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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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