it is equivalent to series switching circuit.

it has no equivalence to switching circuit.

it is equivalent to parallel switching circuit.

it is a mixture of series and parallel switching circuit.

MCQ Questions for Class 12 Engineering Physics Semiconductor Electronics: Materials,Devices And Simple Circuits Quiz 5

The load voltage is approximately constant when a zener diode is

0%
Forward-biased
0%
Reverse-biased
0%
Operating in the breakdown region
0%
Unbiased

What is the order of magnitude of the resistance of a dry human body?

0%
$10\Omega$
0%
$10^4 \Omega$
0%
$10$ M $\Omega$
0%
$10\mu\Omega$

Explanation

It is known that the resistance of a dry human body is

$10 k \Omega = 10^{4} \Omega$ .

If the series resistance decreases in an unloaded zener regulator, the zener current

0%
Decreases
0%
Stays the same
0%
Increases
0%
Equals the voltage divided by the resistance

For detecting the light_______________.

0%
The photodiode has to be forward biased
0%
The photodiode has to be reversed biased
0%
The LED has to connected in forward bias mode
0%
The LED has to be connected in reverse bias mode

A semi conductor device with both active and passive electronic elements diffused into a silicon water to form a functional circuit is called.

0%
Thin-film circuit
0%
Integrated circuit
0%
Hybrid-circuit
0%
Thick-film circuit

Explanation

Integrated circuits or

$I.C$ 's are those semi-conductor device with both active and passive electronic elements diffused into a silicon water to form a functional circuit.

The n-side of the depletion layer of a p-n junction.

0%
Always has same width as of the p-side
0%
Has no bound charges
0%
Is negatively charged
0%
Is positively charged

Which of the following gates will have an output of

$1$ ?

0%
$D$
0%
$A$
0%
$B$
0%
$C$

Explanation

The logic gate

$(C)$ is a NAND gate for which the Boolean expression is

$X = \overline {0.1}$ .
We have

$X = \overline {0} = 1$ .
Hence option D is correct.

The breakdown in a reverse biased p-n junction diode is more likely to occur due to.

0%
large velocity of the minority charge if the doping concentration is small.
0%
large velocity of the minority charge carriers if the doping concentration is small.
0%
strong electric field in a depletion region if the doping concentration is small.
0%
strong electric filed in the deplention region if the doping conentration is large.

Which of the following devices can be used as a square law device for modulation.

0%
p-n-p transistor in active region
0%
p-n junction diode near threshold voltage.
0%
Superconductor
0%
Inductor

Explanation

$p-n$ junction diode near threshold voltage is used as a square law device for modulation.

The logic symbols shown here are logically equivalent to :-

0%
$AND$ and $OR$ gate
0%
$NOR$ and $NAND$ gate
0%
$OR$ and $AND$ gate
0%
$NAND$ and $NOR$ gate

Zener breakdown takes place if:

0%
Doped impurity is low
0%
Doped impurity is high
0%
Less impurity in N-part
0%
Less impurity in P-type

On increasing the temperature of a semi-conductor material

0%
Density of charge-carrier as well as their mobilities both increases
0%
Density of charge-carriers increases, but their mobilities decreases
0%
Density of charge-carries decreases, but their mobilities increases
0%
Both density of charge-carriers as well as their mobilities decreases

Conventional flow of current will always be from -

0%
p type $\rightarrow$ n type
0%
n type $\rightarrow$ p type
0%
n Type $\rightarrow$ n type
0%
None of these

When a p-n junction diode is connected in forward bias its barrier potential

0%
decreases and less current flows in the circuit
0%
decreases and more current flows in the circuit
0%
increases and more current flows in the circuit
0%
decreases and no current flows in the circuit

The process by which ac is converted into dc is known as

0%
Purification
0%
Amplification
0%
Rectification
0%
Current Amplification

Semi-conductor are formed if the bonds are

0%
van der Walls
0%
ionic
0%
metallic
0%
covalent

The nature of binding for a crystal with alternate and evenly spaced positive and negative ions is:

0%
metallic
0%
covalent
0%
dipolar
0%
ionic

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.

0%
A and B are correct.
0%
A and B are wrong.
0%
A is correct but B is wrong.
0%
A is wrong but B is correct.

Explanation

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.

There is a sudden increase in current in zener diode is

0%
Due to rupture of bonds
0%
Resistance of deplection layer becomes less
0%
Due to high doping
0%
Due to less doping

A p-n junction diode cannot be used

0%
as a rectifier.
0%
for converting light energy to electric energy.
0%
for getting light radiation.
0%
for increasing the amplitude of an AC signal.

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.

0%
(a) and (b) are correct
0%
(b) and (c) are correct
0%
(a) only is correct
0%
(d) only is correct

AND gate

0%
it has no equivalence to switching circuit.
0%
it is equivalent to series switching circuit.
0%
it is equivalent to parallel switching circuit.
0%
it is a mixture of series and parallel switching circuit.

In an insulator, the energy gap between conduction band and valence band is about

0%
$0\ eV$
0%
$6\ eV$
0%
$1\ eV$
0%
$0.6\ eV$

Explanation

Atoms are neutral in the insulator, thus there are no valence electrons in the outer orbit. The electrons are tightly bound with the nucleus hence, at room temperature thermal energy is not enough to push the electrons into conduction band and hence, no electrons are available for conduction. The energy required for electron to escape out from orbit and to over come the energy gap is thus of the order of

$6\ eV$ .

Consider the following statements A and B and identify the correct answer:
A) germanium is preferred over silicon in the construction of Zener diode.
B) germanium has high thermal stability than silicon in the construction of Zener diode.

0%
Both (A) and (B) are true.
0%
Both (A) and (B) are false.
0%
(A) is true but (B) is false.
0%
(A) is false but (B) is true.

Explanation

Germanium diode does not have a sharp breakdown in reverse bias and hence, the voltage does not remain constant at

$5\ V$ after breakdown while silicon diode is better in this respect. Hence, silicon is preferred in Zener diode. Also, thermal stability of silicon is better than germanium.

The following is NOT equal to

$0$ in the Boolean algebra is

0%
$\overline{\overline A.0}$
0%
$A.\bar{A}$
0%
$A.0$
0%
$\overline{A+\overline A}$

Explanation

(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

$A = 1$ ,

$A.\bar A = 1. \bar 1 = 1.0 = 0$

$A = 0$ ,

$A.0 = 0.0 = 0$
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

$A = 1$ ,

$\overline{A+\bar A} = \overline{1+\bar 1} = \overline{1+0} = \bar 1 = 0$

Digital circuit can be made by repetitive use of :

0%
OR gates
0%
AND gates
0%
NOT gates
0%
NAND gates

NAND and NOR gates are called universal gates because they

0%
are universally available.
0%
can be combined to produce OR, AND and NOT gates.
0%
are widely used in the integrated circuits.
0%
can be easily manufactured.

Explanation

The truth table for AND gate is as shown in the figure. Thus, the output of an AND gate is

$1$ if and only if both inputs 'A' and 'B' are

$1$ .

In the Boolean algebra:

$\overline{A+B}$ is equal to:

0%
A + B
0%
$\bar{\bar{A}}$ + $\bar{\bar{B}}$
0%
$\overline{\overline{A.B}}$
0%
$\bar{A}$ . $\bar{B}$

Explanation

Clearly, as shown in the truth table,

$\overline{A+B}=\bar{A}.\bar{B}$

The minimum number of gates required to realise the expression

$Z = DABC + D\overline{ABC}$ is

0%
One
0%
Two
0%
Eight
0%
Five

Explanation

The given expression is same as

$Z = D$ .
Hence, one gate is sufficient to realise this expression.

Boolean expression

$Y = A.\bar{B} + B.\bar{A}$ is given. If

$A = 1, B = 1$ then

$Y =$

0%
0
0%
1
0%
11
0%
10

Explanation

$Y=A.\bar B+B.\bar A$
For

$A=1, B=1$

$Y=1.\bar 1+1.\bar 1$

$Y=1.0+1.0 = 0+0 = 0$

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

0%
OR
0%
NOT
0%
AND
0%
NAND

A diode made of silicon has a barrier potential of

$0.7\ V$ and a current of

$20\ mA$ passes through the diode when a battery of emf

$3\ V$ and a resistor is connected to it. The power dissipated across the resistor and diode are

0%
$0.046\ W, \ 0.014\ W$
0%
$4.6\ W,\ 0.14\ W$
0%
$0.46\ W,\ 0.14\ W$
0%
$46\ W, \ 14\ W$

Explanation

Given that:

$V_D = 0.7\ V$

$I = 20\ mA = 0.02\ A$

$V_B = 3\ V$

Now,

$P = (V_B - V_D){I}$
or

$P=(3-0.7)0.02=0.046\ W$

And power across

$R_D$ is

$P_{R_D} = (V_D)I$

$P_{R_D} = 0.7\times 0.02 = 0.014\ W$

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

0%
$0.5$
0%
$0.9$
0%
$0.7$
0%
$1.1$

Explanation

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,

$E = \dfrac{1242}{2480} = 0.5\ eV$

Consider a two-input AND gate of figure below. Out of the four entries for the Truth Table given here, the correct ones are

0%
All
0%
1 and 2 only
0%
1, 2 and 3 only
0%
1, 3 and 4 only

From the adjacent circuit, the output voltage is

0%
$10 V$
0%
$100 V$
0%
$90 V$
0%
zero

Explanation

Here, Zener diode is connected parallel to the input voltage source. Zener diode is known as voltage regulating diode hence, the output voltage will be equal to the voltage across Zener diode i.e.

$10 V$ .

The logic expression which is not equal to

$1$ in Boolean algebra is

0%
$A+1$
0%
$A . \bar{A}$
0%
$A+\bar{A}$
0%
$\overline{\overline {A+A}}$

Explanation

$A+1$ is addition of

$1$ to value of

$A$ , which can be any of

$0$ and

$1$ . This expression would always given '

$1$ ' as the output since addition of

$1$ to anything gives

$1$ .

$A.\bar{A}$ gives

$0$ always because whatever value

$A$ assumes,

$\bar{A}$ assumes the other and they give

$0$ under AND operation.

$A+\bar{A}$ would always give value

$1$ because whatever value

$A$ assumes,

$\bar{A}$ assumes the other and they give

$1$ under OR operation.

$\overline{\overline{A.A}}=A.A=A$ which is might be

$1$ for value of

$A=1$ .

From the circuit shown below, the maximum and minimum value of zener diode current are :

0%
$6\ mA,\ 5\ mA$
0%
$14\ mA,\ 5\ mA$
0%
$9\ mA,\ 1\ mA$
0%
$3\ mA,\ 2\ mA$

Explanation

The Zener diode acts as the voltage regulator in the circuit and the voltage supply is variable hence, excess of voltage over

$50\ V$ will be across

$5 K\Omega$ resistance.

Thus, minimum voltage across

$5 K\Omega$ resistance is

$80\ V-50\ V = 30\ V$

Current through

$5 K\Omega$ resistance is

$I_1$ (say)

$I_1 = \dfrac{30}{5000} = 6\ mA$

Also, maximum voltage across

$5 K\Omega$ resistance is

$120\ V-50\ V = 70\ V$

Current through

$5 K\Omega$ resistance is

$I'_1$ ...(say)

$I'_1 = \dfrac{70}{5000} = 14\ mA$

Now, voltage across load resistance of

$10 K\Omega$ is

$50\ V$

Thus, current through

$10 K\Omega$ will be

$I_2 = \dfrac{50}{10000} = 5\ mA$

Hence, maximum current through Zener diode is

$I_z = I'_1 - I_2 = 14 - 5 = 9\ mA$

Minimum current through Zener diode is

$I_z = I_1 - I_2 = 6 - 5 = 1\ mA$

Boolean expression for the gate circuit shown below is

0%
$A + \bar{A} =1$
0%
$A+1=1$
0%
$A+A=A$
0%
$A+0=A$

Explanation

The input

$A$ directly goes into the OR gate and also first passes through NOT gate, giving another input of

$\bar{A}$ to the OR gate.

$A+\bar{A}=1$ since atleast one of

$A$ and

$\bar{A}$ is

$1$ .

In the given circuit, two input wave forms

$A$ and

$B$ are applied simultaneously. The resultant wave form at

$Y$ is

Explanation

The input waveform indicates the input signals as:
For

$A$ ,

$(1,0,1,0)$ and for

$B, (1,0,0,1)$ . Truth table for given logic is shown below.

Input	Output
$(1,1)$	$0$
$(0,0)$	$1$
$(1,0)$	$1$
$(0,1)$	$1$

Logic gate having an output of

$1$ is

Explanation

Logic gate OR is used for addition of the input signals and Logic gate AND is used for multiplication of the input signals. While, logic gates NOR and NAND are used to reverse the addition and multiplication of the input signals respectively. Hence, input signal in NAND gate (C) will yield output

$1$ .

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

0%
A-0 B-1 C-0
0%
A-1 B-0 C-0
0%
A-1 B-0 C-1
0%
A-1 B-1 C-0

The input of

$A$ and

$B$ for the Boolean expression

$(\overline{A+B}).(\overline{A.B})=1$ is

0%
$0,0$
0%
$0, 1$
0%
$1,0$
0%
$1,1$

Explanation

Given equation is:

$(\overline {A+B})(\overline {A.B})$
A] For

$(0,0)$

$(\overline {A+B})(\overline {A.B}) = (\overline {0+0})(\overline {0.0})$

$(\overline {A+B})(\overline {A.B}) = (1)(1) = 1$

B] For

$(0,1)$

$(\overline {A+B})(\overline {A.B}) = (\overline {0+1})(\overline {0.1})$

$(\overline {A+B})(\overline {A.B}) = (0)(1) = 0$

C] For

$(1,0)$

$(\overline {A+B})(\overline {A.B}) = (\overline {1+0})(\overline {1.0})$

$(\overline {A+B})(\overline {A.B}) = (0)(1) = 0$

D] For

$(1,1)$

$(\overline {A+B})(\overline {A.B}) = (\overline {1+1})(\overline {1.1})$

$(\overline {A+B})(\overline {A.B}) = (0)(0) = 0$

The output Y of the gate circuit shown in the fig is

0%
$\bar{A}+\bar{B}$
0%
$\overline{\bar A + \bar B}$
0%
$\overline{A+B}$
0%
A + B

Explanation

Here, OR gate is followed by NOT gate hence, the addition of inputs will be reversed in output i.e.

$Y = \overline {A+B}$

The combinations of the NAND gates shown hereunder are equivalent to

0%
an OR gate and an AND gate respectively
0%
an AND gate and a NOT gate respectively
0%
an AND gate and OR gate respectively
0%
an OR gate and an NOT gate respectively

Explanation

Truth tables for logic (1) and (2) are shown below:

Input	Output
$(0,0)$	$0$
$(1,0)$	$1$
$(0,1)$	$1$
$(1,1)$	$1$

Input	Output
$(0,0)$	$0$
$(1,0)$	$0$
$(0,1)$	$0$
$(1,1)$	$1$

The truth tables are equivalent to OR and AND gate.

Given above are four logic gate symbols. Those for OR, NOR and NAND are respectively

0%
a, d, c
0%
d, a, b
0%
a, c, d
0%
d, b, a

A Truth table is given below. The logic gate having following truth table is
A B Y
0 0 1
1 0 0
0 1 0
1 1 0

0%
NAND gate
0%
NOR gate
0%
AND gate
0%
OR gate

For the given combination of gates, if the logic states of inputs A, B, C are as follows

$A=B=C=0$ and

$A=B=1, C=0$ then the logic states of output D are

0%
0,0
0%
0,1
0%
1, 0
0%
1,1

The expression of

$Y$ in the above circuit is

0%
$ABCD$
0%
$A+BCD$
0%
$A +B+C+D$
0%
$AB+CD$

Explanation

Here, to build the logic three OR gates are used hence, all the input signals will be added to each other.

Hence, the boolean expression is:

$A+B+C+D = Y$

In the given circuit, if A and B represent two inputs and C represents the output, the circuit represents

0%
NOR gate
0%
AND gate
0%
gate
0%
OR gate

Explanation

The image given shows a diode OR circuit.

$R$ is connected from the output to ground to provide bias current for the diodes. Any positive voltage will represent a logic

$1$ state and the summing of currents through multiple diodes does not change the logic level.

The output

$Y$ of the gate circuit shown in the figure below is

0%
$\overline{A.B}$
0%
$\bar{A}.\bar{B}$
0%
$\overline{\overline{A+B}}$
0%
$\bar{A}+\bar{B}$

Explanation

Here, the inputs are reversed first and then multiplied with each other hence the boolean expression is:

$\bar A.\bar B = Y$

CBSE Questions for Class 12 Engineering Physics Semiconductor Electronics: Materials,Devices And Simple Circuits Quiz 5 - MCQExams.com

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Practice Class 12 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Physics Semiconductor Electronics: Materials,Devices And Simple Circuits Quiz 5 - MCQExams.com

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Practice Class 12 Engineering Physics Quiz Questions and Answers

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