MATERIALS, DEVICES AND SIMPLE

SEMICONDUCTOR ELECTRONICS:

CIRCUITS

CHAPTER – 14

SEMICONDUCTOR ELECTRONICS: MATERIALS, DEVICES AND SIMPLE CIRCUITS

The word "electronics' is derived from electron + dynamics Due to this a large number of electrons are available for
which means the study of the behavior of an electron under electrical conduction and therefore its resistivity is low ( = 10–2
different conditions of externally applied field. This field of – 10–8U–m) and conductivity is high [ =102 – 108 (–m)–1].
science deals with electronic devices and their utilization. An Such materials are called conductors. For example, gold, silver,
electronic device is a device in which conduction takes place by copper, etc.
the movement of electron - through a vacuum, a gas or a Insulator
semiconductor. Main application of electronic is computer In some solids energy gap is large (Eg > 3eV). So, in conduction
which is used in every field. All electronics equipment required band there are no electrons and so no electrical conduction is
D.C. supply for operation (not A.C. supply). possible. Here energy gap is so large that electrons cannot be
Energy Band easily excited from the valence band to conduction band to
The energy levels of an isolated atom are clearly defined. conduction band by any external energy (electrical, thermal or
However, these individual energy levels overlap and undergo optical)
substantial modification when numerous such atoms combine Such materials are called as "insulator". Their U > 1011 –m
to create a real solid. and  < 10–11 (Q–m)–1
The energy values of electrons are not discrete but rather fall Semiconductor
within a range. An energy band is thought to be formed by the In some solids a finite but small band gap exists (Eg < 3eV). Due
accumulation of these densely grouped energy levels. Valence to this small band gap some electrons can be thermally excited
Band and Conduction Band are two terms used to describe these to "conduction band".
types of bands that form in solids. The Valence Band is made up These thermally excited electrons can move in conduction band and
of filled energy levels, while the Conduction Band is made up of can conduct current. Their resistivity and conductivity both are in
partially filled or unfilled energy levels. A space known as the medium range, ≃10–5 – 106 –m and  ≃10–6 – 105 –m–1
energy gap or forbidden gap typically separates the two bands. Example of semiconducting materials
Classification Of Solids According to Energy Band Theory Elemental semiconductor: Si and Ge
Compound semiconductor
According to energy band theory, solids are conductor,
Inorganic: CdS, GaAs, CdSe, InP etc.
semiconductor and insulator:
Organic: Anthracene, Doped pthalocyanines etc.
Conductor Organic Polymers: Poly pyrrole, Poly aniline, polythiophene
In some solids conduction band and valence band are Properties Of Semiconductor
overlapped so there is no band gap between them, it means Eg Negative temperature coefficient (a.), with increase in
= 0. temperature resistance decreases. Crystalline structure with
covalent bonding [Face centered cubic (FCC)]. Conduction bonded electrons become free), vacancy of free e– creates in
properties may change by adding small impurities valence band. The electron vacancy called as "hole" which has
Position in periodic table - IV group (Generally} same charge as electron but positive. This positively charged
Forbidden energy gap (0.1 eV to 3 eV) vacancy move randomly in semiconductor solid.
Charge carriers: electron and hole.
Properties of holes
There are many semiconductors but few of them have practical
➢ It is missing electron in valence band.
application in electronics.
➢ It acts as positive charge carrier.
Holes ➢ Its effective mass is more than electron.
Due to external energy (temperature or radiation) when ➢ Its mobility is less than electron.
electron goes from valence band to conduction band (i.e.,

Holes acts as virtual charge, although there is not physical charge on it.

Q. Find the maximum wavelength of electromagnetic radiation, which can create a hole-electron pair in germanium. Given
that forbidden energy gap in germanium is 0.72eV.
Sol. Here, Eg = 0.72eV = 0.72 × 1.6 × 10−19
ℎ𝑐
The maximum wavelength of radiation, which can create a hole-electron pair in germanium is given by 𝐸𝑔 =
𝜆
Or
hc 6.62×10−34 ×3×108
𝜆= = = 1.724 × 10−6 m
Eg 0.72×1.6×10−19

Intrinsic Semiconductor properties of the semiconductor and makes it more suitable for
Pure semiconductors are in which the conductivity is caused due electronic devices such as diodes and transistors. While adding
to charge carriers made available from within the material are impurities, a small amount of suitable impurity is added to pure
called intrinsic semiconductors. There are no free charge carriers material, increasing its conductivity by many times. Extrinsic
available under normal conditions. However, when the semiconductors are also called impurity semiconductors or
temperature is raised slightly, some of the covalent bonds in the doped semiconductors. The process of adding impurities
material get broken due to thermal agitation and few electrons deliberately is termed as doping and the atoms that are used as
become free. In order to fill the vacancy created by absence of an impurity are termed as dopants. The impurity modifies the
electron at a particular electrical properties of the semiconductor and makes it more
location, electron from suitable for electronic devices such as diodes and transistors.
other position move to this The dopant added to the material is chosen such that the original
location and create a lattice of the pure semiconductor is not distorted. Also, the
vacancy (absence of dopants occupy only a few of the sites in the crystal of the
electron) at another place original semiconductor, and it is necessary that the size of the
called hole. The dopant is nearly equal to the size of the semiconductor atoms.
movement/shifting of
electrons and holes within
the material results in
conduction.
An intrinsic semiconductor behaves as a perfect insulator at
temperature 0 K
Extrinsic Semiconductors
Extrinsic semiconductors are semiconductors that are doped
with specific impurities. The impurity modifies the electrical

Q. Distinguish between intrinsic and extrinsic semiconductors?

Sol. A semiconductor free from all types of impurities is called an intrinsic semiconductor. At room temperature, a few
covalent bonds break up and the electrons come out. In the bonds, from which electrons come out, vacancies are
created. These vacancies in covalent bonds are called holes. In an intrinsic semiconductor, holes and electrons are equal
in number and they are free to move about in the semiconductor. On the other hand, a semiconductor doped with a
suitable impurity (donor or acceptor) so that it possesses conductivity much higher than that of pure semiconductor is
called an extrinsic semiconductor. The extrinsic semiconductor may be of n-type or p-type.
 Q. Why is a semiconductor damaged by a strong current?
Sol. A strong current, when passed through a semiconductor, heats up the semiconductor and the covalent bonds break up.
It results in a large number of free electrons. The material then, behaves just as a conductor. As now the semiconductor
no longer possesses the property of low conductor it is said to be damaged.

n-type semiconductor Formation of P-N Junction

When a pure semiconductor (Si or Ge) is doped by pentavalent As we know, if we use different semiconductor materials to
impurity (P, As, Sb) then four electrons out of the five valence make a P-N junction, there will be a grain boundary that would
electrons of impurity take part in covalent bonding, with four inhibit the movement of electrons from one side to the other by
silicon atoms surrounding it and the fifth electron is set free. scattering the electrons and holes and thus, we use the process
These impurity atoms which donate free e– for conduction are of doping. We will understand the process of doping with the
called as Donor’s impurity (N0). Here free e– increases very much help of this example. Let us consider a thin p-type silicon
so it is called as "N" type semiconductor. Here impurity ions semiconductor sheet. If we add a small amount of pentavalent
known as "Immobile Donor positive Ion". "Free e–" called as impurity to this, a part of the p-type Si will get converted to n-
"majority" charge carriers and "holes" called as "minority" type silicon. This sheet will now contain both the p-type region
charge carriers. and the n-type region and a junction between these two regions.
The processes that follow after forming a P-N junction are of two
types – diffusion and drift. There is a difference in the
concentration of holes and electrons at the two sides of a
junction. The holes from the p-side diffuse to the n-side, and the
electrons from the n-side diffuse to the p-side. These give rise to
a diffusion current across the junction.

p- type semiconductor
When a pure semiconductor (Si or Ge) is doped by trivalent
impurity (B, Al, In) then the outermost three electrons of the
valence band of impurity, take part in covalent bonding with four
silicon atoms surrounded by it. This shows that there remains a
Direction of diffusion current: P to N side and drift current: N to
vacancy in the band. To fill this vacancy, an electron is accepted
P side
from the neighboring atom leaving a hole from its own site.
If there is no biasing then | diffusion current | = | drift current
Thus, an extra hole is formed. These impurity atoms accepting
|So total current is zero. In junction N side is at high
bonded e– from valence band are called as Acceptor impurity
potential relative to the P side. This potential difference
(NA). Here holes increases very much so it is called as "P" type
tends to prevent the movement of electron from the N
semiconductor. Here impurity ions known as "Immobile
region into the P region. This potential difference is called
Acceptor negative Ion", Free e– are called as minority charge
Barrier potential.
carries and holes are called as majority charge carriers.

p-n junction diode under forward bias

In this arrangement the positive terminal of battery is connected
to p-end and negative terminal to n-end of the crystal, so that an
P-N Junction external electric field E is established directed from p to n-end
A P-N junction is an interface or a boundary between two to oppose the internal field, Ei. Thus, the junction is said to
semiconductor material types, namely the p-type and the n- conduct.
type, inside a semiconductor.
 During the first half (positive) of the input signal, S1 is at positive
and S2 is at negative potential. So, the PN junction diode D is
forward biased. The current flows through the load resistance RL
and output voltage is obtained across the RL.

Under this arrangement the holes move along the field E from p-
region to n-region and electrons move opposite to field E from
n-region to p-region; eliminating the depletion layer. A current
is thus set up in the junction diode. The following are the basic
features of forward biasing. Within the junction diode the
current is due to both types of majority charge carriers but in
external circuit it is due to electrons only. The current is due to
diffusion of majority charge carriers through the junction and is
of the order of milliamperes.

During the second half (negative) of the input signal, S1 is at

negative potential and S2 is at positive potential. The PN junction
diode will be reversed biased. In this case, practically no current
would flow through the load resistance. So, there will be no
output across the RL.
p-n junction diode under reverse bias
Thus, corresponding to an alternating input signal, we get a
In this type of biasing we apply a potential difference such that
unidirectional pulsating output called rectified output.
P-side is at low potential and N-side is at high potential as shown
in the diagram. Full Wave Rectifier
It rectifies both the cycles of input ac wave. It is of two types
The applied voltage is same side of to the junction barrier
(fundamentally).
potential. Due to this effective potential barrier increased,
junction width also increased, so no majority carriers will be Centre tape rectifier: Figure shows the experimental
allowed to flow across junction. arrangement for using diode as full wave rectifier. When the
alternating signal is fed to the transformer, the output signal
Only minority carriers are drifted. It means the current flow in
appears across the load resistance RL.
Principledly due to minority charge carries and is very small (UA)
called as reverse current.

The current under reverse bias is essentially voltage independent

up to a critical reverse bias voltage, known breakdown voltage (Ei). During the positive half of the input signal: S1 positive and S2
When V = Ei, the diode reverses current increases sharply. Even a negative. In this case diode D1 is forward biased and D2 is reverse
slight increase in the bias voltage causes large change in the current. biased. So only D1 conducts and hence the flow of current in the
This phenomenon is known as Breakdown. load resistance RL is from A to B.
Application of Junction Diode During the negative half of the input signal: S1 is negative and S2
Rectifier is positive. So D1 is reverse-biased and D2 is forward biased. So
It is device which is used for converting alternating current into only D2 conduct and hence the current flows through the load
direct current. resistance RL again from A to B.
(i) Half wave rectifier: It is dear that whether the input signal is positive or negative,
It rectifies only half of the ac input wave. the current always flows through the load resistance in the same
direction and thus output is called full wave rectified.
Zener Diode that the Zener diode is reverse biased. If the input voltage
It is a special purpose diode, designed to operate under the increases, the current through RS and Zener diode also increases.
reverse bias in the breakdown region and used in voltage This increases the voltage drop across RS without any change in
regulation. Symbol of Zener diode is the voltage across the Zener diode. This is because in the
breakdown region, Zener voltage remains constant even though
the current through the Zener diode changes. Similarly, if the
input voltage decreases, the current through RS and Zener diode
In reverse bias of Zener diode after the breakdown voltage Vz, a also decreases.
large change in the current can be produced by almost
insignificant change in the reverse bias voltage. In other words,
Zener voltage remains constant, even though current through
the Zener diode varies over a wide range. This property of the
Zener diode is used for regulating voltage.

The voltage drop across RS decreases without any change in the

voltage across the Zener diode. Thus, any increase/decrease in
the input voltage results in, increase/decrease of the voltage
Zener diode as a voltage regulator drop across RS without any change in voltage across the Zener
The unregulated de voltage (filtered output of a rectifier) is diode. Thus, the Zener diode acts as a voltage regulator.
connected to the Zener diode through a series resistance RS such

Q. A Zener diode of voltage VZ (= 6 V) is used to maintain a constant voltage across a

load resistance R L (= 1000Ω) by using a series resistance 𝑅𝑆 (= 100Ω ). If the e.m.f.
of source is 𝐸(= 9 V), calculate the value of current through series resistance, Zener
diode and load resistance. What is the power being dissipated in Zener diode.

Sol. Here, E = 9 V; VZ = 6; R L = 1000Ω and R S = 100Ω,

Potential drop across series resistor 𝑉 = 𝐸 − 𝑉𝑍 = 9 − 6 = 3𝑉
V 3
Current through series resistance R S is I = = = 0.03 A
R 100
VZ 6
Current through load resistance R L is IL = = = 0.006 A
RL 1000
Current through Zener diode is IZ = I − IL = 0.03 − 0.006 = 0.024amp.
Power dissipated in Zener diode is 𝑃𝑧 = 𝑉𝑧 𝐼𝑧 = 6 × 0.024 = 0.144 Watt

Optoelectronic Junction Devices

Photodiode
It is a special purpose junction diode used to sense and measure
incident light. It is operated under reverse bias.
Its symbol is-

When light of energy "hv" falls on the photodiode (Here hv > Light Emitting Diode
energy gap) more electrons move from valence band to It is a heavily doped P-N junction which under forward bias emits
conduction band, due to this current in circuit of photodiode in spontaneous radiation. Its symbol is
"Reverse bias'', increases. As light intensity is increased, the
photo current goes on increasing. So, photo diode is used "to
detect light intensity". Example used in "Video camera".
when LED is forward biased then electrons move from N→P and If a thin layer of P-type semiconductor is sandwiched between
holes move from PN. At the junction boundary these are two thick layers of N-type semiconductor, then it is known as
recombined. On recombination, energy is released in the form NPN transistor.
of photons of energy equal to or slightly less than the band gap.
When the forward current of the diode is small, the intensity of
light emitted is small. As the forward current increases, intensity
of light increases and reaches a maximum. Further increase in
the forward current results in decrease of light intensity. LEOs
are biased in such a way that the light emitting efficiency should
be maximum.
In case of Si or Ge diodes, the energy released in recombination
• P-N-P Transistor
lies in infra-red region. Therefore, to form LED, such
If a thin layer of N-type of semiconductor is sandwiched
semiconductors are to be used which have band gap from 1.8 eV
between two thick layer of P-type semiconductor, then it is
to 3 eV. Hence GaAs1–x Px is used in forming LED.
known as PNP transistor.
Solar Cell
A p-n junction which generates emf when solar radiation falls on
it, called solar cell. It works on the same principle (photovoltaic
effect) as the photodiode, except that no external bias is applied
and the junction area is kept much larger for solar radiation to
be incident because we are interested in more power.

Each transistor has three terminals and these are:

(i) Emitter: It is the left most part of the transistor which
emits the majority carriers towards base. It is highly doped
and medium in size.
(ii) Base: It is the middle part of transistor which is sandwiched
by emitter (E) and collector (c). It is lightly doped
and very thin in size.
(iii) Collector: It is right part of the transistor which collects the
majority carriers which is emitted by emitter. It has large
size and moderate doping.
Every transistor has following two junctions
(i) The junction between emitter and base is known as emitter-
base junction (JEB).
When light falls on, emf generates due to the following three (ii) The junction between base and collector is known as base-
basic processes: generation, separation and collection- collector junction (JBC).
(i) generation of e-h pairs due to light (with hv > Eg) in junction
region. Working of Transistor
(ii) separation of electrons and holes due to electric field of the Working of NPN Transistor
depletion region. Electrons are swept to n-side and holes to The emitter base junction is forward biased and base collector
p-side by the junction field. junction is reversed biased to study the behavior of transistor. It
(iii) On reaching electrons at n-side and holes on at p-side. Thus is called active state of transistor. N-P-N transistor in circuit and
n-side becomes negative and p-side becomes positive symbolic representation is shown in figure.
potential and giving rise to photovoltage
Transistor
Transistor is a three terminal device which transfers a signal
from low resistance circuit to high resistance circuit. It is formed
when a thin layer of one type of extrinsic semiconductor (P or N
type) is sandwiched between two thick layers of other type of
extrinsic semiconductor.
Transistors are of two types
• N-P-N Transistor
In active state of n-p-n transistor majority electrons in emitter
are sent towards base.
The barrier of emitter base junction is reduced because of
forward bias therefore electrons enter into the base.
About 5% of these electrons recombine with holes in base region
results very small current (IB) in base.
The remaining electron (  95%) enters into the collector region
because these are attracted towards the positive terminal of
battery results collector current (IC).
The base current is the difference between IE and lC and
Input Characteristics
proportional to the number of electron hole recombination in
The variation of base current (IB) (input) with base emitter
the base.
voltage (VEB) at constant collector emitted voltage (VCE) is called
IE = I B + I C
input characteristic.
We also see IE ≃IC because IB is very small.
Working of PNP Transistor
When emitter-base junction is forward biased, holes (majority
carriers) in the emitter are repelled towards the base and diffuse
through the emitter base junction. The barrier potential of
emitter-base junction decreases and hole enters into then-
region (i.e. base). A small number of holes (i) Kept the collector-emitter voltage (VCE) constant (say VCE =
( 5%) combine with electrons of base-region resulting small 10V)
current (IB). The remaining holes (ii) Now change emitter base voltage VBE in steps of 0.1 volt and
( 95%) enter into the collector region because these are note the corresponding values of base current (IB).
attracted towards negative terminal of the battery connected (iii) Plot the graph between VBE and IB.
with the collector-base junction. These holes constitute the Output characteristics
collector current (IC). The variation of collector current IC (output) with collector-
emitter voltage (VCE) at constant base current (IB) is called output
characteristic.

(i) Keep the base current (IB) constant

(say IB = 10 A)
(ii) Now change the collector-emitter voltage (VCE) and not the
corresponding values of collector current (IC) .
8/As one hole reaches the collector, it is neutralized by the (iii) Plot the graph between VCE and IC.
battery. As soon as one electron and a hole is neutralized in (iv) A set of such curves can also be plotted at different fixed
collector, a covalent bond is broken in emitter region and an values of base current (say 20 A, 30 A etc.)
electron hole pair is produced. The released electron enters the
positive terminal of battery and holes moves towards the Transistor as a switch
collector. So IE = IB + IC When a transistor is used in the cut off (off state) or saturation
Configurations Of A Transistor And Its Characteristics state (on state) only, it acts as a switch. To study this behavior,
The transistor is connected in either of the three ways in circuit. we understand base biased CE transistor circuit.
(i) Common base configuration
(ii) Common emitter configuration
(iii) Common collector configuration
In these three, common emitter is widely used and common
collector is rarely used.
Common emitter transistor characteristics
 we fix the value of VBB corresponding to a point in the middle of
the linear part of the transfer curve then the de base current IB
would be constant and corresponding collector current IC will
also be constant. The dc voltage VCE = VCC – ICRC would also
remain constant. The operating values of VCE and IB determine
the operating point, of the amplifier.
If a small sinusoidal voltage with amplitude i is superposed in
series with the VBB supply, then the base current will have
Applying Kirchhoff’s voltage rule to the input and output sides of
sinusoidal variations superimposed on the value of IB. As a
this circuit we get
consequence, the collector current also will have sinusoidal
Vi = IBRB + VBE (Vi = dc input voltage)
variations superimposed on the value of IC producing in turn
and VO = VCC – IC RC (VO = dc output voltage)
corresponding change in the value of V0.
Now we can analyse how VO changes as Vi increase from zero
Mathematical Analysis:
onwards. In case of Silicon transistor, if Vi is less than 0.6 V, IB will
be zero, hence IC will zero and transistor will be said to be in cut- From KVL equation of base biased CE transistor circuit
off state, and VO = VCC. When Vi become greater than 0.6 V, some Vi = IBRB + VBE
IB flows, so some IC flows (transistor is in active state now) and  Vi = (IB)RB + VBE VBE = 0
output VO decreases as the term IC RC increase. With increase in  Vi = (IB)RB
Vi the IC increase almost linearly and so VO decreases linearly till Similarly Vo = VCC – ICRC
its value becomes less than about 1.0 volt.  Vo = VCC – (IC)RC
VCC = 0
 Vo = –(IC)RC
So voltage gain of CE amplifier
Vo −(IC ) R C R
AV = = = − C
Vin (IB ) R B RB
The negative sign represents that output voltage is opposite
Beyond this, the change becomes nonlinear and transistor goes in phase with the input voltage.
into saturation state. With further increase in Vi the output Power gain (Ap) = current gain × voltage gain = ac × AV  AP
voltage is found to decrease further forwards zero (however, it >1
may never become zero).
Feedback amplifier
It we draw the VO versus Vi curve called transfer characteristic When some part of output signal is fed back to the input of
(see figure), we see that between cut off state and active state amplifier then this process is known as feedback. Feedback of
and also between active state and saturation state there are
two types:
regions of non-linearity showing that the transition from cut-off
• Positive feedback
state to active state and from active state to saturation state are
not sharply defined. When input and output are in the same phase then positive
feedback is there. It is used in oscillators. Voltage gain after
Transistor as an amplifier
A
The process of increasing the amplitude of input signal without feedback Af =
distorting its wave shape and without changing its frequency is 1 − A
known as amplification. • Negative feedback
A device which increases the amplitude of the input signal is If input and output are out of phase and some part of that
called amplifier. is feedback to input then it is known as negative feedback.
It is used to get constant gain amplifier.
A
Voltage gain after feedback Af =
1 + A
Oscillator is device which delivers ac output wave form of
desired frequency without any external input wave form.
The electric oscillations are produced by. L–C circuit (i.e.,
tank circuit containing inductor and capacitor). These
oscillations are damped one i.e., their amplitude decrease
with the passage of time due to the small resistance of the
inductor. In other words, the energy of the L-C oscillations
To operate the transistor as an amplifier it is necessary to fix its decreases. If this loss of energy is compensated from
operating point somewhere in the middle of its active region. If
 outside, then undamped oscillations (of constant memory, or microprocessor. ICs can be made very compact,
amplitude) can be obtained. having up to several billion transistors and other electronic
components in an area the size of a fingernail. The most widely
used technology is the Monolithic Integrated Circuit. The word
monolithic is a combination of two Greek words, monos mean
single and lithos means stone. This, in effect means that the
entire circuit is formed on a single silicon crystal (or chip). The
chip dimensions are as small 1 mm × 1 mm or it could even be
smaller.
Depending upon the level of integration (i.e., the number of
circuit components or logic gates), the ICs are termed as small
This can be done by using feedback arrangement and a transistor integration, SSI (logic gates < 10); Medium Scale Integration, MSI
amplifier in the circuit. (logic gates< 100); Large Scale Integration, LSI (logic gates <
1
Oscillating frequency of oscillator is given by 𝑓 = 1000) and very Large-Scale integration, VLSI (logic gates > 1000).
2𝜋√𝐿𝐶
The technology of fabrication is very involved but large-scale
Advantages of semiconductor devices over vacuum tubes industrial production has made them very inexpensive.
Advantage
Basic Logic Gates
• Semiconductor devices are very small in size as compared
There are three basic logic gates. They are (a) OR gate (b) AND
to the vacuum tubes. Hence the circuits using
gate, and (c) NOT gate.
semiconductor devices are more compact.
• In vacuum tubes, current flows when the filament is heated The OR gate: The output of an OR gate attains the state 1 if one
an starts emitting electrons. So, we have to wait for some or more inputs attain the state1.
time for the operation of the circuit. One the other hand, in Logic symbol of OR gate
semiconductor devices no heating is required and the circuit
begins to operate as soon as it is switched on.
• Semiconductor devices required low voltage for their
operation as compared to the vacuum tube. So a lot of
electrical power is saved.
B, read as C equals A OR B.
• Semiconductor devices do not produce any humming noise
which is large in case of vacuum tube. Truth table of a two-input OR gate
• Semiconductor devices have longer life than the vacuum
tube. Vacuum tube gets damaged when its filament is burnt.
• Semiconductor devices are shock proof.
• The cost of production of semiconductor-devices is very
small as compared to the vacuum tubes.
• Semiconductor devices can be easily transported as
compared to vacuum tube.
Disadvantages
• Semiconductor devices are heat sensitive. They get
damaged due to overheating and high voltages. So they The AND gate
have to be housed in a controlled temperature room. The output of an AND gate attains the state 1 if and only if all the
• The noise level in semiconductor devices in very high. inputs are in state 1.
• Semiconductor devices have poor response in high Logic symbol of AND gate
frequency range.
Integrated circuit (IC)

The Boolean expression of AND gate is C = A.B.

It read as C equals A AND B.
Truth table of a two-input AND gate

An integrated circuit (ICs), sometimes called a chip or microchip,

is semiconductor wafer on which thousands or millions of tiny
resistors, capacitors and transistors are fabricated. An IC can
function as an amplifier, oscillator, timer, counter, computer
 Logic symbol of NAND gate

The Boolean expression of NAND gate is 𝐶 = 𝐴. 𝐵

Truth table of a NAND gate
The NOT gate
The output of a NOT gate attains the state 1 if and only if the
input does not attain state 1.
Logic symbol of NOT gate

The NOR gate

Logic symbol of NOR gate
The Boolean expression is 𝐶 = 𝐴̄, read as C equals NOT A.
Truth table of NOT gate
The Boolean expression of NOR gate is 𝐶 = 𝐴 + 𝐵
Truth table of a NOR gate

Combination Of Gates
The three basis gates (OR, AND and NOT) when connected in
various combinations give us logic gates such as NAND, NOR
gates, which are the universal building blocks of digital circuits.
The NAND gate
• Intrinsic Semiconductor: I = eA(ne ve + nh vh)
The pure semiconductors in which the electrical where ne and nh are the electron and hole densities,
conductivity is totally governed by the electrons excited and ve and vh are their drift velocities, respectively.
from the valence band to the conduction band and in which (b) If µh are the electron and hole mobilities, then
no impurity atoms are added to increase their conductivity the conductivity of the semiconductor will be,
are called intrinsic semiconductors and their conductivity is ρ = e(ne µe + nh µh) and the resistivity will be,
called intrinsic conductivity. Electrical conduction in pure 1
=
e ( ne  + nh h )
semiconductors occurs by means of electron-hole pairs. In
an intrinsic semiconductor,
ne = n h = n i (c) The conductivity of an intrinsic semiconductor
where ne = the free electron density in conduction band, nh increases exponentially with temperature as,
= the hole density in valence band, and n i = the intrinsic  Eg 
carrier concentration.  =  0 exp  − 
• Extrinsic Semiconductors:  2k BT 
A Semiconductor doped with suitable impurity atoms so as • Forward Biasing of a pn-junction:
to increase its conductivity is called an extrinsic If the positive terminal of a battery is connected to the p-
semiconductor. side and the negative terminal to the n-side, then the pn-
• Types of Extrinsic Semiconductors: junction is said to be forward biased. Both electrons and
Extrinsic semiconductors are of two types holes move towards the junction. A current, called forward
i) n-type semiconductors current, flows across the junction. Thus, a pn-junction offers
ii) p-type semiconductors a low resistance when it is forward biased.
• n-type semiconductors: • Reverse Biasing of a pn-junction:
The pentavalent impurity atoms are called donors because If the positive terminal of a battery is connected to the n-
they donate electrons to the host crystal and the side and negative terminal to the p-side, then pn-junction is
semiconductor doped with donors is called n-type said to be reverse biased. The majority charge carriers move
semiconductor. In n-type semiconductors, electrons are the away from the junction. The potential barrier offers high
majority charge carriers and holes are the minority charge resistance during the reverse bias. However, due to the
carriers. Thus, minority charge carriers a small current, called reverse or
ne >> nh leakage current flows in the opposite direction. Thus,
• p-type semiconductors: junction diode has almost a unidirectional flow of current.
The trivalent impurity atoms are called acceptors because • Action of a transistor:
they create holes which can accept electrons from the When the emitter-base junction of an npn-transistor is
nearby bonds. A semiconductor doped with acceptor type forward biased, the electrons are pushed towards the base.
impurities is called a p-type semiconductor. In p-type As the base region is very thin and lightly doped, most of the
semiconductor, holes are the majority carriers and electrons cross over to the reverse biased collector. Since
electrons are the minority charge carriers Thus, few electrons and holes always recombine in the base
nh >> ne
region, so the collector current Ic is always slightly less then
• Holes:
emitter current IE.
The vacancy or absence of electron in the bond of a
IE = I C + I B
covalently bonded crystal is called a hole. A hole serves as a
Where IB is the base current.
positive charge carrier.
• Three Configurations of a Transistor:
• Mobility:
A transistor can be used in one of the following three
(a) The drift velocity acquired by a charge carrier in a unit
configurations:
electric field is called its electrical mobility and is
(a) Common-base (CB) circuit.
denoted by µ.
(b) Common-emitter (CE) circuit.
Vd
= (c) Common-collector (CC) circuit.
E • Current Gains of a Transistor:
(b) The mobility of an electron in the conduction band is Usually low current gains are defined:
greater than that of the hole (or electron) in the valence (a) Common base current amplification factor or ac current
band. gain α:
• Electrical conductivity of a Semiconductor: It is the ratio of the small change in the collector current
(a) If a potential difference V is applied across a conductor to the small change in the emitter current when the
of length L and area of cross-section A, then the total collector-base voltage is kept constant.
current I through it is given by,
 𝛿𝐼𝐶 It is a shorthand method of describing the function of a logic
𝛼=[ ]
𝛿𝐼𝐸 𝑉 gate in the form of an equation or an expression. It also
𝐶𝐵=Constant
(b) Common emitter current amplification factor or ac current relates all possible combination of the inputs of a logic gate
gain β: to the corresponding outputs.
It is the ratio of the small change in the collector current to • Positive and Negative Logic:
the small change in the base current when the collector If in a system, the higher voltage level represents 1 and the
emitter voltage is kept constant. lower voltage level represent 0, the system is called a
𝛿𝐼𝐶 positive logic. If the higher voltage represents 0 and the
𝛽=[ ] lower voltage level represents 1, then the system is called a
𝛿𝐼𝐵 𝑉
𝐶𝐸=Constant negative logic.
• Relations between α and β: • OR Gate:
The current gains α and β are related as, An OR gate can have any number of inputs but only one
output. It gives higher output (1) if either input A or B or
 
= and = both are high (1), otherwise the output is low (0).
1+  1+  A+B=Y
• Transistor as an amplifier: Which is read as ‘A or B equals Y’.
An amplifier is a circuit which is used for increasing the • AND gate:
voltage, current or power of alternating form. A transistor An AND gate can have any number of inputs but only one
can be used as an amplifier. output. It gives a high output (1) if inputs A and B are both
high (1), or else the output is low (0). It is described by the
• AC Current Gain: Boolean expression.
AC current gain is defined as, A. B = Y
Which is read as ‘A and B equals Y’.
 I 
 ac or Ai =  C  • NOT Gate:
  I B V CE = constant A NOT gate is the simplest gate, with one input and one
• DC Current Gain: output. It gives as high output (1) if the input A is low (0),
DC current gain is defined as, and vice versa.
Whatever the input is, the NOT gate inverts it. It is described
 IC 
 dc =   by the Boolean expression:
 I B V 𝐴̅ = Y
CE = constant
Which is read as ‘not A equal Y’.
• Voltage Gain of an Amplifier:
• NAND (NOT+AND) gate:
It is defined as,
𝑉 𝐴smallchangeinoutputvoltage It is obtained by connecting the output of an AND gate to
𝐴𝑣 = 0 = the input of a NOT gate. Its output is high if both inputs A
𝑉𝑖 𝐴smallchangeininputvoltage

V and B are not high. If is described by the Boolean expression.

AV = CE A.B = Y or AB = Y
 VBE Which is read as ‘A and B negated equals Y’.
Rout • NOR (NOT+OR) Gate:
AV =  ac . = Ai . Ar It is obtained by connecting the output of an OR gate to the
Rin input of a NOT gate. Its output is high if neither input A nor
i.e., Voltage gain = Current gain X Resistance gain input B is high. It is described by the Boolean expression.
• Power Gain of an Amplifier: A+B=Y
It is defined as, Which is read as ‘A and B negated equals Y’.
Outputpower • XOR or Exclusive OR gate. The XOR gate gives a high output
𝐴𝑝 = = current x voltage gain
Inputpower if either input A or B is high but not when both A and B are
or high or low. It can be obtained by using a combination of
Rout two NOT gates, two AND gates and one OR gate. It is
A p = Ai . Av =  ac2 . described by Boolean expression:
Rin Y = AB + AB
• Logic Gate: The XOR gate is also known as difference gate because its
A logic gate is a digital circuit that has one or more inputs output is high when the inputs are different.
but only one output. It follows a logical relationship • Integrated Circuits:
between input and output voltage. The concept of fabricating an entire circuit (consisting of
• Truth Table: many passive components like R and C and active devices
This table shows all possible input combination and the like diode and transistor) on a small single block (or chip) of
corresponding output for a logic gate. a semiconductor has revolutionized the electronics
• Boolean Expression: technology. Such a circuit is known as Integrated Circuit (IC).
 PRACTICE EXERCISE
Q9. In the half wave rectifier circuit operating from 50 Hz
MCQ mains frequency, the fundamental frequency in the ripple
would be
Q1. In a p-n junction having depletion layer of thickness (a) 25 Hz (b) 50 Hz
10−6 m the potential across it is 0.1 V. The electric field is (c) 70.7 Hz (d) 100 Hz
107 V 10−6 V Q10. Addition of trivalent impurity to a semiconductor creates
(a) (b)
m m
105 V 10−5 V many
(c) (d) (a) hole (b) bound electrons
m m
Q2. In a reverse biased diode when the applied voltage (c) free electrons (d) valence electrons
changes by 1 V, the current is found to change by 0.5𝜇A. Q11. In a full wave rectifier circuit operating from 50 Hz mains
The reverse bias resistance of the diode is frequency, the fundamental frequency in the ripple would
(a) 2 × 105 Ω (b) 2 × 106 Ω be
(c) 200Ω (d) 2Ω. (a) 25 Hz (b) 50 Hz
(c) 70.7 Hz (d) 100 Hz
Q3. When the forward bias voltage of a diode is changed from
0.6 V to 0.7 V, the current changes from 5 mA to 15 mA. Q12. What bonds are present in a semiconductor?
Then its forward bias resistance is (a) Monovalent (b) Bivalent
(a) 0.01Ω (b) 0.1Ω (c) Trivalent (d) Covalent
(c) 10Ω (d) 100Ω Q13. What happens to the forbidden energy gap of a
Q4. A diode having potential difference 0.5 V across its semiconductor with the fall of temperature?
junction which does not depend on current, is connected (a) Decreases
(b) Increases
in series with resistance of 20Ω across source. If 0.1 A
(c) Unchanged
current passes through resistance then what is the
(d) Sometimes decreases and sometimes increases
voltage of the source?
(a) 1.5 V (b) 2.0 V Q14. In a p-type semiconductor, the current conduction is due
(c) 2.5 V (d) 5 V to
(a) Holes (b) Atoms
Q5. If the forward bias on p-n junction is increased from zero
(c) Electrons (d) Protons
to 0.045 V, then no current flows in the circuit. The
contact potential of junction i.e. 𝑉𝐵 is Q15. What is the main function of a transistor?
(a) zero (b) 0.045 V (a) Simplify (b) Amplify
(c) more than 0.045 V (d) less than 0.045 V (c) Rectify (d) All of the above
Q15. In a semiconductor, what is responsible for conduction?
Q6. The peak voltage in the output of a half-wave diode
(a) Electrons only
rectifier fed with a sinusoidal signal without filter is 10V.
(b) Holes only
The d.c. component of the output voltage is
(c) Both electrons and holes
(a) 20/ 𝜋 S V (b) 10/ 2 V
(d) Neither electrons nor holes
(c) 10/ 𝜋 V (d) 10V
Q17. What happens to the resistance of semiconductors on
Q7. Two ideal diodes are connected to a battery as shown in heating?
the circuit. The current supplied by the battery is (a) Increases
(b) Decreases
(c) Remains the same
(d) First increases later decrease
Q18. In intrinsic semiconductors at room temperature, the
number of electrons and holes are
(a) Unequal (b) Equal
(a) 0.75 A (b) zero
(c) Infinite (d) Zero
(c) 0.25 A (d) 0.5 A
Q19. Which of the following is not a universal gate?
Q8. Hole in semiconductor is (a) NOT (b) AND
(a) an anti – particle of electron (c) OR (d) NAND
(b) a vacancy created when an electron leaves a covalent Q20. Electric conduction in a semiconductor takes place due to
bond (a) electrons only
(c) absence of free electrons (b) holes only
(d) an artificially created particle (c) both electrons and holes
(d) neither electrons nor holes
 ASSERTION AND REASONING
NUMERICAL TYPE QUESTIONS
Directions: Each of these questions contain two statements,
Assertion and Reason. Each of these questions also has four Q1. A P type semiconductor has acceptor level 57 meV above
alternative choices, only one of which is the correct answer. You the valance band. What is maximum wavelength of light
have to select one of the codes (a), (b), (c) and (d) given below. required to create a hole?
(a) Assertion is correct, reason is correct; reason is a correct Q2. A silicon specimen is made into a p-type semiconductor
explanation for assertion. by doping on an average one indium atom per 5 × 107
(b) Assertion is correct, reason is correct; reason is not a correct silicon atom. If the number density of atoms in the silicon
explanation for assertion specimen is 5 × 1028 atoms/ m3; find the number of
(c) Assertion is correct, reason is incorrect acceptor atoms in silicon per cubic centimetre.
(d) Assertion is incorrect, reason is correct Q3. Pure Si at 300 K has equal electron (ne) and hole (nh)
Q1. Assertion: NAND or NOR gates are called digital building concentrations of 1.5 × 1016 m–3. Doping by indium nh
blocks. Reason: The repeated use of NAND (or NOR) gates increases to 3 × 1022 m–3. Calculate ne in the doped Si.
can produce all the basis or complicated gates. Q4. What will be conductance of pure silicon crystal at 300 K
temperature? If electron hole pairs per cm3 is 1.075 × 1010
Q2. Assertion: A transistor amplifier in common emitter
at this temperature, e = 1350 cm2 / volt-s & h = 480 cm2
configuration has a low input impedance. / volt-s.
Reason: The base to emitter region is forward biased.
Q5. What is the value of current I in given circuits?
Q3. Assertion: A p-n junction diode can be used even at ultra-
high frequencies.
Reason: Capacitive reactance of p-n junction diode
increases as frequency increases.
Q4. Assertion: The diffusion current in a p-n junction is from
the p-side to the n-side.
Reason: The diffusion current in a p-n junction is greater
than the drift current when the junction is in forward
biased. Q6. In a transistor, the value of  is 50. Calculate the value of
.
Q5. Assertion: When the temperature of a semiconductor is
Q7. In an NPN transistor 1010 electrons enter the emitter in
increased, then its resistance decreases.
10−6 s and 2% electrons recombine with holes in base,
Reason: The energy gap between valence and conduction then find the value of α and β.
bands is very small for semiconductors.
Q8. The transfer ratio of a transistor is 50. The input
resistance of the transistor when used in the common-
SHORT ANSWER QUESTIONS emitter configuration is 1 kΩ. Then find the peak value for
an A.C input voltage of 0.01 𝑉 peak.
Q1. In a transistor, reverse bias is quite high as compared to
Q9. Three photo diodes 𝐷1 , 𝐷2 and 𝐷3 are made of
the forward bias. Why? A transistor is a temperature
semiconductors having band gaps of 2.5eV, 2eV and 3eV,
sensitive device. Explain. respectively. Which ones will be able to detect light of
Q2. The use of a transistor in common-emitter configuration wavelength 6000Å A?
is preferred over the common-base configuration. Explain Q10. The number of silicon atoms per m3 is 5 × 1028 . This is
why? doped simultaneously with 5 × 1022 atoms per m3 of
Q3. A transistor is a temperature sensitive device. Explain Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate
the number of electrons and holes.
Q4. Explain why the emitter is forward biased and the
Q11. A photodetector is made from a semiconductor with E g =
collector is reverse biased in a transistor?
0.75 eV. Calculate the maximum wavelength which it can
Q5. Why a transistor cannot be used as a rectifier? detect?
 HOMEWORK EXERCISE
(c) becomes bad conductor
MCQ (d) has maximum conductivity

Q1. When a semiconductor is heated, its resistance Q9. The intrinsic semiconductor becomes an insulator at
(a) decreases (a) 0°C (b) –100°C
(b) increases (c) 300 𝐾 (d) 0 𝐾
(c) remains unchanged Q10. Energy bands in solids are a consequence of
(d) nothing is definite (a) Ohm’s Law
Q2. In the case of constants α and β of a transistor (b) Pauli’s exclusion principle
(a) α = β (c) Bohr’s theory
(b) β < 1, α > 1 (d) Heisenberg’s uncertainty principle
(c) α = β2 Q12. Choose the correct statement
(d) β > 1, α < 1 (a) When we heat a semiconductor, its resistance
Q3. The forbidden energy band gap in conductors, increases
semiconductors and insulators are EG1 , EG2 and EG3 (b) When we heat a semiconductor, its resistance
respectively. The relation among them is decreases
(a) EG1 = EG2 = EG3 (c) When we cool a semiconductor to 0 𝐾 then it becomes
(b) EG1 < EG2 < EG3 super conductor
(c) EG1 > EG2 > EG3 (d) Resistance of a semiconductor is independent of
(d) EG1 < EG2 > EG3 temperature
Q4. Which statement is correct? Q13. Two 𝑃𝑁-junctions can be connected in series by three
(a) 𝑁-type germanium is negatively charged and 𝑃-type different methods as shown in the figure. If the potential
germanium is positively charged difference in the junctions is the same, then the correct
(b) Both 𝑁-type and 𝑃-type germanium is neutral connections will be
(c) 𝑁-type germanium is positively charged and 𝑃-type
germanium is negatively charged
(d) Both 𝑁-type and 𝑃-type germanium is negatively
charged
Q5. Let 𝑛ℎ and 𝑛𝑒 be the number of holes and conduction
electrons respectively in a semiconductor. Then
(a) nh > ne in an intrinsic semiconductor (a) In the circuit (1) and (2)
(b) nh = ne in an extrinsic semiconductor (b) In the circuit (2) and (3)
(c) nh = ne in an intrinsic semiconductor (c) In the circuit (1) and (3)
(d) ne > nh in an intrinsic semiconductor (d) Only in the circuit (1)
Q6. Wires 𝑃 and 𝑄 have the same resistance at ordinary Q14. The approximate ratio of resistances in the forward and
(room) temperature. When heated, resistance of 𝑃
reverse bias of the 𝑃𝑁-junction diode is
increases and that of 𝑄 decreases. We conclude that
(a) 102 : 1 (b) 10−2 : 1
(a) 𝑃 and 𝑄 are conductors of different materials −4
(c) 1: 10 (d) 1: 104
(b) 𝑃 is 𝑛-type semiconductor and 𝑄 is 𝑝-type
semiconductor Q15. The dominant mechanisms for motion of charge carriers
(c) 𝑃 is semiconductor and 𝑄 is conductor in forward and reverse biased silicon 𝑃-𝑁 junctions are
(d) 𝑃 is conductor and 𝑄 is semiconductor (a) Drift in forward bias, diffusion in reverse bias
Q7. In an NPN transistor circuit, the collector current is 10 mA. (b) Diffusion in forward bias, drift in reverse bias
If 90% of the electrons emitted reach the collector, the (c) Diffusion in both forward and reverse bias
emitter current (iE ) and base current (iB ) are given by (d) Drift in both forward and reverse bias
(a) 𝑖𝐸 = –1mA, 𝑖𝐵 = 9mA In reverse biasing diffusion becomes more difficult so net
(b) 𝑖𝐸 = 9mA, 𝑖𝐵 = –1mA current (very small) is due to the drift.
(c) 𝑖𝐸 = 1mA, 𝑖𝐵 = 11mA Q16. In a triclinic crystal system
(d) 𝑖𝐸 = 11mA, 𝑖𝐵 = 1mA (a) 𝑎 ≠ 𝑏 ≠ 𝑐, α ≠ β ≠ γ
Q8. At zero Kelvin a piece of germanium (b) 𝑎 = 𝑏 = 𝑐, α ≠ β ≠ γ
(a) becomes semiconductor (c) 𝑎 ≠ 𝑏 ≠ 𝑐, α ≠ β = γ
(b) becomes good conductor (d) 𝑎 = 𝑏 = 𝑐, α = β = γ
 Q5. Why are elemental dopants for Silicon or Germanium
ASSERTION AND REASONING usually chosen from group 13 or group 15?

Directions: Each of these questions contain two statements, CASE STUDY BASED QUESTIONS
Assertion and Reason. Each of these questions also has four
alternative choices, only one of which is the correct answer. You Q1. A student performs an experiment for drawing the static
have to select one of the codes (a), (b), (c) and (d) given characteristic curve of a triode valve in the laboratory.
below. The following data were obtained from the linear portion
(a) Assertion is correct, reason is correct; reason is a correct of the curves:
explanation for assertion. Grid voltage Vg (volt) –2.0 –3.5 –2.0
(b) Assertion is correct, reason is correct; reason is not a correct Plate voltage Vp (volt) 180 180 120
explanation for assertion Plate current Ip (mA) 15 7 10
(c) Assertion is correct, reason is incorrect (i). Calculate the plate resistance rp of the triode valve?
(d) Assertion is incorrect, reason is correct (a) 0.12 × 104 ohm (b) 1.2 × 104 ohm
4
Q1. Assertion: A pure semiconductor has negative (c) 1.3 × 10 ohm (d) 1.4 × 104 ohm
temperature coefficient of resistance. (ii). Calculate the mutual conductance g m of the triode
Reason: In a semiconductor on raising the temperature, valve?
more charge carriers are released, conductance increases (a) 5.33 × 10−3 ohm−1 (b) 53.3 × 10−3 ohm−1
−3 −1
and resistance decreases. (c) 4.32 × 10 ohm (d) 5.00 × 10−3 ohm−1
(iii). Calculate the amplification factor μ, of the triode
Q2. Assertion: A p-type semiconductors is a positive type
valve?
crystal.
(a) 64 (b) 52
Reason: A p- type semiconductor is an uncharged crystal.
(c) 54 (d) 62
Q3. Assertion: When two semiconductors of p and n type are
brought in contact, they form p-n junction which act like NUMERICAL TYPE QUESTIONS
a rectifier.
Reason: A rectifier is used to convent alternating current Q1. Calculate the emitter current for which IB = 20 A,  = 100
into direct current. Q2. The base current is 100 A and collector current is 3 mA.
Q4. Assertion: The number of electrons in a 𝑃-type silicon (a) Calculate the values of , IE and .
semiconductor is less than the number of electrons in a (b) A charge of 20 A in the base current produces a
pure silicon semiconductor at room temperature. change of 0.5 mA in the collector current. Calculate
Reason: It is due to law of mass action. a.c..
Q5. Assertion: We can measure the potential barrier of a PN Q3. In a transistor connected in common emitter mode R 0 = 4
junction by putting a sensitive voltmeter across its k, Ri = 1 k, IC = 1 mA and IB = 20 A. Find the voltage
terminals. gain.
Reason: The current through the PN junction is not same
Q4. In the figures below, circuit symbol of a logic gate and two
in forward and reversed bias
input waveform ‘A’ and ‘B’ are shown.
Q6. Assertion: The resistivity of a semiconductor decreases
with temperature.
Reason: The atoms of a semiconductor vibrate with larger
amplitude at higher temperature there by increasing its
resistivity.

SHORT ANSWER QUESTIONS

Q1. Why we prefer transistor over the vacuum tubes in the (a) Name the logic gate & Write its Boolean expression.
portable radio receivers? (b) Write its truth table.
(c) Give the output wave from.
Q2. How is a light emitting diode fabricated? Briefly state it’s
working. Write any two important advantages of LEDs Q5. Write down the equivalent function performed by given
over the conventional incandescent low power lamps. circuit. Explain your answer.
Q3. How is forward biasing different from reverse biasing in a
p-n junction diode?
Q4. Explain why elemental semiconductor cannot be used to
make visible LEDs.
Q6. In a half wave rectifier, what is the frequency of ripple in Q9. A p-n photodiode is fabricated from a semiconductor with
the output if the frequency of input ac is 50 Hz ? What is a bandgap of 2.8eV. Can it detect a wavelength of
the output ripple frequency of a full wave rectifier? 6000 nm ?
Q7. The number of silicon atoms per m3 is 5 × 1028 . This is
Q10. The number of silicon atoms per m3 is 5 × 1028 . This is
doped simultaneously with 5 × 1022 atoms per m3 of
Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate doped simultaneously with 5 × 1022 atoms per m3 of
the number of electrons and holes. Given that 𝑛1 = Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate
1.5 × 1016 m−3 . Is the material 𝑛-type or 𝑝-type? the number of electrons and holes. Given that 𝑛𝑖 =
Q8. A 𝑝 − 𝑛 photodiode is fabricated from a semiconductor 1.5 × 1016 per m3 . Is the material 𝑛-type or 𝑝-type?
with band gap of 2.8eV. Can it detect a wavelength of
6000 nm ?
 SOLUTIONS MULTIPLE CHOICE

MCQ S11. (d) In full wave rectifier, we get the output for the
positive and negative cycle of input a.c. Hence the
frequency of the ripple of the output is twice than
V 0.1
S1. (c) E= = = 105 V/m that of input a.c. i.e. 100 Hz
d 10−6
S2. (b) S12. (d) Covalent Covalent bonds are present in a
Reverse resistance semiconductor.
ΔV 1 6
= = −6 = 2 × 10 Ω
S13. (b) Increases The forbidden energy gap of a
ΔI 0.5×10
semiconductor increases with the fall of
S3. (c)
ΔV temperature.
Forward bias resistance = S14. (a) Holes In a p-type semiconductor, the current
ΔI
(0.7−0.6)V 0.1
= = = 10Ω conduction is due to holes.
(15−5)mA 10×10−3

S4. (c) S15. (b) Amplify

V ′ = V + IR = 0.5 + 0.1 × 20 = 2.5 V The main function of a transistor is to amplify.
S5. (c) S16. (c) Both electrons and holes
When no current flows at the junction plane, then Both electrons and holes are responsible for
contact potential of junction plane is equal to the conduction in a semiconductor.
forward voltage applied = 0.045 V S17. (b) decreases
S6. (c) The resistance of semiconductors decreases on
V 10
V= o= V heating.
𝜋 𝜋
S18 (b) Equal
S7. (d) Here 𝐷1 is in forward bias and 𝐷2 is in reverse bias so,
In intrinsic semiconductors, the number of electrons
𝐷1 will conduct and 𝐷2 will not conduct. Thus, no
current will flow through 𝐷𝐶. and holes are equal.
𝑉 5 1 S19. (d) NAND
𝐼= = = Amp.
𝑅 10 2 NAND is not a universal gate.
S8. (b) a vacancy created when an electron leaves a S20. (c) Electric conduction, in a semiconductor occurs due to
covalent bond. Because- anti particle of electron is both electrons & holes.
positron ( i.e. e+ ), there is no presence of free
electron in semiconductor and hole is not created. It ASSERTION AND REASONING
is just the absence created when an electron jumps
to the conduction band from the valence band S1. (a) These gates are called digital building blocks because
leaving an empty spot in the semiconductor. Electron using these gates only (either NAND or NOR) we can
breaks the covalent bond and a blank is created compile all other gates also (like OR, AND, NOT, XOR)
which is called hole. S2. (a) Input impedance of common emitter configuration.
S9. (b) In half wave rectifier, we get the output only in one Δ𝑉𝐵𝐸
=| |
half cycle of input a.c. therefore, the frequency of the Δ𝑖𝐵 𝑉
𝐶𝐸= constant
ripple of the output is same as that of input a.c. i.e., where Δ𝑉𝐵𝐸 = voltage across base and emitter (base
50 Hz emitter region is forward biased)
S10. (a) holes S3. (c) Here Assertion is correct but Reason is wrong. As
In a pure semiconductor, each atom is surrounded by capacitive reactance, 𝑋𝑐 = 1/(𝜔𝐶) = 1/(2𝜋𝑣𝐶). It
four atoms and is engaged in chemical bond with all decreases as v increases.
of them. This is because a pure semiconductor has
four valence electrons. A trivalent impurity has 3 S4. (b) Diffusion current is due to the migration of holes and
valence electrons. When it is added, one atom of it electrons into opposite regions, so it will be from p-
forms bonds with three atoms of the semiconductor. side to n-side. Also, in forward bias it will increases.
Only three atoms are engaged in bonding. There is an S4. (a) Assertion is right because with increase in
empty space between the atom of impurity and the temperature, the resistance of semiconductor
fourth atom of pure semiconductor. Such an empty decreases. Due to small gap between valence band
space is called as hole which is available for and conduction band, electrons jump into
conduction. Thus a semiconductor with trivalent
conduction band and increases conduction.
impurity has many holes.
 S3. For a doped semi-conductor in thermal equilibrium
SHORT ANSWER QUESTIONS
n e n h = n i2 (Law of mass action)
S1. In a transistor, charge carriers (electrons or holes)
(1.5 1016 ) 2
n i2
move from emitter to collector through the base. The ne = = = 7.5 109 m −3
reverse bias on collector is made quite high so that it nh 3 10 22

may exert a large attractive force on the charge

carriers to enter the collector region. These moving S4.  = niee + nieh = nie(e + h)
carriers in the collector constitute a collector current. = 1.072 × 1010 × 1.6 × 10–19 × (1350 + 480)
= 3.14 × 10–6 mho/cm
S2. The current gain and voltage gain in the common-
27−0.7
emitter configuration is more one. So maximum S5. 𝐼= = 2 𝑚𝐴
1×103
power gain in common emitter configuration 𝛼 𝛼
S3. In a transistor, conduction is due to the movement of S6. 𝛽=  50 =
1−𝛼 1−𝛼
50
current carriers’ electrons and holes. When 50 –50 =   = = 0.98
51
temperature of the transistor increases, many
Ne 1010 ×1.6×10−19
covalent bonds may break up resulting in the S7. Emitter current Ie = = = 1.6 mA
t 10−6
formation of more electron, and holes. Thus, the 2
Base current Ib = × 1.6 = 0.032 mA
current will increase in the transistor. This current 100
gives rise to the production of more heat energy. The But, Ie = Ic + Ib
excess heat causes complete breakdown of the ∴ Ic = Ie − Ib = 1.6 − 0.032 = 11.568 mA
Ic 1.568 Ic 1.568
transistor. ∴α= = = 0.98 and β = = = 49
Ie 1.6 Ib 0.032
S4. In a transistor, the charge carriers move from
S8. β = 50, 𝑅 = 1000Ω, 𝑉𝑖 = 0.01𝑉
emitter to collector. The emitter sends the charge 𝑖𝑐 𝑉𝑖 0.01
carriers and collector collects them. This can happen β= and 𝑖𝑏 = = = 10−5 𝐴
𝑖𝑏 𝑅𝑖 103
−5
only if emitter is forward biased and the collector is Hence 𝑖𝑐 = 50 × 10 𝐴 = 500 μA
reverse biased so that it may attract the carriers.
S9. Energy of incident light photon,
S5. If transistor is to be used as a rectifier, then either ℎ𝑐 6.6 × 10−34 × 3 × 108
emitter-base or base-collector has to use as diode. 𝐸 = ℎ𝜈 = = = 𝟐. 𝟎𝟔𝐞𝐕
𝜆 6 × 10−7 × 1.6 × 10−19
For equated working of the said set of diodes, the For the incident radiation to be detected by the
number density of charge carriers in emitter and base photodiode, energy of incident radiation photon
or base and collector must be approximately same.
should be greater than the band gap. This is true only
As base is lightly doped and comparatively thin, so
for 𝐷2 . Therefore, only 𝐷2 will detect this radiation.
transistor cannot work as a rectifier
S10. Following values are given in the question:
NUMERICAL TYPE QUESTIONS Number of silicon atoms, N = 5 × 1028 atoms /m3
Number of arsenic atoms, 𝑛𝐴𝑆 = 5 × 1022 atoms
hc /m3
S1. E= Number of indium atoms, 𝑛ln = 5 × 1022 atoms /m3

𝑛𝑖 = 1.5 × 1016 electrons /m3
hc 6.62 10−34  3 108
 = = = 217700 Å 𝑛𝑒 = 5 × 1022 − 1.5 × 1016 = 4.99 × 1022
E 57 10−3 1.6 10−19 Let us consider the number of holes to be 𝑛ℎ
S2. The doping of one indium at6m in silicon In the thermal equilibrium, 𝑛𝑒 𝑛ℎ = 𝑛𝑖2
semiconductor will produce one acceptor atom in p- Calculating, we get
type semiconductor. Since one indium atom has 𝑛h = 4.51 × 109
been clopped per 5 × 107 silicon atoms, so number Here, 𝑛𝑒 > 𝑛ℎ , therefore the material is a 𝑛-type
density of acceptor atoms in silicon semiconductor.
5 1028 𝒉𝒄 𝟔.𝟔𝟐×𝟏𝟎−𝟑𝟒 ×𝟑×𝟏𝟎𝟖
= = 1021 atom/m3 = 1015 atom/cm3 S11. 𝝀
𝟎.𝟕𝟓×𝟏.𝟔×𝟏𝟎−𝟏𝟗
5 10 7 𝑬𝒈
𝒎𝒂𝒙
 HOMEWORK EXERCISE SOLUTIONS
reverse biased. This feature of the junction diode
MULTIPLE CHOICE QUESTIONS enables it to be used as a rectifier
S1. (a) With temperature rise conductivity of S4. (a) According to law of mass action, 𝑛𝑖2 = 𝑛𝑒 𝑛ℎ . In
semiconductors increases. intrinsic semiconductors 𝑛𝑖 = 𝑛𝑒 = 𝑛ℎ and for 𝑃-
S2. (d) α is the ratio of collector current and emitter current type semiconductor 𝑛𝑒 would be less than 𝑛𝑖 , since
while β is the ratio of collector current and base 𝑛ℎ is necessarily more than 𝑛𝑖 .
current. S5. (c) We cannot measure the potential barrier of a PN-
S3. (b) In insulators, the forbidden energy gap is very large, junction by connecting a sensitive voltmeter across
in case of semiconductor it is moderate and in its terminals because in the depletion region, there
conductors the energy gap is zero. are no free electrons and holes and in the absence of
forward biasing, PN-junction offers infinite
S4. (b) Both P-type and N-type semiconductors are neutral
resistance.
because neutral atoms are added during doping.
S6. (d) Resistivity of semiconductors decreases with
S5. (c) In intrinsic semiconductors, the creation or liberation
temperature. The atoms of a semiconductor vibrate
of one free electron by the thermal energy creates
with larger amplitudes at higher temperatures there
one hole. Thus, in intrinsic semiconductors 𝑛𝑒 = 𝑛ℎ
by increasing its conductivity not resistivity.
S6. (d) Conductor has positive temperature coefficient of
resistance but semiconductor has negative SHORT ANSWER QUESTIONS
temperature coefficient of resistance.
90 S1. This is because of two reasons:
S7. (d) 𝑖𝐶 = × 𝑖𝐸 ⇒ 10 = 0.9 × 𝑖𝐸 ⇒ 11𝑚𝐴
100 (a) Transistor is compact and small in size than the
Also iE = iB + iC ⇒ iB = 11 − 10 = 1mA vacuum tube.
S8. (c) At zero Kelvin, there is no thermal agitation and (b) Transistor can operate even at low voltage which can
therefore no electrons from valence band are able to be supplied with two or three dry cells.
shift to conduction band. S2. LED is fabricated by
S9. (d) At 0K temperature semiconductor behaves as an (i) heavy doping of both the p and n regions.
insulator, because at very low temperature electrons (ii) providing a transparent cover so that light can come
cannot jump from the valence band to conduction out. Working: When the diode is forward biased,
band electrons are sent from n→ p and holes from p→ n.
S10. (b) Formation of energy bands in solids are due to Pauli’s At the junction boundary, the excess minority
exclusion principle. carriers on either side of junction recombine with
majority carriers. This releases energy in the form of
S12. (b) With rise in temperature, conductivity of
photon hν = Eg.
semiconductor increases while resistance decreases.
GaAs (Gallium Arsenide): Band gap of
S13. (b) Because in case (1) N is connected with N. This is not semiconductors used to manufacture LED’s should
a series combination of transistor be 1.8 eV to 3 eV. These materials have band gap
S14. (d) Resistance in forward biasing 𝑅𝑓𝑟 ≈ 10Ω and which is suitable to produce desired visible light
resistance in reverse biasing wavelengths.
𝑅𝑓𝑟 1 Advantages-
𝑅𝑅𝑤 ≈ 105 Ω ⇒ =
𝑅𝑅𝑤 104 (i) Low operational voltage and less power
S15. (b) In forward biasing the diffusion current increases and consumption.
drift current remains constant so not current is due (ii) Fast action and no warm-up time required
to the diffusion. S3. Forward Bias:
S16. (a) In a triclinic crystal 𝑎 ≠ 𝑏 ≠ 𝑐 and α ≠ β ≠ γ ≠ 90° (i) Within the junction diode the direction of applied
voltage is opposite to that of built-in potential.
ASSERTION AND REASONING (ii) The current is due to diffusion of majority charge
carriers through the junction and is of the order of
S1. (a) In semiconductors, by increasing temperature, milliamperes.
covalent bond breaks and conduction hole and (iii) The diode offers very small resistance in the forward
electrons increase. bias.
S2. (d) There is no charge on P-type semiconductor, because Reverse Bias:
each atom of semiconductor is itself neutral. (i) The direction of applied voltage and barrier potential is
S3. (b) Study of junction diode characteristics shows that same.
the junction diode offers a low resistance path, when
forward biased and high resistance path when
 (ii) The current is due to leakage of minority charge In full wave rectifier, the output ripple frequency is
carriers through the junction and is very small of the twice of input frequency of ac i.e., 2 × 50 = 100 Hz
order of nA S7. The following values are given in the question:
(iii) The diode offers very large resistance in reverse bias. Number of silicon atoms, N = 5 × 1028 atoms /m3
S4. Elemental semiconductor’s band-gap is such that Number of arsenic atoms, nAS = 5 × 1022 atoms
electromagnetic emissions are in infrared region. /m3
Number of indium atoms, nln = 5 × 1022 atoms/m3
S5. The size of dopant atoms should be such as not to
𝑛𝑖 = 1.5 × 1016 electrons /m3
distort the pure semiconductor lattice structure and
𝑛𝑒 = 5 × 1022 − 1.5 × 1016 = 4.99 × 1022
yet easily contribute a charge carrier on forming
Let us consider the number of holes to be nh
covalent bonds with Si or Ge.
In the thermal equilibrium, 𝑛𝑒 𝑛ℎ = 𝑛𝑖2
CASE STUDY BASED QUESTIONS Calculating, we get
𝑛ℎ = 4.51 × 109
S (i). (b) Here, ne > nh , therefore, the material is an n-type
∆𝑉𝑝 (180−120)
𝑟𝑝 = = (15−10)×10−3 = 1.2 × 104 𝑜ℎ𝑚 semiconductor.
∆𝐼𝑝
S (ii). (a) S8. Energy corresponding to wavelength 6000 nm is
ℎ𝑐
gm =
∆Ip (15−7)×10−3
= (−2.0)−(−3.5) = 5.33 × 10−3 ohm−1 E==
𝜆
∆Vg 6.6×10−34 ×3×108
S (iii). (a) = joule = 3.3 × 10−20 J
6000×10−9
μ = rp × g m = (1.2 × 104 ) × (5.33 × 10−3 ) = 64 3.3×10−20
= = 0.2Ev
1.6×10−19

NUMERICAL TYPE QUESTIONS The photon energy ( 𝐸 = 0.2eV) of given wavelength

is much less than the band gap (𝐸𝑔 = 2.8eV), hence
S1. IC =  IB = 100 × 20 × 10–6 = 2000 A it cannot detect the given wavelength.
IE =  IB + IC = 20 + 2000 = 2020 A = 2.02 × 10–3 S9. No, the photodiode cannot detect the wavelength of
A = 2.02 mA 6000 nm because of the following reason:
IC The energy bandgap of the given photodiode, E𝑔 =
3×10−3
S2. (a) = =
100×10−6
= 30 2.8eV
IB The wavelength is given by 𝜆 = 6000 nm = 6000 ×
𝛼=
𝛽
=
30
=
30
= 0.97 and 𝐼𝐸 =
𝐼𝐶
=
3×31
= 10−9 m
1+𝛽 1+30 31 𝛼 30 We can find the energy of the signal from the
3.1 𝑚𝐴 following relation:
(b) IB = 20 A = 0.02 mA,
𝐸 = ℎ𝑐/𝜆
IC = 0.5 mA In the equation, h is Planck's constant = 6.626 ×
𝛥𝐼 0.5
 𝛽𝑎𝑐 = 𝐶 = = 25 10−34 J and c is the speed of light = 3 × 108 m/s
𝛥𝐼𝐵 0.02

S3. Voltage gain Substituting the values in the equation, we get

𝑅0 𝐼𝐶 𝑅0 1 × 10−3 4 𝐸 = (6.626 × 10−34 × 3 × 108 )/6000 × 10−9 =
𝐴𝑉 = 𝛽 ( ) = ( ) ( ) = ( −6
)( ) 3.313 × 10−20 J
𝑅𝑖 𝐼𝐵 𝑅𝑖 20 × 10 1 But, 1.6 × 10−19 J = 1eV
= 200 Therefore, E = 3.313 × 10−20 J = 3.313 × 10−20 /
S4. (a) NAND gate: Y = A.B 1.6 × 10−19 = 0.207eV
(b) Truth table The energy of a signal of wavelength 6000 nm is
𝐼𝑛𝑝𝑢𝑡 𝐴 𝐼𝑛𝑝𝑢𝑡 𝐵 𝑂𝑢𝑡𝑝𝑢𝑡 𝑌 0.207eV, which is less than 2.8eV - the energy band
0 0 1 gap of a photodiode. Hence, the photodiode cannot
0 1 1 detect the signal.
1 0 1 S10. Arsenic is 𝑛-type impurity and indium is 𝑝-type
1 1 0 impurity.
(c) Output waveform
Number of electrons, 𝑛𝑒 = 𝑛𝐷 − 𝑛𝐴
= 5 × 1022 − 5 × 1020 = 4.95 × 1022 m−3
Also 𝑛𝑖2 = 𝑛𝑒 𝑛ℎ
Given 𝑛𝑖 = 1.5 × 1016 m−3
S5. AND gate, 𝐴 = 𝐴 ⋅ 𝐵 + 𝐵 ⋅ 𝐶 = 𝐴𝐵 ⋅ 𝐵𝐶 = 𝐴𝐵𝐶 = 𝑛𝑖2
2
(1.5×1016 )
𝐴𝐵𝐶 (∵ 𝑋̄ + 𝑌̄ = 𝑋⋅ 𝑌 Number of holes, 𝑛ℎ = =
𝑛𝑒 4.95×1022

S6. In half wave rectifier, the output ripple frequency is ⇒ 𝑛ℎ = 4.54 × 109 m−3
50 Hz. As 𝑛𝑒 > 𝑛ℎ ; so the material is an 𝑛-type
semiconductor