Topic Topics Name No of Page

No MCQs No

152 03
1 Semiconductor Devices and Applications: Introduction to P-N junction
Diode and V-I characteristics, Half wave and Full-wave rectifiers,
capacitor filter.

2 Zener diode and its characteristics, Zener diode as voltage regulator 20 17

3 Introduction to BJT, it’s input-output and transfer characteristics, BJT 37 19

as a single stage CE amplifier, Frequency response and bandwidth

4 Regulated power supply 34 28

5 Field effect transistor (JFET, MOSFET, CMOS) 40 32

6 A/D & D/A Converter 18 37

7 Microprocessor 50 39

MECHTEST GPSC 1 @mechtest1

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MECHTEST GPSC 2 @mechtest1

 1.7. In a rectifier, larger the value of shunt capacitor filter
(A) Larger the peak-to-peak value of ripple voltage
(B) Larger the peak current in the rectifying diode
(C) Longer the time that current pulse flows through the
diode
(D) Smaller the dc voltage across the load
[EE-GWSSB {03/202122}]
1.1. In intrinsic semiconductor, the Fermi level exists 1.8. The basic reason why a full wave rectifier has a twice
(A) just below the conduction band the efficiency of a half wave rectifier is that
(B) just above the valance band (A) it makes use of transformer
(C) centre of the forbidden energy band (B) its ripple factor is much less
(D) deep in the conduction band (C) it utilizes both half-cycle of the input
[AE-GMB {20/202122}] (D) its output frequency is double the line frequency
1.2. The potential which exists in a PN junction to cause [EE-GWSSB {03/202122}]
drift of charge carriers is called 1.9. Drift current in semiconductors depends upon
(A) Contact potential (B) diffusion potential (A) Only the electric field
(C) ionisation potential (D) threshold potential (B) Only the carrier concentration gradient
[DEE-GWSSB {42/202122}] (C) Both the electric field and the carrier concentration
1.3. Under low level injection assumption, the injected (D) Both the electric field and the carrier concentration
minority carrier current for an extrinsic semiconductor is gradient
essentially the [EE-GWSSB {03/202122}]
(A) Diffusion current (B) Drift current 1.10. In a P-type Si sample the hole concentration is 2.25
(C) Recombination current (D) Induced current × 1015/cm3. The intrinsic carrier concentration is 1.5 ×
[EE-GWSSB {03/202122}] [GATE’06] 1010/cm3 the electron concentration is
1.4. The impurity commonly used for realizing the base (A) Zero (B) 1010/cm3
region of a silicon n-p-n transistor is (C) 105/cm3 (D) 1.5 × 1025/cm3
(A) Gallium (B) Indium [EE-GWSSB {03/202122}] [ESE’14]
(C) Boron (D) Phosphorus 1.11. Crystal diode has _______
[EE-GWSSB {03/202122}] (A) one P-N junction (B) two P-N junctions
1.5. Special types of diodes in which transition time and (C) three P-N junctions (D) none of the above
storage time are made small are called [AE-GWSSB {02/202122}]
(A) Snap diodes (B) Rectifier diodes 1.12. The junction is said to be biased in the forward
(C) Storage diodes (D) Memory diodes direction when the positive battery terminal is
connected to _______
[DEE-GWSSB {42/202122}]
(A) P-type region (B) N-type region
1.6. The figure shows a half-wave rectifier. The diode D is
(C) both P&N type region (D) none of the above
ideal. The average steady-state current (in Amperes)
[AE-GWSSB {02/202122}]
through the diode is approximately ___
1.13. A crystal diode is used as _______
(A) an amplifier (B) a rectifier
(C) an oscillator (D) a voltage regulator
[AE-GWSSB {02/202122}]
(A) 2 A (B) 1 A (C) 0.1A (D) A 1.14. If the temperature of a crystal diode increases, then
[EE-GWSSB {03/202122}] [GATE’14] leakage current _______.
(A) remains the same (B) decreases

MECHTEST GPSC 3 @mechtest1

 (C) increases (D) becomes zero (C) extraction and, and subsequent diffusion and generation
[AE-GWSSB {02/202122}] of minority carriers
1.15. As the load current increases, the zener current ___ (D) extraction, and subsequent drift and recombination of
(A) decreases (B) remains same minority carriers.
(C) increases (D) none of the above [GATE’13]
[AE-GWSSB {02/202122}] SOLUTION:
1. 16. The disadvantage of a half-wave rectifier is that the Potential barrier of the pn junction is lowered when a
(A) components are expensive forward bias voltage is applied, allowing electrons and
(B) diodes must have a higher power rating holes to flow across the space charge region (Injection)
(C) output is difficult to filter when holes flow from the p region across the space
(D) none of the above charge region into the n region, they become excess
[AE-GWSSB {02/202122}] minority carrier holes and are subject to diffuse, drift and
1.17. The ripple factor of a half-wave rectifier is _______. recombination processes.
(A) 2 (B) 1.21 (C) 2.5 (D) 0.48 i. holes will be injected from p-side.
[AE-GWSSB {02/202122}] [ISRO’11] [PGCIL’20] ii. holes will be diffused in n-side.
SOLUTION: iii. Recombination will take place.
Ripple
‘Ripple’ is the unwanted AC component remaining when
converting the AC voltage waveform into a DC
waveform. Even though we try our best to remove all AC
components, there is still some small amount left on the
output side which pulsates the DC waveform. This
1.19. Reverse saturation current in a Silicon PN junction
undesirable AC component is called a ‘ripple’.
diode nearly doubles for every
Ripple factor
(A) 2° rise in temperature (B) 5° rise in temperature
The ripple factor is the ratio between the RMS value of
(C) 6° rise in temperature (D) 10° rise in temperature
the AC voltage (on the input side) and the DC voltage (on
the output side) of the rectifier. [EE-GWSSB {03/202122}]
1.20. Leakage current doubles for every
(A) 5°C rise in temperature (B) 10°C rise in temperature
(C) 15°C rise in temperature (D) 20°C rise in temperature
[AE-GMB {20/202122}]
SOLUTION:
CIRCUIT DIAGRAM Ripple factor
Half-Wave Rectifier 1.21
Center-Tap Full Wave
0.48
Rectifier
Bridge-Type Full Wave
0.48
Rectifier
.

1.18. In a forward-biased PN junction diode, the sequence

of events that best describes the mechanism of current
flow is 1.21. For every 10°C increase in temperature, the reverse

(A) injection, and subsequent diffusion and recombination saturation current of a p-n junction will be increased by

of minority carriers (A) 10 times (B) 2 times

(B) injection and subsequent drift and generation of (C) 4 times (D) remains same

minority carriers [ISRO’18]

MECHTEST GPSC 4 @mechtest1

 SOLUTION: [EE-GWSSB {03/202122}] [GATE’90]
1.27. A thin P-type silicon sample is uniformly illuminated
with light which generates excess carriers. The
recombination rate is directly proportional to
(A) The minority carrier mobility
(B) The minority carrier recombination lifetime
(C) The majority carrier concentration
(D) The excess minority carrier concentration
[EE-GWSSB {03/202122}] [GATE’14]
1.22. Which of the following metal is used in making SOLUTION:
electrical resistance wire for electric furnaces and Recombination rate, R = B(nno + nn) (Pno + Pn)
heating elements? nno and Pno = Electron and hole concentrations
(A) Babbit metal (B) Monel metal respectively under thermal equilibrium .
(C) Nichrome (D) Phosphor bronze nn and pn =Excess elements and hole concentrations
[AE-GMB {20/202122}] respectively.
1.23. The relation between mobility (μ), drift velocity (v)
1.28. An ideal rectifier should have
and electric field intensity (ε) is
(A) Efficiency = 100%; Vac = 0; Transformer Utilization Factor
(A) μ = vε (B) ε = μv (C) v = με (D) μvε = 1
[AE-GMB {20/202122}] = 1 and Total harmonic distortion = 0

1.24. N-type semiconductor is (B) Efficiency = 90%; Vac = 0.1; Transformer Utilization Factor
(A) negatively charged (B) positively charged = 0.9 and Total harmonic distortion = 0
(C) no charge at all (D) electrically neutral (C) Efficiency = 90%; Vac = 0.1; Transformer Utilization Factor
[AE-GMB {20/202122}] = 0.9 and Total harmonic distortion = 5%
1.25. A PN junction with a 100 Ω resistor is forward biased (D) Efficiency = 100%; Vac = 0.2; Transformer Utilization
so that a current of 100 mA flows. If the voltage across Factor = 0.8 and Total harmonic distortion = 5%
this combination is instantaneously reversed to 10 volts [UPRVUNL AE’16]
at t = 0, the reverse current that flows through the diode SOLUTION:
at t = 0 is approximately given by An ideal rectifier should have:
(A) 0 mA (B) 100 mA (C) 200 mA (D) 50 mA  Efficiency = 100%
[EE-GWSSB {03/202122}] [ISRO’11]  AC components (Vac) = 0
SOLUTION:  Transformer Utilization Factor = 1
 Power Factor = 1

1.29. For ideal Rectifier and filter circuits, % regulations

must be
(A) 1% (B) 0.1% (C) 5% (D) 0%
[DEE-GWSSB {42/202122}]
1.30. An ideal rectifier should have transformer
utilization factor (TUF) of 1. If the actual TUF is 3.5, it
1.26. In a uniformly doped abrupt PN junction, the
shows that
doping level of the N-side is four times the doping level
(A) Diode is under-loaded
of the P-side. Then the ratio of the depletion layer width
(B) The transformer must be 3.5 times larger
is
(C) The transformer should be only 1/3.5 times the ideal size
(A) Wn/Wp = 1/8 (B) Wn/Wp = 1/6
(D) The ripple factor is low
(C) Wn/Wp = 1/4 (D) Wn/Wp = 1/2
[ISRO’17]

MECHTEST GPSC 5 @mechtest1

 SOLUTION: SOLUTION:
The concentration of minority carriers in an extrinsic
semiconductor is given by the law of mass action,
according to which:
n  p = n i2
n = concentration of electrons in the conduction band
p = concentration of holes in the valence band
ni = intrinsic carrier concentration
Cases:
In an n-type semiconductor, the minority hole
concentration is given by:
1.31. The value of current that flows through RL in a ‘π’
section filter circuit at no load is
ND = Concentration of Donor impurity
(A) ∞ (B) 0.1 mA
In a p-type semiconductor, the minority electron
(C) 0 (D) few mA
concentration is given by:
[DEE-GWSSB {42/202122}]
1.32. Which performance parameter of a regulator is
defined as the change in regulated load voltage due to NA = Concentration of Acceptor impurity
variation in line voltage in a specified range at a constant Observations:
load current? Thus the minority carriers concentration is:
1. inversely proportional to the doping concentration
(A) Load regulation (B) Line regulation
2. directly proportional to the square of intrinsic
(C) Temperature stability factor (D) Ripple rejection
concentration and not to the intrinsic concentration.
[DEE-GWSSB {42/202122}]
1.33. Voltage divider bias or universal bias circuit is also 1.36. Select the CORRECT statement(s) regarding semi-
known as conductor devices.
(A) self-bias circuit (B) collector bias circuit 1. Electrons and holes are of equal density in an intrinsic
(C) collector to base circuit (D) Fixed bias circuit semiconductor at equilibrium.
[DEE-GWSSB {42/202122}] 2. Collector region is generally more heavily doped than

1.34. Silicon PN junction at a temperature of 200° C has a Base region in a BJT.

reverse saturation current of 10 pA. The reverse 3. Total current is spatially constant in a two terminal

saturation current at 400° C for the same bias is electronic device in dark under steady state

approximately condition.

(A) 20 pA (B) 30 pA 4. Mobility of electrons always increases with

(C) 40 pA (D) 80 pA temperature in Silicon beyond 300 K.

[EE-GWSSB {03/202122}] (A) 1 & 2 (B) 2 & 3 (C) 3 & 4 (D) 1 & 3
[GATE’22]
1.35. The concentration of minority carriers in an extrinsic
SOLUTION:
semiconductor under equilibrium is
Statement (1): An intrinsic semiconductor is an undoped
(A) Directly proportional to the intrinsic concentration
semiconductor. For an intrinsic semiconductor,
(B) Inversely proportional to the intrinsic concentration
the concentration of electrons in the conduction band
(C) Directly proportional to the doping concentration
is equal to the concentration of holes in the valence band
(D) Inversely proportional to the doping concentration
at thermal equilibrium.
[ESE’14] [GATE’06] • Hence, st (1) is correct.

MECHTEST GPSC 6 @mechtest1

 Statement (2): Collector region is generally lightly
1.39. A bar of Gallium Arsenide (GaAs) is doped with
doped than base region in BJT
Silicon such that the Silicon atoms occupy Gallium and
• Hence, st (2) is incorrect.
Arsenic sites in the GaAs crystal. Which one of the
Statement (3): The total current due to injected
minority carrier (diffusion current) and majority carrier following statements is true?
(drift current) is spatially constant in two terminal (A) Silicon atoms act as p-type dopants in Arsenic sites and
electronic device, however they individually will be n-type dopants in Gallium sites
spatially different under dark and steady state (B) Silicon atoms act as n-type dopants in Arsenic sites and
condition. p-type dopants in Gallium sites
(C) Silicon atoms act as p-type dopants in Arsenic as well as
Gallium sites
(D) Silicon atoms act as n-type dopants in Arsenic as well as
Gallium sites
[GATE’17]
1.40. A region of negative differential resistance is
observed in the current voltage characteristics of a
• Hence, st (3) is correct. silicon PN junction if
Statement (4): Mobility of electrons always decreases (A) Both the P-region and the N-region are heavily doped
with temperature in Silicon beyond 300 K. (B) The N-region is heavily doped compared to the P-region
(C) The P-region is heavily doped compared to the N-region
(D) An intrinsic silicon region is inserted between the P-
region and the N-region
[GATE’15]
SOLUTION:
A region of negative differential resistance is observed in
the current voltage characteristics of a silicon PN
• Hence, st (4) is incorrect.
junction if both the P-region and the N-region are heavily
1.37. A diode which is formed by using lightly doped doped. Under normal forward bias operation as voltage
GaAs or silicon with metal is called begins to increase, electrons at first tunnel through the
(A) Zener diode (B) Schottky diode very narrow p-n junction barrier and fill electron states in
(C) Varactor diode (D) tunnel diode the conduction band on the n-side, which becomes
[DEE-GWSSB {42/202122}] aligned with empty valence band hole states on the p-
1.38. The bonding forces in compound semiconductors, side of the p-n junction. As voltage increases further,
such as GaAs, arise from these states become increasingly misaligned and the
(A) ionic bonding current drops. This is called negative resistance becomes
(B) metallic bonding current decreases with increasing voltage. p-n diode
(C) covalent bonding with high doping on both sides is called tunnel diode.
(D) combination of ionic and covalent bonding Hence, the correct option is (A).
[ESE-2002]
1.41. The primary reason for the widespread use of Silicon
SOLUTION:
in semiconductor device technology is
In semiconductors like GaAs the elements Ga and As are
(A) abundance of Silicon on the surface of the Earth.
in covalent bond. Arsenic has five valence electrons while
gallium has three valence electrons. Therefore, in GaAs (B) larger bandgap of Silicon in comparison to Germanium.
crystal Arsenic contribute one electron to Gallium and it (C) favorable properties of Silicon - dioxide (SiO2)
is shared between both atoms in a covalent bond. (D) lower melting point

MECHTEST GPSC 7 @mechtest1

 [GATE’05] SOLUTION:
1.42. The Miller effect in the context of a Common Peak inverse voltage is the maximum voltage that
Emitter amplifier explains appears across a diode when it is under reverse bias.
(A) an increase in the low-frequency cutoff frequency
(B) an increase in the high-frequency cutoff frequency
(C) a decrease in the low-frequency cutoff frequency
(D) a decrease in the high-frequency cutoff frequency During the positive half cycle at Vin the capacitor (C) is
charged to Vin (as Diode is forward bias and hence
[GATE’17]
considered as a short circuit).
SOLUTION:
During the negative half cycle of Vin, the circuit will be –
The miller effect causes increase in the Input
(Diode will be reverse bias and act as open circuit).
capacitance increase in the Input capacitance of
Common – Emitter Amplifier
Upper cut-off frequency,

Therefore, there is decrease in upper-cut off frequency Therefore, by applying KVL, the negative voltage across
& Bandwidth of CE Amplifier. the diode is
Vd = Vin + Vin = 2Vin
1.43. Consider avalanche breakdown in a silicon p + n ⇒ PIV = 2Vm
junction. The n-region is uniformly doped with a donor
1.45. A semiconductor differs from a conductor in that it
density ND. Assume that breakdown occurs when the
has
magnitude of the electric field at any point in the device
(A) only one path for the free electrons in the valence band.
becomes equal to the critical field Ecrit. Assume Ecrit to be
(B) only one path for holes in the conductance band.
independent of ND. If the built-in voltage of the p + n
(C) two paths followed by free electrons and holes, one an
junction is much smaller than the breakdown voltage,
ordinary path in the conduction band and the other one
VBR, the relationship between VBR and ND is given by
an extraordinary path in the valence band, respectively.
(A) VBR × √𝑁𝐷 = constant (B) 𝑁𝐷 × √𝑉𝐵𝑅 = constant (D) two paths followed by free electrons and holes, one an

(C) 𝑁𝐷 × VBR = constant (D)

𝑁𝐷
= constant extraordinary path in the conduction band and the other
𝑉𝐵𝑅
one an ordinary path in valence band, respectively
[GATE’16] [ESE’16]
SOLUTION: SOLUTION:
 Conductors, semiconductors and insulators can
be distinguished with respect to their conduction
band and valence band
 For a material to conduct electricity the electrons
from the valence band must shift to the conduction
band
 In the case of conductors like metals, the valence
band and conduction band overlap each other and
1.44. If Vm is the peak value of an applied voltage in a half
almost all the charge carriers are found in the
wave rectifier with a large capacitor across the load, then
conduction band; This is the reason they are good
the peak inverse voltage will be
conductors of electricity
(A) 0.5 Vm (B) Vm
 In the case of semiconductors, there is a gap
(C) 1.5 Vm (D) 2.0 Vm
between the valence band and the conduction
[ESE’16] band and only a small quantity of charge carriers are

MECHTEST GPSC 8 @mechtest1

 found in the conduction band; Hence, they are not (D) inversely proportional to intrinsic concentration.
very good conductors of electricity [ESE’14]
 In the case of insulators, the gap between the SOLUTION:
valence band and the conduction band is very
large, and no charge carriers are found in the
conduction band; Thus, they do not conduct
electricity

1.49. For an n-type semiconductor having any doping

level, which of the following hold(s) good:
A conductor has only one path for the free electrons in 1. pn ND = 𝒏𝟐𝒊 2. PP ND = 𝒏𝟐𝒊
the valence band whereas a semiconductor has two 3. nn ND = 𝒏𝟐𝒊 4. pn nn = 𝒏𝟐𝒊
paths followed by free electrons and holes, one Select the correct answer using the code given below:
an ordinary path in the conduction band and the other (A) 1 and 4 (B) 2 and 4
one an extraordinary path in the valence band, (C) 3 and 4 (D) Only 4
respectively. [ESE’06]
1.46. What is the type of breakdown that occurs in a 1.50. The doping material for an N-type semiconductor is
Zener diode having breakdown voltage (6 V)? (A) Pentavalent (B) Tetravalent
(A) Avalanche breakdown only (C) Trivalent (D) Bivalent
(B) Zener breakdown only [UPSSSC JE’15]
(C) Avalanche breakdown where breakdown voltage is SOLUTION:
below 6 V and Zener breakdown otherwise 1. N-Type Semiconductor: This type of semiconductor is
(D) Zener breakdown where breakdown voltage is below 6 formed by added pentavalent impurity
V and Avalanche breakdown otherwise 2. P-Type Semiconductor: This type of semiconductor is
[ESE-2016] formed by added Trivalent impurity.

1.47. Consider the following statements: The intrinsic 1.51. Match List-I (Parameter) with List-II (Variation) and
carrier concentration of a semiconductor: select the correct answer using the code given below the
1. Depends on doping lists:
2. Increases exponentially with decrease of band gap of List-I
the semiconductor A. Electron mobility around room temperature
3. Increase non-linearity with increase of temperature B. Energy gap
4. Increase linearly with increase of temperature C. Intrinsic carrier concentration
Which of the above statements are correct? D. Mole density (gm/mole)
(A) 1, 2 and 3 (B) 1 and 2 only List-II
(C) 2 and 3 only (D) 2 and 4 only 1. Increases with temperature
[ESE’14] 2. Decreases with temperature
1.48. The concentration of minority carriers in an extrinsic 3. Remains constant as temperature is varied
semiconductor under equilibrium is (A) A-2, B-1, C-1, D-1 (B) A-1, B-2, C-1, D-3
(A) directly proportional to doping concentration (C) A-2, B-2, C-1, D-3 (D) A-2, B-2, C-1, D-1
(B) directly proportional to intrinsic concentration [ESE’06]
(C) inversely proportional to doping concentration.

MECHTEST GPSC 9 @mechtest1

 1.52. The concentration of minority carriers in an extrinsic
semiconductor under equilibrium is
(A) directly proportional to the doping concentration.
(B) inversely proportional to the doping concentration.
(C) directly proportional to the intrinsic concentration.
(D) inversely proportional to the intrinsic concentration.
[ESE’12] • P-type and N-type materials are NOT positively
1.53. If n is the number of electrons per unit volume of and negatively charged.
the semiconductor and vd is the drift velocity of the o because atoms are electrically neutral.
electrons, then the current flowing through a • N-type materials have electrons or negative charges
semiconductor is given by that are majority carriers and holes are minority
(A) i = n/vd (B) i = n vd carriers.
• So it has mainly electrons or negative charge carriers
(C) i = vd /n (D) i = n vd1/2
that can move freely, but it is still neutral because
[ESE’12]
the fixed donor atoms, having donated electrons,
1.54. A p-type semiconductor carries net:
are positive.
(A) Positive charge (B) Negative charge • Thus we can say that the n-type of semiconductor
(C) Zero charge (D) Depends upon biasing is neutral.
Hence option 3 is correct.
SOLUTION:
• Semiconductor: A solid substance that has a 1.55. In a semiconductor, holes exist in:
conductivity between an insulator and metals. (A) the conduction band
Semiconductors have conductivity due to many (B) the forbidden energy band
factors like the addition of an impurity or because (C) the valence band
of temperature effects. (D) both (1) and (2)
• N-type semiconductors: An extrinsic semicondu-
ctor where the dopant atoms provide extra 1.56. Semiconductors have ______ conduction band and
conduction electrons to the host material like ______ valence band.
Phosphorus P in Silicon Si. (A) A lightly filled; a moderately filled
This creates an excess of negative (n-type) electron (B) an almost filled; a moderately filled
charge carriers that are able to move freely.
(C) an almost empty; an almost filled
(D) an almost filled; an almost empty

1.57. Temperature coefficient of resistance in a pure

semiconductor is ______.
(A) zero
(B) positive
• P-type semiconductors: A semiconductor, when (C) negative
the impurity with trivalency is added to (D) dependent on size of specimen
pure semiconductors, then it is known as a p-type
semiconductor. 1.58. A p-type semiconductor is _______.
o Impurities with trivalency such as Boron (B), (A) positively charged
Gallium (Ga), Indium(In), Aluminium(Al), etc are (B) negatively charged
called acceptor impurity. (C) electrically neutral
(D) not used in semiconductor devices

MECHTEST GPSC 10 @mechtest1

 1.59. Match the following: C. poor D. reverse
a P-N Junction i
1.63. If the temperature of a crystal diode increases, then
diode
leakage current ______
b Zener diode ii
A. remains the same B. decreases
C. increases D. becomes zero
c Schottky diode iii

1.64. The d.c. resistance of a crystal diode is ______ its

d Tunnel diode iv
a.c. resistance
A. the same as B. more than
(A) a-iii, b-iv, c-ii, d-i (B) a-iii, b-ii, c-i, d-iv C. less than D. none of the above
(C) a-i, b-ii, c-iii, d-iv (D) a-ii, b-iii, c-iv, d-i
1.65. For the same secondary voltage, the output voltage
SOLUTION: from a centre-tap rectifier is ______ than that of bridge
P-N Zener Schottky Tunnel rectifier
Junction Diode Diode Diode A. twice B. thrice
Diode C. four time D. one-half
Allows Allows Allows Allows
current current current current 1.66. The disadvantage of a half-wave rectifier is that
flow only flow in flow only flow in the______
in one both in one both A. components are expensive
direction directions. direction direction B. diodes must have a higher power rating
Very Slow Low High Ultra-High C. output is difficult to filter
Switching Switching Switching Switching D. none of the above
Speed Speed Speed. Speed.
1.67. A crystal diode is used as ______
V-I V-I V-I V-I
A. an amplifier B. a rectifier
Characteri Characteri Characteri Characteri
C. an oscillator D. a voltage regulator
stics do stics do stics do stics
not showa not show not show shows
1.68. A zener diode is used as ______
negative a negative a negative negative
A. an amplifier B. a voltage regulator
resistance resistance resistance resistance
C. a rectifier D. a multivibrator
region. region. region. region
1.69. A half-wave rectifier has an input voltage of 240 V
r.m.s. If the step-down transformer has a turns ratio of
1.60. If the arrow of crystal diode symbol is positive w.r.t. 8:1, what is the peak load voltage? Ignore diode drop.
bar, then diode is _______ biased. A. 27.5 V B. 86.5 V C. 30 V D. 42.5 V
A. forward B. reverse
C. either forward or reverse D. none of the above 1.70. A zener diode has _______ breakdown voltage
A. undefined B. sharp
1.61. In the breakdown region, a zener didoe behaves C. zero D. none of the above
like a ______ source.
A. constant voltage B. constant current 1.71. A zener diode is always ______ connected.
C. constant resistance D. none of the above A. reverse B. forward
C. either reverse or forward D. none of the above
1.62. When the crystal current diode current is large,
the bias is _______. 1.72. A crystal diode is a _______ device
A. forward B. inverse A. non-linear B. bilateral C. linear D. none

MECHTEST GPSC 11 @mechtest1

 1.73. A crystal diode has OR 1.82. The most widely used rectifier is ______
A zener diode has ______ A. half-wave rectifier
A. one pn junction B. two pn junctions B. centre-tap full-wave rectifier
C. three pn junctions D. none of the above C. bridge full-wave rectifier
D. none of the above
1.74. A 10 V power supply would use ______ as filter
capacitor.
1.83. If the doping level in a crystal diode is increased,
A. paper capacitor B. mica capacitor
the width of depletion layer______
C. electrolytic capacitor D. air capacitor
A. remains the same B. is decreased
C. in increased D. none of the above
1.75. If the a.c. input to a half-wave rectifier is an r.m.s
𝟒𝟎𝟎
value of volts, then diode PIV rating is ______
𝟒𝟎𝟎
√𝟐 1.84. A series resistance is connected in the zener circuit
A. V B. 400 V to______
√𝟐
C. 400 √2 V D. None of the above A. properly reverse bias the zener
B. protect the zener
1.76. When a crystal diode is used as a rectifier, the most C. properly forward bias the zener
important consideration is ______ D. none of the above
A. forward characteristic B. doping level
C. reverse characteristic D. PIC rating
1.85. The forward voltage drop across a silicon diode is
about _______
1.77. The PIV rating of each diode in a bridge rectifier is
A. 2.5 V B. 3 V C. 10 V D. 0.7 V
______ that of the equivalent centre-tap rectifier
A. one-half B. the same as
C. twice D. four times 1.86. The ratio of reverse resistance and forward
resistance of a germanium crystal diode is about ______
1.78. When the graph between current through and A. 1 : 1 B. 100 : 1 C. 1000 : 1 D. 40000 : 1
voltage across a device is a straight line, the device is
referred to as ______ 1.87. The PIV rating of a crystal diode is ______ that of
A. linear B. active equivalent vacuum diode
C. nonlinear D. passive
A. the same as B. lower than
C. more than D. none of the above
1.79. The knee voltage of a crystal diode is
approximately equal to ______
1.88. If the doping level of a crystal diode is increased,
A. applied voltage B. breakdown voltage
the breakdown voltage_____
C. forward voltage D. barrier potential
A. remains the same B. is increased
C. is decreased D. none of the above
1.80. A crystal diode utilises _______ characteristic for
rectification
1.89. A crystal diode has forward resistance of the order
A. reverse B. forward
of _______
C. forward or reverse D. none of the above
A. kΩ B. Ω C. MΩ D. none

1.81. The maximum efficiency of a half-wave rectifier is

1.90. The reverse current in a diode is of the order of ___
____
A. kA B. mA C. μA D. A
A. 40.6% B. 81.2% C. 50% D. 25%

MECHTEST GPSC 12 @mechtest1

 1.91. Mains a.c. power is converrted into d.c. power for 1.100. If the PIV rating of a diode is exceeded, _______
_____ A. the diode conducts poorly
A. lighting purposes B. heaters B. the diode is destroyed
C. using in electronic equipment D. none of the above C. the diode behaves like a zener diode
D. none of the above
1.92. There is a need of transformer for ______
A. half-wave rectifier 1.101. A zener diode is destroyed if it________
B. centre-tap full-wave rectifier A. is forward biased
C. bridge full-wave rectifier B. is reverse biased
D. none of the above C. carrier more than rated current
D. none of the above
1.93. A zener diode is ___________ device
A. a non-linear B. a linear 1.102. The resistivity of a semiconductor ..............
C. an amplifying D. none of the above conductors and insulators.
A. is more than that of
1.94. The leakage current in a crystal diode is due to B. lies between that of
______ C. is less than that of
A. minority carriers B. majority carriers D. none of the above
C. junction capacitance D. none of the above
1.103. A semiconductor is formed by ............... bonds.
1.95. A zener diode utilizes _____ characteristics for its A. covalent B. electrovalent
operation. C. co-ordinate D. none of the above
A. forward B. reverse
C. both forward and reverse D. none of the above 1.104. The most commonly used semiconductor is
...............
1.96. ______ rectifier has the lowest forward resistance A. germanium B. carbon
A. solid state B. vacuum tube C. sulphur D. silicon
C. gas tube D. none of the above
1.105. In a semiconductor, the energy gap between
1.97. The doping level in a zener diode is ___________ valence band and conduction band is about ...............
that of a crystal diode A. 5 eV B. 10 eV
A. the same as B. less than C. 15 eV D. 1 eV
C. more than D. none of the above
1.106. A semiconductor has ............... temperature co-
1.98. The ___________ filter circuit results in the best efficient of resistance.
voltage regulation A. negative B. positive
A. choke input B. capacitor input C. zero D. none of the above
C. resistance input D. none of the above

1.107. A semiconductor generally has ............... valence

1.99. An ideal crystal diode is one which behaves as a electrons.
perfect _______ when forward biased. A. 2 B. 3 C. 4 D. 6
A. conductor B. insulator
C. resistance material D. none of the above

MECHTEST GPSC 13 @mechtest1

 1.108. When a pure semiconductor is heated, its 1.116. When the temperature of an extrinsic semi-
resistance ............... conductor increases, the pronounced effect is on ..........
A. goes down B. goes up A. majority carriers B. minority carriers
C. remains the same D. none of the above C. junction capacitance D. none of the above

1.109. The strength of a semiconductor crystal comes 1.117. With forward bias to a pn junction, the width of
from depletion layer ...............
A. forces between nuclei B. forces between protons A. decreases B. increases
C. electron-pair bonds D. none of the above C. remains the same D. none of the above

1.110. The impurity level in a semiconductor is about 1.118. The leakage current in a pn junction is of the order
.......... of pure semiconductor. of ...............
A. 10 atoms for 10 atoms
8
B. 1 atom for 10 atoms
8
A. A B. mA C. kA D. µA
C. 1 atom for 104 atoms D. 1 atom for 100 atoms
1.119. The difference between an insulator and a
1.111. As the doping increases, the bulk resistance of a semiconductor is
semiconductor ............... A. wider forbidden gap B. the number of free electrons
A. decreases B. remains the same C. the atomic structure D. all of the above
C. increases D. none of the above
1.120. Holes in n-type semiconductor are
1.112. The battery connections required to forward bias a A. minority carriers produced by thermal energy
pn junction are ............... B. minority carriers produced by doping
A. +ve terminal to p and –ve terminal to n C. majority carriers that are thermally produced
B. –ve terminal to p and +ve terminal to n D. majority carriers that are produced by doping
C. –ve terminal to p and –ve terminal to n
D. none of the above 1.121. The depletion region consists of
A. nothing but minority carriers
1.113. A reversed biased pn junction has ............... B. positive and negative ions
A. very narrow depletion layer C. no majority carriers
B. almost no current D. B. and C.
C. very low resistance
D. large current flow 1.122. The term bias means
A. ratio of majority to minority carriers
1.114. A pn junction acts as a ............... B. amount of current across pn junction
A. unidirectional switch B. bidirectional switch C. d.c. voltage applied to electronic device
C. controlled switch D. none of the above D. none of the above

1.115. The leakage current across a pn junction is due to 1.123. When a voltmeter is connected across a forward-
............... biased diode, it will read a voltage approximately equal to
A. majority carriers B. minority carriers A. bias battery voltage
C. junction capacitance D. none of the above B. 0V
C. diode barrier potential
D. none of the above

MECHTEST GPSC 14 @mechtest1

 1.124. When the positive lead of an ohmmeter is 1.131. The process of adding impurity to an intrinsic
connected to the anode of a diode and the negative lead semiconductor is called
to the cathode, the diode is A. recombination B. ionisation
A. reverse biased B. open C. atomic modification D. doping
C. forward biased D. none of the above
1.132. In an intrinsic semiconductor,
1.125. When voltage is applied to a semiconductor, holes A. there are no free electrons
will flow B. there are only holes
A. away from negative terminal C. free electrons are thermally produced
B. toward negative terminal D. cannot say
C. in the external circuit
D. none of the above 1.133. When a pn junction is forward biased, the current
in the external wires is
1.126. At room temperature, a conductor has how many A. zero
holes? B. by free electrons
A. many C. by holes
B. none D. by both free electrons and holes
C. only those produced by thermal energy
D. same number as free electrons 1.134. When a pn junction is reversed biased, the current
in the external wires is
1.127. In an intrinsic semiconductor, the number of free A. zero
electrons B. by holes
A. equals the number of holes C. by both free electrons and holes
B. is greater than the number of holes D. by free electrons
C. is less than the number of holes
D. none of the above 1.135. At room temperature, the charge carries in an
intrinsic semiconductor are
1.128. At absolute zero temperature (–273°C), an intrinsic A. free electrons
semiconductor has B. holes
A. a few free electrons B. many holes C. free electrons and holes
C. no holes or free electrons D. many free electrons D. holes and ions

1.129. The number of free electrons and holes in an 1.136. The ratio of number of holes and number of free
intrinsic semiconductor increases when the temperature electrons in an intrinsic semiconductor is
A. decreases B. increases A. 1 B. more than 1
C. stays same D. none of the above C. less than 1 D. none of the above

1.130. Suppose an intrinsic semiconductor has 1 billion 1.137. Doping of silicon with arsenic leads to
free electrons at room temperature (25ºC). If the A. conductor
temperature changes to 75ºC, how many holes are there? B. insulator
A. less than 1 billion B. more than 1 billion C. p-type semiconductor
C. 1 billion D. impossible to say D. n-type semiconductor

MECHTEST GPSC 15 @mechtest1

 1.138. A piece of germanium is cooled from room 1.146. The energy gap between conduction band and
temperature to 100 K. Its conductivity will valence band of a substance is of the order of 0.07 eV. It
A. increase is
B. decrease A. an insulator B. an alloy
C. remain unchanged C. a semiconductor D. a conductor
D. first increase and then decrease
1.147. A semiconductor is cooled from T1K to T2K. Its
1.139. If the temperature of a piece of germanium resistance
increases, its conductance A. will increase
A. increases B. decreases B. will decrease
C. remains unchanged D. becomes zero C. will remain unchanged
D. will first increase and then decrease
1.140. Which type of charge carrier has the greatest
mobility? 1.148. A full-wave rectifier is ............... as effective as a
A. positive ions B. negative ions halfwave rectifier.
C. free electrons D. holes A. twice B. thrice
C. two and half-times D. four times
1.141. A 0 K, an intrinsic semicondcutor behaves as a
A. superconductor B. semiconductor 1.149. The a.c. component in the output of a full-wave
C. perfect insulator D. perfect conductor rectifier is ............... the d.c. component.
A. equal to B. more than
1.142. The current is the external wire for p-type C. less than D. none of the above
semiconductor is by
A. free electrons B. holes 1.150. The PIV of each diode in a bridge circuit is ...............
C. positive ions D. negative ions that of the equivalent centre-tap circuit.
A. one-half B. the same as
1.143. Ohm’s law is not obeyed by C. twice D. four times
A. conductors B. semiconductors
C. both A and B D. none of the above 1.151. In a capacitor filter the rectifier circuit, ______ the
capacitance value, higher will be the power factor.
1.144. In a semiconductor crystal, if the current flows due A. Lower B. Higher
to breakage of crystal bonds, then the semiconductor is C. Unaffected D. Equal
called
A. acceptor 1.152. The function of the filter capacitor in a rectifier
B. donor circuit is to reduce
C. intrinsic semiconductor A. Frequency B. Ripple
D. extrinsic semiconductor C. Harmonics D. Amplitude

1.145. A p-type semiconductor is

A. uncharged B. positively charged
C. negatively charged D. none of the above

MECHTEST GPSC 16 @mechtest1

 • At reverse breakdown voltage (typically around 5V),
there is a sharp breakdown at a reverse voltage as
shown in the figure

2.1. In the circuit shown below, the Zener diode is ideal

and Zener voltage is 6V. The output voltage V0 (in volts)
is

(A) 6 V (B) 5 V (C) 10 V (D) 4 V

[EE-GWSSB {03/202122}] [GATE’15]
SOLUTION:

2.4. A zener diode is used as _______.

(A) an amplifier (B) a voltage regulator
(C) a rectifier (D) a multivibrator
[AE-GWSSB {02/202122}]
2.5. A zener diode is always _______ connected.
(A) reverse (B) forward
(C) either reverse or forward (D) none of the above
2.2. In a Zener diode, [AE-GWSSB {02/202122}]

(A) Only P-region is heavily doped 2.6. Switching regulators are series type regulators,
(B) Only N-region is heavily doped which has ______ power dissipation & ______ efficiency.
(C) Both P and N-regions are heavily doped (A) increased, increased (B) increased, reduced
(D) Both P and N-regions are lightly doped (C) reduced, increased (D) reduced, reduced
[EE-GWSSB {03/202122}] [GATE’89] [DEE-GWSSB {42/202122}] [ISRO’16]

2.3. A Zener diode, when used in voltage stabilization SOLUTION:

Voltage Regulator:
circuits, is biased in
1. It is an electronic circuit that gives a smooth output
(A) reverse bias region below the breakdown voltage
voltage. Smooth voltages are most important for the
(B) reverse breakdown region
working of digital circuits.
(C) forward bias region
2. An IC-based voltage regulator can be classified in
(D) forward bias constant current mode different ways.
[EE-GWSSB {03/202122}] [ISRO’18] 3. The 3 terminal IC voltage regulator block diagram is
SOLUTION: shown below:
• The forward voltage rating of Zener diode is the
same as a normal diode
• A zener diode is used as a voltage regulator in
reversed biased where voltage across it remains
constant as shown in the figure
• The slope (V/I) is positive

MECHTEST GPSC 17 @mechtest1

 2.10. Zener voltage regulator will cease to act as a
voltage regulator if Zener current becomes ______
(A) Less than load current (B) Zero
Switching regulator:
(C) More than load current (D) None of the above
• It is a voltage regulator that uses a switching element
to transform the incoming power supply into a pulsed
2.11. If the load resistance decreases in a Zener regulator,
voltage, which is then smoothed using capacitors,
then Zener current _______
inductors, and other elements.
(A) Decreases (B) Stays the same
• Power is supplied from the input to the output by
(C) Increases (D) None of the above
turning ON a switch (MOSFET) until the desired
voltage is reached. 2.12. A Zener regulator _____ in the power supply.
• Once the output voltage reaches the predetermined (A) Increases the ripple
value the switch element is turned OFF and no (B) Decreases the ripple
input power is consumed. (C) Neither increases nor decreases the ripple
• Repeating this operation at high speeds makes it (D) Data insufficient
possible to supply voltage efficiently and with less
heat generation. Hence series regulator has 2.13. Zener diode are generally made of ________
reduced power dissipation and increased (A) Germanium (B) Silicon
efficiency. (C) Carbon (D) None of the above
There are three types of Switching voltage regulators:
Step up, Step down and Inverter voltage regulators. 2.14. A 30 V Zener will have depletion layer width
______ that of 10 V Zener
2.7. A Zener voltage regulator is used for ______ load
(A) More than (B) Less than
currents
(C) Equal to (D) None of the above
(A) High (B) Very high
(C) Moderate (D) Small
2.15. Zener diode is used as the main component in dc
power supply for
2.8. A 10-V dc regulator power supply has a regulation of
(A) Filteration (B) Conversion from AC to DC
0.005 per cent. Its output voltage will vary within an
(C) Rectification (D) Voltage regulation
envelope of
[UPPSC’22]
(A) ± 5 mV (B) ± 0.05 mV (C) ± 0.5 mV (D) ± 2.5 mV
SOLUTION:
[EE-GWSSB {03/202122}]
Zener Diode:
2.9. The power dissipated by a transistor approximately
equals the collector current times
(A) Base emitter voltage (B) Collector emitter voltage
(C) Base supply voltage (D) 0.7V  The Zener diode is a PN junction diode that operates
[EE-GWSSB {03/202122}] in the reverse biased condition.
SOLUTION:  During the reverse biased condition, if the voltage
across the Zener diode is greater than the breakdown
voltage, then the Zener diode conducts.

MECHTEST GPSC 18 @mechtest1

  This breakdown voltage is known as the Zener
breakdown voltage.
The Zener diode is used as the main component in
the dc power supply for voltage regulation.

2.16. A Zener diode that has very narrow depletion layer

will breakdown by mechanism
(A) Avalanche (B) Zener
(C) Both avalanche and Zener (D) None of the above 3.1. A bipolar junction transistor is used as power control
switch by biasing it in the cut-off region (OFF state) or in
2.17. In a Zener voltage regulator, the changes in load the saturation region (ON state). In the ON state, for the
current produce changes in _____ . BJT
(A) Zener current (A) Both the base-emitter junction and base-collector
(B) Zener voltage junctions are reverse biased
(C) Zener voltage as well as Zener current (B) The base-emitter is reverse biased, and the base-
(D) None of the above collector junction is forward biased
(C) The base-emitter junction is forward biased, and the
2.18. Two similar 15 V Zeners are connected in series. base-collector junction is reverse biased
What is the regulated output voltage? (D) Both the base-emitter and base-collector junctions are
(A) 15 V (B) 5 V (C) 30 V (D) 45 V forward biased
[EE-GWSSB {03/202122}] [GATE’04]
2.19. A Zener diode _____
3.2. As temperature increases, the β value of a BJT
(A) Is a battery
(A) decreases
(B) Acts like a battery in the breakdown region
(B) increases
(C) Has a barrier potential of 1 V
(C) does not change
(D) Is forward biased
(D) can’t be determined
[AE-GMB {20/202122}]
2.20. Another name for Zener diode is _______ diode
(A) Breakdown (B) Voltage 3.3. The condition for saturation in a BJT is
𝐼 𝐼𝐶
(C) Power (D) Current (𝐴)𝐼𝐵 ≤ | 𝐶 | (B) 𝐼𝐵 ≤ | |
𝛼 𝛽
𝐼𝐶 𝐼𝐶
(C) 𝐼𝐵 ≥ | | (D) 𝐼𝐵 ≥ | |
𝛼 𝛽
[DEE-GWSSB {42/202122}]
SOLUTION:
For transistor to be in saturation
βIB ≥ ICsat

3.4. A Bipolar Junction Transistor saturation point may

be defined as
(A) The point where the collector current exceeds
maximum value
(B) Large base current is present
(C) Point above which the increase in base current does not
increase the collector current significantly
(D) None of the other options
[UPRVUNL AE’16]

MECHTEST GPSC 19 @mechtest1

 SOLUTION: (C) reverse active mode (D) forward saturation mode
For a BJT operating in active mode, the base current (I b) [GATE’96]
and collector current (IC) are related as Ic = β Ib where β is SOLUTION:
the current gain. If a transistor is operating with both junctions forward
At saturation the voltage across collector – emitter biased, then it is operating in saturation region. Now if
junction (VCE) is constant (∼ 0.2V) and collector current is the transistor operating in saturation mode has collector
independent of base current.
base forward bias greater than the emitter base forward
3.5. When the input is symmetrical, to operate the BJT in bias, then it is operating in reverse saturation mode.
active region, the quiescent point is chosen So, reverse saturation mode is correct.
(A) at the top edge of the load line
3.8. When a bipolar junction transistor is operating in the
(B) at the bottom edge of the load line
saturation mode, which one of the following statements
(C) at the centre of the load line
is TRUE about the state of its collector-base (CB) and the
(D) can be chosen anywhere on the load line
base-emitter (BE) junctions?
[DEE-GWSSB {42/202122}]
(A) The CB junction if forward biased and the BE junction is
3.6. What is the biasing condition of junctions in bipolar
reverse biased.
junction transistor to work as an amplifier?
(B) The CB junction is reversed and the BE junction is
(A) Reverse biased base to emitter junction and reverse
forward biased.
biased base to collector junction
(C) Both the CB and BE junctions are forward biased.
(B) Forward biased base to emitter junction and reverse
(D) Both the CB and BE junctions are reverse biased.
biased base to collector junction
[GATE’15]
(C) Forward biased base to emitter junction and forward
SOLUTION:
biased base to collector junction
In Bipolar transistor, saturation mode occurs when
(D) Reverse biased base to emitter junction and forward
1. Collector–base (CB) is forward biased.
biased base to collector junction
2. Base-emitter (BE) is also forward biased.
[ESE’14]
Hence, for the saturation condition, bath the junction
SOLUTION:
should be in forward-bias, as shown in the figure below.
BJT Amplifier:
 Transistors biasing is done to keep stable DC
operating conditions needed for its functioning as an
amplifier.
 A properly biased transistor must have its Q-point
(DC operating parameters like IC and VCE) at the
3.9. The Ebers-Moll model is applicable to
center of saturation mode and cut-off mode i.e. active
(A) Bipolar junction transistors
mode.
(B) MOS transistors
 In the active mode of transistor operation, the base-
(C) Unipolar Junction transistors
emitter junction is forward biased and the base-
(D) Junction field effect transistors
collector junction is reverse biased.
[EE-GWSSB {03/202122}]
3.7. If a transistor is operating with both of its Junctions 3.10. The Ebers-Moll model of a BJT is valid
forward biased, but with the collector base forward (A) only in active mode
biased, but with the collector base forward bias greater (B) only in active and saturation modes
than the emitter-base forward bias, then it is operating (C) only in active and cut-off modes
in the (D) in active, saturation and cut-off modes
(A) forward active mode (B) reverse saturation mode [GATE’16]

MECHTEST GPSC 20 @mechtest1

 SOLUTION: SOLUTION:
Ebers-Moll Model:
It is an ideal model for BJT, that can be used in
forward or reverse active mode of operation, as well
as in both saturation and cut-off mode also.
The Model contains two diodes and two currents source
shown below:

3.13. The breakdown voltage of a transistor with its base

open is BVCEO that with emitter open is BVCBO then
Analysis:
• The two diodes represent Base emitter and Base (A) BVCEO = BVCBO
collector diodes. (B) BVCEO> BVCBO
• The current sources quantify the transport of (C) BVCEO < BVCBO
minority carriers through the base region. These (D) BVCEO not related to BVCBO
current sources depend on the current through the [EE-GWSSB {03/202122}] [GATE’95]
diode. SOLUTION:
• IE,S, and IC,S are the saturation currents of the base- The given voltage ratings are reverse breakdown
emitter and base-collector diodes. voltages.
• αF & αR are forward and reverse transport factors
BVCEO : Voltage between the collector and emitter with
respectively.
base open
• By applying KCL at Node a,
BVCBO : Voltage from collector to base with emitter open
The mechanism involved for such breakdown is due to
The Ebers-Moll Parameters are related by: Avalanche.
Equation relating these breakdowns is BVCEO= BVCBO/(β)1/N
3.11. A BJT is said to be operating in the cutoff region, if This shows that voltage in open base configuration is
(A) Both the junctions are reverse biased smaller by (β)1/N
(B) Base emitter junction is in reverse biased, and base
collector junction is forward biased
(C) Base emitter junction is in forward biased, and base
collector junction is reverse biased
(D) Both the junctions are forward biased
[EE-GWSSB {03/202122}] [GATE’95]
3.12. The relation between ICEO, ICBO and α is ICEO =

(A)
ICBO
(B)
ICBO
3.14. h-parameters are valid over a ________ frequency
α 1+α
range
ICBO ICBO
(C) (D) (A) R.F. (B) For DC only
(1+α)2 1−α
(C) Audio frequency range (D) upto 1 MHz
[DEE-GWSSB {42/202122}]
[DEE-GWSSB {42/202122}]

MECHTEST GPSC 21 @mechtest1

 3.15. What are the benefits of h-parameters? Output Characteristics:
1. They are easy to measure A graph showing the variation of collector current
2. Convenient to use in circuit analysis and design (IC) with collector-emitter voltage (VCE) at a constant
3. They are real numbers at audio frequencies base current (IB) is called output characteristics, it is
given by
4. They vary with widely temperature.
(A) Only 1 (B) Only 2 and 3
(C) Only 1,2 & 3 (D) 1, 2, 3 & 4
[ESE’07] [UPRVUNL AE’14]
SOLUTION:
• The h- parameters are dependent on the
conductivity parameters of the transistor which
vary with intrinsic carrier concentration.
• Since intrinsic carrier concentration varies with
temperature the h-parameters vary with From above it is clear that the input resistance(ri) is
temperature. defined as the ratio of the small change in the base-
• Every linear circuit having input and output can be emitter voltage(ΔVBE) to the corresponding small
analyzed as two-port networks. In these networks, change in base current (ΔIB), it is given by
there are four parameters called hybrid or h-
parameter.
 Important Points
• Out of these four parameters, one is measured in
In Common emitter configuration (CE) emitter is the
ohm, one in mho, and the other two are
common terminal.
dimensionless.
• Since these parameters have mixed dimensions, so
they are called hybrid parameters.
hi = input resistance with output shorted (Ω)
hr = Reverse voltage gain with input open (V)
hf = Forward current gain with output shorted
ho = Output conductance with input open
Advantages of h-parameter:  In common emitter configuration, the emitter is
1. Easy to measure common to both base and collector.
2. Convenient to use in circuit analysis and design  Emitter base junction is always forward bias as
3. Real numbers at audio frequencies a positive terminal of the battery is connected to
4. Can be obtained from the transistor static the p-side of the transistor and the negative
characteristics curves terminal of the battery is connected to the n-side of
5. Most of the transistor manufacturers specify the h- the transistor.
parameters.  The behaviour of the common emitter transistor can
be explained on the basis of three different types of
3.16. In CB configuration, the output V-I characteristics of characteristics.
the transistor are drawn by taking  This provides good overall performance and as
(A) VCB vs. IC for constant IB (B) VCB vs. IE for constant IC such, it is often the most widely used
(C) VCB vs. IE for constant IB (D) VCB vs. IB for constant IE configuration.
[EE-GWSSB {03/202122}] [ISRO’16]
SOLUTION: Input Characteristics: A graph showing the variation of
In CE configuration, the output V-I characteristics are the base current (IB) with base-emitter voltage (VBE), at
drawn by taking VCE versus IC for the constant value of IB constant collector-emitter Voltage (VCE) is called input
characteristics of a transistor.

MECHTEST GPSC 22 @mechtest1

 SOLUTION:
Common collector transistor configurations have the
highest input impedance
The given circuit diagram is a common-collector
amplifier, i.e.

The input resistance(ri) is defined as the ratio of the

small change in the base-emitter voltage(ΔVBE) to the
3.20. The most widely used amplifier configuration is
corresponding small change in base current (ΔIB), it is
(A) Common base
given by
(B) Common emitter
(C) Common collector
3.17. Common base transistor acts as a (D) Combination of common base and common emitter
(A) current control voltage source [ISRO’18]
(B) voltage control voltage source 3.21. If the base width in a bipolar junction transistor is
(C) voltage control current source doubled, which one of the following statements will be
(D) current control current source TRUE?
[AE-GMB {20/202122}] (A) Current gain will increase
3.18. Which of the following transistor configuration (B) Unity gain frequency will increase
circuits is much less temperature dependent? (C) Emitter base junction capacitance will increase
(A) Common base (B) Common emitter (D) Early voltage will increase
(C) Common collector (D) None of the above [EE-GWSSB {03/202122}] [GATE’15]
[EE-GWSSB {03/202122}] SOLUTION:
SOLUTION: • A large collector base reverse bias decreases the
There are three basic transistor configurations. One is effective base of the width as the depletion region
common base configuration in common base penetrates more into the base region. This leads to
configuration collector current IC = αIE + ICO. Here there is an increase of collector current and the phenomenon
temperature dependent component. Other is common is known as the Early effect.
emitter configuration in this collector current is IC = βIB + • The collector current increases with the collector
(1+ β) ICO. Here also temperature dependent component voltage rather than staying constant.
present. The option (a) common base is not an answer • The slope introduced by the Early effect is almost
because IC is more dependent on ICO in common base. linear with IC and the common emitter characteristics
The option (b) common emitter is also not correct extrapolate to an intersection with the voltage axis
answer because IC is more dependent on ICO in common VA, called the Early voltage.
emitter as compared to common collector. • Now, as the base width is increased, the base current
will also increase, reducing the collector current.
3.19. Which of the following transistor configurations has • The collector-emitter characteristics will be more flat
the highest input impedance? which implies that the early voltage will increase.
(A) Common collector (B) Common Emitter This can be explained through characteristic as
(C) Common Base (D) All of the above shown below;
[ISRO’18]

MECHTEST GPSC 23 @mechtest1

 3.24. In a multi-stage RC-coupled Amplifier the coupling
capacitor
(A) Limits the low frequency response
(B) Limits the high frequency response
(C) Does not affect the frequency response
(D) Blocks the dc components without effecting the
frequency response
[EE-GWSSB {03/202122}]
Hence the Early voltage increases on increasing base 3.25. An npn bipolar junction transistor (BJT) is operating
width. in the active region. If the reverse bias across the base-
Important Point: collector junction is increased, then
• The punch-through breakdown occurs if the
(A) the effective base width increases and common-emitter
collector-base reverse bias voltage is increased far
current gain increases
enough to the extent that the depletion region fills
(B) the effective base width increases and common-emitter
the entire base.
current gain decreases
• In this condition holes in the p-n-p transistor are
swept directly from the emitter region to the (C) the effective base width decreases and common-
collector and the normal transistor action is lost. emitter current gain increases
• Punch-through is undesirable and hence it is avoided (D) the effective base width decreases and common-
in circuit design. emitter current gain decreases
[GATE’17]
3.22. In large signal analysis of amplifiers
SOLUTION:
(A) The swing of the input signal is over a wide range around
the operation point
(B) Operating point swings over large range
(C) stability factor is large
(D) power dissipation is large.
[DEE-GWSSB {42/202122}]
3.23. The CE amplifier circuit are preferred over CB As the reverse-biased base-collector voltage is
amplifier circuit because they have increased, the transition (or) the depletion region of the
base-collector junction is increased resulting in the
(A) Lower amplification factor
extension of this region more towards P-side (or base
(B) Larger amplification factor
side), thereby reducing the effective base width.
(C) High input resistance and low output resistance
This reduction in the effective base-width results in
(D) None of these
the increase in current gain as well because of the
[EE-GWSSB {03/202122}] increase in the slope of concentration gradient
SOLUTION: variation from the emitter side to the collector side.
CE amplifier -This transistor configuration is probably the
most widely used. The circuit provides a medium input 3.26. If for a silicon npn transistor, the base-to-emitter
and output impedance levels. Both current and voltage voltage (VBE) is 0.7 V and the collector-to-base voltage
gain can be described as medium, but the output is the (VCB) is 0.2 V, then the transistor is operating in the:
inverse of the input, i.e. 180o phase change. This provides (A) Normal active mode (B) Saturation mode
a good overall performance and as such it is often (C) Inverse active mode (D) Cut-off mode
thought of as the most widely used configuration or can [ISRO’13]
be said as that we get a good amplification factor here in
CE amplifier compared to CB amplifier.

MECHTEST GPSC 24 @mechtest1

 SOLUTION: • The slope introduced by the Early effect is almost
The modes of operation of an NPN BJT are as follow: linear with IC and the common-emitter characteristics
Emitter base Collector Base extrapolate to an intersection with the voltage axis
Mode
junction junction VA, called the Early voltage.
Reverse bias Reverse bias Cutoff • This is explained with the help of the following

Reverse bias Forward bias Reverse active VCE (Reverse voltage) vs IC (Collector current) curve:

Forward bias Reverse bias Active

Forward bias Forward bias Saturation

VB = 0.7V (Forward-biased) and

VBC = -VCB = -0.2V (Reverse-biased)
hence, the transistor operates in normal active mode.

3.27. The effect of reduction in effective base width due

3.30. An increase in the base recombination of a BJT will
to an increase in reverse voltage of BJT is
increase
(A) Hall effect (B) Early effect
(A) the common-emitter dc current gain β
(C) Zener effect (D) Miller effect
(B) the breakdown voltage BVCEO
[ESE’19]
(C) the unity gain cut-off frequency f T
3.28. The early effect in a bipolar junction transistor is
(D) the transconductance gm
caused by
[GATE’14]
(A) Fast turn - on
3.31. α-cut off frequency of a bipolar junction transistor
(B) Fast turn off
increases ______.
(C) Large collector – base reverse bias
(A) with the increase in base width
(D) Large emitter – base forward bias
(B) with the increase in collector width
[ISRO’20] [GATE’95]
(C) with the increase in temperature
3.29. The early effect in BJT is related to
(D) with the decrease in base width
(A) Base narrowing (B) Avalanche breakdown
[SSC JE’17] [GATE’93]
(C) Zener breakdown (D) Thermal runaway
3.32. The phenomenon known as “Early Effect” in a
[EE-GWSSB {03/202122}] [UPPCL JE’18]
bipolar transistor refers to a reduction of the effective
SOLUTION:
base – width caused by
Early Effect:
(A) Electron – hole recombination at the base
• A large collector base reverse bias is the reason
behind the early effect manifested by BJTs. (B) The reverse biasing of the base – collector junction
• As reverse biasing of the collector to base junction (C) The forward biasing of emitter – base junction
increases, the depletion region penetrates more into (D) The early removal of stored base charge during
the base, as the base is lightly doped. saturation – to – cutoff switching
• This reduces the effective base width and hence the [ESE’14] [GATE’06]
concentration gradient in the base increases. SOLUTION:
• This reduction in the effective base width causes For the transistor in common – emitter configuration the
less recombination of carriers in the base region which collector current for a given IB is expected to be
results in an increase in collector current. This is known
independent of VCE. This is true when base width is
as the Early effect.
constant, but when the base collector voltage (reverse
• The decrease in base width causes ß to increase and
bias) is increased the base width will be reduced. This
hence collector current increases with collector
reduced base width causes the minority carries to
voltage rather than staying constant.
increase, which causes increase in diffusion current. As a

MECHTEST GPSC 25 @mechtest1

 result, β will be increased i.e. IC increases with VCE. This 3. The emitter doping concentration to base doping
deviation is known as Early effect or base width concentration ratio is reduced
modulation. An extrapolation of collector currents gives 4. The base doping concentration is increased keeping
interaction with VCE axis, which is called Early voltage the ratio of the emitter doping concentration to base
(VA) doping concentration constant
5. The collector doping concentration is reduced
(A) A – 2, B – 3, C – 1 (B) A – 2, B – 5, C – 1
(C) A – 2, B – 3, C – 4 (D) A – 4, B – 3, C – 1
[GATE’94]
SOLUTION:
• As the base width of the BJT is reduced then the
3.33. Consider the following statements S1 and S2 recombination current (base current IB) decreases as a
S1: The β of a bipolar transistor reduces if the base width resuly colletor current (IC) increases. So, the current
is increased. gain of the BJT increases.
S2: The β of a bipolar transistor increases if the doping α=IC/IE
concentration in the base is increased. • If the emitter doping concentration to base doping
Which one of the following is correct? concentration ratio is reduced then the emitter
(A) S1 is FLASE and S2 is TRUE injection efficiency decreases, so the current
(B) Both S1 and S2 are TRUE gain (α) of BJT reduces.
(C) Both S1 and S2 are FLASE • If the collector doping concentration is increased then
(D) S1 TRUE and S2 is FALSE the breakdown (VBR) of a BJT will be reduced.
[GATE’98] [ISRO’17]
3.35. In a bipolar transistor at room temperature, if the
SOLUTION:
emitter current is doubled, the voltage across its base-
The β of a bipolar transistor reduces if the base width emitter junction
is increased because (A) It increases by about 20 mV
(B) It doubles
(C) Its halves
and base current Ib can be reduces by increasing base (D) It decreases by about 20 mV
width or decreasing doping concentration in the [GATE’97] [UPRVUNL JE’14]
base, resulting increase in β of bipolar transistor. SOLUTION:
However, the β of a bipolar transistor reduces if the
doping concentration in the base region is increased.
Therefore statement 1 is TRUE and 2 is FALSE.

3.34. Match the following:

List-I
A. The current gain of a BJT will be increased
B. The current gain of a BJT will be reduced
C. The break-down voltage of a BJT will be reduced
List-II 3.36. In a uniformly doped BJT, assume that NE, NB and
1. The collector doping concentration is increased NC are the emitter, base and collector doping in
2. The base width is reduced atoms/cm3, respectively. If the emitter injection
efficiency of the BJT is close to unity, which one of the
following conditions is TRUE?

MECHTEST GPSC 26 @mechtest1

 (A) NE = NB = NC (B) NE ≫ NB and NB < NC
(C) NE = NB and NB < NC (D) NE < NB < NC
[GATE’10]
SOLUTION:
The emitter injection efficiency is a measure of how
efficient the emitter is in injecting electrons into the
base.
It is important because the entire transistor action is due
to the electrons injected into the base.
The holes, which are simultaneously injected into the
emitter, are useless from this point of view.
In BJT the doping profile are related as-
NE> NC> NB
Now, as emitter injection efficiency of the BJT is close to
unity we can say
NE≫ NB
Thus, NE≫ NB and NB < NC

3.37. A small percentage of impurity is added to an

intrinsic semiconductor at 300 K. Which one of the
following statements is true for the energy band
diagram shown in the following figure?

(A) Intrinsic semiconductor doped with pentavalent atoms

to form an n-type semiconductor
(B) Intrinsic semiconductor doped with trivalent atoms to
form an n-type semiconductor
(C) Intrinsic semiconductor doped with pentavalent atoms
to form a p-type semiconductor
(D) Intrinsic semiconductor doped with trivalent atoms to
form a p-type semiconductor
[GATE’16]
SOLUTION:
 The new energy level is close to the edge of the
conduction band (Ec).
 Thus it is an n-type semiconductor that is formed by
doping a pure or intrinsic semiconductor with a
pentavalent atom.
Note:
Fermi level of an intrinsic semiconductor is as shown:

MECHTEST GPSC 27 @mechtest1

 4.1. The value of voltage V0 (given figure) will be

All pass filter:

Passes all frequencies without attenuation but provides
a predictable phase shift.
(A) 10 V (B) 0.7 V (C) 10.7 V (D) 30 V Band rejects filter:
[AE-GMB {20/202122}] It rejects only one certain band of frequency.
4.2. The following circuit will act as a

(A) low pass filter (B) high pass filter

(C) band pass filter (D) band reject filter Bandpass filter:
[AE-GMB {20/202122}] It allows only a certain band of frequency.

4.3. Which of the following electrical components can be

used as a high pass filter?
1. Capacitor 2. Inductor
3. Transistor 4. Amplifier
(A) 1, 2 and 3 (B) Only 1
(C) Only 3 and 4 (D) Only 3
[UPSSSC JE’21] Function Type of filter
SOLUTION: Passes low frequency signals Low pass filter
Electric filters are used to select certain bands of Passes high frequency signals High pass filter
frequencies to pass along or accept and another band to
Passes signals between two
stops or reject. The frequency at which the transition Band-pass filter
frequency limits
between passing and rejecting input signal occurs is
Rejects signals between two
called the cut-off frequency (fc) Band reject filter
frequency limits
High pass filter:
It allows only high-frequency components (above cut-  Lead Compensator is analogous to High Pass Filter
off frequency).  Lag Compensator is analogous to Low Pass Filter
 Lead - Lag Compensator is analogous to Band Pass
Filter
 Lag - Lead Compensator is analogous to Band
Stop Filter
NOTE:
Remember that current in a circuit always takes the path
Low pass filter: of least resistance. Since inductors offer such high
It allows only low-frequency components. (below cut off resistance to high-frequency signals, current signals of
frequency) high frequency will not go through the inductor of this

MECHTEST GPSC 28 @mechtest1

 circuit. They will take an alternate path and go through SOLUTION:
another part of the circuit that offers lesser resistance. In Voltage regulation is a measure commonly used to
RL circuit, instead of the high-frequency signals going describe the percentage voltage difference between no-
through the inductor and down to the ground, they go load and full load voltages to the full load voltage for
through to the output. And this is why this circuit is a electrical machines.
high-pass filter circuit. Low-frequency signals, however,
will go through the inductor, because inductors offer
• The ideal voltage regulation for any machine
very low resistance to low-frequency, or Dc, signals.
should be zero
Therefore, low-frequency current will take the path of
• Zero voltage regulation indicates that there is no
going through the inductor to the ground.
difference between its no-load voltage and its
4.4. The rise time of the given circuit will be full-load voltage
• This is not practically possible and is only valid
theoretically for ideal machines
• Therefore, for the practical application, it should
be as low as possible for the proper operation of
(A) 1 msec (B) 1.1 msec the electrical devices
(C) 2 msec (D) 2.2 msec
4.9. In an unregulated power supply, if load current
[AE-GMB {20/202122}]
increases, the output voltage ________
4.5. The ratio of AC signal power delivered to the load to
(A) Remains the same (B) Decreases
the DC input power to the active device as a percentage
(C) Increases (D) None of the above
is called
(A) conversion (B) Rectification η
4.10. If the load current drawn by unregulated power
(C) power η (D) utilisation factor supply increases, the d.c. output voltage ________
[DEE-GWSSB {42/202122}] (A) Increases (B) Decreases
4.6. A DC power supply has a no-load voltage of 30 volts, (C) Stays the same (D) None of the above
and a full load voltage of 25 volts at a full load current of
one amp. Its output resistance and load regulation, 4.11. An ideal regulated power supply is one which has
respectively are voltage regulation of ___________
(A) 5 Ω and 20 % (B) 25 Ω and 20 % (A) 0% (B) 5% (C) 10% (D) 1%
(C) 5 Ω and 16.7 % (D) 25 Ω and 16.7 %
[EE-GWSSB {03/202122}] 4.12. If the input a.c. voltage to regulated or ordinary
power supply increases by 5% what will be the
4.7. When step signal is applied to the input of high pass
approximate change in d.c. output voltage?
RC network, the transient output will be
(A) 10% (B) 20% (C) 15% (D) 5%
(A) step signal (B) ramp signal
(C) impulse (D) zero
4.13. The load voltage is approximately constant when a
[AE-GMB {20/202122}]
Zener diode is _______
4.8. The % load regulation of a power supply should be (A) Forward biased (B) Unbiased
ideally _____ & practically _____. OR (C) Reverse biased
A power supply should possess an ideal voltage (D) Operating in the breakdown region
regulation which is equal to _____ and practical voltage
regulation measuring ______. 4.14. A power supply has a voltage regulation of 1%. If
(A) zero, small (B) small, zero the no-load voltage is 20 V, what is the full-load
(C) zero, large (D) large, zero voltage?
[DEE-GWSSB {42/202122}] [UPPCL JE’18] (A) 19.8 V (B) 7 V (C) 6 V (D) 2

MECHTEST GPSC 29 @mechtest1

 4.15. In a regulated power supply, two similar 15 V zeners regulation ______
are connected in series. The input voltage is 45 V d.c. If A. of 10% B. of 15%
each Zener has a maximum current rating of 300 mA, C. of 25% D. within 1%
what should be the value of the series resistance?
A. 10 Ω B. 50 Ω C. 25 Ω D. 40 Ω 4.25. If the doping level is increased, the breakdown
voltage of the Zener _______
4.16. As the junction temperature increases, the voltage A. Remains the same B. Is increased
breakdown point for Zener mechanism _______ C. Is decreased D. None of the above
A. Is increased B. Is decreased
C. Remains the same D. None of the above 4.26. The current in a Zener diode is limited by ______
A. External resistance B. Power dissipation
4.17. As the junction temperature increases, the voltage C. Both (A) and (B) D. None of the above
breakdown point for avalanche mechanism ______
A. Remains the same B. Decrease 4.27. The Zener current will be minimum when ______
C. Increases D. None of the above A. Load current is maximum
B. Load current is minimum
4.18. For increasing the voltage rating, zeners are C. Load current is zero
connected in ____ D. None of the above
A. Parallel B. Series-parallel
C. Series D. None of the above 4.28. Which of these is the best description for a Zener
diode?
4.19. When load current is zero, the Zener current will be A. It is a diode
A. Zero B. Minimum C. Maximum D. None B. It is a constant current device
C. It is a constant-voltage device
4.20. The rupture of covalent bonds will occur when the D. It works in the forward region
electric field is _____.
A. 100 V/cm B. 0.6 V/cm 4.29. In a 15 V Zener diode, the breakdown mechanism
C. 1000 V/cm D. More than 105 V/cm will occur by _____
A. Avalanche mechanism
4.21. What is true about the breakdown voltage in a B. Zener mechanism
Zener diode? C. Both Zener and avalanche mechanism
A. It decreases when load current increases D. None of the above
B. It destroys the diode
C. It equals current times the resistance 4.30. A Zener diode utilises ____ characteristic for
D. It is approximately constant voltage regulation
A. Forward B. Reverse
4.22. A Zener diode is used as a ______ voltage C. Both forward and reverse D. None of the above
regulating device
A. Shunt B. Series 4.31. In a loaded Zener regulator, which is the largest
C. Series-shunt D. None of the above Zener current?
A. Series current B. Zener current
4.23. A power supply which has voltage regulation of C. Load current D. None of the above
______ is unregulated power supply.
A. 0% B. 5% C. 10% D. 8% 4.32. A power supply can deliver a maximum rated
current of 0.5 A at full-load output voltage of 20 V. What
is the minimum load resistance that you can connect
4.24. Commercial power supplies have voltage
across the supply?

MECHTEST GPSC 30 @mechtest1

 A. 10 Ω B. 20 Ω C. 15 Ω D. 40 Ω

4.33. A certain regulator has a no-load voltage of 6 V and

a full-load output of 5.82 V. What is the load regulation?
A. 3.09 B. 2.87
C. 5.72 D. None of the above

4.34. In an unregulated power supply, if input a.c.

voltage increases, the output voltage.
A. Increases B. Decreases
C. Remains the same D. None of the above

MECHTEST GPSC 31 @mechtest1

 SOLUTION:

Sr. BJT FET

1. BJT is a bipolar device. FET is a unijunction
transistor.
5.1. Field Effect transistor is: 2. Operation depends on Operation depends on
A. three junction both majority and majority charge
B. neither unijunction nor bijunction transistor minority charge carriers (holes or
C. unijunction transistor carriers. electrons).
D. Bijunction 3. Input impedance is Input impedance is
SOLUTION: very low and has less very large.
output impedance
than FET.
4. It is a current- It is a voltage-
controlled device. controlled device.
5. It is cheaper. It is costlier than BJT.
6. It is bigger in size than It is smaller in size
• The field-effect transistor (FET) is a type of transistor FET. than BJT.
that uses an electric field to control the flow of
current in a semiconductor. 5.4. A JFET has high input impedance because _____
• FETs are devices with three terminals: source, gate, A. it is made of semiconductor material
and drain. FETs control the flow of current by the B. input is reverse biased
application of a voltage to the gate, which in turn C. of impurity atoms
alters the conductivity between the drain and the D. none of the above
source.
• Field effect transistor has a large input impedance. 5.5. The gate voltage in a JFET at which drain current
• FETs naturally contain two p-n junctions, hence it becomes zero is called ______ voltage
is a Bi-junction device. A. saturation B. pinch-off
• FETs use either electrons (n-channel) or holes (p- C. active D. cut-off
channel) as charge carriers in their operation, but
not both. 5.6. The output characteristics of a JFET closely resemble
• Field effect transistors generally display very high the output characteristics of a ______ valve
input impedance at low frequencies. A. pentode B. tetrode
• In a FET, there are TWO pn junctions at the sides C. triode D. diode

5.2. Field effect transistor has

5.7. A MOSFET differs from a JFET mainly because ____
A. large input impedance
A. of power rating
B. large output impedance
B. the MOSFET has two gates
C. large power gain
C. the JFET has a pn junction
D. small voltage gain
D. none of the above

5.3. As compared to FET, BJT has _____ input impedance

5.8. If the reverse bias on the gate of a JFET is increased,
and ____ output impedance :
then width of the conducting channel _______
A. low, low B. low, high
A. is decreased B. is increased
C. high, high D. high, low
C. remains the same D. none of the above

MECHTEST GPSC 32 @mechtest1

 5.9. If the gate of a JFET is made less negative, the width A. 1 B. 11.4 C. 8.75 D. 3.2
of the conducting channel ______
A. remains the same B. is decreased 5.13. The input control parameter of a JFET is ______
C. is increased D. none of the above A. gate voltage B. source voltage
C. drain voltage D. gate current
5.10. What can be the effect on the width of the
conductor channel in a JFET, if the gate to source voltage 5.14. The input impedance of a JFET is ___________ that
is reverse biased? of an ordinary transistor
A. grows B. has no effect A. equal to B. less than
C. remains stable D. lacks C. more than D. none of the above
SOLUTION:
• A JFET ( Junction Field Effect Transistor) is a three 5.15. A JFET can operate in–
terminal semiconductor device in which the current A. Depletion mode only
conduction is by either holes or electrons. B. Enhancement mode only
• The current conduction is controlled by means of an C. Depletion and enhancement modes
electric field between the gate and the conducting D. Neither enhancement nor depletion mode
channel of the device.
• The width and hence resistance of this channel can 5.16. A JFET has three terminals, namely _______
be controlled by changing the input voltage (gate to A. cathode, anode, grid
source voltage)VGS. B. emitter, base, collector
• The greater the gate to source voltage VGS, the wider C. source, gate, drain
will be the depletion channel and the narrow will be D. none of the above
the conducting channel.
• That is, the conducting channel offers high resistance
5.17. The pinch-off voltage in a JFET is analogous to
and hence source to drain current decreases. _______ voltage in a vacuum tube
• When VGS decreases the conducting channel A. anode B. cathode
increases offering low resistance. C. grid cut off D. none of the above
• From the above points, it is concluded that if the
gate to source voltage is reverse biased the width
5.18. At cut-off, the JFET channel is ____
of the conducting channel decreases.
A. at its widest point
• The gate of a JFET is reverse biased.
B. completely closed by the depletion region
• JFET has the lowest noise-level.
C. extremely narrow
• Transconductance of a JFET ranges from 0.5 to 30
D. reverse baised
mA/V
• The pinch-off voltage of a JFET is about 5 V.
5.19. In a p-channel JFET, the charge carriers are _______
A. electrons
5.11. If the cross-sectional area of the channel in n-
B. holes
channel JEFT increases, the drain current ______
C. both electrons and holes
A. is increased B. is decreased
D. none of the above
C. remains the same D. none of the above

5.20. The pinch-off voltage of a JFET is about ______

5.12. In a certain common source D-MOSFET amplifier,
A. 5 V B. 0.6 V C. 15 V D. 25 V
Vds = 3.2 V r.m. and Vgs = 280 mV r.m.s. The voltage gain
is _____

MECHTEST GPSC 33 @mechtest1

 5.21. The two important advantages of a JFET are ___ 5.23. IGBT and MOSFET are ___ and ___ controlled
A. high input impedance and square-law property devices
B. inexpensive and high output impedance A. voltage and voltage B. voltage and current
C. low input impedance and high output impedance C. current and voltage D. current and current
D. none of the above
5.24. When drain voltage equals the pinch-off-voltage,
5.22. Which of the following statements is INCORRECT? then drain current ___ with the increase in drain voltage
A. Similar to the MOSFET, the IGBT has a low impedance A. decreases B. increases
gate. C. remains constant D. none of the above
B. Similar to the MOSFET, the IGBT has a high impedance
gate. 5.25. A properly biased JFET will work as OR JFET in
C. Like the BJT, the IGBT has a small on- state voltage. properly biased condition acts as a
D. Similar to the GTO, the IGBT can be designed to block A. current controlled current source
negative voltage. B. voltage controlled voltage source
SOLUTION: C. voltage controlled current source
D. current controlled voltage
Sr. BJT MOSFET IGBT

1. Bipolar device Unipolar Bipolar device 5.26. A MOSFET uses the electric field of a _____ to
device control the channel current

2. Low on-state High on-state Low on-state A. capacitor B. battery C. generator D. none

voltage drop voltage drop voltage drop SOLUTION:

• MOSFET is a voltage-driven/controlled device.
3. Low on-state High on-state Low on-state • By providing a positive voltage to the gate, with
conduction conduction conduction respect to the source, current will be made to flow in
power loss power loss power loss the drain Reducing the voltage to zero will turn the
4. High Low Low drain current off.
switching switching switching • voltage controls the current through the two

power loss power loss power loss terminals at the third terminal (gate) It is a unipolar
device (current conduction is only due to one type of
5. Low input High input High input majority carrier either electron or hole) It has a high
impedance impedance impedance input impedance.
6. Current Voltage Voltage • MOSFET is a unipolar device hence current through it

control device control device control device is only because of Majority carriers but not of
Minority carriers.
7. Usage: UPS Usage: SMPS Usage: • A power MOSFET is a unipolar, voltage controlled
charging Inverter, and majority carrier device.
battery Chopper • A MOSFET is also known as Insulated-Gate JFET.

8. Negative Positive Positive • The input impedance of a MOSFET is of the order of

temperature temperature temperature several MO.

coefficient of coefficient of coefficient of • The Gate terminal of a MOSFET is isolated from the

resistance resistance resistance semiconductor by a thin layer of SiO2.

• A MOSFET can be operated with positive as well as
negative gate voltage.

MECHTEST GPSC 34 @mechtest1

 5.27. Which is the power semiconductor device having 5.34. In a common source JFET amplifier, the output
highest switching speed? OR Which of the following voltage is _______. OR What is the position of the input
transistors can be used in enhancement mode and output signals of CE amplifier?
A. SCR B. IGBT C. MOSFET D. GTO A. 180° out of phase with input
B. In phase with input
5.28. MOSFET can be used as a : C. 90° out of phase with input
A. Current controlled capacitor D. None of the above
B. Voltage controlled capacitor SOLUTION:
C. Current controlled inductor Common Source configuration:
D. Voltage controlled inductor • In the Common Source configuration (similar to
common-emitter), the input is applied to the Gate
5.29. A MOSFET is also known as Insulated-Gate _____. and its output is taken from the Drain as shown.
A. GTO B. BJT C. JFET D. Diode • This is the most common mode of operation of the
FET due to its high input impedance and good
5.30. The gain of MOSFET amplifiers at high frequency voltage amplification and as such Common Source
reduces due to the effect of: amplifiers are widely used.
A. Coupling capacitors B. Parasitic capacitors • The common source mode of FET connection is
C. Oxide capacitors D. Bypass capacitors generally used audio frequency amplifiers and in
SOLUTION: high input impedance pre-amps and stages.

• In the low frequency operation there will be no • Being an amplifying circuit, the output signal is 180o

internal feedback in the MOSFET and therefore it will “out-of-phase” with the input.

provide the best performance.

• In the high frequency operation there will be inter
electrode capacitors (also called as stray) or (parasitic
capacitors) and due to this there will be a negative
feedback in FET and MOSFET therefore performance
of the device is badly affected and MOSFET gain
decreases. Common Gate configuration:
• In the Common Gate configuration (similar to
5.31. For the operation of enhancement only Nchannel
common base), the input is applied to the Source and
MOSFET, value of gate voltage has to be :
its output is taken from the Drain with the Gate
A. Zero B. Low positive
connected directly to ground (0v) as shown.
C. High positive D. High negative
• The high input impedance feature of the previous
connection is lost in this configuration as the
5.32. A MOSFET has _____ terminals
common gate has a low input impedance, but a high
A. two B. five C. four D. three
output impedance.
• This type of FET configuration can be used in high-
5.33 Choose the correct statement. frequency circuits or impedance matching circuits
A. Both BJT and MOSFET are voltage controlled devices were a low input impedance needs to be matched to
B. Both BJT and MOSFET are current controlled devices high output impedance. The output is “in-phase” with
C. MOSFET is a voltage controlled device and BJT is a the input.
current controlled device
D. MOSFET is a current controlled device and BJT is a
voltage controlled device

MECHTEST GPSC 35 @mechtest1

 5.35. Which of the following is correct
A. JFET has zero offset value
B. Compared with MOSFETs, JFET are easier to fabricate
C. The drain resistance of MOSFET is higher than that of
JFET.
D. The input resistance of MOSFET is higher than that of
JFET

5.36. Which power semiconductor device is used for

speed control of small single-phase induction motor
based domestic fans?
A. single SCR B. BJT
C. Triac D. MOSFET

5.37. Which is the important factor in the steady state

characteristics of a MOSFET?
A. Current gain B. Transconductance
C. Output resistance D. Drain source voltage

5.38. The disadvantage of a typical MOSFET as compared

to BJT is
A. Increased power-handling levels
B. Reduced power-handling levels
C. Increased voltage-handling levels
D. Reduced voltage-handling levels

5.39. For the most FET configurations and for common-

gate configurations, the input impedances are
respectively
A. High and high B. High and low
C. Low and low D. Low and high

5.40. A depletion-type MOSFET can be operated in an

enhancement mode where negative charges are
induced into n-type channel by applying
A. Positive Gate Voltage B. Negative Gate Voltage
C. Positive Drain Voltage D. Negative Drain Voltage

MECHTEST GPSC 36 @mechtest1

 A. A 8-bit converter B. A 10-bit converter
C. A 12- bit converter D. A 16- bit converter

6.7. The settling time for DAC with a resolution of 10 mV

6.1. What is the number of comparators required for a 3- will be:
bit flash Analog to Digital Converter? A. 6 mV B. 5 mV C. 2 mV D. 10 Mv
A. 5 B. 9 C. 7 D. 3
SOLUTION: 6.8. A 4-bit R/2R digital-to-analog (DAC) converter has a
n = 3-bit reference of 5V. What is the analog output (in V) for the
Number of comparators = 2n–1 = 23–1 = 8 – 1 = 8 input code 1010?
A. 1.125 V B. 1.725 V
6.2. A 16-bit flash (parallel) ADC needs_____ C. 2.125 V D. 3.125 V
comparators.
A. 64,000 B. 65,530 6.9. What is the frequency of a signal (in Hz) If a five-
C. 65, 536 D. 65,535 digit decimal counting Assembly (DCA) reads 08225 and
SOLUTION: the time base is set to 1 ms
n = 16-bit A. 8825000 Hz B. 8250000 Hz
Number of comparators = 2 –1 = 2 –1 = 65536 – 1 = 65535
n 16 C. 8225000 Hz D. 8290000 Hz
SOLUTION:
6.3. Number of bits needed to code 512 operations is n=5
A. 4 B. 7 C. 8 D. 9 Counting assembly reads = 08225
SOLUTION: t = 1mSec
The number of bits needed to code 512 operations is 9 f=?
2n = 512 f = Reading/Time = 08225/(1 × 10−3)
2 n = 29 f = 8225000 Hz
N=9
6.10. Which of the following types of ADC is also known
6.4. The term digitalization refers to as continuous conversion type ADC?
A. Conversion of analog into digital A. Counter-type ADC
B. conversion of digital into analog B. Dual-slope ADC
C. use of analog form of electricity C. Successive approximation ADC
D. a form of changing physical quantities D. Tracking-type ADC

6.5. In which of the following types of A/D converter 6.11. Settling time is normally defined as the time taken
does the conversion time almost double for every bit by DAC to settle within
added to the device? A. LSB of its final value, when a change occurs in the input
A. Successive approximation type A/D converter code
B. Single slope integrating type A/D converter B. ±1/4 LSB of its final value, when a change occurs in the
C. Counter type A/D converter input code
D. Tracking type A/D converter C. ±1/2 LSB of its final value, when a change occurs in the
input code

6.6. The resolution of D/A converter is approximately D. 1 LSB of its final value, when a change occurs in the input

0.4% of its full-scale range. It is : code

MECHTEST GPSC 37 @mechtest1

 6.12. In which of the following types of A/D converter C. Successive approximation A/D converter
does the conversion time almost double for every bit D. Extended parallel type A/D converter
added to the device?
A. Counter type A/D converter 6.15. The analog-to-digital converters are employed in:
B. Tracking type A/D converter A. Voltmeter B. Wattmeter
C. Single-slope integrating type A/D converter C. Energy meter D. Digital multimeter
D. Successive approximation type A/D converter
SOLUTION: 6.16. What is the value of LSB of an 8-bit DAC for 0–12.8V
For n-bit conversion, the conversion time for different range?
ADC are: A. 1.6V B. 50 mV C. 0.625V D. 1.28V
Successive approx. time ADC: n Tclk SOLUTION:
Flash type ADC: Tclk Least Significant Bit (LSB) value of a DAC:
Dual slope ADC: (2n+1 – 1) Tclk
Counter type ADC: (2n – 1) Tclk ≈ 2n Tclk
he successive approximation A/D converter has a shorter
conversion time compared to the counter ramp A/D
converter, i.e the correct sequence will be: TD < TC< TA<
TB
• Counter type ADC and successive approximate ADC
uses DAC
• Counter type ADC uses linear search and successive
approximation type ADC uses binary search 6.17. The reference voltage and the input voltage are
• A ring counter is used in successive approximation sequentially connected to the integrator with the help
type ADC of a switch in a
• Flash type ADC is fastest ADC A. Successive approximation A/D converter
• Flash type ADC requires no counter B. Dual slope integration A/D converter
• For an n-bit ADC, flash type ADC requires (2n – 1) C. Voltage to time converter
comparators D. Voltage to frequency converter
• Dual slope ADC is most accurate

6.18. An 8-bit A/D converter is used over a span of 0 to

6.13. A 12-bit A/D converter has a full-scale analog input 2.56V. The binary representation of a 1.0V signal is:
of 5 V. Its resolution is A. 01100100 B. 01110001 C. 10100101 D. 10100010
A. 1.22 mV B. 2.44 mV C. 3.66 mV D. 4.88 Mv SOLUTION:
SOLUTION: To find the digital output (binary) of an ADC:

6.14. An A-to-D converter in which one sub-circuit is a D-

to-A converter is
A. Parallel A/D converter Now convert 100 to 8-bit binary:
B. Dual slope A/D converter 100 = 01100100

MECHTEST GPSC 38 @mechtest1

 7.8. A 512 MB RAM is connected to a microprocessor with
data bus length of 8. The size of memory that will remain
unutilized is:
A. 256 MB B. 128 MB
7.1. The number of status flags in 8085 microprocessor is: C. Zero D. 192 MB
A. 6 B. 4 C. 3 D. 5
7.9. Which of the interrupts of 8085 microprocessor has
7.2. In 8085 microprocessor, the general-purpose highest priority?
registers (GPR) are: A. INTR B. TRAP
A. A B C D E F B. A B C D H F C. RST 7.5 D. RST 5.5
C. B C D H L M D. B C D E H L
7.10. In 8085 microprocessor, stack works on
7.3. In 8085 microprocessor system, the direct addressing A. LILO B. LIFO
instruction is – C. FIFO D. None of these
A. MOV A, B B. MOV B, 0AH
C. MOV C, M D. STA addr 7.11. What are the sets of commands in a program which
are not translated into machine instructions during
7.4. In 8086 microprocessor the following has the highest assembly process, called?
priority among all type interrupts. A. Mnemonics B. Directives
A. DIV 0 B. TYPE 255 C. Identifiers D. Operands
C. OVER FLOW D. NMI
7.12. Memory-mapped I/O scheme for address allocation
7.5. In a microprocessor, the address of the next to memories and I/O devices is used for
instruction to be executed is stored in: A. small systems B. large systems
A. Stack pointer C. both D. very large systems
B. Address latch
C. Program counter 7.13. The interfacing device used for generating accurate
D. General purpose register time delay is:
A. Intel-8251 B. Intel-8257
7.6. A programmable ROM has a decoder at the input and C. Intel-8253 D. Intel-8259
OR gates at the output with
A. both these blocks being fully programmable 7.14. Which of the following is a two-byte instruction in
B. both these blocks being partially programmable 8085?
C. only the latter block being programmable A. MOV B. CMA
D. only the former block being programmable C. ADD D. MVI

7.7. Which one of the following is not correct for the AAA 7.15. Which of the following statements about RAM is
instruction in 8086 microprocessors? NOT correct?
A. It works only on the AL register. A. RAM = Random Access Memory
B. It updates AF and CF. B. It is read/write memory
C. It checks the result for correct unpacked BCD. C. Info is lost when power is off
D. It updates all the flags. D. Binary contents are stored during manufacturing

MECHTEST GPSC 39 @mechtest1

7.16. The address bus width of 8085 microprocessor is: 7.25. 8254 programmable timer has how many 16-bit
A. 32-bit B. 8-bit C. 16-bit D. 64-bit counters?
A. 2 B. 3 C. 4 D. 5
7.17. How many and which types of machine cycles are
needed to execute PUSH PSW by Intel 8085 7.26. Handshake signal in data transfer is sent:
microprocessor? A. Along with data bits B. Before data transfer
A. 2; Fetch and Memory write C. After data transfer D. Any of the above
B. 3; Fetch and 2 Memory write
C. 3; Fetch and 2 Memory read 7.27. Total number of memory & I/O devices addressable
D. 3; Fetch, Memory read and Memory write with 16-bit address bus using memory-mapped I/O:
A. 64K memory + 256 I/O
7.18. The total addressable memory size of 8085 is: B. 256 I/O + 65279 memory
A. 16 kb B. 32 kb C. 64 kb D. 128 kb C. 64K memory + 0 I/O
D. 64K either memory or I/O
7.20. When RET instruction is executed by a subroutine
then: 7.28. In Intel 8085, the interrupt enable flip-flop is reset
A. Top of stack is popped into PC by:
B. Program resumes at next instruction after CALL A. DI instructions only
C. PC is incremented B. System RESET only
D. All of the above C. Interrupt acknowledgement only
D. Either DI or system RESET or interrupt acknowledgement
7.21. 8086 has maximum clock frequencies ranging from:
A. 5 MHz - 8 MHz B. 6 MHz - 10 MHz 7.29. A 4k × 8 bit RAM starts at address AA00H. What is
C. 5 MHz - 15 MHz D. 5 MHz - 10 MHz the last byte address?
A. 0FFFH B. 1000H
7.22. 1 MB ROM can be obtained from 64 KB ROM ICs. C. B9FFH D. BA00H
How many are needed?
A. 16 ICs in a row 7.30. The Interrupt Vector Table (IVT) of 8086 contains:
B. 16 ICs in a column A. CS/IP of main program address where interrupt occurred
C. 8 ICs in a column and 2 in a row B. CS/IP where control returns after routine
D. None of the above C. CS/IP of interrupt service routine
D. Starting address of IVT
7.23. A direct memory access (DMA) transfer implies:
A. Memory ↔ Accumulator 7.31. Which of the following does NOT happen when 8085
B. Memory ↔ I/O without microprocessor is reset?
C. Microprocessor registers A. Gives reset out signal B. Resets PC to FFFFH
D. Fast memory ↔ microprocessor C. Disables interrupt system D. Tri-states buses

7.24. Opcode fetch cycle is: 7.32. Which one of the following is the software interrupt
A. Last part of instruction cycle of 8085 microprocessor?
B. First part of instruction cycle A. RST 7.5 B. RST 7
C. Intermediate part C. TRAP D. INTR
D. Data reception through bus

MECHTEST GPSC 40 @mechtest1

7.33. Which one of the following statements 7.40. In mode '0' operation of 8255, the ports can be used
corresponding to execution of SIM instruction is not as:
correct? A. A as input port only
A. It will selectively mark all the interrupts of 8085 B. B as output port only
B. D7 appears on SOD pin only if b6 = 1 in accumulator C. A as output port only
C. RST 7.5 can reset without executing ISR for RST 7.5 D. A as input or output port
D. It can handle interrupts and serial I/O
7.41. On receiving an interrupt, the CPU of an 8085
7.34. Correct sequence of steps for executing an microprocessor:
instruction during CPU processing: A. Completes the current instruction and then goes to the
A. Fetch → Read data → Decode → Store → Execute ISR
B. Decode → Read data → Execute → Fetch next → Store B. Branches off to ISR immediately
C. Decode → Decode next → Fetch → Execute → Store C. Hands over buses to interrupting device
D. Fetch → Decode → Read operands → Execute → Store D. Goes to HALT state temporarily

7.35. Three devices P, Q, and R are connected to 8085. P 7.42. Which value will the program counter have when
has the highest priority, R the lowest. Choose correct the non-maskable interrupt of the 8085 microprocessor
interrupt assignment: is serviced?
A. P → TRAP, Q → RST 5.5, R → RST 6.5 A. 0004H
B. P → RST 5.5, Q → RST 6.5, R → RST 7.5 B. 0014H
C. P → RST 7.5, Q → RST 6.5, R → RST 5.5 C. 0024H
D. P → RST 5.5, Q → RST 6.5, R → TRAP D. 0034H
SOLUTION:
7.36. In microprocessor architecture, the flag indicates: Interrupt Triggering Vectored Maskable
A. Number of microprocessor B. Manufacturer name Address / Non-
C. Internal status of CPU D. Bit size of microprocessor maskable
TRAP Edge and 0024H Non-
7.37. How many memories of 16×4 size are needed to Level maskable
build 64×8 memory? RST 7.5 Edge 003CH Maskable
A. 2 B. 4 C. 6 D. 8 RST 6.5 Level 0034H Maskable
RST 5.5 Level 002CH Maskable
INTR Level Non- Maskable
7.38. Output device at address 1000H is memory-mapped
vectored
to 8085. Which instruction sequence is correct to move
data from accumulator?
7.43. Which one of the following statements is not
A. LXIH, 1000H MOV A, M B. LXI H, 1000H MOV M, A
correct?
C. LHLD 1000H MOV A, M D. LHLD 1000H MOV M, A
A. CMPA is a single byte instruction and CMA is not an
instruction
7.39. The program counter in an 8085 microprocessor is a
B. The instruction SUB A sets the zero flag
16-bit register, because:
C. Bus is a group of wires
A. It counts 16 bits at a time
D. Instruction INR does not affect carry flag
B. There are 16 address lines
C. It facilitates the user storing 16-bit data temporarily
D. It has to fetch two 8-bit data at a time

MECHTEST GPSC 41 @mechtest1

7.44. In a microprocessor-based system, DMA facility is
required to increase the speed of data transfer between
the:
A. Microprocessor and I/O memory
B. Microprocessor and I/O devices
C. Memory and the I/O devices
D. Memory and register

7.45. Sixty-four 256×1 bit RAM ICs are arranged in 8 rows

and 8 columns. What is the total memory?
A. 1 kB B. 2 kB
C. 4 kB D. 8 kB

7.46. In a 5 × 7 dot matrix format

A. 64 bits are required to store 64 alphanumeric characters.
B. 560 bits are required to store 64 alphanumeric characters.
C. 1120 bits are required to store 64 alphanumeric characters.
D. 2240 bits are required to store 64 alphanumeric
characters.

7.47. Direct memory access channel (DMA) facilitates

data to move in and out of the system
A. on first-come first serve basis
B. with equal time delay
C. without a sub-routine
D. without program intervention

7.48. To interface a slow memory, wait states are added

by
A. extending the time of the chip select logic
B. causing READY signal to go low
C. causing READY signal to go high
D. by increasing the clock frequency

7.49. A semiconductor ROM is preferred to a

semiconductor RAM because-
A. ROM is cheaper than RAM
B. ROM is faster
C. ROM does not require power supply for their operation
D. Program stored in the ROM cannot be altered

7.50. The number of output pins of a 8085 microprocessor

are
A. 40 B. 27 C. 21 D. 19

MECHTEST GPSC 42 @mechtest1

 CONCEPT: A BJT has four modes for operation depending on polarities of emitter base junction and collector base
junction.
Base-emitter (BE) Junction Collector-base (CB) junction
Saturation F.B F.B
Active mode F.B R.B
(functioning as an amplifier)
Reverse Active R.B F.B
Cut off R.B R.B
*F.B: Forward biased & R.B: reverse biased
MCQS:

3.6. What is the biasing condition of junctions in bipolar 3.1. A bipolar junction transistor is used as power control
junction transistor to work as an amplifier? [ESE’14] switch by biasing it in the cut-off region (OFF state) or in
(A) Reverse biased base to emitter junction and reverse the saturation region (ON state). In the ON state, for the
biased base to collector junction BJT
(B) Forward biased base to emitter junction and reverse (A) Both the base-emitter junction and base-collector
biased base to collector junction junctions are reverse biased
(C) Forward biased base to emitter junction and forward (B) The base-emitter is reverse biased, and the base-
biased base to collector junction collector junction is forward biased
(D) Reverse biased base to emitter junction and forward (C) The base-emitter junction is forward biased, and the
biased base to collector junction base-collector junction is reverse biased
3.8. When a bipolar junction transistor is operating in the (D) Both the base-emitter and base-collector junctions are
saturation mode, which one of the following statements forward biased
is TRUE about the state of its collector-base (CB) and the [BEE-GWSSB {25/202223}] [EE-GWSSB {03/202122}]
base-emitter (BE) junctions? [GATE’04]
(A) The CB junction if forward biased and the BE junction is SELF QUE 1: When both the junctions of bipolar junction
reverse biased.
transistor (BJT) are in forward biased then in which
(B) The CB junction is reversed and the BE junction is
region BJT will operate?
forward biased.
(A) Saturation Region (B) Active Region
(C) Both the CB and BE junctions are forward biased.
(C) Ohmic Region (D) Cut-off Region
(D) Both the CB and BE junctions are reverse biased.
[SSC’17]
[GATE’15]
SELF QUE 2: A BJT is in inverse active mode if
3.11. A BJT is said to be operating in the cutoff region, if
(A) E-B junction is reverse biased and C-B junction is
(A) Both the junctions are reverse biased
forward biased.
(B) Base emitter junction is in reverse biased, and base
(B) Both E-B junction and C-B junction are forward biased
collector junction is forward biased
(C) Both-E-B junction and C-B junction are reverse biased.
(C) Base emitter junction is in forward biased, and base
(D) Operation of the device is very poor.
collector junction is reverse biased
[UPRVUNL AE EC 2014]
(D) Both the junctions are forward biased
[EE-GWSSB {03/202122}] [GATE’95]

MECHTEST GPSC 43 @mechtest1