JEE Physics Semiconductor Electronics MCQs Set 11

Answer: (b) the conduction and valence bands overlap

Explanation:
1. In a metal the highest occupied band is either partly filled or overlaps the next band.
2. So there is no forbidden gap at all, and electrons can move freely under a small field.

Teacher's Note:
1. Conductors have zero energy gap because their bands overlap.
2. Insulators and semiconductors have a filled valence band and an empty conduction band.

Answer: (b) EG1 < EG2 < EG3

Explanation:
1. A conductor has no gap (overlapping bands), a semiconductor has a small gap (about 1 eV) and an insulator has a large gap (more than about 5 eV).
2. So EG1 < EG2 < EG3.

Teacher's Note:
1. Rank materials by how hard it is to excite electrons into the conduction band.
2. Typical values: Ge 0.7 eV, Si 1.1 eV, diamond about 5.5 eV.

Answer: (c) Resistance of a conductor decreases with increase in temperature

Explanation:
1. (a) is true: more carriers are generated on heating.
2. (b) is true: holes drift along the field, electrons against it.
3. (c) is false: in a metal, more frequent collisions at higher temperature increase the resistance.
4. (d) is true: the donor atoms are neutral overall, so the crystal remains electrically neutral.

Teacher's Note:
1. Metals have a positive and semiconductors a negative temperature coefficient of resistance.
2. Doping does not charge a crystal.

Answer: (b) intrinsic semiconductors

Explanation:
1. In a pure semiconductor each electron that reaches the conduction band leaves behind one hole.
2. So ne = nh, which defines an intrinsic semiconductor.

Teacher's Note:
1. Doping makes one type of carrier dominant.
2. In an intrinsic semiconductor ne = nh = ni.

Answer: (b) In an n-type semiconductor the number of electrons increases in the valence band

Explanation:
1. Doping controls the carrier concentration, so (a) is right.
2. In n-type material the extra electrons are in the conduction band (donor electrons), not in the valence band, so (b) is wrong.
3. (c) and (d) are true.

Teacher's Note:
1. Donor electrons enter the conduction band; acceptors create holes in the valence band.
2. The intrinsic Fermi level lies in the middle of the gap.

Answer: (a) in the middle of the forbidden gap

Explanation:
1. In an intrinsic semiconductor the electron and hole concentrations are equal.
2. This is possible only when the Fermi level lies (almost exactly) at the middle of the forbidden gap.

Teacher's Note:
1. Doping shifts the Fermi level: towards the conduction band for n-type, towards the valence band for p-type.
2. The intrinsic position is the reference.

Answer: (a) ne increases and vd decreases

Explanation:
1. Heating generates more electron-hole pairs, so ne increases.
2. Lattice vibrations increase, scattering is more frequent, mobility falls, and so does the drift velocity for a given field.

Teacher's Note:
1. Carrier number rises rapidly (exponentially) with temperature.
2. Mobility falls only slowly, so conductivity still rises.

Answer: (a) n-type and its resistivity is 0.34 ohm-metre

Explanation:
1. Electrons outnumber holes, so the semiconductor is n-type.
2. \( \sigma=e(n_e\mu_e+n_h\mu_h)=1.6\times10^{-19}(8\times10^{18}\times2.3+5\times10^{18}\times0.01)=2.95\ \text{S/m} \).
3. \( \rho=\frac1\sigma=0.34\ \Omega\,\text{m} \).

Teacher's Note:
1. The majority carrier gives the type.
2. Add both carriers' contributions to conductivity.

Answer: (a) 1.0 × 10⁶ V/m

Explanation:
1. \( E=\frac{V}{d}=\frac{0.50}{5.0\times10^{-7}}=1.0\times10^{6}\ \text{V/m} \).

Teacher's Note:
1. Treat the field as uniform across the depletion layer.
2. Field = potential difference divided by width.

Answer: (a) is zero because the number of charge carriers flowing on both sides is the same

Explanation:
1. Even without bias, carriers diffuse across the junction (diffusion current) and minority carriers drift back (drift current).
2. These two currents are equal and opposite, so the net current is zero.

Teacher's Note:
1. It is a dynamic equilibrium, not an absence of motion.
2. Net current zero does not mean no carrier crosses.

Answer: (b) production of electron-hole pairs due to the electric field

Explanation:
1. In a heavily doped diode the depletion region is very thin, so even a modest reverse voltage produces a very strong field (about 10⁶ V/m or more).
2. This field breaks covalent bonds directly and creates electron-hole pairs, which gives the large reverse current.
3. (Heavy doping makes this possible, but the direct cause is the field-induced pair production.)

Teacher's Note:
1. Zener breakdown: field ionisation in a thin, heavily doped junction.
2. Avalanche breakdown: collision ionisation in a thicker, lightly doped junction.

Answer: (c) a p-n junction

Explanation:
1. A single piece of p-type, n-type or intrinsic material obeys Ohm's law and conducts equally in both directions.
2. Only a p-n junction conducts well in one direction (forward bias) and hardly at all in the other.

Teacher's Note:
1. One-way conduction is the signature of a junction diode.
2. Uniform semiconductors are symmetric.

Answer: (a) In forward bias the voltage across R is V

Explanation:
1. An ideal diode has zero resistance in forward bias, so the whole battery voltage appears across R: VR = V.
2. In reverse bias it has infinite resistance, no current flows, and the voltage across R is zero (not V or 2V).

Teacher's Note:
1. Ideal diode: zero drop when forward biased, open circuit when reverse biased.
2. The voltage across R can never exceed the battery emf in a series circuit.

Answer: (b) 5 Ω

Explanation:
1. Maximum current \( I=\frac{P}{V_d}=\frac{0.1}{0.5}=0.2\ \text{A} \).
2. The resistor takes the rest of the battery voltage: \( R=\frac{1.5-0.5}{0.2}=5\ \Omega \).

Teacher's Note:
1. Diode power = drop x current.
2. Apply Kirchhoff's voltage law to the loop to get R.

Answer: (a) 8 × 10¹⁵ per cm³

Explanation:
1. \( n=\frac{\sigma}{e\mu}=\frac{5}{1.6\times10^{-19}\times3900}=8.0\times10^{15}\ \text{cm}^{-3} \).

Teacher's Note:
1. n = σ/(eμ) for a single dominant carrier type.
2. Use consistent cm units.

Answer: (c) 100 Hz

Explanation:
1. A full-wave rectifier produces one output pulse for every half cycle of the input.
2. So the ripple has twice the input frequency: 2 x 50 = 100 Hz.

Teacher's Note:
1. Half-wave rectifier ripple has the same frequency as the input.
2. Full-wave ripple is double the input frequency.

Answer: (a) 99

Explanation:
1. \( \beta=\frac{\alpha}{1-\alpha}=\frac{0.99}{0.01}=99 \).

Teacher's Note:
1. β = α/(1 - α), and α = β/(1 + β).
2. α is always slightly less than 1.

Answer: (d) 75

Explanation:
1. \( \beta=\frac{\Delta I_C}{\Delta I_B}=\frac{16.5-9}{(200-100)\times10^{-3}}=\frac{7.5\ \text{mA}}{0.1\ \text{mA}}=75 \).

Teacher's Note:
1. AC current gain is the ratio of the changes at constant VCE.
2. Convert 100 μA to 0.1 mA before dividing.

Answer: (a) I and III are correct

Explanation:
1. In a common-base amplifier the output voltage follows the input voltage with no phase reversal, whatever the type (p-n-p or n-p-n).
2. Hence statements I and III are correct.

Teacher's Note:
1. The common-emitter amplifier gives a phase reversal of 180°; the common-base does not.
2. The phase relation does not depend on the transistor type.

Answer: (d) 0.98

Explanation:
1. Electrons reaching the collector are 98% of those entering the emitter.
2. Current transfer ratio \( \alpha=\frac{I_C}{I_E}=0.98 \).

Teacher's Note:
1. α is the fraction of emitter current that reaches the collector.
2. Only the fraction lost in the base (2%) becomes base current.

Answer: (a) 9.1 mA

Explanation:
1. \( I_E=I_B+I_C=0.1\ \text{mA}+9\ \text{mA}=9.1\ \text{mA} \).

Teacher's Note:
1. The emitter current is the sum of the base and collector currents.
2. Keep both currents in the same unit.

Answer: (a) 0.93

Explanation:
1. Voltage gain = current gain x resistance gain: \( A_V=\alpha\times\frac{R_{out}}{R_{in}} \).
2. \( \alpha=\frac{2800}{3000}=0.93 \).

Teacher's Note:
1. Resistance gain is the ratio of output to input resistance.
2. Voltage gain is the product of current gain and resistance gain.

Answer: (b) OR

Explanation:
1. The output is 1 whenever at least one input is 1, and 0 only when both are 0.
2. This is the OR gate. (XOR would give 0 for A = B = 1.)

Teacher's Note:
1. OR gives 0 only for input 0, 0; AND gives 1 only for input 1, 1.
2. XOR differs from OR in the last column.

Answer: (b) Common emitter

Explanation:
1. Power gain = current gain x voltage gain.
2. The common-emitter circuit has both a large current gain (β) and a large voltage gain, so its power gain is the highest.

Teacher's Note:
1. Common base has current gain below 1; common collector has voltage gain below 1.
2. Only the common-emitter circuit has large gain in both.