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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , Introduction:At absolute zero a pure, perfect crystal of the most of the semiconductors will be an, insulator. Semiconductors are electronic conductors with electrical resistivity value, generally in the range of, , 102 to 109 ohm-cm at room temperature, intermediate, , between good conductors ( 106 ohm-cm) and insulators ( 1014 to 1022 ohm-cm). The, electrical resistivity of semiconductors dependent on temperature. Devices based on, the properties of semiconductors include transistors, rectifiers, modulators, detectors,, thermostats, and photocells. Broadly speaking we can have two types of, semiconductors i.e. intrinsic semiconductors and extrinsic semiconductors., , INTRINSIC SEMICONDUCTORS:1. A pure semiconductor is called intrinsic semiconductor. At absolute zero, the valence, band of an intrinsic semiconductor is completely filled. For this reason at absolute zero, they behave as insulators., , 2. However, as the temperature is gradually increased some of the electrons of the, valence band get excited to conduction band. The excitation of electrons from the, valence band to the conduction band leaves an equal number of holes in the valence, band., 3. Both electron in the conduction band and holes in the valence band serve as charge, , Vignana Bharathi Institute of Technology, , Page 1

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , carriers and contribute to the electrical conductivity. Pure form of the germanium and, silicon are the best examples of intrinsic semiconductors., , EXTRINSIC SEMI-CONDUCTORS: If a small amount of a pentavalent (or) trivalent impurity is introduced into a pure, germanium (or) silicon crystal, the conductivity at the crystal increases appreciably and, the crystal becomes an “Extrinsic” Semi-conductor., Extrinsic semi-conductors are of two types: n- type and p- type., , n- Type semiconductors: 1. When a pentavalent (antimony, phosphorus (or) arsenic) atom replaces a Ge, atom in the crystal lattice. Four of its five valence electrons form covalent bonds, with neighboring Ge (or) Si atoms., 2. The fifth electron free at room temperature to move about in the crystal and acts, as a charge carrier. The crystal is now called an “n- type semi-conductor”, because it has an excess of negative charge carries. The impurity is called a, “donor” because it donates the conducting electrons., 3. In the energy level diagram, the impurity atoms introduce discrete energy levels, for the electrons just below the conduction band. These are called “donor, impurity levels”., Vignana Bharathi Institute of Technology, , Page 2

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , 4. Therefore, at room temperature, the ‘fifth’ electrons of donor atoms are, thermally excited from the donor levels into conduction band., 5. At ordinary temperature, almost all the electrons in the conduction band come, from the donor levels. Only a few come from the valence band. Therefore, the, main charge carriers responsible for conduction are electrons contributed by the, donors. Thus, in an n-type semi-conductor the electrons are the “majority, carriers” and the holes are “minority carriers”., , Generation and recombination:, When the covalent bond is break it leads to the electron and hole formation. The, electron is in the conduction band and the hole is in the valence band., This electron -hole formation is known as generation., , In the electron-hole formation the number of electrons are generated would be equal to, the number of holes produced., , Vignana Bharathi Institute of Technology, , Page 3

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Engineering physics, , UNIT –II SEMICONDUCTOR PHYSICS, , In the recombination process the electron in the conduction band may lose its energy, and fall into the valence band and it merges with hole. This transfer of the free electron, into the valence band and forms the covalent bond. In this process electron-hole pair, disappears and energy is emitted. Tis process is known as recombination., , Carrier transport:, Intrinsic Semiconductor:, At any temperature T the number of electrons generated will be equal to the number of, the holes produced. At this temperature T the charge carriers possess an average, velocity vth., , Drift:, If a potential difference is applied to the solid, the charge carriers will move. This, movement of the electrons and holes is the known as drift and the corresponding, velocity of the electrons is known as drift velocity (vd), , The applied potential difference is V  E  l, , vd , , And the drift velocity is, , l, t, , ---------- (1), ---------- (2), , We know that the total charge on the material is equal to Q is, , Q  it, We know that the charge can be given as, ,  i, , Q, t, , Q  ( A l) n e, , Where n is the number of electrons and e is the charge of the electron., , i, , Q, An, t, , Vignana Bharathi Institute of Technology, , l , e , t, Page 4

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, ,  i  A n e vd ---------- (3), We know that J , ,  J  n e vd, According to the Ohm’s law, Where,, , , , J E, , i, is current density, A, , --------------- (4), , -------------- (5), , is conductivity of the charges., , We know that the drift velocity is directly proportional to the applied electric field E, Therefore, , vd  E, ,  vd = E ---------------- (6),  is mobility of the charges., , From the eq (6) & (4), ,  J e  n e e E, , is the current density of the electrons., , And J p  p e  p E is the current density of the holes., ,  e and  p are motilities of electron and holes respectively, and P is number of the, holes., Therefore the total density is, , J  Je  J p, ,  J   n e e E    p e  p E , ,  J  eE  n e + p  p  ------------- (7), Comparing eq (7) and (5) we get, ,   e  n e + p  p , In intrinsic semiconductors number of electron generated is equal to the number of, holes produced. Therefore, , n  p  ni, ,    e n i  e +  p  ----------------- (8), Vignana Bharathi Institute of Technology, , Page 5

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Engineering physics, , UNIT –II SEMICONDUCTOR PHYSICS, , This represents the carrier transport in intrinsic semiconductor., , Extrinsic Semiconductor:, In extrinsic semiconductor the carrier transport is the sum of the carrier transport of, n-type and p-type semiconductors., Hence the carrier transport of n-type semiconductor can be given as, ,   e  n e + p  p , Where nn and pn are number of electrons and holes in n-type semiconductor., We know that electrons are the majority charge carriers the number of electrons in ntype is very much greater than number of holes in n-type., i.e. nn    pn, Therefore, , pn, , can be neglect and, , nn  nD, , where, , Hence the transport of n-type semiconductor is, , nD is number of donor atoms, ,  n  e nD e, , Similarly we can write the transport of p-type semiconductor is, Where, , nA, ,  p  e nA  p, , is number of Acceptor atoms., , Therefore the total transport in extrinsic semiconductor is, ,    e  nD  e  n A  p , Diffusion:, The process by which, charge carriers (electrons and holes) in a semiconductor moves, from a region of higher concentration to a region of lower concentration without any, external voltage is known as diffusion., Or, Without external voltage there is a movement of charges provided a concentration, gradient of charges is exist, this is known as diffusion., The current density due to the diffusion is directly proportional to the concentrate, gradient., Therefore current density for n-type and p-type semiconductors are given as, Vignana Bharathi Institute of Technology, , Page 6

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Engineering physics, , UNIT –II SEMICONDUCTOR PHYSICS, , Jn  e, , dn, dx, ,  J n = Dn e, , dp, dx, , Jp  e, dn, dx, , Therefore the total current density is, , J n = - Dp e, , dp, dx, , J  J  Drift   J  Diffusion, , dn , ,  J n   n e e E    Dn e, , dx , , , dn , , J n  e  n e E  Dn, , dx , , , and J p  p e  p E  D p e, , dp, dx, , dp , , J p  e  p  p E  Dp, , dx , , Carrier concentration in Intrinsic Semiconductor:Calculation of density of electrons:Let dn be the number of electrons available between the energy ranges E and E  dE in, the conduction band is, dn  g ( E ) F ( E )dE ---------------------------- (1), , Where g ( E )dE is the density of electrons in the energy interval E and E  dE .we know, that, , 3, 1, 4, 2, g ( E )dE  3 (2m) E 2 dE, h, , Since electron is moving in a periodic potential, its mass has to be replaced by its, effective mass me* and E starts at the bottom of the conduction band Ec , then, 3, 1, 4, * 2, g ( E )dE  3 (2me ) ( E  Ec ) 2 dE, h, 1, 2, ,  c( E  Ec ) dE ------------------------------- (2), , Vignana Bharathi Institute of Technology, , Page 7

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Engineering physics, , UNIT –II SEMICONDUCTOR PHYSICS, 3, 4, * 2, Where c  3 (2me ), h, , In equation (1) F ( E ) is the probability occupancy of the electrons for a given energy, level E. i.e., , 1, ------------------------- (3),  E  EF , 1  exp, ,  KT , , F (E) , , Now from equation (1),(2) and (3), 1, , dn  c( E  Ec ) 2 F ( E )dE, Now to obtain total number of electrons in the conduction band at any temperature we, have integrate equation (1) from Ec to Etop ., Etop, ,  c( E  E ), , n, , c, , Ec, , 1, 2, , 1,  E  EF, 1  exp ,  KT, , , , , , dE, , For all possible temperatures E  EF  KT , so that the term unity in the denominator is, neglected. Then, Etop, , , , n, , c( E  Ec ), , Ec, , Etop, , , , , , Ec, , 1, 2, , 1,  E  EF, exp ,  KT, ,  E  EF, c( E  Ec ) 2 exp  , KT, , 1, , , , , , dE, , ,  dE, , , In the above integral the upper limit is taken as infinity. Since the probability of, electrons occupying upper level of conduction band is zero at infinity., , ,  E  EF, n   c( E  Ec ) 2 exp  , KT, , Ec, 1, , To solve this integral let us assume that, , ,  dE ---------------------- (4), , , E  Ec, x, KT, ,  E  Ec  kT x, dE  kT dx, , Vignana Bharathi Institute of Technology, , Page 8

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Engineering physics, , UNIT –II SEMICONDUCTOR PHYSICS, f E  Ec then x  0, and E   then x  , , Now equation (4) becomes, , , 1,  E  Ec  Ec  EF, n  c  ( E  Ec ) 2 exp  , KT, , Ec, , ,  dE, , , , , 1,  E  Ec ,  E  EF , n  c  ( xKT ) 2 exp  , exp   c, dx ( KT ), , KT , KT , , , 0, ,  E  EF , n  c( KT ) exp   c, KT , , , , ,  E  EF , n  c( KT ) exp   c, KT , , , , , 3, 2, , 3, 2, , x, , 1, 2, , 0, , x, , 1, 2, ,  E  Ec , exp  , dx, KT , , e  x dx --------------------------- (5), , 0, , Simplifying above integral, we get, , , 1, , 2, x, x, e, dx, , 0, 2, 1, 2, , Now equation (5) becomes, 1, ,  E  EF   2, n  c( KT ) exp   c, KT  2, , 3, 2, , , ,  12, 4, , n  3 (2m ) ( KT ) ,  2, h, , , 3, * 2, e, , 3, 2, , 3, 2, ,  ,  e, , , , , ,  2 m e kT   , n  2,  e, 2, h, , , *, , Ec  EF , KT , , Ec  E F , KT , , i.e. the number of electrons per unit volume of the material is given by, 3, 2, , , ,  2  m e * k T   , n  2,  e, 2, h, , , , Ec  E F , , KT, , , ------------------------- (6), , Calculation of density of holes:Let dp be the number of holes available between the energy ranges E and E  dE in the, valence band is, dp  g ( E ) 1  F ( E )  dE ------------------------------ (7), Vignana Bharathi Institute of Technology, , Page 9

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , Where g ( E )dE is the density of holes in the energy interval E and E  dE . We know that, g ( E )dE , , 3, 1, 4, 2, 2, (2, m, ), E, dE, h3, , Since hole is moving in a periodic potential, its mass has to be replaced by its effective, mass mh* and Ev is the energy of the top of the valence band, g ( E )dE , , 3, 1, 4, * 2, 2, (2, m, ), (, E, , E, ), dE, h, v, h3, 1, ,  c( Ev  E ) 2 dE ------------------------------- (8), , Where c , , 3, 4, * 2, (2, m, ), h, h3, , In equation (7) 1  F ( E )  is the probability of existence of holes in the valence band., For all possible temperatures EF  E  KT , so that the term unity in the denominator is, neglected. Then, 1  F (E) , , 1, E E, exp  F,  1,  KT , ,  E E, 1  F ( E ) = exp   F, , KT , , , Now equation (7) becomes, 1, , dp  c( Ev  E ) 2 1  F ( E )  dE, 1,  E E, p  c( Ev  E ) 2 exp   F,  dE, KT , , , Now to obtain total number of holes in the valence band at any temperature we have, integrate equation (1) from Ev to  ., , p, , Ev, ,  c( E, , v, , , ,  E  EF,  E ) exp ,  KT, 1, 2, , ,  dE ------------------------ (9), , , Now equation (9) becomes, , Vignana Bharathi Institute of Technology, , Page 10

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Engineering physics, , UNIT –II SEMICONDUCTOR PHYSICS, ,  2 KT , ni 2  4 , , 2,  h, , , 3, , m, , *, h, , me, , *, , , , 3, 2, , e, ,  EC  EV , , , KT , , , Where Ec  Ev  Eg is the forbidden energy gap., , Hence, ,  2 KT , ni  2 , , 2,  h , , 3, 2, , m, , h, , *, , 3, 4, , me*  e, , Eg , ,  , , 2, KT , , , FERMI ENERGY:Since n = p in intrinsic semiconductor,, 3, 2, ,  2 me kT , 2,  e, h2, , , *, , m , *, , e, , e, , 3, 2, , e, ,  Ec  EF ,   KT , , , , e, , e, , e, , , ,  2 mh KT   ,  2,  e, h2, , , ,  Ec  EF ,   KT , , , , Ec  EF , , , , KT, , , , 3, 2, , *, ,   mh, ,  E F  EV , , , KT, , , , 3, * 2, , , , e, ,  EF  EV ,   KT , , , ,  m ,   h* ,  me , *, ,  E F  EV  E c  E F , , , KT, , , , m ,   h* ,  me , , EV  E c , ,  , , KT, , , , m ,   h* ,  me , ,  2 EF ,  KT , , , , e, , EF  EV , KT , , *, , *, , 3, 2, , 3, 2, , 3, 2, , Apply the log on both sides, , e, ,  2 EF ,  KT , , , , e, , EV  E c , ,  , , KT, , , , m ,   h* ,  me , *, , 3, 2, ,  m , E  Ec, 2EF,  V,  lo g  h * , KT, KT,  me , *, , Vignana Bharathi Institute of Technology, , 3, 2, , Page 12

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UNIT –II SEMICONDUCTOR PHYSICS, ,  m h*, 3, 2 E F  K T log  *, 2,  me, , Engineering physics, , ,    EV  E c , , , If we assume that me*  mh*, ,  E  Ec , EF   v, , 2, , , Thus Fermi level is located half way between the valence band and conduction bands, and its position is independent of temperature., , Carrier concentration in Extrinsic Semiconductor:, Carrier concentration in n- type semi-conductor: Consider an extrinsic semi-conductor doped with donor atoms. The donor atoms, give rise to donor levels Ed close to the conduction band Ec as shown in figure. Let N D, be the number of impurity atoms gives rise to a single electrons state at Ed . Then the, number of electrons in the energy level Ed would be, nd  g e  Ed  Fe  Ed  ----------------------------------- (1), , But, g e ( Ed )  N D represents the density of states and, 1, , Fe ( ED ) , e, , Ed  EF, kT, , --------------------------------------- (2), 1, , Substituting the above values in equation (1) we get, , 1, , nd  N D, e, , Ed  EF, kT, , 1, , ------------------------------ (3), , Total number of energy states per unit volume at Ed is N D . The total number of filled, energy state per unit volume at Ed is nd . The total number of vacant energy states per, unit volume at Ed is N D   N D  nd ., From equation (3), , ND  ND , , ND,  E  EF , 1  exp  d, ,  KT , , Vignana Bharathi Institute of Technology, , Page 13

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , , , , , 1, , N D   N D 1 ,  1  exp  Ed  EF  , , , ,  KT  , , ,  Ed  EF  ,  exp  KT  , ,  ,  ND , E, , E,  1  exp  d, F , , , , KT, , , , ND, ------------------------------ (4), ND ,  EF  Ed , exp ,  1,  KT , , The density of electrons per unit volume in the conduction band is given by,  2 me* KT , ne  2 , , h2, , , , 3/2, , e, , EF  Ec, kT, , ---------------------------------- (5), , Where EF, , =, , Energy of the Fermi level, , EC, , =, , Energy corresponding to the bottom of the conduction band., , Density of empty donors (or) number of vacancies per unit volume in the donor, level of energy Ed is, , N, , ND , e, , D, ( EF  Ed ), KT, , 1, , ---------------------------------------- (6), , The concentration of electrons in conduction band must be equal to the number, of vacancies per unit volume in the donor level. Therefore, equating equations (5) and, (6), ,  2 me* KT , 2, , h2, , , , 3/2, , e, , EF  Ec, KT, , N, , , e, , D, ( EF  Ed ), KT, , -------------------------------- (7), , 1, , If we suppose EF  Ed  kT . We can neglect 1 from the denominator of R.H.S. of, equation (7). Hence,, ,  2 me* KT , 2, , h2, , , , 3/2, , e, , EF  Ec, KT, , ND, , , , Vignana Bharathi Institute of Technology, , e, , EF  Ec, KT, , Page 14

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Engineering physics, , UNIT –II SEMICONDUCTOR PHYSICS, ,  2 me* KT , 2, , h2, , ,  2 me* KT , 2, , 2, h, , , , 3/2, , e, , EF  Ec, KT, , ND, , , e, , 3/2, , e, , EF  Ec, KT, , EF  Ed, KT, ,  N D .e, , ( Ed  EF ), KT, , Taking log on both sides, we get,  2 me* KT  EF  EC, E  EF, log 2 ,  log N D  d, , 2, h, KT, KT, , , , 2 EF  Ec  Ed, ND,  log, 3/2, KT,  2 me* KT , 2, , h2, , , ND, 2 EF  Ec  Ed  KT .log, 3/2,  2 me* KT , 2, , h2, , , ND, 2 EF  Ec  Ed  KT .log, 3/2,  2 me* KT , 2, , h2, , , , EF , , EC  Ed KT, ND, , .log, 3/2 ----------------- (8), 2, 2,  2 me* KT , 2, , h2, , , , Putting the value of EF from equation (8) in equation (5), we get, , ND,  Ed  Ec KT, , .log,  Ec, 3/2, , *, 2, 2,  2 me KT , , 2, 3/2, , *, , h2,  2 me KT , , , ne  2 ,  exp , 2, h, kT, , , , , , , ,  2 me* KT , , = 2, h2, , , , 3/2, , , , , , , , , , , , , , , , EC , ND,  Ed  Ec 1, exp ,  log, , 3/2, *, 2 KT, 2, KT , , , 2, , m, KT, e, , , 2, , 2, , , h, , , , , , Vignana Bharathi Institute of Technology, , Page 15

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Engineering physics, , UNIT –II SEMICONDUCTOR PHYSICS, ,  2 me* KT , , h2, , , , 3/2, , = 2, ,  2 me* KT , ne  2 , , h2, , , ,  ne   2 N D , , 1/2, , , , , , ND,  Ed  Ec 1, , exp ,  log, 3/2, , 2 KT, 2,  2 me* KT  , , 2,  , , h2, ,  , , 3/2, , 1/2, ,  ND , , ,  2 , , 1,  2 me* KT , , , h2, , , ,  2 me* KT , , , h2, , , , 3/4, , e, , , , E, 2 KT, , .e, 3/4, , Ed  Ec, 2 KT, , ------------------------- (9), , When E  Ec  Ed ionization energy of the donor., The equation (8) at T = 0 k becomes, EF , , Ed  Ec, 2, , This shows that the Fermi level lies exactly halfway between the donor levels and the, bottom of conduction band as shown in figure. But when the temperature is increased, the Fermi level falls below the donor level and at high temperature the value of EF, becomes, , Ec  Ev, ., 2, , P- Type semi-conductor: 1. When a trivalent impurity (boron, aluminum, gallium (or) indium) atom replaces a Ge, atom in the crystal lattice. Only three valence electrons are available to form covalent, bands with the neighboring Ge atoms., 2. This results into an empty space (or) a positive hole around the impurity atom. An, electron bound to a neighbouring Ge (or) Si atom drops into the hole, when an external, field is applied, thereby creating a new hole. This phenomenon continues and the hole, moves in the crystal lattice, acting as a positive charge – carrier. The crystal is called a ptype semi-conductor because it has an excess of positive charge carriers. The impurity is, called an ‘acceptor’ because the impurity atoms create holes which accept electrons., In the energy level diagram the trivalent impurity atoms introduce vacant discrete levels, just above the top of the valence band. There are called “acceptor impurity levels”., , Vignana Bharathi Institute of Technology, , Page 16

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , At room temperature, electrons are easily excited from the valence band into the, acceptor levels. The corresponding holes created in the valence band are the main, charge carries in the crystal when an electric field is applied., Thus, in a p- type semi-conductor the holes are the majority carriers and few, electrons, thermally excited from the valence band into conduction band are minority, carriers., Carrier Concentration p-type semi-conductor: Consider a semi-conductor doped with N A number of acceptor atoms per unit, volume. The acceptor atoms give rise to acceptor levels Ea slightly above the valence, band energy Ev as shown in figure., If some of the atoms, let na get ionized by accepting electrons, then those atoms, are becoming negatively charged., The number of such electrons na occupying energy level Ea would be, na  g e ( Ea ) Fe ( Ea )  N A  ----------------------------------- (1), , But g e ( Ea )  N A represents the density of states and, Fe ( E ) , , 1,  E  EF , 1  exp  a, ,  KT , , na , , NA,  N A  ------------------------------------ (2),  E  EF , 1  exp  a, ,  KT , , Vignana Bharathi Institute of Technology, , Page 17

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Engineering physics, , UNIT –II SEMICONDUCTOR PHYSICS, , Since Ea  EF  KT , e, , Ea  EF, KT, , is a large quantity and thus 1 can be neglected from, , denominator.,  Number of electrons per unit volume in acceptor level, ,  N Ae, , EF  Ea, kT, , ---------------------------------------------- (3), , Since we have neglected the effect of conduction band, the number of electrons in the, acceptor level is equal to the number of holes in valence band., The density of holes per unit volume in the valence band is given by,  2 mh* KT , nh  2 , , h2, , , , 3/2, , e, , Ev  EF, KT, , ----------------------------------- (4), , Now, equating equation (3) and (4), ,  2 mh* KT , 2, , h2, , , , 3/2, , e, , Ev  EF, KT, ,  N A .e, , EF  Ea, KT, , ------------------------------- (5), , Taking log of both sides (5) we get,  2 mh* KT , log 2 , , h2, , , , 3/2, , , , Ev  EF, E  Ea,  log N A  F, KT, KT, ,  2 mh* KT , Ev  EF  EF  Ea,  log N A  log 2 , , KT, h2, , , ,  2 mh* KT , 2 EF  ( Ea  Ev ),   log N A  log 2 , , KT, h2, , , 2 EF  ( Ea  Ev ), NA,   log, 3/2, KT,  2 mh* KT , 2, , h2, , , NA, 2 EF  Ea  Ev  KT log, 3/2,  2 mh* KT , 2, , h2, , , EF , , Ea  Ev KT, NA, , log, 3/2, 2, 2,  2 mh* KT , 2, , h2, , , , 3/2, , 3/2, , --------------------------------- (6), , Putting the values of EF from (6) in (4) we get, , Vignana Bharathi Institute of Technology, , Page 18

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Engineering physics, , UNIT –II SEMICONDUCTOR PHYSICS, , Ea  Ev KT, NA, , 3/2,  Ev  2  2 log,  2 mh* KT , , 2, 3/2, , , h2,  2 mh* KT , , , nh  2 ,  exp , 2, h, KT, , , , , , , ,  2 mh* KT , nh  2 , , h2, , , , 3/2, ,  2 mh* KT ,  2, , h2, , , , , , , , , , , , , , , , , , Ev  Ea 1, NA, , , exp ,  log, 3/2, , *, 2kT, 2, , , 2, , m, KT, h, , 2,  , 2, , h, ,  , 3/2, ,  NA , , ,  2 , , 1/2, , e, , Ev  Ea, 2 KT, , 1,  2 mh* KT , 2, , h2, , , , *, 1/2  2 mh KT , nh  (2 N A ) , , h2, , , , 3/4, , e, , 3/4, , Ev  Ea, 2 KT, , If we put E  Ev  Ea we have, ,  2 mh* KT , nh  (2 N A ) , , h2, , , , 3/ 4, , 1/ 2, , e, , E, 2 KT, , -------------------------- (7), , This represents the carrier concentration in P-type semiconductor., At T = 0k, the form of equation (6) becomes, , EF , , Ea  Ev, 2, , Hence, in this case, the Fermi level lies exactly halfway between the acceptor, levels and the top of valence band., But when T increases the Fermi level rises., At high temperature the value of EF becomes, , Ea  Ev, i.e., EF approaches the center of, 2, , band gap and crosses the acceptor level which makes the substance an intrinsic semiconductor., , Vignana Bharathi Institute of Technology, , Page 19

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , It is clear from equation (7) that density of the holes in the valence band is proportional, to the square root of the acceptor concentration., , HALL EFFECT:, When a current carrying material or semiconductor is placed in a transverse magnetic, field then a potential difference is developed across the semiconductor which is, mutually perpendicular to both the electric and magnetic fields. This phenomenon is, known as Hall Effect., The voltage produced in this phenomenon is known as Hall Voltage (VH), , From the Lorentz force FL  e E, , --------------- (1), , E ---- is Applied electric field, , e ---- is charge of the electron or hole, The magnetic force acting on the charges is given as, , FB  B e vd, , B ---- is magnetic field, , ------------- (2), , vd, , is drift velocity, , Vignana Bharathi Institute of Technology, , Page 20

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UNIT –II SEMICONDUCTOR PHYSICS, , From eq (1) and (2) we get, , Engineering physics, , e EH  B e vd, ,  EH  B vd --------- (3), The applied electric field EH is written as, , EH , , VH, d, , ----------- (4), , VH is Hall Voltage and d is thickness of the sample., From the above equations we get, , VH,  B vd, d, ,  VH  B vd d ---------- (5), Due to the Hall field there will be development of current density J, , J  n e vd,  vd , , J, --------- (6), ne, , Therefore VH  B, , J, d --------- (7), ne, , But we know the current density J in terms of current is J , , I, I, J, ------- (8), A, dw, , Where A  d w, , d, , is thickness of the sample and, , w, , is width of the sample., , Substitute the equations (6) and (7) in (5) we get, , VH  B, , VH , , I  1 , , d, d w n e, , BI 1 , ,  -------- (9), w n e, , This equation represents the Hall Voltage., , Vignana Bharathi Institute of Technology, , Page 21

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Engineering physics, , UNIT –II SEMICONDUCTOR PHYSICS, , Hall Coefficient (RH):, It can be defined as it is the ration of the electric field across the semiconductor to the, current density per unit volume per unit magnetic field., , i.e., RH , , EH, J B, , V  1, i.e., RH   H ,  d J B, , From equation (4), , Substitute the value of VH from eq (7) in the above equation, , RH  B, , J, 1, d, d ne J B, ,  RH , , 1, ne, , This equation represents the Hall coefficient., , Applications of Hall Effect:, Hall Effect is a very useful phenomenon and helps to, , Determine the Type of Semiconductor:, By knowing the direction of the Hall Voltage, one can determine that the given sample is, whether n-type semiconductor or p-type semiconductor. This is because Hall coefficient, is negative for n-type semiconductor while the same is positive in the case of p-type, semiconductor., , if RH  ve it is n-type semiconductor and if RH  ve it is p-type semiconductor., Calculate the Carrier Concentration:, The expressions for the carrier concentrations of electrons (n) and holes (p) in terms of, Hall coefficient are given by n , , 1, 1, and p , e RH, e RH, , Determine the Mobility (Hall Mobility):, Mobility expression for the electrons (μn) and the holes (μp), expressed in terms of Hall, , Vignana Bharathi Institute of Technology, , Page 22

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UNIT –II SEMICONDUCTOR PHYSICS, , coefficient is given by,, , Engineering physics, ,  n   n RH and  p   p RH, , Where, σn and σp represent the conductivity due to the electrons and the holes,, respectively., , Measure Magnetic Flux Density:, This equation can be readily deduced from the equation of Hall voltage and is given by, , VH , , VH w, BI 1 , , B, , , , RH I, w n e, , Further, there are many commercially available types of equipment based on the, principle of Hall effect including Hall-effect sensors and Hall-effect probes., , Formation a PN junction: A P-N junction is formed by joining P-type and N-type semiconductors to, gather in very close contact. The term junction refers to the boundary, interface where two regions of the semiconductor meet. In N- type the, “donor” ion is represented by plus sign because, after this impurity atom, donates an electron it becomes a positive ion and in P-type acceptor ion is, indicated by minus sign because, after this atom accepts an electron, it, becomes a negative ion., , Vignana Bharathi Institute of Technology, , Page 23

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , Space-charge region: There is a large concentration of holes in the P-type material and a large, concentration of electrons in the N-type material. Because there is density, gradient across the junction, holes will initially diffuse to the right across the, junction, and electron to left. The flow of holes from P-region results a, negatively charge acceptor ions and flow of electrons from N-region leaves, positively charged donor ions.,  The unneutralized ions in the neighborhood of the junction are referred to as, “uncovered charges”. Since the region of the junction is depleted of mobile, charges, it is called depletion region, the space-charge region or transition, region. The thickness of this region is of the order of the wavelength of visible, light ≈ 0.5 micron., , Energy diagram of the PN junction: The valence and conduction bands in an n-type material are at slightly lower, energy levels than the valence and conduction bands in a p-type material., This is due to differences in the atomic characteristics of the penta valent, and the trivalent impurity atoms., , An energy diagram for a p-n junction at the instant of formation is shown in, Figure. As you can see, the valence and conduction bands in the n region are at, lower energy levels than those in the p region, but there is a significant amount of, , Vignana Bharathi Institute of Technology, , Page 24

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , overlapping., The free electrons in the n region that occupy the upper part of the conduction, band in terms of their energy can easily diffuse across the junction (they do not, have to gain additional energy) and temporarily become free electrons in the lower, part of the p-region conduction band. After crossing the junction, the electrons, quickly lose energy and fall into the holes in the p-region valence band as indicated, in above Figure., As the diffusion continues, the depletion region begins to form and the energy level of, the n- region conduction band decreases. The decrease in the energy level of the, conduction band in the II region is due to the loss of the higher-energy electrons that, have diffused across the junction to the p region. Soon, there are no electrons left in, the n-region conduction band with enough energy to get across the junction to the pregion conduction band., , At this point, the junction is at equilibrium; and the depletion region is complete, because diffusion has ceased. There is an energy gradient across the depletion region, which acts as an "energy hill" that an n-region electron must climb to get to the p, region., , Vignana Bharathi Institute of Technology, , Page 25

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , I-V Characteristics of PN Junction:, The V-I characteristics or voltage-current characteristics of the p-n junction diode is, shown in the graph. X-axis represents the voltage applied across the p-n junction diode, whereas Y-axis represents the current flows in the p-n junction diode., , Forward Bias:, If the positive terminal of the battery is connected to the p-type semiconductor and the, negative terminal of the battery is connected to the n-type semiconductor, the diode is, said to be in forward bias. In forward biased p-n junction diode, VF represents the, forward voltage whereas IF represents the forward current., , When the external voltage applied on the diode, then it starts allowing large electric, current through it. At this point, a small increase in voltage increases the electric current, rapidly., The forward voltage at which the diode starts allowing large electric current is called, cut-in voltage., , Reverse biasing:If the negative terminal of the battery is connected to the p-type semiconductor and the, positive terminal of the battery is connected to the n-type semiconductor, the diode is, said to be in reverse bias., In reverse biased p-n junction diode, VR represents the reverse voltage whereas IR, represents the reverse current., Vignana Bharathi Institute of Technology, , Page 26

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , If the external reverse voltage applied on the p-n junction diode is increased, the free, electrons from the n-type semiconductor and the holes from the p-type semiconductor, are moved away from the p-n junction. This increases the width of depletion region., Then it does not allow the majority charge carriers but it allows the minority charge, carrier current., , The free electrons (minority carriers) in the p-type semiconductor and the holes, (minority carriers) in the n-type semiconductor carry the electric current. The electric, current, which is carried by the minority charge carriers in the p-n junction diode, is, called reverse current., In n-type and p-type semiconductors, very small number of minority charge carriers is, present. Hence, a small voltage applied on the diode pushes all the minority carriers, towards the junction. Thus, further increase in the external voltage does not increase, the electric current. This electric current is called reverse saturation current., Vignana Bharathi Institute of Technology, , Page 27

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , Zener Diode:, In normal PN junction diode, when the reverse voltage reaches the break down voltage,, the current and the power dissipation through the junction will be high. In such case the, diode gets damage. Hence the diodes can be designed with adequate capability to, operate in breakdown region. Such diodes are known as Zener diode., The schematic symbol for the Zener diode is shown as below., , Zener diode is heavily doped than the ordinary diodes., Zener diode is always operated in reverse bias., The operation of the Zener diode is same as that of ordinary PN diode under forward, biased condition. But under reverse biased condition breakdown of junction occurs., The breakdown voltage depends upon the amount of doping. If the Zener diode is, heavily doped the depletion region will be thin and consequently breakdown occurs at, lower reverse voltage and further the breakdown voltage id sharp., , V-I Characteristics of the Zener Diode, Vignana Bharathi Institute of Technology, , Page 28

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , The sharp increase in the current under breakdown conditions are due to two, mechanisms., 1. Avalanche breakdown, 2. Zener breakdown, , Avalanche Breakdown:,  As the applied reverse bias is increases, the field across the junction increases, correspondingly.,  As a results thermally generated charge carriers velocity is also increases.,  These fast moving charges colliding with immovable ions and hence they rupture, the covalent bonds and they generate electron-hole pairs.,  This process is cumulative in nature and results generation of avalanche of charge, carriers with in a short time.,  This mechanism is known as avalanche multiplication and the diode is said to be, in the region of avalanche breakdown., , Zener breakdown:,  When P, N regions are heavily doped, direct rupture of covalent bonds takes, place because of the strong electric fields at the junction of PN diode.,  Therefore the new electron–hole pairs so created increases the reverse current in, reverse biased PN diode.,  This process of charge creation is known as Zener breakdown.,  The reverse bias typically below 6V for heavily doped diode and hence the, depletion region is very small.,  Zener breakdown takes place at very high fields across the depletion region, become very high of the order 107 V/m and for an applied field 6V or less.,  This is the condition for the Zener breakdown.,  For heavily doped Zener diodes Zener breakdown voltage become high. Though, Zener breakdown occurs at lower breakdown voltage and avalanche breakdown, , Vignana Bharathi Institute of Technology, , Page 29

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , occurs for higher breakdown.,  Zener diodes can be used as a voltage regulators., , Bipolar Junction Transistors:, In bipolar junction transistor the current is due to majority and minority charge, carriers. It is of two types., 1. NPN transistor, 2. PNP transistor., It is three terminal device with two junctions. The three terminals are emitter,, base and collector., , Emitter: Emitter is heavily doped and has moderate region. It supplies or provides, the majority carriers i.e holes in P-region and electrons in N- region., Base: it is always thin (narrower) and lightly doped. Only recombination current, flow through it with the order of  A . It controls the majority charge carriers., Collector: it has got wider region (high region) and is moderately doped. It collects the, majority charge carriers from the emitter after recombination in Base region., Schematic Representation:, , Vignana Bharathi Institute of Technology, , Page 30

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , 1. Always emitter is indicated by an arrowhead, 2. It always shows direction of the conventional current flow i.e P to N, 3. The arrowhead shows the type of the transistor., 4. It always shows the emitter location., Formation of Depletion layer:, P-N-P Transistor:, When no voltage is applied to the transistor then it is said to be unbiased. The, charge carriers always try to move from higher concentrate region to lower, concentrate region. Therefore, the holes from the p-region (both Collector and, Emitter) will be move towards n-region (Base), During this process, hole will meet the free electrons in n-region and recombine, with each other., As a result depletion region (consisting of +ve and –ve ions) is formed at the EB, Junction and BC Junction., , At the EB junction the depletion region is penetrated more towards Base side and, similarly at the BC Junction the depletion layer is penetrated more towards Base, side., This is because at emitter to junction the emitter is heavily doped and the base is, lightly doped. Similarly at the Base Collector junction the collector is heavily, doped and the base is lightly doped., , Vignana Bharathi Institute of Technology, , Page 31

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , N-P-N Transistor:, When no voltage is applied to the transistor then it is said to be unbiased. The, charge carriers always try to move from higher concentrate region to lower, concentrate region. Therefore, the electrons from the n-region (both Collector, and Emitter) will be move towards p-region (Base), During this process, free electrons will meet the holes in p-region and recombine, with each other., As a result depletion region (consisting of +ve and –ve ions) is formed at the EB, Junction and BC Junction., , At the EB junction the depletion region is penetrated more towards Base side and, similarly at the BC Junction the depletion layer is penetrated more towards Base, side., This is because at emitter to junction the emitter is heavily doped and the base is, lightly doped. Similarly at the Base Collector junction the collector is heavily, doped and the base is lightly doped., , Vignana Bharathi Institute of Technology, , Page 32

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , Operation of BJT:, P-N-P transistor:, One of the P-region in P-N-P transistor is connected in forward bias at EB Junction, and another is in connected in reverse bias at CB Junction., From emitter (p region) a lot of holes crossover to the base region (n region)., As base lightly doped with n-type, hence few of holes only recombine with, electrons and causes IB current is very small., , 95% of remaining holes cross over into the collector region to constitute a, collector current IC., ,  I E  I B  IC, N-P-N transistor:, One of the N-region in N-P-N transistor is connected in forward bias at EB, Junction and another is in connected in reverse bias at CB Junction. From emitter, (N- region) a lot of electrons crossover to the base region (P-region)., As base lightly doped with P-type, hence few of electrons only recombine with, holes and causes IB current which is very small., , Vignana Bharathi Institute of Technology, , Page 33

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UNIT –II SEMICONDUCTOR PHYSICS, , Engineering physics, , 95% of remaining electrons cross over into the collector region to constitute a, collector current IC., ,  I E  I B  IC, In BJT, the junction consisting of forward bias has low resistance due to narrow, junction, and the junction consisting of reverse bias has high resistance due to, wider junction., Hence in the transistors the transfer of the resistance is takes place., In BJT, the current flow through the junctions is due to both majority and minority, charge carriers, hence it is known as bipolar transistors., , Vignana Bharathi Institute of Technology, , Page 34