CBSE Class 12 Physics Semiconductor Electronics Materials Devices and Simple Circuits MCQs Set 06

Answer: (c) E_g(C) > E_g(Si) > E_g(Ge)

Explanation:
1. Diamond: about 5.4 eV; silicon: about 1.1 eV; germanium: about 0.7 eV.
2. So the order is E_g(C) > E_g(Si) > E_g(Ge).

Teacher's Note:
1. All three have the same crystal structure, but the band gap falls as we go down group 14.
2. The large gap of diamond is why it is an insulator.

Answer: (b) 5 × 10⁹ m⁻³

Explanation:
1. Mass action relation: n_e n_h = n_i²
2. n_h = \( \frac{(1.5 \times 10^{16})^2}{4.5 \times 10^{22}} \) = \( \frac{2.25 \times 10^{32}}{4.5 \times 10^{22}} \)
3. n_h = 5 × 10⁹ m⁻³

Teacher's Note:
1. Doping raises the majority carriers and lowers the minority carriers below n_i.
2. Square n_i carefully: (1.5)² = 2.25 and (10¹⁶)² = 10³².

Answer: (a) 5 × 10²² m⁻³

Explanation:
1. N_D = \( \frac{5 \times 10^{28}}{10^{6}} \)
2. N_D = 5 × 10²² m⁻³

Teacher's Note:
1. 1 part per million means divide by 10⁶, not multiply.
2. Each arsenic atom gives one free electron, so n_e ≈ 5 × 10²² m⁻³ too.

Answer: (d) decreases

Explanation:
1. Heating breaks more covalent bonds, so more electron–hole pairs are produced.
2. With more charge carriers the conductivity rises and the resistance decreases.

Teacher's Note:
1. Semiconductors have a negative temperature coefficient of resistance.
2. In metals the number of carriers is fixed, so extra collisions make resistance rise instead.

Answer: (b) just below the bottom of the conduction band

Explanation:
1. The fifth electron of the donor atom is very loosely bound (about 0.05 eV in Si).
2. So its level lies just below the bottom of the conduction band, and a small energy lifts it into the band.

Teacher's Note:
1. Acceptor levels, in contrast, lie just above the valence band.
2. This small gap is why almost all donors are ionised at room temperature.

Answer: (c) electrically neutral

Explanation:
1. Each trivalent atom that creates a hole becomes a fixed negative ion when it accepts an electron.
2. The total positive charge of the holes equals the total negative charge of the ions, so the crystal is electrically neutral.

Teacher's Note:
1. Do not confuse 'p' (majority carriers are holes) with a net positive charge.
2. Doping adds neutral atoms, so no net charge is added to the crystal.

Answer: (a) only immobile ions

Explanation:
1. Near the junction, electrons and holes diffuse across and recombine.
2. What is left is a layer of fixed donor and acceptor ions with no mobile carriers, so it has only immobile ions.

Teacher's Note:
1. 'Depletion' means depleted of mobile charge carriers.
2. The ions are fixed in the crystal lattice and cannot move.

Answer: (d) 21.5 mA

Explanation:
1. Voltage across the resistor = 5 − 0.7 = 4.3 V
2. I = \( \frac{4.3}{200} \) = 0.0215 A = 21.5 mA

Teacher's Note:
1. 25 mA is what you get if you forget the 0.7 V drop across the diode.
2. Treat the forward-biased diode as a fixed 0.7 V drop for such problems.

Answer: (b) 5 Ω

Explanation:
1. r_d = \( \frac{\Delta V}{\Delta I} \) = \( \frac{0.75 - 0.70}{(20 - 10) \times 10^{-3}} \)
2. r_d = \( \frac{0.05}{0.01} \) = 5 Ω

Teacher's Note:
1. Use changes in V and I, not V/I at one point.
2. Convert mA to A before dividing.

Answer: (a) the depletion region widens and the barrier height increases

Explanation:
1. In reverse bias the applied voltage adds to the barrier, so the effective barrier becomes V₀ + V.
2. More carriers are pulled away from the junction, so the depletion region widens and the barrier height increases.

Teacher's Note:
1. Forward bias does exactly the opposite.
2. This is why a reverse-biased diode allows only a tiny current.

Answer: (c) during the half cycle in which its anode is positive with respect to its cathode

Explanation:
1. A diode conducts only when it is forward biased, i.e. its p-side (anode) is at a higher potential.
2. So it conducts during the half cycle in which its anode is positive, and output appears only in that half.

Teacher's Note:
1. In the other half the diode is reverse biased and acts as an open switch.
2. Its output frequency is the same as the input frequency.

Answer: (b) 2 A

Explanation:
1. D₁ points from the + rail to the − rail, so it is forward biased and acts as a closed switch.
2. D₂ points the other way, so it is reverse biased and its branch carries no current.
3. I = \( \frac{6}{3} \) = 2 A

Teacher's Note:
1. 3 A would be the answer only if both branches conducted.
2. For ideal diodes, first decide which are forward biased, then remove the others from the circuit.

Answer: (d) reduce the ripple and give a smoother dc output

Explanation:
1. The capacitor charges as the output rises and slowly discharges through R_L as it falls.
2. This fills the gaps between pulses, so it reduces the ripple and gives a smoother dc output.

Teacher's Note:
1. A large C R_L makes the output steadier.
2. This arrangement is called a capacitor filter.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A hole is a vacancy with an effective charge +e and it does carry current. The Assertion is true.
2. Bound electrons jumping into vacancies make the vacancy move the opposite way, like a positive charge. The Reason is true.
3. The Reason explains how holes conduct, so it is the correct explanation.

Teacher's Note:
1. Hole current is really a movement of bound valence electrons.
2. Free electrons move in the conduction band; holes move in the valence band.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. At 0 K there is no thermal energy to lift electrons across the 1.1 eV gap. The Assertion is true.
2. So the valence band is full and the conduction band is empty. The Reason is true.
3. With no free carriers there can be no current, so the Reason explains the Assertion.

Teacher's Note:
1. A full band cannot conduct, and neither can an empty one.
2. Conductivity appears only as the temperature rises.

Answer: (d) A is false, but R is true.

Explanation:
1. The free electrons are balanced by fixed positive donor ions, so the crystal is neutral. The Assertion is false.
2. Electrons are indeed the majority carriers in n-type material. The Reason is true.

Teacher's Note:
1. 'n' refers to the type of majority carrier, not to the net charge.
2. Doping adds neutral atoms, so the total charge stays zero.

Answer: (c) A is true, but R is false.

Explanation:
1. In forward bias the barrier falls to V₀ − V and a large current flows, so the resistance is low. The Assertion is true.
2. In forward bias the depletion region becomes narrower, not wider. The Reason is false.

Teacher's Note:
1. A wider depletion region happens in reverse bias.
2. Low resistance goes with a thin depletion layer.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The current through a diode does not increase in proportion to the voltage. The Assertion is true.
2. Its V–I curve is non-linear and differs for forward and reverse bias. The Reason is true.
3. A non-linear V–I graph is exactly what failing to obey Ohm's law means, so the Reason explains the Assertion.

Teacher's Note:
1. Because the graph is curved, we use dynamic resistance ΔV/ΔI.
2. A device that obeys Ohm's law has a straight-line V–I graph.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. One diode conducts in each half cycle, both sending current through R_L in the same direction. The Assertion is true.
2. A capacitor across the load smooths the output, so the Reason is true.
3. Output in both halves is due to the two diodes, not the capacitor, so the Reason does not explain the Assertion.

Teacher's Note:
1. The full-wave output works even without any filter.
2. Choose (b) when both statements are true but unrelated.

Answer: (a) 5 Ω

Explanation:
1. r_d = \( \frac{\Delta V}{\Delta I} \) = \( \frac{0.8 - 0.7}{(30 - 10) \times 10^{-3}} \)
2. r_d = \( \frac{0.1}{0.02} \) = 5 Ω

Teacher's Note:
1. 27 Ω comes from wrongly using V/I = 0.8/0.030 at one point.
2. Dynamic resistance falls as the forward current increases.

Answer: (c) 0.7 V

Explanation:
1. Below about 0.6 V the current is tiny; it starts rising sharply from about 0.7 V.
2. So the knee voltage is about 0.7 V, as expected for silicon.

Teacher's Note:
1. For germanium the knee voltage is lower, about 0.2–0.3 V.
2. The knee is where the curve turns from flat to steep.

Answer: (b) drift of minority carriers across the junction

Explanation:
1. In reverse bias the higher barrier stops majority carrier diffusion.
2. Only the few minority carriers are swept across by the junction field, so the current is due to drift of minority carriers.

Teacher's Note:
1. The number of minority carriers depends on temperature, not on the applied voltage.
2. That is why this current hardly changes with voltage.

Answer: (d) 20 mA

Explanation:
1. Voltage across the resistor = 2 − 0.7 = 1.3 V
2. I = \( \frac{1.3}{65} \) = 0.02 A = 20 mA

Teacher's Note:
1. 30.8 mA comes from ignoring the diode drop.
2. Always subtract the diode's forward drop from the supply voltage first.

Answer: (a) about 1 μA, practically zero

Explanation:
1. The diode is now reverse biased by only 2 V, far below breakdown.
2. Only the reverse saturation current of about 1 μA flows, which is practically zero.

Teacher's Note:
1. A diode in reverse bias behaves almost as an open switch.
2. Breakdown needs a much larger reverse voltage.

Answer: (c) 100 Hz

Explanation:
1. In a full-wave rectifier there is one output pulse in each half cycle.
2. Output frequency = 2 × 50 = 100 Hz

Teacher's Note:
1. A half-wave rectifier would give 50 Hz.
2. More pulses per second make the output easier to smooth.

Answer: (d) D₁ conducts and D₂ does not

Explanation:
1. When A is positive with respect to C, D₁ is forward biased.
2. At the same time B is negative with respect to C, so D₂ is reverse biased.
3. So D₁ conducts and D₂ does not.

Teacher's Note:
1. In the next half cycle the roles of the diodes swap.
2. In both cases the current through the load is in the same direction.

Answer: (a) work as a half-wave rectifier with output frequency 50 Hz

Explanation:
1. D₁ still conducts in the half cycles when A is positive.
2. No current flows in the other half cycles, so the circuit becomes a half-wave rectifier with output frequency 50 Hz.

Teacher's Note:
1. The output is still unidirectional, just with gaps.
2. The output is not ac, because current still flows only one way through the load.

Answer: (c) much larger than the time between output peaks

Explanation:
1. Between peaks the capacitor discharges through R_L.
2. If C R_L is much larger than the time between peaks, the voltage falls very little before the next peak recharges it.

Teacher's Note:
1. A small time constant lets the capacitor discharge almost fully, leaving large ripple.
2. This is why chargers use capacitors of large capacitance.

Answer: (b) 10 ms

Explanation:
1. Output frequency = 100 Hz, so the time between peaks = \( \frac{1}{100} \) s
2. Time interval = 10 ms

Teacher's Note:
1. 20 ms is the period of the 50 Hz input, which applies to a half-wave rectifier.
2. Time = 1/frequency.