has sharp breakdown at low reverse voltage.

is operated only in forward bias.

There are no free electrons at $$0K$$

There are no free electrons at any temperature

The number of free electrons is more than that of the conductor.

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 Medical Physics Semiconductor Electronics: Materials, Devices And Simple Circuits Quiz 12

For a transistor amplifier power gain and voltage gain are

$7.5$ and

$2.5$ respectively. The value of the current gain will be:

0%
$0.33$
0%
$0.66$
0%
$0.99$
0%
$3$

In a

$NPN$ transistor working as a common base amplifier, current gain is

$0.96$ and emitter current is

$7.2\ mA$ . The base current is

0%
$0.4\ mA$
0%
$0.2\ mA$
0%
$0.29\ mA$
0%
$0.35\ mA$

Explanation

$\alpha=\dfrac{i_{c}}{i_{e}}=0.96$ and

$i=7.2\ mA$

$\Rightarrow i_{c}=0.96\times i_{e}=0.96\times 7.2=6.91\ mA$

$\therefore i_{e}=i_{c}+i_{b}\Rightarrow 7.2=6.91+i_b \Rightarrow i_b=0.29mA$ .

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

The magnitude of the force between a pair of conductors, each of length 110 cm, carrying a current of 10 A and separated by a distance of 10 cm is

0%
$55 \times 10 ^ { - 5 } N$
0%
$44 \times 10 ^ { - 5 } N$
0%
$33 \times 10 ^ { - 5 } N$
0%
$22 \times 10 ^ { - 5 } N$

Explanation

$\begin{array}{l} B=\frac { { { \mu _{ o } }i } }{ { 2\pi r } } \\ F=\frac { { { \mu _{ o } }i } }{ { 2\pi r } } \times iL \\ =\frac { { 4\pi \times { { 10 }^{ -7 } }\times 10\times 10\times 110 } }{ { 2\pi .10 } } \\ =22\times { 10^{ -5 } } \\ =2.2\times { 10^{ -4 } }N \\ Hence, \\ option\, \, D\, \, is\, \, correct\, \, answer. \end{array}$

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%
$0.03 \mathrm { mA }$

Which of the following is not a rectifier circuit ?

0%
0%
0%
0%
none of these

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%
Increase
0%
decrease
0%
remains same
0%
None of these

The maximum efficiency of full wave rectifier is:

0%
$100$ %
0%
$25.20$ %
0%
$42.2$ %
0%
$81.2$ %

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 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

The current gain of a common emitter amplifier is

$69$ . If the emitter current is

$7.0\ mA$ , collector current is

0%
$69\ mA$
0%
$6.9\ mA$
0%
$0.69\ mA$
0%
$9.6\ mA$

Explanation

Given,

$\beta=69$

$I_E=7.0mA$

Current gain,

$\beta=\dfrac{I_C}{I_B}$

$I_C=\beta I_B$ . . . . . . . . . . . (1)

$I_E=I_B+I_C=I_B(1+\beta)$

$I_B=\dfrac{I_E}{1+\beta}$

$I_B=\dfrac{7mA}{1+69}=0.1mA$

From equation (1),

$I_C=\beta I_B$

$I_C=69\times 0.1$

$I_C=6.9mA$

The correct option is B.

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

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,

$y = 0$

$x = 1$ .

Which of the following elements is added to turn artificial diamond into semi conductors?

0%
Beryllium
0%
Boron
0%
Magnesium
0%
Calcium

Explanation

Synthetic diamond has potential uses as a semiconductor

$,$ because it can be doped with impurities like boron and phosphorus

$.$

Hence,

option

$(B)$ is correct answer.

The current gain of a transistor in common base mode is 0.If the change in emitter current is 5 mA then the change in collector current is

0%
4..95 mA
0%
5.4 mA
0%
5.99 mA
0%
0.198 mA

Explanation

In common base transistor, current gain is,

$a=\dfrac{I_c}{I-e}$

Here

$I_c$ represents current through collector and

$I_e$ represents current through emitter.

In the question it is given,

$a=0.99$

$I_e=5\,mA$

Therefore,

$I_c=a\times I_e$

$I_c=0.99\times 5\,mA=4.95\,mA$

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.

In a semiconductor

0%
There are no free electrons at any temperature
0%
The number of free electrons is more than that of the conductor.
0%
There are no free electrons at $0K$
0%
None of these

Explanation

In semiconductor, there are fewer electrons than that of a conductor.

But, at

$0K$ , i.e., at the absolute zero temperature, there can not exist any free electrons. This is also true in the case of any conductor.

The current gain of a transistor in a common base arrangement is

$0.98$ . Find the changes in collector current corresponding to a change of

$5.0$ mA in emitter current. What would be the change in base current?

0%
$4.9 mA, 0.1 mA$
0%
$4.9 mA, 0.2 mA$
0%
$5.9 mA, 0.3 mA$
0%
$5.9 mA, 0.8 mA$

Explanation

Given:

Current gain of a transistor,

$\alpha = 0.98$

Change in the emitter current,

$\Delta {i_e} = 5.0{\text{ mA}}$

The current gain of a transistor (

$\alpha$ ) is the ratio of collector current (

$\Delta {i_c}$ ) to the emitter current (

$\Delta {i_e}$ ) of a transistor

$\Rightarrow \alpha = \dfrac{{\Delta {i_c}}}{{\Delta {i_e}}}$

Substituting the given values,

$\Rightarrow 0.98 = \dfrac{{\Delta {i_c}}}{{5.0}}$

$\Rightarrow \Delta {i_c} = 0.98 \times 5.0 = 4.9{\text{ mA}}$

The change in collector current is 4.9 mA

The change in base current can be found by the formula

$\Delta {i_b} = \Delta {i_e} - \Delta {i_c}$

Substituting the values,

$\Rightarrow \Delta {i_b} = 5.0 - 4.9 = 0.1{\text{ mA}}$

The change in base current is 0.1 mA

Therefore, Option (A) will be the right answer.

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.

In the circuit, if the forward voltage drop for the diode is

$0.5{ V },$ the current will be :

0%
$3.4{ mA }$
0%
$2{ mA }$
0%
$2.5{ mA }$
0%
$3{ mA }$

Explanation

The voltage drop across resistance

$=8-0.5=7.5\ V$

$\therefore$ Current

$i=\dfrac {75}{2.2\times 10^3}=3.4\ mA$

Symbolic representation of photodiode is

The output of the given circuit in the figure.

0%
would be zero at all times
0%
would be like a half-wave rectifier with positive cycles in output
0%
would be like a half -wave rectifier with negative cycles in output
0%
would be like that of a full wave rectifier

Explanation

In this given figure,

The output of the given circuit in the figure would be like

a half- wave rectifier with negative cycles in output.

Option

$C$ is 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

Efficiency of half wave rectifier is:

0%
40.6 %
0%
4.06 %
0%
0.406 %
0%
0.0406 %

Two constant parallel forces act on a wheel provided in a transistor for tunning as shown in the figure. If the wheel rotates then the work done per rotation will be

0%
0.05 joule
0%
0.0015 joule
0%
0.031 joule
0%
0.062 joule

Explanation

$\begin{array}{l} \tau =r\times F \\ =\left( { 2.5\times 1.0 } \right) \times 2 \\ =0.05Nm \\ \omega =\tau .\theta \\ =0.05\times 2\pi \\ =0.1\times 3.14 \\ =0.031\, joule \\ Hence, \\ option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$

The

$PN$ junction diode is used as

0%
An amplifier
0%
A rectifier
0%
An oscillator
0%
A modulator

Explanation

It is used to convert

$ac$ into

$dc$ ( rectifier)
1743482_1820598_ans_a3879a443d4f4d24954aaef26c74a7f5.png

In the circuit below, the value of current is

0%
$0\ amp$
0%
$10^{-2}\ amp$
0%
$10^2\ amp$
0%
$10^{-3}\ amp$

Explanation

Current flow is possible and

$i=\dfrac VR =\dfrac{(4-1)}{300}=10^{-2}A$

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

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

Find current through the ideal diode.

0%
Zero
0%
20A
0%
0.05 A
0%
0.15 A

Explanation

The current should flow from

$10{\text{ Volt}}$ to

${\text{5 Volt}}$ , but the diode is connected in the reverse bias. Hence, the current through the circuit is Zero.

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.

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

Explanation

Output of OR gate

$C=A+B$

$y=(A+B).A$

In a full wave rectifier in which input voltage is represented by

$V=V_M\sin \omega t$ , then peak inversion voltage of non conducting diode will be?

0%
$-V_M$
0%
$V_M/2$
0%
$2V_M$
0%
$0$

Explanation

$=2V_M$ .

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

Thermistor is a semiconductor device whose temperature coefficient of resistance is

0%
Large and negative
0%
Small and negative
0%
Large and positive
0%
Small and positive

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

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

For transistor, the current ratio

$'\beta_{dc}'$ is defined as the ratio of?

0%
Collector current to emitter current
0%
Collector current to base current
0%
Base current to collector current
0%
Emitter current to collector current

Explanation

For Transistor

$(\beta_{dc})$ is the current gain and is defined as the ratio of collector current to base current.

$(\beta_{dc}) = \dfrac{I_c}{I_B}$

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

A thin plate or board that contains electronic components is called.....

0%
RAM
0%
ROM
0%
Circuit Board
0%
All the above

The impurity added in germanium crystal to make

$n-$ type semi-conductor is:

0%
aluminium
0%
gallium
0%
iridium
0%
phosphorus

The valence of an impurity added to germanium crystal in order to convert it into a

$P-$ type semi conductor is

0%
$6$
0%
$5$
0%
$4$
0%
$3$

The given transistor operates in saturation region then what should be the value of

$V_{BB}: (R_{out}=200\Omega, R_{in}=100 K\Omega, V_{CC}=3$ volt,

$V_{BE}=0.7$ volt,

$V_{CE}=0, \beta =200)$ .

0%
$4.1$ volt
0%
$7.5$ volt
0%
$8.2$ volt
0%
$6.8$ volt

Explanation

We know that,

$V_{CE}=V_{CC}−I_{C}R_{O}$

$0=3−I_{C}\times200$

$I_{C}=\dfrac{3}{200}=15\ mA$

$\beta=200$

$\beta=\dfrac{I_{C}}{I_{B}}$

$I_{B}=\dfrac{I_{C}}{I_{β}}=\dfrac{15mA}{200}=75\ \mu A$

$V_{BE}=V_{BB}−I_{B}R_{in}$

$V_{BB}=0.7+75\times10−6\times100\times10^3$

$V_{BB}=0.7+7.5=8.2\ volt$

So, The value of

$V_{BB}$ is

$8.2\ volt$ .

Hence, C is correct option.

Assertion: Gallium arsenide phosphide is used in red

$L.E.D$
Reason: Its work function lies b/w

$1.65ev$

0%
If both assertion and reason are true and reason is the correct explanation of assertion.
0%
If both assertion and reason are true but reason is not correct explanation of assertion.
0%
If assertion is true but reason is false.
0%
If both assertion and reason are false.

In an insulator, band gap of the order of?

0%
$0.1eV$
0%
$1eV$
0%
$5eV$
0%
$100eV$
0%
$1MeV$

Explanation

In an insulator the forbidden band gap is of the order of

$5eV$ .

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

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Practice Class 12 Medical Physics Quiz Questions and Answers

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

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Practice Class 12 Medical Physics Quiz Questions and Answers

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