E vs k diagram - INAOE - Pjmolina/2010_Part3-06-08_Bandas de e… · E vs k diagram m k m p E 2 2!2...
Transcript of E vs k diagram - INAOE - Pjmolina/2010_Part3-06-08_Bandas de e… · E vs k diagram m k m p E 2 2!2...
E vs k diagram
m
k
m
pE
22
222
The parameter k is called the crystal momentum and
is a parameter that results from applying Schrödinger
wave equation to a single-crystal lattice.
Electrons traveling in different directions encounter different potential patterns
and therefore different k-space boundaries. The E vs k diagrams are in general
a function of the k-space direction in a crystal.
111Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
22
2
/*
dkEdm
*
22
2)(
e
Cm
kEkE
When the conduction band edge occurs at k = 0, we
can represent the band-structure as a simple parabola,
1
2
2
2
* 1
dk
Edme
where:
EC = minimum of the conduction band energy
me* = effective mass of an electron,
The effective mass of an electron in a band
with a given (E, k) relationship
Effective mass of carriers – electrons
mdk
Ed 2
2
2
Narrow parabola
small effective mass
GaAs me* = 0.063m0
Ge me* = 0.55m0
Si me* = 1.09m0
112Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Curso propedéutico de Electrónica INAOE 2008 Dr. Joel Molina & Dr. Pedro Rosales
*
22
2h
Vm
kEE
Similarly, the energy momentum relation
for the valence band can be written…
where:
EV = maximum of the valence band energy
mh* = effective mass of a hole.
There are actually two bands near the top
of the valence band of different widths,
leading to heavy holes and light holes.
GaAs mhh* = 0.45m0 mlh* = 0.08m0
Si mhh* = 0.49m0 mlh* = 0.16m0
113
Effective mass of carriers – holes
A perfect semiconductor crystal with no
impurities or lattice defects is called an intrinsic
semiconductor. In such material there are no
charge carries at T = 0 K.
At higher temperatures EHPs are generated as
valence band electrons are thermally excited
across the bandgap to the conduction band.
If a steady state carrier concentration is
maintained, there must be recombination of
EHPs at the same rate at which they are
generated ri = gi
114Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Intrinsic material
In addition to the intrinsic carriers generated thermally … When a group V or III
atom (As, B) is substituted into the Si lattice an electron is donated or accepted and
the semiconductor becomes n-type or p-type respectively.
In an extrinsic semiconductor at any
temperature the carriers concentration have
two contributions:
1. Thermal
2. Doping [ ND or NA ]
For a n-type semiconductor:
n = ND and p = ni2 / ND
For a p-type semiconductor:
p = NA and n = ni2 / NA
[ND]
[NA]
The crystal is extrinsic when the
doping is such that:
(n0, p0) ni
115Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Extrinsic material
At 0K the extra electrons associated with the donor atoms are ‘fixed’ to
the donor sites at an energy level Ed.
As the temperature increases there is enough thermal energy to ionize
the donor atoms i.e. for an electron to make the transition into the
conduction band which is only an energy jump of Ed where Ed <<Eg.
To create holes in the valence band in a p-type semiconductor, electrons
need only an energy of Ea to reach the acceptor level where Ea <<Eg.
116Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
25 meV for Si
7 meV for GaAs
50 meV for Si, GaAsVt= kBT/q at 300 K = 26 meV
In reality, different dopants have different ionization levels and deep levels (|E| > 3kBT), which can be important, but this simple model gives the correct order of magnitude.
He
S
D Em
mE
0
*2
0
hydrogen
energy levels
donor levels
acceptor levels
117Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Approximate energy required to excite the 5th electron of a donor atom into
the conduction band:
Bohr Model
Ionization level of dopants
Ionization level of dopants
118Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
He
S
D Em
mE
0
*2
0
119Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
A quantum well laser (QWL) is a laser diode in which the active region of the device is so narrow that quantum confinement occurs.
The wavelength of the light emitted by a QWL is determined by the width of the active region rather than just the bandgap of the
material from which it is constructed. This means that much shorter wavelengths can be obtained from QLW than from conventional
laser diodes using a particular semiconductor material. The efficiency of a QLW is also greater than a conventional laser diode due to
the stepwise form of its density of states function.
Electrons and holes in Quantum Wells
To obtain the carrier density per unit volume we must first calculate the
number of allowed states (including spin) per energy range per unit volume.
For electrons in the conduction band
where the E-k relation is of the form,
*
22
2 e
Cm
kEE
21
23
2
*24 E
h
mEN e
The density of states is given by:
See appendix IV of Streetman
*
22
2 h
Vm
kEE
Similarly, for holes in the valence band
where the E-k relation is of the form,
21
23
2
*24 E
h
mEN h
The density of states is given by:
23
*23
*23
*
hhlhh mmmAccounting for the
contribution from both
light and heavy holes
120Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Density of states function
The probability that an electron occupies an electronic state with energy E
is given by the Fermi-Dirac distribution function:
TkEE BFEF
/exp1
1)(
The distribution of
electrons over a range of
allowed energy levels at
thermal equilibrium.
For E=EF then F(E)=0.5The Fermi energy is the energy for which the
probability of occupation by an electron is exactly ½
121Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Fermi-Dirac distribution function
Curso propedéutico de Electrónica INAOE 2008 Dr. Joel Molina & Dr. Pedro Rosales
TkEE
BFBFeEFTkEE
3
TkEE
BFBFeEFTkEE
13
The Fermi distribution function is simplified
for an electron in the conduction band since,
and for a hole in the valence band since,
For Si at 300 K:
ni=pi 1010 cm-3
Density of available states at Ev and Ec:
1019 cm-3
Because of the relatively large density of states in
each band, small changes in f(E) can result in
significant changes in carrier concentration.
122
The concentration of
the electrons in the
conduction band is:
Ec
dEENEfn )()(
Ec
dEENEfp )()](1[
The concentration of
the holes in the
valence band is:
The distribution of electrons in the conduction band is given by the density of allowed
quantum states times the probability that a state is occupied by an electron.
123Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
toptop EE
dEEFENdEEnn00
taking the bottom of the conduction band to be E=0
The electron density in the conduction band is given by,
Taking the previous simplified expression for F(E),
dETk
EEE
h
mn
B
Fe
exp2
40
2
*
21
23
124Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Electron concentration
dxexTk
ETk
h
mn x
B
FB
e
0
2
*
21
23
23
exp2
4
Let x = E / kT
Tk
E
h
Tkmn
B
FBe exp2
2
23
2
*
The effective density of states in the conduction band, NC
125Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Electron concentration
For Si (300 K) NC = 2.8 x 1019 cm-3
For GaAs (300 K) NC = 4.7 x 1017 cm-3
Taking the bottom of the conduction band as EC rather than E=0,
Tk
EENn
B
FCC exp
126Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Electron concentration
Where:
2/3
2
*22
h
TkmNc Be
For Si (300 K) NV = 1.04 x 1019 cm-3
For GaAs (300 K) NV = 7 x 1018 cm-3
Tk
EENp
B
VFV exp
127Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Hole concentration
Similarly for holes in the valence band,
2/3
2
*22
h
TkmNv Bh
Where:
Tk
ENNnnp
B
g
VCi exp2
Tk
ENNn
B
g
VCi2
exp
This expression is independent of EF and is valid for extrinsic (doped) semiconductors too.
For Si (300 K) ni = 9.65 x 109 cm-3
For GaAs (300 K) ni = 2.25 x 106 cm-3
128Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Law of Mass Action
Tk
EEnn
B
iF
iexp
Tk
EEnp
B
Fii exp
The product np is then,
Tk
EEn
Tk
EEnnp
B
iFi
B
Fii expexp
2
innp
So the law of mass action holds also for extrinsic semiconductors.
129Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Law of Mass Action. Intrinsic Semiconductors
Calculation of extrinsic Fermi level
Tk
EENn
B
FCC exp
Tk
EENN
B
FCCD exp
D
CBFC
N
NTkEE ln
At 300 K there is usually enough thermal energy to completely ionize the
dopant atoms, so for n-type semiconductor n = ND (donor concentration)
So as the concentration of donor atoms increases the Fermi
level moves closer to the bottom of the conduction band
130Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
So as the concentration of acceptor atoms increases the Fermi
level moves closer to the top of the valence band
A
VBVF
N
NTkEE ln
Tk
EENN
B
VFVA exp
Tk
EENp
B
VFV exp
Similarly for p-type semiconductors, p = NA (acceptor concentration)…
131Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Calculation of extrinsic Fermi level
It is often useful to express the carrier density in terms of the intrinsic
carrier concentration and the intrinsic Fermi level…
Similarly for holes,
Tk
EE
Tk
EENn
B
Fi
B
iCC expexp
Tk
EEnp
B
Fii exp
For electrons,
132Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Calculation of extrinsic Fermi level
Tk
EEnn
B
iFi exp
Intrinsic carrier densities in
Si, Ge and GaAs as a
function of the reciprocal of
temperature.
133Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Intrinsic carrier densities ni as a function of temperature
Tk
ENNn
B
g
VCi2
exp
For Si (300 K) ni = 9.65 x 109 cm-3
For GaAs (300 K) ni = 2.25 x 106 cm-3
Eg (Ge) = 0.66 eV
Eg (Si) = 1.12 eV
Eg (GaAs) = 1.42 eV
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
0 200 400 600 800 1000 1200
Temperature (K)
En
erg
y B
an
dg
ap
(eV
)
Eg (GaAs) = 1.42 eV
Eg (Si) = 1.12 eV
Eg (Ge) = 0.66 eV
The temperature dependence of the energy bandgap, Eg, has been
experimentally determined yielding the following expression for Eg as a
function of the temperature, T:
where Eg(0), a and b are the
fitting parameters. These fitting
parameters are listed for
germanium, silicon and gallium
arsenide in Table.
134Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Energy gap Eg as a function of temperature
At low temperatures the thermal
energy is insufficient to ionize all
donor atoms so n < ND
At higher temperatures the thermal
energy is sufficient to ionize all
donor atoms so n = ND
At some temperature the intrinsic
carrier density becomes
comparable to the donor
concentration and beyond this point
the semiconductor becomes
intrinsic.
135Curso propedéutico de Electrónica INAOE 2010 Dr. Joel Molina & Dra. Claudia Reyes
Electron density n0 as a function of temperature