Semiconductor Electronics

JKBOSE/CBSE
Semiconductor Electronics Class 12 Notes
Semiconductor Electronics - Notes

Semiconductor Electronics

This chapter introduces the behavior of semiconductors, p-n junctions, and their applications in electronic devices.

1. Classification of Solids

  • Metals: Very high conductivity, low resistivity (10-2 to 10-8 Ωm). Band gap is zero.
  • Insulators: Very low conductivity, high resistivity (1011 to 1019 Ωm). Large band gap (> 3 eV).
  • Semiconductors: Conductivity intermediate (10-5 to 106 Ωm). Small band gap (< 3 eV).

2. Intrinsic Semiconductors

  • Pure semiconductors (e.g., Si, Ge).
  • Number of electrons (ne) = Number of holes (nh) = ni.
  • Current I = Ie + Ih.

3. Extrinsic Semiconductors

Formed by adding small amount of impurity (Doping).

  • n-type: Doped with Pentavalent impurities (P, As, Sb). ne >> nh.
  • p-type: Doped with Trivalent impurities (B, Al, Ga, In). nh >> ne.
  • Mass Action Law: ne nh = ni2.

4. p-n Junction

Formed when p-type and n-type materials are joined.

  • Depletion Region: Region near junction devoid of mobile charge carriers.
  • Barrier Potential: Built-in electric field that opposes further diffusion of carriers.

5. Biasing of p-n Junction

  • Forward Bias: P connected to positive, N to negative. Depletion layer width decreases. Current flows easily.
  • Reverse Bias: P connected to negative, N to positive. Depletion layer width increases. Very small leakage current.

6. Semiconductor Diode as Rectifier

Device used to convert AC into DC.

  • Half-wave Rectifier: Conducts only during one half cycle of AC. Efficiency ≈ 40.6%.
  • Full-wave Rectifier: Conducts during both half cycles. Efficiency ≈ 81.2%.

7. Special Purpose Diodes

  • Zener Diode: Operates in reverse breakdown region. Used as Voltage Regulator.
  • Photodiode: Operated in reverse bias. Current increases when light falls on it.
  • LED: Forward biased. Emits light when electrons recombine with holes.
  • Solar Cell: Generates emf when solar radiation falls on it. No external bias.
Semiconductors - Numericals

Numericals: Semiconductor Electronics

Q1: In an intrinsic semiconductor, the number of conduction electrons is 7 × 1019 per cubic meter. Find the total number of current carriers in a piece of size 1 cm × 1 cm × 1 mm.

Given: ne = 7 × 1019 m-3, V = 10-2 × 10-2 × 10-3 = 10-7 m3
Formula: Total carriers = (ne + nh) × V; For intrinsic ne = nh.
Solution:
Total carrier density = 2 × 7 × 1019 = 1.4 × 1020 m-3.
Total carriers = 1.4 × 1020 × 10-7 = 1.4 × 1013.
Answer: 1.4 × 1013
Q2: A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?

Given: Eg = 2.8 eV, λ = 6000 nm = 6000 × 10-9 m
Formula: E = hc / λ
Solution:
E = (6.63 × 10-34 × 3 × 108) / (6000 × 10-9 × 1.6 × 10-19) eV
E ≈ 0.207 eV.
Since E < Eg (0.207 < 2.8), the photodiode cannot detect this wavelength.
Answer: No
Q3: In a p-n junction diode, the current I can be expressed as I = I0 [ exp(eV / 2kBT) - 1 ]. If I0 = 5 × 10-12 A and T = 300 K, what will be the forward current at a forward voltage of 0.6 V?

Given: I0 = 5e-12 A, V = 0.6 V, T = 300 K
Solution:
eV / kBT ≈ 0.6 / 0.026 ≈ 23.07 (using kBT/e ≈ 26 mV at 300K).
I = 5 × 10-12 [ exp(23.07 / 2) - 1 ]
I = 5 × 10-12 [ exp(11.535) - 1 ] ≈ 5 × 10-12 × 1.02 × 105 ≈ 5.1 × 10-7 A.
Answer: 0.51 μA
Q4: Find the maximum frequency of light emitted by a GaAs LED with a bandgap of 1.42 eV.

Given: Eg = 1.42 eV
Formula: Eg = hν
Solution:
ν = Eg / h = (1.42 × 1.6 × 10-19) / (6.63 × 10-34)
ν ≈ 3.43 × 1014 Hz.
Answer: 3.43 × 1014 Hz
Q5: For a Zener diode voltage regulator, given Vin = 20 V, Vz = 10 V, Rs = 100 Ω. Find the Zener current if load resistance RL is 1 kΩ.

Given: Vin = 20, Vz = 10, Rs = 100, RL = 1000
Formula: Is = (Vin - Vz) / Rs; IL = Vz / RL; Iz = Is - IL
Solution:
Is = (20 - 10) / 100 = 0.1 A = 100 mA.
IL = 10 / 1000 = 0.01 A = 10 mA.
Iz = 100 - 10 = 90 mA.
Answer: 90 mA
Q6: Conductivity of intrinsic semiconductor increases with? (Answer: Temperature)
Answer: Temperature
Q7: Majority carriers in p-type semiconductor. (Answer: Holes)
Answer: Holes
Q8: Ripple frequency of full wave rectifier with 50 Hz input. (Answer: 100 Hz)
Answer: 100 Hz
Q9: Barrier potential of Silicon diode. (Answer: 0.7 V)
Answer: 0.7 V
Q10: Ratio of ne/nh in n-type semiconductor. (Answer: >> 1)
Answer: Much greater than 1
Q11: Depletion layer width in forward bias? (Answer: Decreases)
Answer: Decreases
Q12: Band gap of Silicon at 300 K. (Answer: 1.1 eV)
Answer: 1.12 eV
Q13: Mobility of electrons vs holes. (Answer: μe > μh)
Answer: Electrons have higher mobility
Q14: Dynamic resistance rd formula.
Answer: ΔV / ΔI
Q15: Which bias is used for Zener diode as regulator? (Answer: Reverse Bias)
Answer: Reverse Bias
Q16: Pentavalent impurity example. (Answer: Phosphorus, Arsenic)
Answer: Phosphorus (P)
Q17: If ni = 1016 and ne = 1020, find nh.
Formula: ni2 / ne = 1032 / 1020 = 1012
Answer: 1012 m-3
Q18: Knee voltage is also called? (Answer: Cut-in/Barrier voltage)
Answer: Cut-in Voltage
Q19: Efficiency of Half wave rectifier. (Answer: 40.6%)
Answer: 40.6%
Q20: Solar cells work on which principle? (Answer: Photovoltaic effect)
Answer: Photovoltaic effect
Notes

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