Semiconductor Electronics

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Exercise 14.1Semiconductors, p-n Junction and Logic Gates

In an n-type silicon, which of the following statement is true: (a) Electrons are majority carriers and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants.

Solution

Answer: (c) Holes are minority carriers and pentavalent atoms are the dopants. Explanation: • n-type semiconductor is doped with pentavalent atoms (e.g., As, P, Sb) • Pentavalent atoms donate one extra electron, increasing the free electron concentration • Electrons are majority carriers (large concentration) • Holes are minority carriers (small concentration from thermal excitation) Options (a) is wrong because trivalent atoms create p-type, not n-type. Options (b) is wrong because electrons are majority, not minority carriers in n-type. Option (d) describes p-type semiconductor.

Which of the statements given in Exercise 14.1 is true for p-type semiconductors?

Solution

For p-type semiconductors, statement (d) is true: Holes are majority carriers and trivalent atoms are the dopants. Explanation: • p-type semiconductor is doped with trivalent atoms (e.g., B, Al, Ga, In) • Trivalent atoms accept one electron from silicon, creating holes • Holes are majority carriers • Electrons are minority carriers • Dopants are trivalent (3 valence electrons, need 4 to bond with silicon)

Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (E_g)C, (E_g)Si and (E_g)Ge. Which of the following statements is true? (a) (E_g)Si < (E_g)Ge < (E_g)C (b) (E_g)C < (E_g)Si < (E_g)Ge (c) (E_g)C > (E_g)Si > (E_g)Ge (d) (E_g)C = (E_g)Si = (E_g)Ge

Solution

Answer: (c) (E_g)C > (E_g)Si > (E_g)Ge Actual values: • Carbon (diamond): E_g ≈ 5.4 eV (insulator) • Silicon: E_g ≈ 1.1 eV (semiconductor) • Germanium: E_g ≈ 0.7 eV (semiconductor) The band gap decreases as we go down Group 14 (C → Si → Ge) because: • Atomic size increases • Valence electrons are farther from nucleus • Weaker bonding → smaller band gap Ge has the smallest band gap, making it a better conductor than Si at room temperature.

In an unbiased p-n junction, holes diffuse from the p-region to n-region because: (a) free electrons in the n-region attract them (b) they move only from high to low potential (c) hole concentration in p-region is more as compared to n-region (d) All the above.

Solution

Answer: (c) hole concentration in p-region is more as compared to n-region. Explanation: Diffusion occurs due to a concentration gradient — particles move from higher concentration to lower concentration region. In a p-n junction: • p-region has high hole concentration and low electron concentration • n-region has low hole concentration and high electron concentration Holes diffuse from p to n because of the concentration gradient (c). Option (a) is incorrect: free electrons don't attract holes across the junction in this way. Option (b) is incorrect: holes move toward lower potential only in drift, not diffusion.

When a forward bias is applied to a p-n junction, it (a) raises the potential barrier (b) reduces the majority carrier current to zero (c) lowers the potential barrier (d) None of the above.

Solution

Answer: (c) lowers the potential barrier. Explanation: In an unbiased p-n junction, a built-in potential barrier (~0.7 V for silicon, ~0.3 V for germanium) exists at the junction. When forward bias is applied: • The positive terminal is connected to p-side and negative to n-side • External voltage opposes the built-in potential • The potential barrier is reduced • Majority carrier current increases significantly Forward bias does NOT eliminate majority carrier current — it actually greatly increases it once the barrier is overcome.

For transistor action, which of the following statements are correct? (a) Base, emitter and collector regions should have similar size and doping concentrations. (b) The base region must be very thin and lightly doped. (c) The emitter junction is forward biased and collector junction is reverse biased. (d) Both the emitter junction as well as the collector junction are forward biased.

Solution

Correct statements: (b) and (c) (b) The base region must be very thin and lightly doped: • Thin base allows electrons (in npn) injected from emitter to diffuse through without too much recombination • Lightly doped base means fewer holes to recombine with injected electrons • Only ~2–5% recombine; ~95–98% reach the collector (c) Emitter junction forward biased, collector junction reverse biased: • Forward bias at emitter-base junction injects minority carriers into base • Reverse bias at collector-base junction sweeps those carriers into collector • This is the fundamental operating condition for a transistor (a) is wrong: emitter is heavily doped, base is lightly doped and thin, collector is moderately doped. (d) is wrong: both junctions forward biased is the saturation condition, not active/amplifier operation.

For a transistor amplifier, the voltage gain (a) remains constant for all frequencies (b) is high at high and low frequencies and constant in the middle frequency range (c) is low at high and low frequencies and constant at mid frequencies (d) None of the above.

Solution

Answer: (c) is low at high and low frequencies and constant at mid frequencies. Explanation: A transistor amplifier has a frequency-dependent voltage gain: • At low frequencies: Coupling capacitors (at input and output) have high impedance (X_C = 1/2πfC), which increases as f decreases. This reduces the signal transmission → lower gain. • At mid frequencies (midband): Coupling capacitors are effectively short circuits. Gain is nearly constant (flat response region). • At high frequencies: Internal transistor capacitances (junction capacitances) become significant. These short-circuit the signal → gain falls.

In half-wave rectification, what is the output frequency if the input frequency is 50 Hz? What is the output frequency of a full-wave rectifier for the same input frequency?

Solution

Half-wave rectifier: • Only one half of the AC cycle is allowed through • For each complete input cycle, there is one output pulse • Output frequency = Input frequency = 50 Hz Full-wave rectifier: • Both halves of the AC cycle are rectified (both half-cycles appear as positive pulses) • For each complete input cycle, there are two output pulses • Output frequency = 2 × Input frequency = 2 × 50 = 100 Hz

A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm? Justify.

Solution

Band gap E_g = 2.8 eV, λ = 6000 nm = 6000 × 10⁻⁹ m Energy of incident photon: E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(6000 × 10⁻⁹) = 1.988 × 10⁻²⁵/6 × 10⁻⁶ = 3.31 × 10⁻²⁰ J = 3.31 × 10⁻²⁰/(1.6 × 10⁻¹⁹) eV = 0.207 eV Since E_photon (0.207 eV) < E_g (2.8 eV), the photon does NOT have enough energy to create electron-hole pairs. No, the photodiode cannot detect a wavelength of 6000 nm. The photon energy is too small to excite electrons across the band gap.

Q10

The following table gives the values of work function for a few sensitive metals: If each of these metals is exposed to electromagnetic radiations of wavelength 3300 Å, which of them will not emit photo-electrons and why? (Na: 1.92 eV, K: 2.15 eV, Mo: 4.17 eV, Ni: 5.15 eV)

Solution

λ = 3300 Å = 3300 × 10⁻¹⁰ m = 3.3 × 10⁻⁷ m Energy of incident photon: E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(3.3 × 10⁻⁷) = 1.988 × 10⁻²⁵/3.3 × 10⁻⁷ = 6.024 × 10⁻¹⁹ J = 6.024 × 10⁻¹⁹/1.6 × 10⁻¹⁹ eV = 3.77 eV Photoemission occurs only if E_photon > Work function: • Na (1.92 eV): 3.77 > 1.92 → emits ✓ • K (2.15 eV): 3.77 > 2.15 → emits ✓ • Mo (4.17 eV): 3.77 < 4.17 → does NOT emit ✗ • Ni (5.15 eV): 3.77 < 5.15 → does NOT emit ✗ Molybdenum (Mo) and Nickel (Ni) will not emit photoelectrons because their work functions exceed the photon energy.

Q11

Identify the equivalent gate for the following circuit and write its truth table: NOT-AND (NAND) gate circuit where input A is given to a NOT gate, input B is given to a NOT gate, and both outputs are fed into an OR gate.

Solution

The circuit: A → NOT → Ā; B → NOT → B̄; Ā + B̄ → OR → Output Y Y = Ā + B̄ By De Morgan's theorem: Ā + B̄ = ̄(A·B) = NAND(A,B) This is equivalent to a NAND gate. Truth Table: A | B | Ā | B̄ | Y = Ā + B̄ 0 | 0 | 1 | 1 | 1 0 | 1 | 1 | 0 | 1 1 | 0 | 0 | 1 | 1 1 | 1 | 0 | 0 | 0 This is the NAND gate truth table: output is 0 only when both inputs are 1.

Q12

Write the truth table for a NAND gate connected as given in figure (both inputs connected together — single input A). What effective gate does this represent?

Solution

When both inputs of a NAND gate are connected together (A = B): Output Y = NAND(A, A) = NOT(A·A) = NOT(A) = Ā Truth Table: A | Y = NAND(A,A) 0 | 1 1 | 0 This represents a NOT gate (inverter). This is why NAND is called a universal gate: it can implement any logic function. A NOT gate can be constructed from just one NAND gate by connecting both inputs together.

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