Transcript Solid State Physics II - West Virginia University

Fire Eating Demo
While I don’t encourage this at home, if you do decide to try this, here are some tips
that may help prevent an accident.
Breathe in before (prevents from: inhaling fire = bad)
Wet your lips (prevents from: burning lips = bad)
Extinguish quickly (don’t want wick hot, burning mouth = bad)
Pull long hair back (burning hair = bad)
Clothing: Synthetic fabrics are less likely to catch fire.
Adding Constant Heat to H2O
What happens to slope
near phase change?
Heat capacity related to slope
Big changes at phase boundaries
Many solids conduct electricity
There are electrons that are not bound but are able to move through the crystal.
 = n e2  / m
( = 1/)
Conducting solids fall into two main classes; metals and semiconductors.
 ( RT )metals ;106  108   m
and increases by the addition of small
amounts of impurities (semiconductors do opposite).
In a metal, the conductivity  (1/resistivity )
normally gets smaller (or resistivity
increases) at higher temperatures. Why?
Why smaller?
Semiconductors
A small fraction of electrons are
thermally excited into the
conduction band. These electrons
carry current just as in metals.
The smaller the gap the more electrons in the conduction
band: fraction ~ e-Eg/2kBT
Resistivity decreases with temperature (more excited e-s)
Insulator
Semiconductor @ low temp
Semiconductor @ high
Impurties have a
big effect on
semiconductors
Why?
Impurities can add states
within the band gap that
makes excitation (and
thus conduction) much
more likely (convert)
Optical Properties
• If a semiconductor or insulator does not have many
impurity levels in the band gap, photons with energies
smaller than the band gap energy can’t be absorbed
• What applications might this be useful for?
• Sunglasses or Si filters (transmission of infra red light with
simultaneous blocking of visible light)
• This is why many insulators are transparent to visible light
(e.g. glass), whereas semiconductors (Si, GaAs) are not
Photoconductivity
• Charge carriers (electrons or
holes or both) created in the
corresponding bands by
absorbed light can also
participate in current flow,
and thus should increase the
current for a given applied
voltage, i.e., the conductivity
increases
• This effect is called
Photoconductivity
•
•
•
•
If want conductivity to be controlled by light. Need what?
So want few carriers in dark → semiconductor
But want light to be absorbed, creating photoelectrons
→ Band gap of intrinsic photoconductors should be smaller
than the energy of the photons that are absorbed
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
10
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”
11
Indirect Gap Semiconductors
Indirect gap semiconductors: the top of the valence band and the
bottom of the conduction band occur at different k-vector.
Direct bandgap
0.8eV
Indirect bandgap
@ k =0, is 0.66eV
Germanium (Ge)
12
Indirect Optical Absorption
Conduction Band
Energy
hnphoton
~Eg
kphonon
k
How would you expect
the optical absorption
to look this time?
Valence Band
A transition across an indirect band gap
requires a photon to be absorbed and a
phonon (thermal excitation) to be absorbed
or emitted.
Ge: Indirect Optical Absorption
Indirect gap semiconductor: no sharp onset of absorption
0.80
0.88
0.73
0.66
For T = 300K Eg (indirect gap) = 0.66 eV and EG1 (direct gap) = 0.8 eV
Why bigger at low temp?
How should this change with lower temperature?
For T = 77K Eg (indirect gap) = 0.73 eV and EG1 (direct gap) = 0.88 eV
Conductivity in Semiconductors
• What do we need to know in order to predict
the conductivity for semiconductors.
• For metals?
 = n e2  / m
• How did we find n, the density of conduction
electrons, for metals?
15
Electron and Hole Densities
• To determine the number of carriers in each
band, we will modify what we derived for free
electrons in metals

n
 g ( ) f ( )d

• Electron Fermi-Dirac distribution function
means what?
-1


E- 
fe (E) =  exp
 + 1 
 kB T 


f ( )
Chemical potential or Fermi level
At a temperature T the probability that an available electron state is
occupied is given by the Fermi-Dirac function


E- 
fe (E) =  exp
 + 1 
 kB T 


-1
The chemical potential,  , is the energy for which f = ½
(50% likely to be occupied).
Fermi energy: all energy states are occupied below EF at T = 0 (metal).
If intrinsic,  is within the energy gap. In the conduction band
Therefore
E-
exp
  1
 kB T 
  (E - ) 
fe (E)  exp

 kB T 
µ
Then, how should f(E) change with temperature in a semiconductor?
Number of electrons in conduction band
Silicon
log plot
@ 300K
n ~ 2x1016 m-3
Note Units are
cm-3
Electron density
increases ~
exponentially
with temperature
(exponential would be
straight line in log plot)
  (E - ) 
fe (E)  exp

 kB T 
Electron and Hole Densities
• To determine the number of carriers in each
band, we will modify what we derived for free

electrons
n
 g ( ) f ( )d

• Fermi-Dirac distribution function
f ( )
for c and v
  (E - ) 
fe (E)  exp

 kB T 
• What dependence on energy should the DOS have?
Density of States
Conduction
Band
Assume: bottom of conduction band
and the top of valence band parabolic
conduction band
valence band
E = Ec + 2k2/2me*
E = Ev - 2k2/2mh*

1
Conduction g e (E) =
2 3 2me
2 
band
Valence
band
g h (E) =
1
2
2

* 3/2

2m 

* 3/2
h
3
Ec = Eg
( E  Eg )
( E )
1
2
1
2
Ev = 0
Valence
Band
D(E)
Conduction
Band
Valence
Band
0
Eg
E
g (E) 
V1
2 2
3/ 2 1/ 2
(2
m
)
E
3
Electron density in conduction band
Density of states
Distribution function
To find the electron
density of occupied
states:
1  2me

g e (E) =
2
2   2
*
3/2
1

 ( E  Eg ) 2


  (E - ) 
fe (E)  exp

 kB T 

Ne 
 g ( E ) f (E)dE
e
e
Eg
3
 me * k BT  2
N e  2
 exp (   E g ) / k BT ) 
2
 2 
Ne  Nc exp(  Eg ) / kBT ) where
m k T 

N c  2
 2 
*
e B
2
3
2
Number of holes and electrons
Exactly same argument for holes in the valence band gives
Distribution
function
  (  E) 
f h (E)  1  f e ( E )  exp

 kBT 
Total density of
hole states
Nh 
Ev  0
g
v
( E ) f v (E)dE

3
m k T  2
 exp(  ) / k BT   N v exp(  ) / k BT 
N h  2
 2 
3
*
h B
2
Ne  Nc exp(  Eg ) / kBT )
3
 me*k BT 

N c  2
2 
2




2
Group: Find the product NeNh.
 k BT  * * 3 2
Ne N h  4
(me mh ) exp( Eg / k BT )
2 
 2 
Independent

of

As Ne=Nh(intrinsic) we can say: Ne  Nc Nv exp  Eg / 2k BT )
This result is important; known as law of mass
action; find carrier density without µ.
True for both intrinsic & extrinsic semiconductors.

Chemical Potential
For intrinsic SCs, determine an approach that
will find  (the energy for which f = ½).
When is  = Eg/2?
Results of recent discussion:
3
m k T  2
 exp(  ) / k BT   N v exp(  ) / k BT 
N h  2
 2 
*
h B
2
Ne  Nc exp(  Eg ) / kBT )
m k T 

N c  2
 2 
*
e B
2
3
2
Intrinsic semiconductors Ne=Nh
In pure “intrinsic” semiconductors the electrons and holes arise
only from excitation across the energy gap.
Therefore Ne = Nh
m  exp (  E
*
e
3
2
g
) / k T )   m 
*
h
B
3
2
exp   / k BT 
*

1
3
mh 
  E g  k B T n  * 
2
4
 me 
Silicon Eg=1.15eV
m*e = 0.2me & m*h = 0.8me @ 300K.
Gives
E

g
2
 0.026
eV
Chemical potential
near middle of gap
Doping
• Semiconductors can be easily doped
• Doping is the incorporation of
[substitutional] impurities into a
semiconductor in a controlled manner
• In a doped material, Ne is not equal to Nh
For example, impurities change the
conductivity of the material so that it can
be fabricated into a device
Extrinsic Semiconductors
• Electrical Properties of Semiconductors can
be altered drastically by adding very small
amounts of suitable impurities to the pure
crystals
• Impurities: Atoms of the elements different
from those forming solid
– Substitutional: “foreign” atoms occupying the
sites of host atoms
– Interstitial: “foreign” atoms “squeezed”
between regular sites crystal sites
Donors
• Let’s use Silicon (Si) as an example
– Substitute one Si (Group IV) atom with a
Group V atom (e.g. As or P)
– Si atoms have four valence electrons that
participate in covalent bonding
– When a Group V atom replaces a Si atom, it
will use four of its electrons to form the
covalent bonding
– What happens with the remaining electron?
Donors
• The remaining electron will not be
very tightly bound, and can be easily
ionized at T > 0K
• Ionized electron is free to conduct
– In terms of the band structure, this
electron is now in the conduction band
• Such Group V impurities are called
Donors, since they “donate” electrons
into the Conduction Band
– Semiconductors doped by donors are
called n-type semiconductors (extra
electrons with negative charge)
This crystal has been doped with a pentavalent impurity.
The free electrons in n type silicon support the flow of current.
Donors: Energy Levels
•
Such impurities create
“shallow” levels - levels within
the band gap, close to the
conduction band
• A donor is similar to a hydrogen
atom (called hydrogenic donors)
– A positive charge with a
single electron within its
potential
– the small ionization energy
means a sizable fraction of
donor atoms will be ionized
at room temperature 
Another View: Hydrogenic Donors
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 the positive
charge on the donor atom.
-e
+ve
ion
Conduction
Band
DE
Ec =
Eg
ED
The electron bound to the positive
ion is in an energy state ED = Eg- DE
where DE is the binding energy.
How much is DE?
Valence
Band
Ev = 0
Treat Similar to Hydrogen
• Employ the solution for hydrogen atom
– Consider the energy of the bottom of the
conduction band to be zero – “free” electron
– Substitute the effective mass for the electron mass
– Charge shielded in a solid so modify the
Coulomb interaction by the dielectric constant of
the solid (dielectric constant for free space, κ = 1)
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 (= 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  ~15.
13.6 me*
EB  2
eV
 me
0.53  r
a0  *
me / me
~ 50 Å
~10 meV.
33
Examples
• Ge: me* = 0.04m0; κ = 16  ED = -2.1 meV
• GaAs: me*= 0.067m0; κ = 13  ED = -5.4 meV
• Si: me* = 0.26m0; κ = 12  ED = -25 meV
• ZnSe: me* = 0.21m0; κ = 9  ED = -35 meV
Chemical Potential in n-type
At low temp, the chemical
potential must lie
somewhere between the
donor levels and the
conduction band.
f ( )  1 / 2
At higher temp, when the
donor level is depleted of
electrons, the contribution
from intrinsic electrons to
the electrical conductivity
becomes more substantial
and the chemical potential
tends towards ~Eg/2.
Learning Objectives
By the end of this section you should be able to:
• Distinguish between direct and indirect
semiconductors and know the difference
between their optical absorption
• Calculate carrier density, chemical
potential, and donor binding energies
Alternate detailed source: Pierret’s Advanced
Semiconductor Fundamentals, Ch. 4