JEE Physics Semiconductor Electronics MCQs Set 12

Answer: (c) 217100 Å

Explanation:
1. \( \lambda=\frac{hc}{E}=\frac{6.6\times10^{-34}\times3\times10^{8}}{57\times10^{-3}\times1.6\times10^{-19}} \).
2. \( =2.17\times10^{-5}\ \text{m}=217100\ \text{Å} \).

Teacher's Note:
1. A hole is created when an electron jumps from the valence band to the acceptor level.
2. Convert meV to joule with e = 1.6 x 10^-19 C.

Answer: (b) 2 × 10⁷ V/m

Explanation:
1. Energy gained over a path l is \( eEl=1\ \text{eV} \).
2. \( E=\frac{1\ \text{V}}{5\times10^{-8}\ \text{m}}=2\times10^{7}\ \text{V/m} \).

Teacher's Note:
1. Work done by the field is eEl.
2. 1 eV corresponds to a potential difference of 1 V.

Answer: (c) 283 V

Explanation:
1. The capacitor charges to the peak value of the supply voltage: \( V_0=\sqrt2\times200=283\ \text{V} \).

Teacher's Note:
1. Rms value and peak value are related by V0 = √2 Vrms.
2. A smoothing capacitor holds the output near the peak voltage.

Answer: (a) 11.7 V

Explanation:
1. Two parallel diodes share the same voltage, so the one with the smaller threshold (germanium) conducts and clamps the voltage across the pair at 0.3 V.
2. Voltage across the resistor \( V_0=12-0.3=11.7\ \text{V} \).

Teacher's Note:
1. Parallel branches have equal voltage; the lower-threshold diode turns on first.
2. The resistor receives the rest of the supply voltage.

Answer: (a) 11.3 V and 0.3 V

Explanation:
1. The silicon diode is forward biased, so V01 = 12 - 0.7 = 11.3 V.
2. The germanium diode to earth is forward biased and drops 0.3 V, so V02 = 0.3 V (the rest of the voltage is across the 4.7 kΩ resistor).

Teacher's Note:
1. A forward-biased diode holds a fixed voltage across itself.
2. Subtract the diode drop from the source voltage step by step.

Answer: (b) copper strip decreases and that of germanium increases

Explanation:
1. Copper is a metal: lower temperature means fewer lattice collisions, so its resistance decreases.
2. Germanium is a semiconductor: at low temperature far fewer electron-hole pairs exist, so its resistance increases.

Teacher's Note:
1. Metals have a positive temperature coefficient, semiconductors a negative one.
2. Cooling reverses the direction of change compared with heating.

Answer: (c) electrons move from base to collector

Explanation:
1. In an npn transistor the emitter injects electrons into the base, and most of them cross the thin base into the collector.
2. So the electrons move from the base into the collector (the conventional current is opposite).

Teacher's Note:
1. In an npn transistor the emitter-base junction is forward biased and the collector-base junction is reverse biased.
2. Collector current is carried mainly by electrons that have crossed the base.

Answer: (b) Pauli's exclusion principle

Explanation:
1. When N atoms come together, each atomic level splits into N levels.
2. By Pauli's exclusion principle no two electrons can occupy the same state, so the levels spread into a band instead of overlapping at a single energy.

Teacher's Note:
1. Band formation follows from the exclusion principle applied to many closely spaced atoms.
2. The other principles do not force the splitting of levels.

Answer: (a) both the depletion region and the barrier height are reduced

Explanation:
1. Forward bias opposes the built-in potential, so the potential barrier decreases.
2. A lower barrier lets more majority carriers cross, and the depletion region becomes narrower.

Teacher's Note:
1. Forward bias lowers the barrier; reverse bias raises it.
2. Barrier height and width change together.

Answer: (c) 0

Explanation:
1. In the common-base circuit, an increase in emitter current increases the collector current and the output voltage across the load in step with the input.
2. So the output is in phase with the input: phase difference 0. (Common emitter gives π.)

Teacher's Note:
1. Only the common-emitter configuration inverts the signal.
2. The common-base amplifier has voltage gain but no phase reversal.

Answer: (a) 5/4

Explanation:
1. Current is proportional to concentration times drift velocity: \( I=nevA \).
2. \( \frac{I_e}{I_h}=\frac{n_e}{n_h}\cdot\frac{v_e}{v_h}\Rightarrow\frac74=\frac75\cdot\frac{v_e}{v_h} \).
3. \( \frac{v_e}{v_h}=\frac74\times\frac57=\frac54 \).

Teacher's Note:
1. The same charge magnitude and area appear for both carriers, so they cancel in the ratio.
2. Divide the current ratio by the concentration ratio.

Answer: (c) 49

Explanation:
1. \( \alpha=\frac{I_C}{I_E}=\frac{5.488}{5.60}=0.98 \).
2. \( \beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{0.02}=49 \).

Teacher's Note:
1. First find α from the common-base currents, then convert to β.
2. Equivalently, IB = 5.60 - 5.488 = 0.112 mA and β = IC/IB = 49.

Answer: (d) covalent binding

Explanation:
1. Conductivity that rises with temperature is characteristic of a semiconductor.
2. Semiconductors such as Si and Ge have covalent bonding, and their small band gap makes them opaque to visible light.

Teacher's Note:
1. Metals have falling conductivity with temperature; ionic and Van der Waals solids are insulators.
2. Elemental semiconductors are covalent solids.

Answer: (d) The number of free electrons for conduction is significant only in Si and Ge but small in C

Explanation:
1. Carbon (diamond) has a band gap of about 5.4 eV, so almost no electrons are excited across it at room temperature.
2. Silicon (1.1 eV) and germanium (0.7 eV) have small gaps, so a significant number of carriers is thermally excited.

Teacher's Note:
1. Smaller band gap means more thermally generated carriers.
2. Diamond behaves as an insulator at room temperature.

Answer: (d) Ec and Ev increase, but Eg decreases

Explanation:
1. Decreasing the lattice constant brings the atoms closer together, so the overlap of their wave functions increases.
2. Stronger overlap spreads the bands wider (Ec and Ev increase) and therefore reduces the gap between them (Eg decreases).

Teacher's Note:
1. Closer atoms mean broader bands.
2. Broader bands encroach on the gap.

Answer: (a) variation of the number of charge carriers with temperature

Explanation:
1. In a metal the number of free electrons is nearly fixed, and resistance increases with temperature due to increased scattering.
2. In a semiconductor the number of carriers rises very rapidly with temperature, which dominates over the scattering effect and lowers the resistance.

Teacher's Note:
1. Carrier density is nearly constant in metals but exponential in semiconductors.
2. Scattering increases with temperature in both.

Answer: (d) copper decreases and germanium increases

Explanation:
1. Copper's resistance falls on cooling because lattice vibrations decrease.
2. Germanium has fewer thermally generated carriers at 77 K, so its resistance rises.

Teacher's Note:
1. Metals and semiconductors respond oppositely to temperature change.
2. The effect is the same as for cooling to 80 K.

Answer: (b) if either or both inputs are 1

Explanation:
1. OR gives 1 for 0 1, 1 0 and 1 1 and 0 only for 0 0.

Teacher's Note:
1. OR corresponds to Boolean addition Y = A + B.
2. AND needs both inputs high.

Answer: (a) positive feedback

Explanation:
1. An oscillator feeds part of its output back to the input in phase with it (positive feedback) so that oscillations sustain themselves without any external input.
2. Negative feedback reduces gain and stabilises an amplifier instead.

Teacher's Note:
1. The Barkhausen condition requires the loop gain to be 1 and the feedback to be in phase.
2. Positive feedback is the basis of oscillators.

Answer: (a) n-type

Explanation:
1. An impurity level just below the conduction band is a donor level; electrons are easily excited from it into the conduction band.
2. Hence the semiconductor is n-type.

Teacher's Note:
1. Donor levels lie close to the conduction band, acceptor levels close to the valence band.
2. Thermal energy at room temperature is about 26 meV.

Answer: (b) displaced towards the conduction band

Explanation:
1. Extra electrons from donor atoms increase the electron concentration in the conduction band.
2. To hold this larger concentration the Fermi level moves from the middle of the gap upwards towards the conduction band.

Teacher's Note:
1. p-type: Fermi level moves towards the valence band.
2. The shift grows with the doping density.

Answer: (d) Some potential gap develops across the p-n junction when it is formed

Explanation:
1. When the junction forms, diffusion of carriers leaves uncovered ions on both sides, creating a barrier potential (about 0.7 V for silicon).
2. Forward bias allows current and reverse bias blocks it, which is why rectification is possible, so (a), (b), (c) are false.

Teacher's Note:
1. The built-in potential exists without any external battery.
2. One-way conduction makes a junction a rectifier.

Answer: (d) the numbers of both electrons and holes increase

Explanation:
1. Heating breaks more covalent bonds, creating additional electron-hole pairs.
2. So both the electron and the hole concentrations increase (electrons remain the majority carriers).

Teacher's Note:
1. Thermal generation adds electrons and holes in equal numbers.
2. The majority carriers are still the donor electrons.

Answer: (c) 2ν

Explanation:
1. A full-wave rectifier delivers one current pulse in every half cycle of the input.
2. So the output (ripple) frequency is twice the input frequency, 2ν.

Teacher's Note:
1. Half-wave rectifier: output frequency equals the input frequency.
2. Full-wave: twice the input frequency.