Semiconductor Electronics: Materials, Devices and Simple Circuits Test 9

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 9

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Question 1
1 / -0

The truth table for NOT gate is

A

B

C

D

Solution

NOT gate gives the reverse of the input as output. Truth table for NOT gate is:
Input Output
$$1$$ $$0$$
$$0$$ $$1$$
Question 2
1 / -0

The logic expression which is NOT true in Boolean algebra is

A
$$[\bar{1}+\bar{1}].1=0$$

B
$$[\bar{1}+0].1=0$$

C
$$[\bar{1}+0].\bar{1}=0$$

D
$$[1+1].1=0$$

Solution

$$[\bar 1+\bar 1].1= [0+0].1 = 0$$
$$[\bar 1+0].1= [0+0].1 = 0$$
$$[\bar 1+0].\bar 1= [0+0].0 = 0$$
$$[1.1].1 = 1.1 = 1 \neq 0$$
Question 3
1 / -0

In the following figure, the diodes which are forward biased are :

A
(a), (b) and (d)

B
(c) only

C
(c) and (a)

D
(b) and (d)

Solution

c) and a) are forward biased diodes since in these cases, p region is connected to higher potential as compared to n-side.
Question 4
1 / -0

In a PN junction : -

A
High potential at N side and low potential at P side

B
High potential at P side and low potential at N side

C
P and N both are at same potential

D
Undetermined

Solution
Question 5
1 / -0

Which one of the following statement is false

A
Work is a state finction

B
temperature is a state function

C
Change in the state is completely defined when the initial and final states are specified

D
Work appears at the boundary of the system

Solution

Majority carrier in a $$n-$$ type semiconductor are holes.This statement is false.
Since,in $$n-$$type semiconductor, the pentavalent impurity atoms donate electrons o the host crystal and the semiconductor doped with donars (pentavalent impurity) is called $$n-$$type semiconductor.
Therefore majority carrier in a $$n-$$type semiconductor are electrons
Question 6
1 / -0

Boolean Expression for the gate circuit shown in the figure is

A
$$A. 1 = A$$

B
$$A.\bar{A}=0$$

C
$$A.A = A$$

D
$$A.0 = 0$$

Solution

The input $$A$$ directly goes into the AND gate and also first passes through NOT gate, giving another input of $$\bar{A}$$ to the AND gate.
$$A.\bar{A}=0$$ since atleast one of $$A$$ and $$\bar{A}$$ is $$0$$.
Question 7
1 / -0

NAND gate in the following is

A

B

C

D

Solution

logic gate NAND is used to reverse the multiplication of the input signals. Hence, by connecting NOT gate after AND gate we get the NAND gate.
Question 8
1 / -0

The logic gate having following truth table is
A B Y
0 0 1
0 1 1
1 0 1
1 1 0

A
XOR

B
OR

C
AND

D
NAND

Solution

From truth table it is clear that output is high for either one or two low inputs, hence it is the reverse of the condition for AND gate i.e. output is low for either one or two high inputs . Hence, given truthtable is for NAND gate.
Question 9
1 / -0

The output of OR gate is 1 :-

A
If either one or both inputs are 1

B
Only if both inputs are 1

C
If either input is zero

D
If both inputs are zero

Solution

At least one input required to activate $$or -gate$$
Option A
Question 10
1 / -0

Semiconductors are generally made up of which substance?

A
Silicon

B
Carbon

C
Phosphorus

D
Boron

Solution

Semiconductors are made of Si, generally. Because Si is the second most abundant element in earth's crust after oxygen and is less expensive than other semiconductors used in intrinsic semiconductors.