**SEMICONDUCTORS: MATERIALS AND CLASSIFICATION**
**Conductivity order:** Metals (σ ~ 10⁶ S/m) >> Semiconductors (σ ~ 10⁻⁵ to 10⁵ S/m) >> Insulators (σ ~ 10⁻¹⁹ S/m)
**Energy Bands:**
**Intrinsic Semiconductors:**
**Extrinsic Semiconductors:**
**N-type:** Doped with Group V (donors: P, As, Sb)
**P-type:** Doped with Group III (acceptors: B, Al, Ga)
**Note:** Doping does NOT change Eg; it only shifts Fermi level and carrier concentrations.
**P-N JUNCTION DIODE**
**Depletion layer (space-charge region):**
**Forward bias (P-side positive):**
**Reverse bias (P-side negative):**
**V-I Characteristic:**
**TRANSISTOR AS SWITCH & AMPLIFIER (BJT)**
**Structure:** Emitter (E), Base (B), Collector (C) — NPN or PNP
**Active Mode:**
**Saturation Mode (Switch ON):**
**Cutoff Mode (Switch OFF):**
**RECTIFIERS**
**Half-wave:** 1 diode in series → conducts for half AC cycle → pulsating DC
**Full-wave (center-tap):** 2 diodes, center-tapped transformer → both half-cycles conducted
**Full-wave bridge:** 4 diodes in bridge → both half-cycles conducted, no transformer needed → smoother output
**LOGIC GATES (Transistor-based)**
**AND:** Transistors in series → output HIGH only if ALL inputs HIGH
**OR:** Transistors in parallel → output HIGH if ANY input HIGH
**NOT:** Single transistor inverter → output is opposite of input
**NAND, NOR:** Combinations of AND/OR with NOT
Q1. At absolute zero, a pure silicon crystal contains 4N valence electrons. How many empty energy states are available in the conduction band?
Answer: B — The conduction band has 4N empty energy states (same as the 4N filled states in the valence band), since each atom contributes 4 valence electrons and has space for 8 total in the outer shell.
Q2. When a silicon diode is forward-biased with an applied voltage of 0.5 V (which is less than the knee voltage of 0.7 V), the current through it is approximately:
Answer: B — Below the knee voltage, the exponential part of the V-I curve is still very steep but not yet in the linear region; current increases exponentially with voltage but remains relatively small.
Q3. Which of the following statements about N-type and P-type semiconductors is correct?
Answer: A — Doping does not change the energy gap Eg of the semiconductor material itself; it only shifts the Fermi level and changes the concentration of majority and minority carriers.
Q4. A P-N junction is reverse-biased. What happens to the width of the depletion layer compared to the equilibrium (no bias) condition?
Answer: C — Reverse bias applies a negative voltage to the P-side, which strengthens the internal electric field and pulls more charge carriers away from the junction, widening the depletion layer.
Q5. In a silicon p-n junction diode at room temperature (25°C), the reverse saturation current is approximately 10⁻¹² A. If the temperature increases to 75°C, the reverse saturation current will approximately:
Answer: D — Reverse saturation current (due to thermal generation of minority carriers) doubles approximately every 5-10°C temperature increase, so a 50°C rise causes an 8-32 fold increase.
Q6. A BJT in active mode has a base current Ib = 20 μA and current amplification factor β = 100. What is the collector current Ic?
Answer: B — In active mode, Ic ≈ β × Ib = 100 × 20 μA = 2000 μA = 2 mA.
Q7. A transistor is used as a switch. When the base current is made zero (Ib = 0), the transistor enters _______ mode and acts as an open switch.
Answer: C — Cutoff mode occurs when Ib = 0, both junctions are reverse-biased, Ic ≈ 0, and Vce equals the full supply voltage, so the transistor blocks current flow like an open switch.
Q8. Which statement about full-wave and half-wave rectifiers is NOT correct?
Answer: C — A full-wave bridge rectifier does NOT require a center-tapped transformer; instead, it uses 4 diodes in a bridge configuration to rectify both half-cycles.
Q9. Consider the following statements about the P-N junction in equilibrium (no external bias): (I) The depletion layer has a width determined by diffusion of carriers across the junction. (II) The internal electric field opposes further diffusion and maintains equilibrium. Which of the following is correct?
Answer: A — Both statements are correct: initially, carriers diffuse and create a depletion region; the resulting internal field then prevents further diffusion, establishing equilibrium.
Q10. A group V atom (with 5 valence electrons) is substituted for a silicon atom (with 4 valence electrons) in a silicon crystal. The 5th electron of the dopant atom is bound to the dopant nucleus by a small binding energy of about 0.05 eV. At room temperature (kT ≈ 0.025 eV), this electron is most likely to be:
Answer: B — The binding energy (0.05 eV) is comparable to thermal energy at room temperature (~2.5 kT), so the electron can easily escape from the donor and become a free carrier contributing to conductivity.
What is the valence band in a semiconductor?
The energy band containing the outermost (valence) electrons of atoms in the crystal at absolute zero, completely filled with 4N electrons in Si or Ge.
Define the conduction band.
The energy band above the valence band containing 4N empty energy states at absolute zero, into which electrons can jump to become mobile charge carriers.
What is the energy gap (Eg) in a semiconductor?
The energy difference between the lowest level of the conduction band and the highest level of the valence band; electrons must gain at least this much energy to conduct.
What is an N-type semiconductor and how is it created?
A semiconductor doped with Group V atoms (P, As) having 5 valence electrons; the extra electron becomes a free carrier (donor impurity), making electrons the majority carriers.
What is a P-type semiconductor and how is it created?
A semiconductor doped with Group III atoms (B, Al) having 3 valence electrons; the missing electron is a hole (acceptor impurity), making holes the majority carriers.
Explain the formation of the depletion layer in a P-N junction.
Electrons from N-region diffuse into P-region and holes from P-region into N-region, creating a charge-depleted layer near the junction with an internal electric field opposing further diffusion.
What happens when a diode is forward-biased?
The applied positive voltage on the P-side reduces the depletion layer width and the internal electric field, allowing charge carriers to diffuse across the junction and large current to flow.
What happens when a diode is reverse-biased?
The applied negative voltage on the P-side widens the depletion layer and strengthens the internal field, blocking diffusion and allowing only tiny leakage current from minority carriers.
In a BJT, what is the relationship between collector current and base current in active mode?
The collector current is approximately β times the base current: Ic ≈ β × Ib, where β is the current amplification factor.
What are the differences between saturation and cutoff modes of a transistor as a switch?
Saturation: Ib is large, Ic is maximum, Vce is low (~0), transistor is ON; cutoff: Ib = 0, Ic ≈ 0, Vce = full supply voltage, transistor is OFF.
Define the terms 'valence band' and 'conduction band' in a semiconductor. Why are there empty states in the conduction band of a pure semiconductor at absolute zero temperature? [2 marks]
Valence band contains outer-shell electrons of atoms (4N states, all filled at T=0K); conduction band consists of 4N available energy states above the valence band for mobile electrons. Empty because thermal energy at 0K is insufficient for electrons to jump across the energy gap Eg; electrons populate lowest available states first (Pauli exclusion).
Explain the formation of a depletion layer in a P-N junction and how the internal electric field established in this layer reaches equilibrium (no further net charge flow). Why does the width of the depletion layer increase under reverse bias? [5 marks]
Formation: electrons diffuse from N-region into P-region, holes diffuse from P-region into N-region, leaving behind fixed positive ions (donors) on N-side and fixed negative ions (acceptors) on P-side, creating a charge-depleted region. This charge separation creates an internal electric field pointing from N to P. Equilibrium: the electric field opposes further diffusion; at equilibrium, drift current (due to field) equals diffusion current. Reverse bias: external field adds to internal field, pulling more charges away from junction, widening depletion layer.
Derive the condition for a BJT to act as a switch in saturation mode, and explain why a transistor in saturation has minimal power dissipation. Also, state the condition for cutoff mode and compare the circuit behavior in both modes. [6 marks]
Saturation: base current Ib is made sufficiently large (Ib >> Ic,max / β) so that both BE and BC junctions are forward-biased; Ic reaches its maximum (limited by external circuit, not β), and Vce ≈ 0V (transistor acts as closed switch with nearly zero voltage drop). Power dissipation P = Ic × Vce ≈ Ic × 0 ≈ 0. Cutoff: Ib = 0 (no base current), both junctions reverse-biased, Ic ≈ 0 (only leakage), Vce ≈ supply voltage (transistor blocks current, acts as open switch). Saturation is ON state (conducts); cutoff is OFF state (blocks). In switching applications, transistor spends most time in saturation (conducting full current, low loss) or cutoff (blocking, low loss), and transitions quickly between them, minimizing time spent in active mode (where power dissipation is high).
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