Subject: ELECTRONIC DEVICES

PART – I
TYPICAL QUESTIONS & ANSWERS

OBJECTIVE TYPES QUESTIONS

Each Question carries 2 marks
Choose correct or the best alternative in the following:
Q.1 A resistor with colour bands Red, Violet, Green and Black will have a value
(A) (A) 27 K  10% K (B) 2.7 M  20% K
(C) 270 K  5% K (D) 2.7 K  2% K
Ans: B
Q.2 In an n-type semiconductor, as temperature T increases, the Fermi level EF
(A) moves towards conduction band
(B) moves towards middle of forbidden energy gap
(C) does not vary
(D) may or may not shift depending upon the concentration of donor atoms
Ans: A
Q.3 In a BJT with β = 100, α equals
(A) 99 (B) 0.99
(C) 1.0 (D) 1.01
Ans: C
Q.4 In integrated circuits, npn construction is preferred to pnp construction because
(A) npn construction is cheaper
(B) to reduce diffusion constant, n-type collector is preferred
(C) npn construction permits higher packing of elements
(D) p-type base is preferred
Ans: B
Q.5 Pinch-off voltage VP for an FET is the drain voltage at which
(A) significant drain current starts flowing
(B) drain current becomes zero
(C) all free charges get removed from the channel
(D) avalanche break down takes place
Ans: C
Q.6 In SCR, the turn ON time
(A) is independent of Vg
(B) decreases with increase of Vg
(C) varies as Vg
2
(D) varies as Vg
Ans: A
Q.7 Avalanche breakdown results basically due to
(A) impact ionisation
(B) strong electric field across the junction
(C) emission of electrons
(D) rise in temperature
Ans: A
Q8 Dynamic plate resistance of pentode is of the order of
(A) 1 K Ω (B) 10 K Ω
(C) 100 K Ω (D) 1M Ω
Ans: D
Q.9 At room temperature, the current in an intrinsic semiconductor is due to
(A) holes (B) electronics
(C) ions (D) holes and electronics
Ans: D
Q.10 The varactor diode is usually
(A) Forward biased (B) reverse biased
(C) Unbiased (D) holes and electronics
Ans: B
Q.11 The diode in which impurities are heavily doped is
 (A) Varactor diode (B) PIN diode
(C) Tunnel diode (D) Zener diode
Ans: C
Q.12 A transistor in common emitter mode has
(A) a high input resistance and low output resistance
(B) a medium input resistance and high output resistance
(C) a very low input resistance and a low output resistance
(D) a high input resistance and a high output resistance
Ans: B
Q.13In an SCR the holding current is
(A) more than latching current (B) less than latching current
(C) equal to latching current (D) very small
Ans: B
Q.14 The negative potential at the control grid in a vacuum triode that causes plate current Zero is
called
(A) cut off bias (B) cut in voltage
(C) reverse blocking voltage (D) forward blocking voltage
Ans: A
Q.15 A constant current source supplies a current of 300 mA to a load of 1 Kohm. When the Load is
changed to 100 ohm, the load current will be
(A) 3 Amp (B) 300 mAmp
(C) 30 mAmp (D) 600 mAmp
Ans: B
Q.16 An Op-amp as a voltage follower has a voltage gain of
(A) Infinity (B) Zero
(C) Unity (D) less than unity
Ans: C
Q.17 A resistor used in colour TV has the following colour bands: yellow, violet, orange and silver. Its
nominal value is
(A) 4.7 K Ω  10 % (B) 4.7 K Ω  5 %
(C) 47 K Ω  10 % (D) 470 K Ω  5 %
Ans: C
Q.18 Ratings on a capacitor are given 25  F, 12 V. Also a plus sign is written near one of its
terminals. The capacitor is
(A) mica capacitor (B) ceramic capacitor
(C) electrolytic capacitor (D) paper capacitor
Ans: C
Q.19 An ideal voltage source of 12 V provides a current of 150 mA to a load connected across it. If
the load impedance is halved, the new load current will be
(A) 0.3 A (B) 0.15 A
(C) 0.6 A (D) 1.2 A
Ans: A
Q.20 An intrinsic semiconductor at the absolute zero temperature
(A) behaves like a metallic conductor
(B) behaves like an insulator
(C) has a large number of holes
(D) has a large number of electrons
Ans: B
Q.21 Which of the following diodes is operated in reverse bias mode ?
(A) P-N junction (B) Zener
(C) Tunnel (D) Schottky
Ans: B
Q.22 Compared to bipolar transistor, a JFET has
(A) lower input impedance
(B) higher voltage gain
(C) higher input impedance and high voltage gain
(D) higher input impedance and low voltage gain
Ans: D
Q.23 The minimum gate current which can turn on SCR is called
(A) trigger current (B) holding current
(C) junction (D) break over current
Ans: A
Q.24 A virtual ground
(A) is a ground for voltage
(B) is a ground for both voltage and current
(C) is ground for current
(D) is a ground for voltage but not for current
Ans: D
Q.25 Which of the following doping will produce a p-type semiconductor
(A) Germanium with phosphorus (B) Silicon with Germanium
(C) Germanium with Antimony (D) Silicon with Indium
Ans: D
Q.26 The majority charge carriers in the emitter of an NPN transistor are
(A) pentavalent atoms (B) trivalent atoms
(C) electrons (D) holes
Ans: C
Q.27 An ideal differential amplifier has CMRR equaling
(A) Unity (B) – 1 (minus unity)
(C) Infinity (D) Zero
Ans: C
Q.28 Which of the following is an active device
(A) an electric bulb (B) a diode
(C) a BJT (D) a transformer
Ans: C
Q.29 Which configuration has unity voltage gain (ideal)
(A) a Common Collector (CC) (B) a Common Emitter (CE)
(C) a Common Base (CB) (D) CE followed by CB
Ans: A
Q.30 JFET is a
(A) Current controlled device with high input resistance
(B) Voltage controlled device with high input resistance
(C) Current Controlled Current Source (CCCS)
(D) Voltage Controlled Voltage Source (VCVS)
Ans: B
Q.31 The depletion region in a Junction Diode contains
(A) only charge carriers (of minority type and majority type)
(B) no charge at all
(C) vacuum, and no atoms at all
(D) only ions i.e., immobile charges
ANS:D
NUMERICALS
Q.1
-7
The current flowing in a certain PN junction at room temperature is 2 X 10 A, when a large
reverse bias voltage is applied. Calculate the current when a forward voltage of 0.1 V is applied
across the junction.Marks (4)


Ans: The diode current is given as I = I O e
V VT
− 1
From given data:
Large reverse bias current I≈IO = 0.2 µA
Applied Voltage V = 0.1 V
Volt-equivalent of temperature,
VT = 26 mV or 0.026 V at room temperature

so, the current flowing through the diode, I = 0.2 ×10

−6
e 0.1 0.026

− 1 = 29.38 µ A

Q.2
Determine the range of V for obtaining a regulated voltage shown in Fig.5(a)
for the data 0 ≤ IL ≤ 4 mA
2 ≤ IZ ≤ 8 mA
Marks (7)

Ans:
Determination of range of V for the given circuit diagram shown in fig.6(a):
 In the fig.6(a), the voltage across the series or current limiting resistor equals the
difference between the source voltage and zener voltage. Therefore, the current through
the series resistor RS is given by
V − V
I S= S Z

RS
The zener diode and the load resistor are in parallel. So, the sum of their currents has
to equal the total current, which is the same as the current through the series resistor.
i.e., IS = IZ + IL

From the given data the load current is in between 0 mA (I LMIN) and 4 mA (ILMAX) and
the Zener current is in between 2 mA (IZMIN) and 8 mA (IZMAX).

(i) The minimum value of Source Current is given by

I =I ++I i.e.,
s(min) z(min) L(min)
IS (MIN) = 2 mA + 0 mA = 2 mA, and
The minimum value of source current is given by
V −V
S (MIN) Z
I )=
S ( MIN

RS
From the fig.3.1, VZ = 50 V and RS = 5 K
VS MIN −
2 X 10 −3 = 50( )

5 X
3
10
Therefore, the minimum value of source voltage is given by
VS (MIN) = 50 + 10 = 60 V

(ii) The maximum value of source current is given by

I =I +I
SMAX ZMAX LMAX
= 8 mA + 4 mA = 12 mA and
The maximum value of source current is given by
V −V
S ( MAX ) Z
I =
S ( MAX )
R
S
VS MAX
− 50
12 X 10 −3 = ( )
3
5 X 10
Therefore, the maximum value of source voltage is given by
VSMAX= 50 + 60 = 110 V
The value of source voltage is 50 ≤ VS ≤ 110 V

Q.3 For the circuit shown in fig.7(a), draw the waveform of output voltage Vo. Assume ideal
diode D and lossless capacitor C.
 Ans:

The name of the circuit given in fig 7.(a) is Positive Clamper. A circuit that shifts the
signal in the positive side in such a way that the negative peak of the signal falls on the
zero level, is called a Positive Clamper.

Working:

During negative half-cycle of the input voltage, the diode conducts heavily and acts like a
closed switch see fig 7.c(i). The capacitor C is charged to 5 V (V m) at the negative peak
of the signal with the polarity as marked. Slightly beyond the negative peak, the diode
stops conduction through it and behaves as an open switch (see fig.7c(ii)). The charged
capacitor (Vm = 5V) just behaves as a battery whose voltage adds to the signal voltage,
see fig7.c(ii).During positive half-cycle of the signal, the diode is reverse biased and acts
as an open switch. The resultant output voltage coming across the load resistor R L will be
as shown in Fig.7(b): By writing KVL to the circuit shown in fig5.a (iii)., we get 5V + 5V
– Vo = 0 i.e.,
Output Voltage (Vo) = 10 V.(at positive peak)
 DESCRIPTIVE TYPE QUESTIONS
Q1 Describe the phenomenon of avalanche and zener breakdown. Marks (4)

Ans: Avalanche Breakdown: The minority carriers, under reverse biased conditions, flowing
through the junction acquires a kinetic energy, which increases with the increase in
reverse voltage. At a sufficiently high reverse voltage (say 5v or more), the kinetic energy
of minority carriers becomes so large that they knock out electrons from the covalent
bonds of the semiconductor material. As a result of collision, the liberated electrons in
turn liberate more electrons and the current becomes very large leading the breakdown
of the crystal structure itself. This phenomenon is called Avalanche Breakdown.

Zener Breakdown: Under a very high reverse voltage, the depletion region expands and
7
the potential barrier increases leading to a very high electric field (3X10 V/m) across the
junction. The electric field will break some of the covalent bonds of the semiconductor
reverse current. This is called the Zener effect. The breakdown occurs at a particular and
constant value of reverse voltage called the breakdown voltage.

Q.2 Sketch V-I characteristics of a zener diode. How are they determined in the laboratory ?
Marks (6)
Ans: Determination of Forward and Reverse Characteristics of a Zener Diode in the
laboratory:
Forward Characteristic of Zener diode: Connect the circuit shown in fig 7(a) using the
5.1 V zener diode. For forward characteristic, reverse the connection of the zener diode.
Switch on the variable dc power supply. Observe and record the voltage across
the diode for various values of currents say, 1 mA, 10 mA and 30 mA. The current can be
varied by varying the power supply voltage. Observe and record the corresponding
voltage too. It should be evident that the forward characteristic is similar to that of a
junction rectifier diode as shown in below fig.7 (b)
 Reverse Characteristic of a Zener diode: Reset the voltage to zero. Connect zener
diode as shown in fig.7(c) and the reverse characteristics are shown in fig. 7(d)

Raise the voltage slowly and observe the current meter. Set the value of the zener
current of 1 mA, 2 mA, 5 mA, 10 mA, 20 mA, 30 mA and record the corresponding
voltage drop across the diode and plot the points on the graph.

The graph should show that very little current flows until the voltage has risen to a
value near the nominal voltage (5.1 V) of the zener diode. However, a very slight
increase in voltage causes a large current to flow. Once an appreciable current is flowing
in the diode, any alteration in that current will make little difference to the voltage. The
terminals to which voltmeter is connected will look like a source of voltage having a low
source resistance. This voltage may be referred to as the ‘zener voltage’ or the
avalanche voltage, (this being a reference to the physical process inside the
semiconductor).

Q.3 Can an ordinary diode be used as a zener diode? Justify your answer. Marks (4)
Ans: Ordinary diode cannot be used as a Zener diode, because the ordinary diodes are not
operated in the reverse bias breakdown region to avoid them from damaging. These
diodes are normally operated in forward region and never operated in reverse region,
since small signal and rectifier diode are never operated in breakdown (or reverse
region). Zener diode is a specially designed silicon diode, which is optimized to operate in
the breakdown region.
Q.4 What is meant by tunneling phenomenon? Marks (4)
Ans: Tunneling Phenomenon: A tunnel diode is a high conductivity two-terminal P-N junction
doped heavily about 1,000 times higher than a conventional junction diode. Because of
heavy doping depletion layer width is reduced to an extremely small value of the order of
-5
10 nm, reverse breakdown voltage is also reduced to a very small value (approaching
zero) resulting in appearance of the diode to be broken for any reverse voltage, and a
negative resistance section is produced. It is found that the reduced depletion layer can
result in carriers ‘punching through’ the junction with the velocity of light even when they
do not possess enough energy to overcome the potential barrier. The result is that large
forward current is produced at relatively low forward voltage (less than 100 mV). Such a
mechanism of conduction in which electrons (possessing very little energy) punch
through a barrier directly instead of climbing over it is called tunneling phenomenon.

Q.5 Draw the energy band diagrams of a tunnel diode for different biasing conditions and
Describe how tunneling takes place. Marks (6)
Ans: Band Diagrams of a Tunnel diode: The energy band diagram for a heavily doped
unbiased P-N junction is shown in fig. 10(a) Note that the depletion region is very narrow
and the filled levels on the N-side are exactly opposite those on the P-side. In this
condition, no tunneling occurs, as there are no empty lower energy levels to which
electrons from either side might cross the depletion region. It is also to be noted that the
conduction and the valence bands on p- side are (negatively) higher than those on the n-
side, this is due to the formation of the depletion region and barrier potential as a result of
electrons crossed from the n-region to the p-region. The n-region has lost negative
charge and p-region has gained them
 Fig. 10
When the junction is reversed biased, the p-region moves up with respect to the n-region as a result,
filled energy levels on the p- side become opposite empty energy levels on the n-side. At this stage
electrons tunnel through the narrow space charges region from the higher energy levels on the p-
side to the lower energy levels on the n- side. Despite the fact that the junction is reverse biased
significant current flows. With the increase in reverse bias, more and more electrons tunnel from the
p-side to n-side and a larger current flow. Thus the reverse characteristic of a tunnel diode is liner,
just like that of a resistor.
When the tunnel diode is forward biased, its initial behaviour is similar to that when it is
reverse biased. Now, some of the filled energy levels on the n-side shift to a high energy level than
empty level on the p-side electron tunneling now occurs, from the n-side to the p-side. With the
increase in forward bias, more and more electrons tunnel from the N-side to the P-side.

Q.6 Explain the formation of a potential barrier in a p-n junction and show the polarity of the
Barrier potential.
Marks (6) Ans: Formation of Potential Barrier in a P-N junction: The two types of
extrinsic semiconductors p-type and n-type are shown in fig. 13(a). The p-type semiconductor is
having negative acceptor ions and positively charged holes. Whereas, the n-type

semiconductor is having positive donor ions and negatively charged electrons.

P-TYPE
N-TYPE
SEMICONDUCTOR
HOLE SEMICONDUCTOR

- - - - + + + +
-
ELECTRON
- - - - + + + +

POSITIVE
NEGATIVE DONOR IONS
ACCEPTOR IONS

Fig. 13(a)
 When these two pieces are joined together and suitably treated, they from a pn
junction. The moment they form a pn junction, some of the conduction electrons from n-
type material diffuse over to the p-type material and undergo electron-hole recombination
with the holes available in the valence band. Simultaneously holes from p-type material
diffuse over to the n-type material and undergo hole-electron combination with the
electrons available in the conduction band. This process is called diffusion. In this
process, some of the free electrons move across the junction from n-type to p-type,
leaving behind positive donor ions as they are robbed of the free electrons. This
establishes a positive charge on the n-side of the junction.

Simultaneously, the free electrons which cross over the junction recombine with the
holes of p-type and uncover some of the negative acceptor ions as shown in fig. 13(b)
This establishes a negative charge on the p-side of the junction. This process of diffusion
continues till a sufficient number of donor and acceptor impurity ions are uncovered and
establish a requisite potential difference (i.e. nearly 0.3 V in case the pn junction is
formed of germanium semiconductor and 0.7 V for silicon pn junction). After this, further
diffusion is prevented because now positive charge on n-side repels holes to cross from
p-type to n-type and negative charge on p-side repels free electrons to enter from n-type
to p-type. Thus, a potential difference created across the junction acts as a barrier which
restricts further movement of charge carriers i.e. holes and electrons. This is called a
Potential Barrier or Junction Barrier Vo

NEGATIVELY POSITIVELY
CHARGED CHARGED
IONS V
P - o + N IONS
- - - - + + + +

-
- - - + + + +

DEPLETION
LAYER

Vo

.
Fig. 13(b)

Q.7 Explain physically how a p-n junction functions as a rectifier. Marks (4)

Ans: When a P-N junction diode is forward-biased and the applied voltage is increased from
zero, hardly any current flows through the device in the beginning. It is so because the
external voltage is being opposed by the internal barrier voltage V B whose value is 0.7 V
for Silicon and 0.3 V for Ge. As soon as VB is neutralized, current through the diode
increases rapidly with increasing applied battery voltage. So, in forward bias condition,
the P-N junction diode behaves like a closed switch. When a P-N junction diode is
reverse biased, majority carriers are blocked and only a small current (due to minority
carriers) flows through the diode. It is of the order of nanoamperes (nA) for Si and
microamperes (A) for Ge. For all practical purposes, this current is almost negligible.
 So, in Reverse bias condition, the P-N junction diode behaves like an open switch. So,
the above discussion shows that the P-N junction functions as a rectifier.

Q.8 Compare CB and CE transistor configurations with regard to AC input and output resistance.
Marks (6)
Ans: AC Input Resistance in Common Base Configuration: The ratio of change in emitter-
base voltage ( VEB) to the resulting change in emitter current ( IE) is known as input
resistance, i.e., at constant VCB
V
EB
Ri = at constant VCB
IE
The value of input resistance Ri in CB configuration is very low. Its value further
decreases with the increase in collector-base voltage VCB. The typical value of input
resistance varies from a few ohms to 100 ohms.

AC Input Resistance in Common Emitter Configuration: The ratio of change in base-

emitter voltage ( VBE ) to the resulting change in base current ( IB) at constant collector-
emitter voltage (VCE) i.e.,
V
BE
R = at constant VCE
i
IB

In CE configuration, the typical value of input resistance is of the order of a few hundred
ohms

AC Output Resistance in Common Base Configuration: The ratio of change in

collector-base voltage ( VCB) to the resulting change in collector current ( IC) at constant
emitter current (IE). i.e.,
V
CB
R = at constant IE
O
IC
The Output Resistance of CB configuration is very high, of the order of several tens of
kilo-ohms (≅ 1 MΩ).

AC Output Resistance in Common Emitter Configuration: The ratio of change in

collector-emitter voltage ( VCE) to the change in collector current ( IC) at constant base
current IB, i.e.

V
CE at constant IB
RO =
IC

The output resistance of CE configuration is less than the CB configuration. Its value is of
the order of 50 KΩ.

Q.9 Describe the operation of a transistor amplifier in CE configuration. Marks(4)

Ans: Operation of a Transistor Amplifier in CE configuration: A Common Emitter NPN

transistor is shown in fig. 16(a). When a signal is applied in the emitter-base junction,
during positive half cycle, the forward bias across this junction increases. This increases
the flow of electrons from emitter to collector via base and thus increases the collector
current. The increased collector current produces more voltage drop across the collector
load resistor RC. However, during negative half cycle of the signal, the forward bias
 voltage across the emitter-base junction decreases. This decreases the collector current
which consequently decreases the voltage drop across the collector load resistor R C.
Hence, an amplified signal appears across the collector load resistor.

Fig. 16(a)

β
dc
Q.10 Derive the relation : α dc  Marks (4)
1  β dc

Ans: Relationship between α and β: The Beta factor (β) is the current gain factor of a common
emitter circuit is defined as the ratio of change in collector current (I C) to the change in
base current (IB) i.e.,

I C  β dc I B  I CO
I C ≅ β dc I B assuming ICO is very small
IC
or β dc 
IB

and the alpha factor (α) is also called the current amplification factor and is defined as the
ratio of change in collector current (IC) to the change in emitter current (IE) i.e.,
I
C 1
α dc = =
IE 1+ IB
I
C

1 β
 dc
1 =
1  1 + β
β dc

dc

Q.11 Discuss the similarities and differences between JFET and MOSFET with regard to
their construction and applications. Marks (7)

Ans: Similarities in Construction

(i) JFET is a three terminal semiconductor device(source, gate and drain) in which
current conduction is by majority carriers (electrons or holes), whereas MOSFET is also
a three terminal semiconductor device in which current conduction is by majority carriers.
(ii) JFET is Voltage Controllable Device (i.e., the gate terminal is used for controlling the
drain current whereas MOSFET is also a Voltage Controllable Device.

Differences in Construction:
(1) An N-channel JFET is shown in fig. 18(a). It consists of a thin N-type silicon bar with
two islands of P-type semiconductor material embedded in the sides, thus forming
two PN junctions. Whereas in N-channel Depletion MOSFET, there is only one p-
region instead of two. This region is known as substrate. The two P-regions are
embedded in it as shown in fig. 18(b)
 Fig. 18(a)

M ETALLIZATION
LAYER
SOURCE GATE DRAIN
S G D

SiO DIELECTRIC LAYER

2
+ +
N N N

DIFFUSED CHANNEL
P-TYPE SUBSTRATE

Fig. 18(b) N-CHANNEL DEPLETION MODE STRUCTURE

(2) In N-channel JFET, the two p regions are connected with each other (externally or
internally) and are called gate (G). Ohmic contacts are made at the two ends of the N-
type semiconductor bar. One terminal known as the source (S) through which the
majority carriers (electrons in this case) enter the bar. The other terminal as the drain (D)
through which these majority carriers leave the bar. But in MOSFET, over the left side of
the channel, a thin layer of metal oxide (usually silicon dioxide) is deposited. A metallic
Gate is deposited over the silicon dioxide .The gate is insulated from the semiconductor
since silicon dioxide is an insulator. That is why MOSFET is also known as Insulated
Gate FET. Like JFET, ohmic contacts are made at the two ends (Drain and Source) of
the N-type semiconductor bar.
(3) Since the gate is insulated from the channel by a thin layer of silicon dioxide, the
4 9
input impedance of MOSFET is very high (of the order of 10 to 10 MΩ) whereas the
input impedance of JFET is 100MΩ as for a reverse biased pn junction.
(4) Unlike the JFET, a MOSFET has no gate diode, it forms a capacitor. The capacitor
has gate and channel as electrodes and the oxide layer is dielectric. Because of this
property, the device can be operated with negative as well as positive gate voltages.

Applications of JFET:
FET is used as a
(1) Buffer amplifier
(2) Low Noise Amplifier
(3) Cascaded Amplifier
(4) Analog Switch
(5) Chopper
(6) Phase Shift Oscillator circuits
(7) Voltage Variable Resistors in Operational Amplifiers and tone controls
etc.,
(8) For Mixer operation on FM and TV receivers
 Applications of MOSFET: MOSFETs can also be used for most applications where
JFET is used. MOSFETs have become very popular for digital logic circuits due to high
density of fabrication and low power dissipation.
(1) MOSFET is used in Sample and Hold circuit as a switch.
(2) P-MOSFET and N-MOSFET are used in digital logic circuits
(3) C-MOSFET is very popular in fabricating of MSI and LSI technology.
Q 12 Explain the principle of operation of any one type of MOSFET. Marks (6)

Ans: Operating Principle of N-channel Enhancement MOSFET: Operating principle of N-

channel Enhancement MOSFET is shown in fig. 9(a). When drain is applied with positive
voltage with respect to source and no potential is applied to the gate, two N-regions and
one P-substrate form P-N junctions connected back to back with a resistance of the P-
substrate. So a very small drain current i.e., a reverse leakage current flows. If the P-type
substrate is now connected to the source terminal, there is zero voltage across the
source substrate junction, and the drain-substrate junction remains reverse biased.

Fig. 9(a)
When the gate is made positive with respect to the source and the substrate, negative
(i.e. minority) charge carriers within the substrate are attracted to the positive gate and
accumulate close to the surface of the substrate. As the gate voltage is increased, more
and more electrons accumulate under the gate. Since these electrons cannot flow across
the insulated layer of silicon dioxide to the gate, so they accumulate at the surface of the
substrate just below the gate. These accumulated minority charge carriers make N-type
channel stretching from drain to source. When this occurs, a channel is induced by
forming what is termed an inversion layer (n-type). The n-type drain and gate are now
connected by the n-type channel and the current can flow from drain to source.

Q.13 Draw the three configurations in which a transistor may be connected showing battery
Connections for each. Marks (4)

Ans:Three Configurations (CB, CC, CE) :

IE IC IE IC

INPUT INPUT
SIGNAL
R C OUTPUT SIGNAL
RC OUTPUT
IB IB
_ + _ + _ _

V V V V
EE CB EE CB

Common Base circuit of npn transistor Common Base circuit of pnp transistor

25
 ELECTRONIC DEVICES

I E I E

RC RC
OUTPUT OUTPUT
_
IB IB
_
INPUT INPUT
SIGNAL I SIGNAL I
C C
+ _ _ +

Common collector circuit of npn transistor Common collector circuit of pnp transistor

I C
IC

IB
IB R RC
C OUTPUT OUTPUT

INPUT INPUT
SIGNAL SIGNAL
I E IE
+ _ _ + _

_ V V
V BB CC V BB CC

Common emitter circuit of npn transistor Common emitter circuit of pnp transistor
Q14 Explain how a Zener diode can be used as voltage regulator. Marks (8)

Ans: Zener diode as a Voltage Regulator: It provides a constant voltage to the load from a
source whose voltage may vary over sufficient range. Fig.31(a) shows the circuit
arrangement. The zener diode of zener voltage Vz is reverse connected across the load
RL across which constant voltage (Vo = Vz) is desired. A resistor R is connected in series
with the circuit which absorbs the input voltage fluctuations so as to maintain constant
voltage (Vo) across the load. Let a variable voltage Vin be applied to the regulator. When
the value of Vin is less than Zener voltage Vz of the zener diode, no current flows through
it and the same voltage appears across the load. When the input voltage V in is more than
VZ, this will cause the Zener diode to conduct a large current IZ.

+ R I
+

IZ
DC
(V ARIABLE) V
SUPPLY
V
in V RL o
Z

_
_

Fig. 31(a)
 Consequently, more current flows through series resistor R which increases the voltage
drop across it. Thus, the input voltage excess of Vz (i.e. Vin – Vz_) is absorbed by the
series resistor. Hence a constant voltage V0 (=Vz) is maintained across the load RL.
When the load resistance RL decreases for constant input voltage Vin, load current IL
increases. This additional current is not supplied from the source of supply but the
demand of additional load current is met by decrease in Zener current Iz.
This keeps the voltage drop across series resistance R constant and also the output
voltage Vout. The diode current Iz attains its maximum Izmax when load current IL
becomes zero. Thus the value of series resistor R can be determined from the equation

V −V
in out
R=
I
z max

Q.15 Draw the circuits to obtain the input and output characteristics of an NPN transistor
in CE configuration. Marks (10)

Ans: Characteristics of NPN Transistor in Common-Emitter configuration: To determine the

characteristics of a NPN transistor in CE configuration, the circuit arrangement as shown
in fig. 32(a)
Milli Am m eter

IC
Micro Am m eter _ +

IB
+ _
+
V
+ CE V
V
+ _ CC
V V _
V
BB BE

_ _ IE

Fig. 32(a)
Input Characteristics: In CE configuration, the curve plotted between base current I B
and base-emitter voltage VBE at constant collector-emitter voltage VCE is called Input
Characteristics.
To draw the input characteristic, note down the readings of Ammeter (IB)
connected in the base circuit for various values of V BE at constant VCE. Plot the curve on
the graph taking IB along y-axis and VBE along x-axis as shown in fig.32(b) The following
points are noted from these characteristics:
 
IB ( A)
V =2V
CE
V =6V
CE
70
V = 10 V
CE
60 I
B P
50

40
30
20 V
BE
10

0 V (VOLTS)
0.5 1.0 BE

Fig.32(b)

(i) These curves are similar like a forward diode characteristics. IB increases less
rapidly with increase in VBE.

(ii)The change in VCE does not result in a large deviation of the curves and hence, the
effect of change in VCE on the input characteristic is ignored for all practical purposes.

Output Characteristics: In CE configuration, the curve plotted between collector current

IC and collector-emitter voltage VCE at constant base current IB is called output
characteristic.

(i) In the active region, Ic increases slightly as V CE increases. The slope of the curve
is little bit more than the characteristics of CB configuration. Hence, the output
resistance (ro) of this configuration is less compared to CB configuration.

(ii)Since the value of Ic increases with the increase in VCE at constant IB, the value
of β also increases since β = IC/IB

(iii) When VCE falls below the value of VBE (i.e. below a few tenths of a volt.), Ice
decreases rapidly. In fact, at this stage, the collector-base junction is also forward
biased and the transistor works in the saturation region. In the saturation region, Ic
becomes independent and it does not depend upon the input current I B.

(iv) In the active region, IC = β IB. Hence, a small change in base-current IB produces
a large change in output current (IC)
 (v) When input current IB = 0, the collector current IC is not zero but its value is equal
to the reverse leakage current ICEO (i.e. collector emitter current when base is open).

Q.17 Mention the applications of CE, CB and CC configurations of BJT’s Marks(4)

Ans: Applications of CE configuration of BJT:

(i) CE amplifier is used in audio frequency applications like Radio receivers, Tape
recorders, TV receivers, FM receivers, etc.
(ii) CE amplifier is used in Oscillators.

Applications of CB configuration of BJT:

(i) To provide Voltage gain without Current gain: CB amplifier provides
voltage gain without increasing the value of ciruit current gain
(ii) For impedance matching in High Frequency Applications: It matches the low
impedance source and high impedance load in order to have a maximum power
transfer between source &load
Applications of CC configuration of BJT:
(i) To provide current amplification with no voltage gain: In digital
Electronics, where an increase in current is required but no increase in voltage is
needed. In such situation CC configuration can be used.
(ii) Impedance Matching: It matches the high impedance source and
low impedance load in order to have a maximum transfer between source and load.

Q.18 Differentiate between FET and BJT transistors. Marks (4)

Ans Differences between FET and BJT Transistors:

(i) FET operation depends upon the flow of majority carriers only, it is
therefore, a uni-polar (current conduction is due to one type of carriers either
electrons or holes) device. On the other hand, the current conduction in BJT is due to
both majority and minority carriers, it is therefore a bipolar device.
(ii) FET has high input impedance (of the order of 100 Megaohms), because its input
circuit (gate to source) is reverse biased, and so permits high degree of isolation
between the input and output circuits. On the other hand, BJT has low input
impedance, because its input circuit is forward biased.
(iii) FET carries very small current because of reverse biased gate and input voltage
controls the output current. This is the reason that FET is essentially a voltage driven
device. BJT carries high current because of forward biased and it is a current
operated device, since input current controls the output current.
(iv) BJT uses a current into its base for controlling a large current between collector
and emitter whereas in a FET, voltage on the gate terminal is used for controlling the
drain current (current between drain and source. Thus an ordinary transistor, gain is
characterized by current gain whereas the FET gain is characterized as the
transconductance.
(v) FET has no junction like BJT and the conduction is through bulk material current
carriers (N-type or P-type semiconductor material) that do not cross junctions. Hence
the inherent noise of BJT (owing to junction transitions) is not present in FET
(vi) FET has small gain-bandwidth product, low voltage gain in comparision to BJT.

Q.19 Explain the constructional features, principle of operation and characteristics of N-

channel JFET. Marks (10)

Ans: Constructional Features of N- Channel JFET: In a N-channel JFET, an 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, as illustrated in fig.35.1
 Fig. 35.1
The two P-N junctions forming diodes or gates are connected internally and a common
terminal, called the gate terminal is brought out. Ohmic contacts (direct electrical
connections) are made at the two ends of the channel – one lead is called the source
terminal (S) through which the majority carriers (electrons) enter the bar and the other
drain terminal (D) through which the majority carriers leave the bar which is shown in
fig. 35(b)

Fig. 35.2

Principle of Operation of N-Channel JFET:

The circuit diagram of an N-channel JFET with normal polarities is shown in fig.35.3

Fig. 35.3
When a voltage VDS is applied across the drain and source terminals and voltage
applied across the gate and source VGS is zero (i.e. gate circuit is open) as shown in
above fig.3 5.3(a), the two PN junctions establish a very thin depletion layer. Thus a large
amount of electrons will flow from source to drain through a wide channel formed
between the two depletion layers. When a reverse voltage VGS is applied across the gate
and source as shown in above fig. 35.3(b), the width of the depletion layer is increased.
This reduces the width of the conducting channel thereby decreasing the conduction
(flow of electrons) through it. Thus the current flowing from source to drain depends upon
the width of the conducting channel which depends upon the thickness of depletion layer.
The thickness of depletion layer established by the two PN junctions depends upon the
voltage applied across the gate-source terminal. Hence, it is clear that the current from
source to drain can be controlled by the application of potential (i.e. electric field) on the
gate. That is why, the device is called Field Effect Transistor.
 Characteristics of N-Channel JFET: A curve drawn between drain current (ID) and
drain-source voltage (VDS) of a JFET at constant gate-source voltage (VGS) is known as
output characteristics of the JFET are shown in below fig.35.4
(i) At the initial stage, the drain current ID increases rapidly with the increase in drain
source voltage VDS but then becomes almost constant. This voltage is called pinch-
off voltage.
(ii) After pinch off voltage, the depletion layers almost touch each other and the
conduction channel becomes very narrow. Therefore, the increase in drain current ID
is very small with the increase in drain-source voltage VDS. Thus, the drain current
almost becomes constant.

(iv) Further increase in drain-source voltage VDS eventually causes the avalanche
breakdown across the reverse-biased gate junction and the current ID shoots to
a very high value.

Fig. 35.4
Q 20 Explain the construction, working and characteristics of depletion mode MOSFET.
Marks (10)

Ans: Construction of N-Channel Depletion MOSFET: Fig. 42.1 shows the construction of N-
channel depletion MOSFET. It consists of a highly doped P-type substrate into which two
blocks of heavily doped N-type material are diffused forming the source and drain. An N-
channel is formed by diffusion between the source and drain. The type of impurity for the
channel is the same as for the source and drain. Now a thin layer of S i02 dielectric is
grown over the entire surface and holes are cut through the SiO 2 (silicon-dioxide) layer to
make contact with the N-type blocks (Source and Drain). Metal is deposited through the
holes to provide drain and source terminals, and on the surface area between drain and
source, a metal plate is deposited. This layer constitutes the gate. SiO 2 layer results in an
10 15
extremely high input impedance of the order of 10 to 10 ohms for this area.

M ETALLIZATION
LAYER
SOURCE GATE DRAIN
S G D

SiO DIELECTRIC LAYER

2
+
N N N +

DIFFUSED CHANNEL
P-TYPE SUBSTRATE

Fig. 42.1 N-CHANNEL DE-MOSFET STRUCTURE

 Working: When negative gate-to-source voltage is applied, the N-channel MOSFET
operates in the depletion mode.
When VGS = 0, electrons can flow freely from source to drain through the
conducting channel which exists between them. When gate is given negative voltage, it
depletes the N-channel of its electrons by including positive charge in it as shown in fig.
42.2. Greater the negative voltage on the gate, greater is the reduction in the number of
electrons in the channel and consequently, lesser its conductivity. In fact, too much
negative gate voltage called VGS(off) can cut-off the channel. Hence, with negative gate
voltage, a Depletion MOSFET behaves like a JFET.
.

V
_ DD +

V
GG _
+
S G D METALLIZATION
LAYER
_ _ _ _
-----
+++++++++++
SiO DIELECTRIC LAYER
----------------
+ ++ + + 2
+
++
+ +
N
N
N-CHANNEL

P-TYPE SUBSTRATE

Fig 42.2 DEPLETION MODE OPERATION

Characteristics: Typical drain characteristics, for various levels of gate-source voltage,

of N-channel depletion MOSFET are shown in fig.42.3. The upper curves are for positive
VGS and the lower curves are for negative VGS. The bottom drain curve is for VGS =
VGS(OFF). For a specified drain-source voltage VDS, VGS(OFF) is the gate-source voltage at
which drain current reduces to a certain specified negligibly small value as shown in the
figure. For VGS between VGS(OFF) and zero, the device operates in depletion mode.

Fig. 42.3

Q.21 Distinguish between majority and minority carriers in a semiconductor. Define

mobility of charge carriers. Marks (5)

Ans: Distinguish between majority and minority carriers:

(i) Majority Carriers are the charge carriers which are maximum in number. For example
in n-type semiconductor the majority charge carriers are electrons and in p-type
 semiconductor, the majority charge carriers are holes as shown in fig.50.1, whereas
Minority Carriers are the charge carriers which are minimum in number. For example in
n-type semiconductor, the minority charge carriers are holes and in p-type
semiconductor, the minority charge carriers are electrons as shown in fig.50.2.

MAJORITY CARRIERS MAJORITY CARRIERS

(FREE ELECTRONS) (HOLES)

MINORITY CARRIERS MINORITY CARRIERS

N-TYPE P-TYPE
(HOLES) (FREE ELECTRONS)

Fig. 50.1 Fig.50.2

(ii) Most of the current conduction in a semiconductor is due to majority carriers, whereas,
the current due to minority carriers is very small, so, it almost negligible.

Mobility of charge carriers: When a steady electric field E volt/meter is applied to a

metal, the electrons move towards the positive terminal of the applied voltage. In their
way, they continuously collide with other free electrons and rebound in a random fashion.
Each collision being inelastic i.e., the electrons lose some kinetic energy.
After the collision, the electrons are accelerated by the electric field and gain certain
component of velocity in the direction opposite to that of electric field (- E). They lose
some of their energy at the next collision. Thus the applied electric field does not stop
collision and random motion but makes the electrons to drift towards the positive
terminal. So, the drift velocity v is proportional to the applied electric fied E, i.e., V = e E
Where e is called mobility which is defined as the Average Particle Drift velocity per unit
2
electric field and is expressed in m /Vs.

Q.22 Discuss how a depletion layer is formed in a P-N diode and how does it vary with
biasing ? Draw V-I characteristics of P-N junction diode. Marks (7)

Ans: Formation of Depletion Layer in a P-N junction Diode: The two types of extrinsic
semiconductors are p-type and n-type as shown in fig.51(a) The p-type semiconductor is
having negative acceptor ions and positively charged holes. Whereas, the n-type
semiconductor is having positive donor ions and negatively charged electrons.

P-TYPE
N- TYPE
SEMICONDUCTOR
HOLE SEMICONDUCTOR

- - - - + + ++
-
ELECTRON
- - - - + + + +

POSITIVE
NEGATIVE DONOR IONS
ACCEPTOR IONS

Fig. 51(a)
When these two semiconductors are joined together and suitably treated, they from a pn
junction. The moment they form a pn junction, some of the conduction electrons from n-
type material diffuse over to the p-type material and undergo electron-hole recombination
with the holes available in the valence band. Simultaneously holes from p-type material
diffuse over to the n-type material and undergo hole-electron combination with the
electrons available in the conduction band. In this process, some of the free electrons
move across the junction from n-type to p-type, leaving behind positive donor ions as
they are robbed of the free electrons. This establishes a positive charge on the n-side of
the junction. Simultaneously, the free electrons which cross over the junction recombine
with the holes of p-type and uncover some of the negative acceptor ions as shown in
fig.51(d) This establishes a negative charge on the p-side of the junction. This process of
diffusion continues till a sufficient number of donor and acceptor impurity ions are
uncovered and establish a requisite potential difference (i.e. nearly 0.3 V in case the pn
junction is formed of germanium semiconductor and 0.7 V for silicon pn junction). After
this, further diffusion is prevented because now positive charge on n-side repels holes to
cross from p-type to n-type and negative charge on p-side repels free electrons to enter
from n-type to p-type.

A potential difference created across the junction acts as a barrier which restricts further
movement of charge carriers (i.e. holes and electrons) is called a Potential Barrier or
Junction Barrier Vo. Thus, a region around the junction from which the charge carriers
(free electrons and holes) are depleted is called Depletion Layer.

NEGATIVELY POSITIVELY
CHARGED CHARGED
IONS V
P - o + N IONS
-- - - + + + +

-
- - - + + + +

DEPLETION
LAYER

Vo

Fig. 51(b)

Behaviour of a PN junction Diode under Biasing:

When a pn junction is connected across an electric supply (potential difference),
the junction is said to be under biasing.

Forward Biasing: When the positive terminal of a d.c source or battery is connected to
p-type and negative terminal is connected to n-type semiconductor of a pn junction as
shown in fig.51(c), the junction is said to be in forward biased. In this circuit, the holes on
the P-side being positively charged particles are repelled from the positive bias terminal
and driven towards the junction. Similarly, the electrons on the N-side are repelled from
the negative bias terminal and driven towards the junction. The result is that the depletion
region is reduced in width and the barrier potential is also reduced. Once the potential
barrier is eliminated by the forward voltage, a conducting path is established for flow of
current.
 + -

P N

DEPLETION
LAYER WITHOUT DEPLETION
EXTERNAL FIELD LAYER WITH

EXTERNAL FIELD

NO EXTERNAL
EXTERNAL
FIELD
FIELD

Fig. 51(c)

Reverse Biasing: When the positive terminal of a d.c. source or battery is connected to
n-type and negative terminal is connected to p-type semiconductor of a pn junction, as
shown in fig.51 (d), the junction is said to be in reverse biased. In this circuit
arrangement, the electrons form the N-side are attracted to the positive bias terminal and
holes form P-side are attracted from the negative bias terminal. Thus, the depletion
region is widened and the barrier potential is increased by the magnitude of the applied
bias. With the increased barrier potential, there is no possibility of majority carrier current
flow across the junction. Thus the P-N junction is in non-conducting state.

P N

DEPLETION
LAYER WITH DEPLETION
EXTERNAL FIELD LAYER WITHOUT
EXTERNAL FIELD

EXTERNAL
NO EXTERNAL FIELD
FIELD
 Fig. 51 (d)

V-I characteristics of P-N junction diode are shown in fig. 51(e)

Fig. 51(e)

Q.23 What is cut-in voltage in semiconductor ? Marks (2)

Ans: Cut-in voltage: The forward voltage (0.3V for Ge and 0.7V for Si diodes) at which the
current through the diode or pn junction starts rising abruptly is known as cut-in voltage or
knee voltage of a diode.

Q.24 What is zener diode and how does it regulate the voltage ? What happens to the
series current, load current and zener current when the d.c. input voltage of a
zener regulator increases ? Marks (6)

Ans: Zener Diode: Zener diode is specially designed silicon diode which is optimized to operate
in the breakdown region.Zener diode is used as a voltage regulator: Fig.53.1 shows the
circuit arrangement. The zener diode of zener voltage Vz is reverse connected across the
RL across which constant voltage (Vo = Vz) is desired. A resistor R is connected in series
with the circuit which absorbs the output voltage fluctuations so as to maintain constant
voltage (Vo) across the load.

+ R I
+

IZ
DC
(VARIABLE)
SUPPLY
V
in V RL Vo
Z

_
_

Fig. 53.1
When the supply voltage VS increases, the current through both the zener diode and
load resistance RL increases.

At the same time, however, the current through the diode increases more than
proportionately. As a result, a greater voltage drop will occur across the series resistor RS
 and the output voltage VOUT (voltage across the diode or load resistance RL) will become
very close to the original value. When the supply voltage VIN is less than zener voltage
VZ of the zener diode, no current flows though the diode, because the diode is reverse
biased and the same voltage appears across the load. When d.c. input voltage of a zener
regulator increases, series current I, load current IL and zener current IZ increases.

Q.25 What is LED ? Give its principle of working and applications. Marks (4)

Ans: LED: LED (Light Emitting Diode) is a Forward Biased P-N junction diode which emits light
under the influence of an electric field. The structure, schematic diagram and symbol of
LED are shown in fig.54(i), 54(ii) and 54(iii).

Principle of working: When P-N junction diode is forward-biased, the potential barrier is
lowered. The conduction band free electrons from n-region cross the barrier and enter
the p-region. As these electrons enter the p-region, they fall into the holes lying in the
valence band. Hence they fall from a higher level to a lower energy level. In the process,
they radiate energy. In the ordinary diodes (rectifier and signal diodes) this energy

radiates in the form of heat, as these diodes are made of silicon or germanium which are
opaque material and block the passage of light.

__
P N +

(i) Structure (ii) Schematic Diagram (iii) Schematic Symbol

Fig. 54

LED’s are made from gallium arsenide phosphide (GaAsP) and gallium
phosphide (GaP) and these radiate energy in the form light (or photons) and hence they
glow.

Applications: LEDs that emit visible light (such as red, green, blue etc.) find their
applications in the following areas
(i)Instrument Display panel
indicator (ii)Digital watches
(iii)Calculators
(iv)Multimeters
(v)Intercom
(vi)Telephone switch boards
Infrared LED’s find applications in remote control schemes, object detectors etc.

Q.26 What is varactor diode ? On what factor its operation is based ? Give its applications.
Marks (4)

Ans: Varactor diode: The varactor diode [also called the varicap, voltcap, epicap, VVC
(voltage-variable capacitance) or tunning] is a voltage-dependent, variable capacitor.
Basically, it is a reverse-biased P-N junction diode whose mode of operation is based on
its transition capacitance.

Operation based factor: The P- and N-regions (away from the space-charge region) are
essentially low resistance areas because of high concentration of majority carriers. The
space-charge or depletion region, which is depleted of majority carriers, serves as a
dielectric. The P- and N-regions act as the plates of the capacitor and the depletion
region acts as an insulating dielectric. The reverse biased P-N junction thus possesses
junction capacitance, also called the transition capacitance C T. The junction capacitance
is given as
 CT ∈
A
W
Where ∈ is the permittivity of the semiconductor material. A is the area of the PN junction
and W is the width of the space-charge region. With increase in reverse-bias potential,

The width of the space charge region increases, which in turn reduces the Transition
Capacitance CT as shown in fig.55.1

C (0) 80
T

60

40 C T IN pF

20

-14 -12 -10 -8 -6 -4 -2

REVERSE BIAS IN VOLTS

Fig.55.1

Applications: Varactor diodes are used in

(i) FM Modulators
(ii) High Frequency Control Devices
(iii) Adjustable Band pass Filters
(iv) Parametric Amplifiers.
Q 27 Distinguish between forward active, saturation, cut-off and reverse active operation
of a transistor. Marks (4)

Ans: I.Forward Active: In this condition, base-emitter junction is forward biased, whereas the
base-collector junction is reverse biased. When the transistor is an active region,
collector current is proportional to the emitter current. Generally, transistor is operated in
this region for amplification,
II.Saturation: In this condition, both base-emitter and base-collector junctions are
forward biased. When the transistor is in saturation, the collector current becomes
independent of the base current and is limited by external circuitry. The transistor acts
like a closed switch.
III.Cut-Off: In this condition, both junctions are reverse biased. The emitter does not emit
carriers into the base and no carriers are collected by the collector (except a little
thermally generated minority carrier). Thus, the transistor acts like an open switch.
IV Reverse Active: In this condition, the base-emitter junction is reverse biased,
whereas the base-collector junction is forward biased. As the collector is not doped to the
extent as the emitter is therefore, it cannot supply (emit) as many majority carriers to the
base. Hence, in this case very poor transistor action is achieved.
Q.28 Sketch the CE-configuration transistor output characteristics of a transistor and
explain thesignificance of these curves. Indicate the active, cut-off and saturation
regions. Marks (7)

Ans: Output Characteristics of a Transistor in CE-Configuration: In CE configuration, the

curve plotted between collector current IC and collector-emitter voltage VCE at constant
base current IB is called output characteristic, shown in fig. 59.1.

Fig. 59.1

(i) In the active region, Ic increases slightly as VCE increases. The slope of the curve is
little bit more than the characteristics of CB configuration. Hence, the output resistance
(ro) of this configuration is less as compared to CB configuration.
(ii) Since the value of IC increases with the increase in VCE at constant IB, the value of β

β 
I
Q

also increases
C

I
B

(iii) When VCE falls below the value of VBE (i.e. below a few tenths of a volt.), Ic decreases
rapidly. In fact, at this stage, the collector-base junction is also forward biased and the
transistor works in saturation region, where Ic becomes independent of input current IB.
(iv) In the active region, Ic = β Ib, hence a small change in base-current IB produces a
large change in output current (IC)
(v) When the input current IB = 0, the collector current IC is not zero but its value is equal
to the reverse leakage current ICEO (i.e. collector emitter current when base is
open).

Q.29 Prove that in a transistor operating in CE-configuration, the active region, collector
current IC is given by IC= β IB + (β + 1) ICO Marks (3)

Ans: Proof: In CE configuration, the input current is IB and the output current IC. These
currents are related by the equations:
IE = IB + IC and IC = α IE + ICBO = α (IB + IC) + ICBO
α 1 I
Or IC (1 - α) = α IB + ICBO IC = IB + CBO
1−α 1−α
In CE configuration, if the base circuit is open (i.e., I B = 0), the collector current will be
the current to the emitter. This current is abbreviated as I CEO that means collector-emitter
current with base open.
I
1 I
1
Hence, CEO = CBO =  β + 1I CBO Q = β+1
(1 − α ) 1−α
and the basic current amplification factor β in terms of α is given by the equation

β= α
 1−α
By substituting these values in the above equation, we get total collector current as
IC = β IB + (β + 1) ICBO Hence Proved.
Q 30 Explain the conduction of current in a good conductor. Why does a conductor has
low resistance ? Marks (7)

Ans: Conduction of current in a good conductor: The substances (like copper, aluminum,
silver etc.) which allow the passage of current through them are known as conductors. In
case of conducting materials, there is no forbidden energy gap (i.e., the energy gap
between valence band and conduction band) and the valence and conduction bands
overlap, as shown in fig.72.1

ConductionBandBand

Energy
Level in eV

Valence Band
Valence Band

Fig.72.1
 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 throughout the solid. This is the reason that electrons in the
conduction band are called the free electrons. Due to overlapping of valence and conduction bands, a
slight potential difference across the conductor causes the free electrons to constitute heavy flow of
electric current through them. Conductor has low resistance, because its valence and conduction
band overlaps. So, its conductivity becomes high.

Q.31 What is intrinsic semiconductor. How do we make it extrinsic semiconductor, and why so ?
Marks (7)

Ans: Intrinsic Semiconductor: An intrinsic semiconductor is one which is made of the semiconductor
material in its extremely pure form.

Making of Extrinsic Semiconductor from Intrinsic Semiconductor: Although an intrinsic

semiconductor is capable to conduct a little current even at room temperature but as it is, it is not
useful for the preparation of various electronic devices. However, the electrical conductivity of intrinsic
semiconductor can be increased many times by adding very small amount of impurity (of the order of
one atom per million atoms of pure semiconductor) to it in the process of crystallization. This process
is called doping and the doped material is called impurity or extrinsic semiconductor .

The purpose of adding impurity in the semiconductor crystal is to increase the number of free
electrons or holes to make it conductive. If a small amount of pentavalent impurity (having 5 valence
electrons), such as arsenic, antimony, bismuth or phosphorus etc. are added to a pure semiconductor
a large number of free electrons will exist in it and it is called N-type (donor type) extrinsic
semiconductor. If a small amount of trivalent impurity (having 3 valence electrons), such as boron,
gallium, indium or aluminium etc.

are added to a pure semiconductor, a large number of holes will exist in it and it is called
the P-type (acceptor) semiconductor.

Q.32 Explain the Zener and Avalanche effects. Give differences between them.
Marks (7)

Ans: Avalanche Breakdown: The minority carriers, under reverse biased conditions, flowing
through the junction acquire a kinetic energy which increases with the increase in reverse
voltage. At a sufficiently high reverse voltage (say 5v or more), the kinetic energy of
minority carriers becomes so large that they knock out electrons from the covalent bonds
of the semiconductor material.
As a result of collision, the liberated electrons in turn liberate more electrons and the
current becomes very large leading the breakdown of the crystal structure itself. This
phenomenon is called Avalanche Breakdown.

Zener Breakdown: Under a very high reverse voltage, the depletion region expands and
7
the potential barrier increases leading to a very high electric field (3X10 V/m) 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 increases the
reverse current. This is called the Zener effect. The breakdown occurs at a particular and
constant value of reverse voltage called the breakdown voltage.

Differences between Zener and Avalanche Breakdown:

(i) Diode junctions that breakdown below 5 V are caused by Zener effect whereas
Junctions that experience breakdown above 5 V are caused by Avalanche effect.
(ii) The Zener breakdown occurs in heavily doped junctions (P-type semiconductor
moderately doped and N-type heavily doped), which produce Narrow depletion
layers, whereas Avalanche breakdown occurs in lightly doped junctions, which
produce wide depletion layers.
(iii) With the increase in junction temperature, Zener breakdown voltage is reduced
while the Avalanche breakdown voltage is increases.
 (iv) The zener diodes have a negative temperature coefficient while Avalanche
diodes have a positive temperature coefficient.

Q.33 With the help of a neat diagram, explain the operation of a Bridge Rectifier. What
is PIV for the diode used here Marks (7)

Ans: Operation of a Full-Wave Bridge Rectifier: Fig (75.1) shows the circuit of a Full-Wave
Bridge Rectifier. The circuit consists of four diodes D 1, D2, D3 and D4 connected to form
a bridge. When a.c. supply is switched on, the alternating voltage V in appears across the
terminals AB of secondary winding of transformer.
During positive half cycle of secondary voltage, the end A becomes positive and
end B is negative (see fig.75.3). This makes diodes D 1 and D3 forward biased and diodes
D2 and D4 reverse biased. Therefore, diodes D1 and D3 conduct while diodes D2 and D4
do not conduct. Thus, current (I) flows through diode D 1, load resistor RL (from M to L),
diode D3 and the transformer secondary as shown in fig (75.3). The wave shape of
current through RL is shown in fig.75.2. (b)

Fig. (75.1) Fig. (75.2)

During negative half cycle, the end A becomes negative and end B positive (see Fig.
(75.4). This brings diodes D2 and D4 under forward bias and diodes D1 and D3 under
reverse bias. Therefore, diodes D2 and D4 conduct while diodes D1 and D3 do not. Thus,
current (i) flows through diode D2, load resistor R L (from M to L), diode D4 and the
transformer secondary as shown in fig. (75.4). The wave shape of current is shown in
fig.75.2 (b)

Fig.(75.3) Fig (75.4)

The current flowing through load resistor R L is in the same direction (M to L) during both
the half cycles. Hence, rectified output voltage Vout is obtained across the load resistor
RL. The wave shape of output voltage is shown in fig.75.2(c)
Peak Inverse Voltage: The maximum possible value of reverse bias voltage coming
across the diode is called the Peak Inverse Voltage.

Fig (75.5)
Fig (75.5) shows the instant when secondary voltage attains its maximum value and
terminal A is positive and B is negative. At this instant, diodes D 1 and D3 are forward
biased and conducting current. Therefore, terminal M attains the same voltage as that of
A’ or A whereas, L attains the same voltage as that of B’ or B. Hence, the diode D 2 and
D4 are reverse biased and the voltage across both of the diodes is Vm
Thus, the P.I.V. required for diodes in a Bridge Rectifier is at least Vm.
P.I.V = Vm

88
 ELECTRONIC DEVICES

Q.35 Explain the functioning of a capacitor filter used with the rectifiers. Marks (7)

Ans: Function of a Capacitor Filter used with the Rectifiers: A Shunt Capacitor filter is shown
in fig. 76.1 In this case, a capacitor C of large value is connected in parallel with the load
resistance and hence the name Shunt Capacitor Filter. The capacitor offers a low reactance
path to a.c. component and allows it to pass through but it acts as an open circuit to d.c.
component. Hence, all the d.c.current passes through the load.

Fig.76.1 Fig.76.2

Working: The working of a shunt capacitor filter can be explained with the help of a wave
diagram shown in Fig.76.2. The dotted pulsating wave shows the output of a full-wave
rectifier. When the rectifier voltage is increasing, the capacitor is charged to +V m (see curve
AB). At point B, the rectifier voltage tries to fall but the charged capacitor immediately tries to
send the current back to the rectifier. In this process, the rectifier diodes are reverse biased
and stop giving supply to the load. Thus, the capacitor discharges (B to C) through the load.
The capacitor continues to discharge until the source voltage (the dotted curve) becomes
more than the capacitor voltage (point C). The diode starts conducting and the capacitor is
again charged to peak value +Vm (point D). During this time (C to D), the rectifier supplies the
charging current ic and load current iL. So, capacitor not only removes the a.c. component
from the output but it also improves the output voltage. The smoothness and magnitude of
output voltage depends upon the time constant CRL. The longer the time period, the steadier
is the output voltage. This can be achieved by using a large value of capacitor .