A $p - n\;$ junction has acceptor impurity concentration of ${10^{17}}{\rm{c}}{{\rm{m}}^{ - 3}}$ in the $P$ side and donor impurity concentration of ${10^{16}}$ ${\rm{c}}{{\rm{m}}^{ - 3}}$ in the $N$ side. What is the contact potential at the junction? ($kT = $ thermal energy, intrinsic carrier concentration ${n_i} = 1.4 \times {10^{10}}$ ${\rm{c}}{{\rm{m}}^{ - 3}}$ )

$\left( {\frac{{kT}}{e}} \right)$ ln ($2.5 \times {10^{23}}$)

$\left( {\frac{{kT}}{e}} \right)$ ln (${10^{23}}$)

$\left( {\frac{{kT}}{e}} \right)$ ln (${10^9}$)

A $p$-type material is electrically …….

Depends on the concentration of $p$ impurities

In a common emitter amplifier, using output resistance of $5000\;{\rm{\Omega }}$ and input resistance of $2000\;{\rm{\Omega }}$, if the peak value of input signal voltage is 10 mV and ${\rm{\beta }} = 50$, then peak value of output voltage is

$5 \times {10^{ - 6}}{\rm{\;V}}$

$12.50 \times {10^{ - 6}}{\rm{\;V}}$

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NEET • PHYSICS • SEMI CONDUCTOR DEVICES

Semi conductor devices practice · Mixed difficulty

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1 Medium

1 of 108

Analyzing +4 | -1

Following is the relation between current and charge

I = A{T^2}{e^{q/{V_L}}}

, then value of

{V_L}

will be

View answer and explanation

Correct option: C

\frac{{kT}}{V}

Option C (

\frac{{kT}}{V}

) is correct because by comparing the given equation with the standard form,

{V_L}

is identified as

\frac{{kT}}{V}

2 Medium

2 of 108

Understanding +4 | -1

To measure light intensity we use

View answer and explanation

Correct option: D

Photodiode with reverse bias

Option D (Photodiode with reverse bias) is correct because photodiodes are used in reverse bias to measure light intensity by detecting changes in the reverse current.

3 Easy

3 of 108

Application +4 | -1

The ratio of RMS output voltage of full wave rectifier and half wave rectifier is

\sqrt{n}.

The value of

n

is? (The input voltage is sine wave).

View answer and explanation

Correct option: A

Option A (2) is correct because the ratio of RMS output voltage of a full-wave rectifier to a half-wave rectifier is

\sqrt{2}

4 Medium

4 of 108

Understanding +4 | -1

The truth table shown in figure is for
A    0   0   1   1
B    0   1   0   1
Y    1   0   0   1 

View answer and explanation

Correct option: C

XNOR

Option C (XNOR) is correct because the output Y is 1 when both inputs A and B are the same (both 0 or both 1), which is the characteristic of an XNOR gate.

5 Medium

5 of 108

Remembering +4 | -1

The average value of output direct current in a half wave rectifier is

View answer and explanation

Correct option: A

{I_0}/\pi

Option A (

{I_0}/\pi

) is correct because the average value of the output direct current in a half-wave rectifier is calculated by integrating the current over a full cycle and dividing by the period.

6 Hard

6 of 108

Application +4 | -1

p - n\;

junction has acceptor impurity concentration of

{10^{17}}\mathrm{c}{\mathrm{m}^{ - 3}}

in the

P

side and donor impurity concentration of

{10^{16}}

\mathrm{c}{\mathrm{m}^{ - 3}}

in the

N

side. What is the contact potential at the junction? (

kT =

thermal energy, intrinsic carrier concentration

{n_i} = 1.4 \times {10^{10}}

\mathrm{c}{\mathrm{m}^{ - 3}}

)

View answer and explanation

Correct option: A

\left( {\frac{{kT}}{e}} \right)

ln (

4 \times {10^{12}}

)

Option A (

\left( {\frac{{kT}}{e}} \right)

ln (

4 \times {10^{12}}

)) is correct because the contact potential is calculated using the formula

V_0 = \frac{{kT}}{e} \ln \left( {\frac{{{N_A}{N_D}}}{{n_i^2}}} \right)

7 Medium

7 of 108

Application +4 | -1

When the forward bias voltage of a diode is changed from 0.6 V to 0.7 V, the current changes from 5 mA to 15 mA. Then its forward bias resistance is

View answer and explanation

Correct option: C

\mathrm{\Omega }

Option C (10

\mathrm{\Omega }

) is correct because the forward bias resistance is calculated using Ohm's law:

R = \frac{{\Delta V}}{{\Delta I}}

8 Medium

8 of 108

Understanding +4 | -1

p

-type material is electrically …….

View answer and explanation

Correct option: C

Neutral

A p-type material is neutral because the number of protons in the nuclei equals the number of electrons orbiting the nuclei.

9 Medium

9 of 108

Analyzing +4 | -1

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

Assertion A: Diffusion current in a p-n junction is greater than the drift current in magnitude if the junction is forward biased.

Reason R: Diffusion current in a p-n junction is form the n-side to the p-side if the junction is forward biased.

In the light of the above statements, choose the most appropriate answer from the options given below.

View answer and explanation

Correct option: B

A is correct but R is not correct

Option B is correct because Assertion A is correct: diffusion current is greater than drift current in forward bias. However, Reason R is incorrect: diffusion current is from p-side to n-side in forward bias.

10 Hard

10 of 108

Understanding +4 | -1

Given below are three statements.

A. At absolute zero, Si acts as a perfect insulator.

B. When p-n junction diode is forward biased, both the thickness of depletion region and the barrier height are increased.

C. Order of energy band gap of C, Si and Ge is

(\mathrm{E})_{\mathrm{C}}>(\mathrm{E})_{\mathrm{Si}}>(\mathrm{E})_{\mathrm{Ge}}.

Choose the correct answer from the options given below.

View answer and explanation

Correct option: D

Only A & C are correct

Only A & C are correct because statement B is incorrect; forward biasing decreases the depletion region and barrier height.

11 Medium

11 of 108

Analyzing +4 | -1

Assertion: Silicon is preferred over germanium for making semiconductor devices.

Reason: The energy gap for germanium is more than the energy gap of silicon.

View answer and explanation

Correct option: C

Assertion is true but reason is false

Option C (Assertion is true but reason is false) is correct because silicon is preferred due to its wider temperature range and the energy gap of germanium is smaller than silicon.

12 Medium

12 of 108

Remembering +4 | -1

The temperature coefficient of resistance of a semiconductor

View answer and explanation

Correct option: B

Is always negative

Option B (Is always negative) is correct because the resistance of a semiconductor decreases as temperature increases, indicating a negative temperature coefficient.

13 Medium

13 of 108

Remembering +4 | -1

The electrical circuit used to get smooth DC output from a rectifier circuit is called

View answer and explanation

Correct option: A

Filter

Option A (Filter) is correct because a filter circuit is used to smooth the DC output from a rectifier.

14 Easy

14 of 108

Understanding +4 | -1

The depletion layer of a p-n junction:

View answer and explanation

Correct option: B

Acts like an insulating zone under reverse bias

Option B (Acts like an insulating zone under reverse bias) is correct because the depletion region is devoid of free charge carriers and thus acts as an insulator, especially under reverse bias which widens the region.

15 Medium

15 of 108

Application +4 | -1

In a silicon transistor, the base current is changed by

20\mu A

. This results in a change of 2 mA in the collector current. If the input resistance and load resistance of an amplifier are

1k\Omega

and

5k\Omega

respectively, what is the trans conductance of the transistor in SI units?

View answer and explanation

Correct option: B

0.1

Option B (0.1) is correct because transconductance is the change in collector current divided by the change in base current.

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16 Medium

16 of 108

Understanding +4 | -1

Formation of covalent bonds in compounds exhibits

View answer and explanation

Correct option: A

Wave nature of electron

Option A (Wave nature of electron) is correct because covalent bonds are formed by the sharing of electrons, which is best explained by considering the wave nature of electrons and their ability to form standing waves between atoms.

17 Medium

17 of 108

Analyzing +4 | -1

Statement-I: The temperature coefficient of resistance is positive for metals and negative for semiconductors.

Statement-II: On raising the temperature, in metals drift velocity increases but in semiconductors more charge carriers are released.

View answer and explanation

Correct option: A

If both statements are correct.

Option A (If both statements are correct) is correct because metals have a positive temperature coefficient of resistance, while semiconductors have a negative one due to increased carrier concentration with temperature.

18 Medium

18 of 108

Understanding +4 | -1

Any digital circuit can be realised by repetitive use of only

View answer and explanation

Correct option: D

NOR gate

Option D (NOR gate) is correct because a NOR gate can be used to implement any other logic gate, making it a universal gate.

19 Medium

19 of 108

Remembering +4 | -1

The phenomenon of thermionic emission was discovered by

View answer and explanation

Correct option: D

Thomas Edison

Option D (Thomas Edison) is correct because Thomas Edison discovered the phenomenon of thermionic emission.

20 Easy

20 of 108

Understanding +4 | -1

When a

\mathbf{P}-\mathbf{N}

junction is reverse biased :

View answer and explanation

Correct option: A

Holes and electrons move away from the junction

Option A (Holes and electrons move away from the junction) is correct because when a p-n junction is reverse biased, the applied voltage increases the width of the depletion region, causing charge carriers to move away from the junction.

21 Medium

21 of 108

Application +4 | -1

In a common emitter transistor amplifier

\mathrm{\beta } = 60,\;{R_0} = 5000\mathrm{\Omega }

and internal resistance of a transistor is

500\mathrm{\Omega }

. The voltage amplification of amplifier will be

View answer and explanation

Correct option: C

600

Option C (600) is correct because it is the voltage amplification calculated using the given parameters in a common emitter transistor amplifier.

22 Medium

22 of 108

Application +4 | -1

Two amplifiers are connected one after other in series. The first amplifier has voltage gain of 15 and second has voltage gain of 20. If input signal voltage is 0.3 V, the output signal voltage is

View answer and explanation

Correct option: C

90 V

Option D (9V) is correct because the total voltage gain is the product of the individual gains, and the output voltage is the input voltage multiplied by the total gain.

23 Medium

23 of 108

Application +4 | -1

In a p-n junction, the depletion region is 400nm wide and an electric field of

5 \times {10^5}V/m

exists in it. What should be the minimum kinetic energy of a conduction electron which can diffuse from the inside to the p-side?

View answer and explanation

Correct option: C

0.2 eV

Option C (0.2 eV) is correct because the minimum kinetic energy required is equal to the potential difference across the depletion region, which can be calculated as

E = qV = qEd

, where

E

is the electric field and

d

is the width of the depletion region.

24 Medium

24 of 108

Understanding +4 | -1

Avalanche breakdown of p-n junction diode is due to

View answer and explanation

Correct option: A

Collision of electrons with atoms

Option A (Collision of electrons with atoms) is correct because avalanche breakdown occurs when accelerated electrons gain enough kinetic energy to ionize atoms in the depletion region through collisions, creating more charge carriers.

25 Medium

25 of 108

Understanding +4 | -1

A piece of semiconductor is connected in series in an electric circuit. On increasing the temperature, the current in the circuit will

View answer and explanation

Correct option: C

Increase

Option C (Increase) is correct because increasing the temperature of a semiconductor increases the number of charge carriers, leading to increased current.

26 Medium

26 of 108

Remembering +4 | -1

At room temperature, a

P

-type semiconductor has

View answer and explanation

Correct option: A

Large number of holes and few electrons

Option A (Large number of holes and few electrons) is correct because P-type semiconductors are doped with acceptor impurities, creating an excess of holes.

27 Medium

27 of 108

Analyzing +4 | -1

\overline{A \cdot \bar{B}+\bar{A} \cdot B}

is equivalent to

View answer and explanation

Correct option: B

(A+\bar{B})+(\bar{A}+B)

Option B,

(A+\bar{B})+(\bar{A}+B)

, is correct because it represents the complement of the XOR operation, which is equivalent to the given expression.

28 Medium

28 of 108

Analyzing +4 | -1

A semi conductor has an electron concentration of

8 \times 10^{13}

per

\mathrm{cm}^3

and a hole concentration of

5 \times 10^{12}

per

\mathrm{cm}^3

. The electron mobility is

25000 \mathrm{~cm}^2 \mathrm{~V}^{-1} \mathrm{~s}^{-1}

and the hole mobility is

100 \mathrm{~cm}^2 \mathrm{~V}^{-1} \mathrm{~s}^{-1}

. Then,

View answer and explanation

Correct option: A

The semiconductor is n-type.

Option A (The semiconductor is n-type) is correct because the electron concentration is significantly higher than the hole concentration.

29 Easy

29 of 108

Application +4 | -1

The acceptor level of a p-type semiconductor is

6 \mathrm{~eV}

. The maximum wavelength of light which can create a hole would be : Given

\mathrm{hc}=1242 \mathrm{~eV} \mathrm{~nm}

View answer and explanation

Correct option: C

207 nm

Option C (207 nm) is correct because the maximum wavelength corresponds to the minimum energy required to create a hole, which is equal to the acceptor level energy. Using the formula

E = \frac{hc}{\lambda}

, we can solve for

\lambda

30 Medium

30 of 108

Application +4 | -1

In a common emitter amplifier, using output resistance of

5000\;\mathrm{\Omega }

and input resistance of

2000\;\mathrm{\Omega }

, if the peak value of input signal voltage is 10 mV and

\mathrm{\beta } = 50

, then peak value of output voltage is

View answer and explanation

Correct option: C

1.25 V

Option C (1.25 V) is correct because it is the calculated peak value of the output voltage using the given parameters in a common emitter amplifier.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 108 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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