MCQ Questions for Class 12 Medical 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 correct relationship between the two current gains

$\alpha \quad and\quad \beta$ in a transistor is

0%
$\beta =\dfrac { \alpha }{ 1+\alpha }$
0%
$\alpha =\dfrac { \beta }{ 1-\beta }$
0%
$\alpha =\dfrac { \beta }{ 1+\beta }$
0%
$\alpha =\dfrac { 1+\beta }{ \beta }$

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

0%
phosphorus
0%
boron
0%
antimony
0%
arsenic

The part of a transistor which is heavily doped to produce a large number of majority carries is

0%
base
0%
emitter
0%
collector
0%
none of these

Explanation

The emitter is highly doped and used to produce large no. Of charge carriers.

In case of npn transistor it produces holes as majority charge carriers.

And in case of pnp it produces electrons as majority charge carriers.

Option

$\textbf B$ is the correct answer.

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

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

0%
drift in forward bias, diffusion in reverse bias
0%
diffusion in forward bias, drift in reverses bias
0%
diffusion in both forward and reverse bias
0%
drift in both forward and reverse bias

Which is the correct diagram of a half-wave rectifier?

In a common base amplifier circuit, calculate the change in base current if that in the emitter current is

$2$ mA and

$\alpha =0.98$ .

0%
$0.04$ mA
0%
$1.96$ mA
0%
$0.98$ mA
0%
$2$ mA

Explanation

We know emitter current

${I}_{E}={I}_{C}+{I}_{B}$ -----(1) where

${I}_{C}$ =connector current and

${I}_{B}$ = base current

Also we know

$\alpha=\dfrac{{I}_{C}}{{I}_{E}}$

${I}_{C}=\alpha{I}_{E}$ -------(A)

Putting the value of 1 in A we get,

${I}_{B}={I}_{E}-\alpha{I}_{E}$

${I}_{B}=2-1.96=0.04 mA$

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

0%
$\rho ' > \rho$
0%
$\rho ' < \rho$
0%
$\rho ' = \rho$
0%
there is no simple reaction between $\rho '\& \rho$

The current gain of a transistor in a common-emitter circuit isThe ratio of emitter current to base current is

0%
40
0%
41
0%
42
0%
43

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

0%
zero
0%
low
0%
high
0%
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 an unbiased PN-junction,

0%

The junction current at equilibrium is zero as charges do not cross the junction
0%
The junction current reduces with rise in temperature
0%
The junction current at equilibrium is zero as equal but opposite carriers are crossing the junction
0%
The junction current is due to minority carriers only

Consider an NPN transistor amplifier in common-emitter configuration. The current gain of the transistor is

$100$ . If the collector current changes by

$1$ mA, what will be the change in emitter current?

0%
$1.1$ mA
0%
$1.01$ mA
0%
$0.01$ mA
0%
$10$ mA

Explanation

We know that in

$NPN$ transistor,

Current gain

$\beta =\dfrac{\Delta i_{c}}{\Delta i_{b}}\Rightarrow \Delta i_{b}=\dfrac{1\times 10^{-3}}{100}=10^{-5}A=0.01\ mA.$

By using

$\Delta i_{e}=\Delta i_{b}+\Delta i_{c}\Rightarrow \Delta i_{e}=1.01+1=1.01\ mA.$

In NPN transistor the collector current is

$10$ mA. If

$90\%$ of electrons emitted reach the collector, then?

0%
Emitter current will be $9$ mA
0%
Emitter current will be $11.1$ mA
0%
Base current will be $0.1$ mA
0%
Base current will be $0.01$ mA

Explanation

90% emitted elctrons reach the collector,

Hence,

${\dfrac{I_c}{I_e}} = 0.9$

Since,

${ I }_{ c }=10$ mA,

Therefore

${ \quad I }_{ e }=10/0.9=11.11$ mA

Hence, the emitter current will be 11 mA.

A crystal diode is a

0%
Non-linear device
0%
Amplifying device
0%
Linear device
0%
fluctuating device

The value of form factor in case of half wave rectifier is:

0%
$1.11$
0%
$1.57$
0%
$1.27$
0%
$0.48$

Explanation

Form factor of half wave rectifier is $=\dfrac{{{I}_{rms}}}{{{I}_{av}}}.....(1)$

RMS value of current is

${{I}_{rms}}=\sqrt{\dfrac{1}{2\pi }\int\limits_{0}^{\pi }{{{i}^{2}}d\theta }}$

$=\sqrt{\dfrac{1}{2\pi }\int\limits_{0}^{\pi }{I_{\max }^{\pi }{{\sin }^{2}}\theta d\theta }}$

$=\dfrac{{{I}_{\max }}}{\sqrt{2\pi }}\sqrt{\dfrac{1}{2\pi }\int\limits_{0}^{\pi }{{{\sin }^{2}}\theta d\theta }}$

$=\dfrac{{{I}_{\max }}}{\sqrt{4\pi }}\sqrt{\left[ \theta -\dfrac{1}{2}\sin 2\theta \right]}_{0}^{\pi }$

$=\dfrac{{{I}_{\max }}}{\sqrt{4\pi }}\sqrt{\pi }$

$=\dfrac{{{I}_{\max }}}{2}$

Average value of current,

${{I}_{av}}=\dfrac{1}{2\pi }\int\limits_{0}^{\pi }{id\theta }$

$=\dfrac{1}{2\pi }\int\limits_{0}^{\pi }{{{I}_{\max }}\sin \theta d\theta }$

$=\dfrac{{{I}_{\max }}}{\pi }$

So, put above values in equations (1) and (2)

Form factor,

$=\dfrac{\dfrac{{{I}_{\max }}}{2}}{\dfrac{{{I}_{\max }}}{\pi }}$

$=\dfrac{\pi }{2}$

$=1.57$

The following combination of gates is equivalent to :

Select the correct statement from the following

0%
a diode can be used as a rectifier
0%
a diode cannot be used as a rectifier
0%
the current in a diode is always proportional to the applied voltage
0%
None of these

The combination of the gates shown will produce

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

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 -

0%
$( i ) < ( i i ) < ( i i i )$
0%
$\text { (iii) } < \text { (ii) } < \text { (i) }$
0%
$\text { (ii) } = \text { (iii) } < \text { (i) }$
0%
$\text { (i) } = \text { (iii) } < \text { (ii) }$

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

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

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:

0%
$1$
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:

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

A transistor has an

$\alpha = 0.95$ . It has change in emitter current of

$100$ milliampere. Then the change in collector current is:-

0%
$100$ milliampere
0%
$100.95$ milliampere
0%
$99.05$ milliampere
0%
$95$ milliampere

In He-Ne laser, metastable state exists in:

0%
He
0%
Ne
0%
Both (1) & (2)
0%
Neither He nor Ne

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%
$5\times 10^{22}m^{-3}$

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

0%
Rectifier
0%
Zero resistance
0%
Amplifier
0%
None of these

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

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

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 the following common emitter circuit if

$\beta=100,V_{CE}=7V,V_{BE}=Negligible,R_C=2k\Omega,$ then

$I_B=?$

0%
$0.01mA$
0%
$0.04mA$
0%
$0.02mA$
0%
$0.03mA$

Explanation

We know that,

$V=V_{CE}+I_cR_L$

$\Rightarrow 15=7+1\times 2\times 10' \Rightarrow i=4mA$

$\therefore \beta=\dfrac{i_C}{i_B} \Rightarrow i_B=\dfrac{4}{100}=0.04mA$

In a resistor circuit shown here the base current is

$35mA.$ The value of the resistor

$R_b$ is

0%
$300KW$
0%
$450KW$
0%
$257KW$
0%
$500KW$

Explanation

$\begin{array}{l} { V_{ b } }={ i_{ b } }{ R_{ b } } \\ \Rightarrow { R_{ b } }=\frac { 9 }{ { 35\times { { 10 }^{ -6 } } } } =257\, KW \end{array}$

Hence,

option

$(C)$ is correct answer.

Transistor with

$CE$ configuration can be used as

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

Zener diode is used as-

0%
Half wave rectifier
0%
Full wave rectifier
0%
A.C. voltage stabllizer
0%
D.C. voltage stabilizer

When a diode is heavily doped the

0%
Zener voltage will be low
0%
Avalanche voltage will be high
0%
Depletion region will be thin
0%
Leakage current will be low

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

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

The mobility of hole in a semiconductor depend on

0%
Electric field
0%
Potential difference
0%
current
0%
Mass

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

0%
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

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.

In the circuit shown in figure, the current gain

$\beta$ = 100 for a npn transistor. The bias resistance

$R_B$ so that

$V_{CE}$ = 5V is (

$V_{BE}$ < < 10 V)

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

Explanation

$\begin{array}{l} \beta =100 \\ { V_{ CE } }=5V \\ \beta =\dfrac { { { I_{ C } } } }{ { { I_{ B } } } } ={ I_{ B } }=\dfrac { { { 5_{ m } } } }{ { 100 } } \\ { I_{ B } }=0.05mA \\ 10-{ I_{ o } }RB-{ V_{ BE } }=0......\left( 1 \right) \\ 10-1K\times { I_{ C } }-{ V_{ CE } }0........\left( 2 \right) \\ { I_{ C } }=\dfrac { 5 }{ { 1K } } =5mA \\ from\, \, equation \\ if\, { V_{ BE } }<<10\approx 0\approx { V_{ BE } } \\ Hence, \\ { R_{ B } }=\dfrac { { 10 } }{ { { I_{ B } } } } =2\times { 10^{ 5 } }\Omega \end{array}$

Therefore, option

$C$ is correct.

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

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

A voltage amplifier operated from a

$12$ volt battery has a collector load

$6\ k\ \Omega$ . Calculate the maximum collector current in the circuit.

0%
$0.5\ mA$
0%
$1\ mA$
0%
$3\ mA$
0%
$2\ mA$

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 emitter base junction of a transistor is_____biased while the collector base junction is ____biased.

0%
Reverse,forward
0%
Reverse,reverse
0%
Forward,forward
0%
Forward,reverse

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.

In a transistor the base is very lightly doped as compared to the emitter because by doing so

0%
The flow across the base region is mainly because of holes.
0%
The flow across the base region is mainly because of electrons.
0%
Recombination is decreased in the base region.
0%
Base current is high.

A common emitter amplifier circuit is shown in the figure below.For the transistor used in the circuit the current amplification factor,

${\beta _{dc}} = 100.other\;parameters\;are\;mentioned\;in\;the\;figure.$
we find that:-

0%
${v_{BE}} = + 18.2\;V,\;{V_{BC}} = - 3.45\;V\;and\;amplifier\;is\;working.$
0%
${v_{BE}} = + 18.2\;V,\;{V_{BC}} = - 2.85\;V\;and\;amplifier\;is\;working.$
0%
${v_{BE}} = + 18.2\;V,\;{V_{BC}} = - 3.75\;V\;and\;amplifier\;is\;working.$
0%
none

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

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.