Semiconductor Devices
Kerala Board · Class 12 · Physics
Flashcards for Semiconductor Devices — Kerala Board Class 12 Physics. Quick Q&A cards covering key concepts, definitions, and formulas.
What are the three categories of materials based on resistivity values? Give their typical resistivity ranges.
Answer
Based on resistivity (ρ), materials are classified as: 1. **Metals**: ρ = 10⁻² – 10⁻⁸ Ωm (high conductivity) 2. **Semiconductors**: ρ = 10⁻⁵ – 10⁶ Ωm (intermediate conductivity) 3. **Insulators**: ρ =
Define valence band, conduction band, and forbidden gap in energy band theory.
Answer
**Valence Band**: Energy band formed by grouping energy levels of valence electrons (outermost orbit electrons). Present below conduction band. **Conduction Band**: Energy band formed by grouping ene
How do the energy band diagrams differ for metals, semiconductors, and insulators?
Answer
**Metals**: Conduction and valence bands overlap → electrons easily move to conduction band → high conductivity **Semiconductors**: Small band gap (Eg ≈ 1 eV) → some electrons can jump to conduction
What is the difference between intrinsic and extrinsic semiconductors?
Answer
**Intrinsic Semiconductor**: - Pure silicon or germanium - No impurities added - Electrons and holes always equal in number - Limited practical use due to high resistivity **Extrinsic Semiconductor**
Explain n-type and p-type semiconductors with examples of dopants.
Answer
**n-type Semiconductor**: - Doped with pentavalent (5 valence electrons) elements - Dopants: Phosphorus (P), Arsenic (As), Antimony (Sb) - Extra electron becomes free → electrons are majority carriers
How is a p-n junction formed and what is the depletion region?
Answer
**p-n Junction Formation**: - n-type and p-type regions created on same crystal - Electrons diffuse from n-region to p-region - Holes diffuse from p-region to n-region - Recombination occurs at juncti
What happens when a p-n junction is forward biased and reverse biased?
Answer
**Forward Bias** (p-side +ve, n-side -ve): - Reduces barrier potential - Large current flows (mA range) - Low resistance (10-30 Ω) - Knee voltage: ~0.7V (Si), ~0.3V (Ge) **Reverse Bias** (p-side -ve,
Calculate the dc voltage and current for a half-wave rectifier if peak AC voltage is 10V and load resistance is 100Ω.
Answer
**Given**: Vm = 10V, RL = 100Ω **For Half-Wave Rectifier**: DC Voltage: Vdc = Vm/π Vdc = 10V/π = 10/3.14 = 3.18V DC Current: Idc = Vdc/RL Idc = 3.18V/100Ω = 0.0318A = 31.8 mA **Note**: Half-wave r
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Sources & Official References
- Kerala Board of Public Examinations — keralapareekshabhavan.in
- National Education Policy 2020 — education.gov.in
Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.
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