Transcript Slide 1

Semiconductors:
Some theory and application
By: Bob Buckley
Let’s continue where we left off in class
Only partly filled electronic band can contribute to conductivity:
Where does this partially filled band come from for semiconductors?
E
conduction band
EC
Eg
EF
EV
valence band
core electrons
Metal
Semiconductor
Insulator
What is the requirement for electrical conduction in a solid?
Conduction occurs when an electronic band is partially filled. Why?
Due to the symmetry of a full band structure, for every electron with wavevector
K, the band also contains an electron with wavevector –K.
With a filled band, there is no way to continuously change the momentum of
any of the electrons. This means they cannot change their motion.
Of course a band that is empty of electrons cannot carry current.
So, how does a semiconductor do it?
Remember Density of States:
Free electron parabola
 2k x
2m
2
1  2m 
D(E)  2  2 
2   
dE
dk x 
2
L
[E, E  dE]
3/ 2
E
kx
What about for a semiconductor?
1  2mn* 
DC ( E )  2  2 
2 

E
k
*
1  2m p 
Dv ( E )  2  2 

2 

3/ 2
E  EC
( E  EC )
Ev  E
( E  Ev )
3/ 2
( Ev  E  Ec isforbidden)
So what do the actual concentrations look like?

As before, Density:
n   D(E)f (E, T )dE
0
We care about partially filled bands with respect to electrical conductivity.

Density of electrons in conduction band: nC 
 D( E ) f ( E , T )dE
EC
Density of holes in valence band:
pV 
EV
 D( E) 1  f ( E,T ) dE
0
Typically:: Eg>>kTRoom
What does this mean for the
concentrations, n and p
First of all, since Eg>>kT, the Boltzmann approximation for the
Fermi distribution is good for the allowed energies.
1
f ( E,T ) 
e
For Eg>>kT
E  EF
k BT
f ( E, T )  e
1
 ( E  EF )
kBT
So, using the Boltzmann approximation:
1  2m 
n 2 2 
2 

*
n
3/ 2
e
N
3/ 2
C
eff
nN e
C
eff

E
kT
E  EC e dE
EC
 2 mn*kT 
n  2

2
h


Substituting
Ef 
kT
e
 2 m kT 
 2

 h

 ( EC  EF )
kT
3/ 2
 E f Ev
e kT
3/ 2
and
N
V
eff

3/ 2
e
pN e
Ev  Ee dE
( EV  EF )
kT
 2 m kT 
 2

h


*
p
2
V
eff
E
kT

 2 m*p kT 
p  2

2
h


 ( EC  EF )
kT
*
n
2
*
1  2m p 
p 2 2 

2 

3/ 2
( EV EF )
kT
Let’s try something: What is n*p
C
np  Neff
e
 ( EC  EF )
kT
V
Neff
e
( EV  EF )
kT
C
V
 Neff
Neff
e
EV EC
kT
Notice: This result no longer depends on the Fermi energy
 Eg
3
 kT 
* * 3/ 2 kT
np  4 
(
m
e
nmp )
2 
 2 
For an intrinsic semiconductor, n=p for charge balance
3/ 2
 kT 
ni  pi  2 
2 
2



 Eg
(mn*m*p )3/ 4 e 2kT
How is the Fermi energy calculated for an intrinsic semiconductor:
 2 m kT 
n  p  2

h


*
n
2
With some algebra:
3/ 2
e
 ( EC  EF )
kT
 2 m kT 
 2

 h

*
p
2
3/ 2
e
( EV  EF )
kT
 m*p 
EC  EV 3
EF 
 kT ln  * 
m 
2
4
 n
Effective Mass
Electrons in crystal are bound by the crystal and do not act like free electrons.
1 
1  2E

Remember: 
 2

 m ij  k ik j
Effective mass depends on the
curvature of E(k)
The more the curvature of E(k), the lighter the effective mass.
For free electron:
E
2 2
p2
k
E

2m 2m
k
2
2E

2
k
m
Properties of holes:
•
•
•
•
Kh = -Ke :: The wavevector of a hole is opposite to the wave vector of an eThe hole wavevector is the absence of the electron wavevector
Eh(Kh) = - Ee(Ke) :: Assuming zero energy at band gap
Conduction band is concave up and valence band is concave down
Vh = Ve
mh = -me :: Curvature of e- valence band near top of curve is negative
meaning hole mass near the top is positive
E
What
about
this
k
Typical Semiconductor Values (300 K)
Comp.
ML/Me
MT/Me
Eg (eV)
.19
ni
cm^-3
1.5E10
Si
.98
Ge
1.57
.082
2.4E13
.67
GaAs
.07 Me
.12 MLH 5E7
.68 MHH
1.12
1.43
From table (and since Eg>>kT), it is evident that the concentrations of holes
and conduction electrons are very small (ni)
n  p  1010 conductionelectrons / cm ^ 3
Compared to approx. 1023 atoms/cm^3
For example, actual resistivity of undoped GaAs is greater than 1E7 ohm-cm,
this includes the effects due to defects such as unintentional doping
Conductivity:
j  e(nn  p p )
Where μn and μp are the mobilities of
the electrons and holes respectively
e   (k )v 2 (k ) 
n  *
Where  is the relaxation time and v is the velocity.
2
mn  v (k ) 
What can we do to increase n and p in order to increase conductivity?
Semiconductor doping
By increasing the concentration of either the conduction electrons or the
holes in the valence band, we can increase the conductivity.
This is done by substituting atoms in the semiconductor lattice that do
not have four valence electrons, typically a valence 3 or 5 atom.
N doping:
A valence 5 atom contains one more electron than the intrinsic semiconductor.
This extra electron, when excited goes into the conduction band.
P doping:
A valence 3 atom contains one less electron: (Or valence 4 plus one hole).
When this hole is excited, it goes into the valence band.
How exactly does this affect the semiconductor as a whole?
Picture a hydrogen atom superimposed inside a semiconductor lattice
For H atom in free space:
mee4
1
E 
where n=principal Q #
2
2
2(40 ) n
H
n
h2
r  0
 mee2
The semiconductor “background” must be accounted for:
mn*e4
1
E 
2(4 s0 )2 n2
H
n
Example, For Si:
E
and
  s0 and me  mn*
h2
r  s0
 mn*e2
Si  11.7 and mn*  .3me
30meV and r
20 A
Bound e- is smeared over
approximately 10^3 lattice sites
So, for a doped material, it takes a relatively small amount of energy to excite the
extra electron (or hole) into the conduction band (valence band) of the material
Remember np=constant for a specific type of semiconductor. Our
assumptions of np still hold even with doped semiconductors
3
E
g
 kT 
* * 3/ 2 kT
2
np  4 
(
m
m
)
e

n
n p
i  Constant
2 
 2 
Total charge is also conserved:
n  N A  p  ND
Total concentration of doped atoms = Non-ionized atoms + ionized atoms:
For n doped : ND  ND0  ND
Typically, ND>>ni>p, so:
Also: n  ND  p
n  N D
Process is basically the same for a p doped semiconductor
It turns out statistically that the density of non-ionized dopant is the following:
For n: N D 
0
ND
1 e
n  ND  ND  ND0
( ED  EF )
kT

1
n  N D 1 
( ED  EF )

 1  e kT





ND

n  2ND 1  1  4 C e

Neff

e
Ed
kT
For low enough temperatures:
For high enough temperatures:




EF
kT
n EkTC
 C e
Neff
1
C
n  ND / Neff
e
n  ND
 Ed
2 kT
ln(n)
Carrier concentration vs. 1/T
Intrinsic
slope = –E g /2k
ln(Nd )
Extrinsic
Ti
Ts
Ioniz ation
slope = –E/2k
ni(T)
1/T
Conductivity vs. T
Conductivity vs. 1/T
All of this theory can be interesting, but we all know semiconductors
are crucial to electronic devices today
Let’s look at a practical application: The p-n junction
What is a p-n junction:
A single crystal in which there is an abrupt change from a
n-type to a p-type semiconductor
p
n
What happens when we have a joined p-n semiconductor?
First of all, the electrochemical potential (EF) must be constant throughout
the entire crystal for equilibrium
In a p semiconductor, EF lies near the valence band. (It is lowered
because of the decrease in electron density of the semiconductor)
In a n semiconductor, EF lies near the conduction band. (The more
number of electrons, the greater the largest energy at ground state)
Since EFn=EFp, EV and EC are each shifted between the p and n regions
This shift of both EV and EC creates a “band bending”
This band bending is created by a macropotential which lies in the transition region
This occurs because conduction electrons near the transition in the n region
diffuse to the p region. Holes go the opposite way.
The macropotential is created by static dopant ions when their electrons or holes
diffuse into the part of the semiconductor doped oppositely
Differential Maxwell Equation for Gauss’ Law


2
E 
and E=  V so  V= 
0
0
Due to symmetry, V and ρ only depend on one dimension
 2V ( x)
 ( x)
So,

2
x
0
Due to EV and EC shift and the resulting macropotential, we can say:
eVD  ( EVn  EVp )
We can say:
And using previous equations for n, p, and np,
 p p nn 
eVD  kT ln  2 
 ni 
Where pp is the density of holes in the pp region and nn is the density of
conduction electrons in the n region, AKA majority carriers
n
ρ(x)
p
V(x)
E(x)
+
-xp
-
xn
x
x
Static charges are due to ionized doping atoms.
=-eVD
=Constant
x
Two types of charge motion are occurring simultaneously when this system is in
equilibrium
1. Due to the holes’ and conduction electrons’ mobility and the fact that there is a
concentration gradient at the junction, electrons diffuse from the n region to the p
region while holes diffuse from the p region to the n region
2. Due to the macropotential voltage at the junction, electrons are pulled from the
p region into the n region while holes are pulled from the n region into the p region.
Due to the fact that there is no net
current, jdiff=jdrift
For electrons: Dn
For holes:
Dp
n
V ( x )
 n n
x
x
p
V ( x)
  p p
x
x
Dn
n
V ( x )
 n n
x
x
C
n( x)  N eff
e
 ( ECp eV ( x ) EF )
kT
because E(x)=(E Cp  eV ( x))
+
-xp
kT
Dn 
n
e
Is Einstein’s relation between D and µ
Using Poisson equation, one can derive:
eN D
V ( x)  Vn () 
( xn  x)2
20
Also:
1/ 2
 2 0VD N A / N D 
xn  

N A  ND 
 e
0  x  xn
n
ρ(x)
p
-
xn
x
Let’s look at the non-steady state by applying a voltage across the junction:
p
n
+
-xp
-
+U-
n
ρ(x)
p
xn
x
1/ 2
VD  VD  U
 U 
xn (U )  xn (0) 1  
 VD 
The width of the static charge region and the magnitude of VD
are voltage dependent.
2 types of current:
Jdrift
or
Igen
Originates when an electron in the p region or hole in the n region
comes near the junction. This is quite “rare” since these are both
minority carriers. Requires a thermal excitation.
This is basically independent of VD and therefore U since a
thermal excitation must start the process. Remember Eg>>kT
I
Jdiffusion
or
Irec
gen
(U ,T )  I
gen
(0,T )
Originates when an electron in the n region or hole in the p
region is able to diffuse against (across) the potential barrier
between the n and p regions. Majority carriers: n  N n p  N p
Depends strongly on temperature because of the change in V
I rec (U  0)  I gen (U  0)  e
I
rec
I
eU
gen kT
e
 e (VD U )
kT
I gen (U ,T )  I gen (0,T )
I rec  I
eU
gen kT
e
I net (U , T )  I rec (U , T )  I gen (U ,T )  I 0gen (T )*(e
eU
kT
1)
By definition of Igen and Irec, positive current means majority
carriers are moving to minority carriers
p
n
Current is highly asymmetric
U  0 : I  I 0e
eU
kT
Forward
bias
+U-
I  I 0 (e
eU
kT
U  0 : I  I0
 1)
Reverse
bias
U  0 : I  I 0e
eU
kT
U  0 : I  I0
Forward
bias
I
Reverse
bias
U
Semiconductor
breakdown