Semiconductor Electronics: Materials, Devices and Simple Circuits Test 39

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Semiconductor Electronics: Materials, Devices and Simple Circuits Test 39

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Question 1
1 / -0

The output of a OR gate is $$1$$ :

A
If either input is zero

B
If both inputs are zero

C
Only if both inputs are $$1$$

D
If either or both inputs are $$1$$

Solution

The output of a OR gate is $$\gamma =A+B$$.
Question 2
1 / -0

A rod of a certain metal is 1.0 m long and 0.6 cm in diameter. Its resistance is $$\displaystyle 3.0\times { 10 }^{ -3 }\Omega $$. Another disc made of the same metal is 2.0 cm in diameter and 1.0 mm thick. What is the resistance between the round faces of the disc?

A
$$\displaystyle 1.35\times { 10 }^{ -8 }\Omega $$

B
$$\displaystyle 2.70\times { 10 }^{ -7 }\Omega $$

C
$$\displaystyle 4.05\times { 10 }^{ -6 }\Omega $$

D
$$\displaystyle 8.10\times { 10 }^{ -5 }\Omega $$

Solution

As, $$R=\rho l/A=\rho l/(\pi d^2/4)$$
Resistance for a given material is directly proportional to the length and inversely proportional to the square of diameter of round faces.

So for the second rod, diameter is $$2.0/0.6=10/3$$ times diameter of the first one.
and, length is $$1.0mm/1m=1/1000$$ times the length of first one.

Thus resistance of the second rod is, $$\dfrac{1/1000}{(10/3)^2}=9\times10^{-5}$$ times the resistance of the first one.

So resistance of the second rod is $$3\times10^{-3}\times9\times10^{-5}=2.70\times10^{-7}\Omega$$

Option B is correct.
Question 3
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What is the voltage gain in a common emitter amplifier, when input resistance is 3 $$\Omega$$ and load resistance is 24 $$\Omega$$ with $$\beta = 60$$?

A
$$480$$

B
$$2.4$$

C
$$4.8$$

D
$$8.4$$

Solution

Given : $$\beta = 60$$
Input resistance $$R_1 = 3\Omega$$
Output resistance $$R_2 = 24\Omega$$
Voltage gain $$A_v = \beta \times \dfrac{R_2}{R_1} = 60 \times \dfrac{24}{3} = 480$$
Question 4
1 / -0

For the combination of gates shown here, which of the following truth table part is not true?

A
$$A = 0, B = 1, C = 1$$

B
$$A = 0, B = 0, C = 0$$

C
$$A = 1, B = 1, C = 1$$

D
$$A = 1, B = 0, C = 1$$

Solution

Key Point Draw the truth table of the given operation and match the output with given options.
The combination of gates is given
For the given operation, equivalent truth table can be written as
$$A$$ $$B$$ $$A\cdot B$$ $$A + A\cdot B$$
$$0$$ $$1$$ $$0$$ $$0$$
$$1$$ $$0$$ $$0$$ $$1$$
$$0$$ $$0$$ $$0$$ $$0$$
$$1$$ $$1$$ $$1$$ $$1$$
Out of the given option, option (a) can't justify the given truth table. So, option (a) is correct.
Question 5
1 / -0

The input resistance of a common emitter transistor amplifier, if the output resistance is 500 k$$\Omega$$, the current gain $$\alpha = 0.98$$ and the power gain is $$6.0625 \times 10^6$$, is

A
$$198 \Omega$$

B
$$300 \Omega$$

C
$$100 \Omega$$

D
$$400 \Omega$$

Solution

Power gain = Current gain $$\times$$ volume gain .....(1)
Voltage gain $$A_v = \beta \dfrac{R_2}{R_1}$$
Output resistance $$R_2 = 500\times 10^3 \ \Omega$$
Also, current gain $$\beta = \dfrac{\alpha}{1 - \alpha} = \dfrac{0.98}{1 - 0.98} = 49$$
$$\therefore \ A_v = (49) \left( \dfrac{500 \times 10^3}{R_1}\right )$$
Power gain $$ = 6.0625 \times 10^6$$ (Given)
From (1), we get
$$ 6.0625 \times 10^6$$ $$= 49 \times \left( \dfrac{500 \times 10^3}{R_1} \right ) \times 49$$
$$\Rightarrow R_1 = 198 \Omega$$
Question 6
1 / -0

The following truth table corresponds to the logic gate
A B X
0 0 0
0 1 1
1 0 1
1 1 1

A
NAND

B
OR

C
AND

D
XOR

Solution

When both the inputs are shared, the logic gate obtained is the OR gate. OR gate gives output signal as 1 when either of its input signals are 1.
Question 7
1 / -0

Which one of the following represents correctly the truth table of the configuration shown in figure?

A

B

C

D

Solution

In this case
$$\displaystyle y = \bar{y} = \overline{A.B}$$
So, it is satisfied in option (c).
Question 8
1 / -0

What the circuit represents from the equivalent circuit?

A
NAND gate

B
OR gate

C
AND gate

D
NOR gate

Solution

Output of NOR gate
$$ y' = \overline{A+B} $$
Output of NAND gate
$$ y'' = \overline{(A+B).(A+B)}$$
$$ = \overline{(A+B)} \overline{(A+B)}$$
$$ = (A+B) (A+B)$$
$$ A+B $$
Output of NOT gate
$$ y"" = \overline{A+B} $$
$$ \therefore $$ The Boolean expression is for NOR gate.
Question 9
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Identify the mismatched pair from the following

A
zener diode : voltage regulator

B
germanium doped with phosphorous : n-type semiconductor

C
semiconductor : band gap > 3 eV

D
p-n junction diode : rectifier

E
silicon doped with aluminium : p-type semiconductor

Solution

For semiconductor band gap is $$<3eV$$ so given option is incorrect.
When zener dioode is operated in breakdown region, voltge across it remains constant even if current through it changes by large amount.
Germanium when doped with phosphorous forms n-type semiconductor and silicon when doped with aluminum forms p-type.
p-n junction can be used as half wave or full wave rectifier.
Question 10
1 / -0

In a Zener diode regulated power supply, unregulated DC input of $$10V$$ is applied. If the resistance$$(R_s)$$ connected in series with a Zener diode is $$200\Omega$$ and the Zener voltage $$V_z=5V$$, the current across the resistance $$R_s$$ is:

A
$$15mA$$

B
$$10mA$$

C
$$20mA$$

D
$$5mA$$

E
$$25mA$$

Solution

Voltage drop across $$R_s$$ is
$$(10-5)=5V$$
$$\therefore I=\displaystyle\frac{V}{R_s}=\frac{5}{200}=0.025$$A
$$=25mA$$.