14.

SEMICONDUCTOR DEVICES
INTERIORS OF TEXTBOOK
1. Give the classification of materials based on their resistivity or conductivity.
On the basis of the relative values of electrical conductivity (σ) or resistivity (ρ = 1/
σ), the solids are broadly classified as:
➢ Metals: They possess very low resistivity (or high conductivity).
ρ~ 10-2– 10-8 Ωm
σ~ 102 – 108 S m-1
➢ Semiconductor: They have resistivity or conductivity intermediate
to metals and insulators.
ρ ~ 10-5 – 106 Ωm
σ~ 105 – 10-6 S m-1
➢ Insulators: They have high resistivity (or low conductivity).
ρ ~ 1011 – 1019 Ωm
σ~ 10-11 – 10-19 S m-1
2. What are the types of semiconductors?
Types of semiconductors
➢ Elemental semiconductors: Si and Ge
➢ Compound semiconductors: Examples are:
Inorganic: CdS, GaAs, CdSe, InP, etc.
Organic: anthracene, doped pthalocyanines, etc.
Organic polymers: polypyrrole, polyaniline, polythiophene, etc.
3. What valence levels? What are valence electrons?
The partially filled outermost levels in an atom are called valence levels. The
electrons in the outermost level are called valence electrons.
4. Differentiate between valence band and conduction band.
Valence Band Conduction Band
Valence band can never be empty. It Conduction band can be empty. It will
will always be either completely filled always be either partially filled or
or partially filled. remain empty.
Electrons in the valence band are Electrons in the conduction band are
called valence electrons. called conduction electrons.
The electrons in this band are bound to The electrons in this band are not
the nucleus. bound to the nucleus.
5. What is energy band gap?
The gap between the top of the valence band and bottom of the conduction band is
called the energy band gap
6. What is an intrinsic semiconductor? Give an example.
A semiconductor which is pure and contains no impurity is known as an intrinsic
semiconductor. In an intrinsic semiconductor, the number of free electrons and
holes are equal. Common examples of intrinsic semiconductors are pure
germanium and silicon.
7. Write a short note on hole.
Hole is a vacant space which was preoccupied by electrons in the valence band. It is
created when an electron jumps from valence band to conduction band. It is
considered to be positively charged.
8. What is intrinsic carrier concentration?
In intrinsic semiconductors, the number of free electrons, ne is equal to the number
of holes, nh. That is ne = nh = ni where ni is called intrinsic carrier concentration.
9. What is meant by doping?
The Process of addition of a very small amount of impurity into an intrinsic
semiconductor is called as doping.
10. What is an extrinsic semiconductor?
When a small amount, say, a few parts per million (ppm), of a suitable impurity is
added to the pure semiconductor, the conductivity of the semiconductor is
increased manifold. Such materials are known as extrinsic semiconductors or
impurity semiconductors.
11. What are dopants?
The impurity atoms which are added to an intrinsic semiconductor are called as
dopants.
12. What are the types of dopants?
There are two types of dopants
➢ Pentavalent (valency 5) or donor atoms: like Arsenic (As), Antimony (Sb),
Phosphorous (P), etc.
➢ Trivalent (valency 3) or acceptor atom: like Indium (In), Boron (B), Aluminium
(Al), etc.
13. What is a bias?
The process of applying an external voltage to a device is called biasing.
14. Distinguish between N-type and P-type semiconductors.
N-Type semiconductor P-Type semiconductor
N-type semiconductors are obtained P-type semiconductors are obtained
when intrinsic semiconductors are when intrinsic semiconductors are
doped with pentavalent impurities. doped with trivalent impurities.
In N-type semiconductors electrons In P-type semiconductors holes are
are the majority carriers and holes the majority carriers and electrons
are the minority carriers. are the minority carriers.
15. What is meant by a PN junction diode?
When one side of a single crystal of pure semiconductor (Ge or Si) is doped with
acceptor impurity atoms and the other side is doped with donor impurity atoms, a
PN junction is formed
16. How does the d.c. current gain of a transistor change, if the width of the base
region is increased?
It will decrease due to decrease in collector current.
17. The a.c. current gain of a transistor is 120. What is the change in the collector
current in the transistor whose base current changes by 100 µA?
∆𝐼
β = 𝛥𝐼𝐶 , ∆𝐼𝐶 = β∆𝐼𝐵 = 120 × 100 × 10–6 A
𝐵
= 12 × 10−3 A
18. Name the type of biasing of a p – n junction diode so that the junction offers
very high resistance.
Reverse biasing.
19. Name one impurity each, which when added to pure Si, produces (i) n – type,
and (ii) p – type semiconductor.
(i) For n – type, arsenic. (ii) For p – type, Indium
20. Draw energy band diagram of a p – type semiconductor.

21. The signals A and B are not used as two inputs of NOR gate. Sketch the output
wave form. Draw its logic symbol.

22. Which of the diodes is (i) forward biased, and (ii) reverse biased in the
following circuits? Justify your answer.

(a) Diode is reverse biased because p – type of the diode is at lower potential.
(b) Diode is reverse biased because p – type of the diode is at lower potential.
(c) Diode is forward biased because p – type of the diode is at higher potential
(d) Diode is forward biased because p – type of the diode is at higher potential.
23. Why is the conductivity of n – type semiconductor greater than that of the p –
type semiconductor even when both of these have same level of doping?
In n – type semiconductor charge carriers are electrons and mobility of electrons is
more than that of holes.
24. Name two factors on which electrical conductivity of a pure semiconductor at
a given temperature depends.
1. Band gap 2. Biasing
25. Two signals A and B shown in the given figure are used as two inputs of a
NAND gate. Draw its output wave form. Give the logic symbol of NAND gate.

26. Give the ratio of the number of holes and number of conduction electrons in
an intrinsic semiconductor.
1:1
27. Is the ratio of number of holes and number of conduction electrons in a p –
type semiconductor more than, less than or equal to 1?
It’s more than one.
28. In a semiconductor, the concentration of electrons is 8 × 1013 cm–3 and that of
holes is 5 × 1012 cm–3 . Is it a p – type or n – type semiconductor?
p – type
29. How does the energy gap of an intrinsic semiconductor vary, when doped with
a trivalent impurity?
Energy gap decreases.
30. How does the width of the depletion region of a p – n junction vary, if the
reverse bias applied to it decreases?
Width of depletion region increases.
31. How does the width of the depletion region of a p – n junction vary, if the
reverse bias applied to it increase?
Width of depletion region increases.
32. What is a solar cell? How does it work? Give its one use?
Solar cell is a device for converting solar energy into electricity. It is basically a p – n
junction operating in a photovoltaic mode. When light photons fall at the junction,
electron-hole pairs are generated. These move in opposite
33. How does the thickness of the depletion layer in a p – n junction vary with
increase in reverse bias?
On increasing reverse bias, the depletion layer increases.
34. How does the collector current change in a junction transistor, if the base
region has larger width?
If base is made wider, the recombination increases in base region. Hence IB
increases and as IE = IC + IB, IC decreases.
35. How does the conductance of a semiconducting material change with rise in
temperature?
Increases.
36. Give the ratio of the number of holes and the number of conduction electrons
in an intrinsic semiconductor.
In an intrinsic semiconductor, the ratio of the number of holes to the number of
conduction electrons is unity.
37. In the given diagram, is the diode D forward or reverse biased?

The diode D is reverse biased.

38. Write the truth table for the combination of gates shown here.
 A B F
0 0 1
1 0 1
0 1 1
1 1 0
39. What is a hole in a semiconductor?
The hole is introduced into the semiconductor by the acceptor impurity having
energy which is 0.04 eV above the highest energy level of the valence band.
40. What is the charge on a hole?
Hole carries + 1.6 × 10-19 C charge.
41. Which, out of Ge and Si, is a better conductor at room temperature?
Germanium is a better conductor as forbidden energy gap E g is
0.72𝑒𝑉 𝑓𝑜𝑟 𝐺𝑒
{
1.1 𝑒𝑉 𝑠𝑜 𝑆𝑖
42. Which one will you prefer for the rectification of low AC voltages – a Si diode
or a Ge diode? Why?
A Si – diode is preferred over Ge – diode because of its poor conductivity when it’s
reverse biased. Hence, it results in better rectification action.
43. The barrier potential of a silicon p – n junction is 0.7 V. Can we measure it
using a voltmeter?
No, it cannot be measured by voltmeter.
44. Frequency of input voltage to a half – wave rectifier is 50 Hz. What will be the
frequency of the output voltage?
The frequency of output waveform will be 50Hz.
45. Draw the output wave forms for the gates shown below:

The input and output waveforms are as shown.

46. Name any two intrinsic semiconductors.

Germanium and Silicon.
47. On what factors does conductivity of intrinsic semiconductor depend?
The conductivity of intrinsic semiconductor depends only on temperature.
48. How can pure Ge be converted into (i) a p – type semiconductor? (ii) a n – type
semiconductor ?
(i) A pure Ge is converted into an n – type semiconductor by doping with trivalent
impurities. (ii) A pure Ge is converted into an n – type semiconductor by doping
with pentavalent impurities.
49. What is the phase relation between input signal and output signal in the case
of transistor amplifier in (a) common – emitter configuration (b) common –
base configuration?
a) In CE the input and output voltages are out of phase by 180°
b) In CB the input and output voltages are in phase.
50. The inputs of a NAND gate are logic 0 and logic 1. What will be the output ?
The output will be 1.
51. Show how a NAND gate can be used as a NOT gate.
The diagram as shown:
52. What happens to the width of the depletion region in a p – n junction when it is
(i) forward biased (ii) reverse biased?
(i) The depletion width decreases. (ii) The depletion width increases.
53. How is an n – p – n transistor represented symbolically?
Symbol of n – p – n transistor:

54. How is a p – n – p transistor represented symbolically?

55. Draw the logic symbol of an AND gate.

Logic symbol of an AND gate:

56. Give one important difference between crystalline solids and amorphous
solids.
Cystalline solids have very sharp melting points, whereas an amorphous solid melts
over a wide range of temperatures.
57. Why a transistor cannot be used as a rectifier?
A rectifier device should conduct only when forward biased and not when reverse
biased. But a transistor conducts either way.
58. Write down the truth – table of NAND gate.
Truth – table of NAND gate:
A B Y
0 0 1
0 1 1
1 0 1
1 1 0
59. Write the truth – table of an AND gate.
Truth table of an AND gate:
A B Y = A.B
0 0 0
1 0 0
0 1 0
1 1 1
60. What is depletion region in a p – n junction ?
Depletion region in a p – n – junction, is a region around the junction which does not
have any free charge carriers due to electron – hole recombination.
61. What is meant by depletion region of a PN junction diode?
A region which does not have any mobile charges and present very close to the
junction is called depletion region. This region is devoid of any charges.
62. What is meant by forward biasing a diode?
When the positive terminal of the battery is connected to P-side and negative
terminal to the N-side, so that the potential difference acts in opposite direction to
the barrier potential then the PN junction diode is said to be forward biased.
63. What is meant by reverse biasing a diode?
When the positive terminal of the battery is connected to the N-Side and negative
terminal to the P-side so that the applied potential difference is in the same
direction as that of barrier potential the junction is said to be reverse biased.
64. What is cut in voltage or knee voltage?
The voltage at which the current in a diode starts to increases rapidly when
connected in reverse bias is known as cut-in voltage or knee voltage of the diode.
65. What is half wave rectifier?
If an alternating voltage is applied across a diode in series with a load, a pulsating
voltage will appear across the load only during the half cycles of the ac input during
which the diode is forward biased. Such rectifier circuit is called a half-wave
rectifier.
66. What is zener diode?
Zener diode is a reverse biased heavily doped semiconductor (silicon or germanium
) PN junction diode which is operated exclusively in the breakdown region.
Symbol:

67. Mention the types of optoelectronic devices.

Types of optoelectronic devices
➢ Photodiodes used for detecting optical signal (photodetectors).
➢ Light emitting diodes (LED) which convert electrical energy into light.
➢ Photovoltaic devices which convert optical radiation into electricity(solar cells).
68. What are the advantages of LED’s over conventional lamps?
LEDs have the following advantages over conventional incandescent low power
lamps:
➢ Low operational voltage and less power.
➢ Fast action and no warm-up time required.
➢ The bandwidth of emitted light is 100 Å to 500 Å or in other words it is nearly (but
not exactly) monochromatic.
➢ Long life and ruggedness.
➢ Fast on-off switching capability.
69. What are the conditions for the selection of a material for solar cell
fabrication?
The important criteria for the selection of a material for solar cell fabrication are
➢ band gap (~1.0 to 1.8 eV)
➢ high optical absorption (~104 cm–1)
➢ electrical conductivity
➢ availability of the raw material, and
➢ Cost.
70. In a transistor emitter and collector cannot be interchanged, why?
In a transistor the emitter region is heavily doped since emitter has to supply
majority carriers. The base is lightly doped. The collector region is lightly doped.
Since it has to accept majority charges carriers, it is physically larger in size. Hence
emitter and collector cannot be interchanged.
71. What are the types of transistors?
Types of transistors
➢ n-p-n transistor: Here two segments of n-type semiconductor (emitter and
collector) are separated by a segment of p-type semiconductor (base).
➢ p-n-p transistor: Here two segments of p-type semiconductor (termed as emitter
and collector) are separated by a segment of n-type semiconductor (termed as
base).
72. What are the two segments of a transistor?
Segments of a transistor are given below:
➢ Emitter: This is the segment on one side of the transistor. It is of moderate size and
heavily doped. It supplies a large number of majority carriers for the current flow
through the transistor.
➢ Base: This is the central segment. It is very thin and lightly doped.
➢ Collector: This segment collects a major portion of the majority carriers supplied by
the emitter. The collector side is moderately doped and larger in size as compared to
the emitter.
73. Define current amplification factor.
The current amplification factor or current gain of a transistor is the ratio of output
current to the input current. If the transistor is connected in common base mode,
the current gain α = IC/IE and if the transistor is connected in common emitter mode,
the current gain β = IC/IB.
74. Define input impedance of a transistor.
The input impedance of the transistor is defined as the ratio of small change in base
– emitter voltage to the corresponding change in base current at a given VCE. The
input impedance of the transistor in CE mode is very high.
∆V
Formula: Input impedance ri = ∆IBE
B
Unit: ohm
75. Define output impedance of a transistor.
The output impedance ro is defined as the ratio of variation in the collector emitter
voltage to the corresponding variation in the collector current at a constant base
current in the active region of the transistor characteristic curves. The output
impedance of a transistor in CE mode is low.
∆V
Formula: Output impedance ro = ∆ICE
C
Unit: ohm
76. Why transistor is called a current amplification device?
In a transistor, if its emitter base junction is forward biased and collector base
junction is reverse biased, for a small input current Ib there is a large output current
Ic since in forward bias a small amount of voltage produces a large number of charge
carriers. Hence transistor can be called a current amplifying device.
77. Why is CE configuration preferred over CB configuration?
In CE configuration a transistor has,
➢ High input impedance
➢ Low output impedance
➢ High current gain
➢ High voltage gain
Due to these characteristics CE configuration is the most preferred.
78. What is meant by feedback? Name its different types.
Feedback is said to be a process in which a fraction of the output signal is returned
or fed back to the input and combined with the input signal.
If the magnitude of the input signal is reduced by the feedback, the feedback is
called negative or degenerative.
If the magnitude of the input signal is increased by the feedback, such feedback is
called positive or regenerative.
79. What is an oscillator?
An Oscillator may be defined as an electronic circuit which converts energy from a
d.c. source into a periodically varying output.
80. Distinguish between analog signal and digital signal.
Analog Signal Digital Signal
The continuous time varying voltage or A digital signal is discrete in nature. It
current signal(sinusoidal) are called possess two discrete levels, ‘High’ and
analog signals. ‘low’.
These signals are processed using linear These signals are processed using
ICs. digital ICs.
81. What are logic gates?
Circuits which are used to process digital signals are called logic gates. They are
binary in nature. Gate is a digital circuit with one or more inputs but with only one
output. The output appears only for certain combination of input logic levels. Logic
gates are the basic building blocks from which most of the digital systems are built
up.
82. How materials are classified based on energy bands?
Classification of materials based on energy bands:
Case I: This refers to a situation, as shown in Fig. (a). One can have a metal either
when the conduction band is partially filled and the valance band is partially empty
or when the conduction and valance bands overlap. When there is overlap electrons
from valence band can easily move into the conduction band. This situation makes a
large number of electrons available for electrical conduction. When the valence
band is partially empty, electrons from its lower level can move to higher level
making conduction possible. Therefore, the resistance of such materials is low or the
conductivity is high.

Case II: In this case, as shown in Fig. (b), a large band gap Eg exists (Eg > 3 eV).
There are no electrons in the conduction band, and therefore no electrical
conduction is possible. Note that the energy gap is so large that electrons cannot be
excited from the valence band to the conduction band by thermal excitation. This is
the case of insulators.
Case III: This situation is shown in Fig.(c). Here a finite but small band gap (Eg < 3
eV) exists. Because of the small band gap, at room temperature some electrons from
valence band can acquire enough energy to cross the energy gap and enter the
conduction band. These electrons (though small in numbers) can move in the
conduction band. Hence, the resistance of semiconductors is not as high as that of
the insulators.
83. Explain in detail about intrinsic semiconductor.
Intrinsic semiconductors are pure form of semi conductors. Some examples for
intrinsic semiconductors are Si, Ge etc.
Si and Ge have four valence electrons. In its crystalline structure, every Si or Ge atom
tends to share one of its four valence electrons with each of its four nearest
neighbour atoms, and also to take share of one electron from each such neighbour.
These shared electron pairs are referred to as forming a covalent bond or simply a
valence bond.

As the temperature increases, more thermal energy becomes available to these

electrons and some of these electrons may break–away (becoming free electrons
contributing to conduction). The thermal energy effectively ionises only a few
atoms in the crystalline lattice and creates a vacancy in the bond as shown in Fig.
(a). The neighbourhood, from which the free electron (with charge –q) has come out
leaves a vacancy with an effective charge (+q ). This vacancy with the effective
positive electronic charge is called a hole. The hole behaves as an apparent free
particle with effective positive charge. In intrinsic semiconductors, the number of
free electrons, ne is equal to the number of holes, nh. That is ne = nh = ni where ni is
called intrinsic carrier concentration.
Semiconductors posses the unique property in which, apart from electrons, the
holes also move. Suppose there is a hole at site 1 as shown in Fig. (a). The
movement of holes can be visualised as shown in Fig. (b). An electron from the
covalent bond at site 2 may jump to the vacant site 1 (hole). Thus, after such a jump,
the hole is at site 2 and the site 1 has now an electron. Therefore, apparently, the
hole has moved from site 1 to site 2. Note that the electron originally set free [Fig.
(a)] is not involved in this process of hole motion. The free electron moves
completely independently as conduction electron and gives rise to an electron
current, Ie under an applied electric field. Under the action of an electric field, these
holes move towards negative potential giving the hole current, Ih. The total current, I
is thus the sum of the electron current Ie and the hole current Ih:
I = Ie + Ih ------------------------ (1)
It may be noted that apart from the process of generation of conduction electrons
and holes, a simultaneous process of recombination occurs in which the electrons
recombine with the holes. At equilibrium, the rate of generation is equal to the rate
of recombination of charge carriers. The recombination occurs due to an electron
colliding with a hole.

An intrinsic semiconductor will behave like an insulator at T = 0 K as shown in Fig.

(a). It is the thermal energy at higher temperatures (T > 0K), which excites some
electrons from the valence band to the conduction band. These thermally excited
electrons at T > 0 K, partially occupy the conduction band. Therefore, the energy-
band diagram of an intrinsic semiconductor will be as shown in Fig. (b). Here, some
 electrons are shown in the conduction band. These have come from the valence
band leaving equal number of holes there.
84. Explain how extrinsic semiconductors are produced.
The conductivity of an intrinsic semiconductor depends on its temperature, but at
room temperature its conductivity is very low. As such, no important electronic
devices can be developed using these semiconductors. Hence there is a necessity of
improving their conductivity. This can be done by making use of impurities. When a
small amount, say, a few parts per million (ppm), of a suitable impurity is added to
the pure semiconductor, the conductivity of the semiconductor is increased
manifold. Such materials are known as extrinsic semiconductors or impure
semiconductors. The deliberate addition of a desirable impurity is called doping and
the impurity atoms are called dopants. Such a material is also called a doped
semiconductor. The dopant has to be such that it does not distort the original pure
semiconductor lattice. It occupies only a very few of the original semiconductor
atom sites in the crystal. A necessary condition to attain this is that the sizes of the
dopant and the semiconductor atoms should be nearly the same.
There are two types of dopants used in doping the tetravalent Si or Ge:
(i) Pentavalent (valency 5); like Arsenic (As), Antimony (Sb), Phosphorous (P), etc.
(ii) Trivalent (valency 3); like Indium (In), Boron (B), Aluminium (Al), etc.
We shall now discuss how the doping changes the number of charge carriers (and
hence the conductivity) of semiconductors. Si or Ge belongs to the fourth group in
the Periodic table and, therefore, we choose the dopant element from nearby fifth or
third group, expecting and taking care that the size of the dopant atom is nearly the
same as that of Si or Ge.
85. What are n – type semiconductors?

Suppose we dope Si or Ge with a pentavalent element as shown in Fig. When an

atom of +5 valency element occupies the position of an atom in the crystal lattice of
Si, four of its electrons bond with the four silicon neighbours while the fifth remains
very weakly bound to its parent atom. This is because the four electrons
participating in bonding are seen as part of the effective core of the atom by the fifth
electron. As a result the ionisation energy required to set this electron free is very
small and even at room temperature it will be free to move in the lattice of the
semiconductor. For example, the energy required is ~ 0.01 eV for germanium, and
0.05 eV for silicon, to separate this electron from its atom. This is in contrast to the
energy required to jump the forbidden band (about 0.72 eV for germanium and
about 1.1 eV for silicon) at room temperature in the intrinsic semiconductor. Thus,
the pentavalent dopant is donating one extra electron for conduction and hence is
known as donor impurity. The number of electrons made available for conduction
by dopant atoms depends strongly upon the doping level and is independent of any
increase in ambient temperature. The number of free electrons (with an equal
number of holes) generated by Si atoms, increases weakly with temperature. In a
doped semiconductor the total number of conduction electrons ne is due to the
electrons contributed by donors and those generated intrinsically, while the total
number of holes nh is only due to the holes from the intrinsic source. But the rate of
 recombination of holes would increase due to the increase in the number of
electrons. As a result, the number of holes would get reduced further. Thus, with
proper level of doping the number of conduction electrons can be made much larger
than the number of holes. Hence in an extrinsic semiconductor doped with
pentavalent impurity, electrons become the majority carriers and holes the minority
carriers. These semiconductors are, therefore, known as n-type semiconductors. For
n-type semiconductors, we have, ne >> nh
86. What are p – type semiconductors?
p – type semiconductor is obtained when Si or Ge is doped with a trivalent impurity
like Al, B, In, etc. The dopant has one valence electron less than Si or Ge and,
therefore, this atom can form covalent bonds with neighbouring three Si atoms but
does not have any electron to offer to the fourth Si atom. So the bond between the
fourth neighbour and the trivalent atom has a vacancy or hole as shown in Fig. Since
the neighbouring Si atom in the lattice wants an electron in place of a hole, an
electron in the outer orbit of an atom in the neighbourhood may jump to fill this
vacancy, leaving a vacancy or hole at its own site. Thus the hole is available for
conduction. Note that the trivalent foreign atom becomes effectively negatively
charged when it shares fourth electron with neighbouring Si atom. Therefore, the
dopant atom of p-type material can be treated as core of one negative charge along
with its associated hole as shown in Fig. (b). It is obvious that one acceptor atom
gives one hole. These holes are in addition to the intrinsically generated holes while
the source of conduction electrons is only intrinsic generation. Thus, for such a
material, the holes are the majority carriers and electrons are minority carriers.
Therefore, extrinsic semiconductors doped with trivalent impurity are called p-type
semiconductors. For p-type semiconductors, the recombination process will reduce
the number (ni)of intrinsically generated electrons to ne. We have, for p-type
semiconductors

nh >> ne The crystal maintains an overall charge neutrality as the charge of

additional charge carriers is just equal and opposite to that of the ionised cores in
the lattice. In extrinsic semiconductors, because of the abundance of majority
current carriers, the minority carriers produced thermally have more chance of
meeting majority carriers and thus getting destroyed. Hence, the dopant, by adding
a large number of current carriers of one type, which become the majority carriers,
indirectly helps to reduce the intrinsic concentration of minority carriers. The
semiconductor’s energy band structure is affected by doping. In the case of extrinsic
semiconductors, additional energy states due to donor impurities (ED) and acceptor
impurities (EA) also exist. In the energy band diagram of n-type Si semiconductor,
the donor energy level ED is slightly below the bottom EC of the conduction band and
electrons from this level move into the conduction band with very small supply of
energy. At room temperature, most of the donor atoms get ionised but very few
(~10–12) atoms of Si get ionised. So the conduction band will have most electrons
coming from the donor impurities, as shown in Fig. (a). Similarly, for p-type
semiconductor, the acceptor energy level EA is slightly above the top EV of the
 valence band as shown in Fig. (b). With very small supply of energy an electron from
the valence band can jump to the level EA and ionise the acceptor negatively.
(Alternately, we can also say that with very small supply of energy the hole from
level EA sinks down into the valence band. Electrons rise up and holes fall down
when they gain external energy.) At room temperature, most of the acceptor atoms
get ionised leaving holes in the valence band. Thus at room temperature the density
of holes in the valence band is predominantly due to impurity in the extrinsic
semiconductor. The electron and hole concentration in a semiconductor in thermal
equilibrium is given by nenh = ni2
87. Explain about the formation of PN junction diode.
Consider a thin p-type silicon (p-Si) semiconductor wafer. By adding precisely a
small quantity of pentavalent impurity, part of the p-Si wafer can be converted into
n-Si. There are several processes by which a semiconductor can be formed. The
wafer now contains p-region and n-region and a metallurgical junction between p-,
and n- region. Two important processes occur during the formation of a p-n
junction: diffusion and drift. In an n-type semiconductor, the concentration of
electrons (number of electrons per unit volume) is more compared to the
concentration of holes. Similarly, in a p-type semiconductor, the concentration of
holes is more than the concentration of electrons. During the formation of p-n
junction, and due to the concentration gradient across p-, and n- sides, holes diffuse
from p-side to n-side (p → n) and electrons diffuse from n-side to p-side (n → p).
This motion of charge carries gives rise to diffusion current across the junction.
When an electron diffuses from n → p, it leaves behind an ionized donor on n-side.
This ionised donor (positive charge) is immobile as it is bonded to the surrounding
atoms. As the electrons continue to diffuse from n → p, a layer of positive charge (or
positive space-charge region) on n-side of the junction is developed.

Similarly, when a hole diffuses from p → n due to the concentration gradient, it

leaves behind an ionised acceptor (negative charge) which is immobile. As the holes
continue to diffuse, a layer of negative charge (or negative space-charge region) on
the p-side of the junction is developed.
This space-charge region on either side of the junction together is known as
depletion region as the electrons and holes taking part in the initial movement
across the junction depleted the region of its free charges (Fig.). The thickness of
depletion region is of the order of one-tenth of a micrometre. Due to the positive
space-charge region on n-side of the junction and negative space charge region on p-
side of the junction, an electric field directed from positive charge towards negative
charge develops. Due to this field, an electron on p-side of the junction moves to n-
side and a hole on n-side of the junction moves to p-side. The motion of charge
carriers due to the electric field is called drift. Thus a drift current, which is opposite
in direction to the diffusion current (Fig.) starts.
 Initially, diffusion current is large and drift current is small. As the diffusion process
continues, the space-charge regions on either side of the junction extend, thus
increasing the electric field strength and hence drift current. This process continues
until the diffusion current equals the drift current. Thus a p-n junction is formed. In
a p-n junction under equilibrium there is no net current. The loss of electrons from
the n-region and the gain of electron by the p-region causes a difference of potential
across the junction of the two regions. The polarity of this potential is such as to
oppose further flow of carriers so that a condition of equilibrium exists. Figure
shows the p-n junction at equilibrium and the potential across the junction. The n-
material has lost electrons, and p material has acquired electrons. The n material is
thus positive relative to the p material. Since this potential tends to prevent the
movement of electron from the n region into the p region, it is often called a barrier
potential.
88. Explain the working of a PN junction diode connected in forward bias.

When an external voltage V is applied across a semiconductor diode such that p-side
is connected to the positive terminal of the battery and n-side to the negative
terminal [Fig. (a)], it is said to be forward biased. The applied voltage mostly drops
across the depletion region and the voltage drop across the p-side and n-side of the
junction is negligible. This is because the resistance of the depletion region is very
high compared to the resistance of n-side and p-side. The direction of the applied
voltage (V) is opposite to the built-in potential V0. As a result, the depletion layer
width decreases and the barrier height is reduced [Fig. (b)]. The effective barrier
height under forward bias is (V0 – V). If the applied voltage is small, the barrier
potential will be reduced only slightly below the equilibrium value, and only a small
number of carriers in the material—those that happen to be in the uppermost
energy levels—will possess enough energy to cross the junction. So the current will
be small. If we increase the applied voltage significantly, the barrier height will be
reduced and more number of carriers will have the required energy. Thus the
current increases. Due to the applied voltage, electrons from n-side cross the
depletion region and reach p-side. Similarly, holes from p-side cross the junction
and reach the n-side. This process under forward bias is known as minority carrier
injection. At the junction boundary, on each side, the minority carrier concentration
increases significantly compared to the locations far from the junction. Due to this
concentration gradient, the injected electrons on p-side diffuse from the junction
edge of p-side to the other end of p-side. Likewise, the injected holes on n-side
diffuse from the junction edge of n-side to the other end of n-side (Fig.). This motion
of charged carriers on either side gives rise to current. The total diode forward
current is sum of hole diffusion current and conventional current due to electron
diffusion. The magnitude of this current is usually in mA.
89. Explain the working of a PN junction diode connected in reverse bias.
When an external voltage (V) is applied across the diode such that n-side is positive
and p-side is negative, it is said to be reverse biased [Fig.(a)]. The applied voltage
mostly drops across the depletion region. The direction of applied voltage is same as
the direction of barrier potential. As a result, the barrier height increases and the
depletion region widens due to the change in the electric field. The effective barrier
height under reverse bias is (V0 + V), [Fig. (b)]. This suppresses the flow of electrons
from n → p and holes from p → n. Thus, diffusion current, decreases enormously
compared to the diode under forward bias.
 The electric field direction of the junction is such that if electrons on p-side or holes
on n-side in their random motion come close to the junction, they will be swept to
its majority zone. This drift of carriers gives rise to current. The drift current is of
the order of a few μA. This is quite low because it is due to the motion of carriers
from their minority side to their majority side across the junction. The drift current
is also there under forward bias but it is negligible (μA) when compared with
current due to injected carriers which is usually in mA.
The current under reverse bias is essentially voltage independent upto a critical
reverse bias voltage, known as breakdown voltage (Vbr). When V = Vbr, the diode
reverse current increases sharply. Even a slight increase in the bias voltage causes
large change in the current. If the reverse current is not limited by an external
circuit below the rated value the p-n junction will get destroyed. Once it exceeds the
rated value, the diode gets destroyed due to overheating. This can happen even for
the diode under forward bias, if the forward current exceeds the rated value.
81. Differentiate between forward bias and reverse bias in a diode.
Forward bias Reverse bias
When the positive terminal of the When the positive terminal of the
battery is connected to P-side and battery is connected to the N-Side and
negative terminal to the N-side, so negative terminal to the P-side so that
that the potential difference acts in the applied potential difference is in
opposite direction to the barrier the same direction as that of barrier
potential then the PN junction diode is potential the junction is said to be
said to be forward biased. reverse biased.
The potential barrier as well as the The depletion region becomes wider
width of the depletion region is and the potential barrier is increased.
reduced.
The current in forward bias is due to The current in reverse bias is due to
majority carriers. minority carriers.
The forward current does not depend The reverse current depends upon
upon temperature. temperature.
The forward current is in mA. The reverse current is in μA.
82. Distinguish between intrinsic and extrinsic semiconductor.
Intrinsic Semiconductor Extrinsic Semiconductor
It is a pure semiconductor material. It is obtained by doping a small
amount of impurity to a pure
semiconductor.
The number of free electrons is The number free electrons and holes
equal to number of holes. are never equal.
Its electrical conductivity is low. Its electrical conductivity is high.
Its electrical conductivity is Its electrical conductivity depends
function of temperature alone. upon temperature as well as on the
quantity of impurity atoms depends
on the structure.
Examples are crystalline form of Examples are silicon and germanium
pure silicon and germanium. with impurity atoms of arsenic,
antimony, phosphorous etc. or
indium, boron, aluminium etc.
83. Explain the forward bias characteristics.
Forward bias characteristics
The circuit for the study of forward bias characteristics of PN junction diode is
shown in fig a. The voltage between P-end and N-end is increased from zero in
suitable equal steps and the corresponding currents are noted down. Fig b shows
the forward bias characteristic curve of the diode. Voltage is the independent
variable. Therefore, it is plotted against Y-axis. From the characteristic curve, the
following conclusions can be made. (i) the forward characteristic is not a straight
line. Hence the ratio V/I is not a constant (i.e) the diode does not obey ohm’s law.
This implies that the semiconductor diode is a non linear conductor of electricity.
(ii) It can be seen from the characteristic curve that initially, the current is very
small. This is because, the diode will start conducting only when the external voltage
overcomes the barrier potential (0.7V for silicon diode). As the voltage is increased
to 0.7V large number of free electrons and holes start crossing the junction. Above
0.7V the current increases rapidly. The voltage at which the current starts to
increases rapidly is known as cut-in voltage or knee voltage of the diode.
Diagram:

84. Explain the reverse bias characteristics.

Reverse bias characteristics
The circuit for the study of reverse bias characteristics of PN junction diode is
shown in fig a. The voltage is increased from zero in suitable steps. For each voltage,
the corresponding current readings are noted down. Fig b shows the reverse bias
characteristic curve of the diode. From the characteristic curve, it can be concluded
that as voltage is increased from zero, reverse current (in the order of
microamperes) increases and reaches the maximum value at a small value of the
reverse voltage. When the voltage is further increased, the current is almost
independent of the reverse voltage upto a certain critical value. This reverse current
is known as the reverse saturation current or leakage current. This current is due to
the minority charge carriers, which depends on junction temperature.
Diagram:
85. Explain the working of a half wave rectifier.

When an alternating voltage is applied across a diode in series with a load, a

pulsating voltage will appear across the load only during the half cycles of the ac
input during which the diode is forward biased. Such rectifier circuit, as shown in
Fig., is called a half-wave rectifier. The secondary of a transformer supplies the
desired ac voltage across terminals A and B. When the voltage at A is positive, the
diode is forward biased and it conducts. When A is negative, the diode is reverse-
biased and it does not conduct. The reverse saturation current of a diode is
negligible and can be considered equal to zero for practical purposes. Therefore, in
the positive half-cycle of ac there is a current through the load resistor RL and we get
an output voltage, as shown in Fig. (b), whereas there is no current in the negative
halfcycle. In the next positive half-cycle, again we get the output voltage. Thus, the
output voltage, though still varying, is restricted to only one direction and is said to
be rectified. Since the rectified output of this circuit is only for half of the input ac
wave it is called as half-wave rectifier.
86. Explain the working of a full wave rectifier.
 The circuit using two diodes, shown in Fig. (a), gives output rectified voltage
corresponding to both the positive as well as negative half of the ac cycle. Hence, it is
known as full-wave rectifier. Here the p-side of the two diodes are connected to the
ends of the secondary of the transformer. The n-side of the diodes are connected
together and the output is taken between this common point of diodes and the
midpoint of the secondary of the transformer. So for a full-wave rectifier the
secondary of the transformer is provided with a centre tapping and so it is called
centre-tap transformer. As can be seen from Fig. (c) the voltage rectified by each
diode is only half the total secondary voltage. Each diode rectifies only for half the
cycle, but the two do so for alternate cycles. Thus, the output between their common
terminals and the centre tap of the transformer becomes a full-wave rectifier output.
Suppose the input voltage to A with respect to the centre tap at any instant is
positive. It is clear that, at that instant, voltage at B being out of phase will be
negative as shown in Fig.(b). So, diode D1 gets forward biased and conducts (while
D2 being reverse biased is not conducting).
Hence, during this positive half cycle we get an output current (and a output voltage
across the load resistor RL) as shown in Fig. (c). In the course of the ac cycle when
the voltage at A becomes negative with respect to centre tap, the voltage at B would
be positive. In this part of the cycle diode D1 would not conduct but diode D2 would,
giving an output current and output voltage (across RL) during the negative half
cycle of the input ac. Thus, we get output voltage during both the positive as well as
the negative half of the cycle. The rectified voltage is in the form of pulses of the
shape of half sinusoids.
87. What is zener diode? What is its significance?
 Zener diode is a special purpose semiconductor diode, named after its inventor C.
Zener. It is designed to operate under reverse bias in the breakdown region and
used as a voltage regulator. The symbol for Zener diode is shown in Fig. (a). Zener
diode is fabricated by heavily doping both p-, and n- sides of the junction. Due to
this, depletion region formed is very thin (<10–6 m) and the electric field of the
junction is extremely high (~5×106 V/m) even for a small reverse bias voltage of
about 5V. The I-V characteristics of a Zener diode is shown in Fig.(b). It is seen that
when the applied reverse bias voltage(V) reaches the breakdown voltage (Vz) of the
Zener diode, there is a large change in the current. Note that after the breakdown
voltage Vz, a 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 supply voltages so that they are
constant. The reverse current is due to the flow of electrons (minority carriers) from
p → n and holes from n → p. As the reverse bias voltage is increased, the electric
field at the junction becomes significant. When the reverse bias voltage V = Vz, then
the electric field strength is high enough to pull valence electrons from the host
atoms on the p-side which are accelerated to n-side. These electrons account for
high current observed at the breakdown. The emission of electrons from the host
atoms due to the high electric field is known as internal field emission or field
ionisation. The electric field required for field ionisation is of the order of 106 V/m.
88. How does zener diode act as a voltage regulator?

When the ac input voltage of a rectifier fluctuates, its rectified output also fluctuates.
To get a constant dc voltage from the dc unregulated output of a rectifier, we use a
Zener diode. The circuit diagram of a voltage regulator using a Zener diode is shown
in Fig. The unregulated dc voltage (filtered output of a rectifier) is connected to the
Zener diode through a series resistance Rs such that the Zener diode is reverse
biased. If the input voltage increases, the current through Rs and Zener diode also
increases. This increases the voltage drop across Rs without any change in 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 also
decreases. 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 drop across Rs without any change in voltage
across the Zener diode. Thus the Zener diode acts as a voltage regulator. The Zener
diode has to be selected according to the required output voltage and accordingly
the series resistance Rs.
89. What is a photodiode?

A Photodiode is again a special purpose p-n junction diode fabricated with a

transparent window to allow light to fall on the diode. It is operated under reverse
bias. When the photodiode is illuminated with light (photons) with energy (hν)
greater than the energy gap (Eg) of the semiconductor, then electron-hole pairs are
generated due to the absorption of photons. The diode is fabricated such that the
generation of e-h pairs takes place in or near the depletion region of the diode. Due
to electric field of the junction, electrons and holes are separated before they
recombine. The direction of the electric field is such that electrons reach n-side and
holes reach p-side. Electrons are collected on n-side and holes are collected on p-
side giving rise to an emf. When an external load is connected, current flows. The
magnitude of the photocurrent depends on the intensity of incident light. It is easier
to observe the change in the current with change in the light intensity, if a reverse
bias is applied. Thus photodiode can be used as a photodetector to detect optical
signals.
90. Diagrammatically represent the V-I characteristics of a photodiode.

91. Explain briefly about light emitting diode.

Light emitting diode is a heavily doped p-n junction which under forward bias emits
spontaneous radiation. The diode is encapsulated with a transparent cover so that
emitted light can come out. When the diode is forward biased, electrons are sent
from n → p and holes are sent from p → n. At the junction boundary the
concentration of minority carriers increases compared to the equilibrium
concentration (i.e., when there is no bias). Thus at the junction boundary on either
side of the junction, excess minority carriers are there which recombine with
majority carriers near the junction. On recombination, the energy is released in the
form of photons. Photons with energy equal to or slightly less than the band gap are
emitted. 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. LEDs are biased such that the light emitting efficiency is maximum.
The V-I characteristics of a LED is similar to that of a Si junction diode. But the
threshold voltages are much higher and slightly different for each colour. The
reverse breakdown voltages of LEDs are very low, typically around 5V. So care
should be taken that high reverse voltages do not appear across them.
92. Mention the applications of LED.
➢ LEDs that can emit red, yellow, orange, green and blue light are commercially
available. The semiconductor used for fabrication of visible LEDs must at least have
a band gap of 1.8 eV (spectral range of visible light is from about 0.4 μm to 0.7 μm,
i.e., from about 3 eV to 1.8 eV).
➢ The compound semiconductor Gallium Arsenide – Phosphide (GaAs1–xPx) is used for
making LEDs of different colours. GaAs0.6 P0.4 (Eg ~ 1.9 eV) is used for red LED. GaAs
(Eg ~ 1.4 eV) is used for making infrared LED.
➢ These LEDs find extensive use in remote controls, burglar alarm systems, optical
communication, etc.
➢ Extensive research is being done for developing white LEDs which can replace
incandescent lamps.
93. What are the advantages of LED’s over conventional power lamps?
LEDs have the following advantages over conventional incandescent
low power lamps:
(i) Low operational voltage and less power.
(ii) Fast action and no warm-up time required.
(iii) The bandwidth of emitted light is 100 Å to 500 Å or in other words it
is nearly (but not exactly) monochromatic.
(iv) Long life and ruggedness.
(v) Fast on-off switching capability.
94. Explain the construction and working of a solar cell.

A solar cell is basically a p-n junction which generates emf when solar radiation falls
on the p-n junction. 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.
A simple p-n junction solar cell is shown in Fig. A p-Si wafer of about 300 µm is
taken over which a thin layer (~0.3 µm) of n-Si is grown on one-side by diffusion
process. The other side of p-Si is coated with a metal (back contact). On the top of n-
Si layer, metal finger electrode (or metallic grid) is deposited. This acts as a front
contact. The metallic grid occupies only a very small fraction of the cell area (<15%)
so that light can be incident on the cell from the top. The generation of emf by a
solar cell, when light falls on, it is due to the following three basic processes:
generation, separation and collection—
(i) Generation of e-h pairs due to light (with hν> Eg) close to the junction; (ii)
Separation of electrons and holes due to electric field of the depletion region.
Electrons are swept to n-side and holes to p-side; (iii) The electrons reaching the n-
side are collected by the front contact and holes reaching p-side are collected by the
back contact. Thus p-side becomes positive and n-side becomes negative giving rise
to photovoltage. When an external load is connected as shown in the Fig. (a) a
photocurrent IL flows through the load.
95. Draw the I – V characteristics of a solar cell.
A typical I-V characteristics of a solar cell is shown
 in the Fig. (b).

Note that the I – V characteristics of solar cell is

drawn in the fourth quadrant of the coordinate
axes. This is because a solar cell does not draw
current but supplies the same to the load.

96. What are the criterions for the selection of a material for solar cell
fabrication?
The important criteria for the selection of a material for solar cell fabrication are (i)
band gap (~1.0 to 1.8 eV), (ii) high optical absorption (~104 cm–1), (iii) electrical
conductivity, (iv) availability of the raw material, and (v) cost.
97. What is a transistor? Mention its types.
A transistor has three doped regions forming two p-n junctions between them.
There are two types of transistors, as shown in Fig.

(i) n-p-n transistor: Here two segments of n-type semiconductor (emitter and
collector) are separated by a segment of p-type semiconductor (base).
(ii) p-n-p transistor: Here two segments of p-type semiconductor (termed as emitter
and collector) are separated by a segment of n-type semiconductor (termed as
base).
98. In a transistor emitter and collector cannot be interchanged, why?
In a transistor the emitter region is heavily doped since emitter has to supply
majority carriers. The base is lightly doped. The collector region is lightly doped.
Since it has to accept majority charges carriers, it is physically larger in size. Hence
emitter and collector cannot be interchanged.
99. Mention the biasing conditions of a transistor.
For a transistor to work the biasing to be given are as follows:
➢ The emitter-base junction is forward biased, so that majority charge carriers are
repelled from the emitter and the junction offers very low resistance to the
current.
➢ The collector-base junction is reverse biased so that it attracts majority charge
carries and this junction offers a high resistance to the
100. Explain the working of a pnp transistor.
Working of a PNP transistor
A PNP transistor is like two PN junction diodes which are placed back to back. At
each junction there is a depletion region which gives rise to a potential barrier. The
external biasing of the junction is provided by the batteries VEE and Vcc as shown in
fig a. The emitter base junction is forward biased and the collector base junction is
reverse biased.
Diagram:

Since the emitter-base junction is forward biased, a large number of holes cross the
junction and enters the base. At the same time, very few electrons flow from the
base to the emitter. These electrons when reach emitter, recombine with an equal
 number of holes in the emitter. The loss of total number of holes in the emitter is
made by flow of an equal number of electrons from the emitter to the positive
terminal of the battery. The flow of holes from the emitter to base gives rise to
emitter current IE. In the emitter, IE is due to the flow of holes. But in the external
circuit the current is due to the flow of electrons from the emitter to the positive
terminals of the battery VEE. The holes diffuse through the base. These holes take a
very small time to flow through this region before they reach the depletion region.
During this time a very small number of holes recombine with an equal number of
electrons in the base. Because the base is lightly doped and very thin this number is
very small. The loss of total number of electrons per second is made up by the flow
of an equal number of electrons from the negative terminal of VEE into the base. The
flow of these electrons contribute the base current IB.
The remaining number of holes, which do not undergo recombination process in the
base, reach the collector. These are neutralized by an equal number of electrons
flowing from the negative terminal of the battery VCC into the collector. At the same
time an equal number of electrons flows from the negative terminal of VEE and reach
the positive terminal of VCC. The flow of holes per second from the base to the
collector. In the external circuit, it is due to the flow of electrons from the negative
terminal of the battery Vcc into the collector.
Applying Kirchoff’s current law to the circuit, the emitter current is the sum of
collector current and base current.
i.e IE = IB + IC
This equation is the fundamental relation between the currents in a transistor
circuit.
This equation is true regardless of transistor type or transistor configuration.
The action of NPN transistor (fig b) is similar to that of PNP transistor.
101. What are the types of transistor configurations available?
In a transistor, only three terminals are available, viz., Emitter (E), Base (B) and
Collector (C). Therefore, in a circuit the input/output connections have to be such
that one of these (E, B or C) is common to both the input and the output.
Accordingly, the transistor can be connected in either of the following three
configurations: Common Emitter (CE), Common Base (CB), Common Collector (CC)
The transistor is most widely used in the CE configuration.
102. Explain the input and output characteristics of a transistor in CE configuration
When a transistor is used in CE configuration, the input is between the base and the
emitter and the output is between the collector and the emitter. The variation of the
base current IB with the base-emitter voltage VBE is called the input characteristic.
Similarly, the variation of the collector current IC with the collector-emitter voltage
VCE is called the output characteristic. The output characteristics are controlled by
the input characteristics. This implies that the collector current changes with the
base current. The input and the output characteristics of an n-p-n transistors can be
studied by using the circuit shown in Fig.

To study the input characteristics of the transistor in CE configuration, a curve is

plotted between the base current IB against the base-emitter voltage VBE. The
collector-emitter voltage VCE is kept fixed while studying the dependence of IB on
VBE. The collector-emitter voltage VCE is kept large enough to make the base
collector junction reverse biased. Since VCE = VCB + VBE and for Si transistor VBE is 0.6
to 0.7 V, VCE must be sufficiently larger than 0.7 V. Since the transistor is operated as
an amplifier over large range of VCE, the reverse bias across the base collector
junction is high most of the time. Therefore, the input characteristics may be
obtained for VCE somewhere in the range of 3 V to 20 V. Since the increase in VCE
appears as increase in VCB, its effect on IB is negligible. As a consequence, input
characteristics for various values of VCE will give almost identical curves. Hence, it is
enough to determine only one input characteristics. The input characteristic of a
transistor is as shown in Fig. (a). The output characteristic is obtained by observing
the variation of IC as VCE is varied keeping IB constant. It is obvious that if VBE is
increased by a small amount, both hole current from the emitter region and the
electron current from the base region will increase. As a consequence both IB and IC
will increase proportionately. This shows that when IB increases IC also increases.
The plot of IC versus VCE for different fixed values of IB gives one output
characteristic. So there will be different output characteristics corresponding to
different values of IB as shown in Fig. (b). The linear segments of both the input and
output characteristics can be used to calculate some important ac parameters of
transistors as shown below.

(i) Input resistance (ri): This is defined as the ratio of change in base emitter
voltage (ΔVBE) to the resulting change in base current (ΔIB) at constant collector-
emitter voltage (VCE). This is dynamic (ac resistance) and as can be seen from the
input characteristic, its value varies with the operating current in the transistor:
𝛥𝑉
ri = ( 𝛥𝐼𝐵𝐸 )
𝐵 𝑉𝐶𝐸
The value of ri can be anything from a few hundreds to a few thousand ohms.
Output resistance (ro): This is defined as the ratio of change in collector-emitter
voltage (ΔVCE) to the change in collector current (ΔIC) at a constant base current IB.
𝛥𝑉
ro = ( 𝛥𝐼𝐶𝐸 )
𝐶 𝐼𝐵
The output characteristics show that initially for very small values of VCE, IC
increases almost linearly. This happens because the base-collector junction is not
reverse biased and the transistor is not in active state. In fact, the transistor is in the
saturation state and the current is controlled by the supply voltage VCC (=VCE) in this
part of the characteristic. When VCE is more than that required to reverse bias the
base-collector junction, IC increases very little with VCE. The reciprocal of the slope of
the linear part of the output characteristic gives the values of ro. The output
resistance of the transistor is mainly controlled by the bias of the base collector
junction. The high magnitude of the output resistance (of the order of 100 kΩ) is due
to the reverse-biased state of this diode. This also explains why the resistance at the
initial part of the characteristic, when the transistor is in saturation state, is very
low.
103. Explain the transfer characteristic and hence explain the working of the
transistor as a switch.
Transistor as a switch
We shall try to understand the operation of the transistor as a switch by analysing
the behaviour of the base-biased transistor in CE configuration as shown in Fig. (a).

Applying Kirchhoff’s voltage rule to the input and output sides of this circuit, we
get
VBB = IBRB + VBE ----------------------- (1)
And
VCE = VCC – ICRc -------------------- (2)
We shall treat VBB as the dc input voltage Vi and VCE as the dc output voltage
VO. So, we have
Vi = IBRB + VBE and V0 = VCC - ICRC
Let us see how Vo changes as Vi increases from zero onwards. In the case of Si
transistor, as long as input Vi is less than 0.6 V, the transistor will be in cut off state
and current IC will be zero. Hence Vo = VCC When Vi becomes greater than 0.6 V the
transistor is in active state with some current IC in the output path and the output Vo
decrease as the term ICRC increases. With increase of Vi , IC increases almost linearly
and so Vo decreases linearly till its value becomes less than about 1.0 V.
Beyond this, the change becomes non linear and transistor goes into saturation
state. With further increase in Vi the output voltage is found to decrease further
towards zero though it may never become zero. If we plot the Vo vs Vi curve, [also
called the transfer characteristics of the base-biased transistor (Fig. (b)], we see that
between cut off state and active state and also between active state and saturation
state there are regions of non-linearity showing that the transition from cutoff state
to active state and from active state to saturation state are not sharply defined. Let
us see now how the transistor is operated as a switch. As long as Vi is low and
unable to forward-bias the transistor, Vo is high (at VCC ). If Vi is high enough to drive
the transistor into saturation, then Vo is low, very near to zero. When the transistor
is not conducting it is said to be switched off and when it is driven into saturation it
is said to be switched on. This shows that if we define low and high states as below
and above certain voltage levels corresponding to cutoff and saturation of the
transistor, then we can say that a low input switches the transistor off and a high
input switches it on. Alternatively, we can say that a low input to the transistor gives
a high output and a high input gives a low output. The switching circuits are
designed in such a way that the transistor does not remain in active state.
104. Explain the working of the transistor as an amplifier in CE configuration.
To operate the transistor as an amplifier it is necessary to fix its operating point
somewhere in the middle of its active region. If we fix the value of VBB corresponding
to a point in the middle of the linear part of the transfer curve then the dc 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 vs is superposed on the dc base bias by
connecting the source of that signal in series with the VBB supply, then the base
current will have sinusoidal variations superimposed on the value of IB. As a
consequence the collector current also will have sinusoidal variations superimposed
on the value of IC, producing in turn corresponding change in the value of VO. We can
measure the ac variations across the input and output terminals by blocking the dc
voltages by large capacitors.
In the description of the amplifier given above we have not considered any ac signal.
In general, amplifiers are used to amplify alternating signals. Now let us
superimpose an ac input signal vi (to be amplified) on the bias VBB (dc) as shown in
Fig. The output is taken between the collector and the ground.

The working of an amplifier can be easily understood, if we first assume that vi = 0.

Then applying Kirchhoff’s law to the output loop, we get
Vcc = VCE + IcRL
Likewise, the input loop gives
VBB = VBE + IB RB
When vi is not zero, we get
VBE + vi = VBE + IB RB + ΔIB (RB + ri )
The change in VBE can be related to the input resistance ri and the change in IB.
Hence
vi = ΔIB (RB + ri )
= r ΔIB
The change in IB causes a change in Ic. We define a parameter βac, which is similar
to the βdc
ΔI i
βac = ΔIC = i c
B b
Which is also known as the ac current gain Ai. Usually βac is close to βdc in the linear
region of the output characteristics.
The change in Ic due to a change in IB causes a change in VCE and the voltage drop
across the resistor RL because VCC is fixed. These changes can be given as
ΔVCC = ΔVCE + RL ΔIC = 0
or ΔVCE = –RL ΔIC
The change in VCE is the output voltage v0.
v0 = ΔVCE = –βac RL ΔIB
The voltage gain of the amplifier is
vo ΔVCE
Av = =
vi rΔIB
βac RL
=−
r
The negative sign represents that output voltage is opposite with phase with the
input voltage. From the discussion of the transistor characteristics you have seen
that there is a current gain βac in the CE configuration. Here we have also seen the
voltage gain Av. Therefore the power gain Ap can be expressed as the product of the
current gain and voltage gain. Mathematically Ap = βac × Av
Since βac and Av are greater than 1, we get ac power gain. However it should be
realised that transistor is not a power generating device. The energy for the higher
ac power at the output is supplied by the battery.
105. Explain the working of transistor as an oscillator.

In an amplifier, we have seen that a sinusoidal input is given which appears as an

amplified signal in the output. This means that an external input is necessary to
sustain ac signal in the output for an amplifier. In an oscillator, we get ac output
without any external input signal. In other words, the output in an oscillator is self-
sustained. To attain this, an amplifier is taken. A portion of the output power is
returned back (feedback) to the input in phase with the starting power (this process
is termed positive feedback) as shown in Fig. (a). The feedback can be achieved by
inductive coupling (through mutual inductance) or LC or RC networks. Different
types of oscillators essentially use different methods of coupling the output to the
input (feedback network), apart from the resonant circuit for obtaining oscillation at
a particular frequency. For understanding the oscillator action, we consider the
circuit shown in Fig. (b) in which the feedback is accomplished by inductive coupling
from one coil winding (T1) to another coil winding (T2). Note that the coils T2 and T1
are wound on the same core and hence are inductively coupled through their
mutual inductance. As in an amplifier, the base-emitter junction is forward biased
while the base-collector junction is reverse biased. Detailed biasing circuits actually
used have been omitted for simplicity. Suppose switch S1 is put on to apply proper
bias for the first time. Obviously, a surge of collector current flows in the transistor.
This current flows through the coil T2 where terminals are numbered 3 and 4 [Fig.
(b)]. This current does not reach full amplitude instantaneously but increases from
X to Y, as shown in Fig. [(c)(i)]. The inductive coupling between coil T2 and coil T1
now causes a current to flow in the emitter circuit (note that this actually is the
‘feedback’ from input to output). As a result of this positive feedback, this current (in
T1; emitter current) also increases from X’ to Y’ [Fig. (c)(ii)]. The current in T2
(collector current) connected in the collector circuit acquires the value Y when the
transistor becomes saturated. This means that maximum collector current is flowing
and can increase no further. Since there is no further change in collector current, the
magnetic field around T2 ceases to grow. As soon as the field becomes static, there
will be no further feedback from T2 to T1. Without continued feedback, the emitter
current begins to fall. Consequently, collector current decreases from Y towards Z
[Fig. (c)(i)]. However, a decrease of collector current causes the magnetic field to
decay around the coil T2. Thus, T1 is now seeing a decaying field in T2 (opposite from
what it saw when the field was growing at the initial start operation). This causes a
further decrease in the emitter current till it reaches Z’ when the transistor is cut-
off. This means that both IE and IC cease to flow. Therefore, the transistor has
reverted back to its original state (when the power was first switched on). The
whole process now repeats itself. That is, the transistor is driven to saturation, then
to cut-off, and then back to saturation. The time for change from saturation to cut-off
and back is determined by the constants of the tank circuit or tuned circuit
(inductance L of coil T2 and C connected in parallel to it). The resonance frequency
(ν) of this tuned circuit determines the frequency at which the oscillator will
oscillate.
1
ν = 2π√LC
106. What is NOT gate?
This is the most basic gate, with one input and one output. It produces a ‘1’ output if
the input is ‘0’ and vice-versa. That is, it produces an inverted version of the input at
its output. This is why it is also known as an inverter. The commonly used symbol
together with the truth table for this gate is given in Fig.
Input Output
A Y
0 1
1 0
(b)
107. How an OR gate works and mention its truth table.
An OR gate has two or more inputs with one output. The logic symbol and truth
table are shown in Fig. The output Y is 1 when either input A or input B or both are
1s, that is, if any of the input is high, the output is high.

108. Explain AND gate with truth table.

An AND gate has two or more inputs and one output. The output Y of AND gate is 1
only when input A and input B are both 1. The logic symbol and truth table for this
gate are given in Fig.

109. Explain NAND gate with truth table.

This is an AND gate followed by a NOT gate. If inputs A and B are both ‘1’, the output
Y is not ‘1’. The gate gets its name from this NOT AND behaviour. Figure shows the
symbol and truth table of NAND gate. NAND gates are also called Universal Gates
since by using these gates you can realise other basic gates like OR, AND and NOT

110. Explain NOR gate with truth table.

It has two or more inputs and one output. A NOT- operation applied after OR gate
gives a NOT-OR gate (or simply NOR gate). Its output Y is ‘1’ only when both inputs
A and B are ‘0’, i.e., neither one input nor the other is ‘1’. The symbol and truth table
for NOR gate is given in Fig.
111. Write short note integrated circuits.
The conventional method of making circuits is to choose components like diodes,
transistor, R, L, C etc., and connect them by soldering wires in the desired manner.
Inspite of the miniaturisation introduced by the discovery of transistors, such
circuits were still bulky. Apart from this, such circuits were less reliable and less
shock proof. The concept of fabricating an entire circuit (consisting of many passive
components like R and C and active devices like diode and transistor) on a small
single block (or chip) of a semiconductor has revolutionised the electronics
technology. Such a circuit is known as Integrated Circuit (IC). The most widely used
technology is the Monolithic Integrated Circuit. The word monolithic is a
combination of two greek words, monos means 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 as 1mm × 1mm or it could even be smaller.
Depending on nature of input signals, IC’s can be grouped in two categories: (a)
linear or analogue IC’s and (b) digital IC’s. The linear IC’s process analogue signals
which change smoothly and continuously over a range of values between a
maximum and a minimum. The output is more or less directly proportional to the
input, i.e., it varies linearly with the input. One of the most useful linear IC’s is the
operational amplifier. The digital IC’s process signals that have only two values.
They contain circuits such as logic gates. Depending upon the level of integration
(i.e., the number of circuit components or logic gates), the ICs are termed as Small
Scale Integration, SSI (logic gates < 10); Medium Scale Integration, MSI (logic gates <
100); Large Scale Integration, LSI (Logic gates < 1000); and Very Large Scale
Integration, VLSI (logic gates > 1000). The technology of fabrication is very involved
but large scale industrial production has made them very inexpensive.
112. Determine the output wave form for the circuit given below, if the input
waveforms are as indicated by A and B.
A
B

Output Y
113. Draw and explain the output wave forms across the load resistor R, if the input
wave form is as shown in the given figure.
The output wave form is given by
The diode starts conducting exactly when it is in
forward biased and does not conduct at all when
it is reversed biases. In case we consider diode as
ideal one total potential applied will appear
across resistor.
114. The following figure shows the input wave forms (A, B) and the output wave
form (Y) of a gate. Identify the gate and write its truth table.

ANS:
115. Justify the output wave form (Y) of the OR gate for inputs (A) and (B) as given
in the following figure.

Truth Table for OR gat is From the figure we get

A B A+B=Y A B Y
0 0 0 0 0 0
0 1 1 1 0 1
1 0 1 1 1 1
1 1 1 0 1 1
0 0 0
1 0 1
0 1 1
Which in accordance with the truth table.
116. Write the function of base region of a transistor. Why this region is made thin
and slightly doped?
Base region in a transistor forms junctions with the emitter region and the collector
region. Base region is made thin and lightly doped so that only a small fraction of
majority carriers that enter into it from the emitter region are able to recombine
with charge carriers of opposite polarity and constitute base current which is a
small fraction of emitter current.
117. What is an ideal diode? Draw the output wave form across the load resistor R,
if the input wave form is as shown in the figure.

An ideal diode is one which offers Zero resistance when forward biased and Infinite
resistance when reverse biased.
Ideal diode is a device that conducts at 0 V i.e. it starts conducting exactly when
forward biased and do not conduct at all when reverse biased. The output wave
form across R is as:

118. Explain how doping a pure semiconductor with pentavalent impurity

increases the conductivity of the semiconductor.
On doping with pentavalent impurity, the number of electrons increases, thereby
resulting in increase in number of charge carriers and hence conductivity increases.
119. Why is n-p-n preferred over p-n-p transistor?
In n – p – n current is mainly carried by electrons whereas in p – n – p, current is
mainly carried by holes. The mobility of electrons is greater than that of holes and
hence, the n – p – n transistor will have faster response and are preferred.
120. Explain transistor action. OR Why is transistor called transistor?
Transistor – action. Transistor simply sends same current from one resistance path
to another, i.e., it only transfer resistance and it’s called transistor.
i.e., TRANSfer + resISTOR = TRANSISTOR
121. When is a transistor said to be in a state of saturation? Explain.
The transistor is said to be in saturation when current cannot exceed its maximum
value even by varying input voltage. This happens when both emitter collector are
forward biased.
122. When is a transistor said to be in a state of cut-off? Explain.
The transistor is in state of cut – off when both the emitter and collector are reverse
biased. In this case output current is always zero.
123. The diagram shows a piece of pure semiconductor, S in series with a variable
resistor R, and a source of constant voltage V. Would you increase or decrease
the value of R to keep the reading of ammeter (A) constant, when
semiconductor S is heated? Give reason.

(a) Value of R should be increased.

(b) With the increase in temperature of semiconductor. Its resistivity decreases.
As a result, circuit resistance decreases and current tends to increase.
124. Name the type of biasing which results in very high resistance of a p-n junction
diode. In the given circuit, a voltmeter ‘V’ is connected across a bulb ‘B’. What
changes would occur in the bulb ‘B’ and voltmeter ‘V’, if the resistor ‘R’ is
increased in value? Give reason for your answer.
(i) p-n junction diode offers very high resistance in reverse bias. (ii) When R is
increased in value, input circuit, i.e., emitter – base circuit get less forward biased.
Reduction in forward biasing means decrease in the collector current. So bulb will
glow dimmer Voltmeter reading will also decrease.
125. Distinguish between intrinsic and extrinsic semiconductors. Explain the
formation of potential barrier and depletion region in a p-n junction.

(i) Formation of depletion region. When a p – type semiconductor is joined with an

n – type semiconductor diffusion of charge carriers takes place across the junction.
The n – type electrons are majority carriers so they diffuse towards p – type where
they are in minority. In the same way, holes diffuse form p – type to n – type. In both
the cases, when an electron meets a hole, the two cancel the effect of each other As a
result, a thin layer is formed at the junction which is devoid of charge carriers. This
layer is called depletion region. It is of the order of 10–6 m.
(ii) Formation of potential barrier. As p – type acquires electrons from n – type; it
becomes slightly negative, while n – type, which loses electrons, becomes slightly
positive. This way a potential difference is developed across the junction. As it
 opposes further diffusion of charge carriers across the junction it is called potential
barrier. It is about 0.3 volt for germanium junction diode and about 0.7 volt for
silicon junction diode generally.
126. A diode is connected to 220V a.c in series with a capacitor as shown. What is
the voltage V across the capacitor?
Diode conducts only for half of the cycle. During half of the cycle capacitor is charged
to the peak value of supply voltage. Thus
V = Erms × √2 = 220 × √2 = 311.1 volt.
127. Can the potential barrier of a p-n junction be measured by connected a
sensitive voltmeter across the junctions?
There are no free charges in the depletion region, so in the absence of forward
biasing it offers infinite resistance. Thus, voltmeter cannot measure the potential
barrier.
128. Find out the reading of the ammeters A1 and A2 shown in the figure. Neglect
the resistance of the meters.

Current will not flow through the ammeter A1 is a diode is connected in reverse bias
its path. Current through the ammeter A2 is
I = 2/10 = 0.2 A
As diode is connected in forward bias with it.
129. Why in a transistor, the forward bias voltage is always smaller than the
reverse bias voltage?
On increasing forward bias voltage, the majority charge will move from the emitter
to the collector through the base with higher velocities. This would produce
excessive heat and transistor might get damaged.
130. Why can’t transistor be used as a rectifier?
As base is thin and lightly doped, the transistor is not fit for the purpose of
rectification.
131. Analyse the following circuit diagram and write the answer to the questions.

i) Choose from the option below the diodes which are in forward bias when
the switch is ON. (a) D2, D4 (b) D1, D3 (c) D1, D3, D4 (d) D2, D3, D4
ii) What criterion have you used to identify the diodes?
iii) Write any one use of a diode.
(i) (d) D2, D3, D4
(ii) when the p – end of a diode is connected to the positive pole and the n – end to
the negative pole of a cell, the diode is to said to be forward biased.
(iii) Diode is used as rectifier and detector.
132. Why are semiconductors doped?
The conductivity of intrinsic semiconductors is so small that it is practically of no
use. The semiconductors are doped so as to increase their conductivity.
133. Distinguish between energy levels and energy bands.
An isolated atom has various energy levels. In a crystal, atoms are so closely packed
that due to interaction between them, splitting of energy levels takes place. The
energy levels, then, form energy bands. The energy band formed by a series of
 energy levels containing the valence electrons is called valence band and the lowest
unfilled energy band formed just above the valence band is called conduction band.
134. An n-type semiconductor has a large number of electrons but still it is
electrically neutral. Explain.
An n-type semiconductor is obtained by doping pure Si or Ge-crystal with a
pentavalent impurity. As the impurity atoms enter into the configuration of the Si-
crystal, its four electrons take part in covalent bonding, while the fifth electron is left
free. Since each atom of the semiconductor as a whole is electrically neutral; the n-
type Ge-crystal, though having a large number of free electrons, is electrically
neutral.
135. Explain the effect of temperature variation on the resistivity of pure
semiconductors.
The variation in the resistivity of a pure semiconductor is mainly due to the change
in carrier concentration. The fraction of the number of electrons raised from the
valence band to the conduction band is given by f ∝ 𝑒 −𝐸𝑔 /𝑘𝑇
It follows that as temperature (T) increases; the value of fraction f also increases i.e.
conductivity of the semiconductor increases. In other words, as temperature of the
semiconductor increases, its resistivity decreases.
136. Can we measure the potential barrier of a p-n junction by putting a sensitive
voltmeter across its terminals?
In the depletion layer, there are no free electrons or holes and in the absence of
forward bias, it offers infinite resistance. Therefore, potential barrier in the p-n
junction cannot be measured with a voltmeter.
137. What is zener breakdown?
When the reverse voltage exceeds the zener voltage, the valence electrons break
free under the influence of the electric field due to the applied voltage and the diode
conducts a large current in the reverse direction. It is called as Zener-breakdown.
138. Can two p-n junction diodes back to back work as a p-n-p transistor?
In a transistor, the base is thin and lightly doped as compared to the collector and
emitter. In case, two diodes are placed back to back, the middle section (base) will
be too thick and will have the same degree of doping as that of collector or emitter.
Hence, two p-n junction diodes placed back to back cannot work as a p-n-p
transistor.
139. State two advantages of semiconductor devices.
a) The semiconductor devices are temperature sensitive. Even a small overheating
may cause damage to them.
b) The semiconductor devices cannot handle as much power as ordinary vacuum
tubes can do.
140. An n-type semiconductor is the one which has excess of free electrons and p-
type semiconductor is deficient in these. When a p-n junction is formed, the
electrons should flow from n to p region. But all the electrons do not do so.
Explain, why?
Though the semiconductors are called n-type or p-type, yet electrically they are
neutral. When p-n junction is formed, electrons flow from n to p region. But as they
do so, p-type becomes increasingly negative, while n-type becomes positive. This
increasing potential difference across the p-n junction set up due to the electron
flow attains a value, which then prevents the further flow of electrons. That is why
all the electrons do not flow from n to p-type semiconductor.
141. Why the base region of a transistor is usually made thin?
Or
Write the function of base region of a transistor. Why this region is made thin
and slightly doped?
Due to the fact that base is doped lightly; the number density of majority carriers
(electrons in p-n-p and holes in n-p-n transistor) is low. When emitter is forward
biased, electron-hole combination takes place in the base region. Since it is thin and
 lightly doped, only a small amount (about 5%) of electron-hole recombination will
take place.
142. In a transistor, forward bias is always small as compared to the reverse bias.
Why?
In a transistor, emitter is forward biased and collector is reverse biased. In case,
forward bias is made large, majority carriers in the emitter drift towards the
collector through the base region with large velocity. Since their number is also very
large, heat produced is so large that it breaks up the covalent bonds and the
transistor gets spoiled. However, if the reverse bias applied to the collector is
comparatively larger, no such effect is produced as the charge carriers drift only in
collector region.
143. Explain the following:
(i) A transistor is a temperature-sensitive device.
(ii) A transistorised circuit starts working immediately after the circuit is
switched on, whereas a vacuum tube circuit takes some time to start working.
(i) A transistor contains free electrons and holes as charge carriers. When properly
biased (emitter forward biased and collector reverse biased), they carry current
through the transistor in external circuit. In case, temperature increases, covalent
bonds break up, resulting in an additional large number of free electrons and holes.
If transistor (when hot) is operated now, the current will be very strong, which may
result in excessive heat and ultimately in a complete breakdown of the
semiconductor device.
(ii) A vacuum tube is based upon thermionic emission. A vacuum tube circuit will
require some time, so that the cathode gets heated and starts emitting electrons. On
the other hand, in a transistor, as there is no filament and hence no heating is
required, the transistorized circuit starts working immediately, when the circuit is
switched on.
144. What is a p-n-p transistor? How does it differ from a n-p-n transistor?
In a p-n-p transistor, a thin p-section is sandwiched between two n-sections. The
holes are charge carriers in a p-n-p transistor.
In a n-p-n transistor, a thin p-section is sandwiched between two n-sections. In a n-
p-n transistor, electrons are the charge carriers.
145. Why a transistor cannot be used as a rectifier?
To use a transistor as a rectifier, either its emitter-base portion or the collector-base
portion has to be used. Since base is thin and lightly doped, either of the two
portions will not work as a p-n junction diode. Hence, a transistor cannot be used as
a rectifier.
146. A square wave (-1 V to +1 V) is applied to p-n junction diode as shown below.
Draw the output waveform..

When input is -1 V, the diode gets reverse biased and no output is obtained. On the
other hand, when input is +1 V, the diode gets forward biased and output obtained
is +1 V. Therefore, corresponding to the given input cycle, the output is of the shape
as shown.

147. What do you understand by truth table of a logic gate?

The truth table of a logic gate shows all possible inputs combination and the
corresponding output combinations for a logic gate. The functioning of a logic gate is
based on its Boolean expression.
148. NAND and NOR gates may be called as digital building blocks. Why?
The repeated use of NAND and NOR gates can produce all the three basic gates i.e.
OR, AND and NOT gates. For this reason, these are called digital building blocks.
149. What is the logic gate?
A logic gate is a digital circuit, which works according to some logical relationship
between input and output voltages.
TEXT BOOK EXERCISES
1. In an n-type silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
An n-type semiconductor is obtained by doping a semiconductor with a pentavalent
impurity. The impurity so added produces free electrons. Therefore, in an n-type
semiconductor, the electrons are majority carriers or holes are minority carriers
and pentavalent atoms are the dopants. Hence, the option (c) is correct.
2. Which of the statements given in Q.no1 is true for p-type semiconductos.
A p-type semiconductor is obtained by doping a semiconductor with a trivalent
impurity. The impurity so added produces holes. Therefore, in a p-type
semiconductor holes are majority carriers and trivalent atoms are the dopants.
Hence, the option (d) is correct.
3. Carbon, silicon and germanium have four valence electrons each. These are
characterised by valence and conduction bands separated by energy band gap
respectively equal to (Eg)C, (Eg)Si and (Eg)Ge. Which of the following statements
is true?
(a) (Eg)Si < (Eg)Ge < (Eg)C (b) (Eg)C < (Eg)Ge > (Eg)Si
(c) (Eg)C > (Eg)Si > (Eg)Ge (d) (Eg)C = (Eg)Si = (Eg)Ge
Carbon is insulator and hence the value of energy band gap is maximum for it. Of the
other two semiconductors, the value of energy band gap for germanium is lesser.
Hence, the option (c) is correct.
4. In an unbiased p-n junction, holes diffuse from the p-region to n-region
because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All the above.
In an unbiased p-n junction, hole concentration in p-region is more as compared to
n-region. Hence, the option (c) is correct.
5. When a forward bias is applied to a p-n junction, it
(a) Raises the potential barrier.
(b) Reduces the majority carrier current to zero.
(c) Lowers the potential barrier.
(d) None of the above.
When a p-n junction is forward biased, the applied voltage opposes the barrier
voltage across the junction. As a result, the potential barrier gets lowered. Hence,
the option (c) is correct.
6. For transistor action, which of the following statements are correct:
(a) Base, emitter and collector regions should have similar size and doping
concentrations.
(b) The base region must be very thin and lightly doped.
(c) The emitter junction is forward biased and collector junction is reverse
biased.
(d) Both the emitter junction as well as the collector junction is forward
biased.
 When emitter is forward biased, electron-hole recombination takes place in the base
region. So that only a small amount (about 5 %) of electron-hole recombination may
take place, the base is made thin and lightly doped. Hence the option (b) is correct.
In a transistor, emitter is always forward biased and collector is reverse biased.
Hence, the option (c) is correct.
7. For a transistor amplifier, the voltage gain
(a) remains constant for all frequencies.
(b) is high at high and low frequencies and constant in the middle frequency
range.
(c) is low at high and low frequencies and constant at mid frequencies.
(d) None of the above.
In a transistor, the voltage gain is high at high and low frequencies and constant at
mid frequencies. Hence, the option (c) is correct.
8. In half-wave rectification, what is the output frequency if the input frequency
is 50 Hz. What is the output frequency of a full-wave rectifier for the same
input frequency?
In half-wave rectifier, the frequency of the output is same as that of the input i.e.
50 Hz.
In full-wave rectifier, the frequency of the output is twice as that of the input i.e.
100 Hz.
9. For a CE-transistor amplifier, the audio signal voltage across the collected
resistance of 2 kΩ is 2 V. Suppose the current amplification factor of the
transistor is 100, find the input signal voltage and base current, if the base
resistance is 1 kΩ.
Here, collector resistance, Rc = 2 k = 2 x 103 ;
Voltage drop across collector resistance = 2 V
voltage across the collector resistance 2
 Ic = = = 10−3 𝐴
collector resistance 2×103
Now, current gain of common emitter amplifier,
I
β = Ic
b
Ic 10−3
or Ib = = = 10−5 A = 𝟏𝟎 𝐀
β 100
Also, input signal voltage,
Vin = Ib x base resistance
Here, base resistance, Rb = 1 k = 103 
 Vin = 10-5 x 103 = 10-2 V = 0.01 V
10. Two amplifiers are connected one after the other in series (cascaded). The
first amplifier has a voltage gain of 10 and the second has a voltage gain of 20.
If the input signal is 0.01 volt, calculate the output ac signal.
When the amplifiers are connected n series, the net voltage gain (Av) is equal to the
product of the gains of the individual amplifiers i.e.
Av = Av’ x Av’’ ….(i)
𝑉𝑜𝑢𝑡
Also, 𝐴𝑣 = 𝑉 ….(ii)
𝑖𝑛
From the equations (i) and (ii), we have
Vout
= A′v × A′′v or Vout = Vin x Av’ x Av’’ = 0.01 x 10 x 20 = 2 V
V in
11. A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV.
Can it detect a wavelength of 6000 nm?
Here, λ = 600 nm = 600 x 10-9 m
Therefore, energy associated with the wavelength,
ℎ𝑐 6.62×10−34 ×3×108
𝐸= = 𝐽
𝜆 600×10−9
6.62×10−34 ×3×108
= 600×10−9×1.6×10−19 = 𝟐. 𝟎𝟕 𝒆𝑽
 As the energy associated with the wavelength is less than the band gap energy of the
semiconductor, it cannot detect this wavelength.
12. 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 the number of electrons and holes. Given that ni = 1.5 × 1016 m–3. Is
the material n-type or p-type?
Arsenic is donar, whereas indium is acceptor impurity.
Here, number of donar atoms added,
ND = 5 x 1022 m-3
And number of acceptor atoms added,
NA = 5 x 1020 m-3
Therefore, number of free electrons created,
ne = ND = 5 x 1022
and number of holes created,
nh = NA = 5 x 1020
Now, ne > nh,
Therefore, net number of free electrons created,
ne' = ne – nh = 5 x 1022 – 5 x 1020
= 4.95 x 1022 m-3
Also, net number of holes created,
2
𝑛2 (1.5×1016 )
𝑛ℎ′ = 𝑛𝑖 ′ = = 𝟒. 𝟓𝟓 × 𝟏𝟎𝟗 𝒎−𝟑
𝑒 4.95×1022
As ne’ > nh’, the resulting material is n-type semiconductor.
13. In an intrinsic semiconductor the energy gap Eg is 1.2eV. Its hole mobility is
much smaller than electron mobility and independent of temperature. What is
the ratio between conductivity at 600K and that at 300K? Assume that the
temperature dependence of intrinsic carrier concentration ni is given by
𝑬𝒈
ni = no exp(− 𝟐𝒌 𝑻) where n0 is a constant.
𝑩
Here, Eg = 1.2 eV; k = 8.62 x 10-5 eV K-1
𝑛𝑖 (𝑎𝑡 600 𝐾) 𝑛0 𝑒 −𝐸𝑔/2𝑘×600
=
𝑛𝑖 (𝑎𝑡 300 𝐾) 𝑛0 𝑒 −𝐸𝑔/2𝑘×300
𝐸𝑔
= 𝑒 2𝑘 ×600
−5
= 𝑒 1.2/2×8.62×10 ×600 = 𝑒 11.6
= 1,09,098 ≈ 105
𝒆𝑽
14. In a p-n junction diode, the current I can be expressed as I = Io exp(𝟐𝒌 𝑻 − 𝟏)
𝑩
where I0 is called the reverse saturation current, V is the voltage across the
diode and is positive for forward bias and negative for reverse bias, and I is
the current through the diode, kB is the Boltzmann constant (8.6×10–5 eV/K)
and T is the absolute temperature. If for a given diode I0 = 5 × 10–12 A and T =
300 K, then
(a) What will be the forward current at a forward voltage of 0.6 V?
(b) What will be the increase in the current if the voltage across the diode is
increased to 0.7 V?
(c) What is the dynamic resistance?
(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?
Here, I0 = 5 x 10-12 A; k = 8.6 x 10-5 eV K-1 = 8.6 x 10-5 x 1.6 x 10-19 J K-1;
(a) Now, I = I0(eeV/2kT – 1)
For V = 0.6 volt, we have
1.6×10−19 ×0.6
𝐼 = 5 × 10 −12
(e 2×8.6×10−5 ×1.6×10−19 ×300 − 1)
= 5 x 10-12 [e23.52 – 1]
= 5 x 10-12 [1.256 x 1010 – 1] = 0.0628 A
 (b) For V = 0.7 volt, we have
1.6×10−19 ×0.6
−12
I = 5 × 10 (e 2×8.6×10−5 ×1.6×10−19 ×300 − 1)
= 5 x 10-12 [e27.132 – 1]
= 5 x 10-12 [6.054 x 1011 – 1] = 3.0271 A
 I = 3.271 – 0.0628 = 2.9643 A
(c) Now, I = 2.9643 A; V = 0.7 – 0.6 = 0.1 V
Therefore, dynamic resistance
𝑉 0.1
= 𝐼 = 2.9643 = 𝟎. 𝟎𝟑𝟑𝟕 
(d) For change in voltage from 1 to 2 V, the current will remain equal to I0 = 5 x 10-12
A. It shows that the diode possesses practically infinite resistance in reverse
biasing.
15. You are given the two circuits as shown in Fig. Show that circuit (a) acts as OR
gate while the circuit (b) acts as AND gate.

Here, the output of NOR gate is connected to NOT gate. Let y’ be the output of NOR
gate and the final output of the combination of two gates be y. The output of a NOR
gate is 1 only when both the inputs are zero, while in NOT gate, the input gets
inverted. Using these facts, the truth table for the given arrangement can be written
as below:
A B y' y A B y
0 0 1 0 0 0 0
1 0 0 1 = 1 0 1
0 1 0 1 0 1 1
1 1 0 1 1 1 1
It is the truth table of OR gate. Therefore, the given circuit acts as OR gate.
Here, the outputs of two NOT gates are connected to NOR gate. Let y1 and y2 be the
ouputs of the two NOT gates and the final output of the combination of three gates
be y. In a NOT gate, the input gets inverted, while the output of a NOR gate is 1 only
when both the inputs are zero. Using these facts, the truth table for the given
arrangement can be written as below:
A B y1 y2 y A B y
0 0 1 1 0 0 0 0
1 0 0 1 0 = 1 0 0
0 1 1 0 0 0 1 0
1 1 0 0 1 1 1 1
It is the truth table of AND gate. Therefore, the given circuit acts as AND gate.
16. Write the truth table for a NAND gate connected as given in Fig.

Hence identify the exact logic operation carried out by this circuit.
The truth table of a NAND gate is as given below:
A B y
0 0 1
1 0 1
0 1 1
1 1 0
 When the two inputs of a NAND gates are joined as shown in Figure above i.e. when
A = B = 0 or A = B = 1, the above truth table of the NAND gate reduces to the one as
given below:
A B y
0 0 1
1 1 0
It is the truth table of a NOT gate. Therefore, the given circuit carries out the logic
operation of a NOT gate.
17. You are given two circuits as shown in Fig., which consist of NAND gates.
Identify the logic operation carried out by the two circuits.

Here, the output of NAND gate is connected to NOT gate (obtained from NAND gate).
Let y’ be the output of NAND gate, and the final output of the combination of two
gates be y. The output of a NAND gate is 0 only when both the inputs are zero, while
in NOT gate, the input gets inverted. Using these facts, the truth table for the given
arrangement can be written as below:
A B y' y A B y
0 0 1 0 0 0 0
1 0 1 0 = 1 0 0
0 1 1 0 0 1 0
1 1 0 1 1 1 1
It is the truth table of AND gate. Therefore, the given circuit acts as AND gate.+
Here, the outputs of two NOT gates (obtained from NAND gates) are connected to
NAND gate. Let y1 and y2 be the outputs of the two NOT gates and the final output of
the combination of three gates be y. In a NOT gate, the input gets inverted, while the
output of a NAND gate is 0 only when both the inputs are zero. Using these facts, the
truth table for the given arrangement can be written as below:

A B y1 y2 y A B Y
0 0 1 1 0 0 0 0
1 0 0 1 1 = 1 0 1
0 1 1 0 1 0 1 1
1 1 0 0 1 1 1 1
It is the truth table of OR gate. Therefore, the given circuit acts as OR gate.
18. Write the truth table for circuit given in Fig.below consisting of NOR gates and
identify the logic operation (OR, AND, NOT) which this circuit is performing.

(Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence
Y=1. Similarly work out the values of Y for other combinations of A and B.
Compare with the truth table of OR, AND, NOT gates and find the correct one.)
Here, the output of NOR gate is connected to another NOR gate, whose inputs are
short-circuited. Let y’ be the output of NOR gate and the final output of the
combination of two gates be y. The output of a NOR gate is 1 only when both the
inputs are zero. Using this fact, the truth table for the given arrangement can be
written as below:
 A B y' y A B y
0 0 1 0 0 0 0
1 0 0 1 = 1 0 1
0 1 0 1 0 1 1
1 1 0 1 1 1 1
It is the truth table of OR gate. Therefore, the given circuit acts as OR gate.
19. Write the truth table for the circuits given in Fig. 14.48 consisting of NOR gates
only. Identify the logic operations (OR, AND, NOT) performed by the two
circuits.

The truth table of a NOR gate is as given below:

A B y
0 0 1
1 0 0
0 1 0
1 1 0
When the two inputs of a NOR gates are joined as shown in Fig. 9.05 (a) i.e. when A =
B = 0 or A = B = 1, the above truth table of the NOR gate reduces to the one as given
below:
A B y
0 0 1
1 1 0
It is the truth table of a NOT gate. Therefore, the given circuit carries out the logic
operation of a NOT gate.
Refer to figure (b): Here, the outputs of two NOT gates (obtained from NOR gates)
are connected to NOR gate. In a NOT gate, the input gets inverted. Let 𝐴̅ and 𝐵̅ be the
outputs of the two NOT gates. Now, the output of a NOR gate is 1 only when both the
inputs are zero. Using this fact, the truth table for the given arrangement can be
written as below:
A B ̅
A ̅
B y A B y
0 0 1 1 0 0 0 0
1 0 0 1 0 = 1 0 0
0 1 1 0 0 0 1 0
1 1 0 0 1 1 1 1
It is the truth table of AND gate. Therefore, the given circuit acts as AND gate.

ALL THE BEST