MCQ 1011 Mark
The phase difference between input and output voltages of a CE circuit is
- A$ 0$
- B$90$
- ✓$180$
- D$ 270$
Answer
View full question & answer→Correct option: C.
$180$
Questions · Page 3 of 7
MCQ
MCQ 1021 Mark
At ordinary temperatures, the electrical conductivity of semi conductors in mho/meter is in the range
- A$10^{-3}$ to $10^{-4}$
- ✓$10^6$ to $10^9$
- C$10^{-6}$ to $10^{-10}$
- D$10^{-10}$ to $10^{-16}$
Answer
View full question & answer→Correct option: B.
$10^6$ to $10^9$
MCQ 1031 Mark
In extrinsic $P$ and $N$-type, semiconductor materials, the ratio of the impurity atoms to the pure semiconductor atoms is about
- A$1$
- B$10^1$
- C$10^4$
- ✓$10^7$
Answer
View full question & answer→Correct option: D.
$10^7$
MCQ 1041 Mark
When the $P$ end of $P-N$ junction is connected to the negative terminal of the battery and the $N$ end to the positive terminal of the battery, then the $P-N$ junction behaves like
- AA conductor
- ✓An insulator
- CA super-conductor
- DA semi-conductor
Answer
View full question & answer→Correct option: B.
An insulator
In this condition $\mathrm{P}-\mathrm{N}$ junction is reverse biased.
MCQ 1051 Mark
The approximate ratio of resistances in the forward and reverse bias of the $P N$-junction diode is
- A$10^2: 1$
- B$10^{-2}: 1$
- C$1: 10^{-4}$
- ✓$1: 10^4$
Answer
View full question & answer→Correct option: D.
$1: 10^4$
Resistance in forward biasing $\mathrm{R}_{\mathrm{fr}} \approx 10 \Omega$ and resistance in reverse biasing
$\mathrm{R}_{\mathrm{Rw}} \approx 10^5 \Omega \Rightarrow \frac{\mathrm{R}_{\mathrm{fr}}}{\mathrm{R}_{\mathrm{Rw}}}=\frac{1}{10^4}$
$\mathrm{R}_{\mathrm{Rw}} \approx 10^5 \Omega \Rightarrow \frac{\mathrm{R}_{\mathrm{fr}}}{\mathrm{R}_{\mathrm{Rw}}}=\frac{1}{10^4}$
MCQ 1061 Mark
For germanium crystal, the forbidden energy gap in joules is
- ✓$1.12 \times 10^{-19}$
- B$1.76 \times 10^{-19}$
- C$1.6 \times 10^{-19}$
- DZero
Answer
View full question & answer→Correct option: A.
$1.12 \times 10^{-19}$
For Ge, $E_g=0.7 \mathrm{eV}=0.7 \times 1.6 \times 10^{-19} \mathrm{~J}=1.12 \times 10^{-19} \mathrm{~J}$
MCQ 1071 Mark
To obtain electrons as majority charge carriers in a semiconductor, the impurity mixed is
- AMonovalent
- BDivalent
- CTrivalent
- ✓Pentavalent
Answer
View full question & answer→Correct option: D.
Pentavalent
MCQ 1081 Mark
A semiconductor is cooled from $T_1 K$ to $T_2 K$. Its resistance
- AWill decrease
- ✓Will increase
- CWill first decrease and then increase
- DWill not change
Answer
View full question & answer→Correct option: B.
Will increase
Will increase
MCQ 1091 Mark
Electronic configuration of germanium is $2,8,18$ and $4$ . To make it extrinsic semiconductor small quantity of antimony is added
- AThe material obtained will be $N-$type germanium in which electrons and holes are equal in number
- BThe material obtained will be $P-$type germanium
- ✓The material obtained will be $N-$type germanium which has more electrons than holes at room temperature
- DThe material obtained will be $N-$type germanium which has less electrons than holes at room temperature
Answer
View full question & answer→Correct option: C.
The material obtained will be $N-$type germanium which has more electrons than holes at room temperature
The material obtained will be $N-$type germanium which has more electrons than holes at room temperature
MCQ 1101 Mark
At zero Kelvin a piece of germanium
- ABecomes semiconductor
- BBecomes good conductor
- ✓Becomes bad conductor
- DHas maximum conductivity
Answer
View full question & answer→Correct option: C.
Becomes bad conductor
At zero Kelvin, there is no thermal agitation and therefore no electrons from valence band are able to shift to conduction band.
MCQ 1111 Mark
Different voltages are applied across a $P-N$ junction and the currents are measured for each value. Which of the following graphs is obtained between voltage and current
- A
- B
- ✓
- D
Answer
View full question & answer→Correct option: C.
$P N$ junction has low resistance in one direction of potential difference $+V$, so a large current flows (forward biasing). It has a high resistance in the opposite potential difference direction $-V$, so a very small current flows (Reverse biasing).
MCQ 1121 Mark
A hole in a $P$-type semiconductor is
- AAn excess electron
- ✓A missing electron
- CA missing atom
- DA donor level
Answer
View full question & answer→Correct option: B.
A missing electron
MCQ 1131 Mark
A semiconductor 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 almost to zero. The device may be
- AA $P$-type semiconductor
- BAn N-type semiconductor
- ✓A $P N$-junction
- DAn intrinsic semiconductor
Answer
View full question & answer→Correct option: C.
A $P N$-junction
When polarity of the battery is reversed, the $P-N$ junction becomes reverse biased so no current flows.
MCQ 1141 Mark
Let $n_P$ and $n_e$ be the number of holes and conduction electrons respectively in a semiconductor. Then
- A(a) $n_{P} ! n_e$ in an intrinsic semiconductor
- B(b) $n_P \quad n_e$ in an extrinsic semiconductor
- ✓(c) $n_P \quad n_e$ in an intrinsic semiconductor
- D(d) $n_{e} ! n_P$ in an intrinsic semiconductor
Answer
View full question & answer→Correct option: C.
(c) $n_P \quad n_e$ in an intrinsic semiconductor
(c) In intrinsic semiconductors, the creation or liberation of one free electron by the thermal energy has created one hole. Thus in intrinsic semiconductors $n=n$.
MCQ 1151 Mark
The energy band gap of $S i$ is
- A$0.70\ eV$
- ✓$1.1\ eV$
- CBetween $0.70\ eV$ to $1.1\ eV$
- D$5\ eV$
Answer
View full question & answer→Correct option: B.
$1.1\ eV$
MCQ 1161 Mark
In a $P N$-junction diode
- ✓The current in the reverse biased condition is generally very small
- BThe current in the reverse biased condition is small but the forward biased current is independent of the bias voltage
- CThe reverse biased current is strongly dependent on the applied bias voltage
- DThe forward biased current is very small in comparison to reverse biased current
Answer
View full question & answer→Correct option: A.
The current in the reverse biased condition is generally very small
In forward biased $P N$-junction, external voltage decreases the potential barrier, so current is maximum. While in reversed biased $P N$-junction, external voltage increases the potential barrier, so the current is very small.
MCQ 1171 Mark
In a triode valve
- AIf the grid voltage is zero then plate current will be zero
- BIf the temperature of filament is doubled, then the thermionic current will also be doubled
- ✓If the temperature of filament is doubled, then the thermionic current will nearly be four times
- DAt a definite grid voltage the plate current varies with plate voltage according to Ohm's law
Answer
View full question & answer→Correct option: C.
If the temperature of filament is doubled, then the thermionic current will nearly be four times
MCQ 1181 Mark
Which is the wrong statement in following sentences? A device in which $P$ and $N$-type semiconductors are used is more useful then a vacuum type because
- APower is not necessary to heat the filament
- BIt is more stable
- CVery less heat is produced in it
- ✓Its efficiency is high due to a high voltage across the junction
Answer
View full question & answer→Correct option: D.
Its efficiency is high due to a high voltage across the junction
MCQ 1191 Mark
With a change of load resistance of a triode, used as an amplifier, from $50$ kilo ohms to $100$ kilo ohms, its voltage amplification changes from $25$ to $30.$ Plate resistance of the triode is
- ✓$25 k \Omega$
- B$75 k \Omega$
- C$7.5 k \Omega$
- D$2.5 k \Omega$
Answer
View full question & answer→Correct option: A.
$25 k \Omega$
$ \text { Voltage amplification } A_v=\frac{\mu}{1+\frac{r_p}{R_L}}$
$ \Rightarrow 25=\frac{\mu}{1+\frac{r_p}{50 \times 10^3}}$
and $30=\frac{\mu}{1+\frac{r_p}{100 \times 10^3}}$an solving equation (i) and (ii), $r_p=25 \mathrm{k} \Omega$.
$ \Rightarrow 25=\frac{\mu}{1+\frac{r_p}{50 \times 10^3}}$
and $30=\frac{\mu}{1+\frac{r_p}{100 \times 10^3}}$an solving equation (i) and (ii), $r_p=25 \mathrm{k} \Omega$.
MCQ 1201 Mark
In a triode, $g_m=2 \times 10^{-3} ohm ^{-1} ; \mu=42$, resistance load, $R=50$ kilo ohm. The voltage amplification obtained from this triode will be
- A$30.42$
- ✓$29.57$
- C$28.18$
- D$ 27.15$
Answer
View full question & answer→Correct option: B.
$29.57$
$ \text { Voltage gain } A_v=\frac{\mu}{1+\frac{r_p}{R_L}} \text { and } \mu=r_p \times g_m$
$ \Rightarrow r_p=\frac{42}{2 \times 10^{-3}}=21000 \Omega$
$ \Rightarrow A_v=\frac{42}{1+\frac{21000}{50 \times 10^3}}=29.57$
$ \Rightarrow r_p=\frac{42}{2 \times 10^{-3}}=21000 \Omega$
$ \Rightarrow A_v=\frac{42}{1+\frac{21000}{50 \times 10^3}}=29.57$
MCQ 1211 Mark
The grid voltage of any triode valve is changed from $-1$ volt to $-3$ volt and the mutual conductance is $3 \times 10^{-4} mho$. The change in plate circuit current will be
- A$0.8 mA$
- ✓$0.6 mA$
- C$0.4 mA$
- D$1 mA$
Answer
View full question & answer→Correct option: B.
$0.6 mA$
$ \text { By using } g_m=\frac{\Delta i_p}{\Delta v_g} \Rightarrow 3 \times 10^{-4}=\frac{\Delta i_p}{-1-(-3)} $
$ \Rightarrow \Delta i_p=6 \times 10^{-4} \mathrm{~A}=0.6 \mathrm{~mA}$
$ \Rightarrow \Delta i_p=6 \times 10^{-4} \mathrm{~A}=0.6 \mathrm{~mA}$
MCQ 1221 Mark
The most commonly used material for making transistor is
- ACopper
- ✓Silicon
- CEbonite
- DSilver
Answer
View full question & answer→Correct option: B.
Silicon
MCQ 1231 Mark
In an experiment, the saturation in the plate current in a diode is observed at $240 V$. But a student still wants to increase the plate current. It can be done, if
- AThe plate voltage is increased further
- BThe plate voltage is decreased
- CThe filament current is decreased
- ✓The filament current is increased
Answer
View full question & answer→Correct option: D.
The filament current is increased
After saturation plate current can be increased by increasing the temperature of filament. It can be done by increasing the filament current.
MCQ 1241 Mark
The valence of the impurity atom that is to be added to germanium crystal so as to make it a $N$-type semiconductor, is
- A$ 6$
- ✓$ 5$
- C$4$
- D$3$
Answer
View full question & answer→Correct option: B.
$ 5$
MCQ 1251 Mark
The amplification factor of a triode is $20.$ Its plate resistance is $10$ kilo ohms. Mutual conductance is
- A$2 \times 10^5 mho$
- B$2 \times 10^4 mho$
- C$500\ mho$
- ✓$2 \times 10^{-3} mho$
Answer
View full question & answer→Correct option: D.
$2 \times 10^{-3} mho$
Using $\mu=r_p \times g_m \Rightarrow g_m=\frac{20}{10 \times 10^3}=2 \times 10^{-3}$.
MCQ 1261 Mark
In a $P$-type semi-conductor, germanium is dopped with
- AGallium
- BBoron
- CAluminium
- ✓All of these
Answer
View full question & answer→Correct option: D.
All of these
Gallium, boron and aluminum are trivalent.
MCQ 1271 Mark
The process of adding impurities to the pure semiconductor is called
- ADrouping
- BDrooping
- ✓Doping
- DNone of these
Answer
View full question & answer→Correct option: C.
Doping
MCQ 1281 Mark
Semiconductor is damaged by the strong current due to
- ALack of free electron
- ✓Excess of electrons
- CExcess of proton
- DNone of these
Answer
View full question & answer→Correct option: B.
Excess of electrons
When a strong current passes through the semiconductor it heats up the crystal and covalent bond are broken. Hence because of excess number of free electrons it behaves like a conductor.
MCQ 1291 Mark
In NPN transistor the collector current is $10\ mA$. If $90 \%$ of electrons emitted reach the collector, then
- AEmitter current will be $9 mA$
- ✓Emitter current will be $11.1 mA$
- CBase current will be $0.1 mA$
- DBase current will be $0.01 mA$
Answer
View full question & answer→Correct option: B.
Emitter current will be $11.1 mA$
MCQ 1301 Mark
In an $N P N$ transistor the collector current is $24 mA$. If $80 \%$ of electrons reach collector its base current in $m A$ is
- A$36$
- B$26$
- C$16$
- ✓$6$
Answer
View full question & answer→Correct option: D.
$6$
Given $i_c=\frac{80}{100} \times i_e \Rightarrow 24=\frac{80}{100} \times i_e \Rightarrow i_e=30 \mathrm{~mA}$
By using $i_e=i_b+i_c \Rightarrow i=30-24=6 \mathrm{~mA}$.
By using $i_e=i_b+i_c \Rightarrow i=30-24=6 \mathrm{~mA}$.
MCQ 1311 Mark
At room temperature, a $P$-type semiconductor has
- ✓Large number of holes and few electrons
- BLarge number of free electrons and few holes
- CEqual number of free electrons and holes
- DNo electrons or holes
Answer
View full question & answer→Correct option: A.
Large number of holes and few electrons
In $P$-type semi conductor, holes are majority charge carriers.
MCQ 1321 Mark
In the given figure, which of the diodes are forward biased ?
- A(a) 1,2,3
- ✓(b) 2, 4, 5
- C(c) $1,3,4$
- D(d) $2,3,4$
Answer
View full question & answer→Correct option: B.
(b) 2, 4, 5
(b) In figure 2,4 and 5. P-crystals are more positive as compared to $N$-crystals.
MCQ 1331 Mark
The emitter-base junction of a transistor is biased while the collector-base junction is biased
- AReverse, forward
- BReverse, reverse
- CForward, forward
- ✓Forward, reverse
Answer
View full question & answer→Correct option: D.
Forward, reverse
The emitter base junction is forward biased while collector base junction is reversed biased.
MCQ 1341 Mark
In a forward biased $P N$-ju nction diode, the potential barrier in the depletion region is of the form ...
- A
- ✓
- C
- D
Answer
View full question & answer→Correct option: B.
Potential across the $P N$ junction varies symmetrically linear, having $P$ side negative and $N$ side positive.
MCQ 1351 Mark
The electrical circuit used to get smooth $d c$ output from a rectifier circuit is called
- AOscillator
- ✓Filter
- CAmplifier
- DLogic gates
Answer
View full question & answer→Correct option: B.
Filter
Filter circuits are used to get smooth $d c \pi$-filter is the best filter.
MCQ 1361 Mark
If $l_1, l_2, l_3$ are the lengths of the emitter, base and collector of a transistor then
- A(a) $l_1=l_2=l_3$
- B(b) $l_3
l_1$ - C(c) $l_3
- ✓(d) $l_3>l_1>l_2$
Answer
View full question & answer→Correct option: D.
(d) $l_3>l_1>l_2$
(d)
MCQ 1371 Mark
In an NPN transistor circuit, the collector current is $10 mA$. If $90 \%$ of the electrons emitted reach the collector, the emitter current ( $i$ ) and base current $(i)$ are given by
- A$i=-1 mA , i=9 mA$
- B$i=9 m A, i=-1 m A$
- C$i=1 mA , i=11 mA$
- ✓$i=11 mA , i=1 mA$
Answer
View full question & answer→Correct option: D.
$i=11 mA , i=1 mA$
$i_C=\frac{90}{100} \times i_E \Rightarrow 10=0.9 \times i_c=11 \mathrm{~mA}$
Also $i_E=i_B+i_C \Rightarrow i_B=11-10=1 \mathrm{~mA}$.
Also $i_E=i_B+i_C \Rightarrow i_B=11-10=1 \mathrm{~mA}$.
MCQ 1381 Mark
The electrical circuits used to get smooth d.c. output from a rectifier circuit is called
- AFilter
- ✓Amplifier
- CFull wave rectifier
- DOscillator
Answer
View full question & answer→Correct option: B.
Amplifier
MCQ 1391 Mark
Least doped region in a transistor
- AEither emitter or collector
- ✓Base
- CEmitter
- DCollector
Answer
View full question & answer→Correct option: B.
Base
In transistor, base is least doped.
MCQ 1401 Mark
In the depletion region of an unbiased $P-N$ junction diode there are
- AOnly electrons
- BOnly holes
- CBoth electrons and holes
- ✓Only fixed ions
Answer
View full question & answer→Correct option: D.
Only fixed ions
MCQ 1411 Mark
The impurity atom added to germanium to make it $N$-type semiconductor is
- ✓Arsenic
- BIridium
- CAluminium
- Dlodine
Answer
View full question & answer→Correct option: A.
Arsenic
For $N$-type semiconductor, the impurity should be pentavalent.
MCQ 1421 Mark
In a good conductor the energy gap between the conduction band and the valence band is
- AInfinite
- BWide
- CNarrow
- ✓Zero
Answer
View full question & answer→Correct option: D.
Zero
The conduction and valence bands in the conductors merge into each other.
MCQ 1431 Mark
When a semiconductor is heated, its resistance
- ✓Decreases
- BIncreases
- CRemains unchanged
- DNothing is definite
Answer
View full question & answer→Correct option: A.
Decreases
Decreases
MCQ 1441 Mark
In a transistor in $CE$ configuration, the ratio of power gain to voltage gain is
- A$\alpha$
- B$\beta / \alpha$
- C$\beta \alpha$
- ✓$\beta$
Answer
View full question & answer→Correct option: D.
$\beta$
For CE configuration voltage gain $=\beta \times R_L / R_i$
Power gain $=\beta^2 \times R_L / R_i $
$\Rightarrow \frac{\text { Power gain }}{\text { Voltage gain }}=\beta$
Power gain $=\beta^2 \times R_L / R_i $
$\Rightarrow \frac{\text { Power gain }}{\text { Voltage gain }}=\beta$
MCQ 1451 Mark
$N P N$ transistor are preferred to $P N P$ transistor because they have
- ALow cost
- BLow dissipation energy
- CCapability of handing large power
- ✓Electrons having high mobility than holes
Answer
View full question & answer→Correct option: D.
Electrons having high mobility than holes
MCQ 1461 Mark
The maximum efficiency of full wave rectifier is
- A$100 \%$
- B$25.20 \%$
- C$40.2 \%$
- ✓$81.2 \%$
Answer
View full question & answer→Correct option: D.
$81.2 \%$
$\text { For full wave rectifier } \eta=\frac{81.2}{1+\frac{r_f}{R_L}} $
$ \Rightarrow n_{\max }=81.2 \% \quad(r \ll R)$
$ \Rightarrow n_{\max }=81.2 \% \quad(r \ll R)$
MCQ 1471 Mark
In intrinsic semiconductor at room temperature, number of electrons and holes are
- AUnequal
- ✓Equal
- CInfinite
- DZero
Answer
View full question & answer→Correct option: B.
Equal
In intrinsic semiconductors, at room temperature $n=n$.
MCQ 1481 Mark
In a $P N$-junction diode not connected to any circuit
- AThe potential is the same everywhere
- BThe $P$-type is a higher potential than the $N$-type side
- ✓There is an electric field at the junction directed from the $N$ type side to the $P$-type side
- DThere is an electric field at the junction directed from the $P$ type side to the $N$-type side
Answer
View full question & answer→Correct option: C.
There is an electric field at the junction directed from the $N$ type side to the $P$-type side
At junction a potential barrier/depletion layer is formed, with $N$-side at higher potential and $P$-side at lower potential.
Therefore there is an electric field at the junction directed from the $N$-side to $P$-side
Therefore there is an electric field at the junction directed from the $N$-side to $P$-side
MCQ 1491 Mark
Which of the following statements is not true
- AThe resistance of intrinsic semiconductors decrease with increase of temperature
- BDoping pure $S i$ with trivalent impurities give $P$-type semiconductors
- ✓The majority carriers in $N$-type semiconductors are holes
- DA $P N$-junction can act as a semiconductor diode
Answer
View full question & answer→Correct option: C.
The majority carriers in $N$-type semiconductors are holes
In $N$-type semiconductor majority charge carriers are electrons.
MCQ 1501 Mark
The dominant mechanisms for motion of charge carriers in forward and reverse biased silicon $P-N$ junctions are
- ADrift in forward bias, diffusion in reverse bias
- ✓Diffusion in forward bias, drift in reverse bias
- CDiffusion in both forward and reverse bias
- DDrift in both forward and reverse bias
Answer
View full question & answer→Correct option: B.
Diffusion in forward bias, drift in reverse bias
In forward biasing the diffusion current increases and drift current remains constant so not current is due to the diffusion.In reverse biasing diffusion becomes more difficult so net current (very small) is due to the drift.