In the case of constants ${\rm{\alpha\; and\; \beta }}$ of a transistor

${\rm{\beta }} > {\rm{ }}1,$${\rm{\alpha }} < 1$

${\rm{\alpha }} = {\rm{\beta }}$

The gain factor of an amplifier is increased from 10 to 12 as the load resistance is changed from $4 \, k\Omega$ to $8 \, k\Omega.$ Calculate (a) the amplification factor and (b) the plate resistance.

(a) 15; (b) $4 \, \mathrm{k} \Omega$

(a) 7; (b) $2 \, \mathrm{k} \Omega$

(a) 17; (b) $6 \, \mathrm{k} \Omega$

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JEE • PHYSICS • SEMI CONDUCTOR DEVICES

Semi conductor devices practice · Mixed difficulty

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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 1-30 of 370 questions

1 Medium

1 of 370

Remembering +4 | -1

A PN-junction has a thickness of the order of

View answer and explanation

Correct option: C

{10^{ - 6}}\,{\text{m}}

Option C (

{10^{ - 6}}\,{\text{m}}

) is correct because the depletion region in a PN junction is typically on the order of micrometers.

2 Medium

2 of 370

Understanding +4 | -1

In a

PNP

transistor the base is the

N

-region. Its width relative to the

P

-region is

View answer and explanation

Correct option: A

Smaller

Option A (Smaller) is correct because the base in a PNP transistor is intentionally made very thin to minimize recombination of carriers.

3 Medium

3 of 370

Understanding +4 | -1

A solid reflects incident light and its electrical conductivity decreases with temperature. The binding in this solid is

View answer and explanation

Correct option: C

Metallic

Option C (Metallic) is correct because metals reflect light due to free electrons and their conductivity decreases with increasing temperature.

4 Medium

4 of 370

Remembering +4 | -1

An alternating current can be converted into direct current by a

View answer and explanation

Correct option: D

Rectifier

Option D (Rectifier) is correct because a rectifier is an electrical device that converts alternating current (AC) to direct current (DC).

5 Medium

5 of 370

Understanding +4 | -1

A N–P–N transistor is used in inverse active mode of operation then its emitter base junction should be in

View answer and explanation

Correct option: B

Reverse biased

Option B (Reverse biased) is correct because in inverse active mode, the emitter-base junction is reverse biased.

6 Medium

6 of 370

Understanding +4 | -1

In the case of constants

\mathrm{\alpha\; and\; \beta }

of a transistor

View answer and explanation

Correct option: D

\mathrm{\beta } > \mathrm{ }1,

\mathrm{\alpha } < 1

Option D (

\mathrm{\beta } > \mathrm{ }1,\mathrm{\alpha } < 1

) is correct because in a transistor, alpha (

\alpha

) is the current gain from emitter to collector, which is always less than 1, and beta (

\beta

) is the current gain from base to collector, which is typically greater than 1.

7 Medium

7 of 370

Understanding +4 | -1

When forward bias is applied to a

P

N

junction, then what happens to the potential barrier

{V_B}

, and the width of charge depleted region

x

View answer and explanation

Correct option: D

{V_B}

decreases,

x

decreases

Option D (

{V_B}

decreases,

x

decreases) is correct because forward bias reduces the potential barrier and depletion region width.

8 Medium

8 of 370

Application +4 | -1

In a common-base mode of a transistor, the collector current is 5.488 mA for an emitter current of 5.60 mA. The value of the base current amplification factor (

\mathrm{\beta }

) will be

View answer and explanation

Correct option: A

Option A (49) is correct because using the formula

\mathrm{\beta } = \frac{{{I_C}}}{{{I_E} - {I_C}}}

with the given values yields 49.

9 Medium

9 of 370

Understanding +4 | -1

Which one of the following statement is not correct in the case of light emitting diodes?
A) It is a heavily doped

p-n

junction.
B) It emits light only when it is forward biased.
C) It emits light only when it is reverse biased.
D) The energy of the light emitted is equal to or slightly less then the energy gap of the semiconductor used.
Choose the correct answer from the options given below:

View answer and explanation

Correct option: B

C

Option B (C) is correct because LEDs emit light only when forward biased, not reverse biased.

10 Medium

10 of 370

Application +4 | -1

Transistor with

\beta=60

is connected to common base configuration. For an emitter current of

5 \, \mathrm{mA}

, the value of collector current will be nearly

View answer and explanation

Correct option: D

4.9 \mathrm{~mA}

Option D (

4.9 \mathrm{~mA}

) is correct because it is calculated using the relationship between

\beta

and

\alpha

in a common base configuration.

11 Medium

11 of 370

Understanding +4 | -1

A full wave rectifier circuit consists of two p-n junction diodes, a centre-tapped transformer, capacitor and a
load resistance. Which of these components remove the ac ripple from the rectified output?

View answer and explanation

Correct option: A

Capacitor

Option A (Capacitor) is correct because a capacitor is used in rectifier circuits to filter out the AC ripple and provide a smoother DC output.

12 Medium

12 of 370

Understanding +4 | -1

Let

{n_e}

and

{n_h}

represent the number density of electrons and holes in a semiconductor. Then

View answer and explanation

Correct option: D

{n_e}

{n_h}

if the semiconductor is intrinsic

Option D (

{n_e}

{n_h}

if the semiconductor is intrinsic) is correct because in an intrinsic semiconductor, the number of electrons equals the number of holes.

13 Medium

13 of 370

Application +4 | -1

26 If

\alpha=0.98

and current though emitter

\mathrm{i}_{\mathrm{e}}=20 \mathrm{~mA}

, the value of

\beta

is -

View answer and explanation

Correct option: B

Option B (49) is correct because it is the value of β calculated using the given value of α and the relation between α and β.

14 Medium

14 of 370

Remembering +4 | -1

Which gate is also called as inverter?

View answer and explanation

Correct option: C

NOT

Option C (NOT) is correct because a NOT gate performs the inversion operation, which is why it is also called an inverter.

15 Medium

15 of 370

Application +4 | -1

A=B=1

, then in terms of Boolean algebra the value of

A \cdot B+A

is not equal to

View answer and explanation

Correct option: D

\bar{A} \cdot B

Option D (

\bar{A} \cdot B

) is correct because when A=B=1,

A \cdot B + A = 1

, but

\bar{A} \cdot B = 0

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16 Medium

16 of 370

Application +4 | -1

In a transistor circuit, the collector current is 50

\,{mA}

and the base current is

1\, {~mA}

. The current gain

\alpha

View answer and explanation

Correct option: C

\frac{50}{51}

Option C (

\frac{50}{51}

) is correct because the current gain

\alpha

is defined as

I_C/I_E

and

I_E = I_C + I_B

17 Medium

17 of 370

Remembering +4 | -1

p

-type semiconductor are

View answer and explanation

Correct option: B

Produced when boron is added as an impurity

Option B is correct. p-type semiconductors are produced when trivalent impurities like Boron are added to intrinsic semiconductors.

18 Medium

18 of 370

Analyzing +4 | -1

Statement I: In a common-emitter transistor amplifier the input current is much less than output current.

Statement II: The common-emitter transistor amplifier has very high input impedance.

View answer and explanation

Correct option: C

Statement I is true; Statement II is false

Option C (Statement I is true; Statement II is false) is correct because in a common-emitter amplifier, the output current is much larger than the input current, but the input impedance is relatively low.

19 Medium

19 of 370

Remembering +4 | -1

In silicon when phosphorus is doped ……. is formed

View answer and explanation

Correct option: B

n

-type semiconductor

Option B (

n

-type semiconductor) is correct because phosphorus is a pentavalent impurity, which donates extra electrons to silicon, creating an

n

-type semiconductor.

20 Medium

20 of 370

Application +4 | -1

The gain factor of an amplifier is increased from 10 to
12 as the load resistance is changed from

4 \, k\Omega

8 \, k\Omega.

Calculate
(a) the amplification factor and (b) the plate
resistance.

View answer and explanation

Correct option: A

(a) 15; (b)

2 \, \mathrm{k} \Omega

Option D (Statement- 1 is False, Statement-2 is True) is correct because the first statement requires zero net force, not just zero external torque, while the second statement accurately reflects momentum conservation in an isolated system.

21 Medium

21 of 370

Application +4 | -1

A transistor is operated in common – emitter configuration at

{V_C} = 2\,V

such that a change in the base current from

100\,\mu A

200\,\mu A

produces a change in the collector current is 5 mA to 10 mA then current gain is

View answer and explanation

Correct option: D

Option D (50) is correct because the current gain is the ratio of change in collector current to the change in base current.

22 Medium

22 of 370

Understanding +4 | -1

In n-type silicon, which of the following statement is true

View answer and explanation

Correct option: C

Holes are minority carriers and pentavalent atoms are the dopants

Option C (Holes are minority carriers and pentavalent atoms are the dopants) is correct because in n-type silicon, pentavalent impurities are added, making electrons majority carriers and holes minority carriers.

23 Hard

23 of 370

Remembering +4 | -1

The emission of electrons from the host atoms due
to the high electric field is known as

View answer and explanation

Correct option: B

internal field emission

Option B (internal field emission) is correct because it describes the process of electron emission from host atoms due to a strong electric field.

24 Medium

24 of 370

Application +4 | -1

In an p-n-p transistor working as a common base amplifier current gain is 0.96 and emitter current is 7.2 mA. The base current is

View answer and explanation

Correct option: B

0.29 mA

Option B (0.29 mA) is correct because it represents the base current calculated using the current gain and emitter current.

25 Medium

25 of 370

Remembering +4 | -1

In LED visible light is produced by

View answer and explanation

Correct option: B

Gallium arsenide

Option A (Gallium phosphide) is correct because Gallium phosphide is a semiconductor material used in LEDs to produce visible light.

26 Medium

26 of 370

Remembering +4 | -1

In order to rectify an alternating current one uses a

View answer and explanation

Correct option: B

Diode

Option B (Diode) is correct because a diode allows current to flow in only one direction, thus rectifying alternating current (AC) to direct current (DC).

27 Medium

27 of 370

Understanding +4 | -1

In a

p - n

junction diode

View answer and explanation

Correct option: A

The current in the reverse biased condition is generally very small

Option A (The current in the reverse biased condition is generally very small) is correct because in reverse bias, the depletion region widens, hindering the flow of majority carriers.

28 Medium

28 of 370

Remembering +4 | -1

In semiconductors the forbidden energy gap between V.B. and C.B. is of the order of

View answer and explanation

Correct option: B

1 \, \mathrm{eV}

Option B (

1 \, \mathrm{eV}

) is correct because the forbidden energy gap in semiconductors is typically around 1 eV.

29 Medium

29 of 370

Understanding +4 | -1

The plate current in a diode is

20 \mathrm{~mA}

when the plate voltage is

50 \mathrm{~V}

60 \mathrm{~V}

. What will be the current if the plate voltage is

70 \mathrm{~V}?

View answer and explanation

Correct option: A

20 \mathrm{~mA}

Option A (

20 \mathrm{~mA}

) is correct because the diode is operating in the saturation region, where the current remains constant despite changes in voltage.

30 Medium

30 of 370

Remembering +4 | -1

The coordination number of body centred crystal is

View answer and explanation

Correct option: B

Option B (8) is correct because in a body-centered cubic crystal, each atom is in contact with eight other atoms.

Revision notes and key takeaways

Review SEMI CONDUCTOR DEVICES in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 370 single-correct-answer MCQs on SEMI CONDUCTOR DEVICES.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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