Transcript Document

Semiconductors
At zero temperature
semiconductors are insulators
with completely filled bands.
Conduction Band
(empty at T=0)
Eg
Valence Band
(full at T=0)
At higher temperatures they conduct at due to the thermal excitation of
electrons across a relatively small band gap.
In a semiconductor the highest energy filled band is called the valence
band and the lowest energy band called the conduction band.
We will consider states near the top of the valence band to be holes
(particles of charge +e) with free electron like dynamics but effective
mass m*h
We will consider states near the bottom of the conduction band to be
electrons with free electron like dynamics but effective mass m*e
1
Direct Gap Semiconductors
Direct gap semiconductors: the top of the valence band
and the bottom of the conduction band occur at the
same k-vector.
Conduction
band
Energy
Gap
Valence
band
k
Gallium Arsenide GaAs
2
Indirect Gap Semiconductors
Indirect gap semiconductors: the top of the valence band and the
bottom of the conduction band occur at different k-vector.
Germanium (Ge)
Indirect bandgap 0.8eV
Direct bandgap, @ k =0,
is 0.66eV
3
Direct Optical absorption
Direct gap semiconductor: sharp
onset of absorption when the photon
energy is equal to the bandgap
Optical absorption in the
direct gap semiconductor
InSb at 4K
Conduction
Band
Energy
k
hn =Eg
Valence
Band
Bandgap of InSb at
4K is 0.23eV
Photon creates an
“electron-hole pair”
4
Indirect Optical absorption
A transition across an indirect band gap requires a photon
to be absorbed and a phonon to be absorbed or emitted.
Conduction Band
Energy
hnphoton
~Eg
kphonon
k
Valence Band
5
Ge: Indirect Optical absorption
Indirect gap semiconductor: no sharp onset of absorption
0.80
0.88
0.73
0.66
For T = 300
K Eg (indirect gap) = 0.66 eV and EG1 (direct gap) = 0.8 eV
For T = 77K Eg (indirect gap) = 0.73 eV and EG1 (direct gap) = 0.88 eV
Note: numerical values for E(k) for Germanium in figure in Kittel, reproduced earlier.
6
Number of electrons in conduction band
Silicon
@ 300K
n ~ 2x1016 m-3
Note Units are
cm-3
Electron density
increases ~
exponentially
with temperature
7
Density of States
Assume: bottom of conduction band
and the top of valence band parabolic
i.e.
conduction band E = Ec + 2k2/2me*
valence band
E = Ev - 2k2/2mh*
1  2me
Conduction De (E) =
2
2
2


band

valence
band
D(E)
Conduction
Band
Valence
Band
0
Eg
Ec =
Eg
3/2

1
 (E  E g )2


3/2
*


1
1  2mh 
2
Dh (E) = 2  2  (  E)
2   
*
Conduction
Band
Ev = 0
Valence
Band
Note. I do not set
Ev = 0 until later in
notes document
E
8
Chemical potential or Fermi level
At a temperature T the probability that an electron state is
occupied is give by the Fermi-Dirac function
-1


E- 

fe (E) =  exp
 + 1 
 kB T 


The chemical potential,  , is the energy for which f = ½ .
Fermi energy: all energy states are occupied below EF at T = 0.
In discussing semiconductors  is often referred to as the Fermi level !
Will find that  is within the energy gap. In the conduction band
Therefore
E-
exp 
  1
 kB T 
  (E - ) 
fe (E)  exp

 kB T 
9
Electrons density in conduction band
3/2
Density of states
1  2me
De (E) = 2  2
2  
Distribution function
  (E - ) 
fe (E)  exp

 kB T 
*

1
 (E  E g )2



Total number density
of occupied states
n
 D ( E ) f (E)dE
e
e
Eg
3
 me * k B T  2
n  2
 exp (  Eg ) / k B T )
2
 2 
n  Nc exp (  E g ) / k B T )
where
“Fermi level”   E g  k BT ln n / N c 
m k T 

N c  2
 2 
Silicon Eg=1.15eV m*e = 0.2me & n~1016 m-3 @ 300K. Gives
*
e B
2
3
2
  E g  0.5eV
Fermi level near middle of gap
10
Number of holes and electrons
Exactly same argument for holes in the valence band gives
  (  E ) 
f h (E)  1  f e ( E )  exp 

 kBT 
Distribution
function
m k T
Total number

p  2
density of
 2 
occupied states
*
h B
2
3
2
exp ( ) / k B T   Nv exp ( ) / k B T 
n  Nc exp (  E g ) / k B T )
3
3
k T
np  4 B 2  (m*e m*h ) 2 exp( E g / k BT)
Independen t
 2 
This last result is particularly important
of

True for both intrinsic and extrinsic semiconductors
11
Intrinsic semiconductors n = p
In pure “intrinsic” semiconductors the electrons and holes arise
only from excitation across the energy gap. Therefore n = p
3
 k BT 
* * 32
np  4
(me mh ) exp(  E g / k BT )  N v N c exp(  E g / k BT )
2 
 2 
3
 k BT  2 * * 3 4
1/ 2
ni  n  p  2
(
m
m
)
exp(

E
/
2
k
T
)

(
N
N
)
exp(  Eg / 2k BT )

e h
g
B
v c
2
 2 
Chemical potential? m
 mh* 
1
3
  Eg  k BT n  * 
2
4
 me 

Eg
2
*
e

3
 0.026
2
 
exp (  Ec ) / k B T )  m
*
h
3
2
exp   / k B T 
Silicon Eg=1.15eV m*e = 0.2me &
m*e = 0.8me @ 300K. Gives
eV
Fermi level near
middle of gap
12
Hydrogenic Donors & Acceptors
An electron added to an intrinsic semiconductor at T=0 would go into
the lowest empty state i.e. at the bottom of the conduction band.
When one adds a donor atom at T=0
the extra electron is bound to positive
charge on the donor atom.
-e
+ve
ion
Conduction
Band
DE
The electron bound to the positive
Ion is in an energy state ED = Eg- DE
where DE is the binding energy.
An electron which moves on to an
acceptor atom has energy EA
+e
-ve
ion
Ec =
Eg
ED
EA
Valence
Band
Ev = 0
13
Magnitude of binding energy
Similar to a hydrogen atom. Ground state wavefunction is
3
1 1
 (r ) 
  exp[ r / a0 ]
  a0 
2
+ve
ion
-e
The Bohr radius, a0 = 4ro2/mee2 determines the spatial extent of
the wavefunction. Hydrogen atom (r = 1 ) a0 = 0.53 Å.
Binding energy of an electron in the ground state of a hydrogen
atom is
4
EB 
e me
 13.6 eV
2
2(40  r )
Typical Semiconductor
me* ~0.15 me and r ~15.
0.53   r
a0  *
me / me
13.6 m*e
EB  2
eV
 r me
~ 50 Å
~10 meV.
14
Number of Electrons in
the conduction band
Consider a semiconductor with ND
donor atoms per m3
ND0 and ND+ – number density of neutral
and ionised donors
Conduction
Band
DE
Ec =
Eg
ED
At T = 0 all electrons in the lowest
available energy states. No electrons are
excited from donor states into
conduction band
n = 0 ; ND0 = ND
Valence
Band
Ev = 0
15
T ~ Room temperature
kBT > ~ (EC -ED) & number of available states in conduction band >>
ND . Therefore almost all the donors will be ionised. n = ND+ ~ ND
Relevant regime for all electronics. Note that the density of electrons
is ~ independent of temperature.
The chemical potential is well below EC and the expression obtained
for n in an extrinsic semiconductor can be used to give
N 
  E g  k B T n  C 
 n 
where
Silicon Nc ~ 2.6x1024 so for n ~ 1022
 m * k BT 
N C  2

2
 2 
3
2
  E g  6k B T  E g  150meV
T >> Room temperature
In this limit number of electrons excited across the bandgap becomes
larger than number of donors. Behaves like an intrinsic semiconductor.
16
n-type semiconductors
n  exp  Eg / 2k B T )
n  Cons tan t
ND
ln(n)
n  exp  (E g  ED ) / k B T )
Number of electrons
in conduction band
ln(T)
Eg
ED
ln()
“Fermi level”
Eg / 2
ln(T)
17
p-type semiconductors
p  exp  Eg / 2k B T )
p  Cons tan t
NA
ln(p)
p  exp EA / kBT)
Number of holes in
valence band
ln(T)
Eg / 2
ln()
“Fermi level”
EA
ln(T)
18
Compensated semiconductor
Compensated semiconductor:
both donors and acceptors present.
Conduction
Band
ND donors per m3 and NA acceptors per m3
For ND > NA have an n-type semiconductor
with n ~ ND - NA for T ~ 300K
Ec =
Eg
ED
NA electrons fall into
acceptor states
For NA > ND have an p-type semiconductor
with p ~ NA - ND for T ~ 300K
EA
Valence
Band
Ev = 0
19
Impurity Bands
Have considered the impurities as isolated atoms. Reasonable as
doping level normally ~ one donor per 106 semiconductor atoms.
At very high donor concentrations, one has substantial overlap
between the donor or acceptor wavefunctions.
f(r)
aB
+
b
+
Above a critical doping level one has an impurity energy band with a
finite conductivity.
Electron density at which this “metal insulator transition” occurs?
aB ~ 50 Å & lattice constant, a ~ 2.5 Å.
Need b ~ aB = 20a . i.e. one donor per 203 = 8000 semiconductor atoms
20
Mobility of semiconductors
E
Both electrons and holes carry current in
the same direction in a semiconductors.
-e
Conductivity: s nee + peh
+e
e: electron mobility, h: holes mobility
In considering scattering of electrons and holes it is important to
consider mobility as the numbers of carriers varies with temperature.
Conductivity
/me*
se = ne2tp /me*
Mean Free Path Le  tpv
Mobility
e = se/ne = etp
so e = e Le /vme* ~ Le /v
The electron and hole distributions are non-degenerate and <E> ~ kBT
<v> ~ < 2E/me* > ~ T1/2.
T > ~ TD number of phonons increases as T.
L ~ T-1 .
 ~ T-3/2
T << TD ionised impurity scattering dominates. Similar to Rutherford
scattering & scattering cross section ~ E-2 ~ T-2. So L ~ T2 and  ~ T3/2
21
Hall Effect in Semiconductors
n-type semiconductors (n>>p) RH ~ -1/ne
p-type semiconductors (p>>n) RH ~ +1/pe.
The Hall effect is used to obtain the carrier densities in semiconductors.
In an electric field electrons and holes drift in opposite directions.
j
Consider case of n = p and e = h.
F
Have no Hall field.
B
-e
B
F
+e
The free carrier Hall coefficient is generally (Hook and Hall p153)
RH = ( p h2 - n e2 )/e(n e + p h )2
22
Semiconductor devices:
Inhomogeneous semiconductors
All solid-state electronic and opto-electronic devices are
based on doped semiconductors.
In many devices the doping and hence the carrier
concentrations are non-homogeneous.
In the following section we will consider the p-n junction
which is an important part of many semiconductor
devices and which illustrated a number of key effects
23
The p-n semiconductor junction:
p-type / n-type semiconductor interface
We will consider the p-n
interface to be abrupt. This is a
good approximation.
n-type ND donor atoms per m3
p-type NA acceptor atoms per m3
Consider temperatures ~300K
Almost all donor and acceptor
atoms are ionised.
p-n interface at x=0.
impurity atoms m-3
ND
NA
p-type
n-type
x=0
xa
ND (x) = ND (x>0) = 0 (x<0)
NA (x) = NA (x<0)
(x>0) = 0 (x>0)
(x<0)
24
p-type semiconductor
Electron and
hole transfer
n-type semiconductor
Electrons
EC

EC

EV
EV
Holes
Consider bringing into contact p-type and n-type semiconductors.
n-type semiconductor: Chemical potential,  below bottom of
conduction band
p-type semiconductor: Chemical potential,  above top of valence
band.
Electrons diffuse from n-type into p-type filling empty valence states.
25
EC
Electrons
Band
Bending
EC

EV
Holes
eDf0
EV
p-type semiconductor
n-type semiconductor
Electrons diffuse from n-type into p-type filling empty valence band states.
The p-type becomes negatively charged with respect to the n-type material.
Electron energy levels in the p-type rise with respect to the n-type material.
A large electric field is produced close to the interface.
Dynamic equilibrium results with the chemical potential (Fermi level)
constant throughout the device.
Note: Absence of electrons and hole close to interface -- depletion region
26
Electrostatic voltage drop, Df0
EC
Electrons
In equilibrium  constant.
Electrostatic voltage
difference, Df0, between nand p- regions.
EC

EV
eDf0
Holes
EV
p-type semiconductor
n-type semiconductor

 Nc 
Nc 

E

k
T

n
c
B



N

N
N
A
 D
 D
For x >> 0
  Ec  k BT n 
for x << 0
 Nv 
 Nv 
  Ev  k BT n 
  Ev  k BT n  
 N A  ND 
 NA 
Since Eg = Ec – Ev
 ND N A 
  E g
eD 0  k BT n
 Nc Nv 
27
Depletion region
Depletion region
Assume the electric field in the
region of the junction removes all
the free carriers creating a
depletion region for –dp<x < dn.
n,p
Electron and
Hole Density
r
The ionised impurities are fixed in
the lattice. So charge density is
r  +eND per m3 for 0 <x <dn
r  –eNA per m-3 for -dp < x < 0.
Net charge
density
E
-dp
0
Depletion region
dn
xa
The total charge in the depletion region must be zero as the number of
electrons removed from the right equals the number of holes removed
from the left
i.e.
NDdn = NAdp.
28
Electric field E(x)
Can calculate the electrostatic potential, f(x) from the Poisson’s equation
 2f ( x)
  r ( x) /  0  r
2
x
Charge density: r(x) = eND for 0 < x < dn r(x) = -eNA for –dp < x < 0
Boundary condition: E = 0 for x > dn and x < –dp
Depletion region
So integration gives
  N A e
E 
( x  d p ) for  d p  x  0
x  0 r
r
Net charge
density
E
  N D e
E 
( x  d n ) for 0  x  d n
x  0 r
Electric Field
(negative)
0
xa
29
Electrostatic potential, f(x)
Integration of E gives the potential (x).
Since   0 for x < –dp and   D0 for x < dn.
eN A
( x) 
(x  d p )2
2 0 r
eN D
( x)  D 0 
(x  dn )2
2 0 r
(x) is continuous at x = 0 so
D 0 
So since NDdn = NAdp
dp  x  0
for
for
0  x  dn
e( N D d n  N A d p )
 2 0  r N A D 0 
dn  

eN
(
N

N
)
D 
 D A
2
2
2 0 r
1
2
 2  N D 0 
dp   0 r D

eN
(
N

N
)
D 
 A A
1
2
Resulting depletion width is ~100nm to 1m. Self consistent
30
Electron and
Hole Density
n,p
r
Net charge
density
E
Electric Field
(negative)
Electrostatic
Potential
Energy of
Conduction
band edge
f(x)
EC
0
Depletion layer
xa
31
p-n junction with a forward bias
Electric Current
Net flow of electrons
EC

EC

eV
EV
e(Df0V
Net flow of holes
p-type biased positive
EV
n-type biased negative
Forward bias: p-type region biased positive with respect to n-type region.
The voltage is dropped across depletion region since the free carrier density
is low and therefore the resistivity is high.
Total potential across the depletion layer is D = D0 – V
32
Generation Current
EC
EC

EV
eDf0
EV
p-type semiconductor
n-type semiconductor
Electron-hole pairs created in the depletion region move apart in
the strong electric field. A generation current, Jgen, in the negative
x-direction results.
Magnitude of the generation current density is
Jgen = A exp(-Eg/2kBT) where A is a constant.
33
Recombination Current
EC
eDf0
Photons Out
EC

EV
eDf0
EV
p-type semiconductor
n-type semiconductor
Electrons with energies greater than eΔ0 can move into the p-type
material where they recombine with holes.
A recombination current, Jrec, in the positive x-direction results
Jrec = B exp(-e Δ0/2kBT) where B is ~ constant.
For direct band gap semiconductors recombination leads to photon emission
34
Current-Voltage Characteristic
At equilibrium, without a bias voltage Jgen + Jrec = 0
With external positive voltage V the Jgen is ~ unchanged, but Jrec becomes
 - (eD 0 - eV) 
J rec (V ) = B exp 

kBT


 eV 
  1)
Total net current density is J  J rec (V )  J gen (0)  J gen (0)(exp 
 kBT 
J
Reverse bias
negative
p
n
positive
Check
Forward bias
positive
p
n
negative
V
-Jgen
35
Applications of p-n junctions
• p-n junction diodes: Excellent diodes, which can be used for
rectification of AC signals.
• Light emitting diodes (LEDs) and lasers: In forward bias
one has an enhanced recombination current. For direct band gap
semiconductors light is emitted.
• Solar cells: If photons with hn>Eg are absorbed in the depletion
region of a p-n junction one has an enhanced generation current.
The energy of the photons can be converted to electrical power in
solar cells based on this mechanism.
• p-n-p junction transistors: Transistors based on the
properties of p-n junctions can also be produced.
See Hook and Hall p184-8.
36