Semiconductor — Page 4 | NEET Physics

Q31

To get output 1 for the following circuit, the correct choice for the input is

A A = 1, B = 1, C = 0

B A = 1, B = 0, C = 1

C A = 0, B = 1, C = 0

D A = 1, B = 0, C = 0

Correct Answer

Option B

Solution

Output of the circuit, Y = (A + B)·C Y = 1 if C = 1 and A = 0, B = 1 or A = 1, B = 0 or A = B = 1

Q32

Consider the junction diode as ideal. The value of current flowing through AB is

A 10

-

1 A

B 10

-

3 A

C 0 A

D 10

-

2 A

Correct Answer

Option D

Solution

Here, the p-n junction diode is forward biased, hence it offers zero resistance. So I AB =

{{4 - \left( { - 6} \right)} \over {1 \times {{10}^3}}}

= 10 -2 A

Q33

The input signal given to a CE amplifier having a voltage gain of 150 is V i = 2cos(15t +

{\pi \over 3}

). The corresponding output signal will be

A 2cos

\left( {15t + {{5\pi } \over 6}} \right)

B 300cos

\left( {15t + {{4\pi } \over 3}} \right)

C 300cos

\left( {15t + {{\pi } \over 3}} \right)

D 75cos

\left( {15t + {{2\pi } \over 3}} \right)

Correct Answer

Option B

Solution

Given, V i = 2cos(15t +

{\pi \over 3}

) and Voltage gain A V = 150 For CE transistor phase difference between input and output signal is $\pi$ = 180 o Using formula, A V =

{{{V_0}} \over {{V_i}}}

$\Rightarrow$ V 0 = A V $\times$ V i = 150 $\times$ 2cos(15t +

{\pi \over 3}

+ $\pi$ ) = 300cos

\left( {15t + {{4\pi } \over 3}} \right)

Q34

Which logic gate is represented by the following combination of logic gates ?

A AND

B NOR

C OR

D NAND

Correct Answer

Option A

Solution

The Boolean expression of this arrangement is Y =

\overline {\overline A + \overline B }

=

\overline{\overline A} + \overline{\overline B}

= A.B Thus, the combination represents AND gate.

Q35

The barrier potential of a p-n junction depends on (1) type of semiconductor material (2) amount of doping (3) temperature Which one of the following is correct ?

A (1) and (2) only

B (2) only

C (2) and (3) only

D (1), (2) and (3)

Correct Answer

Option D

Solution

The barrier potential depends on type of semiconductor (For Si, V b = 0.7 V and for Ge, V b = 0.3 V), amount of doping and also on the temperature.

Q36

The output from a NAND gate is divided into two in parallel and fed to another NAND gate. The resulting gate is a

A AND gate

B NOR gate

C OR gate

D NOT gate

Correct Answer

Option A

Solution

C' =

\overline {A.B}

$\Rightarrow$ C =

\overline{\overline {A.B}}

= A.B Hence the resultant gate is AND gate.

Q37

One way in which the operation of a n-p-n transistor differs from that of a p-n-p

A The emitter junction injects minority carries into the base region of the p-n-p

B The emitter injects holes into the base of the p-n-p and electrons into the base region of n-p-n

C The emitter injects holes into the base of n-p-n

D The emitter junction is reversed biased in n-p-n

Correct Answer

Option B

Solution

In p-n-p transistor holes are injected into the base while electrons are injected into the base of n-p-n transistor.

Emitter-base junction is forward biased.

Q38

In an unbiased p-n junction, holes diffuse from the p-region to n-region because of

A The attraction of free electrons of n-region.

B The higher hole concentration in p-region than that in n-region.

C The higher concentration of electrons in the n-region than that in the p-region.

D The potential difference across the p-n junction.

Correct Answer

Option B

Solution

The higher hole concentration is in p-region than that in n-region.

Q39

In a n-type semiconductor, which of the following statement is true.

A Holes are minority carries and pentavalent atoms are dopants.

B Holes are majority carries and trivalent atoms are dopants.

C Electrons are majority carries and trivalent atoms are dopants.

D Electrons are minority carriers and pentavalent atoms are dopants.

Correct Answer

Option A

Solution

In n-type semiconductor, electrons are majority charge carriers and holes are minority charge carriers and pentavalent atoms are dopants.

Q40

The output (X) of the logic circuit shown ion figure will be

A X = A . B

B X =

\overline {A + B}

C X =

\overline {\overline A } .\overline {\overline B }

D X =

\overline {A.B}

Correct Answer

Option A

Solution

The output of the given logic circuit is X =

\overline{\overline {A.B}}

= A.B