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JEE • PHYSICS • SEMI CONDUCTOR DEVICES

Semi conductor devices practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 301-330 of 361 questions

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301 Medium
301 of 361
Application +4 | -1
When the reverse bias in a junction diode is changed from 5V to 15V, the value of current changes from 38 μA38\,\mu A to 88 μA88\,\mu A. The reverse resistance of the junction diode will be
View answer and explanation
Correct option: B
2×105 Ω2 \times {10^5}\,\Omega
Option B (2×105 Ω2 \times {10^5}\,\Omega) is correct because it is calculated using Ohm's law for the change in voltage and current in the reverse bias.
302 Medium
302 of 361
Understanding +4 | -1
A piece of copper and another of germanium are cooled from room temperature to 77 K, the resistance of
View answer and explanation
Correct option: C
Copper decreases and germanium increases
Option B is correct because both the assertion and the reason are true, but the reason is not the correct explanation of the assertion. Vibrational degrees of freedom become significant at high temperatures, and diatomic gases do have vibrational degrees of freedom.
303 Medium
303 of 361
Remembering +4 | -1
If a transistor, the emitter, base and collector are respectively
View answer and explanation
Correct option: A
Heavily doped, lightly doped, moderately doped
Option A (Heavily doped, lightly doped, moderately doped) is correct because the emitter needs a high carrier concentration, the base needs minimal recombination, and the collector handles a significant current.
304 Medium
304 of 361
Understanding +4 | -1
Statement-I: A p−n\mathrm{p-n} junction with reverse bias can be used as a photodiode to measure light intensity.

Statement-II: In a reverse bias condition the current is small but it is more sensitive to changes in incident light intensity.
View answer and explanation
Correct option: A
If both statements are correct.
Option A is correct because a reverse-biased p-n junction can be used as a photodiode to measure light intensity, and the current in reverse bias is small but highly sensitive to changes in incident light intensity.
305 Medium
305 of 361
Understanding +4 | -1
A piece of copper and another of germanium are cooled from temperature to 80∘ K.80^\circ \,K. The resistance of
View answer and explanation
Correct option: D
Copper decreases and germanium increases
Option D (Copper decreases and germanium increases) is correct because copper is a metal and its resistance decreases with decreasing temperature, while germanium is a semiconductor and its resistance increases with decreasing temperature.
306 Medium
306 of 361
Remembering +4 | -1
Bonding in a germanium crystal (semi-conductor) is
View answer and explanation
Correct option: D
Covalent
Option D (Covalent) is correct because germanium, like silicon, forms a crystal lattice through covalent bonds.
307 Medium
307 of 361
Understanding +4 | -1
Electric conduction in semi-conductor takes place due to
View answer and explanation
Correct option: C
Both electrons and holes
Option C (Both electrons and holes) is correct because both electrons and holes contribute to the electric current in a semiconductor.
308 Medium
308 of 361
Remembering +4 | -1
Let np  and ne{n_p}\,\,{\text{and }}{n_e} be the number of holes and conduction electrons in an intrinsic semiconductor. Then
View answer and explanation
Correct option: B
np=ne{n_p} = {n_e}
Option B (np=ne{n_p} = {n_e}) is correct because, in an intrinsic semiconductor, the number of holes is equal to the number of conduction electrons.
309 Medium
309 of 361
Understanding +4 | -1
In order to forward bias a PNPN junction, the negative terminal of battery is connected to
View answer and explanation
Correct option: C
NN-side
Option C is correct because to forward bias a PN junction, the negative terminal of the battery is connected to the N-side, and the positive terminal to the P-side.
310 Medium
310 of 361
Understanding +4 | -1
In an nn-type semiconductor, the fermi energy level lies
View answer and explanation
Correct option: A
In the forbidden energy gap nearer to the conduction band
Option A (In the forbidden energy gap nearer to the conduction band) is correct because in an n-type semiconductor, the Fermi level is close to the conduction band due to the presence of excess electrons.
311 Medium
311 of 361
Understanding +4 | -1
A LED (light emitting diode) in its proper Operation is
View answer and explanation
Correct option: B
A heavily doped p-n junction in forward biased
Option B (A heavily doped p-n junction in forward biased) is correct because LEDs require heavy doping and forward bias to emit light efficiently.
312 Medium
312 of 361
Understanding +4 | -1
If ne{n_e} and vd{v_d} be the number of electrons and drift velocity in a semiconductor. When the temperature is increased
View answer and explanation
Correct option: A
ne{n_e} increases and vd{v_d} decreases
Option A is correct because in a semiconductor, increasing the temperature increases the number of free electrons (nen_e), but the increased scattering reduces the drift velocity (vdv_d).
313 Medium
313 of 361
Understanding +4 | -1
CC and SiSi both have same lattice structure, having 4 bonding electrons in each. However, CC is insulator where as Si is intrinsic semiconductor. This is because
View answer and explanation
Correct option: C
The four bonding electrons in the case of CC lie in the second orbit, whereas in the case of SiSi they lie in the third
The difference in conductivity between C and Si is due to the energy band gap, which is influenced by the orbital location of bonding electrons. Carbon's electrons are closer to the nucleus, resulting in a larger energy gap.
314 Medium
314 of 361
Understanding +4 | -1
In semiconductors at room temperature: -
View answer and explanation
Correct option: A
The valence band is partially empty and conduction band is partially filled
Option A is correct because, at room temperature, some electrons in semiconductors gain enough energy to move from the valence band to the conduction band, resulting in both bands being partially filled.
315 Medium
315 of 361
Analyzing +4 | -1
A⋅Bˉ+Aˉ⋅B‾\overline{A \cdot \bar{B}+\bar{A} \cdot B} is equivalent to
View answer and explanation
Correct option: B
(A+Bˉ)+(Aˉ+B)(A+\bar{B})+(\bar{A}+B)
Option D ((A+B)⋅(Aˉ+Bˉ)(A+B) \cdot(\bar{A}+\bar{B})) is correct because it represents the simplified form of the given Boolean expression after applying De Morgan's law and simplification.
316 Medium
316 of 361
Understanding +4 | -1
Zener diodes have higher dopant densities as compared to ordinary p-n junction diodes. This
View answer and explanation
Correct option: C
decreases the width of depletion layer but increases the electric field
Option C is correct because higher dopant densities in Zener diodes decrease the depletion layer width and increase the electric field.
317 Medium
317 of 361
Understanding +4 | -1
Statement A: In intrinsic semiconductors, conductivity is mainly due to the breakage of covalent bonds.
Statement B: In extrinsic semiconductors, the conductivity is mainly due to the addition of impurities.
View answer and explanation
Correct option: C
Both A and B are true
Option C is correct because both statements are true: intrinsic semiconductors conduct due to broken covalent bonds, and extrinsic semiconductors conduct due to added impurities.
318 Medium
318 of 361
Understanding +4 | -1
When a P-N junction is forward biased, the flow of current across the junction is mainly due to
View answer and explanation
Correct option: B
Diffusion of charges
Option B (Diffusion of charges) is correct because in forward bias, the majority carriers diffuse across the junction, contributing significantly to the current.
319 Medium
319 of 361
Remembering +4 | -1
For germanium crystal, the forbidden energy gap in joules is
View answer and explanation
Correct option: A
1.12×10−191.12 \times {10^{ - 19}}
Option A (1.12×10−191.12 \times {10^{ - 19}}) is correct because the forbidden energy gap for germanium is approximately 0.7 eV, which is equal to 1.12×10−191.12 \times {10^{ - 19}} joules.
320 Medium
320 of 361
Remembering +4 | -1
For LED’s to emit light in visible region of electromagnetic light, it should have energy band gap in the range of:
View answer and explanation
Correct option: D
1.7 eV to 3.0 eV
Option D (1.7 eV to 3.0 eV) is correct because this range corresponds to the energy of photons in the visible light spectrum.
321 Medium
321 of 361
Understanding +4 | -1
Which of the following statements is not correct when a junction diode is in forward bias?
View answer and explanation
Correct option: D
Electrons on nn-side and holes on pp-side will move away from junction.
Option D is correct because in forward bias, electrons from the n-side and holes from the p-side move towards the junction, not away from it.
322 Medium
322 of 361
Understanding +4 | -1
For a transistor α\alpha and β\beta are given as α=ICIE\alpha = \frac{{{I_C}}}{{{I_E}}} and β=ICIE\beta = \frac{{{I_C}}}{{{I_E}}}. Then, the correct relation between α\alpha and β\beta will be
View answer and explanation
Correct option: B
β=α1−α\beta = \frac{\alpha }{{1 - \alpha }}
Option B is correct because the correct relation between α\alpha and β\beta is β=α1−α\beta = \frac{\alpha}{1 - \alpha}.
323 Medium
323 of 361
Understanding +4 | -1
In forward bias, the width of potential barrier in a PP-NN junction diode
View answer and explanation
Correct option: B
Decreases
Option B (Decreases) is correct because forward bias reduces the potential barrier width by reducing the depletion region.
324 Medium
324 of 361
Understanding +4 | -1
In BJT, maximum current flows in which of the following?
View answer and explanation
Correct option: A
Emitter region
Option A (Emitter region) is correct because the emitter current is the sum of the base and collector currents in a BJT.
325 Medium
325 of 361
Application +4 | -1
Three amplifier stages each with a gain of 10 are cascaded. The overall gain is
View answer and explanation
Correct option: C
1000
Option C (1000) is correct because the overall gain of cascaded amplifiers is the product of the individual gains.
326 Medium
326 of 361
Remembering +4 | -1
The coordination number of CuCu is
View answer and explanation
Correct option: D
12
Option D (12) is correct because copper has an FCC structure, and the coordination number for FCC structures is 12.
327 Medium
327 of 361
Remembering +4 | -1
The logic behind ‘NOR’ gate is that it gives
View answer and explanation
Correct option: A
High output when both the inputs are low
Option A (High output when both the inputs are low) is correct because a NOR gate outputs a high signal only when both inputs are low.
328 Medium
328 of 361
Understanding +4 | -1
Which of the following is not equal to 1 in Boolean algebra?
View answer and explanation
Correct option: B
A⋅AˉA \cdot \bar A
Option B (A⋅AˉA \cdot \bar A) is correct because in Boolean algebra, A⋅AˉA \cdot \bar A always equals 0, while the other options equal 1.
329 Medium
329 of 361
Remembering +4 | -1
The band gap in germanium and silicon in eVeV respectively is
View answer and explanation
Correct option: A
0.7,1.1
Option A (0.7,1.1) is correct because the band gap for germanium is approximately 0.7 eV and for silicon is approximately 1.1 eV.
330 Medium
330 of 361
Analyzing +4 | -1
A semiconducting material has a band gap of 1 eV1 \mathrm{~eV}. Acceptor impurities are doped into it which create acceptor levels 1 meV1 \mathrm{~meV} above the valence band. Assume that the transition from one energy level to the other is almost forbidden if kTk T is less than 1/501 / 50 of the energy gap. Also, if kTk T is more than twice the gap, the upper levels have maximum population. The temperature of the semiconductor is increased from 0 K0 \mathrm{~K}. The concentration of the holes increases with temperature and after a certain temperature it becomes approximately constant. As the temperature is further increased, the hole concentration again starts increasing at a certain temperature. Find the order of the temperature range in which the hole concentration remains approximately constant.
View answer and explanation
Correct option: A
2020 to 230 K230 \mathrm{~K}
Option A (2020 to 230 K230 \mathrm{~K}) is correct because the hole concentration remains approximately constant within this temperature range, determined by the given conditions on kT and the band gap.
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