Semiconductor Electronics: Materials, Devices and Simple Circuits Test 72

Result

Semiconductor Electronics: Materials, Devices and Simple Circuits Test 72

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 diffusion 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 baised

D
from the P-side to the N-side if the junction is forward-biased and in the opposite direction if it is reverse baised

Solution

Diffusion of current in he $${\text{p - n}}$$ junction is a result of electrons moving near the
$${\text{p - }}$$end and holes moving near the $${\text{n - }}$$end of the diode. Because of this current flows from the $${\text{p - }}$$side to the $${\text{n - }}$$side.
Question 2
1 / -0

Zener diode is used as

A
Half wave rectifier

B
Full wave rectifier

C
ac voltage stabilizer

D
dc voltage stabilizer

Solution

A Zener diode is always used in the reverse bias to get the property of breakdown voltage. We use a Zener diode to regulate the voltage of the alternating input signal in various places because of this property. Zener diode is used as the ac voltage stabilizer.
So, option (C) is correct.
Question 3
1 / -0

In the diagram, the input is across the terminals A and C and the output is across B and D. Then, the output is

A
zero

B
same as the input

C
full wave rectifier

D
half wave rectifier

Solution

In the circuit, we can put an Alternating Current source in between the points $$A$$ & $$C$$ and we will connect the points $$B$$ & $$D$$ with an external resistance $$PQ$$ with resistance $$R{\text{ }}\Omega $$.
When the voltage of the point $$A'$$ is more than the point $$C'$$, the direction of current flow will be as follows-
$$A' \to A \to B \to P \to Q \to D \to C \to C'$$
When the voltage of the point $$A'$$ is less than the point $$C'$$, the current will not flow because $${D_3}$$ and $${D_4}$$ both are in reverse bias.
Hence, the output will be a half-wave rectifier.
Question 4
1 / -0

Metals are all good electrical conductors, which of the following statements is correct?

A
Metals are soft and hence allow electrons to pass through

B
Metals posses small value of band gap

C
Metals have large void spaces in their lattices that accommodate electrons

D
Metals are high melting solids, so electrons can translate on their surface

Solution
Question 5
1 / -0

In the given circuit, the current through the resistor $$2 k \Omega $$ is:

A
$$2 m A $$

B
$$6 m A $$

C
$$1 m A $$

D
$$18 m A $$

Solution
Question 6
1 / -0

The figure shown the function of

A
$$OR$$ gate

B
$$NAND$$ gate

C
$$XOR$$ gate

D
$$AND$$ gate

Solution

In the given circuit, $${y_1} = {\text{A}}{\text{.B}}$$ and $${y_2} = {\text{B}}{\text{.C}}$$. So, the final output after passing through the $$NOR$$ gate is-
$${\text{Q}} = {\text{ }}\overline {{\text{A}}{\text{.B}} + {\text{B}}{\text{.C}}} $$
If we do the Boolean algebra, we will get the truth-table as a $$XOR$$ gate.
Question 7
1 / -0

What is the Boolean equation of the networking in figure..

A
ABCD

B
A + BCD

C
A + B + C + D

D
AB + CD

Solution

All the logic gates connected in the circuit are $${\text{OR}}$$ gates. The Boolean algebra for $${\text{OR}}$$ gates is $$Y = A + B$$
Therefore, the resultant output of the whole circuit is $$Y = A + B + C + D$$
Question 8
1 / -0

A zener diode having breakdown voltage equation to 15 V is used in voltage regulator circus shown in figure the current through diode is :

A
100 mA

B
95 mA

C
150 mA

D
195 mA

Solution

The current through the load resistance, $${{I}_{1}}=\frac{{{V}_{zener}}}{{{R}_{Load}}}=\frac{5}{1\times {{10}^{3}}}=5\times {{10}^{-3}}A$$
The current through the $$150\Omega $$ resistance, $${{I}_{2}}=\frac{E-{{V}_{zener}}}{R}=\frac{\left( 20-5 \right)}{150}=\frac{15}{150}=0.1A$$
Clearly, the Zener current, $${{I}_{zener}}={{I}_{2}}-{{I}_{1}}=\left[ 0.1-\left( 5\times {{10}^{-3}} \right) \right]A=95mA$$
Option (B) is correct.
Question 9
1 / -0

In $$n$$ -type semiconductor when all donor states are filled, then the net charge density in the donor states becomes

A
$$1$$

B
$$< 1$$

C
$$< 1 ,$$ but not zero

D
zero

Solution

The major charge carriers in a $${\text{n - }}$$type semiconductor are electrons. When all donor states are filled, then the net charge density in the donor states becomes less than $$1$$ because of all the negative charges,
But, it can not be equal to zero.
Question 10
1 / -0

From the Zener diode circuit shown in figure, the current through the Zener diode is

A
$$34mA$$

B
$$31.5mA$$

C
$$36.5mA$$

D
$$2.5mA$$

Solution

The current through the load resistance, $${I_1} = \frac{{{V_{zener}}}}{{{R_{Load}}}} = \frac{{50}}{{20 \times {{10}^3}}} = 2.5 \times {10^{ - 3}}A$$
The current through the $$5{\text{k}}\Omega $$ resistance, $${I_2} = \frac{{E - {V_{zener}}}}{R} = \frac{{220 - 50}}{{5 \times {{10}^3}}} = 34 \times {10^{ - 3}}A$$
Clearly, the Zener current, $${I_{zener}} = {I_2} - {I_1} = \left( {34 - 2.5} \right) \times {10^{ - 3}}A = 31.5mA$$
Option (B) is correct.