Semiconductor Electronics: Materials, Devices and Simple Circuits Test 32

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 32

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Weekly Quiz Competition

Question 1
1 / -0

The conductivity of a semi conductor can be increased by:

A
Heating the semi conductor

B
Doping the semi conductor

C
Both (a) and (b)

D
None

Solution

The conductivity of a semi-conductor can be increased by releasing the electrons which are bounded by atoms initially by heating the semi-conductor.
The conductivity of semi-conductor can also be increased by adding the trivalent or pentavalent impurity in a semi-conductor which is called as doping in a semiconductor.
Question 2
1 / -0

A Zener diode having break-down voltage $$5.6V$$ is connected in reverse bias with a battery of emf $$10V$$ and a resistance of $$100\Omega$$ in series. The current flowing through the Zener is

A
$$88mA$$

B
$$0.88mA$$

C
$$4.4mA$$

D
$$44mA$$

Solution

A Zener diode offers infinite resistance as long as the reverse voltage across it is less than the break-down value. As soon as the reverse bias voltage cross the break-down level, the residue voltage remains in the circuit with the diode offering zero resistance to it.
Hence here the voltage remaining across the Zener=$$10V-5.6V=4.4V$$
Therefore the current in the circuit is $$\dfrac{4.4V}{100\Omega}$$
$$=44mA$$
Question 3
1 / -0

A NOR gate and a NAND gate are connected as shown in the figure. Two different sets of inputs are given to this set up. In the first case, the input to the gates are A=0, B=0, C=0. In the second case, the inputs are A=1, B=0, C=1. The output D in the first case and second case respectively are

A
0 and 0

B
0 and 1

C
1 and 0

D
1 and 1

Solution

In the first case the output of first NOR gate is $$\overline{A+B}$$
$$=\overline{0+0}=1=E$$
and that of NAND gate is $$\overline{E.C}=\overline{1.0}=1$$
In the second case, the output of first NOT gate is $$\overline{1+0}=0$$
and that of NAND gate is $$\overline{0.1}=1$$
Question 4
1 / -0

When a forward bias is applied to p-n junction it:

A
Raises the potential barrier.

B
Reduces the majority carrier current to zero.

C
Lowers the potential barrier.

D
None of the above.

Solution

When a forward bias is applied to the p-n junction diode, the voltage applied opposes the barrier voltage. Due to this the potential barrier across the is lowered.
Hence option C is correct.
Question 5
1 / -0

In a n - type semi conductors which of the following statement is true:

A
Electrons are majority carrier and trivalent impurity are the dopants.

B
Electrons are minority carrier and pentavalent atoms are dopants.

C
Holes are minority carrier and pentavalent atoms are dopants.

D
Holes are majority carrier and trivalent atoms are the dopants.

Solution

n-type semi-conductor can be produced by doping pentavalent impurity (atoms of valence 5) such as phosphorus.
Because of extra electrons due to pentavalent impurity electrons are majority carriers and holes are minority carriers.
$$\therefore$$ Here holes are minority carriers and pentavalent atoms are dopants. Hence option C is correct.
Question 6
1 / -0

Insulators are materials having an electrical conductivity equal to:

A
Between $$10^{-8}$$ S/cm to $$10^{3}$$ S/cm

B
$$<\ 10^{-8}$$ S/cm

C
Equal to $$ 10^{3}$$ S/cm

D
$$>\ 10^{3}$$ S/cm

Solution

Insulators are the least conducting materials and hence have the least conductivity or the highest resistivity. Therefore, they have a conductivity of $$< 10^{-8}$$ $$S/cm$$.
Conductors have the highest conductivity i.e. $$> 10^3$$ $$S/cm$$ whereas semiconductors have an electrical conductivity in between $$10^{-8}$$ and $$10^3$$ $$S/cm$$.
Question 7
1 / -0

The given truth table is for
Input Input Output
A B Y
$$0$$ $$0$$ $$1$$
$$0$$ $$1$$ $$1$$
$$1$$ $$0$$ $$1$$
$$1$$ $$1$$ $$0$$

A
AND gate

B
OR gate

C
NAND gate

D
NOR gate

Solution

A NAND gate gives output as $$1$$ when either of the input signals is low i.e. $$0$$ and gives output as $$0$$ only when both the input signals are high i.e. $$1$$
Hence the above truth table is for NAND gate.
Question 8
1 / -0

Fill in the blank.
The following truth table with $$A$$ and $$B$$ as inputs is for _________ gate.
A B Output
1 0 1
1 1 0
0 1 1
0 0 0

A
AND

B
OR

C
XOR

D
NOR

Solution

In the given truth table, output is 1 when exactly one of the inputs is 1 and the other is 0. Output is 0 if both inputs are same (0 or 1). This is the characteristic of XOR gate.
$$Y = A\overline B+ \overline A B$$
Question 9
1 / -0

A proper combination of $$3$$ NOT and $$1$$ AND gates is shown. If $$A=0$$, $$B=1$$, $$C=1$$ then the output of this combination is

A
$$1$$

B
Zero

C
Not predictable

D
None of these

Solution

A NOT gate gives an output as opposite to that of its input. An AND gate gives output as 1 only when all of its input signals are 0, else it gives an output as 0.
Thus truth table for the above circuit is shown which suggests that its output (i.e. Y) is zero.
INPUTS OUTPUTS
A B C D E F Y
0 1 1 1 0 0 0
Question 10
1 / -0

The addition of pentavalent impurities ________ of the semiconductor.

A
brings no change to the conductivity

B
decreases the conductivity

C
increases the conductivity

D
reduces the formation of free electrons

Solution

The additional pentavalent impurity added to the semi conductor increases the concentration of free electrons in a semi-conductor.
Due to this increase in number of free electrons in a semi-conductor conductivity of semi-conductor increases.
Hence option C is correct.

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Input	Input	Output
A	B	Y
$$0$$	$$0$$	$$1$$
$$0$$	$$1$$	$$1$$
$$1$$	$$0$$	$$1$$
$$1$$	$$1$$	$$0$$

INPUTS			OUTPUTS
A	B	C	D	E	F	Y
0	1	1	1	0	0	0

Semiconductor Electronics: Materials, Devices and Simple Circuits Test 32

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 32

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SHARING IS CARING

Question 1

Heating the semi conductor

Doping the semi conductor

Both (a) and (b)

None

Solution

Question 2

$$88mA$$

$$0.88mA$$

$$4.4mA$$

$$44mA$$

Solution

Question 3

0 and 0

0 and 1

1 and 0

1 and 1

Solution

Question 4

Raises the potential barrier.

Reduces the majority carrier current to zero.

Lowers the potential barrier.

None of the above.

Solution

Question 5

Electrons are majority carrier and trivalent impurity are the dopants.

Electrons are minority carrier and pentavalent atoms are dopants.

Holes are minority carrier and pentavalent atoms are dopants.

Holes are majority carrier and trivalent atoms are the dopants.

Solution

Question 6

Between $$10^{-8}$$ S/cm to $$10^{3}$$ S/cm

$$<\ 10^{-8}$$ S/cm

Equal to $$ 10^{3}$$ S/cm

$$>\ 10^{3}$$ S/cm

Solution

Question 7

AND gate

OR gate

NAND gate

NOR gate

Solution

Question 8

AND

OR

XOR

NOR

Solution

Question 9

$$1$$

Zero

Not predictable

None of these

Solution

Question 10

brings no change to the conductivity

decreases the conductivity

increases the conductivity

reduces the formation of free electrons

Solution

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