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
nc = Nc exp
kT

c
 m kT 
Nc = 2

2




*
F
)


 (E  E
pv = Nv exp
kT

3
2
e
2
Intrinsic (pure) semiconductor:
F
 m kT 
Nv = 2

2




*
h
2
nc = pv
V
3
2
)


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
Si atoms
P
5th electron is unpaired and
weakly bound to P.
Thermal excitation is usually enough
to break bond between P and unpaired
electron.
This leaves an ionised P (donor atom)
bonded to the Si network, and a free
(donor) electron which can carry charge
eP+
Can represent process on energy level diagram:
conduction
band
valence
band
c
d
v
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 Ed 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 106.
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
between all 4 bonds to neighbouring Si’s]
Can regard as tetrahedrally bonded B
binding a hole.
B
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
bond (mobile or free hole).
B-
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 Ea – 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 ntype 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
 m r 
*
2
e
2
4 r
2
(1)
0
m* - effective mass of electron
 - dielectric constant of semiconductor
Bohr postulate:
angular momentum quantised
in units of 
 m r   n
[n – integer]
Eliminate  between eq’ns (1) and (2) to give
*
2
4 n 
r
em
2
2
(3)
0
2
(2)
*
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
U

4  n 
4 r
Hence total energy En is given by
2
4
*
2
0
En = U + T 
2
2
0
e m
4
24
*
n
2
0
2
2
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
Eb = E 1 - E =
24  
4
*
2
0
2
Example
 = 12
For Si:
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 PV
- IN DOPED SEMICONDUTORS
Doping  large increase in nc (or pv)  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 – ni2 = 0

N  N  4n
nc =
2
d
2
2
d
i
[Take positive root – negative one gives negative nc!]
Hence from eq’n (2)
pv = ni2/nc
So from eq’n (3), if ni << 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
(3)
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
N
where NSi – number of Si atoms per unit volume
6
d
Si
Obtain NSi from
N 
Si
Avagadro’s No.

28  10
3
 6  10
This gives
NSi = 5 × 1028 m-3
So
Nd = 5 × 1022 m-3.
23
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

Obtain ni from:
For Si:
nc = pv = ni
3
2
  
 kT 
ni = 2m m  
 exp

  
 2kT 
*
*
e
h
3
4
g
2
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 EF 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 Eg –  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
[See end Lecture 1]
2k
g
slope =
slope =
 E  E
2k
c
d

2
1
T-1
Analysis in Tanner (pages 127-130) shows:
In region 1 (low T)
 (E  E ) 
exp

2kT 

[nc – conc’n of electrons in conduction band]
nc = ZN

1
2
d
c
d
1
1
N 
F = (Ec + d) + kTln  
2
2
 Z 
d
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
pv = ZN

1
2
a
 (E  E ) 
exp

2kT 

a
v