In an n-p-n common emitter (CE) transistor the collector current changes from 5 $$\mathrm{mA}$$ to $$16 \mathrm{~mA}$$ for the change in base current from $$100~ \mu \mathrm{A}$$ and $$200 ~\mu \mathr

The current gain of a transistor in common emitter configuration is given by: $$\beta = \frac{I_C}{I_B}$$ where $$I_C$$ is the collector current and $$I_B$$ is the base current. In this case, the collector current changes from $$5 \mathrm{~mA}$$ to $$16 \mathrm{~mA}$$ for the change in base current from $$100~ \mu \mathrm{A}$$ to $$200 ~\mu \mathrm{A}$$. Therefore, we have: $$\Delta I_C = 16 \mathrm{~mA} - 5 \mathrm{~mA} = 11 \mathrm{~mA}$$ $$\Delta I_B = 200~ \mu \mathrm{A} - 100~ \mu \mathrm{A

Semiconductor

JEE Physics · 70 questions · Page 6 of 7 · Click an option or "Show Solution" to reveal answer

Q51

Choose the correct statement about Zener diode :

A It works as a voltage regulator in reverse bias and behaves like simple pn junction diode in forward bias.

B It works as a voltage regulator in both forward and reverse bias.

C It works as a voltage regulator only in forward bias.

D It works as a voltage regulator in forward bias and behaves like simple pn junction diode in reverse bias.

Correct Answer

Option A

Solution

Option A is the correct statement about Zener diode.

It works as a voltage regulator in reverse bias and behaves like a simple pn junction diode in forward bias.

When a Zener diode is reverse-biased, it operates in the breakdown region, where a relatively constant voltage is maintained across the diode, regardless of the current flowing through it.

This property makes it useful as a voltage regulator.

In forward bias, the voltage applied across the diode is in the same direction as the normal direction of current flow.

In this condition, the Zener diode behaves like a simple pn junction diode and allows current to flow in the forward direction.

Q52

The effect of increase in temperature on the number of electrons in conduction band (

\mathrm{n_e}

) and resistance of a semiconductor will be as:

\mathrm{n}_{\mathrm{e}}

decreases, resistance increases

B Both

\mathrm{n}_{\mathrm{e}}

and resistance increase

\mathrm{n}_{\mathrm{e}}

increases, resistance decreases

D Both

\mathrm{n}_{\mathrm{e}}

and resistance decrease

Correct Answer

Option C

Solution

As temperature increases $n_{e}$ increases, this results in increase in conductance. $\therefore T$ increases, $n_{e}$ increases and $R$ decreases.

Q53

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 than the energy gap of the semiconductor used. Choose the correct answer from the options given below:

A A

B B

C C and D

D C

Correct Answer

Option D

Solution

The correct answer is C.

It is not correct that a light-emitting diode (LED) emits light only when it is reverse biased.

In fact, an LED emits light only when it is forward biased, which is stated correctly in option B.

Option A is also correct, as an LED is indeed a heavily doped p-n junction.

Option D is also correct, as the energy of the light emitted by an LED is equal to or slightly less than the energy gap of the semiconductor material used in the device.

Therefore, the statement that is not correct in the case of light-emitting diodes is C.

Q54

Statement I : When a Si sample is doped with Boron, it becomes P type and when doped by Arsenic it becomes N-type semi conductor such that P-type has excess holes and N-type has excess electrons. Statement II : When such P-type and N-type semi-conductors, are fused to make a junction, a current will automatically flow which can be detected with an externally connected ameter. In the light of above statements, choose the most appropriate answer from the options given below

A Statement I is incorrect but statement II is correct

B Both Statement I and statement II are correct

C Statement I is correct but statement II is incorrect

D Both Statement I and Statement II are incorrect

Correct Answer

Option C

Solution

Statement I is correct but in statement II we cannot detect the current through ammeter thus the statement II is incorrect.

Q55

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R Assertion A : Photodiodes are used in forward bias usually for measuring the light intensity. Reason R : For a p-n junction diode, at applied voltage V the current in the forward bias is more than the current in the reverse bias for

\mathrm{|{V_z}| > \pm v \ge |{v_0}|}

where

\mathrm{v_0}

is the threshold voltage and

\mathrm{V_z}

is the breakdown voltage. In the light of the above statements, choose the correct answer from the options given below

A Both A and R are true but R is NOT the correct explanation of A

B A is true but R is false

C Both A and R are true and R is the correct explanation of A

D A is false but R is true

Correct Answer

Option D

Solution

Photodiodes are used in reverse bias therefore the assertion is incorrect.

Q56

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R Assertion A : Photodiodes are preferably operated in reverse bias condition for light intensity measurement. Reason R : The current in the forward bias is more than the current in the reverse bias for a

p-n

junction diode. In the light of the above statements, choose the correct answer from the options given below :

A Both A and R are true and R is the correct explanation of A

B A is false but R is true

C A is true but R is false

D Both A and R are true but R is NOT the correct explanation of A

Correct Answer

Option D

Solution

Photodiodes are preferably operated in reverse bias condition for light intensity measurement because it increases the width of depletion layer, therefore both are correct but not the correct explanation.

Q57

In an n-p-n common emitter (CE) transistor the collector current changes from 5

\mathrm{mA}

16 \mathrm{~mA}

for the change in base current from

100~ \mu \mathrm{A}

and

200 ~\mu \mathrm{A}

, respectively. The current gain of transistor is __________.

A 210

B 0.9

C 9

D 110

Correct Answer

Option D

Solution

The current gain of a transistor in common emitter configuration is given by:

\beta = \frac{I_C}{I_B}

where

I_C

is the collector current and

I_B

is the base current. In this case, the collector current changes from

5 \mathrm{~mA}

16 \mathrm{~mA}

for the change in base current from

100~ \mu \mathrm{A}

200 ~\mu \mathrm{A}

. Therefore, we have:

\Delta I_C = 16 \mathrm{~mA} - 5 \mathrm{~mA} = 11 \mathrm{~mA}

\Delta I_B = 200~ \mu \mathrm{A} - 100~ \mu \mathrm{A} = 100~ \mu \mathrm{A}

Therefore, the current gain of the transistor is:

\beta = \frac{\Delta I_C}{\Delta I_B} = \frac{11 \mathrm{~mA}}{100~ \mu \mathrm{A}} = 110

Therefore, the current gain of the transistor is 110.

Q58

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 from the

\mathrm{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

A Both A and R are correct but R is NOT the correct explanation of A

B A is correct but R is not correct

C A is not correct but R is correct

D Both A and R are correct and R is the correct explanation of A

Correct Answer

Option B

Solution

A p-n junction consists of a p-type semiconductor (which has an excess of holes) in contact with an n-type semiconductor (which has an excess of electrons).

In a forward-biased p-n junction, an external voltage is applied such that the positive terminal is connected to the p-side and the negative terminal is connected to the n-side.

This configuration promotes the flow of majority charge carriers (holes from the p-side and electrons from the n-side) across the junction.

There are two types of currents in a p-n junction: drift current and diffusion current.

Drift current is caused by the electric field due to the built-in potential, which opposes the flow of majority charge carriers.

Diffusion current is caused by the concentration gradient of the charge carriers, which promotes the flow of majority charge carriers.

When the p-n junction is forward biased, the applied voltage reduces the potential barrier, allowing more majority charge carriers to flow across the junction.

This results in an increase in the diffusion current.

In a forward-biased p-n junction, the diffusion current is indeed greater than the drift current in magnitude, so Assertion A is correct.

Regarding Reason R: The diffusion current in a p-n junction is actually from the p-side to the n-side, as holes (majority charge carriers in the p-side) move from the p-side to the n-side, and electrons (majority charge carriers in the n-side) move from the n-side to the p-side.

Therefore, Reason R is incorrect.

Q59

A light emitting diode (LED) is fabricated using GaAs semiconducting material whose band gap is

1.42 \mathrm{~eV}

. The wavelength of light emitted from the LED is :

A 1243 nm

B 875 nm

C 650 nm

D 1400 nm

Correct Answer

Option B

Solution

The wavelength of light emitted by a Light Emitting Diode (LED) fabricated using a semiconducting material can be determined by the energy band gap of the material.

The energy of the photon emitted, which corresponds to the band gap energy, is given by the equation:

E = \frac{hc}{\lambda}

Where: $E$ is the energy of the emitted photon (in joules when $h$ and $c$ are in SI units), corresponding to the band gap energy of the material. $h$ is Planck's constant ( $6.626 \times 10^{-34} \, \mathrm{m^2\,kg/s}$ ). $c$ is the speed of light in vacuum ( $3.0 \times 10^{8} \,\mathrm{m/s}$ ). $\lambda$ is the wavelength of the emitted light (in meters).

However, since the energy band gap given is in electronvolts (eV), and we are looking for the wavelength in nanometers (nm), we can use the energy formula directly in terms of eV and then do the unit conversion conveniently.

The conversion between energy (in eV) and wavelength (in nm) without needing to convert eV to Joules is facilitated by the equation:

\lambda(\mathrm{nm}) = \frac{1240}{E(\mathrm{eV})}

Here, 1240 nm·eV is a conversion factor used for directly converting energy in eV to wavelength in nm.

Given the band gap energy of GaAs is $1.42 \mathrm{~eV}$ , the wavelength ( $\lambda$ ) of light emitted can be found as:

\lambda = \frac{1240}{1.42} = 873.24 \mathrm{~nm}

Thus, the wavelength of light emitted from the LED is approximately 873.24 nm, which is closest to: Option B: 875 nm.

Q60

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}

A 407 nm

B 103.5 nm

C 414 nm

D 207 nm

Correct Answer

Option D

Solution

The energy required to create a hole in a p-type semiconductor can be directly related to the acceptor level because this energy level represents the minimum energy required to excite an electron from the valence band into the acceptor level, effectively creating a hole.

The acceptor level is given as

6 \, \text{eV}

To find the maximum wavelength of light that can excite an electron into this level, we use the equation that relates the energy (

E

) of a photon to its wavelength ( $\lambda$ ):

E = \frac{hc}{\lambda}

Where:

E

is the energy in electronvolts (eV),

h

is Planck's constant,

c

is the speed of light, and $\lambda$ is the wavelength in nanometers (nm). The product of

hc

is given as

1242 \, \text{eV nm}

, allowing us to solve for $\lambda$ directly:

\lambda = \frac{hc}{E}

Substituting the given values:

\lambda = \frac{1242 \, \text{eV nm}}{6 \, \text{eV}} = 207 \, \text{nm}

Thus, the maximum wavelength of light that can create a hole in the semiconductor by exciting an electron into the acceptor level is

207 \, \text{nm}

. Therefore, the correct answer is Option D: 207 nm.

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