Get the App Signup / Login
Sign in Open App

31 Year NEET Previous Year Questions: Semiconductor Electronics - 3


25 Questions MCQ Test Physics 31 Years NEET Chapterwise Solved Papers | 31 Year NEET Previous Year Questions: Semiconductor Electronics - 3


Description
Attempt 31 Year NEET Previous Year Questions: Semiconductor Electronics - 3 | 25 questions in 50 minutes | Mock test for NEET preparation | Free important questions MCQ to study Physics 31 Years NEET Chapterwise Solved Papers for NEET Exam | Download free PDF with solutions
QUESTION: 1

The following truth table belongs to which of the following four gates? [1997]

 

Solution:

The given truth table is of
(OR gate + NOT gate) ≡ NOR gate

QUESTION: 2

A semi-conducting device is connected in aseries circuit with a battery and a resistance. Acurrent is found to pass through the circuit. Ifthe polarity of the battery is reversed, the currentdrops to almost zero. The device may be

Solution:

In reverse bias, the current through a p-n
junction is almost zero.

QUESTION: 3

The transfer ratio β of a transistor is 50. Theinput resistance of the transistor when used inthe common emitter configuration is 1 kΩ. Thepeak value of the collector A.C. current for anA.C. input voltage of 0.01 V peak is [1998]

Solution:

QUESTION: 4

Which of the following gates will have an output of 1?

(a)

(b)

(c)

(d)
Solution:

(a) is a NAND gate so output is 

(b) is a NOR gate so output is  

(c) is a NAND gate so output is 

(d) is a XOR gate so output is 

Following is NAND Gate

QUESTION: 5

The truth-table given below is for which gate?   [1998]
Solution:

Relation between A, B and C shows that

QUESTION: 6

The cause of the potential barrier in a p-n diodeis [1998]
Solution:

During the formation of a junction diode,
holes from p-region diffuse into n-region and
electrons from n-region diffuse into p-region.
In both cases, when an electron meets a hole,
they cancel the effect of each other and as a
result, a thin layer at the junction becomes
free from any of charge carriers. This is called
depletion layer. There is a potential gradient
in the depletion layer, negative on the p-side
and positive on the n-side. The potential
difference thus developed across the
junction is called potential barrier.

QUESTION: 7

In forward bias, the width of potential barrier ina p-n junction diode [1999]
Solution:

We know that in forward bias of p-n junction
diode, when positive terminal is connected
to p-type diode, the repulsion of holes takes
place which decreases the width of potential
barrier by striking the combination of holes
and electrons.

QUESTION: 8

A depletion layer consists of [1999]
Solution:

Depletion layer is formed by diffusion of
holes and electrons from p-type
semiconductor to n-type semiconductor and
vice-versa. Hence, donor and acceptor atom
get positive and negative charge leading to
formation of p-n junction. Thus, donor and
acceptor are immobile.

QUESTION: 9

Which of the following when added acts as animpurity into silicon produced n-type semiconductor?    [1999]
Solution:

n-type of silicon semiconductor is formed
when impurity is mixed with pentavalent
atom. Out of given choices only phosphorus
is pentavalent.

QUESTION: 10

In a junction diode, the holes are due to        [1999]
Solution:

Holes are produced due to missing of
electrons.

QUESTION: 11

A gate has the following truth table

The gate is
Solution:

P, Q and R are related as R = P. Q which is
relation of AND gate.

QUESTION: 12

In the study of transistor as amplifier, if  and  , where, IC, IB and IE are the collector, base and emitter currents, then   [2000]
Solution:

As we know that Ie = Ic = Ib
Divide both side by Ie

QUESTION: 13

The forward biased diode is            [2000]
Solution:

As A is at higher potential than B, hence
current will flow from A to B i.e. from p
junction to n junction and diode is forward
biased.

QUESTION: 14

Which gate is represented by the following truth table ?
Solution:

Which is truth table of NAND gate

QUESTION: 15

The intrinsic semiconductor becomes insulator at

[2001]
Solution:

At 0K, motion of free electrons stop. Hence
conductivity becomes zero. Therefore, at 0K
intrinsic semiconductor becomes insulator.

QUESTION: 16

For a common emitter circuit if    then  current gain for common emitter circuit will be   [2001]
Solution:

We have, 

QUESTION: 17

Transmission of light in optical fibre is due to         [2001]
Solution:

Light is trasmitted through optical fibre due
to multiple total internal reflections.

QUESTION: 18

A d.c. battery of V volt is connected to a series combination of a resistor R and an ideal diode D as shown in the figure below. The potential difference across R will be [2002]
Solution:

In forward biasing, the diode conducts. For
ideal junction diode, the forward resistance
is zero; therefore, entire applied voltage
occurs across external resistance R i.e., there
occurs no potential drop, so potential across
R is V in forward biased.

QUESTION: 19

In a p-n junction     [2002]
Solution:

For conduction, p-n junction must be
forward biased. For this p-side should be
connected to higher potential and n-side to
lower potential.

QUESTION: 20

In the case of a common emitter transistoramplifier, the ratio of the collector current to theemitter current Ic /Ie is 0.96. The current gain ofthe amplifier is [2002]
Solution:

QUESTION: 21

If a full wave rectifier circuit is operating from50Hz mains, the fundamental frequency in theripple will be     [2003]
Solution:

In case of full wave rectifier,
Fundamental frequency = 2 × mains
frequency
= 2 × 50 = 100Hz.

QUESTION: 22

An n-p-n transistor conducts when [2003]
Solution:

When the collector is positive and emitter is
negative w.r.t. base, it causes the forward
biasing for each junction, which causes
conduction of current.

QUESTION: 23

Barrier potential of a p-n junction diode doesnot depend on [2003]
Solution:

Barrier potential does not depend on diode
design while barrier potential depends upon
temperature, doping density, and forward
biasing.

QUESTION: 24

Following diagram performs the logic function of         [2003]
Solution:

Y = A. B by Demorgan therorem
∴ This diagram performs the function of
AND gate.

QUESTION: 25

The following circuit represents         [1999]
Solution:

Output of upper AND gate = 

Output of lower AND gate = 

∴ Output of OR gate, Y = 

This is Boolean expression for XOR gate.

Use Code STAYHOME200 and get INR 200 additional OFF
Use Coupon Code
Use Coupon Code

Recommended Content

Recommended Tests

How to Prepare for NEET

Read our guide to prepare for NEET which is created by Toppers & the best Teachers

Download free EduRev App

Track your progress, build streaks, highlight & save important lessons and more!
© EduRev
Follow Us