Semiconductor Electronics: Materials, Devices and Simple Circuits Test 26

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 26

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Question 1
1 / -0

The two ends of a p-n junction are joined by a wire

A
There will not be a steady current in the circuit

B
There will be a steady current from the n-side to the p-side

C
There will be a steady current from the p-side to the n-side

D
There may or may not be a current depending upon the resistance of the connecting wire

Solution

Since the potential difference across p-n junction is 0 in ideal condition, on connecting a wire, no current will flow through it.
Question 2
1 / -0

The drift current in a p-n junction is

A
from the n-side to the p-side

B
from the p-side to the n-side

C
from the n-side to the p-side if the junction is forward-biased and in the opposite direction if it is reverse biased

D
from the p-side to the n-side if the junction is forward-biased and in the opposite direction if it is reverse-biased

Solution

When pn junction is formed, the free holes and the electrons in their respective region diffuse to the other region which results in the formation of the depletion region and an electric field establishes from n-type to p-type semiconductor in the depletion region. Now if an electron from p-type comes in the depleton region then the electric field pushes the electron to the n-type. Similarly holes are drifted towards the p-type. Thus due to these holes and electron drift, a drifting current exists which flows from n-type to p-type.
Question 3
1 / -0

The free electron density is more in .....

A
Conductors

B
Insulators

C
Semi conductors

D
Electrolytes

Solution

Free electron density is more in conductors that is why they are useful in conducting electricity.
Question 4
1 / -0

A substance in which large energy gap separates the valence and conduction band is a

A
metal

B
metalloid

C
semi conductor

D
insulator

Solution

A useful way to visualize the difference between conductors, insulators and semiconductors is to plot the available energies for electrons in the materials. Instead of having discrete energies as in the case of free atoms, the available energy states form bands. Crucial to the conduction process is whether or not there are electrons in the conduction band. In insulators the electrons in the valence band are separated by a large gap from the conduction band, in conductors like metals the valence band overlaps the conduction band, and in semiconductors there is a small enough gap between the valence and conduction bands that thermal or other excitations can bridge the gap. With such a small gap, the presence of a small percentage of a doping material can increase conductivity dramatically.
Question 5
1 / -0

The mobility of free electrons is greater than that of free holes because

A
They carry negative charge

B
They are light

C
They mutually collide less

D
They require high energy to continue their motion

Solution

The mass of an electron is significantly less than that of a hole therefore making it more mobile.
Moreover, $$K.E = \dfrac{1}{2}mv^2$$
Thus if the same energy is given to both, then electron will move faster than the hole because mass of electron is less than that of hole.
Question 6
1 / -0

In the following circuit, the output Y for all possible inputs A and B is expressed by the truth table.

A

B

C

D

Solution

The output of the first NOR gate $$\overline{A+B}=Y'$$ is the input for the second OR gate.
Hence, $$Y=$$ $$\overline{Y'+Y'}$$
$$=\overline{\overline{A+B}+\overline{A+B}}$$
$$=\overline{\overline{A+B}}+\overline{\overline{A+B}} $$
$$=(A+B)+(A+B)=A+B$$
Thus the output is an OR output.
Question 7
1 / -0

To which logic gate does the truth table given below correspond?
A B Y
0 0 0
0 1 1
1 0 1
1 1 0

A
AND gate

B
NAND gate

C
NOR gate

D
XOR gate

Solution

For this truth table, Boolean expression is $$Y = A.\bar{B} + \bar{A}B$$ which is for XOR gate
Question 8
1 / -0

NAND and NOR gates are called universal gates primarily because they

A
are available universally

B
can be combined to produce OR, AND and NOT gates

C
are widely used in Integrated circuit packages

D
are easiest to manufacture

Solution

Combination of NAND & NOR gates can produce OR, AND & NOT gates.
Question 9
1 / -0

The truth-table given below is for which gate?
A B C
0 0 1
0 1 1
1 0 1
1 1 0

A
XOR

B
OR

C
AND

D
NAND

Solution

Relation between A, B and C shows that $$C = \overline{AB}$$
So NAND Gate
Question 10
1 / -0

The truth table given below is
A B Y
0 0 1
0 1 1
1 0 1
1 1 0

A
NOR

B
AND

C
XOR

D
NAND

Solution

The given truth table is for NAND gate.