Resistance of an electrolyte increases with rise of temperature

Resistance of metals increases with increasing temperature

Resistance of semi-conductors decreases with rise of temperature

MCQ Questions for Class 12 Engineering Physics Semiconductor Electronics: Materials,Devices And Simple Circuits Quiz 11

The I -V characteristic of an LED is:

Explanation

LED is working only in forward biased condition.

$\therefore I-V$ characteristics will be forward

$I-V$ characteristics of a diode. therefore option D is correct.

The impurity atoms needed to make a p-type semiconductor are

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phosphorus
0%
boron
0%
antimony
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arsenic

In the depletion layer, the electric field created

0%
is from n-side to p-side
0%
is from p-side to n-side
0%
changes randomly
0%
depends on the biasing

Explanation

Because of diffusion a layer of holes is formed over N-side and a layer of electron is formed over P-side as shown below.

Now, as we see the direction of

$\vec { E }$ is form

$N$ to

$P$ .

$\Rightarrow$ Option (A) is the correct answer.

1104361_1017630_ans_cb2c571501bb47d2b152cd873d58e000.jpg

The dominant mechanisms for motion of charge carriers in forward and reversed biased silicon p-n junctions are :

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Diffusion in forward bias, drift in reverse bias
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Diffusion in both forward and reverse bias
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Drift in both forward and reverse bias
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None of these

Explanation

$\Rightarrow$ In forward bias we apply voltage in a direction opposite to that of barrier potential p-side to positive terminal, n-side to negative terminal of battery. So electrons in the n-side, holes in p side pushed towards the junction. Results in increased diffusion.

$\Rightarrow$ In reverse bias, electrons in n-side, holes in p-side pushed away from function.

Ref. image

n-side to positive terminal p-side to negative terminal.

Which of the following circuits correctly represents the following truth table?

A zener regulated power supply consists of a

$9 V$ battery connected in series with a and a zener diode. The zener diode maintains a constant voltage drop of

$4V$ The current drawn by the load resistance will be :

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$0.025 A$
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$0.050 A$
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$0.01 A$
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$0.015 A$

In which of the following circuits is the diode forward biased?

Explanation

Here,

The pointed part of the triangle in the diode symbol is n type and the other is p type,

Diode conducts in forward bias,

And forward bias means high voltage on p-type and low voltage on the n-type,

In all the above options,

Only option

$D$ has high voltage on p-side and low voltage on low side.

The dominant mechanisms of motion of change carriers in forward and reverse biased silicon

$P-N$ junction are

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drift in forward bias, diffusion in reverse bias
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diffusion in forward bias, drift in reverses bias
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diffusion in both forward and reverse bias
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drift in both forward and reverse bias

A non-conducting device is connected in a series circuit with a battery and a resistance. A current is found to pass through the circuit. If the polarity of the battery is reversed, the current drops to almost zero. The device may be

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a p-n junction diode
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an intrinsic semiconductor
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a p-type semiconductor
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a n-type semiconductor

If the resistivity of an alloy is $\rho '$ and that of constituent metal is $\rho$ then :

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$\rho ' > \rho$
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$\rho ' < \rho$
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$\rho ' = \rho$
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there is no simple reaction between $\rho '\& \rho$

If a p-n junction diode is reverse biased, then the resistance measure by an ohm-meter will be

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zero
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low
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high
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infinite

Explanation

A p-n junction diode conducts negligible (not zero) current under reverse biased mode,

Hence there's high resistance is offered under reverse bias mode.

Option

$\textbf C$ is correct.

In the case of forward biasing of a p-n junction diode,which one of the following figure correctly depicts the direction of conventional current?

A Zener diode is connected to a battery and a load as shown below: The current

$I,I_{Z}$ and

$I_{L}$ are respectively :

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$12.5\ mA,735\ mA,5\ mA$
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$12.5\ mA,5\ mA,7.5\ mA$
0%
$15\ mA,7.5\ mA,7.5\ mA$
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$15\ mA,5\ mA,10\ mA$

Explanation

Given that,

${{R}_{L}}=2\,k\Omega$

$R=4\,k\Omega$

$V=60\,V$

${{V}_{Z}}=10\,V$

The Zener diode in the breakdown region maintain a constant voltage, the voltage across $R_{L}$ is

${V_{{R}_{L}}}={{V}_{L}}=10\,V$

Now, the load current is

${{I}_{L}}=\dfrac{{{V}_{L}}}{{{R}_{L}}}$

${{I}_{L}}=\dfrac{10}{2000}$

${{I}_{L}}=5\,mA$

Now, using Kirchhoff’s law

${{I}_{L}}{{R}_{L}}-60+4000\Omega I=0$

$10-60+4000I=0$

$4000I=50$

$I=\dfrac{50}{4000}$

$I=0.0125$

$I=12.5\,mA$

Now, from Kirchhoff’s current law

$I={{I}_{Z}}+{{I}_{L}}$

${{I}_{Z}}=12.5-5$

${{I}_{Z}}=7.5\,mA$

Hence, the current is $12.5\ mA$ , $5\ mA$ and $7.5\ mA$

In a pure silicon

$()$ crystal at

$300 K, 10^{21}$ atoms of phosphorus are added per cubic meter. The new hole concentration will be:-

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$10^{21} \text{per} \ m^3$
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$10^{19} \text{per} \ m^3$
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$10^{11} \text{per} \ m^3$
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$10^{5} \text{per} \ m^3$

The following combination of gates is equivalent to :

In an unbiased PN-junction,

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The junction current at equilibrium is zero as charges do not cross the junction
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The junction current reduces with rise in temperature
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The junction current at equilibrium is zero as equal but opposite carriers are crossing the junction
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The junction current is due to minority carriers only

For a purely resistive circuit, which of the following statements is incorrect?

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rms current will vary if the frequency of the source is varied
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Power factor is equal to $1$
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Current and voltage are in phase
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The average Power dissipated through the resistor is directly proportional to ${i}_{rms}^{2}$

Which is not true:

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Resistance of metals increases with increasing temperature
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Resistance of semi-conductors decreases with rise of temperature
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Resistance of an electrolyte increases with rise of temperature
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None of the above

The combination of the gates shown will produce

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AND gate
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NAND gate
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NOR gate
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XOR gate

In a P-type semiconductor, the acceptor level is

$57meV$ , above the valence band. The maximum wavelength of light required to produce a hole will be

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$57 \mathrm { A } ^ { o }$
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$57 \times 10 ^ { - 3 } A ^ { o }$
0%
$2.18\times10 ^ { -5 }A^o$
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$11.61 \times 10 \mathrm { A } ^ { \circ }$

In the following circuits, PN-junction diodes Dl, D2 and D3 are ideal for the following potential of AB, the correct increasing order of resistance between A and B will be -

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$( i ) < ( i i ) < ( i i i )$
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$\text { (iii) } < \text { (ii) } < \text { (i) }$
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$\text { (ii) } = \text { (iii) } < \text { (i) }$
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$\text { (i) } = \text { (iii) } < \text { (ii) }$

Find current through the ideal diode.

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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 output Y of the combination of gates shown is equal to:

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

Explanation

Here,

$Y=A.(A+B)$

$Y=A.A+A.B$

$Y=A+A.B$

$Y=A.(1+B)$

$Y=A.1$

$Y=A$

Option

$\textbf A$ is the correct answer

In a silicon diode, the reverse current increases from

$10\mu A$ to

$20\mu A$ . when the reverse voltage change from

$2V$ to

$4V$ . The reverse ac resistance of the diode is

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$2\times 10^{5}\ \Omega$
0%
$1\times 10^{5}\ \Omega$
0%
$3\times 10^{5}\ \Omega$
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$4\times 10^{5}\ \Omega$

Explanation

change in current

$= 20\mu A-10\mu A = 10\mu A$

change in voltage

$= hv-2v = 2v$

reverse ac resistance

$R = \frac{V}{2}$

$= \dfrac{2}{10\mu A} = 2\times 10^{5}\Omega$

1214997_1142987_ans_da93997ab837413d912f0decb29f5aab.JPG

How many NOR gates are required to form NAND gate:

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$1$
0%
$3$
0%
$2$
0%
$4$

A two inputed XOR gate produces an high output only when its both inputs are:

0%
same
0%
different
0%
low
0%
high

When all the inputs of a NAND gate are connected together, the resulting circuit is:

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a NOT gate
0%
an AND gate
0%
an OR gate
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a NOR gate

Explanation

The equation of NAND gate is

$Y=\bar{A}.\bar{B}$

When both the terminals are connected together.

$A=B=A$

$Y=\bar{A}.\bar{A} =\bar{A}. \bar{A}=\bar{A}$

Which is nothing but the equation of NOT gate.

$Y=\bar{A}$

Option

$\textbf A$ is the correct answer

Select the correct statement from the following

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a diode can be used as a rectifier
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a diode cannot be used as a rectifier
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the current in a diode is always proportional to the applied voltage
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None of these

In a zener diode, break down occurs in reverse bias due to

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Impact ionisation
0%
Internal field emission
0%
High doping concentration
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All of these

The probability of electrons to be found in the conduction band of an intrinsic semiconductor at a finite temperature :-

0%
decreases exponentially with increasing band gap
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increases exponentially with increasing band gap
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decreases with increasing temperature
0%
is independent of the temperature and the band gap

Explanation

Solution We know,

$P = \frac{1}{1+e(\frac{E-E_{F}}{KT})}, K\Rightarrow$ Boltzmann's conduction

$P \rightarrow$ Probability of finding to

election in conduction band

P is clearly decreasing

exponentially with increasing band

gap i.e.

$(E-E_{F})$

Option - A is correct.

1156165_1190032_ans_1b63f68c077e4d40a1cf29d83041768a.jpg

A.P.N junction Diode when connected to an AC sourice behaves as an ?

0%
Rectifier
0%
Zero resistance
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Amplifier
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None of these

If electromagnetic rays are incident on a semiconductors its conductivity .

0%
Increases
0%
Increases with increases frequency
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Increases when frequency increases above certain frequency
0%
Increases when wavelength is increases above a certain wavelength

In the circuit given the current through the Zener diode is :- -

0%
$10 \mathrm { mA }$
0%
$6.67 \mathrm { mA }$
0%
$5 \mathrm { mA }$
0%
$3.33 \mathrm { mA }$

In an intrinsic semiconductor, the density of conduction electrons is

$7.07\times 10^{15}m^{-3}$ . When it is doped with indium, the density of holes becomes

$5\times 10^{22}m^{-3}$ . Find the density of conduction electrons in doped semiconductor

0%
$Zero$
0%
$1\times 10^9m^{-3}$
0%
$7\times 10^{15}m^{-3}$
0%
$5\times 10^{22}m^{-3}$

When a diode is heavily doped the

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Zener voltage will be low
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Avalanche voltage will be high
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Depletion region will be thin
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Leakage current will be low

In the adjacent circuit, A and B represent two inputs and C represents the output. The circuit represents.

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

A light emitting diode (LED) has a voltage drop of 2 V across it and passes a current of 10 mA. When it operates with a 6 V battery through a limiting resistor R, the value of R is

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200 $\Omega$
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400 $\Omega$
0%
40 $k\Omega$
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4 $k\Omega$

The mobility of hole in a semiconductor depend on

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Electric field
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Potential difference
0%
current
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Mass

If potential

$V=100\pm0.5$ Volt and current

$I=10\pm0.2$ amp are given to us, then what will be the value of resistance

0%
$10\pm0.7$ ohm
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$5\pm2$ ohm
0%
$0.1\pm0.2$ ohm
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none of these

A zener diode, having breakdown voltage equal to 15 V is used in a voltage regulator circuit shown in figure. The current through the diode is

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20 mA
0%
5 mA
0%
10 mA
0%
15 mA

For detecting intensity of light we use

0%
Photodiode in forward bias
0%
Photodiode in reverse bias
0%
LED in forward bias
0%
LED in reverse bias

The material used in solar cells contains

0%
$C_s$
0%
$Si$
0%
$Sn$
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None of the above

Which of the following statements is true for a p-n-p transistor when used in an amplifier circuit ?

0%
Emitter is connected to negative terminal of a battery.
0%
Base current is always lower than emitter current.
0%
Collector is connected to positive terminal of a battery
0%
Collector current is always more than the emitter current

The number of particles crossing unit area perpendicular to X-axis in unit time is given by

$\dfrac { D\left( n_ 2-n_ 1 \right) }{ \left( x_ 2-x_ 1 \right) }$ where

$n_1 and n_2$ are number of particle per unit volume respectively at and

$x_2$ The dimension of the diffusion constant D are

0%
$\left[ LT^ { -1} \right]$
0%
$\left[ L^2 T^ { -1} \right]$
0%
[LT]
0%
$[L^{-1} T ]$

Explanation

$\begin{array}{l} \left[ n \right] ={ \left[ L \right] ^{ -2 } }\times { \left[ T \right] ^{ -1 } } \\ \left[ { { n_{ 2 } }-{ n_{ 1 } } } \right] ={ \left[ L \right] ^{ -3 } } \\ \left[ { { x_{ 2 } }-{ x_{ 1 } } } \right] =\left[ L \right] \\ \left[ n \right] =\left[ d \right] \times \left[ { { n_{ 2 } }-{ n_{ 1 } } } \right] /\left[ { { x_{ 2 } }-{ x_{ 1 } } } \right] \\ put\, \, the\, \, above\, \, values\, \, in:- \\ { \left[ L \right] ^{ -2 } }\times { \left[ T \right] ^{ -1 } }=\left[ d \right] \times { \left[ L \right] ^{ -3 } }/\left[ L \right] \\ \left[ d \right] ={ \left[ L \right] ^{ -2 } }\times { \left[ T \right] ^{ -1 } }\times \left[ L \right] /{ \left[ L \right] ^{ -3 } } \\ \left[ d \right] ={ \left[ L \right] ^{ 2+1+3 } }\times { \left[ T \right] ^{ -1 } } \\ \left[ d \right] ={ \left[ L \right] ^{ 2 } }\times { \left[ T \right] ^{ -1 } }=\left[ { { L^{ 2 } }{ T^{ -1 } } } \right] \end{array}$

Hence,

option

$(B)$ is correct answer.

The relation between dynamic plate resistance

$\left( r _ { p } \right)$ of a vacuum diode and plate current in the space charge limited region, is

0%
$r _ { p } \propto I _ { p }$
0%
$r _ { p } \propto I _ { p } ^ { 3 / 2 }$
0%
$r _ { p } \propto \dfrac { 1 } { I _ { p } }$
0%
$r _ { p } \propto \dfrac { 1 } { \left( I _ { p } \right) ^ { 1 / 3 } }$

Explanation

The dynamic plate resistance is

$r_p=\dfrac{\Delta V_p}{\Delta i_p}$

Now for a vacuum diode

$i_p=K_p^{-3/2} \Rightarrow =V_p= \left(\dfrac{i_p}{K} \right)^{2/3}$

$\Rightarrow \dfrac{\Delta V_p}{\Delta i_p} =\dfrac{2}{3 K^{2/3}} \times \dfrac{1}{i_p^{\left(\dfrac{2}{3}-1 \right)}}$

$\Rightarrow r_p=(constant) I_p^{-1/3} \Rightarrow r _ { p } \propto \dfrac { 1 } { \left( I _ { p } \right) ^ { 1 / 3 } }$

Which of the following break down of pn junction is reversible?

0%
Avalanche breakdown
0%
Zener breakdown
0%
Dielectric breakdown
0%
All of these

Explanation

If reverse voltage is increase contineously than at a certain voltage the electric field across the junction become so high that the covalent bond inside the position region start breaking

$.$ This process of breaking of covalent bond is known as break down voltage

$.$

These are two types of break down

$:$

$(1)$ Zener

$:-$ due to high electric field

$(2)$ Avalanche

$-$ due to successive collision of carries

$.$

Hence,

option

$(A)$ and

$(B)$ are both correct answer.

A gate in which all the inputs must be low to get a high output is called

0%
NAND gate
0%
An inverter
0%
NOR gate
0%
AND gate

The given figure shows the waveforms for three inputs

$A,B$ and

$C$ and that for the output

$Y$ of a logic circuit. The logic circuit is

0%
$XOR$ Gate
0%
$XNOR$ Gate
0%
$NAND$ Gate
0%
$NOR$ Gate

In PN-junction diode the reverse saturation current is

$10 ^ { - 5 }$ amp at

$27 ^ { \circ } C$ . The forward current for a voltage of 0.2 volt is

0%
$2037.6 \times 10 ^ { - 3 } \mathrm { amp }$
0%
$203.76 \times 10 ^ { - 3 } \mathrm { amp }$
0%
$20.376 \times 10 ^ { - 3 }amp$
0%
$2.0376 \times 10 ^ { 3 }amp$

Explanation

The forward current

$\large i=\;{i_s}\left( {{e^{ev/kT}} - 1} \right)$

$={10^{ - 5}}\left[ {{e^{\cfrac{{1.6 \times {{10}^{ - 19}} \times 0.2}}{{1.4 \times {{10}^{ - 23}} \times 300}} - 1}}} \right]$

$= {10^{ - 5}}\left[ {2038.6 - 1} \right] = 20.376 \times {10^{ - 3}}A$

Hence,

option

$(C)$ is correct answer.

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

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

CBSE Questions for Class 12 Engineering Physics Semiconductor Electronics: Materials,Devices And Simple Circuits Quiz 11 - 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 11 - MCQExams.com

0:0:2

Practice Class 12 Engineering Physics Quiz Questions and Answers

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