Homogeneous Semiconductors
• Dopants
• Use Density of States and Distribution
Function to:
• Find the Number of Holes and Electrons.
Energy Levels in Hydrogen Atom
Energy Levels for Electrons in a Doped Semiconductor
Assumptions for Calculation
Density of States
(Appendix D)
Energy Distribution Functions
(Section 2.9)
Carrier Concentrations
(Sections 2.10-12)
GOAL:
• The density of electrons (no) can be found
precisely if we know
1. the number of allowed energy states in a small energy
range, dE: S(E)dE
“the density of states”
2. the probability that a given energy state will be
occupied by an electron: f(E)
“the distribution function”
no =  S(E)f(E)dE
band
For quasi-free electrons in the conduction band:
1. We must use the effective mass (averaged over all directions)
2. the potential energy Ep is the edge of the conduction band (EC)
*
1

1  2mdse
S ( E )  2  2 E  EC 2
2   
For holes in the valence band:
1. We still use the effective mass (averaged over all directions)
2. the potential energy Ep is the edge of the valence band (EV)
*
1

1  2mdsh
S ( E )  2  2 EV  E 2
2   
Energy Band Diagram
Eelectron
E(x)
conduction band
EC
S(E)
EV
valence band
Ehole
x
note: increasing electron energy is ‘up’, but increasing hole energy is ‘down’.
Reminder of our GOAL:
• The density of electrons (no) can be found
precisely if we know
1. the number of allowed energy states in a small energy
range, dE: S(E)dE
“the density of states”
2. the probability that a given energy state will be
occupied by an electron: f(E)
“the distribution function”
no =  S(E)f(E)dE
band
Fermi-Dirac Distribution
The probability that an electron occupies an
energy level, E, is
f(E) = 1/{1+exp[(E-EF)/kT]}
– where T is the temperature (Kelvin)
– k is the Boltzmann constant (k=8.62x10-5 eV/K)
– EF is the Fermi Energy (in eV)
– (Can derive this – statistical mechanics.)
f(E) = 1/{1+exp[(E-EF)/kT]}
T=0 oK
T1>0
T2>T1
1
f(E)
0.5
0
E
EF
All energy levels are filled with e-’s below the Fermi Energy at 0 oK
Fermi-Dirac Distribution for holes
Remember, a hole is an energy state that is NOT occupied by
an electron.
Therefore, the probability that a state is occupied by a hole
is the probability that a state is NOT occupied by an electron:
fp(E) = 1 – f(E) = 1 - 1/{1+exp[(E-EF)/kT]}
={1+exp[(E-EF)/kT]}/{1+exp[(E-EF)/kT]} 1/{1+exp[(E-EF)/kT]}
= {exp[(E-EF)/kT]}/{1+exp[(E-EF)/kT]}
=1/{exp[(EF - E)/kT] + 1}
The Boltzmann Approximation
If (E-EF)>kT such that exp[(E-EF)/kT] >> 1 then,
f(E) = {1+exp[(E-EF)/kT]}-1  {exp[(E-EF)/kT]}-1
 exp[-(E-EF)/kT] …the Boltzmann approx.
similarly, fp(E) is small when exp[(EF - E)/kT]>>1:
fp(E) = {1+exp[(EF - E)/kT]}-1  {exp[(EF - E)/kT]}-1
 exp[-(EF - E)/kT]
If the Boltz. approx. is valid, we say the semiconductor is non-degenerate.
Putting the pieces together:
for electrons, n(E)
f(E)
T=0 oK
T1>0
T2>T1
1
S(E)
0.5
0
E
EV EF EC
n(E)=S(E)f(E)
E
Putting the pieces together:
for holes, p(E)
fp(E)
T=0 oK
1
T1>0
T2>T1
0.5
S(E)
0
EV EF EC
p(E)=S(E)f(E)
hole energy
E
Finding no and po
Ec (max)
n0 
 S ( E ) f ( E )dE 
Ec
1  2m

2 
2  
 N C exp[ ( EC  E F ) / kT ]
*
dse
2



3/ 2 
E  EC e  E  EF  / kT dE

Ec
 m kT 

...where N C  2
2 
 2 
*
dse
3/ 2
the effective density of states
in the conduction band
1  2m

p0   S ( E ) f p ( E )dE 
2 
2  
Ev (min)
Ev
 NV exp[ ( E F  EV ) / kT ]
*
dsh
2



3 / 2 Ev

EV  E e  EF  E  / kT dE

 m kT 

...where NV  2
2 
 2 
*
dsh
3/ 2
Energy Band Diagram
intrinisic semiconductor: no=po=ni
E(x)
conduction band
EC
n(E)
p(E)
EF=Ei
EV
valence band
x
where Ei is the intrinsic Fermi level
Energy Band Diagram
n-type semiconductor: no>po
n0  NC exp[ ( EC  EF ) / kT ]
E(x)
conduction band
EC
n(E)
EF
p(E)
EV
valence band
x
Energy Band Diagram
p-type semiconductor: po>no
p0  NV exp[ ( EF  EV ) / kT ]
E(x)
conduction band
EC
n(E)
p(E)
EF
EV
valence band
x
A very useful relationship
n0 p0  N C exp[ ( EC  EF ) / kT ]  NV exp[ ( EF  EV ) / kT ]
 N C NV e ( EcEv ) / kT  N C NV e
 Eg / kT
…which is independent of the Fermi Energy
Recall that ni = no= po for an intrinsic semiconductor, so
nopo = ni2
for all non-degenerate semiconductors.
(that is as long as EF is not within a few kT of the band edge)
n0 p0  N C NV e
ni  N C NV e
 Eg / kT
 Eg / 2 kT
 ni2
The intrinsic carrier density
n0 p0  N C NV e
ni  N C NV e
 Eg / kT
n
2
i
 Eg / 2 kT
is sensitive to the energy bandgap, temperature, and m*
 m kT 

N C  2
2 
 2 
*
dse
3/ 2
The intrinsic Fermi Energy (Ei)
For an intrinsic semiconductor, no=po and EF=Ei
N C exp[ ( EC  Ei ) / kT ]  NV exp[ ( Ei  EV ) / kT ]
which gives
Ei = (EC + EV)/2 + (kT/2)ln(NV/NC)
so the intrinsic Fermi level is approximately
in the middle of the bandgap.
Higher Temperatures
Consider a semiconductor doped with NA ionized
acceptors (-q) and ND ionized donors (+q), do not assume
that ni is small – high temperature expression.
positive charges = negative charges
po + ND = no + NA
using ni2 = nopo
ni2/no + ND = no+ NA
ni2 + no(ND-NA) - no2 = 0
no = 0.5(ND-NA)  0.5[(ND-NA)2 + 4ni2]1/2
we use the ‘+’ solution since no should be increased by ni
no = ND - NA in the limit that ni<<ND-NA
Temperature variation of some important “constants.”
Simpler
Expression
Degenerate Semiconductors
impurity band
EC
ED1
+ + + +
Eg(ND)
1. The doping concentration is so
high that EF moves within a few kT
of the band edge (EC or EV).
Boltzman approximation not valid.
2. High donor concentrations cause
the allowed donor wavefunctions to
overlap, creating a band at Edn.
First only the high states overlap, but
eventually even the lowest state
overlaps.
Eg0
This effectively decreases the
bandgap by
EV
for ND > 1018 cm-3 in Si
DEg = Eg0 – Eg(ND).
Degenerate Semiconductors
As the doping conc. increases more, EF rises above EC
DEg
available impurity band states
filled impurity band states
EC (intrinsic)
EF
EC (degenerate) ~ ED
apparent band gap narrowing:
DEg* (is optically measured)
Eg* is the apparent band gap:
an electron must gain energy Eg* = EF-EV
-
EV
Electron Concentration
in degenerately doped n-type semiconductors
The donors are fully ionized: no = ND
The holes still follow the Boltz. approx. since EF-EV>>>kT
po = NV exp[-(EF-EV)/kT]
= NV exp[-(Eg*)/kT]
= NV exp[-(Ego- DEg*)/kT]
= NV exp[-Ego/kT]exp[DEg*)/kT]
nopo = NDNVexp[-Ego/kT] exp[DEg*)/kT]
= (ND/NC) NCNVexp[-Ego/kT] exp[DEg*)/kT]
= (ND/NC)ni2 exp[DEg*)/kT]
Summary
non-degenerate:
nopo= ni2
degenerate n-type:
nopo= ni2 (ND/NC) exp[DEg*)/kT]
degenerate p-type:
nopo= ni2 (NA/NV) exp[DEg*)/kT]