L-28: Semiconductors and Semiconducting Devices
Physics — Class 12 · NIOS Code 312 · Module 8 · Source: 312_Physics_Eng_Lesson28.pdf
Semiconductors and Semiconducting Devices
Solid-state electronics rests on understanding energy bands, doped semiconductors, p-n junction diodes, and bipolar junction transistors. Module 8: Semiconductors Devices and Communication.
28.1 Energy Bands in Solids
In crystals, interacting atoms form quasi-continuous energy bands from discrete levels. Conduction band (CB) — unfilled levels; valence band (VB) — filled levels.
28.2 Intrinsic and Extrinsic Semiconductors
Intrinsic (pure Si/Ge): electrons and holes generated in pairs by thermal energy; equal concentrations; low conductivity at room temperature (Ge: ~2.5×10¹⁹ m⁻³ pairs at 300 K).
Extrinsic (doped): ~1 dopant atom per 10⁸ host atoms. Group V (P, As, Sb) → n-type (donor impurities, excess electrons). Group III (B, Al, Ga) → p-type (acceptor impurities, holes). Neither n-type nor p-type is net charged — ions balance carriers.
Current convention: direction of hole motion = conventional current direction.
28.3 p-n Junction
Joining n-type and p-type: electrons diffuse to p-side, holes to n-side → recombination → depletion region (~0.5 μm) of immobile donor/acceptor ions → barrier potential opposes further diffusion (Si ~0.7 V, Ge ~0.3 V). Symbol: arrow shows conventional current direction (p → n).
28.4–28.5 Forward and Reverse Bias; I-V Characteristics
Forward bias: p to +, n to −. Barrier overcome when V > knee voltage (Si ~0.7 V, Ge ~0.3 V). Low forward resistance (~10–30 Ω); current ~mA. Depletion width decreases.
Reverse bias: p to −, n to +. Majority carriers move away; small reverse saturation (leakage) current from minority carriers (μA for Ge, nA for Si). High resistance. Depletion width increases. At breakdown voltage: avalanche or Zener effect → sharp current rise.
28.6 Special Diodes
- Zener diode: heavily doped; thin depletion; operates in reverse breakdown — constant VZ for voltage regulation
- LED: forward bias; electroluminescence (GaAsP, InP); Group III–V materials
- Photodiode: reverse bias; photovoltaic effect; photocurrent ∝ light intensity
- Solar cell: photovoltaic; generation, separation (depletion field), collection of e⁻–h⁺ pairs; I–V in 4th quadrant (supplies current)
28.7–28.8 Transistors
BJT (1948, Bardeen/Brattain/Shockley): three regions — Emitter (heaviest doping), Base (thinnest, lightest doping), Collector (largest, moderate doping). Types: npn and pnp. Arrow in symbol shows conventional emitter current direction.
Active region biasing: emitter-base junction forward biased; collector-base junction reverse biased. >95% emitter carriers reach collector (thin lightly doped base).
β (common-emitter gain) >> 1; β = α/(1−α)
Example: α = 0.98 → β = 49
Configurations: Common Emitter (CE) — most used (voltage + current gain); Common Base (CB) — constant current source; Common Collector (CC) — impedance matching.
CE characteristics: input (VBE vs IB) like forward diode; output (VCE vs IC at fixed IB). Input resistance Rie ≈ 20–100 Ω (npn) or ~kΩ (pnp CE).
SEMICONDUCTORS — KEY POINTS
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Conductor : CB overlaps VB
Semiconductor : Eg ≤ 3 eV ; insulator Eg > 3 eV
n-type : pentavalent donor (extra e⁻)
p-type : trivalent acceptor (holes)
Barrier pot. : Si 0.7 V ; Ge 0.3 V
Forward bias : low R, knee voltage, depletion shrinks
Reverse bias : high R, leakage current, depletion grows
Zener : reverse breakdown, voltage regulator
Photodiode : reverse bias · Solar cell: photovoltaic
Transistor : E-B forward + C-B reverse (active)
α, β : β = α/(1−α) ; CE config most common
Quick Revision
- Intrinsic: e⁻ and holes in equal numbers; extrinsic: majority carriers from dopants.
- Depletion region = immobile ions; not depleted of charge, but of mobile carriers.
- Diode: unidirectional — rectification, battery protection, detection.
- Two back-to-back diodes ≠ transistor (four doped regions, wrong base).
- Transistor naming: AC125 = Ge AF transistor; BY127 = Si rectifier diode.
Q1. A semiconductor has a forbidden energy gap of about:
Q2. n-type semiconductor is obtained by doping with:
Q3. The barrier potential of a silicon p-n junction is approximately:
Q4. In forward bias, the width of the depletion region:
Q5. A p-n junction diode conducts significantly when:
Q6. A Zener diode is used as a:
Q7. A photodiode is normally operated in:
Q8. In active region, a transistor has:
Q9. If α = 0.98, then β is approximately:
Q10. The most widely used transistor configuration is:
PYQ — Previous Year Questions
Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L28 — Semiconductors and Semiconducting Devices only. Use Model Answer for marking points; Explanation for concept clarity.
Chapter 28 — Semiconductors and Semiconducting Devices (L28)
12 questions · Sections A & B · Sources: 312/TUS/104A, 312/MAY/204A–C, 68/ESS/1-312-A
Section A — Objective (1 mark)
PYQ1. Which of the following devices has its I–V characteristics in the fourth quadrant of the Cartesian coordinate system? (A) Zener diode (B) Photodiode (C) LED (D) Solar cell
Model Answer
(D) Solar cell
Under illumination the solar cell supplies power (acts as a source); its operating I–V curve lies in the fourth quadrant.
Explanation
LED and forward-biased junctions → 1st quadrant; photodiode/Zener in conventional plotting → reverse regions. Solar cell photovoltaic mode → 4th quadrant (L28 §28.2).
Section A — Short Answer (2 marks)
PYQ2. Fill in the blanks: (i) A photodiode is always connected in __________ biasing. (ii) A forward biased p-n junction offers __________ resistance to the flow of electrons.
Model Answer
(i) reverse — photons generate e⁻–h⁺ pairs in depletion region; reverse bias widens depletion layer
(ii) low — barrier height reduced; majority carriers cross junction easily
Explanation
Photodiode/solar cell operate reverse biased for detection; forward bias → low R, heavy conduction (L28 §28.2).
PYQ3. Write TRUE or FALSE: (1) A NOT gate uses two p-n junctions. (2) When a Zener diode is not working on its optimum voltage condition the load draws less power than the power dissipated in the diode.
Model Answer
(1) FALSE — NOT gate is a logic inverter (transistor/diode logic), not two separate p-n junctions.
(2) TRUE — off Zener breakdown/regulation point, excess power is dissipated in the diode; load receives less than diode loss.
Explanation
Zener regulator needs Vi > VZ and minimum IZ; otherwise regulation fails (L28 §28.2).
PYQ4. Match Column-I with Column-II: (i) Use of a p-n junction diode → (a) Rectifier (b) Oscillator (c) Receiver for remote (d) Stabilizer; (ii) Use of a transistor → same options.
Model Answer
(i) → (a) Rectifier — unidirectional conduction converts AC to pulsating DC
(ii) → (c) Receiver for remote (also amplifier/oscillator) — transistor amplifies/detects weak RF signals in receivers
Explanation
Diode rectification is core L28; transistor switching/amplification enables practical circuits (L28 §28.3).
Section B — Short Answer (2 marks)
PYQ5. Name a diode which is used in reverse bias. What is it used for?
Model Answer
Zener diode (also photodiode acceptable)
Operated in reverse breakdown region for voltage regulation/stabilization — maintains constant output VZ across load.
Explanation
Photodiode: reverse bias for light detection. Zener: heavily doped, thin depletion, designed breakdown (L28 §28.2).
PYQ6. Draw a circuit diagram of a stabilized power supply showing a step-down transformer, a full-wave rectifier, a capacitor filter and a Zener diode.
Model Answer
AC mains → step-down transformer → full-wave rectifier (2 diodes, centre-tap or bridge) → capacitor filter across load → Zener regulator (series resistor RS, Zener in reverse across RL).
Label: Vi, VO ≈ VZ, smoothing C, RL.
Explanation
Rectifier + filter give unregulated DC; Zener shunt maintains fixed VO despite load/input variations (L28 §28.2).
PYQ7. Draw a circuit diagram showing an n-p-n transistor amplifier in common-emitter configuration. Write expressions for current gain and voltage gain.
Model Answer
CE circuit: VCC, RC in collector, RE optional, input at base–emitter, output across RC; E-B forward biased, C-B reverse biased.
Current gain β = IC/IB (≈ 50–300)
Voltage gain Av ≈ β × RC/Rin (or −β RC/RE with emitter resistor)
Explanation
CE is most common — high voltage and current gain; 180° phase shift between input and output (L28 §28.3).
PYQ8. Draw a circuit diagram for a half-wave rectifier. Also show its input and output waveforms.
Model Answer
AC source → transformer secondary → single diode D in series with load RL.
Input: sinusoidal AC. Output: pulsating unidirectional — only positive half-cycles appear across RL; negative half blocked (zero current).
Explanation
Diode conducts forward half-cycle only; PIV rating must exceed peak AC voltage (L28 §28.2 / L29 §29.1).
PYQ9. Write the symbol of (a) p-n junction and (b) p-n-p transistor.
Model Answer
(a) p-n junction: triangle/arrow on cathode (n-side arrow) standard diode symbol
(b) p-n-p: two arrows inward on emitter; BJT symbol with E-B-C labelled, arrow on emitter pointing toward base
Explanation
Arrow indicates conventional current direction on emitter; npn arrow out, pnp arrow in (L28 §28.3).
PYQ10. Describe in brief the formation of depletion region in a p-n junction diode with a suitable diagram.
Model Answer
Electrons diffuse n→p, holes p→n → recombination near junction → immobile ions (n⁺, p⁻) left → depletion layer devoid of mobile carriers (~0.5 μm).
Built-in potential opposes further diffusion; region acts as insulating barrier until bias applied.
Explanation
Marking scheme: concentration gradient drives diffusion; recombination creates space charge — core p-n junction physics (L28 §28.1).
Section B — Long Answer (3 marks)
PYQ11. Draw circuit diagrams to plot the characteristics of a p-n junction diode in (a) forward bias and (b) reverse bias. Also draw the characteristic curves in each case.
Model Answer
Forward: microammeter + milliammeter, diode forward, variable V; plot I vs V — threshold ~0.7 V (Si), then exponential rise.
Reverse: reverse connection, µA meter; small reverse saturation current I0 until breakdown.
Curves: forward steep after knee; reverse flat then breakdown at high |V|.
Explanation
Forward bias narrows depletion layer; reverse widens it — asymmetric I–V is basis of rectification (L28 §28.2).
PYQ12. Distinguish clearly between intrinsic and extrinsic semiconductors. (Give any three points.)
Model Answer
- Purity: intrinsic = pure Si/Ge; extrinsic = doped with impurity
- Carriers: intrinsic ne = nh = ni; extrinsic — n-type ne >> nh, p-type nh >> ne
- Conductivity: intrinsic low, increases with T; extrinsic much higher, controlled by dopant
- Dopants: none vs pentavalent (n) / trivalent (p)
Explanation
Extrinsic doping enables practical devices; intrinsic ni sets mass-action law nenh = ni² (L28 §28.1).
Problem Solving — L28 Semiconductors and Devices
Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). Explanations open by default.
Distinguish conductor, insulator, semiconductor by band gap qualitatively.
Solution — step by step with formulas
- Conductor: overlapping/partial band; insulator: large gap; semiconductor: small gap (~1 eV).
Final answer: Small E_g for semiconductors
Textbook formal language
Electrical behaviour follows availability of states and gap between valence and conduction bands.
Working formula set for this problem: (see solution steps). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Metals always have free electrons; glass has a huge jump; silicon has a small jump heat/light can bridge.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Energy bands
Si, Ge classic elemental semiconductors.
Link to chapter notes (L28 — Energy bands): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: (see solution steps). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write (see solution steps) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
What are n-type and p-type semiconductors? Name typical dopants for Si.
Solution — step by step with formulas
- n-type: donor impurities (P, As) extra electrons.
- p-type: acceptor (B, Al) create holes.
Final answer: n: pentavalent donors; p: trivalent acceptors
Textbook formal language
Doping controls majority carrier type and conductivity.
Working formula set for this problem: (see solution steps). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Add phosphorus → extra electrons (n). Add boron → missing electrons/holes (p).
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Doping
n_i² = n p still holds in thermal equilibrium.
Link to chapter notes (L28 — Doping): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: (see solution steps). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write (see solution steps) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Explain depletion layer formation at a p–n junction.
Solution — step by step with formulas
- Diffusion of e⁻ and holes across junction leaves uncovered ions ⇒ depletion + built-in potential.
Final answer: Diffusion creates ion space charge region
Textbook formal language
Built-in field opposes further diffusion at equilibrium.
Working formula set for this problem: (see solution steps). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Electrons wander into p side, holes into n; leftover ions make a quiet charged zone.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — p–n junction
Bias changes barrier: forward shrinks, reverse widens.
Link to chapter notes (L28 — p–n junction): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: (see solution steps). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write (see solution steps) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Why does a p–n diode conduct preferentially in forward bias?
Solution — step by step with formulas
- Forward bias lowers barrier; majority carriers injected; large current. Reverse raises barrier.
Final answer: Forward: low barrier; reverse: high barrier
Textbook formal language
Asymmetric I–V characteristic enables rectification of AC.
Working formula set for this problem: (see solution steps). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
One way street for current—push the right way and it flows; opposite way almost blocks.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Diode rectifier
Half-wave uses one diode; full-wave uses bridge/centre tap.
Link to chapter notes (L28 — Diode rectifier): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: (see solution steps). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write (see solution steps) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
State the main use of a Zener diode and the bias condition for regulation.
Solution — step by step with formulas
- Voltage regulation in reverse breakdown (Zener/avalanche region).
Final answer: Reverse breakdown for voltage reference/regulation
Textbook formal language
Sharp reverse characteristic holds nearly constant V over current range.
Working formula set for this problem: (see solution steps). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Wired backward on purpose to clamp voltage at a fixed value.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Zener diode
Needs series resistor to limit current.
Link to chapter notes (L28 — Zener diode): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: (see solution steps). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write (see solution steps) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
In CE amplifier, what roles do base and collector currents play qualitatively?
Solution — step by step with formulas
- Small base current controls large collector current; current gain β = I_C/I_B.
Final answer: Small I_B controls large I_C (β)
Textbook formal language
BJT uses carrier injection and collection between junctions.
Working formula set for this problem: (see solution steps). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
A tiny tap (base) steers a strong stream (collector)—that’s amplification.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Transistor as amplifier idea
Operating point set by biasing network.
Link to chapter notes (L28 — Transistor as amplifier idea): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: (see solution steps). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write (see solution steps) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).