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					2.2 EXTRINSIC SEMICONDUCTORS

    Brief Summary of Previous Lecture & Intro’

Discussed basic ideas of semiconducting behaviour.

Energy, E
                                                Conduction
                                                Band (CB)

                                                Energy Gap
                                                    Eg


                                                Valence
                                                Band (VB)

nc – conc’n of electrons in conduction band
pv – conc’n of holes in valence band


Discussed how we could calculate nc and pv

             (E  E          )                (E  E          )
nc = Nc exp     c       F
                                  pv = Nv exp     F       V
                                                                  
                kT                              kT             
                      3                                  3

           m kT                             m kT 
              *       2                         *        2

    Nc = 2   e
                                      Nv = 2  h
                                                    
           2                              2 
                  2                                  2




Intrinsic (pure) semiconductor:            nc = pv
Typical values of conductivity  for intrinsic
semiconductors (at room temperature):

          Si:  =  10-4 -1m-1
          Ge:  =  1 -1m-1

Often measure much higher values than this.
Why?     small amounts of impurities.



Deliberate addition of impurities in a controlled way
allows tailoring of charge concentration (and hence )
to desired values.

Process is referred to as doping – essential for device
fabrication.

Doped semiconductors referred to as extrinsic.

Most important impurities (dopants) are from:

          Group V in Periodic Table e.g. P, As

          Group III in Periodic Table e.g. B, Al


[Si, Ge etc. – Group IV]
QUALITATIVE PICTURE OF DOPING

1.   Intrinsic Semiconductor e.g.Si


[2-D representation]

each Si atom has 4 nearest neighbours

strong, covalent bonds – tetrahedral
                         bonding in 3-D


2.   Doping with Group V Impurity e.g. P


P atom has 5 available electrons to form
bonds with neighbouring atoms                       e-

4 go to form bonds with neighbouring            P
Si atoms

5th electron is unpaired and
weakly bound to P.


Thermal excitation is usually enough
to break bond between P and unpaired
electron.                                  e-

This leaves an ionised P (donor atom)           P+
bonded to the Si network, and a free
(donor) electron which can carry charge
Can represent process on energy level diagram:


conduction
band                                      c
                                          d


valence                                   v
band



At T = 0, all donor electrons are weakly bound to P
atoms – represented by level d in energy diagram.

Typically       c – d =  0.01 – 0.1 eV
This is small or comparable with thermal energies
(kT =  0.025 eV at room temperature).

So at room temperature, most of the donor atoms have
given up their excess electron (most excited from E d into
the conduction band)  large increase in nc
(concentration of electrons in conduction band)
 large increase in conductivity .

Semiconductors doped with donors – referred to as
n-type.

Typical doping levels – 1 part in 10 6.
3.   Doping with Group III Impurity e.g. B




B has 3 available electrons to form
with neighbouring atoms.

B adopts tetrahedral bonding but with
broken bond. [In fact, broken bond shared             B
between all 4 bonds to neighbouring Si’s]
Can regard as tetrahedrally bonded B
binding a hole.




B can capture (accept) an electron from
a Si-Si bond elsewhere.

This leaves a negatively ionised B
atom (acceptor), and a broken Si-Si                   B-
bond (mobile or free hole).



Can think of B as donating a free hole.




Process occurs easily at thermal energies (room temp’).
Energy level diagram at T = 0


                                         Ec


                                         Ea
                                         Ev



Energy level diagram at Room Temp’.



                                         Ec


                                         Ea

                                         Ev



Thermal energy large compared with E a – Ev.

Most holes excited down into valence band.
[Alternatively: large numbers of electrons excited from
valence band up into levels at Ea – leaves large numbers
of holes in valence band].

Large increase in pv.   Large increase in .

Acceptor–doped semiconductors – referred to as p-type.
EXTRINSIC SEMICONDUCTORS - SUMMARY SO FAR

n-Type

Doped with Group V donor impurities e.g. P, As

If ionised, impurities donate extra electron (into
conduction band)

Nearly all donors (positively) ionised at Room
Temperature.

Hence:
          large increase in nc at Room Temp.
          large increase in  at Room Temp.


p-Type

Doped with Group III acceptor impurities e.g. B

If ionised, impurities accept extra electron (from valence
band) – leaves extra hole in valence band.

Nearly all acceptors (negatively) ionised at Room Temp.

Hence:
          large increase in pv at Room Temp.
          large increase in  at Room Temp.
QUASI-BOHR MODEL FOR DONORS

Can estimate binding energy for donor electrons in n-
type material using a quasi-Bohr model.

Recall Bohr model for H atom i.e. electron orbitting
positively charged H nucleus (proton). [PA114].


Apply model to n-type semiconductor (doped with P)

     Weakly bound unpaired electron orbits singly
     charged P nucleus.



                                   r               e-
                             +
                         P
                                                        

Electron circles P + under action of central Coulomb force
                                       2
                                   e
                m r 
                     *   2
                                                             (1)
                                 4 r    0
                                               2




m* - effective mass of electron
 - dielectric constant of semiconductor


Bohr postulate:     angular momentum quantised
                         in units of 
               m r   n
                      *   2
                                 [n – integer]                                                             (2)
Eliminate  between eq’ns (1) and (2) to give

                   4 n                        2    2

               r                    2
                                          0
                                              *
                                                                                                           (3)
                     em

Kinetic energy of orbiting electron T given by
                                                           4       *
             1              em
          T  m r  
                  *           2


                        24  n 
                                                                   2           2           2
             2                                                 0




Potential energy U is given by

                          e      e m
                                  2                                    4               *

              U            
                   4 r     4  n 
                                  0                                    0
                                                                               2           2       2



Hence total energy En is given by

                                                           e m        4               *

               En = U + T 
                                                      24               n     2           2       2
                                                                       0




So orbiting electron can occupy a series of levels just
below conduction band edge c.

     n=1       lowest bound state – ground state

     n       En  0 i.e. impurity atom ionised,
               electron goes into conduction band

So binding energy Eb of donor electron is given by

                                e m                               4               *

               Eb = E1 - E =
                              24  
                                                                                   2           2
                                                                           0
Example

For Si:                   = 12

Hence (using) m* = m      Eb = - 0.09 eV

                              r=6Å

          i.e. electron is weakly bound


[Can do similar calculation for hole orbiting around
acceptor in p-type semiconductor]
ELECTRON AND HOLE CONCENTRATIONS – NC AND P V
          - IN DOPED SEMICONDUTORS

Doping  large increase in nc (or p v)  large increase in
conductivity 

Can we calculate nc and pv for a given amount of doping?

n-type semiconductor

Si sample doped with concentration Nd of P donors.

Assumption:
               complete ionisation of donors

Can determine nc and pv using:

          (1) charge neutrality
          (2) law of mass action

Charge neutrality

     Positively charged: ionised donors
                         holes in valence band
     Negatively charged: electrons in conduction band

     So must have: Nd + pv = nc                 (1)

Law of mass action
                     ncpv = ni2                 (2)

     [true for intrinsic and extrinsic semiconductors]
Solve equations (1) and (2) for nc and pv.

              nc2 – Ndnc – n i2 = 0

                    N  N  4n   2       2

              nc =    d         d       i
                                                         (3)
                        2

[Take positive root – negative one gives negative nc!]


Hence from eq’n (2)            pv = n i2/nc


So from eq’n (3), if n i << Nd

                                     nc  Nd

and also (from eq’n (1))
                                     pv << nc

                           i.e. can neglect pv


p-type semiconductor

Similarly analysis for p-type semiconductor with
concentration Na of acceptors gives

                     pv  Na

                     nc << pv
               i.e. can neglect nc
EXAMPLES!

Question 1: A Si sample is doped with 10-4 atomic% of P
           donors. Assuming complete ionisation of
           donors at room temperature, calculate the
           charge carrier concentration and conductivity
           at room temperature.
[For Si:  = 2330 kg m-3, atomic weight = 28, e = 0.15
m2V-1s-1]

Answer 1
                       N
                            10
                            d        6


                       N    Si

where NSi – number of Si atoms per unit volume

Obtain NSi from
                                                 Avagadro’s No.
                                 
                    N                     6  10   23


                           28  10   3
                      Si




This gives          NSi = 5 × 1028 m-3

So                   Nd = 5 × 1022 m-3.

Complete ionisation, n-type semiconductor:

So charge carrier concentration is

                    nc = Nd = 5 × 1022 m-3

                    (neglect pv)
Conductivity                   = ncee
[hole contribution negligible]

So  = 5 × 1022  1.6  10-19  0.15 -1m-1
            = 1200 -1m-1

Question 2: What are the carrier concentrations and
              Conductivity in intrinsic Si?
[For Si: g = 1.1 eV, me* = 0.25me, mh* = 0.5me, e = 0.15
m2V-1s-1, h = 0.05 m2V-1s-1]

Answer 2

Intrinsic          nc = pv = ni
                                             3

                                  kT        
                    ni = 2m m  
                                     3       2

                                       exp     
                             *   *                g
Obtain n i from:                     4

                                         2kT 
                             e   h       2




For Si:        g = 1.1 eV, me * = 0.25me, mh* = 0.5me

So at room temperature (T = 300K):

                    ni = 9.8 × 1015 m-3

Conductivity    = ncee + pveh
            = niee + nieh
               = (2.3  10-4) + (7.8  10-5) -1m-1
               = 3.1  10-4 -1m-1

Comparing answers for Q1 and Q2:
doping Si with 1 part in 106 (of P) has led to an increase
in  of a factor > than 106.
 CONDUCTIVITY OF EXTRINSIC SEMICONDUCTORS
       AS FUNCTION OF TEMPERATURE T


Summary so far

Assuming complete ionisation of donor or acceptor
impurities:

     doping semiconductors can lead to large increases
     in 

     can estimate carrier concentrations and  under
     these conditions


Questions

How does  vary with temperature T in doped
semiconductors?

Does position of Fermi level F change with doping?
Qualitative Picture

n-type semiconductor


 E
                                                   Ec

                                                   Ed

     Eg


                                                   Ev




T = 0K – Insulating!
Valence band full.
Donor electrons all in bound states – energy Ed.
No electrons in conduction band.
So E F must lie between Ed and Ec.

1. As T is raised, weakly bound donor electrons are
promoted into the conduction band –  rises

2. Eventually, nearly all donor electrons have been
excited into conduction band. However, T still not high
enough for excitation across energy gap E g –  levels off.
3. At higher T still, start to get excitation of electrons
across Eg – intrinsic behaviour.  rises rapidly.
Graph of ln vs T-1 for n-type semiconductor


     ln
                                       E
                        slope =           [See end Lecture 1]
                                          g


                                       2k



                                                               E  E        
                                                    slope =       c       d

                                                                  2k
                           2
                                                                      1


                                              T-1

Analysis in Tanner (pages 127-130) shows:

In region 1 (low T)
                                (E  E ) 
                  nc = ZN         
                                   1

                           exp   2
                                          
                                                    c   d


                                   2kT 
                               d



[nc – conc’n of electrons in conduction band]

                      1           1     N 
                  F = (Ec + d) + kTln                     d

                      2           2      Z 

So for an extrinsic n-type semiconductor, F lies between
c and d, and falls with increasing T. [Nd << Z]
[Nd – donor concentration]
[Z – (average) density of states in conduction band]
p-type semiconductor

Similar results for extrinsic p-type semiconductor.

     

                                             c



                                             a
                                             v


Concentration of acceptors Na



For an extrinsic p-type semiconductor, F lies between a
and v, and rises linearly with increasing temperature.



Hole concentration pv (at low T) given by


                                     (E  E ) 
               pv = ZN 
                            1
                            2
                                exp   a    v
                                               
                                       2kT 
                        a

				
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