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					Semiconductor
   Physics
                   Introduction

• Semiconductors are materials whose electronic properties
  are intermediate between those of Metals and Insulators.

• They have conductivities in the range of 10 -4 to 10 +4S/m.

• The interesting feature about semiconductors is that they
  are bipolar and current is transported by two charge
  carriers of opposite sign.

• These intermediate properties are determined by
  1.Crystal Structure bonding Characteristics.
  2.Electronic Energy bands.
• Silicon and Germanium are elemental semiconductors
  and they have four valence electrons which are distributed
  among the outermost S and p orbital's.

• These outer most S and p orbital's of Semiconductors
  involve in Sp3 hybridanisation.

• These Sp3 orbital's form four covalent bonds of equal
  angular separation leading to a tetrahedral arrangement of
  atoms in space results tetrahedron shape, resulting crystal
  structure is known as Diamond cubic crystal structure
Semiconductors are mainly two types

1. Intrinsic (Pure) Semiconductors
2. Extrinsic (Impure) Semiconductors
           Intrinsic Semiconductor

• A Semiconductor which does not have any kind of
 impurities, behaves as an Insulator at 0k and
 behaves as a Conductor at higher temperature is
 known as Intrinsic Semiconductor or Pure
 Semiconductors.

• Germanium and Silicon (4th group elements) are
 the best examples of intrinsic semiconductors and
 they possess diamond cubic crystalline structure.
       Intrinsic Semiconductor
                          Valence Cell


Covalent bonds
                     Si




             Si      Si          Si




                     Si
                Conduction band

                                           KE of
                                   Ec      Electron
                                           = E - Ec
           Ec
    E
           Ef
Electron          Fermi energy level
energy

           Ev
                                           KE of Hole
                                           =
                            Valence band
                                           Ev - E


                    Distance
 Carrier Concentration in Intrinsic Semiconductor

When a suitable form of Energy is supplied to a
Semiconductor then electrons take transition from Valence
band to Conduction band.

Hence a free electron in Conduction band and
simultaneously free hole in Valence band is formed. This
phenomenon is known as Electron - Hole pair generation.


In Intrinsic Semiconductor the Number of Conduction
electrons will be equal to the Number of Vacant sites or
holes in the valence band.
         Calculation of Density of Electrons
Let ‘dn’ be the Number of Electrons available between
energy interval ‘E and E+ dE’ in the Conduction band

         dn  Z ( E )dE F ( E )
              top of the band

         n          z ( E ) F ( E )dE.......... 1)
                    Ec
                                               ......(


Where Z(E) dE is the Density of states in the energy
interval E and E + dE and F(E) is the Probability of
Electron occupancy.
We know that the density of states i.e., the number of energy states
per unit volume within the energy interval E and E + dE is given by


                               4
                                       3   1
                    Z ( E )dE  3 (2m) 2 E 2 dE
                               h

                             4
                                       3   1
                                   
                  Z ( E )dE  3 (2me ) 2 E 2 dE
                             h

     Since the E starts at the bottom of the Conduction band E c

                          4
                                    3            1
                                
               Z ( E )dE  3 (2me ) 2 ( E  Ec ) 2 dE
                          h
Probability of an Electron occupying an energy state E is
given by

                       1
      F (E) 
                        E  Ef
               1  exp(        )
                          kT
      For all possible temperatures E  EF  kT
                        1
      F (E) 
                       E  Ef
                exp(            )
                       kT
                       E  EF          EF  E
      F ( E )  exp (        )  exp(        )
                         kT             kT
Substitute Z(E) and F(E) values in Equation (1)

       top of the band

  n          z ( E ) F ( E )dE
             Ec
       
         4      3            1
                                    E E
  n   3 (2me ) 2 ( E  Ec ) 2 exp( F
             
                                         )dE
      Ec
         h                           kT

                             
     4       3
  n  3 (2me )  ( E  Ec ) 2 exp( E F  E )dE
                            1
            2

     h          Ec                   kT

                                   
     4                                       E
                         3             1
                     E
  n  3 (2me ) 2 exp( F )  ( E  Ec ) 2 exp(
           
                                                 )dE.....(2)
     h               kT Ec                    kT
       To solve equation 2, let us put


E  Ec  x
E  Ec  x
dE  dx
                         
   4                                    E
              3                    1
                  EF
          2
n  3 (2me ) exp(    )  ( E  Ec ) exp(
                                   2
                                            )dE
   h              kT 0                   kT
                        
   4        3
                    EF          1
                                        E x
         
n  3 (2me ) 2 exp(    )  ( x) 2 exp ( c   )dx
   h                kT 0                 kT
                             
   4             EF  Ec
            3                     1
                                           x
          2
n  3 (2me ) exp(         )  ( x) exp (
                                  2
                                                     3
                                             )dx.....( )
   h                kT      0
                                          kT
                                                1
                     1
                         x                3   2
we know that  ( x) exp(
                      2
                            )dE  (kT )     2

             0
                         kT              2

substitute in equation (3)

                                                1
   4             E F  Ec            
            3                       3           2
           2
n  3 (2me ) exp(          ) {(kT ) 2
                                         }
   h                 kT                2
      2me kT 3
          
                       E F  Ec
n2 (     2
               2
              ) exp(            )
        h                 kT

The above equation represents
Number of electrons per unit volume of the Material
               Calculation of density of holes

  Let ‘dp’ be the Number of holes or Vacancies in the
  energy interval ‘E and E + dE’ in the valence band

           dp  Z ( E )dE {1  F ( E )}
                        Ev
            p            z ( E ){1  F ( E )}dE.......... 1)
                 bottom of the band
                                                         ......(


Where Z(E) dE is the density of states in the energy interval
E and E + dE and
1-F(E) is the probability of existence of a hole.
      Density of holes in the Valence band is


             4
                                        3       1
                    
 Z ( E ) dE  3 ( 2mh ) E dE            2       2
             h
Since Ev is the energy of the top of the valence band



            4                 3
                              2
                                                    1
 Z ( E )dE  3 (2m ) ( Ev  E ) dE
                             h
                                                    2
            h
Probability of an Electron occupying an energy state E is
given by
                            1
 1  F (E)  1  {                    }
                             E  Ef
                    1  exp(        )
                               kT
                             E  E f 1
 1  F ( E )  1  {1  exp(        )}
                               kT
 neglect higher order terms in above exp ansion
 f or higher T values
                      E  Ef
 1  F ( E )  exp(            )
                       kT
Substitute Z(E) and 1 - F(E) values in Equation (1)

            Ev
p            z( E ){1  F ( E )}dE
     bottom of the band

          4                        E  EF
     Ev                  3       1

      
                        2
p          3
              (2m ) ( Ev  E ) exp(
                       h
                                 2
                                           )dE
     
          h                           kT

   4             EF
                   3             Ev    1
                                         E
p  3 (2m ) exp(  2
                 h    )  ( Ev  E ) exp( )dE....(2)
                                       2
   h              kT                   kT
             To solve equation 2, let us put

Ev  E  x
E  Ev  x
dE   dx
   4             EF
                3              Ev               1
                                         E
               2
p  3 (2m ) exp(
              h       )  ( Ev  E ) exp( )dE   2
   h              kT                   kT

   4             EF               Ev  x
                3              0        1
               2
p  3 (2m ) exp(
              h       )  ( x) exp(     2
                                           )( dx)
   h              kT                kT
                                    
   4            Ev  E F              x
                3                           1
               2
p  3 (2m ) exp(
              h           )  ( x) exp( )dx 2
   h               kT      0
                                       kT
                                            1
     4            Ev  E F        
                    3                   3   2
                   2
  p  3 (2m ) exp(h         )(kT )      2
     h               kT             2

         2m kT
              
                       Ev  E F
                        3
  p  2(      h
              2
                ) exp(  2
                                )
           h             kT

The above equation represents
Number of holes per unit volume of the Material
Intrinsic Carrier Concentration
In intrinsic Semiconductors n = p
Hence n = p = n i is called intrinsic Carrier Concentration

ni2  np
ni  np
          2me kT 3
              
                        E F  Ec       2mh kT 3
                                           
                                                     Ev  E F
ni  {2 (     2
                  2
                 ) exp(          )}{2(     2
                                               2
                                              ) exp(          )}
            h              kT            h             kT
       2kT               E v  Ec
             3       3
                   4
ni  2( 2 ) (me m ) exp(
             2
                   h               )
        h                   2kT
       2kT 2   4
            3       3
                           Eg
ni  2( 2 ) (me mh ) exp(       )
        h                 2kT
    Fermi level in intrinsic Semiconductors
In intrinsic semiconductors n  p
   2me kT 3
        
                    E F  Ec         2mh kT 3
                                           
                                                       Ev  E F
2(      2
              2
            ) exp(           )  2(        2
                                                 2
                                                ) exp(          )
      h                kT                h               kT
  2me kT 3
      
                   E F  Ec      2mh kT 3
                                       
                                                    Ev  E F
(     2
            2
          ) exp(            )(        2
                                              2
                                             ) exp(          )
    h                 kT             h                 kT
                 
     2 EF       mh 3        Ev  E c
exp(      )  (  ) exp(
                    2
                                     )
      kT        me            kT
taking logarithms on both sides
                 Conduction band


                          Ec

           Ec
    E
                                   mh  me
                                    *    *

           Ef
Electron
energy

           Ev

                    Valence band




                Temperature
               
    2 EF 3    mh       Ev  Ec
         log(  )  (         )
     kT  2    me         kT
                  
         3kT     mh 3      E  Ec
    EF      log(  ) 2  ( v     )
          4      me           2
                                                 
    In intrinsic semiconductor w eknow that me  mh
           Ev  Ec
    EF  (         )
              2


Thus the Fermi energy level EF is located in the
middle of the forbidden band.
      Extrinsic Semiconductors

• The Extrinsic Semiconductors are those in which
  impurities of large quantity are present. Usually,
  the impurities can be either 3rd group elements or
  5th group elements.

• Based on the impurities present in the Extrinsic
  Semiconductors, they are classified into two
  categories.
            1. N-type semiconductors
            2. P-type semiconductors
          N - type Semiconductors


When any pentavalent element such as Phosphorous,
Arsenic or Antimony is added to the intrinsic
Semiconductor , four electrons are involved in covalent
bonding with four neighboring pure Semiconductor
atoms.

The fifth electron is weakly bound to the parent atom.
And even for lesser thermal energy it is released Leaving
the parent atom positively ionized.
N-type Semiconductor

                Free electron

        Si




Si      P        Si




         Si       Impure atom
                    (Donor)
The Intrinsic Semiconductors doped with pentavalent
impurities are called N-type Semiconductors.

The energy level of fifth electron is called donor level.

The donor level is close to the bottom of the conduction
band most of the donor level electrons are excited in to the
conduction band at room temperature and become the
Majority charge carriers.

Hence in N-type Semiconductors electrons are Majority
carriers and holes are Minority carriers.
                Conduction band

                                Ec

           Ec
    E                                  Ed
                        Donor levels
                                            Eg
Electron
energy

           Ev

                 Valence band



                    Distance
  Carrier Concentration in N-type Semiconductor

• Consider Nd is the donor Concentration i.e., the number
  of donor atoms per unit volume of the material and Ed is
  the donor energy level.

• At very low temperatures all donor levels are filled with
  electrons.

• With increase of temperature more and more donor
  atoms get ionized and the density of electrons in the
  conduction band increases.
Density of electrons in conduction band is given by
                   2me kT 3
                       
                                 EF  Ec
            n  2(     2
                           2
                          ) exp(         )
                     h             kT

      The density of Ionized donors is given by

                                               Ed  E F
               N d {1  F ( Ed )}  N d exp(            )
                                                 kT

At very low temperatures, the Number of electrons in the
conduction band must be equal to the Number of ionized
donors.

           2me kT 3
               
                         EF  Ec              Ed  E F
        2(     2
                   2
                  ) exp(         )  N d exp(          )
             h             kT                   kT
         Taking logarithm and rearranging we get

        E F  Ec      Ed  E F                       2me kT 3
                                                        
      (           )(           )  log N d  log 2(        )2
           kT             kT                           h2
                                          Nd
      2 E F  ( Ed  Ec )  kT log
                                        2me kT 2
                                                  3
                                     2(      2
                                                )
                                          h
             ( Ed  Ec ) kT               Nd
      EF                     log
                                       2me kT 2
                                                 3
                   2         2
                                    2(      2
                                                )
                                          h
      at.,0k
             ( E d  Ec )
      EF 
                   2
At 0k Fermi level lies exactly at the middle of the donor level
and the bottom of the Conduction band
   Density of electrons in the conduction band
       2me k T 3
            
                        E  Ec
n  2(      2
                ) 2 exp( F     )
          h                kT
                        ( E  Ec ) k T          Nd
                       { d            log                }  Ec
                                              2me k T 2
                                                       3
                             2      2
                                           2(         )
     E F  Ec                                   h 2
exp(          )  exp{                                           }
        kT                               kT
                                                          1
       E F  Ec         ( E  Ec )                (Nd )   2
                                                               E
exp(            )  exp{ d          log                       c}
                                                 2me k T 2 2
                                                         3 1
          kT               2k T                                kT
                                            [ 2(         ) ]
                                                   h2
                                                          1
       E F  Ec         ( E  Ec )                (Nd )   2
exp(            )  exp{ d          log                      }
                                                 2me k T 2 2
                                                         3 1
          kT               2k T
                                            [ 2(     2
                                                         ) ]
                                                   h
                                1
       E F  Ec            (Nd )2
                                           ( E d  Ec )
exp(            )                     exp
                          2me k T 2 2
                                  3 1
          kT                                   2k T
                     [ 2(         ) ]
                            h2
        2me kT 3
            
                      E F  Ec
 n  2(     2
                2
               ) exp(          )
          h              kT
                                   1
        2m kT
                   3
                              (Nd )2
                                                ( E d  Ec )
 n  2(      e      2
                  ) {                      exp               }
                            2me kT 2 2
                                 
             2                         3 1
            h                                       2kT
                       [ 2(      2
                                      ) ]
                               h
             1
                2me kT 3
                     
                                   ( E d  Ec )
 n  2( N d ) (
             2
                     2
                             4
                            ) exp
                   h                   2kT

Thus we find that the density of electrons in the conduction
band is proportional to the square root of the donor
concentration at moderately low temperatures.
      Variation of Fermi level with temperature

To start with ,with increase of temperature Ef increases
slightly.

As the temperature is increased more and more donor atoms
are ionized.

Further increase in temperature results in generation of
Electron - hole pairs due to breading of covalent bonds and
the material tends to behave in intrinsic manner.

The Fermi level gradually moves towards the intrinsic Fermi
level Ei.
             P-type semiconductors

• When a trivalent elements such as Al, Ga or Indium have
  three electrons in their outer most orbits , added to the
  intrinsic semiconductor all the three electrons of Indium are
  engaged in covalent bonding with the three neighboring Si
  atoms.

• Indium needs one more electron to complete its bond. this
  electron maybe supplied by Silicon , there by creating a
  vacant electron site or hole on the semiconductor atom.

• Indium accepts one extra electron, the energy level of this
  impurity atom is called acceptor level and this acceptor
  level lies just above the valence band.

• These type of trivalent impurities are called acceptor
  impurities and the semiconductors doped the acceptor
  impurities are called P-type semiconductors.
                         Hole
   Co-Valent
    bonds          Si




              Si   In     Si




                    Si

Impure atom
 (acceptor)
                Conduction band
                                Ec

           Ec
    E

Electron                                      Eg
energy                 Acceptor levels
                                         Ea
           Ev


                 Valence band


                temperature
• Even at relatively low temperatures, these
  acceptor atoms get ionized taking electrons
  from valence band and thus giving rise to holes
  in valence band for conduction.

• Due to ionization of acceptor atoms only holes
  and no electrons are created.

• Thus holes are more in number than electrons
  and hence holes are majority carriers and
  electros are minority carriers in P-type
  semiconductors.
• Equation of continuity:

• As we have seen already, when a bar of n-type
  germanium is illuminated on its one face, excess charge
  carriers are generated at the exposed surface.

• These charge carriers diffuse through out the material.
  Hence the carrier concentration in the body of the
  sample is a function of both time and distance.

• Let us now derive the differential equation which governs
  this fundamental relationship.

• Let us consider the infinitesimal volume element of area
  A and length dx as shown in figure.
• If tp is the mean lifetime of the holes, the holes lost
  per sec per unit volume by recombination is p/tp .
• The rate of loss of charge within the volume under
  consideration


                                       p
                                eAdx
                                      tp



                                                             eAdxg
  If g is the thermal rte of generation of hole-electron
  pairs per unit volume, rate of increase of charge wthin
  the volume under consideration
• If i is the current entering
  the volume at x and i + di
  the current leaving the
  volume at x + dx, then
  decrease of charge per
  second from the volume
  under consideration = di                                     dp
                                                       eAdx
• Because of the above                                         dt p

  stated three effects the
  hole density changes with
  time.
• Increase in the number of
  charges per second
  within the volume

                    Increase = generation - loss

                           dp                  p
                    eAdx         eAdxg  eAdx  dI
                           dt p               tp
Since the hole current is the sum of the diffusion current and the drift current

                                 dp
                  I   AeDp         Ape h E
                                 dx


Where E is the electric field intensity within the volume. when no external
field is applied, under thermal equilibrium condition, the hole density
attains a constant value p0.

                                         dp
       under these conditions di  0 and    0
                                         dt
           p0
       g
           tp
       this equation indicates that the rate of generation of
       holes is equal to therate of loss due to recombination
       under equilibriu m conditions.
             combain,.eq s ...3,4 & 5.
             dp    ( p  p0 )     2 p    d ( pE )
                             Dp 2  h
             dt        tp         x         dx

This is called equation of conservation of charge or the continuity equation.


           if p is a function of both t and x,
           partial derivatives should be used.
           p    ( p  p0 )      2 p       ( pE )
                           Dp       h
           t        tp          x 2         x
           if w e are considering holes in the n - type material
           pn    ( p n  p0 n )       2 pn       ( pn E )
                                Dp         h
            t          tp             x 2           x
           if w e are considering electrons in the p - type material
           n p        (n p  n0 p )           2n p           (n p E )
                                      Dn             e
            t              te                x 2               x
Direct band gap and indirect band gap semiconductors:

• We known that the energy spectrum of an electron moving
  in the presence of periodic potential field is divided into
  allowed and forbidden zones.

• In crystals the inter atomic distances and the internal
  potential energy distribution vary with direction of the crystal.
  Hence the E-k relationship and hence energy band
  formation depends on the orientation of the electron wave
  vector to the crystallographic axes.

• In few crystals like gallium arsenide, the maximum of the
  valence band occurs at the same value of k as the minimum
  of the conduction band as shown in below. this is called
  direct band gap semiconductor.
           E   Conduction
                                E     Conduction
               band
                                      band




               Eg
                                              Eg

                        k                          k

Valence band                Valence
                            band
• In few semiconductors like silicon the maximum of the
  valence band does not always occur at the same k value as
  the minimum of the conduction band as shown in figure.
  This we call indirect band gap semiconductor.

• In direct band gap semiconductors the direction of
  motion of an electron during a transition across the
  energy gap remains unchanged.

• Hence the efficiency of transition of charge carriers across
  the band gap is more in direct band gap than in indirect
  band gap semiconductors.
                         Hall effect

When a magnetic field is applied perpendicular to a current carrying
conductor or semiconductor, voltage is developed across the
specimen in a direction perpendicular to both the current and the
magnetic field. This phenomenon is called the Hall effect and voltage
so developed is called the Hall voltage.

Let us consider, a thin rectangular slab carrying current (i) in the x-
direction.
If we place it in a magnetic field B which is in the y-direction.
Potential difference Vpq will develop between the faces p and q which
are perpendicular to the z-direction.
                        Z




+
  VH           +   +   + P    +   +
-                                         Y


          + + +                           B
                  + + +
         + + +++ + + Q + +++
       +     +


       X
                             P – type semiconductor
           i
                             Z




-
                  _ _    _   _ P_     _
    VH
+                                          Y

                 _
                 _ _ _      Q_
                         _ _ _ _ __
                                           B
                    _


         X
                             N – type semiconductor
             i
                  Magnetic deflecting force

                       F  q ( vd  B )
                Hall eclectic deflecting force

                          F  qEH
When an equilibrium is reached, the magnetic deflecting force on
the charge carriers are balanced by the electric forces due to
electric Field.

                       q(vd  B)  qEH
                       E H  ( vd  B )
                       Where vd is drift velocity
 The relation between current density and drift velocity is

                              J
                         vd 
                              ne
Where n is the number of charge carriers per unit volume.

             E H  ( vd  B )
                    J
             EH  (  B)
                    ne
                    1
             E H  (  JB )
                    ne
             E H  RH  JB
                                         1   EH
             RH ( Hall,.coef f icien ) 
                                    t      
                                         ne JB
If VH be the Hall voltage in equilibrium ,the Hall electric field.

                  VH
              EH 
                   d
              Where d is the width of the slab.
                      EH
              RH 
                      JB
                       1 VH
              RH        
                      JB d
              If t is the thicknessof the sample,
              Then its crosssection is dt and current density
                  I
              J
                  dt
              VH  RH JBd
                       I
              VH  RH ( ) B
                       t
                   V t
              RH  H
                    IB
• Since all the three quantities EH , J and B
  are measurable, the Hall coefficient RH and
  hence the carrier density can be found out.

• Generally for N-type material since the Hall
  field is developed in negative direction
  compared to the field developed for a P-
  type material, negative sign is used while
  denoting hall coefficient RH.

				
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