TYPICAL QUESTIONS &
ANSWERS
PART
- I
OBJECTIVE TYPE QUESTIONS
Each
Question carries 2 marks.
Choose
correct or the best alternative in the following:
Q.1 The “Superposition
theorem” is essentially based on the concept of
(A) duality. (B)
linearity.
(C) reciprocity. (D) non-linearity.
Ans: B
Q.2 Cells are connected
in parallel in order to
(A)
increase the voltage available. (B) reduce cost
of wiring.
(C) increase the current available. (D) reduce the
time required to fully
charge them after use.
Ans: C
Q.3 The power factor of a purely resistive
circuit is
(A) zero. (B) unity.
(C)
lagging. (D) leading.
Ans: B
Q.4 The power taken by a 3-phase load is
given by the expression
(A) 
. (B) 
.
(C) 
. (D) 
.
Ans: B
Q.5 Which of the following generating
stations has the minimum running cost?
(A) hydro-electric station. (B) nuclear power station.
(C) thermal power station. (D)
diesel power plant.
Ans: A
Q.6 Which of the following motors has a high
starting torque?
(A)
ac series motor. (B) dc series motor.
(C) induction motor. (D) synchronous motor.
Ans: B
Q.7 A step-up transformer increases
(A) voltage. (B) current.
(C) power. (D) frequency.
Ans: A
Q.8 The effect of increasing the length of
the air gap in an induction motor will be to increase
(A) power factor. (B)
speed.
(C) magnetising current. (D) air-gap flux.
Ans: C
Q.9 The combined
resistance of two equal resistors connected in parallel is equal to
(A) One half the resistance of one resistor.
(B)
Twice the resistance of one
resistor.
(C)
Four times the
resistance of one resistor.
(D)
One fourth the
resistance of one resistor.
Ans: A
Q.10 Superposition
theorem can be applicable only to circuits having _________ elements.
(A)
Non- linear (B) Passive
(C) Resistive (D) Linear bilateral
Ans: D
Q.11 The
Q- factor of a coil is given by
(A) Its power factor cos j.
(B)
Ratio of max. energy stored & energy dissipated per
cycle..
(C)
Reciprocal of its power factor.
(D)
Ratio R/Z.
Ans: C
Q.12 Voltage
equation of a dc motor is
(A) V = Eb + Ia
Ra. (B)
Eb = V + Ia Ra.
(C) V = Eb / Ia Ra. (D) V = Eb
+ Ia 2Ra.
Ans: A
Q.13 The
efficiency of a transformer is maximum when
(A) It runs at half full load. (B) It runs at full load.
(C) Its Cu loss equals iron loss. (D) It runs overload.
Ans: C
Q.14 The
crawling in an induction motor is caused by
(A) Improper design of
the machine (B) Low voltage supply.
(C) High loads. (D) Harmonics developed in the motor.
Ans: D
Q.15 The
starting winding of a single-phase motor is placed in
(A)
Rotor. (B) Stator.
(C) Armature. (D) Field.
Ans: B
Q.16 Reduction
in the capacitance of a capacitor- start motor results in reduced
(A)
Noise. (B)
Speed.
(C) Starting torque. (D) Armature
reaction.
Ans: C
Q.17 In
an ac circuit, the ratio of kW / kVA represents
(A) Power factor. (B) Load
factor.
(C) Form factor. (D) Diversity
factor.
Ans: A
Q.18 The
unit of inductance is
(A) Ohm. (B)
Mho.
(C) Farad. (D)
Henry.
Ans: D
Q.19 Thevenin’s equivalent
circuit consists of _________.
(A) Series combination of RTh, ETh
and RL.
(B)
Series combination of RTh,
ETh.
(C) Parallel combination
of RTh, ETh.
(D) Parallel combination
of RTh, ETh and RL.
Ans:
B
Q.20 In an R – L –C circuit, the phase of the current with respect to the
circuit voltage will be_________.
(A)
Leading. (B) Same.
(C) Lagging. (D) Depends upon the value of Land C.
Ans: D
Q.21 The
frequency of DC supply is__________.
(A) Zero. (B) 16 ⅔ Hz.
(C) 50 Hz. (D) 100 Hz.
Ans: A
Q.22 Load
factor is defined as the ratio of _________.
(A)
Average
Demand / Max. Demand.
(B)
Max. Demand / Average Demand.
(C)
Average Demand / Connected load.
(D) Connected load / Max. Demand.
Ans: A
Q.23 Static
Capacitors are used for__________.
(A) Power improvement. (B) Current improvement.
(C) Voltage improvement. (D) Power factor improvement.
Ans: D
Q.24 The
speed of an induction motor__________.
(A) Decreases too much
with the increase of load.
(B)
Increases with the increase of load.
(C)
Decreases slightly with the increase of load.
(D)
Remains constant with the increase of load.
Ans: C
Q.25 Centrifugal
switch is provided for disconnecting the auxiliary winding in a_______.
(A) Capacitor- start
motor. (B) Capacitor run motor.
(C) Reluctance motor. (D)
Hysteresis motor.
Ans: A
Q.26 Rotating
magnetic field is produced in a ________.
(A) Single- phase induction motor. (B) Three- phase induction motor.
(C) DC series motor. (D) AC series
motor.
Ans: B
Q.27 The
frequency of the secondary voltage of a transformer will be_________.
(A) Less than the frequency of the primary voltage.
(B) Equal to the primary voltage.
(C) Greater than the
frequency of the primary voltage.
(D) Very much greater
than the frequency of the primary voltage.
Ans: B
Q.28 The demand factor for the electrical
system is the ratio of
(A) Maximum demand to
connected load
(B) Maximum demand to
average load
(C) Average power to maximum
power
(D) Relative power to total
power
Ans:
A
Q.29 When a low resistance is
connected in parallel with a high resistance, the combined resistance is
(A) Always more than the high resistance.
(B)
Always less than the low resistance.
(C)
Always between the high resistance & low resistance.
(D)
Either lower or higher than low resistance depending on the
value of high resistance.
Ans: B
Q.30 Q factor of an
inductive coil is given by
(A) R/Z (B)
(C) 
(D)
l
r/l
Ans: B
Q.31 The
r.m.s. value of sinusoidal 100 V peak to peak is _________ volt.
(A)
(B) 
(C) 50 (D) 100
Ans:
B
Q.32 If the readings of the two
wattmeters in the 2-wattmeter method of power measurement are 4.5 kW and 3.5 kW
respectively and the latter reading has been obtained after reversing the
current coil of the wattmeter. What will
be the total power in kW?
(A) 1 (B) 3.5
(C) 4.5 (D)
8
Ans: A
Q.33 A DC series motor is best suited for
driving
(A) Lathes. (B) Cranes and hoists.
(C) Shears and punches. (D) Machine tools.
Ans: B
Q.34 Transformer
cores are built up from laminations rather than from solid metal so that
(A) Oil penetrates the
core more easily.
(B) Eddy current loss is
reduced.
(C)
Less lamination is required for the windings.
(D)
Turn ratio is higher
than voltage ratio.
Ans: B
Q.35 In
a DC series motor increase in load current results in
(A) Decrease in speed (B) Increase in speed
(C) Better commutation (D) Increase in the
back emf.
Ans: A
Q.36 The
starting torque of a 1-phase induction motor is
(A) High. (B) Moderate.
(C) Low. (D)
Zero.
Ans: D
Q.37 An
electric motor in which rotor and stator fields rotate simultaneously is called
a __________ motor.
(A) DC (B) Induction
(C) Synchronous (D) Universal
Ans: C
Q.38 In
India,
electrical power is transmitted by
(A) 1 – phase a.c. system. (B)
3-wire d.c.
system.
(C) 3-phase 3-wire a.c. system. (D) 2-wire d.c. system.
Ans:
C
Q.39 In ac circuit the product of voltage and current is known as
(A) Power. (B) Real power.
(C) Resistive power. (D) Apparent power.
Ans:
D
Q.40 A network that does not
have either voltage or current sources is called
(A) Active network. (B) Passive network.
(C) Resistive network. (D) Dummy network.
Ans:
B
Q.41 The Power- factor at resonance in R-L-C circuit is
(A) Zero. (B) Unity.
(C) 0.5 lagging. (D)
0.5 leading.
Ans: B
Q.42 In an 8 – pole wave connected motor
armature, the number of parallel paths are
(A) 8 (B) 4
(C) 2 (D) 1
Ans:
C
Q.43 Transformer core is
laminated to
(A) Reduce the copper losses. (B) Reduce the core losses.
(C) Reduce the eddy current losses. (D)
None of these.
Ans:
C
Q.44 The relation between
frequency, speed and number of poles is given by
(A) 
(B) 
(C) 
(D) 
Ans: A
Q.45 Star – delta starter
of an induction motor
(A) Inserts resistance in rotor
circuit.
(B) Inserts resistance in stator
circuit.
(C) Applies reduced voltage to
rotor.
(D)
Applies reduced voltage to stator.
Ans: D
Q.46 Stator core of an
induction motor is made of
(A) Laminated cast iron. (B) Mild steel.
(C) Silicon steel stampings. (D)
Soft wood.
Ans:
C
Q.47 Watt hour is the unit
of
(A) Electric power. (B) Electric capacity.
(C) Electric energy. (D) Electric charge.
Ans:
C
Q.48 A
battery is a source of
(A) DC voltage. (B) 1 f AC voltage.
(C) 3 f AC voltage. (D) AC or DC voltage.
Ans: A
Q.49 Which DC
motors has approximately constant speed?
(A) Series motor. (B)
Shunt motor
(C) Cumulatively compound
motor (D) All of the above.
Ans:
B
Q.50 Which of the following bulbs will
have the least resistance?
(A)
220V, 60W (B)
220 V, 100 W
(C) 115 V, 60 W (D) 115V, 100 W
Ans:
D
Q.51 Resistance
of a wire is r ohms. The wire is
stretched to double its length, then its resistance in ohms is
(A) r/2 (B)
4r
(C) 2r (D)
r/4
Ans:
B
Q.52 An electric machine will have high
efficiency when
(A) input/output ratio is
low (B) reactive power is more
(C)
kWh consumption is low (D)
losses are low
Ans:
D
Q.53 Which type of loss is not common to
transformers and rotating machines?
(A) Eddy current loss (B)
Copper loss
(C)
Hysteresis loss (D) Windage loss
Ans:
D
Q.54 The
difference between the synchronous speed and the actual speed of an induction
motor is known as
(A)
Regulation (B)
back lash
(C) slip (D)
lag
Ans:
C
Q.55 In two wattmeter method of
power measurement, if one of the wattmeter shows zero reading, then it can be
concluded that
(A)
Power factor is unity (B)
Power factor is zero
(C)
Power factor is 0.5 lagging (D)
Power factor is 0.5 leading
Ans:
C
Q.56 Which of the following will remain
the same in all parts of a series circuit?
(A) Voltage (B) Current
(C) Power (D)
Resistance
Ans:
B
Q.57 Which single phase motor would you
select for a tape recorder?
(A) Reluctance motor (B) Hysteresis motor
(C) Synchronous motor (D) Universal motor
Ans:
B
Q.58 Under the condition of resonance, RLC
series circuit behaves as a,
(A) Purely resistive circuit. (B) Purely inductive circuit.
(C) Capacitive circuit. (D) Reactive
circuit.
Ans:A
Q.59 During
charging, the electrolyte of a lead acid cell becomes
(A) Stronger. (B) Weaker.
(C) Water. (D)
Diluted.
Ans:D
Q.60 As
compared to shunt and compound motors, series motor have the highest torque
because of its comparatively __________ at the start.
(A) Lower armature resistance. (B) Stronger series field.
(C) Fewer series turns. (D) Larger armature current.
Ans:D
Q.61 The
input of an ac circuit having p.f. of 0.8 lagging is 20 kVA. The power drawn by the circuit is __________
kW.
(A)
12. (B) 20.
(C) 16. (D) 8.
Ans: C
Q.62 The
voltage ratio of the transformer is given as
(A) 
(B) 
(C) 
(D)
Ans:A
Q.63 The
relationship between the frequency of ac wave and the time period is given by
(A) f = T (B) 
(C) 
(D)
Ans: C
Q.64 Which
of the following power plant has the maximum efficiency?
(A) Thermal (B)
Hydroelectric
(C) Nuclear (D)
Diesel
Ans:C
Q.65 Three capacitors of
value 
, 
and 32
are connected in series, the total capacitance will be
(A) 
. (B)
7.32 .
(C)
56 . (D)
32 .
Ans: A
Q.66 The following components are all active
components
(A)
a resistor and an inductor.
(B)
a diode, a BJT and an FET.
(C)
a capacitor, and an inductor.
(D)
an Opamp, a BJT and thermionic triode.
Ans: B
Q.67 In
forward mode NPN BJT, if we increase the voltage 
, the collector current increases
(A)
due to ohm’s law, higher causes higher
current.
(B)
due to base width decrease less carrier recombine in the
base region.
(C)
as the gradient of the minority carriers in the base region
becomes steeper.
(D)
due to both the reasons (B) and (C).
Ans: D
Q.68 The
barrier voltage 
in a junction
diode is the effect of
(A)
the p-side and n-side of the junction forming a battery.
(B)
the emf required to move the holes fast enough to have the
mobility equal to that of the electrons.
(C)
the recombination of charge carriers across the junction
leaving behind the opposite charged ions.
(D) the voltage needed to
make the semiconductor material behave as a conductor.
Ans:C
Q.69 An emitter follower has high
input impedance because
(A)
large emitter resistance is used.
(B)
large biasing resistance is used.
(C)
there is negative feedback in the base emitter circuit.
(D)
the emitter-base junction is highly reverse biased.
Ans: C
Q.70 In a differential
amplifier an ideal CMRR is
(A) infinity. (B)
zero.
(C) –1. (D)
+1.
Ans: A
Q.71 FET is advantageous in
comparison with BJT because of
(A) high input impedance. (B)
high gain-bandwidth product.
(C) its current
controlled behaviour. (D) high noise immunity.
Ans: A
Q.72 The emission of electrons in a vacuum
diode is achieved by
(A)
electrostatic field. (B) magnetic field.
(C) heating. (D) electron bombardment.
Ans: C
Q.73 The colour code of a
resistor of nominal value 
10% is
(A) Red, violet, red and silver. (B) Red, violet, yellow and gold.
(C)
Red, violet, orange
and silver. (D) Red, violet, red and gold.
Ans: A
Q.74 Capacitor
that can have the highest capacitance value is
(A)
Mica (B) Paper
(C) Electrolytic (D) Ceramic
Ans: C
Q.75 The
equivalent current-source representation for a voltage-source with open circuit
voltage 12 V and internal resistance 3 ohms is
(A) a current-source of strength 4A in shunt
with a resistance of 
.
(B) a current –source of
strength 4A in series with a resistance of 
.
(C) a current-source of
strength 4A in shunt with a resistance of 3 ohms.
(D) a current-source of strength 4A in shunt with a resistance
of 36 ohms.
Ans: C
Q.76 An
intrinsic semiconductor at absolute zero temperature
(A) has a large number of holes.
(B)
behaves like an insulator.
(C)
behaves like a metallic conductor.
(D)
has few holes and same number of electrons.
Ans:
A
Q.77 The
current flow through a Ge PN junction diode with a forward bias of 0.22 Volt
and a reverse saturation current of 1 mA at 
is around
(A)
6.3 A (B) 5.22 A
(C) 4 mA (D) 5.1 mA
Ans: B
Q.78 For
the operation of a depletion-type N-MOSFET, the gate voltage has to be
(A)
low positive (B) high positive
(C) high negative (D) zero
Ans: D
Q.79 The
typical operating voltage for LED’s ranges from
(A)
0.2 V to 0.6 V. (B) 6 V to 10 V.
(C) 1.5 V to 2.5 V. (D) 9 V to 10 V.
Ans: C
Q.80 Capacitors
for integrated circuits
(A)
cannot be made using diffusion techniques.
(B)
can be made with very high values of capacitance.
(C)
are always discrete components connected externally.
(D)
can be made using silicon dioxide as the dielectric.
Ans: D
Q.81 The
magnitude of variation in the output voltage for a 10 V regulated dc power
supply of 0.002% regulation will be
(A) 0.2 mV. (B)
0.002 mV.
(C) 0.02 mV. (D) 0.2
.
Ans: A
Q.82 For
the circuit shown in Fig.1, the output voltage is given by
(A) 
.
(B) 
.
(C) 
.
(D) 
.
Ans: C
Q.83 Which one of the
following statements is not true?
(A) Capacitance is a measure of a capacitor’s
capability to store charge.
(B)
A capacitor offers high impedance to ac but
very low impedance to dc.
(C) A capacitor is also
used as bypass capacitor.
(D)
Capacitors are used to couple alternating
voltages from one circuit to another and at the same time to block dc voltage
from reaching the next circuit.
Ans:
B
Q.84 A voltage source having an open-circuit voltage of 100 V and
internal resistance of 50
is equivalent to a current source
(A) 2A in parallel with 50. (B)
2A with 50 in series.
(C) 0.5A in parallel with
50. (D)
2A in parallel with 100.
Ans: A
Q.85 In a Zener diode large
reverse current is due to
(A)
collision. (B)
presence of impurities.
(C) rupture of bonds (D)
lower resistance in reverse biased
region.
Ans: D
Q.86 Ripple factor of a
full-wave rectifier without filter will be
(A) 0.2. (B)
0.48.
(C) 0.24. (D) 1.21.
Ans: B
Q.87 JFET has main drawback of
(A)
having low input impedance.
(B)
having high output impedance.
(C)
being noisy.
(D)
Having small gain-bandwidth product.
Ans: D
Q.88 A
UJT has
(A) stable negative
resistance characteristics.
(B)
low firing current.
(C)
use as a waveform generator.
(D)
all of these characteristics.
Ans: D
Q.89 For
thermionic emission
(A) a material with high
work function is preferable.
(B)
a material with low work function is preferable.
(C)
the work function of the material has no importance.
(D)
None of these is true.
Ans: B
Q.90 Ideal
operational amplifier has input impedance of
(A) 
. (B) infinity.
(C) zero. (D) 
.
Ans: B
Q.91 The
CE configuration amplifier circuits are preferred over CB configuration
amplifier circuits because they have
(A) lower amplification factor.
(B) Larger amplification factor.
(C) high input resistance
and low output resistance.
(D) none of these.
Ans: B
Q.92 The most commonly used type of electron emission in electron tubes
is
(A) Photo-electron emission. (B) Thermionic emission.
(C) Field emission. (D) Secondary emission.
Ans: A
Q.93 The colour band sequence of a
resistor is grey, Blue, gold, and gold.
The range in which its value must lie so as to satisfy the tolerance
specified is between
(A) 
and 
(B) 
and 
(C) 
and 
(D) 
and 
Ans: D
Q.94 A device whose characteristics are very close to that of an ideal
current source is
(A) a gas diode. (B)
a BJT in CB mode.
(C) a BJT in CE mode. (D) a triode.
Ans: C
Q.95 In
an N-type semiconductor, the concentration of minority carriers mainly depends
upon
(A) the doping technique. (B)
the number of donor atoms.
(C) the temperature of the material (D) the quality of the intrinsic material,
Ge or Si.
Ans: B
Q.96 When
forward bias is applied to a junction diode, it
(A) increases the
potential barrier.
(B) decreases the
potential barrier.
(C) reduces the majority-carrier current to zero.
(D) reduces the minority-carrier current to zero.
Ans: B
Q.97 The
theoretical maximum efficiency of a Bridge rectifier circuit is
(A) 48.2%. (B) 81.2%.
(C) 82%. (D) 40.6%.
Ans: B
Q.98 The
input resistance of a common-collector configuration will be of the order
of
(A) 
(B) 
(C) 
(D) 
and above
Ans: D
Q.99 A
switching voltage regulator can be of the following type:
(A) step-down (B) step-up
(C) inverting (D) none of these
Ans: A
Q.100 A
UJT contains
(A) four pn junctions (B) three pn junctions
(C) two pn junctions (D) one pn
junction
Ans: D
Q.101 The foundation on which an IC is built is called
(A) an insulator. (B) a base.
(C) a wafer. (D) a
plate.
Ans: C
Q.102 X-ray tubes make
use of
(A) Thermionic emission.
(B) Secondary emission.
(C) High field emission. (D) Photoelectric
emission.
Ans: C
Q.103 Which of the
following components are all active components?
(A) A resistor and a capacitor.
(B) A microphone, a LCD and a Thyratron.
(C) An electric bulb, a
transformer and a varactor diode.
(D) An SCR, a vacuum
diode and an LED.
Ans: D
Q.104 Doping materials
are called impurities because they
(A)
Decrease the number of charge carriers.
(B) Change the chemical
properties of semiconductors.
(C)
Make semiconductors less than 100 percent pure.
(D)
Alter the crystal structures of the pure semiconductors.
Ans: B
Q.105 Avalanche
breakdown is primarily dependent on the phenomenon of
(A) Collision (B) Doping
(C) Ionisation (D) Recombination
Ans: D
Q.106 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.
Ans: D
Q.107 The
main reason why electrons can tunnel through a P-N junction is that
(A) They have high
energy.
(B) Barrier potential is
very low.
(C)
Depletion layer is extremely thin.
(D)
Impurity level is low.
Ans:
C
Q.108 If a change in base current does not
change the collector current, the transistor amplifier is said to be
(A) Saturated. (B) Cut-off.
(C) Critical. (D) Complemented.
Ans: A
Q.109 The
extremely high input impedance of a MOSFET is primarily due to the
(A) Absence of its
channel.
(B) Negative gate-source
voltage.
(C) Depletion of current
carriers.
(D) Extremely small
leakage current of its gate capacitor.
Ans: A
Q.110 After
firing an SCR, the gating pulse is removed.
The current in the SCR will
(A) Remains the same. (B) Immediately fall to zero.
(C) Rise up. (D) Rise a little and then fall to zero.
Ans: A
Q.111 An
inverting operational amplifier has 
and 
. Its scale
factor is
(A) 1000. (B) 
.
(C) 
. (D) 
.
Ans: B
Q.112 In the context of IC
fabrication, metallisation means
(A) Connecting metallic wires.
(B) Forming
interconnecting conduction pattern and bonding pads.
(C) Depositing 
layer.
(D) Covering with a
metallic cap.
Ans: B
Q.113 The colour band sequence of a resistor is yellow, violet,
orange and gold. The range in which its
value must lie so as to satisfy the tolerance specified is between
(A) 
and 
(B) 
and 
(C) 
and 
(D) 
and 
Ans: D
Q.114 A device whose
characteristics are very close to that of an ideal voltage source is
(A) a vaccum diode. (B) a DIAC.
(C) a zener diode. (D) a FET.
Ans: C
Q.115 The
forbidden energy gap in semiconductors
(A) lies just below the
valance band
(B) lies just above the
conduction band
(C) lies between the
valence band and the conduction band
(D) is the same as the
valence band
Ans: C
Q.116 The
barrier potential for a Ge PN junction is
(A) 0.6V. (B) 0.3V.
(C) 0.1V. (D) 0.5V.
Ans: B
Q.117 The
ripple factor of a power supply is a measure of
(A) its voltage regulation. (B) its diode rating.
(C) purity of power output. (D) its filter efficiency.
Ans: C
Q.118 In
a BJT, if the emitter junction is reverse-biased and the collector junction is
reverse-biased, it is said to operate in
(A)
in active region (B) in saturation region
(C) in cut-off region (D)
none of the above
Ans: C
Q.119 In
the switching type of voltage regulators, the power efficiency will be of the
order of
(A) 50% or less. (B) 60%.
(C) 40% or more. (D) 90% or more.
Ans: D
Q.120 The
resistance between bases of a UJT is typically in the range of
(A) 2 to 3 K
(B) 5 to
10 K
(C) 15 to 20 K (D) 18 to 20 K
Ans: B
Q.121 The
quantity that serves as a figure of merit for a DIFF AMP is
(A) slew rate. (B) bandwidth.
(C) input bias current. (D) CMRR.
Ans: D
Q.122 Practical range of
resistance values obtainable with base diffused resistors is
(A) 10 to 1 K (B) 20 to 30 K
(C) 5 to 3 K (D) 20K to 50 K
Ans: D
Q.123 The colour band sequence of a resistor is Yellow, Violet, Orange and Gold. The range in which its value must lie so as
to satisfy the tolerance specified is between
(A) 
and 
(B) 
and 
(C) 
and 
(D) 
and 
Ans:
B
Q.124 With increasing temperature, the resistivity of an intrinsic
semiconductor decreases. This is because, with the increase of temperature
(A)
The carrier concentration increases but the mobility of
carriers decreases.
(B) Both the carrier
concentration and mobility of carriers decreases.
(C) The carrier concentration decreases, but the mobility of carriers increases.
(D) The carrier concentration remains the same but the mobility of carriers
decreases.
Ans: A
Q.125 At room temperature of 25ºC, the barrier Potential for Silicon is
0.7V. Its value at 0ºC will be
(A)
0.7 V. (B) 0.65 V.
(C) 0.75 V. (D) 0.55 V.
Ans: C
Q.126 Which
of the following is a unipolar device?
(A) P-N junction diode (B) Zener diode
(C) Tunnel diode (D) Schottky diode
Ans: D
Q.127 On
applying a Positive voltage signal to the base of a normally biased N-P-N CE
transistor amplifier
(A) Base current will
fall.
(B) Collector current will fall.
(C) Emitter current will fall.
(D) Collector voltage will become less positive.
Ans: D
Q.128 An N-channel
JFET has Pinch-off Voltage of
VP = – 4V and
given that VGS =
–1V, then the minimum VDS for the device to operate in the Pinch-off
region will be
(A) +1V (B) +3V
(C) +4V (D) +5V
Ans: B
Q.129 The
extremely high input impedance of a MOSFET is Primarily because of
(A) Absence of its channel
(B) Depletion of current carriers
(C) Extremely small
leakage current of its gate capacitor
(D) Negative VGS
Ans: A
Q.130 When
two identical SCRs are placed back-to-back in series with a load and if each is
fired at 90º, then the voltage across the load will be
(A) 
(B) Zero
(C) 
(D) 
Ans: A
Q.131 
In the
differentiating circuit shown in Fig.1, the function of resistor R1
is to
(A)
Enable the circuit to approach ideal differentiator
(B) Maintain high input
impedance
(C) Eliminate high
frequency noise spikes
(D) Prevent oscillations
at high frequencies
Ans:C
PART
– II
DESCRIPTIVES
Q.1 Define the
following and give their units of measurement:-
(i) Resistance. (ii) Electric Potential.
(iii) Electric current. (2 x 3)
Ans:
(i) Resistance: - The opposition offered by a substance to the flow of
electric current. The unit of resistance is ohm and given by the symbol W.
(ii) Electric Potential: -The capacity of a charged body to
do work. The unit of electric potential is volt (V).
(iii) Electric Current: - The flow of free electrons constitutes
electric current. The unit of electric current is called ampere (A).
Q.2 Give reasons, why, starters are required for starting a motor. (6)
Ans:
In case of DC motors, when the motor is at rest, the induced emf in the
armature is zero. Consequently, if full voltagte
is applied across the motor terminals, the armature will draw heavy current
because the armature resistance is relatively small. This heavy starting
current will blow out the fuses and it may also damage the armature winding due
to excessive heating effect. Excessive voltage
drop will occur in the lines to which the motor is connected. To avoid this
havy current at start, a variable resistance is connected in series with the
armature called starting resistance or starter, thus the armature current is
limited to a safe value. Once the motor picks up speed, emf is built up and the
resistance is gradually reduced. The whole resistance is taken out of circuit
when the motor attains normal speed. The starter contains the protective device
as overload protection coil (or relay), which provides necessary protection to
the motor against overloading. In case of induction motors, the current drawn by the
motor from the supply mains depends upon the rotor current. This current is
very large as compared to its full load current. So when squirrel cage motors
are directly connected to the supply mains, it draws very large current from
the mains which effects in two ways – it produces very lare voltage drop in the
distribution lines and affects the voltage
regulation of the supply system. It causes disturbance of the other motors
connected to the same lines. Hence these motors should be started by means of
some starting device known as starter.
Q.3 Why single phase induction motor are not self starting? (8)
Ans:
Single phase induction motor
Single phase induction motors set up
pulsating torque, instead inidirectional and continuous torque. This is because
force experienced by the current carrying conductors depends upon the direction
of current and the magnitude of the flux. As an alternating current, direction
and magnitude is changing so varying force is experienced by the conductors.
Once in one direction, say clockwise, then in other direction, i.e.
anticlockwise. The change is so quick that neither it moves in a clockwise
direction nor in an anticlockwise direction. However, if
the motor is rotated by some means
in any direction it will continue to rotate, even though the starting means
have been withdrawn due to resultant torque in that direction. Hence we can say
single phase motors are not self starting and certain means have to be used for
starting single phase induction
motors.
Q.4 State and explain Maximum power transfer
theorem. Also give its applications. (8)
Ans:
Maximum power transfer theorem deals with transfer of maximum power
from the source to load. This theorem states the relationship between load
resistance and internal resistance of the source for maximum power transfer
from source to load. This condition is also referred to as impedance matching.
Impedance matching is very important in electronic and communication circuits
so as to obtain maximum power. Power transferred in an ac circuit is maximum
when RL (load resistance) = Ri (internal resistance of
the source). This theorem is useful in electronic circuits where maximum power
transfer is usually desirable such as Public address System. Also this theorem
is applicable in starting of car engines.
Q.5 Based on the core
construction, explain the two types of transformer. (8)
Ans:
Two types of core construction are adapted for transformers-core
type and shell type. In the core type of construction, the LV and HV coils are interleaved to reduce the
leakage flux. Half of LV
and half HV are wound on each limb of the core type transformer. For economical
insulation, the LV
coils are placed next to the core and HV coils are placed on the outside. In
the shell type transformer reduced leakage flux is obtained by sandwiching the LV and HV coils. The LV coils are sandwiched
between the sections of the HV coil. Both the coils are placed on the central
limb of the core.
HV LV LV
HV
Windings
Windings Core
CORE TYPE
Sandwiched LV
HV windings
Core
SHELL TYPE
Q.6 Explain the word back emf
used for a dc motor and highlight its significance. (6)
Ans:
The rotating conductors of
the armature between the poles of magnet, in a DC motor, cut the magnetic flux,
thereby developing an induced emf, which opposes the applied / external
voltage. The induced emf set up in the coil of DC motor opposing the current
flowing through the conductor, when the armature rotates, is called back emf.
The value of the back emf depends upon the speed of rotation of the armature
conductors.When the motor starts, the back emf in the beginning is zero.
Consequently, the current flowing through the armature conductors is very
large, since the armature resistance is very small. This current is very large
and may damage the motor. In order to avoid this , additional resistance is
connected in series with the armature to limit the current at starting.
Q.7 Write a note on selection of motors for specific
engineering applications. (8)
Ans:
Selection of motors for different engineering
applications:
Series motors are used in cranes, pumps, trains, trolleys, etc. due to
its very high starting torque and variable speed.
Shunt
motors run practically at constant speed at almost all loads. Such motors are
used in lathes, drills, printing press and for driving pumps.
Cumulative
compounds are used in machine tools, coal cutting machines, punch presser,
crushers, compressor, rolling mills, elevators where very high starting torque
is required and adjustable varying speed is required.
Three
phase induction motors are used for high power applications such as in
industries.
Single-phase
motors are used in most homes, offices and rural areas.
Fractional kilowatt
motors are used in fans, refrigerators, mixers, vacuum cleaners, washing machines, and small farming appliances.
Shaded pole motors are used in small fans,
convectors, vending machines, photocopying machines, advertising displays.
Q.8 Explain the
principle of a transformer. (6)
Ans:
Working Principle of a transformer: The basic principle of a transformer is electromagnetic induction.
It consists of two separate windings placed over the laminated silicon steel
core. The winding to which ac supply is connected is called
primary
winding and the winding to which load is connected is called a secondary
winding. When ac supply of voltage v1 is connected to primary
winding, an alternating
flux
is set up in the core. This alternating flux when links with secondary winding,
an emf is induced in it and is called mutually induced emf. The direction of
this induced emf is opposite to the applied voltage v1. The same
alternating flux also links with the primary winding and produces self-induced
emf e1. Although there is no electrical connection between primary
and secondary winding, but electrical power is transferred from primary circuit
to the secondary circuit through mutual flux.
The induced emf in the primary and secondary winding depends upon the rate of
change of flux linkages (i.e. N df/dt).
The rate of change of flux (df/dt) is same for both primary and
secondary. Therefore, the induced emf in the primary is proportional to number
of turns of the primary winding (e1 µ N1)
and in the secondary it is proportional to the number of turns
of the secondary windings (e2 µ N2).
In case N2 > N1 the transformer is step up and in
case N2 < N1 the
transformer is step down.
Q.9 Explain the term slip in an
induction motor. (4)
Ans:
Induction motor rotor always rotates at a speed less than synchronous
speed. the difference between the flus speed (Ns) and the rotor speed (N) is
called slip. It is usually expressed as a percentage of synchronous speed (Ns)
and represented by the symbol ‘S’.
Q.10 Differentiate
between the ‘squirrel cage’ and ‘phase wound’ rotor types of induction motors. (8)
Ans:
|
Squirrel cage
rotor
|
phase wound rotor
|
|
Almost constant speed but decreases slightly with
increased load.
|
Speed decreases more rapidly than squirrel cage
motor.
|
|
Starting torque is somewhat less, but running
torque is good.
|
Starting torque is about three times the full load
torque. Running torque is also good.
|
|
Starting current is about 5-6 times the full load
current.
|
Starting current is about 2 times the full load
current.
|
|
Speed control is done by changing poles.
|
Speed control is done by changing external
resistance of rotor circuit.
|
|
Power factor is about 0.7 to 0.8.
|
Power factor is about 0.8 to 0.9.
|
|
Cost of fabrication is low.
|
Cost of fabrication is high.
|
|
Maintenance cost is very low.
|
Maintenance cost is high (because of extra
resistance).
|
|
Application- lathes, drills, printing machines,
blowers.
|
Applications – lifts, cranes, where high starting
torque is needed.
|
Q.11 Explain application and advantages of storage batteries? (10)
Ans:
Applications
of storage batteries: Because of the fact that storage batteries are portable,
economical, efficient and reliable source of d.c. supply, they have a number of
commercial applications. Some of the important applications are:-
i)
These are used for starting, ignition and lighting of
automobiles, aircrafts etc.
ii)
For lighting on steam and diesel railways trains.
iii)
As a source of power supply in telephone exchange,
laboratories and broad casting stations.
iv)
Used at generating stations and substations for operation of
protective devices and for emergency lighting.
v)
For emergency lighting at hospitals, banks, rural areas
where electricity supply is not possible.
Advantages of storage batteries
Following are the
advantages of using storage batteries:-
i)
It is the highest and most
efficient device for the storage of energy in portable form.
ii)
The stored energy is available
immediately because there is no lag of time for delivering the stored energy.
iii)
The energy storing in the
battery may be done at any convenient rate and delivered at any other rate.
iv)
It is very reliable source for
supply of energy.
v) The energy can be drawn at a
fairly constant rate.
Q.12 How
does a three-phase synchronous motor differ from a three-phase induction
motor? Give a few applications of
synchronous motors. (8)
Ans:
Synchronous motor is not self-starting and requires starting
devices. It runs only at synchronous speed. So the speed is constant. It has to
be synchronized. It can be operated under a wide range of p.f. both leading and
lagging. The change in the applied voltage, does not cause much effect on its
torque. It is more costly and complicated. The breakdown torque is
approximately proportional to applied voltage. Where as induction motors are
self-starting and do not require any starting devices. Its speed decreases with
load and it has not to be synchronized. It always runs at lagging p.f., whose
value becomes very low at light loads. The change in the applied voltage causes
much effect on its torque. It is simple, rugged and low in cost. The breakdown
torque depends on the square of the applied voltage.
Applications: Synchronous motors are used to
improve power factor of large industries, in substations. It is used to control
the voltage at the end of transmission line by varying their excitation. Also
used in textile mills, cement factories, mining industries and rubber mills for
power applications. They are also used to drive constant speed equipment such
as centrifugal pumps, centrifugal fans, air compressors, motor-generator sets,
and blowers.
Q.13 Explain the different methods
for the starting of a synchronous motor. (6)
Ans:
Starting methods: Synchronous motor can
be started:
(1) by means of an auxiliary
motor: In this case, an auxiliary motor rotates the rotor of synchronous motor.
Then rotor poles are excited due to which the rotor field is locked with the
stator revolving field and continuous rotation is obtained.
( 2) By providing damper winding: In this case, bar conductors are
embedded in the outer periphery of the rotor poles and are short-circuited with
the short-circuiting rings at both sides. The machine is started as a squirrel
cage induction motor first. When it picks up speed, excitation is given to the
rotor and the rotor starts rotating continuously as the rotor field is locked
with stator revolving field.
Q.14 Name
the types of motors used in: vacuum cleaners, phonographic appliances, vending
machines, refrigerators, rolling mills, lathes, power factor improvement and
cranes. (8)
Ans:
Motors used are: -
Vacuum cleaners- Universal motor.
Phonographic appliances – Hysteresis motor.
Vending machines – Shaded pole motor.
Refrigerators – Capacitor split phase
motors.
Rolling
mills – Cummulative motors.
Lathes
– DC shunt motors.
Power
factor improvement – Synchronous motors.
Cranes
– DC series motors.
Q.15 Name
the different types of 1-phase A.C motors. Give some important application of
these motors. (8)
Ans:
Different types of
1-phase AC motors and their applications:
i)
Single phase Induction motor:- These motors are of different
types
1.
Capacitor start single phase induction motor is generally
used for fans, refrigerator, washing machines, blowers and centrifugal pumps
etc.
2.
Split phase induction
motor is used in bench grinder, drills etc.
3.
Shaded pole single phase induction motor is used in electric record
players, slide projectors etc.
ii)
Due to high efficiency and good speed of motor, universal
motor is used for vacuum cleaners, electric type writers etc.
iii)
Reluctance motor is used in electric clocks due to constant
speed.
iv) Hysteresis
motor is used in record player, tape recorders and clocks due to
steady hysteresis torque.
Q.16 With
the help of a neat sketch explain the various parts of a nuclear reactor. (8)
Ans:
Parts of a nuclear reactor: The fission of a nuclear material
is carried out in a nuclear reactor.
Fuels: - used in the reactor have
some components of 238U. In advanced gas cooled reactor enriched
uranium dioxide fuel in pellet form encased in stainless steel cans is used.
The fuel could be in the form of rods enclosed in stainless steel.
Moderators: - are used to slow down
the neutrons. Commonly used moderators are graphite, light water and heavy
water.
Coolants: - these remove the heat
generated in the core by circulation and transfer it outside for raising steam.
Common coolants are light ordinary water, heavy water, CO2 gas and
also metals like sodium or sodium- potassium alloy in liquid form.
Control Materials: - control is
achieved by means of a neutron absorbing material. The control elements are
commonly located in the core in the form of either rods or plates. The most
commonly used neutron absorber is boron.
Reactor Shield: - surrounding the
reactor core with a radiation shield makes provisions for health protection.
This is also called biological shield.
The energy given off in a reactor
appears in the form of heat, which is removed by a gas or liquid coolant. The
hot coolant is then used in a heat exchanger to raise steam. If the coolant is
ordinary water, steam could be raised inside the reactor. This steam runs a
turbo generator for producing electric energy.
Q.17 Define the
following terms:
(i) Diversity Factor. (ii)
Annual Load Factor.
(iii) Capacity Factor. (6)
Ans:
Diversity Factor = S individual maximum demands of consumers
Maximum load on the system
Annual load factor
= Total annual load (Mwh)
Annual peak load (MW) X 8760 h
Capacity factor =
Actual annual generation (Mwh)
Maximum rating (Mw) X 8760
h
Q.18 Write note on Energy storage. (7)
Ans:
Energy storage: - Large-scale storage of energy, which can be quickly converted to
electrical form, can help fast changing loads. The options available are pumped
storage, compressed air storage, heat storage, hydrogen storage and batteries.
Pumped storage: - In areas where sufficient hydrogenation is not
available, peak load may be handled by means of pumped storage. This consists
of upper and lower reservoirs and reversible turbine-generator sets, which can
also be used as motor –pump sets. The upper reservoir has enough storage for
full load generations.
Compressed air storage: - Compressed air can be stored in natural underground caverns or
old mines. The energy stored equals the volume of air multiplied by pressure.
At times of need, this air can be mixed with gas fuel to run a gas turbine.
Heat storage:- Water with good specific and latent heat has been
used. In generating stations, boilers can be kept ready on full steam for the
turbine to pick up fast rising load. Boiler steam, when not in use can heat
feed water for boilers in the station.
Secondary
batteries: - These have
possible use in local fluctuating loads, electric vehicles and back up for wind
and solar power. There are a number of batteries like lead acid cell, nickel
cadmium cell and sodium sulphur cell.
Q.19 State the following:
(i)
Thevenin’s Theorem.
(ii)
Norton’s Theorem.
(iii)
Maximum power transfer theorem.
(iv)
Kirchoff’s laws. (8)
Ans:
(i)Thevenin’s Theorem states that the current flowing
through a load resistance RL connected across any two terminals A
and B of a linear, active bilateral network is given by Voc / (Ri
+ RL) where Voc is the open circuit voltage (ie. the
voltage across the two terminals when RL is removed) and Ri is
the internal resistance of the network as viewed back into the open circuited
network from terminals A and B with all voltage sources replaced by their
internal resistance (if any) and current sources by infinite resistance.
(ii)Norton’s Theorem states that any two – terminal
active network containing voltage sources and resistances when viewed from its
output terminals, is equivalent to a constant current source and parallel
resistance. The constant current is equal to the current which would flow in a
short circuit placed across the terminals and parallel resistance is the
resistance of the network when viewed from these open circuited terminals after
all voltage and current sources have been removed and replaced by their
internal resistances.
(iii)Maximum power
transfer theorem: A
resistive load will abstract maximum power from a network when the load
resistance is equal to the resistance of the network as viewed from the output
terminals, with all energy sources removed leaving behind their internal
resistances.
(iv)Kirchoff’s first law states that the algebraic sum of
all currents meeting at a point is zero.
S I =
0.
Kirchoff’s second law
states that, in a closed circuit, the algebraic sum of all the emf’s plus the
algebraic sum of all the voltage drops (i.e. product of current and
resistances) is zero.
S I R + S
emf = 0.
Q.20 Write short notes
on
(i)
Different losses in transformer. (8)
(ii)
Resonance in R-L-C series circuit. (8)
Ans:
(i) Different losses in transformer
There are two types of
losses occurring in transformer:
1.
Constant losses or Iron losses:These
losses occur in the core, therefore known as core losses or iron losses. There
are two types of iron losses, one is the eddy current loss and other is
hysteresis loss. These losses depend upon the supply voltage,
frequency, core material and its construction. As long as supply voltage and frequency is constant, these losses remain
the same whether the transformer is loaded or not. Hence core losses are known
as constant losses.
2.
Variable losses or copper losses: When the transformer is loaded, current flows in primary and secondary
windings and there is loss of electrical energy due to the resistance of the
primary winding and secondary winding. If current in primary is
I1 amp and in
secondary is I2 amp and primary resistance is r1 and
secondary resistance is r2
ohms then total copper losses are equal to 
. In fact these losses are
winding material losses; therefore, these are known as copper
losses.These losses depend upon the loading conditions of the transformers.
Therefore, these losses are also called as variable losses.
(ii) Resonance in R-L-C
Series circuit: A circuit in which the two components L and C are connected in
series with each other across a variable frequency a.c. source is called a
series resonance circuit as shown in fig.9(a)
XL 
XL
O P f
XC
XC [at 
]
XC
Fig 9(a) Fig 9(b)
If the frequency
of the voltage source is varied, then the value of inductive reactance XL and
capacitor reactance XC at a particular frequency can be given as 
The
total impedance of the circuit will be given as
Where R is the resistance of the
circuit which may be resistance of the coil.
It is clear from
the above equation that XL increases linearly with frequency whereas
XC decreases inversely with frequency as shown in the fig. 9b. There
will be a particular frequency at which XL is equal to XC.
This frequency is called resonance frequency (fr). At this frequency
Z = R and circuit will behave as purely resistive circuit.
At resonant
frequency 
Q.21 What are the
different types of D.C motors? Give their applications? (8)
Ans:
Different type of DC motors
and their applications are as follows:-
1.
Shunt motors: Shunt motor is a fairly
constant speed motor though its starting torque is not very high. Hence it is suitable for constant
speed drive which do not require very high starting torque such as pumps,
blowers, fan, lathe machines, tools’ belt or chain conveyor etc.
2.
Service motors: This motor develops a
high starting torque & its sped is inversely proportional to the loading
conditions i.e. when lightly loaded, the speed is high and
when heavily loaded, it is low. Therefore, motor is used in lifts,
cranes, traction work, coal loader and coal cutter in coal mines etc.
3.
Compound motors: This motor has a variable
speed and high starting torque. It can also run at NIL loads without any
danger. This motor will therefore find its application in loads having high
inertia load or requiring high intermittent torque such as elevators, conveyor,
rolling mill, planes, presses, shears and puches, coal cutter and winding
machines etc.
Q.22 Derive
the emf equation of a transformer. (6)
Ans:
When a sinusoidal voltage is applied to the primary winding of a
transformer, a sinusoidal flux as shown in the fig. is set up in the iron core
which links with the primary and secondary winding. Let j m
= maximum value of flux in wb, f-= supply frequency in Hz. N1= No.
of turns of the primary and N2 = No. of turns of secondary. As shown
in the fig. the flux changes from + jm to - jm in
half a cycle ie. 1/2f seconds.
Average rate of change of flux = jm – (-jm) = 4 jm f wb/s
1/2f
Now, the rate of
change of flux per turn is the average induced emf per turn in volts.
Therefore, average
induced emf / turn = 4 jm f volts.
For a sinusoidal
wave, R.M.S. value / Average value = Form factor = 1.11
Therefore, R.M.S.
value of emf induced / turn, E = 1.11 X 4 jm
f
volts.
Therefore, R.M.S.
value of emf induced in primary, E1 = (emf induced/ turn) X No. of
primary turns. = 4.44 N1 f jm volts.
Similarly R.M.S.
value of emf induced in secondary, E2 = (emf induced/ turn) X No. of
secondary turns. = 4.44 N2 f jm volts.
Q.23 What are the different methods of measurement of power in 3-phase
circuit. Explain two wattmeter method in brief. (8)
Ans:
Following methods are available for measuring power in 3-phase circuit
i)
Three wattmeter method
ii)
Two wattmeter method
iii)
One wattmeter method
Two wattmeter method: In this method
Two wattmeters are used for power measurement. As shown in fig.3a, the current
coils of two wattmeters are inserted in any two line and the
voltage coil of each joined to
the 3rd line. It can be proved that the sum of the instantaneous
power indicated by W1 and W2
gives the instantaneous power absorbed by the three loads L1, L2
& L3.
Fig 3a Fig 3b
This method can be applied to star
connected as well as delta
connected load. Considering star connected load
Instantaneous
current through 
Instantaneous Voltage across 
Instantaneous Power read by 
Instantaneous
current through 
Instantaneous Voltage across 
Instantaneous Power read by 
Therefore 
= 
Now 
by kirchoffs point law
Where p1 is the power absorbed by L1, p2
that absorbed by L2 and p3 that absorbed by L3
W1 + W2
= total power absorbed
Hence in two
wattmeter method the sum of readings of two wattmeters gives the total power
absorbed by 3-Ф circuit.
Q.24 Explain the process of commutation in a dc
machine. Explain what are inter-poles and why they are required in a dc
machine. (8)
Ans:
Commutation: When an armature
coil moves under the influence of one pole- pair, it carries constant current
in one direction. As the coil moves into the influence of the next pole- pair,
the current in it must reverse. This reversal of current in a coil is called
commutation. Several coils undergo commutation simultaneously.
The
reversal of current is opposed by the static coil emf and therefore must be
aided in some fashion for smooth current reversal, which otherwise would result
in sparking at the brushes. The aiding emf is dynamically induced into the
coils undergoing commutation by means of compoles
or interpoles, which are series
excited by the armature current. These are located in the interpolar region of
the main poles and therefore influence the armature coils only when these
undergo commutation.
Q.25 What are the different
network theorems? State Thevenin’s theorem. (6)
Ans:
There are a number of
theorems to solve electrical networks. Some of the important network theorems
are:
i.
Thevenin’s Theorem
ii.
Norton’s Theorem
iii.
Super Position Theorem
iv.
Maximum Power
Transfer Theorem
Thevenin’s Theorem: It states that any two terminal linear networks containing a
number of e.m.f. sources and impedances may be replaced by an equivalent
circuit consisting of a voltage generator (Vth)
in series with an impedance (Rth). This circuit will be called as
Thevenin’s equivalent circuit.
Fig
2a
Where
Vth – Thevenin’s equivalent voltage (open circuit voltage
across terminal AB)
Rth – Thevenin’s equivalent
impedance (Resistance between terminal AB when all
emf sources in the
network are reduced to zero.)
Q.26 Explain the operation of a three phase induction motor. (6)
Ans:
Operation of a 3- phase
induction motor: When the 3- phase supply is given
to the stator of a 3- phase wound induction motor, a rotating field is set-up
in the stator. At any instant the magnetic field set up by the stator is shown
in fig. An arrowhead Fm marks
the direction of
resultant
field. Let this field be rotating in an anti- clockwise direction at an angular
speed of ws
radians per second ie. Synchronous speed. The stationary rotor conductors cut
the revolving field and due to electromagnetic induction an emf is induced in
the rotor conductors. As the rotor conductors are short circuited, current
flows through them in the direction as marked in the fig. Rotor current
carrying conductors set up a resultant field Fr. This tries to come
in line with the stator main field Fm. Due to this an
electromagnetic Te is developed in the anticlockwise direction.
Thus, the rotor starts rotating in the same direction in which stator field is
revolving.
Q.27 Explain
the working principle of operation of a single phase transformer. (6)
Ans:
Working principle of operation of a single
phase transformer: When AC supply is
given to the primary winding, a current will start flowing in the primary. This
will set up flux. This flux is linked with primary and secondary windings.
Hence voltage is induced in both the windings. Now, if load is connected to the
secondary side, then current will start flowing in the load in the secondary
winding, resulting in flow of additional current in the secondary winding.
Hence according to Faraday’s laws of electromagnetic induction, emf will be
induced in both the windings. The voltage induced in the primary winding is due
to its self inductance and known as self induced emf and according to Lenze’s
law it will oppose the cause i.e. supply voltage hence called as back emf. The
voltage induced in secondary coil is known as mutually induced voltage. Hence
transformer works on the principle of electromagnetic induction.
Q.28 Define the following terms:-
Reliability, Maximum
demand, Reserve-generating capacity, Availability (operational). (8)
Ans:
Reliability: It is
measure by the power system’s ability to serve all power demands without
failure over long periods of times.
Maximum Demand: It is the greatest demand of load on the power station during a
given period.
Reserve generating capacity: Modern
generating plants are stressed to limits of temperature and pressure to reduce
the overall power costs. Therefore, extra generation capacity must be installed
to meet the need of scheduled downtimes for preventive maintenance.
Availability: As the
percentage of the time a unit is available to produce power whether needed by
the system or not.
Q.29 What are
the disadvantages of low power factor? How can it be improved? (8)
Ans:
Disadvantages of low power factor:
1)
Line losses are 1.57 times those at unity power factor.
2) Larger generators and
transformers are required.
3) Low lagging power
factor causes a large voltage drop, hence extra regulation equipment is
required to keep voltage drop within prescribed limits.
4) Greater conductor
size: To transmit or distribute a fixed amount of power at fixed voltage, the
conductors will have to carry more current at low power factor. This requires a
large conductor size.
Methods of improving power factor:
1) Static Capacitors: The
static capacitors are connected in parallel with the load operating at lagging
power factor.
2) A synchronous motor
takes a leading current when over excited and therefore behaves like a
capacitor.
3) Phase advancers: Are
used to improve the power factor of induction motors. It provides exciting
ampere turns to the rotor circuit of the motor. By providing more ampere-turns
than required, the induction motor can be made to operate on leading power
factor like an over-excited synchronous motor.
Q.30 Explain
why the following motors are used in the particular applications indicated
against them. Synchronous motors – power-factor improvement, DC shunt motors –
lathes, DC series motors- lifts and cranes, Cumulative compound motor – rolling
mills. (8)
Ans:
Synchronous motors – power factor
improvement- the power factor of the motor can be controlled over a wide range
by adjusting its excitation. Since it can be operated under a wide range of
power actors both lagging and leading by its field current it is used in power
factor improvement.
DC shunt motors – Lathes - shunt motor is
almost constant speed motor. It is used where the speed between no loads to
full load has to be maintained almost constant.
DC series motors- lifts and cranes –
series motor is a variable speed motor. It is used where high torque is
required at the time of starting to accelerate heavy loads.
Cumulative compound motor- rolling mills –
Unlike a series motor, it has a finite no-load speed but speed drops sharply
relieving the peak power drawn from the mains as the billet is passed through
rolls.
Q.31 What are
the advantages and disadvantages of high voltage DC transmission? (8)
Ans:
Advantages of the high voltage DC transmission are:
·
These systems are economical
for long distance bulk power transmission by overhead lines.
·
There is greater power per
conductor and simpler line construction.
·
Ground return is possible.
·
There is no charging current.
·
The voltage regulation problem
is much less serious for DC since only the IR drop is involved (IX =0).
·
There is reversibility and
controllability of power flow through a DC link.
·
The DC line is an asynchronous or flexible
link and it can interconnect two rigid systems operating at different
frequencies.
·
Smaller amount of right of way
is required. The distance between two outside conductors of a 400kV AC line is
normally 20m, whereas the same between a corresponding DC line is roughly half.
·
Line losses are smaller.
·
There is considerable
insulation economy. The peak voltage of the 400 kV AC line is Ö2 X 400 = 564kV. So the AC line requires more insulation between the
tower and conductors, as well as greater clearance above the earth as compared
to corresponding 400 kV DC line.
The disadvantages of high
voltage DC transmission are:
·
The systems are costly since
installation of complicated converters and DC switchgear is expensive.
·
Converters require considerable
reactive power.
·
Harmonics are generated which
require filters.
·
Converters do not have overload
capability.
·
Lack of HVDC circuit breakers
hampers multiterminal or network operation. There is nothing like DC
transformer which can change the voltage level in a simply way.
·
Reactive power required by the
load is to be supplied locally as no reactive power can be transmitted over a
DC link.
Q.32 Explain
the following terms – Busbar, load, system, outage. (8)
Ans:
Busbar – It is a solid electrical connection made of aluminium or
copper bars connecting various power system components like generators,
transformers, lines, loads.
Load – It is a device or devices which draw electrical power
from the busbar to do useful work for the consumers, drive motors and other
processes in industry, in domestic load it is lighting, refrigeration, small
electrical appliances.
System – The complete electrical networks, prime movers,
generators, transformers, lines and loads.
Outage – Removal of a circuit either
deliberately or inadvertently.
Q.33 State a
few applications of solar energy. Also explain the structure of a solar
photovoltaic cell. (2+6)
Ans:
Applications of
solar energy: Solar energy is used in water
heating, solar drying, desalination, industrial process heating and passive /
active heating of buildings. Also solar radiation is used to heat a working
fluid, which runs turbines. Also solar photovoltaic are widely used in
satellites in space, for meeting energy requirements of defence personnel
stationed at remote areas.
The structure of a solar
photovoltaic cell is:
The top layer is glass cover,
transparency 90 –95 %. Its purpose is to protect the cell from dust, moisture
etc. The next is a transparent adhesive layer, which holds the glass cover.
Underneath the adhesive is an antireflection coating
to reduce the reflected sunlight to below 5 %. Then follows a metallic grid,
which collects the charge carriers, generated by the cell under incidence of
sunlight, for circulating to outside load. Under the lower side of the metallic
grid lies a p-layer followed by n-layer forming a pn- junction at their
interface. The thickness of the top p- layer is so chosen that enough photons
cross the junction to reach the lower n-layer. Then follows another metallic
grid in contact with the lower n- layer. This forms the second terminal of the
cell.
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Structure
of a photovoltaic cell
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Q.34 State
the factors, for the choice of electrical system for an aero turbine. Also draw
the block diagram of VSCF wind electrical system. What are the advantages of
VSCF wind electrical system? (2+2+4)
Ans:
The
choice of electrical system for an aero turbine is guided by three factors:
1.
Type of electrical output: dc, variable- frequency ac,
and constant- frequency ac.
2.
Aero turbine rotational speed: constant speed with
variable blade pitch, nearly constant speed with simpler pitch- changing
mechanism or variable speed with fixed pitch blades.
3.
Utilization of electrical energy output: in
conjunction with battery or other form of storage, or interconnection with
power grid.
Block
diagram of VSCF wind electrical system:
VF(variable
frequency), FF (fixed frequency)
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Advantages of VSCF wind electrical system are:
1.
No complex pitch changing mechanism is needed.
2.
Aero turbine always operates at maximum efficiency
point.
3.
Extra energy in the high wind speed region of the
speed – duration curve can be extracted.
4.
Significant reduction in aerodynamic stresses, which
are associated with constant – speed operation.
Q.35 Derive the
equivalent star circuit from a delta circuit. (8)
Ans:
Delta/Star Transformation: Consider
three resistors RAB, RBC, RCA connected in
delta to three terminals A, B, C as shown in the Fig 2 (a). Let the equivalent
star- connected network have resistances RA, RB and RC
(Fig 2(b)). Since the two arrangements are electrically equivalent, the
resistance between any two terminals of one network is equal to the resistance
between the corresponding terminals of the other network.
Fig 2(a)
Fig 2(b)
Consider
the terminals A and B of the network. RAB, RBC, RCA are
connected in delta. Equivalent star connected network has resistances RA,
RB and Rc.
Resistance
between A and B for star = Resistance between A and B for delta
RA+RB = RAB
llel (RBC + RCA )
or 
(i)
Similarly 
(ii)
(iii)
Adding (i),(ii) and (iii)
+
+
By adding,
or 
(iv)
Subtracting (i) from (iv)
(v)
Q.36 Explain
the uses of: shaded – pole motor,
capacitor start motor, DC series motor and DC shunt motor. (8)
Ans:
Shaded pole motors - are used
in small fans, convectors, vending machines, photocopying machines, advertising
displays.
Capacitor start
motors – It have larger starting torque and is used
in machine tools, refrigeration, and air-conditioning.
DC series
motors- lifts and cranes – series motor is a
variable speed motor. It is used where high torque is required at the time of
starting to accelerate heavy loads.
DC
shunt motors – Lathes, drills,
printing press and for driving pumps.- Shunt motor is almost constant
speed motor. It is used where the speed between no loads to full load has to be
maintained almost constant.
Q.37 Explain
the terms real power, apparent power and reactive power for ac circuits and
also the units used. (6)
Ans:
Real Power: is equal to the
product of voltage, current and power factor i.e.
Power = voltage X
current X power factor or P = V I cos j and basic unit of
real power is watt. i.e. Expressed as W or kW.
Apparent power: is equal to the
product of voltage and current
Apparent power =
voltage X current or Apparent power = V I and basic unit of apparent power is volt- ampere. Expressed as VA
or KVA.
Reactive Power: is equal to the
product of voltage, current and sine of angle between the voltage and current i.e.
Reactive power =
voltage X current X sinj or Reactive power =
V I sin j and has no other
unit but expressed in VAR or KVAR.
Q.38 Explain how motors are selected
for specific engineering applications. (8)
Ans:
Selection of motors for different engineering applications:
Series motors are used in cranes, pumps,
trains, trolleys, etc. due to its very high starting torque and variable speed.
Shunt motors runs practically at
constant speed at almost all loads. Such motors are used in lathes, drills,
printing press and for driving pumps.
Cumulative compounds are used in machine
tools, coal cutting machines, punch presser, crushers, compressor, rolling
mills, elevators where very high starting torque is required and adjustable
varying speed is required.
Three phase induction motors are used
for high power applications such as in industries.
Single-phase motors are used in most
homes, offices and rural areas.
Fractional kilowatt motors are used
in fans, refrigerators, mixers, vacuum cleaners, washing machines, and small farming appliances.
Shaded
pole motors are used in small fans, convectors, vending machines, photocopying
machines, advertising displays.
Synchronous motors – power factor
improvement- the power factor of the motor can be controlled over a wide range
by adjusting its excitation. Since it can be operated under a wide range of
power actors both lagging and leading by its field current it is used in power
factor improvement.
Q.39 Explain, the construction, working principle & applications of a
single-phase induction motor. (8)
Ans:
Working : Construction of a single -phase induction motor is similar to that of a
three -phase induction motor except that the stator is provided with a single-
phase winding. Thus, it has a stator with slots, and squirrel cage rotor with a
small air-gap in between.
When it
is connected to single- phase ac supply, alternating current flows in its
stator winding and the polarity of stator poles would alternately be N and S.
The field so produced will be pulsating i.e. polarities will be alternating
with the flux rising and falling in strength. The current induced in the rotor
will tend to turn it in both directions alternately and thus the rotor will be
at standstill due to inertia. If rotor is given a push by hand or by another
means in any direction, it will rotate in the same direction developing
operating torque. Thus a single –phase induction is not self- starting and
requires special starting means.
Applications:
Due to their relatively simple construction, availability in variety of
designs, and characteristics and promoted by economics as well as meeting the
special requirements,
single-phase
induction motors are widely used, particularly where fractional horse power
range is less than 2 H.P. For example motors in 1/8 to 3/4 H.P. ranges are used
in fans, refrigerators, washing machines, blowers, centrifugal pumps, 1/30 to
1/20 H.P. range, are used in toys, hair dryers, vending machines, etc.
Single Phase
Induction Motor
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Q.40 Explain the basic construction and working
principle of a single –phase transformer. (8)
Ans:
Basic Construction and Working Principle of a single
– phase Transformer: A transformer consists of a soft iron or
silicon steel core and two windings placed on it. The windings are insulated from both the core and each other.
The core is built up of thin soft iron or silicon steel laminations to provide
a path of low reluctance to the magnetic flux. The winding connected to the
supply mains is called the primary and that connected to the load circuit is
called the secondary. When the primary winding is connected to an ac supply
mains, current flows through it. Since this winding links with an iron core, so
current flowing through this winding produces an alternating flux in the core.
Since this flux is alternating and links with secondary winding also, it
induces an emf in the secondary winding. The frequency of induced emf in the
secondary winding is the same as that of the flux or that of the supply
voltage. The induced emf in the secondary winding enables it to deliver current
to an external load connected across it. Thus the energy is transformed from
primary winding to the secondary winding by means of electro-magnetic induction
without any change in frequency. The flux of the iron core not only links with
the secondary winding but also with the primary winding, so produces
self-induced emf in the primary winding. This induced emf in the primary
winding opposes the applied voltage and therefore, sometimes it is known as
back emf of primary.
Q.41 How does the three –
phase transformer differ from a single – phase one. Give advantages and disadvantages of a 3 –
phase transformer. (8)
Ans:
Three, single - phase transformers have each a primary winding upon one
leg. These transformers are symmetrically wound and each winding is connected
to one wire of a 3 - phase system. The
three cores are placed 1200 apart so that the empty legs of
the three are in contact. The centre leg formed by these three carries the sum
of the three flux produced by the three phase currents. Since the sum of the
three currents at any instant is zero, the sum of the three fluxes must also be
zero. Any two legs act as the return for third, just as in a 3- phase system any two wire act as the
return for the current in third wire. Like single phase transformers 3 - phase transformers are also of core and
shell type.
Advantages
– 3 – Phase transformers have considerably less
weight,occupy less floor space and cost less than 3 single phase transformers
of equal rating.
Disadvantages
– If one of the phase becomes
defective, then whole of transformer is to be replaced,but in case of 3 -
single phase transformers, if one of the transformer becomes defective, the
sysrem can still be run open delta at reduced capacity or the defective
transformer can be replaced by a single spare.
Q.42 Explain DC
series, shunt and compound motors and their speed torque characteristics. (8)
Ans:
Types of D.C. Motors:
Series – wound motor possesses the field
winding of a few turns of heavy conductor, connected in series with the
armature, i.e. load current flows through both the field and armature. With
increasing load, the speed decreases. Consequently, at no-load, the speed of
the motor is very high. Hence, series-wound motor should never be used without
load. Such motors are used in cranes, pumps, trains, trolleys, etc. due to its
very high starting torque.
Fig (a).
Shunt – wound motor possesses the field
winding of large number of turns, and high resistance, which is connected in
parallel with the armature. Its staring torque is about 2.5 to 3 times greater
than the full-torque. By using shunt regulator the variations of speed of the
motor can be achieved. It runs practically at constant speed at almost all
loads. Such motors are used in lathes, drills, printing press and for driving
pumps. Fig.(b)
Compound – wound motor has
series as well shunt windings. Depending upon the type of field connections, a
compound motor can be one in which series field assists the shunt field windings.
With heavy starting loads, the torque increases. As the load increases, the
speed decreases, and vice-versa, similar to series motor. However, when the
load is suddenly decreased, the shunt prevents the motor from speeding beyond
safe limits.Such motors are used in machine tools,coal cutting machines, punch
presser, crushers, compressor, etc. Fig (c)
Differential compound motor is one
in which the field due to series winding opposes that due to shunt- winding.
Its speed remains constant. However, when such a motor is started, the series
winding requires to be short- circuited; otherwise the series winding would
rise to its full-value before the shunt field does so. If the series winding is
not short circuited at the time of starting, motor starts with high speed, and
that too in wrong direction. Such motors are rarely used since ordinary shunt
motor serves the purpose of providing constant speed.
Fig. (d)
Speed–torque characteristics: Series Motor: Since a
series motor develops a high initial torque
at low speeds; and a low torque at
high speed, so speed-torque characteristic of a series motor is hyperbola. High
initial torque at low speeds enables even a small series motor to start a heavy
load. However when starting friction is overcome the motor begins to
accelerate, counter emf increases, current and torque decreases correspondingly
as the motor speeds up. Fig. (e)
Shunt Motor: The speed – torque
characteristics is similar to speed-armature current characteristics. The flux
is independent of load and remains constant. As the back emf is also
practically constant, speed is a constant. But strictly speaking both back emf
and flux decrease with increasing load. However the back emf decreases slightly
more than flux so that on the whole there some decrease in speed. Hence, the
torque curve is slightly drooping. Fig. (f)
Compound motors: Speed –Torque
characteristic depends on the type of compound motor.
In a
cumulative compound motor Fig.(g), the series excitation helps the shunt
excitation. So, its speed- torque characteristic lies between that of shunt-
motor and series motor.
In a
differential compound motor Fig.(g)the torque increases very slightly with
speed.
Q.43 Define the
following:
(i) Average demand
(ii) Maximum
demand
(iii) Demand
factor.
(iv) Load factor. (8)
Ans:
i)
Average Demand
By
average demand of an installation is meant its average power requirement during
some specified period of time of considerable duration such as a day or month or year giving a daily or monthly or yearly average power
respectively.
ii)
Maximum Demand
The maximum
demand of an installation is defined as the greatest of all the demand which
have occurred during a given period.
It is measured
accordingly to specifications, over a prescribed time interval during a certain
period such as day, a month
or a year.
iii)
Demand Factor
It is defined
as the ratio of actual maximum demand made by the load to the rating of the
connected load.
iv)
Load Factor
It is defined as the ratio
of the average power to the maximum demand. It is necessary that in each case
the time interval over which the maximum demand is based and the period over
which the power is average must be definitely specified.
When applied to a
gereating station annual load factor is
= 
Q.44 Explain the working of a capacitor-start and capacitor-start and-run
single-phase induction motors with suitable diagrams. (8)
Ans:
Capacitor – start
motor – For obtaining the necessary phase
difference in the currents of the two windings a capacitor is placed in series
with the auxiliary winding. While the main winding draws a lagging current Im the
current in the auxiliary winding Ia
is leading and it is possible to make the phase
difference between them 900 at start. During running the auxiliary
winding is cut out so the capacitor is only short – time rated. Such a motor is
known as Capacitor – start motor.
Capacitor – start motor
Capacitor start and run single phase
induction motor: The connection diagram is as shown
in the fig. A larger capacitance (C (run)) and C(start in parallel) is employed
to provide best starting conditions. The phase separation is adjusted to more
than 900. The C (start) is cut out at a certain speed leaving C
(run) in circuit to give best running performance. C (run) also helps to
improve the overall pf of the motor. While C (run) is continuous rated, C
(start) need only the short time rated. This motor is employed for hard to
start loads.
Capacitor start and run single -
phase induction motor
Q.45 Explain, how Biofuels can be used to produce
electricity. Also draw the biomass cycle. (8)
Ans:
Biomass
is the material of all plants and animals. The organic carbon part of this
material reacts with oxygen in combustion and in the natural metabolic
processes. The end product of these processes is mainly CO2 and
heat. This biomass can be transformed by chemical and biological processes into
intermediate products (biofuels) like methane gas, ethanol liquid or charcoal
solid that are used in agro industries, which may be nonpolluting.
Biofuels
can be used to produce electricity in two ways; - By burning in furnace to
produce steam to drive turbines or by allowing fermentation in land fill sites
or in special anaerobic tanks, both of which produce methane gas which is used
as fuel for household stoves and in
spark
ignition engines or gas turbines. The carbon di-oxide produced in this process
may be recycled by cultivating crops or planting trees as CO2 is
absorbed during photosynthesis by plants. Biofuels have a potential to meet about
5 per cent of the electricity requirement of an industrialized country by
exploiting all forms of the household and industrial waste, agricultural waste
etc.
Q.46 Explain
the construction of a lead acid battery and give the equations during the
charging and discharging process. (8)
Ans:
The most common type
of secondary cell used is Lead - acid Accumulator. The electrolyte is a
solution of sulphuric acid (H2SO4) and pure water, and
the electrodes are made from lead.
Initial Charging: The lead acid cell fundamentally has two
electrodes made of pure lead. For charging purposes, the device is connected
across a source of D.C. supply having a voltage approximately 3 volts. When the
circuit is switched on, the current flows inside the cell through ions and
outside the cell through electrons. The acid molecules break into
negative ions represented by (SO4-)
and positive ions given by (H+) which are two in number against each
negative ion. or H2SO4 (SO4-) + 2
(H)+
Each
negative ion has two extra electrons and each positive ion is short of one
electron. The negative ions go towards the positive electrode and vice versa.
Each
negative ion transfers two electrons to the external circuit after coming in
contact with the positive electrode. The ion becomes radical after departing
with its extra electrons. It now reacts with water as follows
SO4+H2O H2SO4 +O
Which
shows formation of sulphuric acid and nascent oxygen.
Two such
oxygen atoms react with lead of the positive electrode of anode forming lead
peroxide on the surface of the electrode.
Pb
+2O PbO2
As the
charging proceeds a layer of brown coloured lead peroxide is formed on the
positive plate .The electron supplied by the negative ions reach the negative
plate / electrode of cathode through the charging circuit. The H+
ions move towards the cathode and receive one electron each, when coming in
contact with the electrode. These ions become hydrogen
atoms.
Two such atoms combine together to form Hydrogen molecules, which escape into
the atmosphere.
Thus
during charging: Lead peroxide is coated on the positive
electrode. Density or specific gravity of the electrolyte improves due to
formation of H2SO4.Hydrogen escapes from the negative
electrode, which remains pure lead during the process of charging. The
electrodes show a potential difference, which reaches a value of 2.6 to 2.7
volts when across the charger.
(a) (b)
It goes
to 2.1 volts when removed from the charger. The larger potential difference is
due to various ions in contact with electrodes, and the indication given by the
voltmeter does not
give the
true voltage or emf developed across the electrode if measured by keeping the
charger circuit on.
Discharging
: For this, the charged cell is connected across some load. This
can be a small resistance as shown in fig (b). Since, the direction of current
through the cell is reversed during discharging the negative ions go towards
the negative electrode and vice versa. This is opposite to the movement of ions
during charging. After passing on two electrons to the external circuit, the
negative ions become SO4 radical, which reacts with lead of negative
electrode so that Pb + SO4 PbSO4 or the negative
electrode is coated with a layer of white coloured lead sulphate.
On the
other hand the H+ ions receive electrons from the external circuit to become
hydrogen atoms. The reaction on the positive electrode is as follows:
PbO2 + 2H+ H2SO4 PbSO4 + 2H2O.
Thus during discharging both the
electrodes are coated with PbSO4 coating which is whitish in colour, formation of water results
in fall in specific gravity of the electrolyte.
Q.47 Give constructional
details of ceramic, mica, film and electrolytic capacitors. Give their typical applications. (7)
Ans:
Mica: Mica
is a transparent, high dielectric strength mineral that is easily formed into
uniform sheets as thin as 0.0025mm.
Mica-capacitors are
built in round, rectangular or irregular-shapes. They are constructed by
sandwiching layers of metal foil and mica. Some times silver is deposited in
lieu of metal foil and encapsulated in a plastic-package.
Applications:
1) Used as a precision capacitors.
2) Employed in high-frequency applications such as oscillator tuning and
filter construction.
Ceramic: Ceramic
capacitors are quite suitable for generation of large-powers at radio
frequencies. The ceramic is a dielectric material made from earth fired under
extreme heat. Titanium oxide or several other types of silicates are used to
obtain very high value of dielectric constant of ceramic material.
Applications:
1) Primarily used as COUPLING and Bypass
portions of radio-frequency circuits.
2) Specially designed ceramic-capacitors are
employed in resonant circuits.
Film: Plastic-film
capacitors are constructed by a thin-sheet of plastic (such as Mylar, Teflon,
or Polyethylene) is employed as dielectric. Thus dielectric improves the
properties of the capacitor by minimizing leakage currents.
Applications: Used
for both dc and ac circuits.
Electrolytic:
Electrolytic capacitors are usually made of aluminium or tantalum because they
form oxides with very high dielectric-strengths. Electrolytic capacitors should
only be connected in a circuit with the
proper polarities.
Applications: 1)
Used in ac-circuits.
Q.48 Differentiate
between an insulator, a semi-conductor and a good conductor. How can we make an intrinsic material to
improve conduction necessary for use in BJTs. (7)
Ans:
(a) For Insulator (b) For Semiconductor (c) For Conductor
In
case of insulators, there is practically no electron in the conduction band,
and valence band is filled. For an insulator, the valence band and conduction
bands are so apart.
For semiconductors at a
temperature of absolute zero the valence band is usually full and there may be
no-electron in the conduction band. However both the bands are so close that
electron can be lifted from the valence band to the conduction band by
imparting some energy to it. This energy must be more than energy gap 
.
In case of
conducting materials there is no forbidden gap, and valence and
conduction bands overlap. The orbits in the conduction band are very-large.
When
the material is heated, electrons break away from their atoms and move from the
valence band to conduction-band. This produces holes in the valence band and
free-electrons in conduction-band. Conduction can then occur by electron
movement and by hole-transfer with the increase in temperature, the rate of
generation of electron-hole pairs is increased. Thus in turn increases the
rate-of recombination.
When the semiconductor
is illuminated, its resistance decreases in the same way as in case of increase
in temperature. The forbidden energy gap also depends somewhat on temperature.
Q.49 For a p n junction diode, draw a
typical V-I characteristic. What is
meant by
a.
forward resistance
b.
static resistance
c.
dynamic resistance of a diode. (7)
Ans:
VI Characteristics of Diode
Forward Resistance: The resistance offered by a diode in the
circuit, when forward biased, is known as the forward-resistance. Thus
resistance is not the same for dc as well as changing-current.
DC or Static Resistance: R is the resistance offered by a diode to the
direct-current. It is the simply the
ration of the dc-voltage across the diode to the direct-current flowing through
it. At any point P on the V-I characteristic of the diode, the voltage across
the diode is OA and corresponding current is OB.
So dc or static resistance, 
Thus at any point
on the V-I characteristic of the diode, the dc or static resistance R is equal
to the reciprocal of the slope of the line joining the operating point to the
origin.
AC or Dynamic Resistance: r is a
resistance offered by a diode to the changing forward-current. It may also
defined as the reciprocal of the slope of the forward characteristic of the
diode.
For 
.
Q.50 With
the help of neat diagram, explain the functioning of a full-wave
rectifier. Clearly explain the
importance of
(i)
PIV
(ii)
Ripple factor
(iii)
Voltage regulation
(iv)
Capacitor filter in the context of a full-wave rectifier
with centre tapped transformer. (14)
Ans:
When the top of the transformer secondary winding is positive the anode
of diode D1 is positive w.r.t cathode, and anode of diode D2
is negative w.r.t cathode. Thus only diode D1 conducts, being
forward biased and current flows from cathode to anode of diode D1,
through load resistance RL and top half the transformer secondary
making cathode end of load resistance RL positive. During the
secondary half-cycle of input voltage the polarity is reversed, making the
bottom of the secondary winding positive w.r.t centre top and thus diode D2
is forward biased and diode D1 is reverse biased.
(i)PIV(Peak Inverse
Voltage): It is the maximum possible-voltage across a diode when it is
reverse-biased.
PIV of diode, 
PIV of diode, 
(ii) Ripple Factor:
(iii) Voltage
Regulation:
= 
(iv) Capacitor filter in context of
transformer :
.
Q.51 Explain the Zener
phenomenon. How does it differ from
Avalanche breakdown? (7)
Ans:
Under
a very high-reverse voltage, the depletion region expands and the potential
barrier increases leading to a very high electric field across the junction.
The electric-field will break some of the covalent-bonds of the semiconductor
atoms leading to a large number of free minority carriers, which suddenly
increase the reverse current. This is also called the Zener-Effect.
Zener-breakdown or
Avalanche breakdown may occur independently or both of these may occur
simultaneously. Diode junctions that breakdown below 5v are caused by Zener
Effect. Junctions that experience breakdown above 5v are caused by
avalanche-effect.
The Zener-breakdown
occurs in heavily doped junctions which produce narrow depletion layers. The
avalanche breakdown occurs in lightly doped junctions, which produce wide
depletion layers.
Q.52 Why
do we require Voltage Regulators? Explain in detail the working of a DC series
Voltage Regulator. Clearly explain the functions of series-pass transistor,
current limiter and error amplifier of such a Voltage Regulator. (10)
Ans:
The
primary function of a voltage-regulator is to maintain a constant dc-output
voltage. However, it also rejects ac-ripple voltage that is not removed by the
filter. The regulator may also include protective functions such as
short-circuit protection, current limiting, thermal shut down, or over-voltage
protection.
Transistor-series
voltage regulator:
Thus
circuit is a series-regulator because collector and emitter-terminals of the
transistor are in series with the load.
Series-pass
transistor: In the above circuit, the transistor Q is termed a series-pass
transistor. The series element controls the magnitude of the input-voltage that
gets to the output.
Current
limiter: If the load resistance RL is reduced or load-terminals
are shorted accidentally, a very large load current will flow in the circuit.
It may destroy the pass-transistor Q1, diode or possibly some other
component. To avoid this situation, a current limiting circuit is added to a
series regulator.
Error
Amplifier: The error amplifier is used to maintain a constant-voltage
through a negative feedback. The internal voltage reference is tightly
controlled during the fabrication of IC.
Q.53 With the help of
neat diagram explain the working of a Voltage Doubler. (4)
Ans:
Voltage-multiplier is a
modified-capacitor filter circuit that delivers a dc-voltage twice or more
times of the peak value (Amplitude) of the input ac-voltage. Such power
supplies are used for high-voltage and low-current devices such as
cathode-ray-tubes.
Half-wave voltage doubler:
During
the positive half-cycle of the ac-input voltage, diode D1 being
forward biased conducts and charges C1 upto peak value of secondary
voltage VSmax. During the negative half-cycle of the input voltage
diode D2 gets forward-biased and conducts charging capacitor C2.
Applying
Kirchoff’s- voltage law to the outer loop we have
–
Or 
= 
= Twice the peak value of the
transformer secondary voltage.
Q.54 Explain the functioning of a bipolar
junction transistor. What is the
a. relation between 
and 
(3)
b.
effect of
variation of on the collector
current (4)
c.
method of biasing the BJT (3)
d.
selection of proper Q-point for linear operation of a BJT amplifier. (4)
Ans:
Operation of transistor: For normal operation the emitter-base junction is always forward biased
while the collector-base junction is always reverse-biased.
The forward bias at the
emitter-base junction reduces the barrier potential and narrows the depletion
region. However, the relatively light doped base and collector-regions produce
a wide depletion region under the reverse-bias. Thus the effective base width Wb
between the two depletion regions is very narrow.
Electrons are injected
into the emitter region by the emitter bias supply VEB. These
conduction band electrons have enough energy to overcome the emitter-base
barrier potential. The injected electrons enter the very thin, lightly doped
base region. Because the base is very lightly doped relative to the emitter
region, only a few of the electrons recombine with the holes doped into the
base.
Injected electrons
diffuse into collector region due to extremely small thickness of base which is
much less than the diffusion length. Most of the electrons cross into the
collector-region. Collector is reverse-biased and creates a strong
electro-static field between base and collector. The field immediately collects
the diffused-electrons which enter the collector junction.
(a) The relation between 
and β:
and 
.
(b) Effect of variation of VCC on the collector current:
- The collector current IC varies
with VCB or VCC only for very low-voltage but transistor
is never operated in this region.
- In active-region collector current IC
is almost equal to IE and appear to remain constant when VCB
is increased.
- The increase in VCB, it conducts
better, although the effect is not very significant. This is because large
reverse-bias voltages causes the depletion layer at the collector-base junction
to penetrate deeper into the base of the transistor, thus reducing the distance
and the resistance between the emitter-base and collector-base regions.
(c) Method of biasing the BJT:
There is a large
number of circuits for biasing of a transistor. These circuits differ so as to
their ability to keep the quiescent point fixed in spite of variations in
transistor characteristics and also effects of temperature variations and
ageing.
A Biasing network
associated with a transistor should fulfil the following requirements:
(1) Establish the
operating point in the middle of the active region of the characteristics, so
that on applying the input-signal the instantaneous operating does not move
either to the saturation region, even at the extreme values of the input
signal.
(2) Stabilize the
collector-current IC against temperature variations.
(3) Make the
operating point independent of transistor parameters so that replacement of
transistor by another of the same type in the circuit does not shift the
operating point.
(d) Methods of different biasing:
1.
Simplest Biasing Circuit,
2.
Fixed bias circuit
3.
Self-bias or Emitter Bias
4.
Potential-divider bias.
5.
Collector-to-base bias.
Q.55 In the cases of CE and CC
configurations of BJT amplifiers, compare:
(i)
their input and output impedances. (3)
(ii)
their Voltage gains and Current gains. (7)
(iii)
their typical uses-give two uses of each case. (4)
Ans:
|
 Configuration
Characteristics
|
Common
Emitter
|
Common
Collector
|
|
Input
Impedance
Output
Impedance
Current
Gain
Voltage
Gain
|
Medium
( )
|
Very
high( )
|
|
High
( )
|
Low( )
|
|
High
( )
|
High
( )
|
|
About
500
|
Less
than unity
|
Applications: CE
– For AF-applications.
CC – For impedance
matching.
Q.56 Explain the principle of operation of Field
Effect Transistors (FET). How does a JFET
and a MOSFET differ in operation? Define
the FET parameters 
Show that 
. (7)
Ans:
Operation: let us consider an
N-channel JFET for
discussing it’s operation
I.
When neither any bias is applied to the gate nor any voltage
to the drain w.r.t source (i.e. when 
), the depletion regions around
the P-N junctions are of equal thickness and symmetrical.
II.
When positive voltage is
applied to the drain terminal D w.r.t source terminal S without connecting
gate-terminal G to supply, the electrons flow from terminal S to terminal D
whereas conventional drain-current ID flows through the channel from
D to S.
Comparison of JFET’s and MOSFET’s
I.
JFET’s can only be operated in
the depletion mode whereas MOSFET’s can
be operated in either depletion or in
enhancement mode. In a JFET, if the gate is forward-biased, excess-carrier
injunction occurs and the gate-current is substantial.
II.
MOSFET’s have input impedance
much higher than that of JFET’s. Thus is due to negligible small leakage
current.
III.
JFET’s have characteristic
curves more flatter than those of MOSFET’s indicating a higher drain
resistance.
IV.
When JFET is operated with a
reverse-bias on the junction, the gate-current IG is larger than it would be in a comparable
MOSFET.
Amplication factor, 
at constant ID.
= 
= a.c. drain
resistance × transconductance.
Amplification factor μ of a JFET may
be as high as 100.
Q.57 How can we use FET
(i)
as an Amplifier.
(ii)
as a Switch. (7)
Ans:
FET
Amplifier:
The
circuit consists of a three independent signal sources (i.e. 
). For a common-source
amplifier 
, and the output VOUT1 is taken at the drain
terminal D.
For common-gate circuit 
, the input-signal voltage is VS with source
resistance RS, and the
output VOUT1 is again taken at drain terminal D.
For common-drain
(or source follower)
, the input-signal voltage is Vin and the
output VOUT2 is taken at the-source terminal.
As a Switch:
When no gate-voltage is applied to
the FET i.e. VGS = 0, FET – becomes saturated and it behaves like a
small-resistance usually of the value less than 100Ω
and, therefore, output-voltage becomes equal to
Since 
is very large in comparison to 
, so 
can be taken
equal to zero.
When a
negative-voltage equal to 
is applied to
gate, the FET operates in the cut-off region and it acts like a very high
resistance usually of some mega-ohm’s. Hence output voltage becomes nearly
equal to input-voltage.
Q.58 Describe
in detail the construction of a triode.
To what use a triode may be put?
How does it differ from a BJT? (7)
Ans:
Construction: It consists of three electrodes namely cathode, anode and control grid.
The cathode is located at the centre of the tube and is surrounded by the
control-grid which in turn is surrounded by the anode (or plate). The grid is
nearer to the cathode than to plate. The control-grid has a mesh-structure so
that electrons emitted by the cathode can pass through it. The whole assembly
of heater filament, cathode, grid and plate is placed inside an evacuated glass
envelope. The connections for grid, plate, and cathode and heater filament are
usually brought-out at the base of the tube.
Applications:
1.
As an amplifier.
2.
Detectors and oscillators at audio
or radio frequencies.
The main difference between BJT and vacuum triode is that the transistor
is a current- controlled
device where as vacuum triode is a voltage-controlled device.
Q.59 Give three uses of a Unijunction Transistor
(UJT). Explain one use in detail. (7)
Ans:
UJT
can be used in variety of applications. A few include oscillators,
pulse-generators, saw-tooth generators, triggering circuits, phase control,
timing circuits, and voltage-or current-regulated-suppliers.
UJT Relaxation
Oscillator:
Basic Circuit Out-put voltage wave-form
across C
The relaxation
oscillator consists of UJT and a capacitor C which is charged through resistor
RE when interbase voltage VBB is switched on. During the
charging period, the voltage across the capacitor increases exponentially until
it attains the peak-point voltage VP. When the capacitor voltage
attains voltage VP, the UJT switches on and the capacitor C rapidly
discharges via B1 and capacitor voltage drops to the value Vv.
The device then cuts off and capacitor commences charging again. The cycle is
repeated continually generating a saw-tooth wave-form across capacitor C.
Q.60 Write short notes on any TWO of the
following:
(i)
An Operational Amplifier as an adder and as a voltage
follower.
(ii)
Differential Amplifier, explain CMRR and the uses of a
differential amplifier.
(iii)
IC Fabrication techniques – for monolithic IC’s.
(iv)
Realization of an Integrator and a Differentiator using
OPAmps. (2 x 7)
Ans:
(i)
An Operational
Amplifier as an adder and as a voltage follower.
This circuit can add
ac or dc-signals. Thus provides an output-voltage proportional to or equal to
the algebraic sum of two or more input voltages multiplied by a constant
gain-factor.
If 
, Then 
.
Voltage follower:
The output-voltage
of the op-amp exactly track the input voltage both in sign and magnitude. This
is the reason that this current is called voltage-follower.
(ii) Differential Amplifier:
Sometimes it necessary to amplify the voltage difference between two
input-lines neither of which is grounded. In this case, the amplifier is called
a differential-amplifier.
This reduces the amount of noise injected into the amplifier, because
any noise appears simultaneously on both input-terminals and the amplifying
circuitry rejects it being a common mode signal.
CMRR: It is defines as the ratio of
differential voltage-gain to common made voltage gain and it is given as
If a differential
amplifier is perfect, CMRR would be infinite because in that case common mode
voltage gain 
would be zero.
(iii) IC-Fabrication
Techniques for-monolithic IC’s:
A monolithic IC is
one in which all circuit components and their interconnections formed on a
single thin wafer, called the substrate. The basic production process for
monolithic IC’s are given below:
1.
A typical P-type or N-type is
grown in dimensions of 250mm length and 25mm diameter. The crystal is then
cut-by a diamond saw into thin slices called wafers. These wafers after being
lapped and polished to mirror –finish serve as the base or substrate on which
hundreds of IC’s are produced.
2.
Epitaxial Growth: On high
resistivity P-type substrate a low resistivity 25μm
thick layer of N-type is epitaxially grown. On this epitaxial-layer all active
and passive components of an IC are formed.
3.
Insulation layer: In order to
prevent the contamination of the epitaxial layer, a thin layer of 
is formed over the entire surface.
4.
Photolithographic Process: The
monolithic technique requires a the selective removal of the to form openings through which imparities may be
diffused, if required.
5.
Isolation Diffusion: layer is removed
from the desired areas using photolithographic etching process. The remaining layer serves as mask for the diffusion of acceptor
imparities. This process results in formation of N-type regions called the
isolation islands.
6.
Base Diffusion: During this
process new layer of is formed over
the wafer. The new pattern of openings is created depending upon the circuit
needs.
7.
Emitter Diffusion: A layer of again formed
over the entire surface and openings in the P-type regions are formed again by
employing masking and etching process.
8.
Aluminium Metallization: For
making electrical connections between various components of the IC, several
windows are opened on a newly created layer.
(iv) Realization of an Integrator and a Differentiator using OPAmps.
An
integrator is a circuit that performs a mathematical operation called
integration.
and 
Differentiator: It’s function is to provide an output voltage proportional
to the rate of change of the input voltage.
.
Q.61 What is a passive circuit element? Name the most commonly used passive circuit
elements. Briefly explain the following:
(i) Thin film resistors.
(ii) Wire-wound resistors. (8)
Ans:
Passive
components by themselves are not capable of amplifying or processing an electrical
signal. Passive components include resistors, inductors and capacitors.
(i)Thin film
resistors- It is constructed by using film deposition techniques of
depositing a thin film of resistive material on to an insulating substrate.
Desired values are obtained by either trimming the layer thickness or by
cutting helical grooves of suitable pitch along its length. During this process
the value of the resistance is monitored closely and cutting of grooves is
stopped as soon as the desired value of resistance is obtained.
(ii) Wire wound
resistors – These resistors are a length of wire wound around an insulating
cylindrical core. Usually wires made of materials such as Constantan and
Manganin which have high resistivity and low temperature coefficients are
employed. The complete wire wound resistor is coated with an insulating
material such as baked enamel.
Q.62 Describe the V-I
characteristic of a practical voltage source.
Find the largest practical value
of load resistance to provide constant current from a current source with 
and 
. Comment on the
variation of current from the short-circuited value.
(8)
Ans:
(a)
DC-voltage source (b) AC-voltage
source V-I
characteristics
An
ideal voltage source is not practically possible. There is no voltage source
which can maintain its terminal voltage constant even when its terminals are
short circuited. An ideal voltage source does not exist in practice. A
practical voltage source can be considered to consist of an ideal voltage
source in series with an impedance. The impedance is called internal impedance
of the source.
Practical DC-voltage
Source Practical
AC-voltage Source
Q.63 What is an N-type semiconductor? Write its energy band diagram. (5)
Ans:
Energy band diagram
for N-Type semiconductor
When a small amount of
pentavalent impurity such as Arsenic, Antimony, Bismuth or Phosphorous is added
to pure semi-conductor crystals during the crystal growth, the resulting
crystal is called N-type extrinsic semi conductor.
Q.64 What is monolithic IC? Explain photolithographic Process in
monolithic IC Production. (8)
Ans :
The word monolithic
is derived from Greek mono meaning ‘single’ and lithus meaning stone. Thus
monolithic circuit is built into a single stone or single crystal ie., in
monolithic IC’s all circuit components and their interconnections are formed
into or on the top of a single chip of silicon.
Photolithographic Process :
The
monolithic technique requires the selective removal of the Sio2 to
form openings through which impurities may be diffused. The Photolithographic
process shown in the figures (a&b) is used for this purpose.
During
the process wafer is coated with a thin layer of photosensitive material (Kodak
photo resist). The negative or stencil of the required dimensions is placed as
a mask over the photo-resist as shown in fig(a). This wafer surface with mask
is exposed to the ultra violet light. Due to UV light the photo-resist below
the transparent portions of the mask becomes polymerised. The mask is now
removed and the wafer is developed by using a chemical like trichloroethylene.
The chemical dissolves the unpolymerised portions of the photo-resist film and
leaves the surface as shown in fig (b). The oxide not covered by polymerised
photo-resist is then removed by immersing the chip in an etching solution HCL.
After etching and diffusion of impurities the resist mask is stripped off with
a chemical solvent like hot sulphuric acid (H2SO4) and by
means of mechanical abrasion process.
Q.65 What is a PN junction? Draw its circuit symbol. What is the convention followed in writing
its symbol? Illustrate its
characteristic and make it self explanatory. (6)
Ans:
The PN junction is produced by
placing a layer of P type semiconductor next to the layer of N type
semiconductor. The contact surface is called PN junction.
(a) Circuit Symbol
(b) Graphical Symbol
The graph plotted between potential
difference across the PN junction and the circuit current is known as
volt–ampere characteristics.
Forward Characteristics: When the external voltage is zero, i.e.,
when the circuit is open, the potential barrier at the junction does not allow
the flow of current and, therefore, the circuit current is zero.
With forward bias to PN junction,
very little current, called the forward current flows until the forward voltage
exceeds the junction barrier potential. As the forward voltage increased to the
knee of characteristics, the potential barrier is completely eliminated,
forward current increases linearly with the increase in forward voltage.
Reverse characteristics: When the reverse bias is applied, the
potential barriers at junction is increased. Therefore, the junction resistance
becomes very high and there is no possibility of a majority carriers flowing
across a reverse-biased junction. But still minority carriers generated on each
side can cross the junction. This results in a very small current which is
known as reverse current.
Q.66 Explain the operation of a two-diode full
wave rectifier circuit. (7)
Ans:
When
the top of the transformer secondary winding is positive, the anode of diode D1
is positive with respect to cathode and anode of diode D2 is
negative with respect to cathode. Thus only diode D1 conducts, being
forward biased and current flows from cathode to anode of diode D1,
through load resistance RL.
During the second
half-cycle of the input voltage polarity is reversed, making the bottom of the
secondary winding positive with respect to centre-tap and thus diode D2
is forward biased and the diode conducts and current flows the load resistance
RL.
Q.67 How
are Zener diodes specified? Define the
important specification factors for the device. (5)
Ans:
Specification of typical Zener
diode at 250C ambient are given below.
VZT : 20 V ± 10%; IZT : 12.5 mA for VZT
= 20 V;
IZK
= 0.25 mA for VZK = 12 V; IZM
: 32 mA; rZK = 22 Ω max;
PZmax
: 1 W; IR = 1 µA for VR : 6 V
A Zener diode is specified by its
breakdown voltage VZ, breakdown current IZK, the maximum
power dissipation PZ(max) and Zener-impedance measured at test
point, ZZT.
Zener
Impedance: Zener impedance ZZ is essentially the dynamic
resistance of a Zener diode. It is defined as the reciprocal of the slope of
the Zener curve
i.e. rZ = 
Where
ΔVZ and ΔIZ are the small
variations in voltage and current respectively.
Zener
Voltage (VZ) and Zener Current (IZ):
When the reverse bias on a crystal
diode is gradually increased, a point is reached when the junction breaksdown
and a reverse current increases abruptly. The breakdown voltage is called Zener
Voltage (VZ) and the sharply increased current is called the Zener
Current (IZ).
Q.68
Establish
the theory of a Zener diode shunt regulator.
(7)
Ans:
Above circuit diagram shows Zener
diode can be used as a voltage regulator to provide a constant voltage from a
source whose voltage may vary appreciably. A resistor RS is
necessary to limit the reverse current through the diode to safer value.
As
long as voltage across the load resistor RL is less than the
break-down voltage VZ, the Zener diode does not conduct, the
resistors RS and RL constitute a potential divider across
VS’. At an increased supply voltage VS, the voltage drop
across load resistor becomes greater than the Zener breakdown voltage. It then
operates in its break down region. The series resistor RS limits the
Zener current IZ from exceeding its rated IZmax because
Zener current is given as IZ= 
So, IS =
IZ + IL
Q.69 What are the three modes in which a transistor can operate? Explain the meaning of each mode of
operation. (9)
Ans:
The
three modes of operations of a transistor are
(1) Common – Base configuration – In
common base configuration, input is connected between emitter base and output
is taken across collector and base.
(2) Common – Emitter configuration – In common emitter configuration, input is connected between
emitter base and output is taken across collector and emitter. This emitter is
common to both input and output circuits.
(3) Common – Collector configuration – In common collector configuration, input is applied between base
and collector while the output is taken across collector and emitter. Thus the
collector forms the terminal is common to both input and output circuits.
Q.70 Draw
the circuits of an NPN and a PNP transistor in CE configuration. Define the following:
(v)
CE dc current gain.
(vi)
CE ac current gain. (4)
Ans:
CE-N-P-N-Transistor CE-P-N-P-Transistor
The output
characteristics used to determine the dc- current gain β and ac current
gain βo is as follows.
DC current gain
β = IC / IB
and AC current
gain, βo = Δ IC / Δ IB
VCE =
constant.
Q.71 What is a field effect transistor
(FET)? Which are the different types of
FET’s available? Draw the circuit
arrangement for obtaining the drain characteristics of a JFET and explain the
procedure for obtaining the above characteristic curves. Illustrate the typical drain characteristic
curves for the device. (13)
Ans:
The
device is called the FET because the drain current is controlled by the effect
of the extension of the field associated with the depletion region developed by
the reverse –bias at the gate.
Types – There
are two major categories of FET namely
(1)
Junction field effect transistor
(2) The Insulated –gate
field effect transistor (MOSFET or MOST’s)
Output
or Drain characteristics- The curve drawn between drain current ID
and drain source voltage VDS with gate- to- source voltage VGS
as the parameters is called the drain or output characteristic.
Circuit Diagram
Initially when VDS is zero, there is no attracting potential
at the drain, so no current flows in spite of the fact that the channel is
fully open. Thus given drain- current ID= 0. For small applied
voltage VDS, the N-type bar acts as a simple semiconductor resistor,
and the drain current ID increases linearly with the increase in VDS,
upto the knee point. This region of the curve is called the Channel ohmic
–region.
With the increase in drain current ID the ohmic voltage drop
between the source and channel region reverse-biases the gate junction. The
reverse biasing of the gate junction is not uniform throughout. The reverse-
bias is more at the drain end than that at the source- end of the channel, so
with the increase in VDS, the conducting portion of the channel
begins to constrict more at the drain- end. Eventually a voltage VDS
is reached at which the channel is pinched off.
The drain current ID no longer increases with the increase in
VDS. It approaches a constant saturation value. The value of voltage
VDS at which the channel is pinched off is called the pinch-off
voltage VP. The pinch off voltage VP, is not too sharply
defined on the curve, where the drain current ID begins to level off
and attains a constant value. From point A to the point B, the drain current ID
increases with the increase in voltage VDS following a reverse
square law. The region of the characteristic in which drain current ID remains
fairly a constant is called the pinch off region. It is also called the
saturation region of the amplifier-region. In this region the JFET operates as
a constant current device, since the drain current remains almost constant. The
drain current in the pinch off region with VGS =0 is referred to the
drain source saturation current IDSS.
Drain current in the pinch off region is given by Shockely’s equation
Where ID is the drain
current at a given gate source voltage VGS, IDSS is the
drain current with gate shorted to source and VGS (off) is gate
source cut-off voltage.
If the drain source
voltage VDS is continuously increased, a stage will come when the
gate channel junction breaksdown. At this point the drain current increases
very rapidly, and the JFET may be destroyed. This happens because the charge
carriers making up the saturation current at the gate channel junction
accelerate to a high velocity and produces an avalanche effect.
Q.72 What
is an unijunction transistor? Compare it
with an ordinary diode & briefly describe its construction. Draw its circuit symbol and equivalent
circuit. (9)
Ans:
Unijunction
transistor is also called the double base diode is a two layer, three terminal
solid state switching device. The device has unique characteristic that when it
is triggered, its emitter current increases regeneratively until it is
restricted by emitter power supply.
The device, because of
one PN junction, is quite similar to a diode but it differs from an ordinary
diode that it has three terminals.
Basic structure Schematic symbol
Construction-
The basic structure of a unijunction transistor is shown in the above fig. It
essentially consists of a lightly doped N-type silicon bar with a small piece
of heavily doped P-type material alloyed to its one side to produce single P-N
junction. The single P-N junction accounts for the terminology unijunction.
Q.73 What is an
integrated circuit? What are its
limitations? (5)
Ans:
An integrated circuit
consists of several interconnected transistors, resistors, capacitors etc., all
contained in one small package with external connecting terminals.
Limitations-
1.In
an IC, the various components are part of a small semiconductor chip and the
individual component or components cannot be removed or replaced, therefore, if
any component in an IC fails, the whole IC has to be replaced by a new one.
2. Limited power rating as it is not possible to manufacture high power.
3. Need of connecting inductors
and transformers exterior to the semiconductor chip as it is
not possible to fabricate inductors
and transformers on the semiconductor chip surface.
4. Operation at low voltage, as
IC’s function at fairly low voltage.
5. High grade P-N-P assembly is not
possible.
6. Low temperature co-efficient is difficult to be achieved.
7. Difficult to fabricate an IC with low
noise.
Q.74 Define the term ‘work-function’ of a metal. What is thermionic emission? (2)
Ans:
The work function of a metal may be
defined as the difference between the energy required to move an electron of a
metal to infinitely large distance and maximum energy an electron can have at
absolute zero of temperature.
A very common method used for
electron emission is by heating the metal piece to a high temterature.
The process of electron emission
from the surface of a metal into the surrounding space by heating the material
to a very high temperature is known as thermionic
emission.
EW= EB–EF
where EW work function of a metal,
EB is the total barrier
an electron has to overcome for coming out of the metal surface,
EF is Fermi level of
energy.
Q.75 List the characteristics of an ideal and a
practical OPAMP. (6)
Ans:
Characteristics
of an ideal and a practical OPAMP
1.
It’s
open loop gain A is infinite.
2.
It’s
input resistance Rin is infinite. It means that the input current is
zero and so it does not load the source.
3.
It’s
output impedance Rout is zero. Output voltage Vout is
independent of the current drawn by the load.
4.
Perfect
balance ie. Differential input voltage Vd = V2-V1
is essentially zero.
5.
Infinite
frequency bandwidth.
6.
Drift
of characteristics with temperature is not.
7.
CMRR
is infinite so that amplifier is free from undesired common mode signals such
as pick-ups thermal noise etc.
8.
Slew
rate is infinite.
9.
Output
voltage is zero when input voltage is zero ie. Offset voltage is zero.
Q.76 Draw the circuit of
an OPAMP V-to-I converter with grounded load and derive the equation for the
current through the load. (6)
Ans:
V to I-Converter
In
industrial electronics, it is necessary to provide a current proportional to
certain voltage, even though the load resistance may vary. A circuit which can
perform this job is called a voltage to current converter.
Q.77 What are active and passive
components? Categorise the following
components into these categories.
Mettalized polyster capacitor, Preset Filter circuits, Audio-frequency
chokes, FET, Vacuum tubes. (4+4)
Ans:
All electronic circuits, however
complicated contain a few basic components – two active and three passive.
Though passive components by themselves are not capable of amplifying or
processing an electrical signal but these components are as important as active
ones.
Active components – FET, Vacuum tubes
Passive components – Metalized polyester capacitors, preset filter
circuits, audio frequency chokes.
Q.78 Draw the energy band diagram of a P-N
junction under open-circuited condition.
Clearly indicate energy levels in P-region, space region and N-region.
How will it be modified if P-N junction is forward biased? (8)
Ans:
If the
external bias voltage were set equal to zero, the P-N junction would be
short-circuited. Under these conditions no current can flow i.e. I = 0 and
electrostatic potential VO remains unchanged and equal to the value
under open circuit conditions.
Suppose forward bias
voltage V is increased until V approaches junction potential VO. If
V were equal to VO, the barrier would disappear and the current
could be arbitrarily large, exceeding diode rating. In practice, barrier cannot
be reduced to zero because, as the current increases without limit, the bulk
resistance of the crystal and the resistance of the ohmic contacts will limit
the current.
Thus it is no longer
possible to assume that all the voltage V appears as change across the P-N
junction.
Forward Biasing
Q.79 Draw input and output characteristics
of common base transistor configuration.
(8)
Ans:
Common-Base
NPN transistor
Input
characteristics for common-base NPN transistor
Output characteristics for common-base
NPN transistor
Q.80 Sketch and explain the basic structure of an
N-channel junction field effect transistor. (8)
Ans:
(a) N-channel JFET N-Channel JFET
In an N-channel JFET a N-type
silicon bar referred to as the channel, has two smaller pieces of P-type
silicon material diffused on the opposite sides of its middle part, forming P-N
junctions. The two P-N junctions forming diodes or gates are connected
internally and common terminal, called the gate terminal is brought out. Ohmic
contacts are made at the two ends of
channel one lead is called the source terminal and the other drain
terminal D.
The
silicon bar behaves like a resistor between two terminals D and S. The gate
terminal is used to control the flow of current from source to drain.
Q.81 Why is a FET known as a unipolar device? How do you compare this
device with BJT? (8)
Ans:
In field effect transistors
current conduction is only by one type of majority carriers (either
by electrons or holes) and
therefore, these are called unipolar transistor.
1) It’s operation depends upon the
flow of majority carriers only. It is, therefore, a unipolar device. In BJT
both majority and minority carriers take part in conduction and therefore BJT
is sometimes called the bipolar transistor.
2) It has high input impedance (@100MΩ)
because its input circuit is reverse biased, and so permits high degree of
isolation between the input and output circuits. However, the input circuit of
an ordinary BJT transistor is forward biased and, therefore, ordinary
transistor has low input impedance.
3) JFET carries very small current
because of reverse biased gate and, therefore, it operates just like a vacuum
tube where control grid carries extremely small current and input voltage
controls the output current. This is the reason that JFET is essentially a
voltage driven device. BJT is a current operated device since input current
controls the output current.
Q.82 Explain the terms “work function” and “threshold frequency” in
connection with electron emission. Name
one material suitable for thermionic emission and one material for
photo-emission. (8)
Ans:
The work function of
a metal may be defined as the difference between the energy required to move an
electron of a metal to an infinitely large distance and maximum energy an electron
can have at absolute zero of temperature.
Threshold
frequency – The minimum frequency which can cause photo emission is
called the threshold frequency and is
given by fo= ef/h where
e = electronic charge, h= Plank’s constant f = work function.
Photo emission materials- Alkaline
material such as sodium, potassium, cesium or rubidium.
Thermionic emission materials- Carbon,
cesium, molybdenum, nickel, platinum.
Q.83
What is
photoelectric emission? How is the
electron emission affected if
(i) the frequency and
(ii) the intensity of
the incident radiations are increased? (8)
Ans:
When the surface
of certain alkaline material such as sodium, potassium, cesium is illuminated
by a beam of light or ultraviolet radiations the electrons are emitted. This
phenomenon is called photoelectric emission.
The work function of
the alkaline materials is very low and therefore when energy of the light
radiations (called photons) or the energy of the ultraviolet radiations (called
quanta) fall on the alkaline material, it gives sufficient energy to the free
electrons of the material to speed up sufficiently to overcome the surface
restraining forces of the metal and hence emission takes place. The electrons
emitted in this way are called photo electrons. The number of electrons emitted
depends upon the intensity of light beam falling upon the emitter surface and
the frequency of radiations. This property is very useful for the measurement
of intensity of illumination.
If
the frequency of incident radiations is greater than fo then the
incident radiations has more energy.
Q.84 Explain the working of a
full-wave rectifier using centre-tapped transformer. (8)
Ans.
In
centre tap rectifier, the ac input is applied through a transformer, the anodes
of the two diodes D1 and D2 are connected to the opposite
ends of the centre tapped secondary winding and two cathodes are connected to
each other and are connected through the load resistance RL and back
to the centre of the transformer as shown in the fig.
Input and Output
waveform
When the top of the
transformer secondary winding is positive, diode D1 is positive with
respect to cathode and anode of diode D2 is negative with respect to
cathode. Thus only diode D1 conducts, being forward biased and
current flows from cathode to anode of diode D1 through load
resistance RL and top half of the transformer secondary , making
cathode end of load resistance RL positive.
During the second half cycle
of the input voltage the polarity is reversed, making the bottom of the
secondary winding positive with respect to centre tap and thus diode D2
is forward biased and diode D1 is reversed biased.
Q.85 Draw the schematic diagram of an op-amp
connected as
(i) an inverter (ii) a scale changer
(iii) a phase shifter and (iv) an adder. (8)
Ans:
Inverting Amplifier
Scale changer – If the ratio
, a real constant, then
amplifier gain Af = –K. Thus the input voltage scale has been
multiplied by a factor –K to give the output voltage scale. The circuit, can
act as negative - scaler or scale changer.
Phase shifter- In the inverting amplifier, resistors Rf and R1 in
the circuit are replaced by Zf and Z1 respectively so
that Zf and Z1 are equal in magnitude but differ in phase
angle, the inverting OP- amp shifts the phase of the sinusoidal input voltage
without making any change in it’s amplitude. Thus any phase-shift from 00
to 3600 can be obtained.
Summing Operational Amplifier
Q.86 Briefly explain the thin-film and thick
–film methods of producing ICs. Discuss
their advantages and limitations. (8)
Ans:
Thin
and thick film IC’s are larger than monolithic IC’s but smaller than discrete
circuits. These IC’s can be used when power requirement is comparatively
higher.
Thin film IC’s are fabricated by
depositing films of conducting material on the surface of a glass or ceramic
base. By controlling the width and thickness of the films and by using
different materials of selected resistivity resistors and conductors are
fabricated.
Thick film IC’s are sometimes referred to
as printed thin film circuits. In their manufacturing process silk screen
printing techniques are used to create the desired circuit pattern on a ceramic
substrate.
IC’s produced by thin or thick film
techniques have the advantages of forming passive components with wider range
and better tolerances, better isolation between their components, greater flexibility
in circuit design and providing better high frequency performance than
monolithic IC’s.
However such IC’s suffer from the
drawbacks of larger physical size, comparatively higher cost and incapability
of fabrication of active components.
Q.87 Differentiate between SSI, MSI,
LSI and VLSI. (8)
Ans:
IC’s can be classified on the basis of their chip size as given
below:
1)
Small Scale integration (SSI) –
3 to 30 gates/chip.
2)
Medium Scale integration (MSI)
– 30 to 300 gates/chip.
3)
Large Scale integration (LSI) –
300 to 3000 gates/chip.
4)
Very Large Scale integration
(VLSI) – more than 3000 gates/chip.
Q.88 Why colour coding system is used to indicate
the value of a resistor? What is the
role of a capacitor in an electronic circuit?
Write a brief note on paper capacitors. (8)
Ans:
Some resistors are large enough in
size to have their resistance values (in Ω) printed on the body. However
there are some resistors, which are too small in size to have their resistance
values printed on them. Hence, a system of colour coding is employed for
indicating their values.
A
capacitor is a physical device which is capable of storing energy by virtue of
a voltage existing across it. The voltage applied across the capacitor sets up
an electric field within it and the energy is stored in the electric field. A
capacitor is basically meant to store electrons (or electrical energy), and
release them when required.
Paper capacitor – Paper capacitors are the most widely used type of
capacitors. Their popularity is due to their low cost and the fact that they
can be built over a wide range of capacitance values. They are designed to
withstand very high voltages. The leakage currents of paper capacitors are high
and their tolerances are relatively poor.
Q.89 Illustrate and
explain the V-I characteristic of a practical current source. Comment on the equivalence between voltage
source and current source. (6)
Ans:
An
ideal current source is not practically possible. There is no current source
which can maintain current supplied by it constant even when its terminals are
open circuited.
A practical current source
can be represented as shown in above fig. A practical current source can be
considered to consist of an ideal current source in parallel with an impedance
Zin. The shunt impedance is called internal impedance of the source
and accounts for the fall in output current with increase in load impedance.
A given voltage source
with a series resistance can be converted into an equivalent current source
with a parallel resistance. Conversely a current source with a parallel
resistance can be replaced by an equivalent voltage source with a series
resistance.
Q.90 What is a semiconductor? Give its important properties. Briefly explain the energy band diagram for a
semiconductor. (10)
Ans:
The
group of materials which are neither good conductors nor good insulators are
called semiconductors. At room temperature such materials have conductivities
considerably lower than those of conductors and much higher than those of
insulators such materials are called semiconductors. The resistivity of various
semiconductor materials lies in a very wide range from 10-4 to about
0.5 -m.
Properties-
1) Their resistance depends
largely on various factors and therefore, it can be controlled.
2) The resistance of
semiconductors decreases with the increase in temperature i.e. temperature
coefficient of semiconductors is negative.
3) Semiconductors are non-linear resistor.
4) The resistivity of
semiconductors changes considerably when even minute amounts of certain other
substances called the impurities are added to them.
Energy Bands-
Within
any given material there are two distinct energy bands in which electrons may
exist. These two energy bands are valence band and conduction band and are
separated by an energy gap in which no electron can normally exist.
The energy band of
interest is the highest band or valence band. If a sufficient amount of energy
is given to an electron in the valence band, the electron is free of the atomic
structure. Such an electron is said to posses enough energy to be in the
conduction band where it can take part in electric current flow. Free electrons
(electrons in conduction band) can move readily under the influence of an
external field.
Q.91
Define the
following as applied to a PN-junction:
(i) Depletion region (ii)
Width of the barrier
(iii) Barrier voltage
Support
your answer with neat illustrations. (6)
Ans:
The
negative potential on the P-side prevents the migration of any more electrons
from the N-type material to the P-type material. Similarly the positive
potential on the N-side prevents any further migration of holes across the
boundary. Thus the initial diffusion of charge carriers creates a Barrier
Potential at the junction.
The
region around the junction is completely ionised. As a result, there are no
free electrons on the N-side nor there are holes on the P-side. Since the
region around the junction is depleted of mobile charges it is called the Depletion
Region. The thickness of the depletion region (or layer) is of the order of
1 micron.
Barrier
voltage depends on doping density, electronic charge and temperature, the first
two factors are fixed thus making barrier potential dependent on temperature.
Q.92 What
do you mean by a voltage regulator?
Distinguish between a linear regulator and a switching regulator. Draw the circuit of a simple emitter-follower
regulator and briefly explain. (12)
Ans:
The primary function of a
voltage regulator is to maintain a constant dc output voltage. However, it also
rejects ac ripple voltage that is not removed by the filters. The regulator may
also include protective functions such as short-circuit protection, current-
limiting, thermal - shut down or over -voltage protection.
Linear voltage regulator-
The main drawback of linear voltage regulator is the power dissipation in the
pass –transistor which is operated in its linear mode. Other drawbacks are
regulated power supplies using these regulators require a step-down transformer
and large sized filter capacitors to reduce the ripple.
Switching Regulators- In
this the transistor is operated either in cut-off region or in the saturation
region. This results in much less power dissipation in the pass-transistor.
Switching regulators can provide large load currents at low voltages.
Emitter-follower
regulator-
This
circuit is called a series regulator because collector and emitter terminals of
the transistor are in series with the load. The unregulated dc- supply is fed
to the input- terminals and regulated output voltage Vout is
obtained across the load resistor RL. Zener diode provides the reference
voltage and the transistor acts as a variable resistor, whose resistance varies
with the operating conditions (Base current IB)
Vout=
VZ- VBE
Q.93 What are the unique features of IC voltage
regulators? (4)
Ans
:
IC voltage regulators are versatile
and relatively inexpensive and are available with features such as programmable
output, current-voltage boasting, terminal short circuit current limiting,
thermal switching and floating operation for high voltage applications.
Q.94 How are BJTs classified? Draw
the circuit symbol for each type. What
are the advantages of transistors over electron tubes? (8)
Ans :
Transistors are of
two types. P-N-P and N-P-N, behave exactly in the same way except change in
biasing and majority carrier. In P-N-P transistors the conduction is by holes
whereas in N-P-N transistors the conduction is by electrons.
P-N-P
Transistor
N-P-N
Transistor
Advantages over vacuum tubes-
Compact size, light weight, Rugged
construction, more resistive to shocks and vibrations, instantaneous operation,
low operating voltage, high operating efficiency and long life with essentially
no ageing effect if operated with in permissible limits of temperature and
frequency.
Q.95 Give a table of comparison between CE and CB configurations with
regard to the important parameters. (4)
Ans :
|
COMMON
BASE
Low input impedance(@100Ω)
Very high output impedance (@500KΩ)
Current gain less than unity.
Voltage gain @150
Very small leakage current.
|
COMMON
EMITTER
Medium
input impedance (@800Ω)
Output impedance high (@50KΩ)
High current gain.
Voltage gain @800
Very large leakage current.
|
Q.96
How does an FET differ from the conventional junction transistor? In the structure of an N-channel JFET, why
the N-type bar is called a channel? Give
the structure of a P-channel JFET. What
is the difference between a JFET and a MOSFET?
(9)
Ans :
JFET’s operation depends upon the
flow of majority carriers only. It is therefore a unipolar device. On the other
hand BJT is sometimes called the bipolar transistor.
JFET has high input impedance,
because its input current is reversed biased. However the input current of a
BJT is forward biased and therefore ordinary transistor has low input
impedance.
JFET is essentially a voltage
driven device. BJT is a current operated device since input current controls
the output current.
Channel – The region between the
source and drain sandwiched between the two gates is called the channel and the
majority carriers move from source to drain through this channel.
P-Channel JFET
Difference
between a JFET and a MOSFET:-
1. JFET’s
can only be operated in the depletion mode whereas MOSFET’s can be operated in
either depletion or enhancement mode.
2.
MOSFET’s have input impedance much higher than that of
JFET’s. This is due to negligibly small leakage current.
3.
JFET’s have characteristic curves
more flatter than those of MOSFET’s indicating
a higher drain resistance.
4.
When JFET is operated with a reverse
bias on the junction, the gate current IG is larger than it would be
in a comparable MOSFET.
Q.97 Write a brief note on DIAC. (7)
Ans :
Basic structure Schematic
Symbols
A DIAC
is an important member of the thyristor and usually employed for triggering
triacs. A DIAC is a two electrode bidirectional avalanche diode which can be
switched from OFF state to the ON state for either polarity of the applied
voltage.
A
DIAC is a P-N-P-N structured four layer two terminal semiconductor device. MT1
and MT2 are the two main terminals of the device. There is no
control terminal in this device.
Q.98 What is an OPAMP? Why is it called so? Briefly explain the following for an OPAMP
(i) Input offset voltage (ii) Input bias current
(iii) CMRR (9)
Ans :
An operational amplifier is
basically a multistage very high gain direct coupled negative feedback
amplifier that uses voltage shunt feedback to provide a stabilized voltage
gain.
An
OPAMP is so called as it was originally designed to perform mathematical
operations.
Input offset voltage – Input bias current Vin(offset) defined
as that voltage which is to be applied between the input terminals to balance
the amplifier.
Input bias current –The OPAMP‘s input is a differential amplifier.
It may be made of BJT’s or FET’s. In either case these transistors are required
to be biased and this takes current.
i.e., IB = 
for VOUT
=0
CMRR – It is defined as
the ratio of differential voltage gain to common mode voltage gain and it is
given as CMRR = Ad/ACM.
Q.99 What do you mean by Passive components? Explain how the variable resistor can be used
as a rheostat and Potentiometer with the help of symbols. Give their applications. (5)
Ans :
Passive components are those components which by themselves are not
capable of amplifying or processing an electrical signal. Passive components
include resistors, inductors and capacitors.
Symbols of Rheostats
Symbols of “POTS”
Variable resistors
usually have three leads, two fixed and one movable. If contacts are made with
to only two leads of the resistor, the variable resistor is being used as a rheostat. Rheostats
are usually employed to limit the current flowing in the circuit branches.
If all three contacts are employed
in a circuit it is termed as a potentiometer or ‘POT’. POT’s are often used as voltage dividers to control or vary
voltage across a circuit branch.
Q.100 What is a Practical Current source; Explain
its V-I characteristics. Convert an a.c.
current source of 2A in parallel with an impedance of 
into its
equivalent voltage source. (6)
Ans
:
Practical Current
source is one in which if the load impedance is very small in comparison to the
internal impedance of the source.
The
current supplied by a source should remain constant irrespective of the load
impedance.
Q.101 Explain what do you understand by intrinsic,
P-type and N-type semiconductors.
Discuss the position of Fermi Level in each case with the help of Energy
Band Diagram.
(7)
Ans :
An
intrinsic semiconductor is one which is made of the semiconductor material in
its extremely pure form.
When a small
amount of pentavalent impurity such as arsenic, antimony or phosphorous is
added to a pure semiconductor crystal during the crystal growth, the resulting
crystal is called the N-type extrinsic semiconductor.
When a small
amount of trivalent impurity, such as baron, gallium, indium or aluminium is added
to a pure semiconductor crystal during the growth the resulting crystal is
called the P-type extrinsic semiconductor.
The Fermi level is simply a reference energy level. It is the energy
level at which the probability of finding electron n energy units above it in
the conduction band is equal to the probability of finding hole n energy units
below it in the valence band.
Average energy level
of Energy
Zero-energy reference
level
In intrinsic
semiconductor the Fermi level lies midway between the conduction and valence
bands.
Q.102 Distinguish between Mobile Charge Carriers and Immobile Ions. (5)
Ans :
The mobility of
electrons is more than that of holes because the probability of an electron
having the energy required to move to an empty state in conduction is much
greater than the probability of an electron having the energy required to move
to the empty state in valence band. The mobility of an electron is double that
of an hole.
Q.103 What is breakdown diode? What is its use? Describe physically how two mechanisms of
breakdown occur in
a p-n junction diode. (7)
Ans
:
Zener diode also sometimes called
the breakdown diode is a PN junction diode specially designed for operation in
the breakdown region in reverse bias condition.
The
diode may use either Zener breakdown mechanism or avalanche breakdown
mechanism.
When
the reverse bias on a crystal diode is gradually increased a point is reached
when the junction breakdown and a reverse current increases abruptly, the
breakdown region is the knee of the reverse characteristic.
The
minority carriers under reverse biased conditions flowing through the junction
acquire a kinetic energy which increases in reverse voltage. At a sufficiently
high reverse voltage the kinetic energy of minority carriers becomes so large
that they knock out electrons from the covalent bonds of semiconductor
material. As a result of collision, the liberated electrons in turn liberate
more electrons and the current becomes very large leading to the breakdown of
the crystal structure itself. This phenomenon is called the “Avalanche
Breakdown”.
Q.104 What is Schottky
diode? Why is it also called Hot-Carrier
diode? How does it differ in
construction from a normal P-N junction diode? (5)
Ans :
Schemaic Symbols of Schottky Diode
(a)
Equivalent Circuit (b)
Approximate Equivalent Circuit
The
reverse recovery time is so short in small signal diodes that its effect cannot
be noticed at frequencies below 10MHZ or so. It becomes very important well
above 10 MHZ. The solution is a special purpose device called Schottky diode.
Such a diode has no depletion layer eliminating the stored charges at the
junction. Due to the lack of charge storage the Schottky diode can switch off
faster than an ordinary diode.
Its
construction is very different from the normal PN junction in which metal
semiconductor junction is developed. On one side of the junction a metal is
used and the other side of the junction N-type doped silicon is used.
In
both materials, the electrons are the majority carriers. In the metal, the
level of minority carriers is insignificant. When diode is unbiased, electrons
on N-side have low energy levels than the electrons in the metal and so the
electrons cannot cross the junction barrier called Schottky barrier. But when
the diode is forward biased the electrons on the N-side gain enough energy to
cross the junction and enter the metal. Since these electrons plung into the
metal with very large energy they are usually called the hot carriers and the
diode is called the hot carrier diode.
Q.105 Draw
the circuit diagram of Four-diode
Full-wave Bridge
Rectifier and explain its operation.
What are its advantages and disadvantages? (6)
Ans :
When the upper
end of the transformer secondary winding is positive, diodes D1 and
D3 are forward biased and current flows through arm AB, enters the
load at positive terminal, leaves the load at negative terminal and returns
back flowing through arm DC. During this half of each input cycle, the diodes D2
and D4 are reverse biased, and so the current is not allowed
to flow in arms AD and BC. The flow of current is indicated by solid arrows in
the figure.
In the
second half of the input cycle the lower end of ac supply becomes positive
diodes D2 and D4 become forward biased and current flows
through arm CB, enters the load at the positive terminal, leaves the load at
the negative terminal and returns back flowing through arm DA. Flow of current
has been shown by dotted arrows in the figure. Thus the direction of flow of
current through the load resistance RL remains the same during both
half cycles of the input supply voltage.
Advantages-
1.
Low
cost, highly reliable and small sized silicon diodes.
2.
No centre tap is required in the transformer secondary so in
case of a bridge rectifier the transformer required is simpler.
3.
The PIV is one half that of centre-tap rectifier. Hence
bridge rectifier is highly suited for high voltage applications.
4.
Transformer utilization factor, in case of a bridge
rectifier is higher than that of a centre tap transformer.
Disadvantages-
It needs four diodes, two of
which conduct in alternate half cycles. Because of this the total voltage drop
in diodes becomes double of that in case of centre tap rectifier.
Q.106 Draw the functional block diagram of Three-Terminal Voltage
Regulator IC and describe its operation. (5)
Ans :
Fundamental
block diagram of a three terminal IC voltage regulator
The
latest generation of IC voltage regulators has devices with only three pins-
one for the unregulated input voltage, one for regulated output voltage and one
for ground.
The
error amplifier is used to maintain a constant voltage through a negative feedback.
The
series pass element is driven by the output of the error amplifier. It acts as
an automatically controlled variable resistor. It’s resistance varies as
required for maintaining the output voltage constant. The series pass element
is typically a BJT that is rated to pass maximum load current.
Q.107 Define a Transistor. Draw the
circuit diagrams of p-n-p and n-p-n transistors with proper biasing
voltages. Also indicate the reference
directions for the currents and the reference polarities for the voltage. (5)
Ans :
The transistor is a
solid state device whose operation depends upon the flow of electric charge
carriers within the solid. The transistor is a current controlled device.
(a) NPN-transistor (b)
PNP-transistor
Q.108 Define Transistor
characteristics? Sketch the output
characteristics of a transistor in its CB mode.
Explain the Active, cut-off and saturation Regions. (7)
Ans
:
The performance of transistors
when connected in a circuit may be determined from their characteristic curves
that relate different dc currents and voltage of a transistor. Such curves are
known as static characteristic curves.
There
are two important characteristics of transistor
1.
Input characteristics, 2.Output
characteristics.
Out put characteristics for Common
Base NPN transistor
The curve drawn between collector
current IC and collector base voltage VCB for a given
value of emitter current IE is known as output characteristics.
In
an active region (emitter is forward biased and collector reverse biased)
collector current IC is almost equal to IE and appears to
remain constant when VCB is increased. In fact, there is very small
increase in IC with increase in VCB. This is because the
increase in VCB expands the collector base depletion region and thus
shortens the distance between the two depletion regions.
In cut-off region (emitter and
collector junctions both are reverse biased) small collector current IC
flows even when emitter current IE =0. This is the collector leakage current ICBO or
ICO.
In saturation region (both emitter
and collector junctions are forward biased) collector current IC
flows even when VCB 
0.
Q.109 What is MOSFET? Why MOSFETs are more widely used than the
JFETs? (4)
Ans :
MOSFET – Metal
Oxide Semiconductor Field Effect transistor is an important semiconductor
device and is widely used in many circuit applications. MOSFET is a three
terminal device (Source, Gate and Drain) and drain current in it is controlled
by gate bias.
These
devices are more useful in electro meter applications than the JFETs. For the
above reasons, and also because MOSFETs are easier to manufacture, they are
widely used than JFETs.
Q.110 Draw the structure of an N-channel JFET and explain its
principle of operation with neat diagrams along with V-I characteristics. Define Pinch-off voltage and mark it on the
characteristics. Explain its
significance in the operation of JFET. (8)
Ans
:
(a) N-Channel JFET Schematic
symbol of
N-Channel
JFET
(b) JFET- Drain Characteristics with Shorted gate
Operation: When neither any
bias is applied to the gate (i.e. When VGS =o) nor any voltage to the drain w.r.t. source (i.e. when VDS
=o) the depletion regions around the P-N junctions are of equal thickness and
symmetrical.
When positive voltage is applied to
the drain terminal D w.r.t. source terminal S, without connecting gate terminal
G to supply, the electrons (which are the majority carriers) flow from terminal
S to terminal D, whereas conventional drain current ID flows through
the channel from D to S .Due to flow of this current, there is a uniform
voltage drop across the channel resistance as we move from terminal D to
terminal S. This voltage drop reverse biases the diode. The gate is more
negative w.r.t. to those points in the channel which are nearer to D to S.
Hence, the depletion layers penetrate more deeply into the channel at points
lying closer to D than to S. Hence the device is called the field effect
transistor because the drain current is controlled by the effect of the
extension of the field associated with the depletion region developed by the
reverse bias at the gate.
Q.111 What is an SCR? Explain the construction, working and V-I
characteristics of an SCR for different gate currents and indicate there-upon
holding current, latching current and break over voltage. (8)
Ans :
The SCR (Silicon controlled rectifier)
is a controlled rectifier constructed of a silicon semiconductor material with
a third terminal for control purposes. Silicon was chosen because of its high
temperature and power capabilities. The third terminal gate, determines when the
rectifier switches from the open circuit to short circuit state.
Schematic Diagram & Symbolic representation of SCR
Construction – SCR is essentially an
ordinary rectifier (PN) and a junction transistor (NPN) combined in one unit to
form PNPN device. It consists of a four layer pellet of P and N type silicon
semiconductor materials. The junctions are diffused or alloyed. The material
which may be used for P- diffusion is aluminium and for N diffusion is
phosphorous. The contact with anode can be made with an aluminium foil through
cathode and gate by metal sheet.
Working - SCR is a switch .Ideally it remains off or
appears to have an infinite impedance until both the anode and gate terminals
have suitable positive voltages with respect to the cathode terminal.
VI Characteristics of SCR
When anode is made positive w.r.t.
the cathode, junction J1 and J3 are forward biased and
junction J2 is reverse biased and only the leakage current will flow
through the device. The SCR is then said to be in the forward blocking state or
in the forward mode or OFF state.
But
when cathode is made positive w.r.t. the anode, junctions J1 and J3
are reverse biased a small reverse leakage current will flow through the SCR
and the SCR is said to be in the reverse blocking or reverse mode.
When the SCR is in forward mode
the SCR conducts when the forward voltage exceeds certain value called the
forward breakover voltage VFBO.
If a positive gate current is
supplied, the SCR can become conducting at a voltage less than forward break
over voltage. The larger the gate current, lower the break over voltage. With
sufficiently large gate current the SCR behaves identical to PN rectifier.
Once the SCR has been switched ON,
it has no control on the amount of current flowing through it. The current
through the SCR is entirely controlled by the external impedance connected in
the circuit and the supplied voltage. There is a very small about 1v, potential
drop across the SCR. The forward current through the SCR can be reduced by
reducing the applied voltage or by increasing circuit impedance. There is,
however, a minimum forward current that must be maintained to keep the SCR in
conducting state. This is called the holding
current rating of SCR. If the
current through the SCR is reduced below the level of holding current, the
device returns to OFF state or blocking state.
Q.112 What is an
integrated circuit? Explain how a
capacitor can be constructed in a monolithic IC. (4)
Ans :
An integrated circuit (IC) consists of several interconnected
transistors, resistors, capacitors, etc. all contained in one small package
with external connecting terminals.
IC-Diffused capacitor
All
P-N junctions have capacitance, so capacitors may be produced by fabricating
suitable junctions. As shown in the above figure P and N regions form the
capacitor plates and depletion region between them is the dielectric.
IC capacitors may also be fabricated by
utilising the Sio2 surface layer as a dielectric. A heavily doped N-
region is diffused to form one plate of capacitor. The other plate is formed by
depositing a film of aluminium on the Sio2 formed on the wafer
surface
.
Q.113 What is photo-electric emission? Explain how is this emission affected if the
frequency and the intensity of the
incident radiations are increased. (4)
Ans :
When the surface
of certain alkaline material such as sodium, potassium, cesium or rubidium is
illuminated by a beam of light or ultraviolet radiations, the electrons are
emitted. The phenomenon is called photoelectric emission.
The
work function of the alkaline materials is very low and therefore when energy
of the ultraviolet radiations (called quanta) fall on the alkaline material, it
gives sufficient energy to the free electrons of the material to speed up
sufficiently to overcome the surface retaining forces of the metal and hence
emission takes place. The electrons emitted in this way are called photo
electrons. The number of electrons emitted depends upon the intensity
(brightness) of light beam falling upon the emitter surface and the frequency
of radiations.
The energy per bundle, is related
to the frequency of light by W=hf = quantum energy, where h is the Planck’s
constant and is equal to 6.626 X 10-34 Joules and f is the frequency
in hertz. The energy required to liberate an electron = eF where e electronic charge = 1.602 x 10-19 and F = work function in electron volt.Hence for a photon to cause
emission
hf ≥ eF ;
f ≥
eF / h
The minimum frequency which can
cause photoemission is called the threshold frequency and is given by fo
= eF / h and fo =C/lo;
lo= C/ fo = Ch / eF
Q.114 Draw the circuit of an OP-AMP Differential
Amplifier and derive an expression for its output voltage. (5)
Ans :
Sometimes
it is necessary to amplify the voltage difference between two input lines
neither of which is grounded. In this case the amplifier is called a
differential amplifier.
Since circuit has two
inputs Vin1 and Vin2 superposition theorem will be used
for determination of voltage gain of the amplifier.
When Vin1 =
0, then Vout = 
…………………………..(1)
When Vin2 = 0, then V1= 
And output due to Vin1
then is
Vout 1= 
Vin1
If R1 = R2 and Rf = R3
\ Vout 1 = Rf /R3 Vin1 ……………………..
2
The net o/p voltage, Vout = Vout 1+ Vout
2
Vout = 
Q.115 Explain the operation of an OPAMP current-to-voltage
converter with the help of circuit diagram. (5)
Ans :
Current
to voltage convert
A device that
produces a voltage proportional to input signal current is called a current to
voltage converter. In this circuit a photocell or photo multiplier tube that
provides output current is connected to the inverting terminal of the op-amp .Rs
is the shunt source resistance.
Vout = - IS
RL
Above equation
indicating that the output voltage is directly proportional to the input
current IS. The capacitor is connected in parallel with resistor RL
for reducing the high frequency noise.
Q.116 Why are resistors, capacitors and inductors
called passive components? Write a note
on moulded-carbon composition resistor.
How are active components are broadly classified? (6)
Ans
:
Passive components are those components which by themselves are not
capable of amplifying or processing an electrical signal. This is the reason R,
L and C is called passive components.
Carbon composition
resistors- This is the most common type of low voltage resistor. The
resistive material is of carbon clay composition and the leads are made of
tinned copper. The resistor is enclosed in a plastic case to prevent the entry
of moisture and other harmful elements from outside. These resistors have
advantages of being cheap and reliable and their stability is high during their
lifetimes, but are highly sensitive to temperature variations. The
power-dissipating capacity of such units ranges from about 0.1 to 2 watts and
the physical size is of the larger units
have diameters less than 10mm.
Active
components – are devices capable of amplifying or processing with the help of
passive components. These active components can be broadly classified into two
categories.
1)
Tube Type – Vacuum tubes, Gas tubes
2)
Semiconductor Devices – BJT, UJT, FET, SCR, Diode, etc.
Q.117 What do you mean by a
constant current source? Write its
symbolic representation. What is the
symbol for a practical current source?
Given an a.c. current source of strength 0.2A and impedance 100 ohms,
write an equivalent voltage-source representation for this source. (8)
Ans
:
Constant current source
– A source that supplies a constant current to a load even if its impedance
varies.
Symbolic
representation of an Ideal-current source
Symbol for a
practical current source
Open circuit voltage
across terminals A and B is given as VOC = ISRin=0.2
x 100 = 20 V
Equivalent-voltage source
Q.118 State Thevenin’s theorem. (2)
Ans :
Thevenin’s
theorem – provides a mathematical technique for replacing a two terminal
network by a voltage source VT and resistance RT
connected in series. The voltage source VT is open circuit voltage
that appears across the load terminals when the load is removed or disconnected
and resistance RT called the Thevenin’s equivalent resistance is
equal to the resistance of the network looking back into the loading terminals.
Thevenin’s Equivalent Circuit
Where I = 
for dc network and I = 
for ac machines.
Q.119 What type of material can conduct electricity
in it? Write the energy band diagrams
for metals and insulators and briefly explain. (8)
Ans :
Conducting
materials (such as silver, copper, aluminium etc.) are good conductors of
electricity and are characterised by a large electrical conductivity and small
electrical resistivity.
Within
any given material there are two distinct energy bands in which electrons may
exist. These two energy bands are valence band and conduction band and are
separated by an energy gap in which no electrons normally exist. This energy
gap is termed the Forbidden gap EG.
For
Insulator For Semiconductor For Conductor
The
energy band of interest is the highest energy band or valence band. If a
sufficient amount of energy is given to an electron in the valence band the
electrons is freed of the atomic structure, such an electron is said to possess
enough energy to be in conduction band , where it can take part in electric
current flow. Free electrons can move readily under the influence of an
external field.
Q.120 Briefly describe the effect
of temperature on the conductivity of instrinsic semiconductors. (3)
Ans
:
At
absolute zero temperature, all the electrons of intrinsic semiconductors are
tightly held by their atoms. The inner orbit electrons are bound to nucleus
whereas, the valence electrons are bound by the forces of covalent bonds. Thus,
at absolute zero temperature no free electron is available in the intrinsic
semiconductor so it behaves like a perfect insulator.
When the material
is heated, electrons break away from their atoms and move from the valence band
to conduction band. Thus produces holes in the valence band and free electrons
in the conduction band. Conduction can then occur by electron movement and hole
transfer. With the increase in temperature, the rate of generation of electron
hole pairs is increased. This intrim increases the rate of recombination. Thus
with the increase in temperature, the concentration of charge carriers
increases. As more charge carriers are made available, the conductivity of a
pure semiconductor increases with the increase in temperature.
Q.121 Explain the operation of a
PN-junction under forward bias condition. (5)
Ans :
Forward Biasing
When an external
field, with P-region connected to positive terminal and N-region connected to
negative terminal of the battery, is applied across the junction, as shown in
the figure, the junction is said to be forward biased.
After forward biased,
barrier is reduced and it is eliminated altogether .The junction offers a low
resistance called the forward resistance, rf, to the flow of current
and current flows in the circuit due to establishment of low resistance path
and magnitude of current depends upon the magnitude of applied forward voltage.
Q.122 Explain the operation of a Voltage Tripler with a suitable
diagram. (6)
Ans :
Voltage Tripler
In operation capacitor C1
is charged through diode D1 to a peak value of transformer secondary
voltage VSmax during first positive half cycle of the ac input
voltage. During the negative half cycle capacitor C2 is charged to
twice the peak voltage 2 VSmax developed by the sum of voltage
across capacitor C1 and transformer secondary winding. The voltage
across capacitors C1 and C3 is 3 VSmax.
Q.123 Mention the effects
of increasing the capacitance of a shunt capacitor filter on the performance of
a rectifier. (3)
Ans :
Larger the filter capacitor the
more charge it can hold and less it will discharge. Hence the peak to peak
value of the ripple will be less and the average dc level will increase.
But larger the capacitance
value, greater is the current required to charge the capacitor to a given
voltage.
Q.124 Why do the regulated supplies include current limiting? Name the building block of the first
generation IC voltage regulators like the 
. What is the
disadvantage of these early IC regulators? (4)
Ans
:
If the load resistance RL
is reduced or load terminals are shorted accidently, a very large load current
will flow in the circuit. It may destroy the pass transistor, diode or possible
some other component. To avoid this situation a current limiting circuit is
added.
Fundamental
Block-Diagram of IC-Regulator
Series regulators are very
popular of our needs. The main drawback of these regulators is the power
dissipation in the pass transistor. Other drawbacks are regulated power
supplies using these regulators require a step-down transformer and alarge
sized filter capacitor to reduce the ripple.
Q.125 What is a transistor? Define
a common-emitter configuration. Show
that for a CE configuration 
with usual
notations. (9)
Ans :
The
transistor is a solid state device, whose operation depends upon the flow of
electric charge carriers within the solid.
In
common emitter configuration input is connected between base and emitter while
the output is taken between collector and emitter.
Common
emitter is commonly used because it’s current, voltage and power gains are
quite high and output to input impedance ratio is moderate.
The
ratio of change in collector current and change in base current is called the
base current amplification factor.ie. β = ΔIC/ΔIB
Q.126 Write the input characteristics of a PNP transistor connected in
common-emitter configuration. (3)
Ans
:
CE-configuration: pnp transistor CE-configuration: Input
characteristics
Q.127 What are the advantages of FETs over
BJTs? Write the structure of an N-channel JFET. What do you mean by PINCH-OFF voltage of a
JFET? (9)
Ans :
Advantages
1.
FET’s
operation depends upon the flow of majority carriers only. It is therefore a
unipolar device.
2.
FET’s are simpler to fabricate, smaller in size and has
higher efficiency.
3.
FET’s have high input impedance(@100
MΩ)
4.
FET’s are relatively immune to noise.
5.
FET’s have very high power gain and therefore the necessity
of employing driver stage is eliminated.
6.
FET’s have negative temperature coefficient of resistance
and therefore has better thermal stability.
N-Channel JFET Schematic Symbol for N-Channel JFET
If the negative voltage at the
gate is increased, depletion layers meet at the centre and the drain current ID
is cut-off completely. The gate to source voltage VGS at which drain
current ID is cut-off completely, is called the PINCH-OFF voltage VP.
Q.128 Write a brief note on UJT. (7)
Ans :
Unijunction transistor
is also called double base diode. It is a two layer, three terminal solid state
switching device. This device has unique characteristic that when it is
triggered its emitter current increases regeneratively until it is restricted
by emitter power supply.
It can be used in a
wide variety of applications including oscillators, pulse generators, saw tooth
generators, triggering circuits, phase control timing circuits and voltage or
current regulated supplies.
The device has only one
junction ie. one P-N junction, which is quite similar to a diode but it differs
from an ordinary diode that it has three terminals.
Base
Structure (UJT) Schematic Symbol (UJT)
Q.129 Write the circuit of the most general form of
a differential amplifier using BJTs and briefly explain. (6)
Ans
:
Differential
Amplifier circuit
An amplifier which is
designed to give the difference between the two input signals is called differential
amplifier.
There are two inputs and two output
as shown. When the input signal drives Q1 there will be more voltage
drop across RC1 and therefore, the collector of Q1 will
be less positive and when the input signal is negative it will turnoff the transistor and so voltage
drop across RC1 will be negligible and collector of Q1
will be more positive.
The amplifier can also be driven
differentially by taking output between the collectors of Q1 & Q2.
The advantage of the differential amplifier is that, hum and noise signal
called common mode signal which is common to both inputs, is cancelled out in
the output.
Q.130 Write the circuit of
an OPAMP non-inverting voltage feedback amplifier and deduce the equation for
its closed-loop gain. (6)
Ans :
The
closed loop gain
So AR1
>>(
) and 
Then 
Q.131 Write the circuit of a current-to-voltage converter using an
OPAMP and explain its operation. (4)
Ans :
A device that produces
a voltage proportional to input signal current is called a current to voltage
converter. There is a virtual ground at the inverting input terminal, current
flowing through RS is zero, and, therefore, the entire input current
IS flows through the load resistor RL resulting in the
output voltage given as Vout = - IS. RL
The above equation
clearly indicates that the output voltage is directly proportional to the input
current IS.
Q.132 Define the following terms as used in IC fabrication:
(i) Chip (ii) Diffusion (iii) Etching. (4)
Ans :
(i) Chip
- An integrated circuit IC is one in which all active and passive components
are automatically part of a small semiconductor chip.
(ii) Diffusion
- is the process of introduction of controlled amount of dopant atoms into the
semiconductor .Diffusion alters the type of conductivity of the semiconductor.
In silicon integrated circuit processing diffusion is used to form bases,
emitters and resistors in bipolar technology and source and drain regions of
MOSFET’s in MOS technology. Commonly used diffusion methods are diffusion from
a chemical source, diffusion from a doped oxide source.
(iii)
Etching – Selective removal of material in silicon IC process is known
etching. The process may be chemical or physical. By physical means etching can
be done by the kinetic energy associated with the bombarding ions in the ion
stream or plasma. Etching can be classified as dry and wet etching. In case of
dry-etching the wafer is bombarded by ions radicals or atoms in the vapour
phase. In wet etching liquid chemicals are used.
Q.133 Briefly explain a Schottky diode. (4)
Ans :
The reverse recovery
time is short in small signal diodes that its effect cannot be noticed at
frequencies below 10MHZ or so. It becomes very important well above 10 MHZ.
The solution is a
special purpose device called a Schottky diode. Such a diode has no depletion layer
eliminating the stored charges at the junction. Due to the lack of charge
storage, the Schottky diode can switch off faster than ordinary diode. It’s
construction is very different from the normal PN junction in that a metal
semiconductor junction is developed.
Q.134 What is an inductor? Explain briefly various
types of fixed inductors employed in electronic industry. What is the role of
variable inductors in radio receiver? (6)
Ans :
An inductor has been defined as a
physical device which is capable of storing energy by virtue of a current
flowing through it.
In case of an inductor current does
not change instantaneously. It offers high impedance to ac but very low impedance to dc. It blocks ac signal but
passes dc signal.
Inductors can be classified into
filter chokes audio frequency chokes and radio frequency chokes.
Filter choke has many turns of fine
wire wound on an iron core made of laminated sheets of E and I shapes and is
used in smoothing the pulsating current produced by rectifying ac into dc.
Audio frequency chokes are used to
provide high impedance to audio frequencies.
Radio frequency chokes are employed
to block the radio frequency.
Variable inductors – Tuning circuits, phase shifting and switching
of bands in amplifier sometimes require a variable inductance.
Q.135 Differentiate
between a current source and a voltage source. Give their graphical
representations. How can they be converted from one another? Determine the
current flowing through 7Ω resistor in the circuit shown in Fig. 2 by
using source transformation technique. (10)
Fig 2
Ans :
Any device that produces voltage
output continuously is known as voltage source. It’s basic purpose is to supply
power to load connected across it.
Dc voltage source ac
voltage source
Ideal current source
Constant
current source, a source that supplies a
constant current to a load even if its impedance varies.
It
should be noted that a voltage source series resistance combination is
equivalent to a current source parallel resistance combination if, and only if
their respective open circuit voltages are equal to respective short circuit
currents are equal.
Source Equivalence
A A
Þ
+
-
B
B
Q.136 “As regards conduction of current in
concerned, a semiconductor is bipolar in nature whereas a metal is
unipolar”-Justify (or) nullify the above statement. (7)
Ans :
In semiconductors both
holes and electrons take part in conduction. This is the reason that these are
bipolar in nature.
In conducting materials
there is no forbidden gap. The orbits in the conduction band are very large. An
electron in the conduction band experiences almost negligible nuclear
attraction. In fact an electron in the conduction band does not belong to any
particular atom but it moves randomly through out the solid.
Q.137 Explain what do you
understand by intrinsic, N-type and P-type semiconductors. Discuss the position
of Fermi level in each case. (9)
Ans :
An intrinsic
semiconductor is one which is made of the semiconductor material in its
extremely pure form.
When a small
amount of pentavalent impurity such as arsenic, antimony or phosphorous is
added to a pure semiconductor crystal during crystal growth the resulting
crystal is called the N-type extrinsic semiconductor.
When a small
amount of trivalent impurity such as boron, gallium, indium or aluminium is
added to a pure semiconductor crystal during the crystal growth, the resulting
crystal is called the P-type extrinsic semiconductor.
Q.138 Discuss
the reasons for the existence of a depletion layer in a P-N junction. Relate it
to the rectifying properties of a P-N junction. (10)
Ans :
On the formation of
P-N junction some of the holes from P-type material tend to diffuse across the
boundary into N-type material and some of the free electrons similarly diffuse
into the P-type material. This happens due to density gradient (as
concentration of holes is higher on P-side than that on N-side and
concentration of electrons is higher on N-side than that on P-side.) This
process is known as diffusion.
As a result of the
displacement of the chargers, an electric field appears across the junction.
Equilibrium is established when the field becomes large enough to restrain the
process of diffusion. The electric charges are confined to the neighbourhood of
the junction, and consists of immobile ions. The initial diffusion of charge
carriers creates a barrier potential at the junction. The region around the
junction is completely ionised. As a result there are no free electrons on the
N-sides nor the holes on the P-side. Since the region around the junction is
depleted of mobile charges it is called the depletion region, the space charge
region or transition region.
Q.139 What is a Zener diode? Explain, with the help of a circuit
diagram. How Zener diode can be used as a voltage regulator? (6)
Ans :
Zener diode also sometimes called
the breakdown diode is a P-N junction diode specially designed for operation in
the breakdown region in reverse bias condition.
Zener diode symbol Zener
diode used as a Voltage regulator
Voltage regulation is a measure of
a circuit‘s ability to maintain a constant output voltage even when either
input voltage or load current varies.
A resistor RS is
necessary to limit the reverse current through the diode to a safer value. The
voltage source VS and resistor RS are selected that the
diode operates in the breakdown region. The diode voltage in this region which
is also the voltage across the load RL is called Zener Voltage VZ
and the diode current is called the Zener current IZ.
As long as voltage across the load
resistor RL is less than the breakdown voltage VZ the zener
diode does conduct. The resistors RS and RL constitute a
potential divider across VS. At an increased supply voltage VS
the voltage drop across load resistor becomes greater than the zener breakdown
voltage. It then operates in the breakdown region. The series resistor RS
limits the zener current IZ from exceeding its rated maximum value
because zener current is given as IZ =
, so IS = IZ +IL
When zener diode operates in its
breakdown region the voltage across it VZ remains fairly constant
even though the current IZ flowing through it may vary considerably.
Q.140 Explain the operation of JFET as an analog
switch. (7)
Ans :
JFET as am Analog Switch
When
no gate voltage is applied to the FET ie. VGS =0, FET becomes saturated
and it behaves like a small resistance usually of the value of less than
100Ω and, therefore, output voltage becomes equal to 
Since
RD>>RDS(ON) so VOUT
can be taken equal to zero.. When a negative voltage equal to VGS(OFF)
is applied to the gate , the FET operates in the cut-off region and it acts
like a very high resistance usually of some mega ohms. Hence output voltage
becomes nearly equal to input voltage.
Q.141 Compare
the characteristics of CB, CE and CC configurations of a transistor. Draw the
circuit of a common collector transistor configuration and explain its
operation. Also derive the relation between 
and α
current amplification factors. (9)
Ans
:
|
Characteristics
|
Common Base
|
Common Emitter
|
Common Collector
|
|
Input impedance
|
Low (@100Ω)
|
Medium (@800Ω)
|
Very high (@750kΩ)
|
|
Output impedance
|
Very high (@500kΩ)
|
High (@50kΩ)
|
Low (@50Ω)
|
|
Current gain
|
Less than unity
|
High(@80)
|
High(@100)
|
|
Voltage gain
|
About 150
|
@500
|
Less than unity
|
|
Leakage current
|
@5µA
|
Very large @500µA
|
Very large
|
Common
collector configuration
CC-NPN Transistor
In this arrangement base current IB
flows in the input circuit and emitter current IE flows in the
output circuit. So, change in emitter current Δ IE to change in base
current Δ IB gives the current amplification factor g
IC= α IE+ ICBO and IE= IB + IC
\ IE=
IB+ α IE+ ICBO
IE(1-
α ) = IB+ ICBO
This
configuration primarily is used for impedance matching.
Q.142 Why are MOSFETs available in both enhancement
and depletion modes, while JFETs operate almost invariably in the depletion
modes. (4)
Ans :
In a JFET, if the gate
is forward biased, excess carrier junction occurs and gate current is
substantial. Thus channel conductance is enhanced to some degree due to excess
carriers but device is never operated with gate forward biased because gate
current is undesirable.
Q.143 Sketch the output characteristics for N-channel JFET with
gate-source voltage shorted (i.e. VGS=0). How Ohmic, Pinch-off and
Breakdown regions are created? (8)
Ans :
JFET-Drain
characteristic with Short-gate
Initially when VDS
is zero, there is no attracting potential at the drain, so no current
flows in spite of the fact that the channel is fully open. Thus ID =
0.
For small supply
applied voltage VDS, the N-type bar acts as a simple semiconductor
resistor, and the drain current ID increases linearly with the
increase in VDS, up to the knee point. This region of the curve is
called the Channel Ohmic Region.
With the increase in
drain current ID, the ohmic voltage drop between the source and
channel region reverse biases the gate junction. The reverse biasing of the
gate junction is not uniform throughout. The reverse bias is more at the drain
end than that at the source end of the channel, the conducting portion of the
channel begins to constrict more at the drain end. Eventually a voltage VDS
is reached at which channel is pinched off, (ie. All the free charges from the
channel gets removed), is called the Pinch-off voltage (VP)
If VDS is continuously
increased, a stage comes when the gate channel junction breaks down. At this
point the drain current increases very rapidly and the JFET may be destroyed.
This is known as Avalanche Effect.
Q.144 Describe the structure, symbol and operation
of SCR with the help of suitable diagrams. (8)
Ans :
Schematic Diagram Symbolic Diagram
The SCR (Silicon controlled rectifier) is a
controlled rectifier constructed of a silicon semiconductor material with a
third terminal for control purposes. The basic operation of SCR is different
from that of an ordinary two layer semiconductor diode in that, the third
terminal gate determines when the rectifier switches from the open
circuit to short circuit state. SCR deice is a switch .Ideally it remains off
or appears to have infinite impedance until both the anode and gate terminals
have suitable positive voltages with respect to the cathode terminal. The
thyristor then switches ON and current flows and continues to conduct without
further gate signals.
Q.145 Explain the following terms as referred to an
operational amplifier
(i) Input offset Voltage
(ii) Input offset Current
(iii) Slew Rate
(vii)
CMRR (12)
Ans :
(i)
Input
offset Voltage
(a) Output offset voltage (b) Elimination of Output
off-set voltage
When
the inputs of the op-amp are grounded, there is almost always an output offset
voltage as shown in fig(a) because the input transistors have different VBE
values.
(ii) Input offset current
(a) Output offset voltage (b) Reduced offset voltage
due to Return-path resistance by equal return resistors
Input offset current Iin
offset – is defined as
the difference between the two currents entering the input terminals of a
balanced amplifier for VOUT = 0
ie. Iin(offset)
= IB1-IB2 for VOUT = 0
(iii)
Slew Rate – The slew rate of an op-amp is defined as the maximum rate at which the
output voltage can change, no matter how large an input signal applied.
SR = d VOUT /dt
Max
This is usually measured in V / µs
(iv)
CMRR- It is defined as the ratio of
differential voltage gain to common mode voltage gain and it is given as CMRR=
Ad / ACM
If the differential amplifier is perfect, CMRR would be infinite because
in that case common mode voltage gain ACM would be zero.
CMRR(log) = 20 log(Ad / ACM) and VOUT=
AdVd
PART
– III
NUMERICALS
Q.1 A, power supply
is having the following loads:-
Type of load Max. demand (kW) Diversity of group Demand factor
Domestic 1500
1.2
0.8
Commercial 2000 1.1 0.9
Industrial
10,000 1.25 1
If the overall system
diversity factor is 1.35, determine the maximum demand and connected load of
each type. (8)
Ans:
The sum of maximum demands of
three types of loads is = 1500 + 10,000+ 2000 = 13,500kW.
As the
system diversity factor is 1.35,
Therefore,
max. demand on the supply = 13,500 / 1.35 = 10,000 kW.
Each type of load has
its own diversity factor among its consumers.
\ connected domestic load = 1500
X 1.2 / 0.8 = 2250 kW.
\ connected commercial load =
2000 X 1.1 / 0.9 = 2444 kW.
\ connected domestic load =
10,000 X 1.25 / 1 = 12, 500 kW.
Q.2 A two-pole alternator runs
at 3000 rpm and supplies power to a 10 –pole single – phase induction motor,
which has full load slip of 5 %. Find the full load speed of the induction
motor and the frequency of its rotor emf due to forward field. (8)
Ans:
where NS =
Synchronous speed and f = frequency of the supply voltage generated by the
alternator, then
= 
so, 
= N = 570 rpm.
If fr is the rotor
emf frequency, then fr = 50 X
0.05 =2.5 Hz.
Q.3 The voltage applied to a
dc shunt motor is 220V. The armature current is 20A. The armature resistance is
0.5 W. The speed is 80 radians per second.
Determine the induced emf, the electromagnetic torque and speed in rpm. (8)
Ans:
Given V= 220V, Ia = 20A where Ia = the
armature current. Ra = Armature resistance = 0.5 W and w = 8o rad. /s.
The
back emf of the motor Eb = V - Ia Ra = 220 – 20 X 0.5 = 210 V.
The
electromagnetic torque Te = Eb
Ia / w = 210 X 20 / 80 = 52.5 N- m.
If N is the speed in rpm, then
total angular distance covered in one minute = 2 p
N radians.
Or
angular distance covered in one second = 2 p
N / 60 rad. / s
Hence
80 = 2 p N / 60 or 
= 764 rpm.
Q.4 For the circuit
shown in Fig.1, find the value of 
for maximum power transfer.
What will be the value of maximum power? (8)
Ans:
Reduce the given circuit to
Thevenin’s circuit
Remove
the load resistance and replace battery
by its internal resistance as shown in Fig d.
Rth =
Calculate Eth = (72 X
24) / (24+12) = 48 V as shown in Fig c.
Fig.c. Fig.d.
For maximum power transfer the internal resistance of the source and load
resistance should be equal. Hence load resistance of circuit is 10 W. Maximum power = Eth2/4RL =
482/ 4 X! 0 = 57.6W.
Q.5 A series circuit
of resistance 250 and inductance 0.25 H is excited from a pulse voltage of
strength 10 V of duration 1 ms. Find the value of the current at 0.5 ms and 2
ms. (8)
Ans:
e= Em
sinω t ; Em = 250V and 2
f = 2 x 3.14 x 50. e =
250 sin314t
When e =125V then sin314t= 125/250 =0.5 ; 314t = sin-10.5= 100t =300
t = (30) /100 x 1800= 1/600 = 1.667ms.
Q.6 Find the average value of a full
wave rectified sine wave shown in Fig.2. (4)
Ans:
t = L / R =
0.25 / 250 = 10-3 s
Consider step voltage of strength 10 V . Reponse is given by
i(t) = 10 / 250 (1 – e –100t)
u(t)
10 P (t, 10-3) = 10
u(t) – 10u( t- 10-3)
Pulse response is given by
i(t) = 0.04 (1 – e –100t) u(t) - 0.04 
u(t
– 10-3)
t = 0.5 ms
i (0.5 x
10-3) = 0.04
-
0.04 (1 – e –0.5) = 0.0157 A
t = 0.5 ms
i (2 x 10-3) = 0.04 
- 0.04
= 0.04 (1 – e –2) - 0.04
( 1- e –1 ) = 0.0093 A
Q.7 The electric mains in a house is
marked as 230 V, 50 Hz. Write down the
equation for instantaneous voltage in sinusoidal form. (4)
Ans:
Vrms
=230V; f = 50Hz Vmax =√2 x Vrms = 325.22 V0lts. w =2 π f = 2x
3.14 x 50
Hence e = 325.22 sin 314 t
Q.8 The secondary of a
750 KVA, 11000/ 400 V, 50 Hz transformer has 160 turns. Determine the primary
number of turns, primary and secondary full load current neglecting losses. If
the area of cross section of the core is
100 cm2, what will be flux density in the core? (8)
Ans:
where N2
and N1 are the number of turns on the secondary and primary
windings.
turns. N1=4400
turns.
= \
= 68.182 A.
= \
= 1875 A.
E1 = 4.44 jmax f N1
volts = 
jmax =
= 0.01126 Wb.
\ Bmax = 
\ Bmax = 1.126 Wb/ m2.
Q.9 A 3 – phase transformer
consisting of three 1 – phase transformers each with turn ratio of 10:1
(primary : secondary) is used to supply a 3 – phase load of 120 kVA at 400 V on
the secondary side. Calculate the
primary line current and voltage if the transformer is connected (i) 
(ii) 
. What is the
line-line transformation ratio in each case? (8)
Ans:
D/Y connection – I = (120 X 1000) /√ 3 X 400 = 173.2 A
Primary line-to line
voltage = a V /Ö3 = 10 X 400/√3 = 2309 V;
where a = turns ratio
Primary
line current = Ö3 X I / a = 1.732 X 173.2 x 1/10 = 30 A;
Line to
line transformation ratio = a V / Ö3 / V =
a / Ö3 = 10 /Ö3
Y/D = Primary line-to line voltage
= Ö3 a V = Ö3
X 10 X 400 = 6928 V;
Primary
line current = 173.2 / Ö3 = = 10 X 1.732 = 10 A;
Line to
line transformation ratio =10 Ö3
Q.10 A
separately excited dc motor is operating at an armature voltage of 300 V. It’s no-load speed is 1200 rpm. When fully loaded it delivers a motor torque
of 350 N-m and its speed drops to 1100 rpm.
What is the full load current and power?
What is the armature resistance of the motor? The motor is now fed with an armature voltage
of 600 V, while its excitation is held fixed as before. If it is once again fully loaded, find the
motor torque, power and speed.
(8)
Ans:
Given Ea
= V = 300V, 300 = (Ka x Φ x 2 π x 1200) / 60 or Ka
x Φ =2.39
Ea = (300 x 1100)/1200 =275 V.
Ka = Z P/ 2 π x A is a
constant.
Ia = 350/2.39 = 146.4 A where Ia
= the armature current.
Mechanical power
developed = Ea Ia = 275 x 146.7 = 40.3 KW.
Ra = (300
– 275) / 146.4 = 0.171W
Armature voltage =
600V
Ia =
350/2.39 = 146.4 A; T = 350 N-m
The back emf of the
motor Eb = V - Ia Ra = 600 – 146.4 X 0.171 = 575 V.
575
= 2.39 x 2 p n / 60; n = 2297
rpm.
Power = Ea
Ra = 575 x 146.4 = 84.2 kW
Q.11 A coil, which has 10W resistance and 50mH inductance
is connected to 230V, 50Hz supply. Calculate the current in the coil. (5)
Ans:
XL
= 2 X 3.14 X 50 X 50 X10-3 = 15.7 W
\ 
= 
= 18.6W.
\
I = 12.37A.
Q.12 A 3-phase induction motor which is
wound for 4-poles, when running on full load, develops a useful torque of 100
Nm; also the rotor emf is observed to make 120-cycles/ min. It is known that the torque lost on account
of friction and core loss is 7 Nm.
Calculate the shaft power output, rotor copper loss, motor input and
motor efficiency. (8)
Ans:
f2= sf ;
120 / 60 = 2 Hz where f2 = rotor frequency
s(slip) = 2.5/ 50
=0.04
n s
(synchronous speed) = 1500 rpm
n = (1 – 0.04) x 1500
= 1440 rpm
w = 2 π
X1440 / 60= 150.7 rad/s
Shaft power
output = 100 x 150.7 = 15.07 KW.
Pm =
(100+7) x 150.7 = 16.12 kW.
Rotor
copper loss = 3 I22R2 = Pm (s /1-s)
= 16.12 X 0.04/ (1- 0.04) = 0.67 kW
Motor input
= 16.12 + 0.67 + 0.7 = 17.49 kW.
h = 15.07 /
17.49 = 86.16 %
Q.13 When
a coil is connected to a 230V, 50Hz supply, it takes a current of 2A and the
power consumption is 150W. Calculate the resistance and inductance of the coil. (5)
Ans:
Where Z is the
impedance, V is the voltage and I the current.
\
= 115W. And P = I2 R; \
= 
= 37.5 W.
\ 
= 
= 108.71W.
Or
= 0.346 H.
Q.14 Three non- inductive
resistances of 5W, 20W, and 25W
are connected in delta. Obtain its equivalent star connected system maintaining
the same phase sequence. (8)
Ans:
Star equivalent
of delta connection can be calculated by using the following expressions
RA = 5 X 25 =
125 = 2.5 W
5 + 20 + 25 50
RB = 5 X 20
= 100 = 2W
5 + 20 + 25 50
RC = 2 5 X 20 =
500 = 10W
5 +
20 + 25 50
Q.15 A power
station has a load cycle as under: 260 MW for 6 hr: 200MW for 8 hr; 160 MW for
4hr; 100MW for 6hr. If the power station is equipped with 4 sets of 75 MW each.
Calculate the load factor and capacity factor from the above data. (8)
Ans:
Daily load
factor = Units actually supplied in a day
Max.
Demand X 24
MWh
supplied per day = (260 X 6) + (200 X 8) + (160 X 4) + (100 X 6) =4,400
\ Station daily load factor =
4,400 = 0.704 or 70.4 %
260 X 24
Capacity factor = Average
demand on station
Installed capacity of the station
No.
of MWh supplied per day = 4,400 \Average power / day = 4,400/24 MW.
Total installed capacity of the station = 75 X 4 = 300 MW.
Capacity factor = 4,400/24 = 0.611 or 61.1 %
300
Q.16 A
generatin
