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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.

Fig 28.1 — Energy Bands in Solids CB overlaps VB Conductor Eg ≤ 3eV Semiconductor Eg > 3eV Insulator At 0 K intrinsic semiconductor acts as insulator · NTC of resistance
Fig 28.1 — Band gap determines conduction type: overlap, small gap, or large gap

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).

Fig 28.6–28.7 — p-n Junction & Depletion Region p-region (holes) n-region (e⁻) depletion ⊖ fixed ions ⊕ fixed ions Unidirectional conduction · diode = one-way for electrons
Fig 28.6–28.7 — Diffusion creates depletion layer and barrier potential at junction

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.

Fig 28.9–28.10 — Forward & Reverse Bias I–V V 0.7V knee Forward: conducts · Reverse: blocks until breakdown
Fig 28.10–28.11 — Diode conducts mainly in forward direction above knee voltage

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).

Fig 28.16–28.19 — npn Transistor (Active Bias) E (n) B (p) C (n) V_BE fwd V_CB rev Cannot make transistor from two separate diodes — base must be single thin shared region
Fig 28.16–28.19 — npn transistor: E-B forward, C-B reverse for amplification
α = IC/IE  |  β = IC/IB  |  IE = IC + IB
α (common-base current gain) ≈ 0.95–0.99, always < 1
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
         ==============================
    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.
20 cards · click any card to flip
Energy bands classification
Conductor: CB overlaps VB. Semiconductor: empty CB, full VB, Eg ≤ 3 eV. Insulator: Eg > 3 eV. Bands form from interaction of ~10²³ atoms/cm³.
Intrinsic semiconductor
Pure Si/Ge. e⁻ and holes generated in pairs thermally. Equal concentrations. Insulator at 0 K. Negative temperature coefficient of resistance (conductivity rises with T).
n-type semiconductor
Doped with pentavalent (Group V) donor: P, As, Sb. Fifth electron free for conduction. Electrons = majority carriers. Material remains electrically neutral.
p-type semiconductor
Doped with trivalent (Group III) acceptor: B, Al, Ga. Creates holes. Holes = majority carriers. More holes than thermally generated e⁻. Electrically neutral overall.
Doping
Adding ~1 impurity atom per 10⁸ host atoms. Replaces lattice sites. Converts intrinsic to extrinsic. Lowers resistivity, increases useful conductivity.
Depletion region
At p-n junction: mobile carriers recombine; immobile donor (+) and acceptor (−) ions remain. ~0.5 μm thick. "Depleted" of mobile charges, not of net charge.
Barrier potential
Built-in potential across depletion layer opposes diffusion. Si ~0.7 V, Ge ~0.3 V. Must be overcome in forward bias for significant current (knee voltage).
Forward bias
p to +, n to −. Depletion region narrows. Current flows when V > knee voltage. Low resistance (~10–30 Ω). Majority carriers cross junction.
Reverse bias
p to −, n to +. Depletion widens. Small leakage current (minority carriers). High resistance. Breakdown at high V: avalanche or Zener effect.
Knee voltage
Forward voltage to start significant conduction. Si ~0.7 V, Ge ~0.3 V. Below this, negligible forward current. I-V curve does not pass through origin.
Zener diode
Heavily doped p-n junction. Operates in reverse breakdown (Zener effect). Maintains nearly constant voltage V_Z. Used for voltage stabilization/regulation.
LED and photodiode
LED: forward bias, electroluminescence, Group III–V (GaAsP). Photodiode: reverse bias, light generates e⁻–h⁺ pairs, photocurrent ∝ intensity. Solar cell: photovoltaic, no external bias needed.
Transistor structure
E (emitter): heaviest doping. B (base): thinnest, lightest doping. C (collector): largest size, moderate doping. Two junctions, three regions. npn or pnp.
Transistor active region bias
Emitter-base: forward biased (small V_BE ~0.6–0.7 V Si). Collector-base: reverse biased (larger V_CB). >95% emitter carriers collected; IC ≈ IE.
Current gains α and β
α = IC/IE (common base), <1. β = IC/IB (common emitter), >>1. IE = IC + IB. Relation: β = α/(1−α). α=0.98 → β=49.
Transistor configurations
CE (common emitter): best voltage/current gain, most used. CB (common base): constant current. CC (common collector): impedance matching. Each has unique input/output characteristics.
Why two diodes ≠ transistor?
Two discrete diodes = 4 doped regions with thick base between. Real transistor needs one thin lightly doped base shared by E and C for carrier transfer and gain.
Avalanche vs Zener breakdown
Avalanche: minority carriers accelerated, collide, create more carriers (chain). Zener: strong field breaks covalent bonds in depletion region. Both cause sharp reverse current increase.
Hole motion and current
Hole shifts when electron jumps from neighbouring bond. Conventional current direction = hole drift direction. In intrinsic material, e⁻ and holes always created in pairs.
Diode applications
Rectification (AC to DC), battery reverse-protection, adapters/chargers, detection. Unidirectional conduction — one-way turnstile for electrons. Basis for all digital electronics with transistors.

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:

Semiconductor: Eg ≤ 3 eV
Conductor: CB overlaps VB
Intrinsic: n_e = n_h
n-type: Group V donor · p-type: Group III acceptor
Barrier: Si 0.7 V · Ge 0.3 V
Forward bias: p → +, n → −
Reverse bias: leakage · high R
α = I_C / I_E
β = I_C / I_B
β = α / (1 − α)
I_E = I_C + I_B
Active: E-B fwd · C-B rev

1. Formulas & Definitions

Full Ch 28 study guide — energy bands, doping, p-n junction, diodes, and transistors.

Semiconductor: Eg ≤ 3 eV

Definition: Band-gap criterion for semiconductors vs insulators.

Derivation

Empty conduction band, filled valence band at 0 K; small gap allows thermal excitation at room T.

Variables

Eg = forbidden energy gap · CB = conduction band · VB = valence band

Why it works

Classifies Si, Ge as semiconductors; distinguishes from insulators (large Eg).

Historical context

Bands form from ~10²³ interacting atoms/cm³ in crystals.

Deep understanding

At 0 K intrinsic semiconductor acts as insulator; NTC — resistance falls as T rises.

2. Diagrams & Visuals

Eg ≤ 3 eV

Color-coded visual · step-by-step breakdown below

  1. Draw CB and VB separated by Eg
  2. Check gap size
  3. Eg ≤ 3 eV → semiconductor
  4. Eg > 3 eV → insulator

3. Solved Examples

Basic

Q: Si band gap?

Solution: ≤ 3 eV

Answer: Semiconductor

Intermediate

Q: Insulator Eg?

Solution: > 3 eV

Answer: > 3 eV

Advanced

Q: 0 K behaviour?

Solution: insulator-like

Answer: No free carriers

Exam

Q: Semiconductor criterion?

Solution: Eg ≤ 3 eV

Answer: Sec 28.1

Conductor: CB overlaps VB

Definition: Metals and conductors — no forbidden gap.

Derivation

Conduction band partially filled or overlaps valence band; electrons free at all temperatures.

Variables

Eg ≈ 0 for conductors

Why it works

Explains high conductivity of metals vs semiconductors.

Historical context

Quasi-continuous bands replace discrete atomic levels in solids.

Deep understanding

Semiconductor at high T can conduct; conductor always conducts.

2. Diagrams & Visuals

CB overlaps VB

Color-coded visual · step-by-step breakdown below

  1. Identify material type
  2. Check if bands overlap
  3. Overlap → conductor
  4. Small gap → semiconductor

3. Solved Examples

Basic

Q: Copper bands?

Solution: overlap

Answer: Conductor

Intermediate

Q: Eg for metal?

Solution: ~0

Answer: Zero gap

Advanced

Q: vs Si?

Solution: Si has gap

Answer: Semiconductor

Exam

Q: Conductor bands?

Solution: CB overlaps VB

Answer: Sec 28.1

Intrinsic: n_e = n_h

Definition: Pure semiconductor — equal electron and hole concentrations.

Derivation

Thermal energy creates e⁻–h⁺ pairs; each broken bond gives one electron and one hole.

Variables

Ge at 300 K: ~2.5×10¹⁹ m⁻³ pairs

Why it works

Baseline before doping; explains low conductivity of pure Si/Ge at room T.

Historical context

Intrinsic = undoped Si or Ge crystal.

Deep understanding

Holes move opposite to electron drift; conventional current follows hole direction.

2. Diagrams & Visuals

n_e = n_h (pairs)

Color-coded visual · step-by-step breakdown below

  1. Start with pure crystal
  2. Thermal excitation creates pairs
  3. Count electrons and holes
  4. Concentrations equal

3. Solved Examples

Basic

Q: Pure Si carriers?

Solution: e⁻ and holes equal

Answer: n_e = n_h

Intermediate

Q: One pair broken?

Solution: 1 e⁻ + 1 hole

Answer: Always paired

Advanced

Q: Doped material?

Solution: not intrinsic

Answer: Extrinsic

Exam

Q: Intrinsic condition?

Solution: n_e = n_h

Answer: Sec 28.2

n-type: Group V donor · p-type: Group III acceptor

Definition: Extrinsic semiconductors from pentavalent and trivalent doping.

Derivation

Donor (P, As, Sb) adds free e⁻; acceptor (B, Al, Ga) creates holes. ~1 dopant per 10⁸ atoms.

Variables

n-type: majority e⁻ · p-type: majority holes · material stays neutral

Why it works

Doping controls conductivity without adding net charge.

Historical context

Replaces lattice site of host Si/Ge atom.

Deep understanding

Neither n-type nor p-type is charged — ions balance carriers.

2. Diagrams & Visuals

n-type (V) p-type (III)

Color-coded visual · step-by-step breakdown below

  1. Identify dopant group
  2. Group V → n-type (donor)
  3. Group III → p-type (acceptor)
  4. Majority carrier type

3. Solved Examples

Basic

Q: Phosphorus in Si?

Solution: n-type

Answer: Donor

Intermediate

Q: Boron in Ge?

Solution: p-type

Answer: Acceptor

Intermediate

Q: Pentavalent?

Solution: n-type

Answer: Extra e⁻

Exam

Q: n-type dopant?

Solution: Group V

Answer: Sec 28.2

Barrier: Si 0.7 V · Ge 0.3 V

Definition: Built-in potential across p-n junction depletion region.

Derivation

Diffusion of e⁻ and holes → recombination → immobile ions → electric field opposes further diffusion.

Variables

Depletion width ~0.5 μm · knee voltage ≈ barrier potential

Why it works

Must be overcome in forward bias for significant diode current.

Historical context

Depletion region depleted of mobile carriers, not net charge.

Deep understanding

Forward bias narrows depletion; reverse bias widens it.

2. Diagrams & Visuals

Si: 0.7 V Ge: 0.3 V

Color-coded visual · step-by-step breakdown below

  1. Identify semiconductor (Si or Ge)
  2. Si barrier ≈ 0.7 V
  3. Ge barrier ≈ 0.3 V
  4. Compare to applied forward V

3. Solved Examples

Basic

Q: Si knee voltage?

Solution: ~0.7 V

Answer: 0.7 V

Intermediate

Q: Ge barrier?

Solution: ~0.3 V

Answer: 0.3 V

Advanced

Q: Below knee?

Solution: negligible I

Answer: No conduction

Exam

Q: Si barrier potential?

Solution: 0.7 V

Answer: Sec 28.3

Forward bias: p → +, n → −

Definition: Diode conducting configuration — external voltage opposes barrier.

Derivation

Majority carriers pushed toward junction; depletion narrows; current flows when V > knee.

Variables

R_fwd ~ 10–30 Ω · current ~ mA

Why it works

Enables rectification, LED emission, transistor E-B forward bias.

Historical context

I-V curve does not pass through origin — offset by knee voltage.

Deep understanding

Diode arrow shows conventional current direction (p → n).

2. Diagrams & Visuals

p (+) n (−)

Color-coded visual · step-by-step breakdown below

  1. Connect p-side to positive terminal
  2. Connect n-side to negative
  3. V > knee voltage
  4. Depletion shrinks, current flows

3. Solved Examples

Basic

Q: p-side connection?

Solution: positive

Answer: + terminal

Intermediate

Q: Depletion width?

Solution: decreases

Answer: Narrows

Advanced

Q: Below 0.7 V Si?

Solution: little current

Answer: Knee not reached

Exam

Q: Forward bias?

Solution: p to +, n to −

Answer: Sec 28.4

Reverse bias: leakage · high R

Definition: Diode blocking configuration — majority carriers move away from junction.

Derivation

Depletion widens; only minority carriers contribute tiny reverse saturation (leakage) current.

Variables

Ge leakage ~ μA · Si leakage ~ nA · breakdown at high |V|

Why it works

Used in photodiodes, Zener regulators; blocks AC negative half-cycle in rectifiers.

Historical context

Breakdown: avalanche (carrier multiplication) or Zener (field ionization).

Deep understanding

At breakdown voltage, reverse current rises sharply.

2. Diagrams & Visuals

p → − · n → + · I ≈ I_s

Color-coded visual · step-by-step breakdown below

  1. Connect p to negative, n to positive
  2. Depletion region widens
  3. Small leakage current flows
  4. Breakdown at high reverse V

3. Solved Examples

Basic

Q: Depletion in reverse?

Solution: widens

Answer: Increases

Intermediate

Q: Si leakage scale?

Solution: nA

Answer: Very small

Advanced

Q: Photodiode bias?

Solution: reverse

Answer: Reverse bias

Exam

Q: Reverse bias effect?

Solution: high R

Answer: Sec 28.5

α = I_C / I_E

Definition: Common-base current gain — fraction of emitter current reaching collector.

Derivation

In active region, >95% emitter carriers cross thin base to collector; small fraction to base.

Variables

α ≈ 0.95–0.99 · always α < 1

Why it works

Describes transistor action in CB configuration.

Historical context

BJT invented 1948 — Bardeen, Brattain, Shockley.

Deep understanding

α close to 1 because base is thin and lightly doped.

2. Diagrams & Visuals

α = I_C / I_E

Color-coded visual · step-by-step breakdown below

  1. Bias E-B forward, C-B reverse
  2. Measure I_E and I_C
  3. α = I_C / I_E
  4. Typically 0.95–0.99

3. Solved Examples

Basic

Q: α range?

Solution: 0.95–0.99

Answer: < 1

Intermediate

Q: α = 0.98?

Solution: 98% to collector

Answer: 0.98

Advanced

Q: α can exceed 1?

Solution: No

Answer: Always < 1

Exam

Q: Common-base gain?

Solution: I_C/I_E

Answer: Sec 28.8

β = I_C / I_B

Definition: Common-emitter current gain — large amplification factor.

Derivation

Small base current controls large collector current in CE configuration.

Variables

β >> 1 · example α=0.98 → β=49

Why it works

CE is most used configuration — voltage and current gain.

Historical context

β relates to α: β = α/(1−α).

Deep understanding

I_B very small because base thin; most I_E becomes I_C.

2. Diagrams & Visuals

β = I_C / I_B

Color-coded visual · step-by-step breakdown below

  1. Set active region bias
  2. Measure I_B and I_C
  3. β = I_C / I_B
  4. Typically >> 1

3. Solved Examples

Basic

Q: β magnitude?

Solution: >> 1

Answer: Large

Intermediate

Q: I_B small?

Solution: yes

Answer: Thin base

Advanced

Q: CE vs CB gain?

Solution: β >> α

Answer: β larger

Exam

Q: Common-emitter gain?

Solution: I_C/I_B

Answer: Sec 28.8

β = α / (1 − α)

Definition: Relation between common-base and common-emitter current gains.

Derivation

From I_E = I_C + I_B and α = I_C/I_E, solve for β = I_C/I_B.

Variables

α = 0.98 → β = 0.98/0.02 = 49

Why it works

Converts between CB and CE gain values in problems.

Historical context

Standard BJT parameter relation in all textbooks.

Deep understanding

As α → 1, β → very large.

2. Diagrams & Visuals

β = α / (1 − α)

Color-coded visual · step-by-step breakdown below

  1. Know α value
  2. Compute 1 − α
  3. β = α / (1 − α)
  4. Check with I_C, I_B if given

3. Solved Examples

Basic

Q: α = 0.98?

Solution: β = 49

Answer: 49

Intermediate

Q: α = 0.95?

Solution: β = 19

Answer: 19

Advanced

Q: α = 0.99?

Solution: β = 99

Answer: 99

Exam

Q: β from α?

Solution: α/(1−α)

Answer: Sec 28.8

I_E = I_C + I_B

Definition: Emitter current equals sum of collector and base currents.

Derivation

Charge conservation at transistor terminals in active region.

Variables

I_E ≈ I_C when β >> 1

Why it works

Fundamental BJT current relation for all configurations.

Historical context

Applies to npn and pnp with proper current directions.

Deep understanding

Two separate diodes cannot replace transistor — need one thin shared base.

2. Diagrams & Visuals

I_E = I_C + I_B

Color-coded visual · step-by-step breakdown below

  1. Identify I_E, I_C, I_B
  2. Add I_C + I_B
  3. Equals I_E
  4. Use with α or β

3. Solved Examples

Basic

Q: I_E = 10 mA, I_C = 9.8 mA?

Solution: I_B = 0.2 mA

Answer: 0.2 mA

Intermediate

Q: β large?

Solution: I_E ≈ I_C

Answer: ≈ equal

Advanced

Q: Missing I_B?

Solution: I_E − I_C

Answer: Subtract

Exam

Q: Transistor currents?

Solution: I_E = I_C + I_B

Answer: Sec 28.8

Active: E-B fwd · C-B rev

Definition: Transistor biasing for amplification — emitter-base forward, collector-base reverse.

Derivation

Forward E-B injects carriers; reverse C-B collects them at collector.

Variables

V_BE ~ 0.6–0.7 V (Si) · V_CB reverse · R_ie ~ 20–100 Ω (npn CE)

Why it works

Required for transistor action; wrong bias → cutoff or saturation.

Historical context

E heaviest doped, B thinnest/lightest, C largest/moderate doping.

Deep understanding

CE input like forward diode (V_BE vs I_B); output I_C vs V_CE at fixed I_B.

2. Diagrams & Visuals

E-B fwd C-B rev

Color-coded visual · step-by-step breakdown below

  1. Forward bias emitter-base junction
  2. Reverse bias collector-base junction
  3. Carriers injected at E
  4. >95% collected at C

3. Solved Examples

Basic

Q: E-B junction bias?

Solution: forward

Answer: Forward

Intermediate

Q: C-B junction?

Solution: reverse

Answer: Reverse

Advanced

Q: Two diodes work?

Solution: No

Answer: Need thin base

Exam

Q: Active region bias?

Solution: E-B fwd, C-B rev

Answer: Sec 28.7

5. Special Features & Extras

Complete study guide for Semiconductors and Semiconducting Devices.

Exam Tips & Tricks

  • Classification: conductor (overlap) · semiconductor (Eg ≤ 3 eV) · insulator (Eg > 3 eV).
  • Doping: Group V → n-type · Group III → p-type · ~1 per 10⁸ atoms.
  • Barrier/knee: Si 0.7 V · Ge 0.3 V — I-V curve offset from origin.
  • Forward bias: p→+, n→− · depletion shrinks · low R (~10–30 Ω).
  • Reverse bias: depletion widens · leakage (nA Si, μA Ge) · breakdown at high V.
  • Special diodes: Zener (regulator) · LED (forward) · photodiode (reverse) · solar cell (photovoltaic).
  • Transistor active: E-B forward + C-B reverse · CE most common.
  • Gains: α = I_C/I_E < 1 · β = I_C/I_B >> 1 · β = α/(1−α).
  • Two diodes ≠ transistor — need one thin shared base.

Common Student Mistakes

  • Confusing Si (0.7 V) and Ge (0.3 V) barrier potentials
  • Thinking n-type or p-type material has net electric charge
  • Reversing forward/reverse bias connections
  • Assuming depletion region has no ions (it has immobile ions)
  • Using β > 1 to mean α > 1 (α is always < 1)
  • Believing two back-to-back diodes make a working transistor

Memory Aids & Mnemonics

Doping: "Pentavalent gives electrons — n-type; Trivalent takes — p-type holes"
Forward bias: "Plus on p, minus on n — diode conducts"
Barrier: "Si seven-tenths, Ge three-tenths"
Transistor active: "Emitter kisses base (fwd), Collector blocks base (rev)"

Which Formula When?

  • Material type? → band gap Eg (≤3 eV = semiconductor)
  • Majority carrier? → Group V = e⁻ (n) · Group III = holes (p)
  • Diode knee voltage? → Si 0.7 V · Ge 0.3 V
  • Depletion width change? → forward shrinks · reverse widens
  • CB gain? → α = I_C/I_E
  • CE gain? → β = I_C/I_B or β = α/(1−α)
  • Current balance? → I_E = I_C + I_B
  • Amplification bias? → E-B forward, C-B reverse

QUICK REFERENCE — Ch 28 Semiconductors & Devices

Semiconductor: Eg ≤ 3 eVConductor: CB overlaps VBIntrinsic: n_e = n_hn-type: Group V donor · p-type: Group III acceptorBarrier: Si 0.7 V · Ge 0.3 VForward bias: p → +, n → −Reverse bias: leakage · high Rα = I_C / I_Eβ = I_C / I_Bβ = α / (1 − α)I_E = I_C + I_BActive: E-B fwd · C-B rev

Diode: forward p→+ n→− · reverse leakage · Zener in breakdown

BJT: α = I_C/I_E · β = I_C/I_B · β = α/(1−α) · I_E = I_C + I_B

Tip: CE configuration gives best voltage + current gain — most common in exams.

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

1 mark · Section A Q16 · 312/TUS/104A

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.

2 marks · Section A Q25 · 68/ESS/1-312-A

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.

2 marks · Section A Q27 · 68/ESS/1-312-A

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.

2 marks · Section A Q28 · 68/ESS/1-312-A

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?

2 marks · Section B Q30 · 312/MAY/204A (also Q29 · 204B, Q31 · 204C)

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.

2 marks · Section B Q37 · 312/MAY/204A

Model Answer

AC mains → step-down transformerfull-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.

2 marks · Section B Q37 · 312/MAY/204B

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.

2 marks · Section B Q37 · 312/MAY/204C

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.

2 marks · Section B Q31 · 312/TUS/104A (OR: n-p-n in CB and CE configurations)

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.

2 marks · Section B Q29 · 68/ESS/1-312-A (Marking Scheme)

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.

3 marks · Section B Q40 · 312/MAY/204A

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.)

3 marks · Section B Q40 · 312/MAY/204B

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.

Question 1 of 6Band

Distinguish conductor, insulator, semiconductor by band gap qualitatively.

Solution — step by step with formulas

  1. 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).
Question 2 of 6Intrinsic/extrinsic

What are n-type and p-type semiconductors? Name typical dopants for Si.

Solution — step by step with formulas

  1. n-type: donor impurities (P, As) extra electrons.
  2. 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).
Question 3 of 6Diode

Explain depletion layer formation at a p–n junction.

Solution — step by step with formulas

  1. 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).
Question 4 of 6Rectifier

Why does a p–n diode conduct preferentially in forward bias?

Solution — step by step with formulas

  1. 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).
Question 5 of 6Zener

State the main use of a Zener diode and the bias condition for regulation.

Solution — step by step with formulas

  1. 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).
Question 6 of 6Transistor

In CE amplifier, what roles do base and collector currents play qualitatively?

Solution — step by step with formulas

  1. 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).