Semiconductor Questions — NEET Physics

Q1

The current / in the circuit shown below is: (All diodes are ideal and identical)

A

\dfrac{5}{3} \mathrm{~A}

B

\dfrac{5}{9} \mathrm{~A}

C

\dfrac{1}{3} \mathrm{~A}

D

\dfrac{15}{2} \mathrm{~A}

Correct Answer

Option D

Solution

For ideal diode, forward resistance $=0$ and reverse biased resistance $=\infty$ $\therefore \quad$ Circuit can be redrawn as,

\begin{aligned} I & =\frac{10}{2}+\frac{10}{4} \\ & =\frac{30}{4}=\frac{15}{2} \mathrm{~A} \end{aligned}

Q2

A full wave rectifier circuit with diodes

\left(D_1\right)

and

\left(D_2\right)

is shown in the figure. If input supply voltage

\mathrm{V}_{\text{in }}=220 \sin (100 \pi t)

volt, then at

t=15 \mathrm{msec}

A

\quad D_1

and

D_2

both are forward biased

B

\quad D_1

and

D_2

both are reverse biased

C

\quad D_1

is forward biased,

D_2

is reverse biased

D

\quad D_1

is reverse biased,

D_2

is forward biased.

Correct Answer

Option D

Solution

To determine the state of the diodes in the full wave rectifier circuit at $t = 15 \, \text{ms}$ , we analyze the input supply voltage given by: $V_{\text{in}} = 220 \sin(100 \pi t)$ Here are the key steps to explain the situation: Substitute the Time : Convert $t = 15 \, \text{ms}$ to seconds: $t = 0.015 \, \text{s}$ Calculate the Angular Frequency : $\omega = 100\pi$ Determine the Period : $\dfrac{2\pi}{T} = 100\pi \quad \Rightarrow \quad T = \dfrac{1}{50} \, \text{s} = 0.02 \, \text{s}$ Calculate the Phase : $t = \dfrac{3T}{4}$ This corresponds to the negative half-cycle of the sine wave, occurring when the input voltage goes negative.

Given the circuit configuration during the negative half-cycle: $D_1$ is reverse biased. $D_2$ is forward biased.

Hence, at $t = 15 \, \text{ms}$ , the diode $D_1$ is reverse biased, and diode $D_2$ is forward biased, allowing current to flow through diode $D_2$ .

Q3

The output (

Y

) of the given logic implementation is similar to the output of an/a ________ gate.

A OR

B NOR

C AND

D NAND

Correct Answer

Option B

Solution

\begin{aligned} & Y_1=\overline{A+B} \\ & Y_2=\overline{A \cdot B} \\ & Y=Y_1 \cdot Y_2 \\ & =\overline{A+B} \cdot \overline{A \cdot B} \\ & =\overline{(A+B)+A \cdot B} \\ & =\overline{A+B(1+A)} \\ & =\overline{A+B} \text{ NOR gate } \end{aligned}

Q4

The I-V characteristics shown above are exhibited by a

A Light emitting diode

B Zener diode

C Photodiode

D Solar cell

Correct Answer

Option D

Solution

The I-V characteristics of solar cell is

Q5

When the output of an OR gate is applied as input to a NOT gate, then the combination acts as a

A NAND gate

B NOR gate

C AND gate

D OR gate

Correct Answer

Option B

Solution

When the output of an OR gate is applied as input to a NOT gate, the combination acts as a NOR gate.

To understand why, let's examine the behavior of these logic gates.

An OR gate outputs a logic high (1) when at least one of its inputs is high.

Mathematically, this can be expressed as:

Y_{\text{OR}} = A + B

Where $A$ and $B$ are the inputs, and $Y_{\text{OR}}$ is the output of the OR gate.

An OR gate followed by a NOT gate inverts this output.

The NOT gate, also known as an inverter, changes a high input to a low output, and vice versa.

So the output of the NOT gate can be expressed as:

Y_{\text{NOT}} = \overline{Y_{\text{OR}}} = \overline{(A + B)}

This is exactly the logic operation performed by a NOR gate.

Therefore, this combination of an OR gate followed by a NOT gate behaves as a NOR gate.

So, the correct answer is Option B: NOR gate.

Q6

The output Y for the inputs A and B of the given logic circuit is:

A

A \cdot B

B

\bar{A} \cdot \bar{B}

C

A+B

D

\bar{A}+\bar{B}

Correct Answer

Option C

Solution

Y=\overline{\bar{A} \bar{B}}=A+B

Q7

A logic circuit provides the output

Y

as per the following truth table : The expression of the output Y is :

A

A \cdot B+\bar{A}

B

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

C

\bar{B}

D B

Correct Answer

Option C

Solution

According to given truth table, output is independent on value of

A

$\therefore$ Output

Y=\bar{B}

Q8

\text{ The output ( } Y \text{ ) of the given logic gate is similar to the output of an/a }

A NAND gate

B NOR gate

C OR gate

D AND gate

Correct Answer

Option D

Solution

\begin{aligned} Y_1 & =\overline{A \cdot A} \\ & =\bar{A} \\ Y_2 & =\overline{B+B} \\ & =\bar{B} \\ Y & =\overline{Y_1+Y_2} \\ & =\overline{\bar{A}+\bar{B}} \\ & =\overline{\bar{A}} \cdot \overline{\bar{B}} \end{aligned}

= A.B is similar to output of AND gate.

Q9

On the basis of electrical conductivity, which one of the following material has the smallest resistivity?

A Germanium

B Silver

C Glass

D Silicon

Correct Answer

Option B

Solution

Electrical conductivity is a measure of a material's ability to conduct an electric current.

The higher the conductivity, the lower the resistivity.

So, if we are looking for the material with the smallest resistivity, we are looking for the material with the highest electrical conductivity.

Among the options given, silver (Option B) is known to have the highest electrical conductivity and thus, the smallest resistivity.

Germanium (Option A) and silicon (Option D) are semiconductors, meaning they have moderate conductivity that can be manipulated, while glass (Option C) is generally a poor conductor, or an insulator.

Therefore, silver is the correct answer.

Q10

A p-type extrinsic semiconductor is obtained when Germanium is doped with:

A Antimony

B Phosphorous

C Arsenic

D Boron

Correct Answer

Option D

Solution

For p type semiconductor trivalent impurity added