carriers and pentavalent atoms are dopants.
   (D) Holes are majority carriers and trivalent atoms are dopants.
2. Rule Used:
   Pentavalent dopants create n-type semiconductors.
3. Final Result:
   Answer: (C) Holes are minority carriers and pentavalent atoms are dopants

Q2. Which Statement Is Correct For p-Type Semiconductors?

Holes are majority carriers, and trivalent atoms are dopants. Trivalent dopants create acceptor levels.

1. Options:
   (A) Electrons are majority carriers and trivalent atoms are dopants.
   (B) Electrons are minority carriers and pentavalent atoms are dopants.
   (C) Holes are minority carriers and pentavalent atoms are dopants.
   (D) Holes are majority carriers and trivalent atoms are dopants.
2. Rule Used:
   Trivalent dopants create p-type semiconductors.
3. Final Result:
   Answer: (D) Holes are majority carriers and trivalent atoms are dopants

Q3. Which Energy Gap Order Is Correct For Carbon, Silicon, And Germanium?

Carbon has the highest energy gap, followed by silicon and germanium. The order is Eg(C) > Eg(Si) > Eg(Ge).

1. Given Data:
   Eg(C) = 5.4 eV
   Eg(Si) = 1.1 eV
   Eg(Ge) = 0.7 eV
2. Formula Used:
   Compare energy gap values.
3. Final Result:
   Answer: Eg(C) > Eg(Si) > Eg(Ge)

Q4. Why Do Holes Diffuse From p-Region To n-Region In An Unbiased p-n Junction?

Holes diffuse because their concentration is higher in the p-region. Diffusion occurs due to concentration gradient.

1. Given Data:
   Hole concentration is high in p-region.
2. Rule Used:
   Carriers diffuse from higher concentration to lower concentration.
3. Final Result:
   Answer: Hole concentration is higher in p-region

Q5. What Does Forward Bias Do To A p-n Junction?

Forward bias lowers the potential barrier. It also reduces the depletion region width.

1. Given Data:
   p-side connects to positive terminal.
   n-side connects to negative terminal.
2. Rule Used:
   Applied voltage opposes the built-in potential.
3. Final Result:
   Answer: It lowers the potential barrier

Q6. What Is The Output Frequency Of A Half-Wave Rectifier For 50 Hz Input?

The output frequency is 50 Hz. Half-wave rectifier gives one output pulse per input cycle.

1. Given Data:
   Input frequency = 50 Hz
2. Formula Used:
   Half-wave output frequency = input frequency
3. Calculation:
   Output frequency = 50 Hz
4. Final Result:
   Half-wave output frequency = 50 Hz

Q7. What Is The Output Frequency Of A Full-Wave Rectifier For 50 Hz Input?

The output frequency is 100 Hz. Full-wave rectifier gives two output pulses per input cycle.

1. Given Data:
   Input frequency = 50 Hz
2. Formula Used:
   Full-wave output frequency = 2 × input frequency
3. Calculation:
   Output frequency = 2 × 50
   Output frequency = 100 Hz
4. Final Result:
   Full-wave output frequency = 100 Hz

Semiconductor Class 12 Questions And Answers On Materials

These semiconductor class 12 questions and answers cover classification by conductivity and energy bands. The NCERT values help identify metals, insulators, and semiconductors.

Q1. How Are Solids Classified On The Basis Of Conductivity?

Solids are classified as metals, semiconductors, and insulators based on conductivity or resistivity values.

1. Metals:
   Resistivity is very low.
   Conductivity is high.
2. Semiconductors:
   Resistivity lies between metals and insulators.
3. Insulators:
   Resistivity is very high.
   Conductivity is very low.
4. Final Result:
   Solids are classified as metals, semiconductors, and insulators

Q2. What Are The Resistivity Ranges Of Metals, Semiconductors, And Insulators?

Metals have the lowest resistivity, semiconductors have intermediate resistivity, and insulators have the highest resistivity.

1. Metals:
   ρ ≈ 10⁻² to 10⁻⁸ Ω m
2. Semiconductors:
   ρ ≈ 10⁻⁵ to 10⁶ Ω m
3. Insulators:
   ρ ≈ 10¹¹ to 10¹⁹ Ω m
4. Final Result:
   Semiconductors have intermediate resistivity

Q3. What Are Elemental And Compound Semiconductors?

Elemental semiconductors contain one element, while compound semiconductors contain two or more elements.

1. Elemental Semiconductors:
   Silicon and germanium.
2. Compound Semiconductors:
   GaAs, CdS, CdSe, and InP.
3. Final Result:
   Si and Ge are elemental semiconductors

Q4. Why Is Carbon An Insulator While Silicon And Germanium Are Semiconductors?

Carbon has a much larger energy gap than silicon and germanium. It gives negligible free electrons for conduction.

1. Given Data:
   Eg(C) = 5.4 eV
   Eg(Si) = 1.1 eV
   Eg(Ge) = 0.7 eV
2. Rule Used:
   Larger energy gap reduces thermal excitation.
3. Final Result:
   Carbon behaves as an insulator

Energy Band Theory Class 12 Important Questions

Energy band theory class 12 important questions explain why metals conduct and insulators do not. Semiconductors conduct when electrons cross a small energy gap.

Q1. What Are Valence Band And Conduction Band?

The valence band contains valence electron energy levels, while the conduction band contains energy levels where electrons can move freely.

1. Valence Band:
   It contains bound valence electrons.
2. Conduction Band:
   It allows free electron movement.
3. Final Result:
   Conduction band electrons contribute to electrical conduction

Q2. How Does Energy Band Gap Classify Materials?

Energy band gap classifies materials by the separation between valence and conduction bands.

1. Metals:
   Eg ≈ 0 or bands overlap.
2. Semiconductors:
   Eg < 3 eV.
3. Insulators:
   Eg > 3 eV.
4. Final Result:
   Small band gap makes a semiconductor

Q3. Why Do Metals Have High Conductivity?

Metals have high conductivity because electrons can easily move into available conduction states. The conduction band is partly filled or overlaps the valence band.

1. Key Condition:
   Conduction band is partially filled.
2. Alternate Condition:
   Conduction and valence bands overlap.
3. Final Result:
   Metals have many free electrons for conduction

Q4. Why Do Semiconductors Conduct At Room Temperature?

Semiconductors conduct at room temperature because some electrons gain thermal energy and enter the conduction band.

1. Given Data:
   Semiconductors have small energy gap.
2. Process:
   Electrons move from valence band to conduction band.
3. Result:
   Holes form in the valence band.
4. Final Result:
   Both electrons and holes contribute to conduction

Intrinsic And Extrinsic Semiconductor Class 12 Important Questions

These intrinsic semiconductor class 12 and extrinsic semiconductor class 12 questions test carrier concentration and doping logic. They form the base of semiconductor device questions.

Q1. What Is An Intrinsic Semiconductor?

An intrinsic semiconductor is a pure semiconductor with equal electron and hole concentrations. It conducts due to thermally generated carriers.

1. Given Data:
   No impurity is added.
2. Formula Used:
   ne = nh = ni
3. Final Result:
   Intrinsic semiconductor has ne = nh = ni

Q2. What Are Holes In Semiconductors?

A hole is an electron vacancy with effective positive charge. It behaves like a positive charge carrier.

1. Cause:
   A valence electron leaves a covalent bond.
2. Result:
   The vacant bond behaves as a hole.
3. Final Result:
   Hole carries effective positive charge

Q3. Why Is Intrinsic Semiconductor Conductivity Low At Room Temperature?

Intrinsic semiconductor conductivity is low because thermally generated carriers are few. Pure silicon or germanium has limited free electrons at room temperature.

1. Given Data:
   Carrier generation depends on thermal excitation.
2. Limitation:
   Few electrons cross the energy gap.
3. Final Result:
   Intrinsic semiconductors have low conductivity

Q4. What Is An Extrinsic Semiconductor?

An extrinsic semiconductor is a doped semiconductor whose conductivity increases due to controlled impurity addition.

1. Process:
   Add a small amount of suitable impurity.
2. Name:
   The impurity is called a dopant.
3. Final Result:
   Doping increases semiconductor conductivity

Q5. What Is The Difference Between Intrinsic And Extrinsic Semiconductor?

Intrinsic semiconductor is pure, while extrinsic semiconductor is doped. Extrinsic semiconductors have higher conductivity.

1. Intrinsic Semiconductor:
   ne = nh = ni
2. Extrinsic Semiconductor:
   One carrier type becomes majority carrier.
3. Final Result:
   Extrinsic semiconductors conduct better than intrinsic semiconductors

n Type Semiconductor Class 12 And p Type Semiconductor Class 12 Questions

These n type semiconductor class 12 and p type semiconductor class 12 questions focus on dopants and majority carriers. Do not confuse dopant charge with mobile carrier charge.

Q1. What Is An n-Type Semiconductor?

An n-type semiconductor is formed by doping silicon or germanium with pentavalent atoms. Electrons become majority carriers.

1. Dopants:
   Phosphorus, arsenic, antimony.
2. Majority Carriers:
   Electrons.
3. Minority Carriers:
   Holes.
4. Final Result:
   In n-type semiconductor, ne >> nh

Q2. Why Are Pentavalent Dopants Called Donor Impurities?

Pentavalent dopants donate one extra electron for conduction. Four electrons form covalent bonds, and the fifth becomes free.

1. Given Data:
   Pentavalent atom has five valence electrons.
2. Bonding:
   Four electrons form bonds with silicon neighbours.
3. Final Result:
   Pentavalent dopants are donor impurities

Q3. What Is A p-Type Semiconductor?

A p-type semiconductor is formed by doping silicon or germanium with trivalent atoms. Holes become majority carriers.

1. Dopants:
   Boron, aluminium, indium.
2. Majority Carriers:
   Holes.
3. Minority Carriers:
   Electrons.
4. Final Result:
   In p-type semiconductor, nh >> ne

Q4. Why Are Trivalent Dopants Called Acceptor Impurities?

Trivalent dopants are acceptor impurities because they create a hole by accepting an electron from a neighbouring bond.

1. Given Data:
   Trivalent atom has three valence electrons.
2. Bonding:
   It forms only three complete covalent bonds.
3. Final Result:
   Trivalent dopants create holes

Q5. What Is The Relation Between Electron And Hole Concentration In Thermal Equilibrium?

The relation is ne nh = ni². It applies in intrinsic and extrinsic semiconductors at thermal equilibrium.

1. Given Data:
   Electron concentration = ne
   Hole concentration = nh
   Intrinsic concentration = ni
2. Formula Used:
   ne nh = ni²
3. Final Result:
   ne nh = ni²

p-n Junction Diode Class 12 Important Questions

These p-n junction diode class 12 questions cover depletion region, diffusion current, drift current, and barrier potential. These concepts explain diode biasing.

Q1. What Is A p-n Junction?

A p-n junction forms when p-type and n-type regions exist inside the same semiconductor crystal. It is the basic unit of diodes.

1. Formation:
   A part of p-type semiconductor converts into n-type.
2. Requirement:
   The junction must form at the atomic level.
3. Final Result:
   p-n junction is the basic building block of semiconductor devices

Q2. Why Can We Not Make A p-n Junction By Physically Joining p-Type And n-Type Slabs?

Physical joining cannot form a proper p-n junction because atomic-level continuous contact is not possible.

1. Reason:
   Slab surfaces have roughness larger than interatomic spacing.
2. Effect:
   The junction behaves as a discontinuity.
3. Final Result:
   A p-n junction must form inside one crystal

Q3. What Is Depletion Region Class 12?

The depletion region is the junction region depleted of free electrons and holes. It contains immobile ionised donor and acceptor cores.

1. Formation:
   Electrons and holes diffuse across the junction.
2. Result:
   Immobile ions remain near the junction.
3. Final Result:
   Depletion region has no mobile charge carriers

Q4. What Is Barrier Potential In A p-n Junction?

Barrier potential is the potential difference across the depletion region. It opposes further diffusion of charge carriers.

1. Cause:
   Ionised donors and acceptors create an electric field.
2. Effect:
   The field opposes further diffusion.
3. Final Result:
   Barrier potential maintains equilibrium

Q5. What Is The Difference Between Diffusion Current And Drift Current?

Diffusion current occurs due to concentration gradient, while drift current occurs due to electric field.

1. Diffusion Current:
   Carriers move from high concentration to low concentration.
2. Drift Current:
   Carriers move due to junction electric field.
3. Final Result:
   At equilibrium, diffusion current equals drift current

Forward Bias And Reverse Bias Class 12 Questions

These forward bias and reverse bias class 12 questions explain diode current. Forward bias gives large current, while reverse bias gives small saturation current.

Q1. What Happens When A p-n Junction Diode Is Forward Biased?

In forward bias, the p-side connects to positive terminal and n-side connects to negative terminal. The barrier height decreases.

1. Given Data:
   p-side to positive terminal.
   n-side to negative terminal.
2. Effect:
   Depletion width decreases.
3. Current:
   Majority carriers cross the junction.
4. Final Result:
   Forward bias gives large current in mA

Q2. What Happens When A p-n Junction Diode Is Reverse Biased?

In reverse bias, the n-side connects to positive terminal and p-side connects to negative terminal. The barrier height increases.

1. Given Data:
   n-side to positive terminal.
   p-side to negative terminal.
2. Effect:
   Depletion width increases.
3. Current:
   Minority carriers cause small reverse current.
4. Final Result:
   Reverse bias gives small current in μA

Q3. Why Is Current Large In Forward Bias?

Forward bias current is large because the barrier potential decreases and majority carriers cross the junction.

1. Applied Voltage:
   It opposes the built-in potential.
2. Barrier Height:
   Effective barrier becomes V0 - V.
3. Final Result:
   Forward bias supports majority carrier flow

Q4. Why Is Reverse Current Almost Constant Before Breakdown?

Reverse current is almost constant because it depends on minority carrier concentration, not applied voltage.

1. Given Data:
   Reverse bias sweeps minority carriers.
2. Limiting Factor:
   Minority carrier concentration limits current.
3. Final Result:
   Reverse current remains nearly voltage independent before breakdown

Q5. What Is Cut-In Voltage Of Silicon And Germanium Diodes?

The cut-in voltage is about 0.7 V for silicon and 0.2 V for germanium.

1. Silicon Diode:
   Cut-in voltage ≈ 0.7 V
2. Germanium Diode:
   Cut-in voltage ≈ 0.2 V
3. Final Result:
   Silicon has higher cut-in voltage than germanium

Semiconductor Diode Class 12 Important Questions

These semiconductor diode class 12 questions focus on V-I characteristics and dynamic resistance. A p-n junction diode allows current mainly in one direction.

Q1. What Is A Semiconductor Diode?

A semiconductor diode is a p-n junction with metallic contacts at both ends. It is a two-terminal device.

1. Structure:
   p-region and n-region.
2. Symbol Meaning:
   Arrow shows conventional current direction in forward bias.
3. Final Result:
   A diode allows current mainly in forward bias

Q2. What Does V-I Characteristic Of A Diode Show?

The V-I characteristic shows how diode current changes with applied voltage in forward and reverse bias.

1. Forward Bias:
   Current increases sharply after cut-in voltage.
2. Reverse Bias:
   Current remains small until breakdown.
3. Final Result:
   V-I graph shows diode conduction behaviour

Q3. What Is Dynamic Resistance Of A Diode?

Dynamic resistance is the ratio of small change in voltage to small change in current. It is written as rd = ΔV/ΔI.

1. Given Data:
   Small voltage change = ΔV
   Small current change = ΔI
2. Formula Used:
   rd = ΔV/ΔI
3. Final Result:
   Dynamic resistance = ΔV/ΔI

Q4. Why Does A Diode Work As A Rectifier?

A diode works as a rectifier because it conducts mainly in forward bias and blocks current in reverse bias.

1. Forward Half Cycle:
   Diode conducts.
2. Reverse Half Cycle:
   Diode blocks current.
3. Final Result:
   A diode converts AC into unidirectional current

Rectifier Class 12 Physics Important Questions With Solutions

These rectifier class 12 physics questions cover half-wave and full-wave rectifiers. CBSE 2026 questions often ask circuit working and output frequency.

Q1. What Is A Half-Wave Rectifier?

A half-wave rectifier converts only one half-cycle of AC into unidirectional output. It uses one diode.

1. Given Data:
   One diode and load resistance.
2. Positive Half Cycle:
   Diode conducts if forward biased.
3. Negative Half Cycle:
   Diode does not conduct.
4. Final Result:
   Half-wave rectifier gives output in alternate half-cycles

Q2. What Is A Full-Wave Rectifier?

A full-wave rectifier converts both half-cycles of AC into unidirectional output. A centre-tap circuit uses two diodes.

1. Given Data:
   Two diodes and centre-tap transformer.
2. Positive Half Cycle:
   One diode conducts.
3. Negative Half Cycle:
   The other diode conducts.
4. Final Result:
   Full-wave rectifier gives output in every half-cycle

Q3. Why Is Full-Wave Rectifier More Efficient Than Half-Wave Rectifier?

A full-wave rectifier is more efficient because it gives output during both half-cycles of AC input.

1. Half-Wave Rectifier:
   Uses only one half-cycle.
2. Full-Wave Rectifier:
   Uses both half-cycles.
3. Final Result:
   Full-wave rectifier gives more continuous output

Q4. What Is The Role Of Capacitor In A Rectifier Circuit?

A capacitor filter smoothens pulsating DC output. It charges during rising voltage and discharges through load during falling voltage.

1. Charging:
   Capacitor charges near peak voltage.
2. Discharging:
   Capacitor discharges through load resistance.
3. Final Result:
   Capacitor reduces AC ripple in rectified output

Q5. What Is The Difference Between Half-Wave And Full-Wave Rectifier Output Frequency?

Half-wave rectifier output frequency equals input frequency. Full-wave rectifier output frequency equals twice the input frequency.

1. Half-Wave:
   fout = fin
2. Full-Wave:
   fout = 2fin
3. Final Result:
   For 50 Hz input, outputs are 50 Hz and 100 Hz

Semiconductor Electronics Class 12 Numerical Questions

These semiconductor electronics class 12 numerical questions use NCERT formulas. Focus on carrier concentration, dynamic resistance, and rectifier frequency.

Q1. A Pure Silicon Crystal Has 5 × 10²⁸ Atoms m⁻³. It Is Doped With 1 ppm Pentavalent Arsenic. Find Electron Concentration.

The electron concentration is 5 × 10²² m⁻³. One donor atom gives one conduction electron.

1. Given Data:
   Silicon atoms = 5 × 10²⁸ m⁻³
   Doping = 1 ppm = 10⁻⁶
2. Formula Used:
   Donor concentration = atom concentration × doping fraction
3. Calculation:
   ND = 5 × 10²⁸ × 10⁻⁶
   ND = 5 × 10²² m⁻³
4. Final Result:
   ne ≈ 5 × 10²² m⁻³

Q2. For The Same Silicon Sample, Find Hole Concentration If ni = 1.5 × 10¹⁶ m⁻³.

The hole concentration is 4.5 × 10⁹ m⁻³. Use ne nh = ni².

1. Given Data:
   ne = 5 × 10²² m⁻³
   ni = 1.5 × 10¹⁶ m⁻³
2. Formula Used:
   ne nh = ni²
3. Calculation:
   nh = ni²/ne
   nh = (1.5 × 10¹⁶)²/(5 × 10²²)
   nh = 2.25 × 10³²/(5 × 10²²)
4. Final Result:
   nh = 4.5 × 10⁹ m⁻³

Q3. A Silicon Diode Current Changes From 10 mA To 20 mA When Voltage Changes From 0.7 V To 0.8 V. Find Dynamic Resistance.

The dynamic resistance is 10 Ω. Use rd = ΔV/ΔI.

1. Given Data:
   ΔV = 0.8 - 0.7 = 0.1 V
   ΔI = 20 mA - 10 mA = 10 mA
2. Formula Used:
   rd = ΔV/ΔI
3. Calculation:
   rd = 0.1/0.010
   rd = 10 Ω
4. Final Result:
   Dynamic resistance = 10 Ω

Q4. A Diode Has Reverse Current 1 μA At 10 V Reverse Bias. Find Reverse Resistance.

The reverse resistance is 1.0 × 10⁷ Ω. Use R = V/I.

1. Given Data:
   V = 10 V
   I = 1 μA = 1 × 10⁻⁶ A
2. Formula Used:
   R = V/I
3. Calculation:
   R = 10/(1 × 10⁻⁶)
   R = 1 × 10⁷ Ω
4. Final Result:
   Reverse resistance = 1.0 × 10⁷ Ω

Q5. Input AC Frequency Is 60 Hz. Find Output Frequency For Half-Wave And Full-Wave Rectifiers.

The half-wave output is 60 Hz, and the full-wave output is 120 Hz.

1. Given Data:
   Input frequency = 60 Hz
2. Formula Used:
   Half-wave output = fin
   Full-wave output = 2fin
3. Calculation:
   Half-wave = 60 Hz
   Full-wave = 2 × 60 = 120 Hz
4. Final Result:
   Half-wave = 60 Hz, Full-wave = 120 Hz

Class 12 Physics Chapter 14 Previous Year Questions

These class 12 physics chapter 14 important questions follow repeated board-style patterns. They test definitions, reasoning, and simple calculations.

Q1. Explain Why A p-n Junction Has No Net Current At Equilibrium.

A p-n junction has no net current because diffusion current equals drift current at equilibrium.

1. Diffusion Current:
   Electrons and holes move due to concentration gradient.
2. Drift Current:
   Electric field drives carriers in the opposite direction.
3. Equilibrium:
   Both currents become equal and opposite.
4. Final Result:
   Net current is zero at equilibrium

Q2. Why Does Forward Bias Reduce Depletion Layer Width?

Forward bias reduces depletion layer width because applied voltage opposes the built-in potential.

1. Given Data:
   p-side connects to positive terminal.
2. Effect:
   Barrier potential decreases.
3. Result:
   Majority carriers cross the junction.
4. Final Result:
   Forward bias narrows the depletion layer

Q3. Why Does Reverse Bias Increase Depletion Layer Width?

Reverse bias increases depletion layer width because applied voltage supports the built-in potential.

1. Given Data:
   n-side connects to positive terminal.
2. Effect:
   Barrier potential increases.
3. Result:
   Majority carrier movement is suppressed.
4. Final Result:
   Reverse bias widens the depletion layer

Q4. Explain Why Semiconductor Conductivity Increases With Temperature.

Semiconductor conductivity increases with temperature because more electrons gain energy and cross into the conduction band.

1. Thermal Energy:
   It breaks some covalent bonds.
2. Carrier Generation:
   Electrons and holes increase.
3. Final Result:
   More carriers increase conductivity

Q5. Why Are Donor Levels Close To The Conduction Band?

Donor levels lie close to the conduction band because the extra electron is weakly bound. Small energy can free it.

1. n-Type Material:
   Pentavalent dopant adds one extra electron.
2. Energy Need:
   Ionisation energy is small.
3. Final Result:
   Donor electrons enter conduction band easily

Q6. Why Are Acceptor Levels Close To The Valence Band?

Acceptor levels lie close to the valence band because valence electrons can easily fill acceptor vacancies. This creates holes.

1. p-Type Material:
   Trivalent dopant creates one hole.
2. Energy Need:
   Small energy transfers an electron to acceptor level.
3. Final Result:
   Acceptor doping creates holes in valence band

Class 12 Physics Chapter List