# Chapter 24 Solids and Semiconductor Devices free study material by TEACHING CARE online tuition and coaching classes

Chapter 24 Solids and Semiconductor Devices free study material by TEACHING CARE online tuition and coaching classes

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# Solids.

It is a state of matter which has a definite shape and a definite volume. The characteristic properties of the solid depends upon the nature of forces acting between their constituent particles (i.e. ions, atoms or molecules). Solids are divided into two categories.

Crystalline solids                                          Amorphous or glassy solids

• These solids have definite external geometrical These solids have no definite external geometrical form.

• Ions, atoms or molecules of these solid are arranged in a definite fashion in all it’s three

Ions, atoms or molecules of these solids are not arranged in a definite fashion.

• Examples : Quartz, calacite, mica, diamond Example : Rubber, plastic, paraffin wax, cement etc.
• They have well defined facets or They do not possess definite facets or faces.
• They are ordered at short range as well as at long These may be short range order, but there is no long range

order.

• They are anisotropic, e. the physical properties like elastic modulii, thermal conductivity, electrical conductivity, refractive index have different values in different direction.

They are isotropic i.e. physical properties are similar in all direction.

• They have sharp melting point. They may not have a sharp melting
• Bond strengths are identical throughout the Bond strengths vary.
• These are considered as true These are considered as pseudo-solids or super cooled

liquids.

• An important property of crystals is their Amorphous solids do not have any symmetry.

# Terms Related With Crystal Structure.

## (1)  Crystal lattice

It is a geometrical arrangement of points in space where if atoms or molecules of a solid are placed, we obtain an actual crystal structure of the solid.

## (2)  Basis

The atoms or molecules attached with every lattice point in a crystal structure is called the basis of crystal structure. Thus,

## (3)  Unit cell

=

Basis                      Crystal

Is defined as that volume of the solid from which the entire crystal structure can be constructed by the translational repetition in three dimensions. The length of three sides of a unit cell (3D) are called primitives or lattice constant they are denoted by a, b, c

## (4)  Primitive cell

A primitive cell is a minimum volume unit cell or the simple unit cell with particles only at the corners is a primitive unit cell and other types of unit cells are called non-primitive unit cells. There is only one lattice point per primitive cell.

## (5)  Crystallographic axis

The lines drawn parallel to the lines of intersection of the faces of the unit cell are called crystallographic axis.

Note : @The location of each atom or molecule in a crystal lattice may be marked as a point which is called lattice point.

All the crystals on the basis of the shape of their unit cells, have been divided into seven crystal systems as shown in the following table.

 S.           System No. No. of lattices Lattice constants Angle between lattice constants Examples (i)       Cubic 3 a = b = c a = b = g = 90o Diamond,    NaCl,    Li,    Ag,    Cu, NH4Cl, Pb etc. (ii)       Tetragonal 2 a = b ¹ c a = b = g = 90o White tin, NiSO4 etc. (iii)      Orthorhombic 4 a ¹ b ¹ c a = b = g = 90o HgCl2, KNO3, gallium etc. (iv)      Monoclinic 2 a ¹ b ¹ c a = g = 90o and b ¹ 90o KclO3, FeSO4 etc. (v)      Triclinic 1 a ¹ b ¹ c a ¹ b ¹ g ¹ 90o K2Cr2O7, CuSO4  etc. (vi)      Rhombo-hedral or Trigonal 1 a = b = c a = b = g ¹ 90o Calcite, As, Sb, Bi etc. (vii)     Hexagonal 1 a = b ¹ c a = b = 90o and g = 120o Zn, Cd, Ni etc.

# Cubic Lattices.

## (1)  Different symmetry

• Centre of symmetry : An imaginary point within the crystal that any line drawn through it intersects the surface of the crystal at equal distances in both

• Plane of symmetry : It is an imaginary plane which passes through the centre of a crystal and divides it into two equal portions such that one part is exactly the mirror image of the

A cubical crystal possesses

Six diagonal plane of symmetry

Three rectangular plane of symmetry

• Axis of symmetry : It is an imaginary straight line about which, if the crystal is rotated, it will present the same appearance more than once during the complete

In general, if the same appearance of a crystal is repeated on rotating through an angle imaginary axis, the axis is called an n-fold axis.

360 o

n

, around an

A cubical crystal possesses in all 13 axis of symmetry.

• Elements of symmetry : The total number of planes, axes and centre of symmetry possessed by a crystal are termed as elements of A cubic crystal possesses a total of 23 elements of symmetry.

Planes of symmetry = (3 + 6) = 9 , Axes of symmetry = (3 + 4 + 6) = 13 , Centre of symmetry = 1. Total number of symmetry elements = 23

## (2)    Different lattice in cubic crystals

There are three lattice in the cubic system. The simple cubic (sc) lattice.

The body-centered cubic (bcc). The face-centered cubic (fcc).

SC                                    BCC                               FCC

The half of the distance between two atoms in contact is defined as atomic radius.

## (4)  Atoms per unit cell

An atom located at the corner of a unit cell of a lattice is shared equally by eight other unit cells in the three dimensional lattice. Therefore, each unit cell has 1/8th share of an atom at its each corner. Similarly, a face of the unit cell is common to the two unit cells in the lattice. Therefore, each unit cell has 1/2 share of an atom at its each face. The atom located at the centre of the unit cell belongs completely to the unit cell.

Let Nc, Nb and Nf be the number of atoms at the corners, centre and face of the unit cell respectively.

Therefore the number of atoms per unit cell is given by N = N

• N f+ Nc

• In sc lattice : Nb = 0, N f   = 0, Nc  = 8 so N = 1

b      2       8

• In bcc lattice :

Nb   = 1, N f

= 0, Nc

= 8 so

N = 2

• In fcc lattice :

Nb  = 0, N f

= 6,

Nc  = 8 so

N = 4

## (5)  Co-ordination number

It is defined as the number of nearest neighbours that an atom has in a unit cell. It depends upon structure.

• Simple cubic structure : Each atom has two neighbours along X-axis, two along Y-axis and two along Z-axis so coordination number =
• Face-centred cubic structure: Every corner atom has four neighbours in each of the three planes XY, YZ, and ZX so coordination number = 12
• Body-centred cubic structure: The atom of the body of the cell has eight neighbours at eight corner of the unit cell so coordination number =

## (6)  Atomic packing fraction (or packing factor or relative packing density)

The atomic packing fraction indicates how close the atoms are packed together in the given crystal structure or the ratio of the volume occupied by atoms in a unit cell in a crystal and the volume of unit cell is defined as APF.

• For sc crystal : Volume occupied by the atom in the unit cell = 4pr 3 = pa 3 . Volume of the unit cell = a 3

3          6

Thus P.F. = pa 3 / 6 = p = 0.52 = 52%

a 3       6

• For bcc : F. =  3p

8

• For fcc : F. = p

3 2

= 68%

= 74%

## (7)  Density of unit cell

Density of unit cell =

Mass of the unit cell     =

Volume of the unit cell

nA  =

NV

nA Na 3

where n = Number of atoms in unit cell (For sc lattice n = 1, for bcc lattice n = 2, for fcc lattice n = 4),

A = atomic weight, N = Avogadro’s number, V = Volume of the unit cell.

## (8)  Bond length

The distance between two nearest atoms in a unit cell of a crystal is defined as bond length.

• In a sc lattice : Bond length = a (ii) In a bcc lattice : Bond length =

Note : @Hexagonal closed packed (HCP) lattice –

Þ a = b ¹ c

Þ Coordination number = 12

Þ P.F. = p 2

6

Þ Number of atoms per unit cell = 2

3a (iii) In a fcc lattice : Bond length = a

2

Þ Magnesium is a special example of HCP lattice structure.

# Types of Binding and Crystals.

• Binding : Bindings among the atoms or molecules are mainly of following

 S.         Binding No. Cause of binding M.P. Electrical conductivity Examples (i)       Ionic Electrostatic      force      between positive and negative ions Very high Very low NaCl, CsCl, LiF etc.   Ge, Si, diamond etc.   H2O     HF, NH3etc.     Cl2, I2, CO2 etc. (ii)       Covalent Sharing of electrons of opposite spins between two neutral atoms High Semi conductor (iii)      Hydrogen Mutual electrostatic interaction between molecules of surface of different electron densities Low Insulator (iv)      Vander Waal Non polar molecules or Vander Waal forces or dipole-dipole interaction Low Normally insulator (v)      Metallic Mutual interaction between electrons and ion lattice — High

## (2)  Three kinds of crystals

• Single crystal : The crystals in which the periodicity of the pattern extends throughout the piece of the crystal are known as single Single crystals have anisotropic behaviour i.e. their physical properties (like mechanical strength, refractive index, thermal and electrical conductivity) are different along different directions. The small sized single crystals are called mono-crystals.
• Poly-crystals : A poly-crystal is the aggregate of the monocrystals whose well developed faces are joined together so that it has isotropic Ceramics are the important illustrations of the poly-crystalline solids.

• Liquid crystals : The organic crystalline solid which on heating, to a certain temperature range becomes fluid like but its molecules remain oriented in a particular directions, showing that they retain their anisotropic properties, is called liquid These crystals are used in a liquid crystal displays (L.C.D.) which are commonly used in electronic watches, clocks and micro-calculators etc.

# Energy Bands.

In isolated atom the valence electrons can exist only in one of the allowed orbitals each of a sharply defined energy called energy levels. But when two atoms are brought nearer to each other, there are alterations in energy levels and they spread in the form of bands.

Energy bands are of following types

## (1)  Valence band

The energy band formed by a series of energy levels containing valence electrons is known as valence band.

At 0 K, the electrons fills the energy levels in valence band starting from lowest one.

• This band is always fulfill by
• This is the band of maximum
• Electrons are not capable of gaining energy from external electric
• No flow of current due to such
• The highest energy level which can be occupied by an electron in valence band at 0 K is called fermi

## (2)  Conduction band

The higher energy level band is called the conduction band.

• It is also called empty band of minimum energy.
• This band is partially filled by the
• In this band the electrons can gain energy from external electric
• The electrons in the conduction band are called the free They are able to move any where within the volume of the solid.
• Current flows due to such

## (3)  Forbidden energy gap (DEg)

Energy gap between conduction band and valence band

• No free electron present in forbidden energy

DEg

= (C.B.)min – (V.B.)max

• Width of forbidden energy gap upon the nature of
• As temperature increases (­), forbidden energy gap decreases (¯) very

# Types of Solids.

On the basis of band structure of crystals, solids are divided in three categories.

 S.No.              Properties Conductors Insulators                 Semiconductors (1)               Electrical conductivity (2)               Resistivity (3)               Band structure         (4)               Energy gap   (5)               Current carries (6)               Condition of V.B. and C.B.         at               ordinary temperature (7)               Temperature co-efficient of resistance (a) (8)               Effect of temperature on conductivity (9)               Effect of temperature on resistance (11)              Examples   (12)              Electron density 102 to 108 Ʊ/m 10–2 to 10–8 W-m (negligible)     C.B.   V.B. Zero or very small Free electrons V.B. and C.B. are completely filled or C.B. is some what empty Positive Decreases Increases Cu, Ag, Au, Na, Pt, Hg etc. 1029/m3 10– 8 Ʊ/m                                10– 5 to 100 Ʊ/m 108 W-m                               105 to 100 W-m   C.B.                                             C.B.   DEg (maximum)                                DEg (less)   V.B.                                             V.B.   Very large; for diamond    For   Ge   Eg    =   0.7 it is 6 eV                                                  for Si Eg = 1.1 eV ––                   Free electrons and holes V.B. – completely filled         V.B. – somewhat empty C.B. – completely unfilled      C.B. – somewhat filled Zero                                     Negative —                                           Increases   —                                          Decreases   Wood,    plastic,   mica,    Ge, Si, Ga, As etc. diamond, glass etc. —                   Ge ~ 1019 /m3 Si ~ 1016 /m3 eV

## Holes in semiconductors

At absolute zero temperature (0 K) conduction band of semiconductor is completely empty and the semiconductor behaves as an insulator.

When temperature increases the valence electrons acquires thermal energy to jump to the conduction band (Due to the braking of covalent bond). If they jumps to C.B. they leaves behind the deficiency of electrons in the valence band. This deficiency of electron is known as hole or cotter. A hole is considered as a seat of positive charge, having magnitude of charge equal to that of an electron.

• Holes acts as virtual charge, although there is no physical charge on
• Effective mass of hole is more than
• Mobility of hole is less than

# Types of Semiconductors.

## (1)  Intrinsic semiconductor

A pure semiconductor is called intrinsic semiconductor. It has thermally generated current carriers

• They have four electrons in the outermost orbit of atom and atoms are held together by covalent bond
• Free electrons and holes both are charge carriers and ne (in B.) = nh (in V.B.)
• The drift velocity of electrons (ve ) is greater than that of holes(vh )
• For them fermi energy level lies at the centre of the B. and V.B.
• In pure semiconductor, impurity must be less than 1 in 108 parts of
• In intrinsic semiconductor n(o) = n(o) = n  = AT 3 / 2e DEg  / 2KT  ; where n(o) =  Electron density in conduction

e                 h                 i                                                                                             e

 h

band, n(o) = Hole density in V.B., ni = Density of intrinsic carriers.

• Because of less number of charge carriers at room temperature, intrinsic semiconductors have low conductivity so they have no practical

## Net current and conductivity

When some potential difference is applied across a piece of intrinsic semiconductor current flows in it due to

both electron and holes i.e. i = ie + ih Þ i = ne eAve      i = eA[neve + nhvh]

Hence conductivity of semiconductor s = e[ne me + nh mh]

where ve

= drift velocity of electron, vh

= drift velocity of holes,

E = Applied electric field

m = ve E

= mobility of e and

m   = vh   =

 n

E

mobility of holes

 e

Note : @ (ni)Ge  ~- 2.4 ´ 1019 / m3 and (ni)Si  ~- 1.5 ´ 1016 / m3

@    At room temperature s Ge > s Si

@     me > mh

@    Conductivity of semiconductor increases with temperature because number density of charge carriers increases.

 i

@    In a doped semiconductor, the number density of electrons and holes is not equal. But it can be

established that

ne nh = n2 ; where ne, nh are the number density of electrons and holes respectively

and ni is the number density of intrinsic curries (i.e. electrons or holes) in a pure semiconductor. This product is independent of donor and acceptor impurity doping.

## (2)  Extrinsic semiconductor

• It is also called impure semiconductor.
• The process of adding impurity is called
• Impurities are of two types :

The elements whose atom has five valance impurities

e.g. As, P, Sb etc. These are also called donor impurities. These impurities are also called donor impurities because they donates extra free electron.

The elements whose each atom has three valance electrons are called trivalent impurities e.g. In, Ga, Al, B, etc. These impurities are also called acceptor impurities as they accept electron.

• The number of atoms of impurity element is about 1 in 108 atoms of the semiconductor.
• ne ¹ nh
• In these fermi level shifts towards valence or conduction energy
• Their conductivity is high and they are practically

 Ge
 Ge

## Types of extrinsic semiconductor

P-N Junction Diode.

When a P-type semiconductor is suitably joined to an N-type semiconductor, then resulting arrangement is called P-N junction or P-N junction diode

## (1)  Depletion region

On account of difference in concentration of charge carrier in the two sections of PN junction, the electrons from N-region diffuse through the junction into P-region and the hole from P region diffuse into N-region.

Due to diffusion, neutrality of both N and P-type semiconductor is disturbed, a layer of negative charged ions appear near the junction in the P-crystal and a layer of positive ions appears near the junction in N-crystal. This layer is called depletion layer

• The thickness of depletion layer is 1 micron = 10–6 m.

• Width of depletion layer µ

1

Dopping

• Depletion is directly proportional to
• The P-N junction diode is equivalent to capacitor in which the depletion layer acts as a

## (2)  Potential barrier

The potential difference created across the P-N junction due to the diffusion of electron and holes is called potential barrier.

For Ge

VB   = 0.3V

and for silicon VB   = 0.7V

On the average the potential barrier in P-N junction is ~ 0.5 V and the width of depletion region ~ 10–6.

So the barrier electric field

Some important graphs

Potential

E = V

d

= 0.5

10 -6

= 5 ´ 105 V / m

Charge density

## (3)  Diffusion and drift current

Because of concentration difference holes/electron try to diffuse from their side to other side. Only these holes/electrons crosses the junction, having high kinetic energy. This diffusion results is an electric current from the P-side to the N-side known as diffusion current (idf)

As electron hole pair (because of thermal collisions) are continuously created in the depletion region. These is a regular flow of electrons towards the N-side and of holes towards the P-side. This makes a current from the N-side to the P-side. This current is called the drift current (idr).

Note : @In steady state idf

= idr

so inet   = 0

@    When no external source is connected, diode is called unbiased.

## (4)  Biasing

Means the way of connecting emf source to P-N junction diode

Forward biasing                                                     Reverse biasing

• Positive terminal of the battery is connected to the P– crystal and negative terminal of the battery is connected
• Positive terminal of the battery is connected to the

N-crystal and negative terminal of the battery is

to N-crystal

E                            connected to P-crystal                                       E

 –  +
 –  +

Eb                                                                                                                                                                 Eb

P                                                       N                                                                                                        P                                                       N

+   –                                                                                                      –  +

• Width of depletion layer decreases (ii) Width of depletion layer increases
• RForward » 10W – 25W (iii) RReverse » 105W

• Forward bias opposes the potential barrier and for V

> VB a forward current is set up across the junction.

• Cut-in (Knee) voltage : The voltage at which the current starts to For Ge it is 0.3 V and for Si it is 0.7 V.
• df – diffusion

dr – drift

• Reverse bias supports the potential barrier and no current flows across the junction due to the diffusion of the majority

(A very small reverse currents may exist in the circuit due to the drifting of minority carriers across the junction)

• Break down voltage : Reverse voltage at which break down of semiconductor For Ge it is 25 V and for Si it is 35 V.

Reverse

P                              N

Knee Forward

Idf Idr Inet

Break

Reverse current

Idf Idr Inet

# Reverse Breakdown and Special Purpose Diodes.

## (1)  Zener breakdown

When reverse bias is increased the electric field at the junction also increases. At some stage the electric field becomes so high that it breaks the covalent bonds creating electron, hole pairs. Thus a large number of carriers are generated. This causes a large current to flow. This mechanism is known as Zener breakdown.

## (2)  Avalanche breakdown

At high reverse voltage, due to high electric field, the miniority charge carriers, while crossing the junction acquires very high velocities. These by collision breaks down the covalent bonds, generating more carriers. A chain reaction is established, giving rise to high current. This mechanism is called avalanche breakdown.

• Special purpose diodes

 Zener diode Light emitting diode (LED) Photo diode Solar cells It is a highly doped p-n junction which is not damaged by high reverse current. The breakdown voltage is made very sharp. In the forward bias, the zener diode acts as ordinary diode. It can be used as voltage regulator N     Specially designed diodes, which give out light radiations when forward biases. LED’S are made of GaAsp, Gap etc. In these diodes electron and hole pairs are created by junction photoelectric effect. That is the covalent bonds are broken by the EM radiations absorbed by the electron in the V.B. These are used for detecting light signals. It is based on the photovoltaic effect. One of the semiconductor region is made so thin that the light incident on it reaches the p-n junction and gets absorbed. It converts solar energy into electrical energy.

# P-N Junction Diode as a Rectifier.

 put gnal + + put gnal D1 –     D2 D1 –     D2 Fluctuating dc

 Ib
 Ib

Note : @ Fluctuating dcconstant dc.

# Transistor.

A junction transistor is formed by sandwiching a thin layer of P-type semiconductor between two N-type semiconductors or by sandwiching a thin layer of n-type semiconductor between two P-type semiconductor.

E – Emitter (emits majority charge carriers)

C – Collects majority charge carriers

B – Base (provide proper interaction between E and C)

Note : @ In normal operation base-emitter is forward biased and collector base junction is reverse biased.

• Working of Transistor : In both transistor emitter – base junction is forward biased and collector – base junction is reverse

Note : @In a transistor circuit the reverse bias is high as compared to the forward bias. So that it may exert a large attractive force on the charge carriers to enter the collector region.

• Characteristics of transistors : A transistor can be connected in a circuit in the following three different
• Common base (CB) (ii) Common emitter (CE)     (iii) Common collector (CC)
• CB characteristics : The graphs between voltages and currents when base of a transistor is common to input and output circuits are known as CB characteristic of a
• CE characteristics : The graphs between voltages an d currents when emitter of a transistor is common to input and output circuits are known as CE characteristics of a
• Transistor as an amplifier : A device which increases the amplitude of the input signal is called

The transistor can be used as an amplifier in the following three configuration

(i) CB amplifier            (ii) CE amplifier         (iii) CC amplifier

## (4)  Parameters of CE/CB amplifiers

• Relation between a and b :

b =   a    or a =    b

1 – a               1 + b

## (6)  Comparison between CB, CE and CC amplifier

 S.No.           Characteristic Amplifier CB CE CC (i)           Input resistance (Ri) » 50 to 200 W low » 1 to 2 kW medium » 150 – 800 kW high (ii)          Output resistance (Ro) » 1 – 2 kW high » 50 kW medium » kW low (iii)         Current gain 0.8 – 0.9 low 20 – 200 high 20 – 200 high (iv)         Voltage gain Medium High Low (v)          Power gain Medium High Low (vi)         Phase difference between input and output voltages Zero 180o Zero (vii)        Used as amplifier for current Power Voltage

Example: 1        What is the coordination number of sodium ions in the case of sodium chloride structure                    [CBSE 1988]

(a) 6                               (b) 8                                  (c) 4                            (d) 12

Solution : (a)       In NaCl crystal Na+ ion is surrounded by 6 Cl

ion, therefore coordination number of Na+ is 6.

Example: 2         A Ge specimen is doped with Al. The concentration of acceptor atoms is ~1021 atoms/m3. Given that the intrinsic concentration of electron hole pairs is ~ 1019 / m3 , the concentration of electrons in the specimen is

[AIIMS 2004]

(a)

1017 / m3

(b)

1015 / m3

(c)

104 / m3

(d)

102 / m3

Solution : (a)

n2 = n n      Þ (1019 )2 = 1021 ´ n    Þ n

= 1017 / m3 .

i               h   e                                                                     e                 e

Example: 3         A silicon specimen is made into a P-type semi-conductor by doping, on an average, one Indium atom per

5 ´ 107

silicon atoms. If the number density of atoms in the silicon specimen is

5 ´ 1028

atoms/m3, then the

number of acceptor atoms in silicon will be                                                                         [MP PMT 1993, 2003]

(a)

2.5 ´ 1030 atoms/cm3        (b)

1.0 ´ 1013 atoms/cm3           (c)

1.0 ´ 1015 atoms/cm3 (d)

• ´ 1036 atoms/cm3

Solution : (c)        Number density of atoms in silicon specimen = 5 ´ 1028 atom/m3 = 5 ´ 1022 atom/cm3

Since one atom of indium is doped in 5 ´ 107 Si atom. So number of indium atoms doped per cm-3 of silicon.

n = 5 ´ 1022

5 ´ 107

= 1 ´ 1015 atom / cm3 .

Example: 4         A P-type semiconductor has acceptor levels 57 meV above the valence band. The maximum wavelength of light required to create a hole is (Planck’s constant h = 6.6 ´ 1034 J-s)                                                  [MP PET 1995]

• 57 Å (b)

57 ´ 103 Å

(c) 217100 Å                      (d)

11.61´ 1033 Å

Solution : (c)

E hc

l

Þ l = hc

E

=   6.6 ´ 10 34 ´ 3 ´ 108   = 217100 Å.

57 ´ 10 3 ´ 1.6 ´ 10 19

Example: 5         A potential barrier of 0.50V exists across a P-N junction. If the depletion region is

5.0 ´ 107 m

wide, the

intensity of the electric field in this region is                                                                                [UPSEAT   2002]

(a)

1.0 ´ 106 V / m

1.0 ´ 105 V / m

2.0 ´ 105 V / m

2.0 ´ 106 V / m

Solution : (a)

E = V

d

=    0.50   = 1 ´ 106 V/m.

5 ´ 10 7

Example: 6 A 2V battery is connected across the points A and B as shown in the figure given below. Assuming that the resistance of each diode is zero in forward bias and infinity in reverse bias, the current supplied by the battery when its positive terminal is connected to A is                                                                                        [UPSEAT 2002]

• 2 A
• 4 A
• Zero
• 1 A

Solution : (a)       Since diode in upper branch is forward biased and in lower branch is reversed biased. So current through

circuit i =     V      ; here rd = diode resistance in forward biasing = 0

R + rd

So       i = V R

= 2 = 0.2A .

10

Example: 7         Current in the circuit will be                                                                                                  [CBSE PMT 2001]

(a)    5 A

40

(b)

5 A

50

• 5 A

10

• 5 A

20

Solution : (b)       The diode in lower branch is forward biased and diode in upper branch is reverse biased

\      i =

5      =  5 A

20 + 30     50

Example: 8         Find the magnitude of current in the following circuit                                                                         [RPMT 2001]

• 0
• 1 amp
• 1 amp
• 2 amp

Solution : (a)       Diode is reverse biased. Therefore no current will flow through the circuit.

Example: 9         The diode used in the circuit shown in the figure has a constant voltage drop of 0.5 V at all currents and a maximum power rating of 100 milliwatts. What should be the value of the resistor R, connected in series with

the diode for obtaining maximum current                                                                                 [CBSE  PMT  1997]

(a) 1.5 W

(b) 5 W

(c) 6.67 W

(d) 200 W

P      100 ´ 10 3

Solution : (b)       The current through circuit i = V =

0.5

= 0.2A

\ voltage drop across resistance = 1.5 – 0.5 = 1 V   Þ R =

1

0.2

= 5 W

Example: 10       For a transistor amplifier in common emitter configuration for load impedance of 1 kW (hfe = 50 and hoe = 25) the current gain is                                                                                                                                  [AIEEE 2004]

(a) – 5.2                          (b) – 15.7                           (c) – 24.8                     (d) – 48.78

Solution : (d)       In common emitter configuration current gain A

=      hfe            =                   50           = – 48.78.

i      1 + hoe RL

1 + 25 ´ 10 6 ´ 103

Example: 11       In the following common emitter configuration an NPN transistor with current gain b = 100 is used. The output voltage of the amplifier will be                                                                                          [AIIMS 2003]

• 10 mV
• 1 V
• 0 V
• 10 V

Solution : (c)        Voltage gain = Output voltage

Input voltage

Þ Vout

= Vin

• Voltage gain

Þ Vout

= Vin

´ Current gain ´ Resistance gain = Vin

´ b ´

RL RBE

= 103 ´ 100 ´ 10 = 1V.

1

Example: 12       While a collector to emitter voltage is constant in a transistor, the collector current changes by 8.2 mA when the emitter current changes by 8.3 mA. The value of forward current ratio hfe is                                              [KCET 2002]

(a) 82                              (b) 83                                 (c) 8.2                          (d) 8.3

 è

æ Dic ö

8.2

Solution : (a)

hfe   = ç Di

÷

b ø Vce

= 8.3 – 8.2 = 82

Example: 13       The transfer ratio of a transistor is 50. The input resistance of the transistor when used in the common-emitter configuration is 1 KW. The peak value for an ac input voltage of 0.01 V peak is                                                   [CBSE PMT 1998]

(a) 100 mA                               (b) 0.01 mA                                 (c) 0.25 mA                        (d) 500 m A

Solution : (d)

ic = bib

= b ´ Vi

Ri

= 50 ´ 0.01 = 500 ´ 10 6 A = 500m A

1000

Example: 14       In a common base amplifier circuit, calculate the change in base current if that in the emitter current is 2 mA

 and a = 0.98 [BHU 1995] Solution : (a) (a) 0.04 mA Dic = a Die = 0.98 ´ 2 = (b) 1.96 mA 196 mA (c) 0.98 mA (d) 2 mA \ Dib = Die – Dic = 2 – 1.96 = 0.04 mA .