has sharp breakdown at low reverse voltage.

is operated only in forward bias.

There is no free electrons at OK

There are no free electrons at any temperature

The number of free electrons is more than in a conductor

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

The intrinsic charge carrier density in germanium crystal at

$300 K is 2.5 \times 10^{13} / cm^3$ . density in an n-type germanium crystal at 300 K be

$5 \times 10^{16} /cm^3$ ,the hole density in this n-type crystal at 300 K would be

0%
$2.5 \times 10^{13} / cm^3$
0%
$5 \times 10^6 / cm^3$
0%
$1.25 \times 10^{10} / cm^3$
0%
$0.2 \times 10^4 / cm^3$

Explanation

$\large \begin{array}{l} nP=n{ i^{ 2 } }\left[ { By\, \, Mass\, \, action\, \, law } \right] \\ n=\frac { { 2.5\times 2.5\times { { 10 }^{ 26 } } } }{ { 5\times { { 10 }^{ 16 } } } } \\ n=1.25\times { 10^{ 10 } }/c{ m^{ 3 } } \\ Hence, \\ option\, \, C\, \, is\, correct\, \, answer. \end{array}$

The combination of gates shown below yields.

0%
NAND gate.
0%
NOR gate.
0%
XOR gate.
0%
OR gate.

A wire of resistance 'R' is cut into four equal parts and they are bundled together side by side to form a thicker wire. The resistance of the bundle is

0%
$\cfrac{R}{16}$
0%
$\cfrac{R}{2}$
0%
$2R$
0%
$4R$

The input signals at NAND gate is given below

Which of the PN-junction is forward biased?

0%
0%
0%
0%
None of these

To get output

$'1'$ and

$R$ , for the given logic gate circuit the input values must be

0%
$X = 0, Y = 1$
0%
$X = 1, Y = 1$
0%
$X = 0, Y = 0$
0%
$X = 1, Y = 0$

Explanation

$p = \overline {x} + y$

$Q = \overline {\overline {y} . x} = y + \overline {x}$

$O/P = \overline {P + Q}$
To make

$O/P$

$P + Q$ must be

$'O'$
So,

$y = 0$

$x = 1$ .

Energy bands in solids are a consequence of

0%
Ohm's Law
0%
Pauli's exclusion principle
0%
Bohr's theory
0%
Heisenberg's uncertainty principle

A PN-junction has a thickness of the order of :

0%
$1\, gm$
0%
$1\, mm$
0%
$1\, \mu m$
0%
$1\, pm$

Explanation

$P-N$ Junction or depletion region is the area between

$P-$ type and

$N-$ type semiconductors; which is free from charge carriers.

This region between the two arts as a barrier and has a very small width

$\approx\ 10^{-6}m$

$1\mu\ m$

The output of an AND gate is connected to both the inputs of a NOR gate, then this circuit will act as a

0%
OR gate
0%
NOR gate
0%
AND gate
0%
NOT gate
0%
NAND gate

In a P-N junction, a potential barrier exists across the function. A hole with kinetic energy of 300 meV *approaches the junction Find the kinetic energy of the hole it crosses the junction it the hole approached the Junction from the p-side and

0%
105 me V
0%
55 me V
0%
50 me V
0%
100 me V

Explanation

Potential barrier 'd'

$=250$ mev

Initial KE of hole

$=300$ mev

KE of the hole decreases when the junction in forward biased and increases when reverse biases in given Px dioide.

(a) Final KE

$=(300-250)$ mev

$=50$ mev

(b) Initial KE

$=(300+250)$ mev

$=550$ mev.

1212740_1288481_ans_1897414b38aa45d1ab22aa6b7d0fd624.JPG

In a p-type semiconductor, the acceptor level lies

0%
near the conduction band
0%
half way between conduction and valence bands
0%
within conduction band
0%
near the valence band

In the following common emitter circuit if

$\beta = 100, V _ { C E } = 7 V , V _ { B E } = \text { negligible, } R _ { C } = 2 \mathrm { k } \Omega$ then

$I _ { B }$ is

0%
$0.01 \mathrm { mA }$
0%
$0.04 \mathrm { mA }$
0%
$0.02 \mathrm { mA }$
0%
$0.03 \mathrm { mA }$

For the circuit show below, the current through the zener diode is :

0%
5 mA
0%
Zero
0%
14 mA
0%
9 mA

Explanation

Assuming zener diode does not undergo breakdown, current in circuit =

$\dfrac{120}{15000} =8mA$
Voltage drop across diode = 80 V > 50 V. The diode undergo breakdown.
Current is

$R_1 = \dfrac{70}{5000} = 14 mA$
Current is

$R_2 = \dfrac{50}{10000} =5A$
current through diode = 9 mA
1144124_1333998_ans_1224e35e86d44fb8b3bbc5ca397e458c.PNG

Find the value of

${ V }_{ 0 }$

0%
5.7 V
0%
5.3 V
0%
6 V
0%
5 V

Explanation

From the given diagram,

$V = 6V$

${V_{Ge}} = 0.3V$

${V_{Si}} = 0.7V$

Germanium starts conducting at 0.3 V and silicon starts conducting at 0.7 V. As both the diodes are in forward biased condition, the current will flow through the diode having less potential difference i.e. through Germanium

Therefore, the value

${V_0}$ will be

${V_0} = 6 - 0.3 = 5.7V$

Hence Option (A) is the correct answer

When PN junction os forward biased , the current in junction

0%
increases
0%
decreases
0%
Remains same
0%
None of these

With the rise of temperature, the specific resistance of semiconductor?

0%
Increases
0%
Remains unchanged
0%
Decreases
0%
None of these

Truth table for the following digital circuit will be:-

The figure shows two

$NAND$ gates follow by a

$NOR$ gate. The system is equivalent to the following logic gate

0%
$OR$
0%
$AND$
0%
$NAND$
0%
None of these

Explanation

Output

$Z$ of single three input gate is that of

$AND$ gate.
1742897_1350943_ans_2da1d6e5bbe84af7982ac8ca299a9ff8.png

The output of the given logic circuit is:

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

Explanation

$Y=\overline { (\bar { A } { + }\bar { B } )\bar { A } }$

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

$=A (\overline {\bar {A} B})$

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

$A+A\bar { B} = A\bar { B}$
1144007_1332828_ans_796100c64edf454f9bf757be010d9f3f.PNG

With the help of the arrangement shown in figure which of the following gates could be realised?

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

A Zener diode

0%
has negative temperature coefficient of resistance.
0%
has sharp breakdown at low reverse voltage.
0%
rectifies alternating voltage.
0%
is operated only in forward bias.

Explanation

A Zener diode has a negative temperature coefficient of breakdown voltage.

The mechanism of Zener breakdown is the quantum tunneling of electrons. In a Zener diode, the depletion region width is very small due to very highly doped n and p regions. This makes the depletion region width very narrow, which creates a high electric field across it on application of reverse bias. On application of a suitable reverse bias across the diode, this high electric field causes the breaking of covalent bonds in the lattice, to liberate electron-hole pairs, and then pulls the carriers in the respective directions of the field.

The concerned electric field is directed from $n$ side to $p$ side, so minority electrons in the $p$ side are drawn by this field to the $n$ side, and minority holes in the $n$ side are drawn to the $p$ side.

As temperature increases, the thermal vibration of the lattice atoms increases, increasing the vibrational energy of the constituent electrons. This additional vibrational energy makes it easier for an electron to escape from the covalent bond, under the influence of the existing electric field.

Thus, with increased temperature, we have a large generation of minority carriers and their subsequent movement in the respective directions, at a lower value of reverse bias voltage. In other words, Zener breakdown voltage decreases with increase of temperature, as temperature facilitates breakdown.

Thus, the temperature coefficient of the Zener breakdown voltage is negative.

Two conducting wires of same material and of equal lengths, resistance and resistivity having ratio

$1:2$ . Their diameter ratio will be

0%
$1:2$
0%
$1:4$
0%
$4:1$
0%
$1:\sqrt{2}$

Explanation

Hence, option

$(D)$ is correct answer.
1380755_1388548_ans_8785bb9b561245df87a1364efef49207.jpg

In a semiconductor

0%
There are no free electrons at any temperature
0%
The number of free electrons is more than in a conductor
0%
There is no free electrons at OK
0%
None of these

Explanation

In semiconductor, no free electron at normal temperature.

Hence,

option

$C$ is correct answer.

With rise in temperature the electrical conductivity of intrinsic semiconductor:

0%
increases
0%
decreases
0%
first increases and then decreases
0%
first decreases and then increases

Wire

$P$ and

$Q$ have the same resistance at a ordinary room temperature .When heated resistance at a ordinary room temperature .When heated resistance of

$P$ increases and that of

$Q$ decreases. we conclude that

0%
$P$ is semiconductor and $Q$ is conductor
0%
$P$ is conductor and $Q$ is semiconductor
0%
$P$ is n-type semiconductor and $Q$ is $p-$ type conductor
0%
None of the above

Explanation

Conductor has positive temperature coefficient of resistance but semiconductor has negative temperature coefficient of resistance. Hence,

$P$ is semiconductor and

$Q$ is conductor.

The width of deplection layer in P-N junction diode is :

0%
$10^-3$ m
0%
$10^-4$ m
0%
$10^-5$ m
0%
$10^-6$ m

Explanation

In general cases, the magnitude width of the depletion region of a

${\text{p - n}}$ junction that is used in experiments and in other places is

$1 \times {10^{ - 5}}m$ .

In a common emitter transistor the current gain isWhat is change in collector current, when change in base current is

$250\mu A$

0%
$9 \times {10^{ - 2}}A$
0%
$8 \times {10^{ - 6}}A$
0%
$2 \times {10^{ - 2}}A$
0%
$3 \times {10^{ - 7}}A$

Explanation

$\begin{array}{l} Current\, \, given=\beta =\frac { { \Delta { i_{ e } } } }{ { \Delta { i_{ b } } } } \\ \Rightarrow \Delta { i_{ e } }=\beta \times \Delta { i_{ b } } \\ =80\times 20\mu A \\ =2\times { 10^{ -2 } }A \end{array}$

Hence,

option

$(C)$ is correct answer.

Symbolic representation of photodiode is

What n - type semiconductor is heated :

0%
number of electrons increases while that of holes decreases
0%
number of holes increases while that of electrons decreases
0%
number of electrons and holes increase equally.
0%
number of electrons and holes increase equally
0%
none of these

Explanation

When the temperature of a

${\text{n - }}$ type semiconductor is increased, the electrons with the minimum energy acquire more energy and jumps to the valence band. This increases the number of electrons, also increasing the number of the ‘Holes.’

Option (D) is the correct answer.

In the forward bias characteristic curve, a diode appears as :

0%
an OFF switch
0%
a high resistance
0%
an ON switch
0%
a capacitor

In a

$\mathrm { n }$ -p-n transistor circuit, the collector current is

$10 \mathrm { mA }$ . If

$95$ percent of the electrons emitted reach the collector, which of the following statements are true?

0%
The emitter current will be 8 $\mathrm { mA }$
0%
The base current will be $2 \mathrm { mA }$
0%
The base current will be $0.53\mathrm { mA }$
0%
None of these

Choose the correct option:
Electrical conductivity of a semi conductor-

0%
decreases with rise in its temperature.
0%
increases with rise in its temperature.
0%
does not changes with temperature.
0%
first decreases and then increases with rise in temperature.

Identity the gate in the figures

0%
AND
0%
NOR
0%
XOR
0%
NAND

Explanation

Truth table

$A$	$B$	$\overline{AB}$
0	0	1
0	1	1
1	0	1
1	1	0

This is the truth table of NAND gate.

The correct option is D.

1495886_1547720_ans_5b2c0795a62d4762908653ea397d932a.png

The Boolean expression for NAND gate is -

0%
$Y = A + B$
0%
$Y = A.B$
0%
$Y = \overline {A + B}$
0%
$Y = \overline {A. B}$

Explanation

The Boolean algebra for

${\text{AND}}$ gate is-

$Y = A.B$

So, the Boolean algebra for

${\text{NAND}}$ gate is-

$Y = \overline {A.B}$

The thickness of the P-N junction is m.

0%
${10^{ - 2}}$
0%
${10^{ - 6}}$
0%
${10^{ - 8}}$
0%
${10^{ - 12}}$

Explanation

${\text{p - n}}$ junction is made of a

${\text{p - }}$ type semiconductor and a

${\text{n - }}$ type semiconductor. In general, he thickness of a

${\text{p - n}}$ junction is in the order of

${10^{ - 6}}m$

Select and write the correct option from the options given in each question:
Optical Detector is:

0%
Diode Laser
0%
Laser
0%
LED
0%
Photo Diode

The arrangement of

$NAND$ gates shown below effectively work as

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

The circuit given is equivalent to

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

Explanation

If we do the Boolean algebra for the whole system, we will get the truth table of the output as follows-

$1,0,0,0$

This is the truth table of the

${\text{NOR}}$ gate.

Option (D) is correct.

Good resistance coils are made of ?

0%
Copper
0%
Manganin
0%
Gold
0%
Silver

The combination of NAND gates shown here under (figure) are equivalent to

0%
An OR gate and an AND gate respectively
0%
An AND gate and a NOT gate respectively
0%
An AND fate and an OR gate respectively
0%
An OR gate and a NOT gate respectively.

Determine the current through zener diode for the circuit shown in figure is: (Given: zener diode break down voltage

$V_z=5.6V$ )

0%
$7mA$
0%
$17mA$
0%
$10mA$
0%
$15mA$

Explanation

For zener brak down potential difference across

$800\Omega$ resistor will be

$5.6V$ 4

$V_z=5.6V$

$i_2=\dfrac{V_z}{800}=\dfrac{5.6}{800}=7mA$

$\Delta V$ across

$200\Omega =9-5.6=3.4V$

$i_1=\dfrac{3.4}{200}=17mA$

$i_1=1_2+i_z$

$i_z=17mA-7mA=10mA$
1242319_1611265_ans_25f97e5cec9444c1b259c1c5cd593d97.PNG

For the following combination of gates, select the correct statement

0%
The output is $1$ when both the inputs are $1$
0%
The output is $0$ when both the inputs are $0$
0%
The output is $0$ when the two inputs differ
0%
The outputs is $1$ when the two inputs differ

A device or power strip designed to protect electronic equipment from power surges and spikes is known as.....

0%
Surge Suppressors
0%
Ping
0%
Gateway
0%
None of these

Logic gate for inpute A an B is given in figure .
Which table is correct for given gates system:

Explanation

Output of OR gate

$C=A+B$

$y=(A+B).A$

LCD stands for ______________.

0%
Low Crystal Display
0%
Less Crystal Display
0%
Liquid Crystal Display
0%
All the above

The reverse breakdown voltage of a Zener diode is

$5.6\ V$ in the given circuit.
The current

$I_{Z}$ through the Zener is

0%
$7\ mA$
0%
$17\ mA$
0%
$10\ mA$
0%
$15\ mA$

Explanation

$9 = V_{Z} + V_{R_{1}}$

$V_{Z} = 5.6\ V$

$V_{R_{1}} = 9 - 5.6$

$V_{R_{1}} = 3.4$

$I_{R_{1}} = \dfrac {V_{R_{1}}}{R} = \dfrac {3.4}{200}$

$I_{R_{1}} = 17\ mA$

$V_{Z} = V_{R_{2}} = I_{R_{2}} (R_{2})$

$\dfrac {5.6}{800} = I_{R_{2}}$

$I_{R_{2}} = 7\ mA$

$I_{Z} = (17 - 7) mA$

$= 10\ mA$ .
1245666_1613806_ans_989192d9c1024ed1993a3e93db728adf.png

The truth table for the circuit given in the fig. is:

0%
$\begin{vmatrix} A&B&Y \\0&0&1\\0&1&1\\1&0&0\\1&1&0 \end{vmatrix}$
0%
$\begin{vmatrix} A&B&Y \\0&0&0\\0&1&0\\1&0&1\\1&1&1 \end{vmatrix}$
0%
$\begin{vmatrix} A&B&Y \\0&0&1\\0&1&0\\1&0&0\\1&1&0 \end{vmatrix}$
0%
$\begin{vmatrix} A&B&Y \\0&0&1\\0&1&1\\1&0&1\\1&1&1 \end{vmatrix}$

Explanation

$C = A + B$
and

$y = \overline{A.C}$

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

For the given circuit shown in figure, the potential of the battery is varied from

$10V$ to

$16V$ . If by zener diode breakdown voltage is

$6V$ , find maximum current through zener diode.

0%
$1.5\ mA$
0%
$3.5\ mA$
0%
$5\ mA$
0%
$6.5\ mA$

Explanation

$i_{i} = \dfrac {6}{4} = 1.5\ mA$
Maximum current will be obtained for battery voltage

$16\ V$

$i = \dfrac {16 - 6}{2} = 5mA\ i_{z}(max) = 5 - 1.5 = 3.5\ mA$ .
1245252_1613181_ans_505916b6a4954ce0b48eb4e43d8ff0f4.png

The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6V and the load resistance is

$R_L = 4 k \Omega$ . The series resistance of the circuit is

$R_i = 1 k \Omega$ . If the battery voltage

$V_B$ varies from

$8V$ to

$16V$ , what are the minimum and maximum values of the current through Zener diode ?

0%
$0.5 \, mA ; \, 6 \, mA$
0%
$0.5 \, mA ; \, 8.5 \, mA$
0%
$1.5 \, mA ; \, 8.5 \, mA$
0%
$1 \, mA ; \, 8.5 \, mA$

Explanation

$V_B = 8V$

$i_L = \dfrac{6 \times 10^{-3}}{4} = 1.5 \times 10^{-3} A$

$i_R = \dfrac{8 - 6 \times 10^{-3}}{1} = 2 \times 10^{-3} A$

$\therefore i_{zener \, diode} = i_R - i_{load}$

$= 0.5 \times 10^{-3} A$
At

$V_B = 16 V$

$i_L = 1.5 \times 10^{-3} A$

$i_R = \dfrac{(16 - 6) \times 10^{-3}}{1} = 10 \times 10^{-3} A$

$\therefore i_{zener \, diode} = i_R - i_L$

$= 8.5 \times 10^{-3} A$

Figure shown a DC voltage regulator circuit, with a Zener diode of breakdown voltage =

$6V$ . If the unregulated input voltage varies between

$10 V$ to

$16 V$ , then what is the maximum Zener current ?

0%
$2.5 \, mA$
0%
$3.5 \, mA$
0%
$7.5 \, mA$
0%
$1.5 \, mA$

Explanation

Maximum current will flow from zener if input voltage is maximum.
When zener diode works in breakdown state, voltage across the zener will remain same.

$\therefore V_{across \,4k\Omega} = 6V$

$\therefore$ Current through

$4K\Omega = \dfrac{6}{4000} A = \dfrac{6}{4}mA$
since input voltage

$= 16V$

$\therefore$ Potential difference across

$2K\Omega = 10V$

$\therefore$ Current difference across

$2K\Omega = \dfrac{}{} = 5mA$

$\therefore$ Current through zener diode

$= (I_S - I_L) = 3.5 \,mA$
1248218_1615164_ans_6236d53f64b7447cb3ba41876052ae88.png

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

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Practice Class 12 Engineering Physics Quiz Questions and Answers

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