Semiconductor Electronics: Materials, Devices and Simple Circuits Test 21

Result

Semiconductor Electronics: Materials, Devices and Simple Circuits Test 21

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The following is NOT equal to $$0$$ in the Boolean algebra is

A
$$\overline{\overline A.0}$$

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

C
$$A.0$$

D
$$\overline{A+\overline A}$$

Solution

(A) For $$A = 0$$,
$$\overline{\bar A.A} = \overline{\bar 0.0} = \overline{1.0} = \bar 0 = 1$$
For $$A = 1$$,
$$\overline {\bar A.A} = \overline {\bar 1.1} = \overline {0.1} = \bar 0 = 1$$

(B) For $$A = 0$$,
$$A.\bar A = 0. \bar 0 = 0.1 = 0$$
For $$A = 1$$,
$$A.\bar A = 1. \bar 1 = 1.0 = 0$$

(C) For $$A = 0$$,
$$A.0 = 0.0 = 0$$
For $$A = 1$$,
$$A.0 = 1.0 = 0$$

(D) For $$A = 0$$,
$$\overline{A+\bar A} = \overline{0+\bar 0} = \overline{0+1} = \bar 1 = 0$$
For $$A = 1$$,
$$\overline{A+\bar A} = \overline{1+\bar 1} = \overline{1+0} = \bar 1 = 0$$
Question 2
1 / -0

Consider the following statements A and B and identify the correct answer.
Statement(A):A Zener diode is always connected in reverse bias to use it as voltage.
Statement(B):The potential barrier of a p-n junction lies between $$0.1$$ to $$0.3\ V$$, approximately.

A
A and B are correct.

B
A and B are wrong.

C
A is correct but B is wrong.

D
A is wrong but B is correct.

Solution

When Zener diode is forward biased, it behaves like ordinary diode and when reverse biased, it will be a voltage stabilizer. Also, at room temperature, the voltage across the depletion layer for silicon is about $$0.6-0.7V$$ and for germanium is about $$0.3-0.35 V$$
Hence, statement A is correct but statement B is wrong.
Question 3
1 / -0

There is a sudden increase in current in zener diode is

A
Due to rupture of bonds

B
Resistance of deplection layer becomes less

C
Due to high doping

D
Due to less doping

Solution

When Zener diode is connected in reverse bias, as applied voltage reaches at Zener breakdown voltage, due to rupture of bonds there is a sudden increase in current in Zener diode.
Question 4
1 / -0

A Zener diode when used as a voltage regulator, is connected
(a) in forward bias.
(b) in reverse bias.
(c) in parallel to the load.
(d) in series to the load.

A
(a) and (b) are correct

B
(b) and (c) are correct

C
(a) only is correct

D
(d) only is correct

Solution

When Zener diode is biased in the forward direction it behaves like a normal signal diode passing the current increasing linearly with voltage, but as soon as a reverse voltage applied across the Zener diode attains the breakdown voltage of the device, a large current starts to flow through the diode. Thus, for a constant voltage large current will flow through it. Hence, it always connected in parallel to the load.
Question 5
1 / -0

The gate that can act as a building block for the digital circuits is

A
OR

B
NOT

C
AND

D
NAND

Solution
Question 6
1 / -0

In the Boolean algebra: $$\overline{A+B}$$ is equal to:

A
A + B

B
$$\bar{\bar{A}}$$+$$\bar{\bar{B}}$$

C
$$\overline{\overline{A.B}}$$

D
$$\bar{A}$$ . $$\bar{B}$$

Solution

Clearly, as shown in the truth table, $$\overline{A+B}=\bar{A}.\bar{B}$$
Question 7
1 / -0

The minimum number of gates required to realise the expression $$Z = DABC + D\overline{ABC}$$ is

A
One

B
Two

C
Eight

D
Five

Solution

The given expression is same as $$Z = D $$.
Hence, one gate is sufficient to realise this expression.
Question 8
1 / -0

Boolean expression $$Y = A.\bar{B} + B.\bar{A}$$ is given. If $$A = 1, B = 1$$ then $$Y =$$

A
0

B
1

C
11

D
10

Solution

$$Y=A.\bar B+B.\bar A$$
For $$A=1, B=1$$
$$Y=1.\bar 1+1.\bar 1$$
$$Y=1.0+1.0 = 0+0 = 0$$
Question 9
1 / -0

The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than $$2480\ nm$$ is incident on it. The forbidden band energy for the semiconductor in $$eV$$ is

A
$$0.5$$

B
$$0.9$$

C
$$0.7$$

D
$$1.1$$

Solution

Given that:
The wavelength of electromagnetic radiation is $$\lambda = 2480\ nm$$
The energy of electromagnetic radiation is given by
$$E = \dfrac{hc}{\lambda}$$
where, $$h$$ is Plank's constant and $$c$$ is velocity of light in nm.
Hence,
$$E = \dfrac{1242}{2480} = 0.5\ eV$$
Question 10
1 / -0

To get an output Y=1 from the circuit above, the input must be

A
A-0 B-1 C-0

B
A-1 B-0 C-0

C
A-1 B-0 C-1

D
A-1 B-1 C-0

Solution

Logic gate OR is used for addition of the input signals and Logic gate AND is used for multiplication of the input signals. Hence, here inputs A = 1, B = 0 and C = 1 will yield the output Y = 1.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Semiconductor Electronics: Materials, Devices and Simple Circuits Test 21

Result

Semiconductor Electronics: Materials, Devices and Simple Circuits Test 21

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\overline{\overline A.0}$$

$$A.\bar{A}$$

$$A.0$$

$$\overline{A+\overline A}$$

Solution

Question 2

A and B are correct.

A and B are wrong.

A is correct but B is wrong.

A is wrong but B is correct.

Solution

Question 3

Due to rupture of bonds

Resistance of deplection layer becomes less

Due to high doping

Due to less doping

Solution

Question 4

(a) and (b) are correct

(b) and (c) are correct

(a) only is correct

(d) only is correct

Solution

Question 5

OR

NOT

AND

NAND

Solution

Question 6

A + B

$$\bar{\bar{A}}$$+$$\bar{\bar{B}}$$

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

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

Solution

Question 7

One

Two

Eight

Five

Solution

Question 8

0

1

11

10

Solution

Question 9

$$0.5$$

$$0.9$$

$$0.7$$

$$1.1$$

Solution

Question 10

A-0 B-1 C-0

A-1 B-0 C-0

A-1 B-0 C-1

A-1 B-1 C-0

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.