Transcript Vibrational Properties of the Lattice

VII. Semiconducting Materials & Devices
A. Band Structure and Terminology
B. Intrinsic Behavior
C. Optical Absorption by Semiconductors
D. Impurity Conductivity
E. Extrinsic Behavior
F. Hall Effect and Hall Mobility
G. The Diode: A Simple p-n Junction
A. Band Structure and Terminology
Semiconductors and insulators have qualitatively similar band structures, with the
quantitative distinction that the band gap Eg > 3.0 eV in insulators.
conduction band
Ec = conduction
band edge
valence band
Energy band diagram in k-space
Ev = valence
band edge
“Flat-band” diagram in real space
Fermi-Dirac Distribution Function
The important external parameter that determines the properties of a semiconductor is
the temperature T, which controls the excitation of electrons across the band gap in a
pure (intrinsic) semiconductor.
Fermi-Dirac
distribution
function
f (E) 
1
e (   ) / kT  1
For nearly all T of interest:
So we can approximate:
The probability for an electron to be in an
energy level  at temperature T
 = chemical potential  EF for T << TF
(   )  kT
f ( E)  e
(  ) / kT
This is the Maxwell-Boltzmann
(classical) limit of Fermi-Dirac
statistics
B. Intrinsic Behavior
“Intrinsic” means without impurities. Electrical conductivity is zero at T = 0, but for
T > 0 some electrons are excited into the conduction band, also creating holes (H+)
in the valence band. In general the conductivity can be written (using the nearly
FEG model):
The conductivity is controlled by the magnitude of n and
2
2
ne  e pe  h
p, which rise exponentially as T increases. The


relaxation times are only dependent on 1/T, and this
me
mh
dependence is often ignored because the exponential
behavior dominates.
  as T  due
to increased
scattering ( )
Experimentally we
find that for a pure
semiconductor:
high-T
(intrinsic)
region
low-T
(extrinsic)
region
Intrinsic Carrier Statistics
It is relatively simple to calculate n(T) and p(T) for the intrinsic region, where the
conductivity is caused by excitation of e- across the energy gap:
e- in the conduction band:
k 2
E  Eg 
2me
h+ in the valence band:
k 2
E
2mh
Ec = Eg

Ev = 0
3/ 2
For parabolic bands the
density of states are:
V  2m 
Nn (E)  2  2e 
2   
Think: why are we
justified in assuming
parabolic E(k) here?
V  2m 
Nh ( E)  2  2h 
2   
E  E 
1/ 2
g
3/ 2
 E 1/ 2
E  0 
Intrinsic Carrier Statistics, cont’d.
Nn (E)
1  2me 
 2 2 
V
2   
We can write the density
of states per unit volume:
3/ 2
Nh ( E)
1  2mh 
 2 2 
V
2   
And now calculate
the carrier
concentration n(T):
E  E 
1/ 2
g
3/ 2
 E 1/ 2

N (E)
1  2me 
n(T )   n
f ( E )dE 
2 
2 
V
2




Eg
3/ 2 
3/ 2

Now rearrange
1  2me 
1/ 2  E  E g
kT  
cleverly and pull out n(T ) 
2 
2 
2   
kT
Eg 
a factor of (kT)1/2:

Which allows:
n(T ) 
1  2me 
2 
2 
2   
3/ 2
 E  E 
g
e
dE
Eg
1/ 2
 (   Eg ) / kT ( E  Eg ) / kT
 e
e
dE

1/ 2 (   E g ) / kT  E  E g
kT  e
E  kT
g

1/ 2  ( E   ) / kT
1/ 2
 ( E  Eg ) / kT
 e
dE

Intrinsic Carrier Statistics, cont’d.
Now make a variable
substitution:
The integral
becomes:
x
E  Eg
 dx 
kT
1  2me 
n(T ) 
2 
2 
2   
3/ 2
dE
 dE  kTdx
kT
kT 
3/ 2
e
(   E g ) / kT

1/ 2  x
x
 e dx
0
3/ 2
So finally:
 m kT  (   E ) / kT
n(T )  2 e 2  e g
 2 
Whew! And next we can do the same calculation for holes to get p(T)!

2
Intrinsic Hole Carrier Statistics
Now for holes in the
valence band:
And since
fh (E)  1  fe (E)  1 
(   E )  kT
we have
1
1

e ( E   ) / kT  1 1  e (   E ) / kT
f h (E)  e( E ) / kT
3/ 2 0
Just as before,
calculate the carrier
concentration p(T):
Nh (E)
1  2mh 
p(T )  
f h ( E )dE  2  2 
V
2   

Now rearrange :
1  2m 
1/ 2
p(T )  2  2 h  e   / kT   E  e (  E ) / kT [d ( E )]
2   

0
3/ 2
1/ 2 ( E   ) / kT



E
e
dE


0
3/ 2

Replace –E with E
1  2mh 
1 / 2  E / kT
  / kT


p
(
T
)

e
E
e
dE


and flip limits due to
2
2
2   
0
minus sign:

Pull out a factor of
(kT)1/2:
1  2mh 
p (T ) 
2 
2 
2   
3/ 2

1/ 2
 E 
(kT )1/ 2 e   / kT    e  E / kT dE
kT 
0
Intrinsic Hole Carrier Statistics, cont’d.
E
kT
Now make the
variable substitution:
x
The integral becomes:
1  2mh 
p(T ) 
2 
2 
2   
 dE  kTdx
3/ 2

(kT ) 3 / 2 e   / kT  x1/ 2 e  x dx
0

3/ 2
 mh kT   / kT
p(T )  2
e
2 
2




So finally:
2
again!
So far these relations for n(T) and p(T) are true for any semiconductor, with or without
impurities. It is very convenient to calculate the product np:
3/ 2
3/ 2
3
 m kT  (   E ) / kT  mh kT   / kT
 kT 
3 / 2  Eg / kT


np  2 e 2  e g
2
e

4
m
m
e



e h
2
2
 2 
 2 
 2 
Intrinsic Carrier Statistics: Results
Now for an intrinsic semiconductor (or in
the intrinsic region of a doped
semiconductor) ni = pi, so:
And equating the earlier
expressions for n and p:
3/ 2
3/ 2
me mh 3/ 4 eE / 2kT
g
3/ 2
 m kT  (   E ) / kT
 m kT 
2 e 2  e g
 2 h 2  e / kT
 2 
 2 
e
This gives an
expression for (T):
 kT 
ni  pi  2
2 
2




( 2   E g ) / kT
m 
  h 
 me 
2  Eg
3/ 2
1
kT  mh 

  Eg 
ln
2
2  me 
kT
3/ 2
m 
 ln h 
 me 
3/ 2
 mh 
1
3

 E g  kT ln
2
4
 me 
So the chemical potential, or Fermi level, has some dependence on T, but if mh and me
are similar, then this is very small.
Carrier Mobility
The total conductivity, including both the
electron and hole contributions, is:
ne2 e pe2 h


me
mh
It is common to define a quantity that expresses the size of the drift speed for each type
of carrier in an electric field E:
Earlier FEG result:
e
v
E
m
(Note: the carrier
mobility is directly
related to the
switching speed of
a solid-state
electronic device)
v e i
Definition of
i 

E mi
carrier mobility:
Now we can rewrite the total conductivity as:   nee  peh
Experiment shows that  has a power-law temperature dependence:   T n  2  n  2
Thus the exponential temperature dependence of n and p
dominates, and we can approximate the intrinsic conductivity
 Eg / 2 kT
 i  Ae
So a plot of vs. 1/T gives a straight line with slope –Eg/2k. Conductivity measurements
allow us to determine Eg!
C. Optical Absorption by Semiconductors
Examine the following calculated 3-D band structures for semiconductors Si and GaAs.
What difference(s) do you see?
Ec
Ev
Ec
Ev
Si
GaAs
Indirect band gap
Direct band gap
(kgap  0)
(kgap = 0)
Optical Absorption and Conservation Laws
Absorption of a photon by a semiconductor can promote an electron from the valence
to the conduction band, but both energy and momentum must be conserved:
Ei  h  E f
  
ki  k  k f
  photon frequency

k  photonwavevector
For semiconductors Eg  1 eV so the photon wavevector can be estimated:
p E/c
1 eV (1.60  1019 J / eV )
6
1
k  


5

10
m


(3.00  108 m / s)(1.05  1034 Js)
But this is utterly tiny compared to a typical BZ dimension:
2
2
10
1


2

10
m
a 3.00  1010 m
 
So essentially we have:
A direct-gap (vertical) transition
ki  k f
kmax 
Direct vs. Indirect Gap Semiconductors
 
But for indirect gap semiconductors it is clear that: ki  k f
So for an indirect gap transition momentum can only be conserved by absorption or
emission of a phonon (lattice vibration)
Ei  h  q  E f q  phonon frequency
  

ki  q  k f
q  phononwavevector
To estimate a typical phonon energy, we know:
q  (1.05  1034 Js)(1014 s 1 )
q  1014 s 1
1 eV
 0.07 eV
19
1.60  10 J
Optical Absorption: Experimental Results
Experimental absorption coefficients () are measured to be:
i
d
Eg
h
Eg  q
 d  Ah  Eg 1/ 2
While for an indirect
gap material with a
direct transition at a
slightly higher energy:
h
 i (T )  A(T )h  Eg 2
i
Eg  q
E2
h
D. Impurity Conductivity in Semiconductors
Consider two types of substitutional impurities in Si:
pure Si:
(each line
represents
an e-)
Si:P
weakly
bound
extra
electron
missing
electron
Si:B
donor energy level
(n-type material)
acceptor energy level
(p-type material)
Donor Impurities in Semiconductors
We can estimate the ionization energy of a pentavalent donor impurity using the Bohr model:
Bohr model for H:
-e, m
Ed
r
+e
4 0  2 2
rn 
n n  1, 2, 3...
m e2
 m e4 1
En 
32 2 02  2 n 2
For an electron orbiting a positive ion inside a semiconductor, what changes must we
make in the Bohr model equations?
dielectric medium:
So in a
semiconductor:
 0   r 0
periodic potential
(effective mass):
m  me
  m e4  1  me  1 
1  me  1 




En  
 2   13.6 eV 2   2 
2 2 2  2 
n  m   r 
 32  0 r   n  m   r 
Donor Ionization Energy
Assuming that the electron is initially in its lowest energy level, the donor ionization
energy Ed is:
Ed  E  E1  0   13.6 eV 
For Si we can use representative
values of the effective mass and
dielectric constant to obtain:
Experimental data reveal
ionization energies (in meV):
The orbital radius
is predicted to be:
1  me  1 
 me  1 





13
.
6
eV
 2 
  2 
2 
1  m   r 
 m   r 
2
 1 
Ed  13.6 eV 0.26
  0.026 eV
11
.
8


Si
P
As
Sb
45
49
39
All within a factor of
two of our rough
estimate!
o
o
4 0  2 2  m 
1 

r1 
1   r   0.529
11.8  24 
2
me
 0.26 
 me 
So this electron moves through a region that includes hundreds of atoms, which
supports the use of the dielectric constant of the bulk semiconductor.
Acceptor Impurities in Semiconductors
What happens when the impurity atom is trivalent?
At 0 K the acceptor level is empty, so a “hole” is
bound to the impurity atom.
Ea
However, the energy Ea is so small (50 meV in Si)
that at room T electrons in the valence band bound to
other Si atoms can be excited into the acceptor level,
leaving behind a mobile hole in the valence band.
Summary: Both donor (P, As) and acceptor (B, Ga) impurities provide an easy way to
increase either n or p even at low T. When such impurity-related carriers dominate the
electrical properties, the semiconductor material displays extrinsic behavior.
Note: If impurity concentration is very large, the Bohr orbits (wavefunctions) of the
donor electrons can overlap and form an “impurity band” that extends throughout the
material. This leads to a so-called insulator-metal transition and causes an abrupt
increase in the conductivity (see problem 10.5).
E. Extrinsic Behavior and Statistics
Let’s consider donor impurities in a semiconductor (n-type):
Nd  Nd   Nd 0
Nd = concentration of donor atoms
Nd+ = concentration of ionized donor atoms
Nd0 = concentration of neutral donor atoms
Now in the presence of a large donor e- concentration, then n >> ni so p must decrease in
order to keep the product np = constant. What physical process causes p to decrease?
Essentially the large number of e- in the
conduction band will be sufficient to fill most
available holes in the valence band, so that:
Eg
Eg-Ed
Ev = 0
n  Nd 
Nd 0  Nd f (Eg  Ed )
Nd 0 
Nd
e
( E g  Ed   ) / kT
1
and
p  n
Ed  0
Extrinsic Carrier Statistics
Now solve for the electron concentration n:
n  Nd 
1


 Nd  Nd 0  N d 1  ( Eg  Ed   ) / kT 
 1
 e
 e ( Eg  Ed   ) / kT  
Nd

n  N d  ( E g  Ed   ) / kT   
(  E g  Ed   ) / kT 
e

1
1

e




Now from our earlier
treatment of intrinsic
behavior:
Equating the
expressions for n:
3/ 2
 me kT  (   Eg ) / kT
(   Eg ) / kT
n  2
e

n
e

0
2
2




n0e
(   E g ) / kT

Nd
e
(   E g  Ed ) / kT
1
You can always use this exact master eqn. to solve for  and thus n,
but you have to do it numerically.
Limits of Low and High Impurity Concentrations
This discussion is relevant to several HW problems in Myers (see 10.4, 10.8, 10.9). It
provides simple approximations for n and  corresponding to very small Nd and very
large Nd.
n  n0e
(   E g ) / kT

Nd
e
(   E g  Ed ) / kT
We will develop approximations to
simplify the solution of this eqn.
1
Now consider two extreme limiting cases:
1.
Nd << n0
In Si at 300K
(high T limit)
1
3
m 
1
  Eg  kT ln h   Eg
2
4
 me  2
 m kT 
n0  2 e 2 
 2 

e
Thus,
n  n0 e
Nd

 Nd
1
3/ 2
 3.4 1024 m3
(   Eg  Ed ) / kT
Since Eg >> 2kT at room temp, this means
(   E g ) / kT
Eg
e
Ev = 0
e
(  12 Eg  Ed ) / kT
(   Eg  Ed ) / kT
And solving
for :

e
(  Eg / 2 kT )
 1
 Nd
 n0
  Eg  kT ln



Limits of Low and High Impurity Concentrations
2. Nd >> n0
e
(   Eg  Ed ) / kT


(low T limit)
e
(  12 Ed  Ed ) / kT
We can neglect the “1”
in the denominator here:
And now
solve for :
e
( 2   2 E g  Ed ) / kT
n  n0e

Does the low T limit make sense?
Substituting into the
above eqn. for n:
Thus,
n  n0 e
e
Ed / 2 kT
(   E g ) / kT
 e1 at 300K
Eg-Ed
 1 for low T

Ev = 0
Nd
e
(   E g  Ed ) / kT
1

Nd
e
(   E g  Ed ) / kT
 Nd
 n0
Nd
n0
  E g  12 Ed  12 kT ln
1
2
  E g  Ed
n  n0 Nd  e Ed / 2kT




N
( E g  12 Ed  kT ln n d   E g ) / kT
 0 
1/ 2
Eg
 Nd
 n0 
 n0
1/ 2
  Ed / 2 kT
 e

Summary of Impurity Semiconductor Behavior
 in  e
Now our schematic
plot of ln  vs. 1/T is
even easier to
understand:
 Eg / 2kT
  constant
 ex  e E
d
/ 2 kT
n  Nd
Extrinsic Behavior for p-Type Semiconductors
Let’s consider acceptor impurities in a semiconductor:
Na  Na   Na0
Na = concentration of acceptor atoms
Na- = concentration of occupied acceptor levels
Na0 = concentration of neutral (unoccupied) acceptor levels
Now in the presence of a large hole concentration, then p >> pi so n must decrease in
order to keep the product np = constant. What physical process causes n to decrease?
Essentially most of e- in the conduction band
will fall down to fill up holes in the valence
band, so that:
Eg
Ea
Ev = 0
p  Na
p  Na  Na f ( Ea )
p  N a 
Na
e ( Ea   ) / kT  1
and
n  p
Ea  0
Extrinsic p-Type Carrier Statistics
Now from our earlier
treatment of intrinsic
behavior:
 m kT 
p  2 h 2  e / kT  p0e / kT
 2 
Equating the
expressions for p:
p0 e   / kT 
3/ 2
Na
e ( Ea   ) / kT  1
Again, you can always use this exact master eqn. to solve for  and
thus p, but you have to do it numerically.
Limits of Low and High Impurity Concentrations
Here we provide simple approximations for p and  corresponding to very small Na
and very large Na.
p0 e
  / kT

Na
e ( Ea   ) / kT  1
We will develop approximations to
simplify the solution of this eqn.
Now consider two extreme limiting cases:
1.
Eg
Na << p0 (high T limit)
As before, we argue that  is near Eg/2 as in the intrinsic case:
1
3
m 
1
  Eg  kT ln h   Eg
2
4
 me  2

Since Eg >> 2kT at room temp, this means
Thus,
p  p0 e
  / kT
Na

 Na
1
e
( Ea  ) / kT
e
(  12 Eg ) / kT

Ev = 0
e
(  Eg / 2 kT )
e( Ea  ) / kT  1
And solving
for :
 p0 

 Na 
  kT ln
Limits of Low and High Impurity Concentrations
2. Na >> p0
e

( Ea   ) / kT
Eg
(low T limit)
e
( Ea  12 Ea ) / kT
e
We can neglect the “1”
in the denominator here:
And now
solve for :
e ( Ea 2  ) / kT 
Thus,
 e1 at 300K
 1 for low T
p  p0 e   / kT 
p  p0 e
Ea

Ev = 0
Na
e ( Ea   ) / kT  1

Na
e ( Ea   ) / kT
 Na 

 p0 
Na
p0
Does the low T limit make sense?
Substituting into the
above eqn. for p:
Ea / 2 kT
  12 Ea  12 kT ln

1
Ea
2

p
(  12 Ea  12 kT ln N0  ) / kT
 a
p   p0 Na  e Ea / 2kT
1/ 2
1/ 2
 p0   Ea / 2 kT
 e
 p0 
 Na 
F. Hall Effect and Mobility
The Hall effect is easier to measure in semiconductors than in metals, since the
carrier concentration is smaller:
When one carrier dominates, we have
a Hall coefficient:
Hall measurements can tell us
whether a semiconductor is n-type
or p-type from the polarity of the
Hall voltage:
RH 
EH
1

JB nq
I
+
+
VH  wEH
B
-+
+
where
-+
-+
-+
n-type
+
w
p-type
When one carrier dominates, we
  nee ( peh )
can write the conductivity:
Measuring RH and  will thus
So the mobility can be written:   RH 
give: sign, concentration, and
mobility of carrier,
General Form of Hall Coefficient
For a semiconductor with significant concentrations of both types of carriers:
ph2  ne2
EH
RH 

JB e(ne  ph )
So if holes predominate (ph > ne ), RH > 0 and the material is said to be p-type,
while if RH < 0 (as for simple metals), the material is said to be n-type.
G. The Diode: A Simple p-n Junction
When p- and n-type materials are fabricated and brought together to form a junction,
we can easily analyze its electronic properties.
p
Near the junction the free electrons and holes
“diffuse” across the junction due to the
concentration gradients there. As this happens,
a contact potential  develops.
n
Na
Nd
p
-
+
+
+
+
E
The field E due to the contact potential inhibits
further flow of electrons and holes toward the
junction, and equilibrium is established at finite .
n
depletion
region
Physics of a Simple p-n Junction
We can also describe the situation in terms of flat band diagrams.
Ecn
Ecp
Initially (before equilibrium)
Evp
Evn
Ecp
e
Ecn
Finally (equilibrium established)
Evp
Evn
Here we see “band bending” in equilibrium. This reflects the potential at the junction
and the equalization of the chemical potential throughout the system.
Physics of a Simple p-n Junction
In this dynamic equilibrium two types of carrier fluxes are equal and opposite:
1. recombination flux: electrons in the n-type region and holes in the p-type region
“climb” the barrier, cross the junction, and recombine with h+/e- on the other side.
2. generation flux: thermally-generated electrons in the p-type region and holes in the
n-type region are “swept” across the junction by the built-in electric field there.
We can picture the carrier fluxes (currents):
In equilibrium at V=0:
Jng
J nr (0)  J ng (0)  0
Jnr
p-type
Jpr
n-type
Jpg
J pr (0)  J pg (0)  0
A Simple p-n Junction With Applied Voltage
Now when an external voltage V is applied to the junction, there are two cases:
1. Forward bias: electrons in the n-type region are shifted upward in energy
e
Generation currents are not affected
since they depend on excitation across
band gap:
eV
J ng (V f )  J ng (0)   J nr (0)
J pg (V f )  J pg (0)  J pr (0)
Recombination currents are increased by a
Boltzmann factor, since they depend on
carriers climbing the potential energy step
at the junction (and Maxwell-Boltzmann
statistics applies):
J nr (Vf )  J nr (0)eeV / kT
J pr (V f )  J pr (0)eeV / kT
Current-Voltage Relation for A Simple p-n Junction
We can now calculate the net current density from both holes and electrons:




J (V f )  J nr (V f )  J ng (V f )  J pr (V f )  J pg (V f ) J nr (0)  J pr (0) eeV / kT  J nr (0)  J pr (0)



J  J nr (0)  J pr (0) eeV / kT 1


I (V )  I s eeV / kT 1
I > 0 so current
flows from p  n
Is = “saturation current”
2. Reverse bias: electrons in the n-type region are shifted downward in energy
e
eV
Here the only difference is that
recombination currents are decreased by a
Boltzmann factor, which just changes the
sign of V in the exponential terms. So the
resulting current is:

I (V )  I s e
We can express both cases in one “ideal
diode equation” if we define forward bias
to be V > 0 and reverse bias to be V < 0:
eV / kT


1
I < 0 so current
flows from n  p
(leakage current)

I (V )  I s eeV / kT 1
Current-Voltage Relation for A Real Diode
What about in
the real world?


I (V )  I s eeV / kT 1 ???
The ideal diode equation is approximately
correct, but we have made some
assumptions that are not rigorously true,
and have neglected other effects, so in a
real diode we see behavior like this:
I
Vbr
V
-Is
Mechanisms for breakdown in reverse bias include:
1. Zener breakdown: a large reverse bias allows tunneling of electrons from valence
band of p-type region to conduction band of n-type region, where they can carry current!
2. Avalanche breakdown: electrons generated in p-type region and swept across the
junction acquire enough kinetic energy to generate other electrons, which in turn
generate more, etc.