Principles of Semiconductor Devices-L8
Principles of Semiconductor Devices-L8
Principles of Semiconductor Devices-L8
www.nanohub.org
EE‐606: Solid State Devices
EE‐606: Solid State Devices
Lecture 8: Density of States
Muhammad Ashraful Alam
alam@purdue.edu
Alam ECE‐606 S09 1
Outline
1) Calculation of density of states
2) Density of states for specific materials
Density of states for specific materials
3) Characterization of Effective Mass
4) Conclusions
Reference: Vol. 6, Ch. 3 (pages 88‐96)
Alam ECE‐606 S09 2
Density of States
E
A single band has total of N‐states
4
O l f ti
Only a fraction of states are occupied
f t t i d
How many states are occupied upto E?
3
Or equivalently…
2
How many states per unit energy ? (DOS)
1
k
Alam ECE‐606 S09 3
Density of States in 1‐D Semiconductors
N atoms
a
Δk
States between E1 +ΔE & E1 = 2 × E
δk
Δk k1+Δk
= 2× k1
2π Na E1+ΔE
E1
Na Δk
States/unit energy @ E1 =
π ΔE k
2π
π
δk = a
Na
Alam ECE‐606 S09 4
1D‐DOS L N atoms
N a Δk a
States/unit energy @ E =
π ΔE
2m* ( E − E0 )
E − E0 =
=2k 2
⇒k= E
2m *
=2
dk m* k1+Δk
=
dE 2= 2 ( E − E0 ) k1
E1+ΔE
L m* E1
States/unit energy @ E =
π 2= 2 ( E − E0 )
Alam ECE‐606 S09 5
1D‐DOS
E E DOS =
1 m*
π 2= 2 ( E − E0 )
k
π DOS
2π a
Δk =
Na
Conservation of DOS
Alam ECE‐606 S09 6
Density of States in 2D Semiconductors
b
a
Alam ECE‐606 S09 7
Density of States in 3D Semiconductors
Macroscopic Sample
States between E1 +ΔE & E1
4 4
π ( k + dk ) − π k 3
3
V H
=3 3 = 2 k 2 Δk
2π 2π 2π 2π
W
L W H
L
V 2 Δk
States/unit energy @ E = 2 k
2π dE 2π/H
2π/W
=2k 2 2m* ( E − E0 ) dk m* 2π/L
E − E0 = ⇒k= ⇒ =
2m *
= 2
dE 2= 2 ( E − E0 )
k
K+dk
States/unit energy/unit volume @ E1
m*
DOS = 2 3 2m* ( E − E0 )
2π =
Alam ECE‐606 S09 8
3D‐DOS
m3* 2m3* ( E − E0 )
DOS =
E E π 2 =3
k
π π DOS
−
a 2π a
Δk =
Na
Conservation of DOS
Alam ECE‐606 S09 9
Outline
1) Calculation of density of states
2) Density of states for specific materials
Density of states for specific materials
3) Characterization of Effective Mass
4) Conclusions
Alam ECE‐606 S09 10
Density of States of GaAs: Conduction/Valence Bands
m*n 2m*n ( E − Ec )
gc ( E ) =
2π 2 =3
⎧ m* 2m* ( E − E )
υ
⎪ hh hh
⎪ 2π 2 =3
⎪
gυ ( E ) = ⎨
⎪ *
lh ( E − Eυ )
*
⎪ lh
m 2 m
⎪⎩ 2π 2 =3
11
Four valleys inside BZ for Germanium
Alam ECE‐606 S09 12
Ellipsoidal Bands and DOS Effective Mass
k2 const. E
t E
=k
2
= k2
2
= k3
2 2 2 2
E − EC = + 1
*
+
2m 2mt*
l 2mt* E=const. ellipsoid
k12 k2 2 k3 2
k1
1= + +
⎡ 2ml ( E − EC ) ⎤ ⎡ 2mt ( E − EC ) ⎤ ⎡ 2mt* ( E − EC ) ⎤
* *
k3
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
⎣ = 2
⎦ ⎣ = 2
⎦ ⎣ = 2
⎦
α2 β 2
Transform into …
⎛4 2⎞ 4 3
Vk = N el ⎜ παβ ⎟ ≡ π keff
⎝3 ⎠ 3
3
2m ( E − Ec ) 2m ( E − Ec ) 2m ( E − Ec ) 4 ⎡ 2meff ( E − Ec ) ⎤
* * * *
4
N el π l t t
≡ π⎢ ⎥
3 =2 =2 =2 3 ⎢ =2 ⎥⎦
⎣
m *
eff =N 23
el (m m )*
l
*2 1 3
t
13
Alam ECE‐606 S09
DOS Effective Mass for Conduction Band
m *
eff =4 23
(m m )
*
l
*2 1 3
t m *
eff =6 23
(m m )
*
l
*2 1 3
t
14
Outline
1) Calculation of density of states
2) Density of states for specific materials
Density of states for specific materials
3) Characterization of Effective Mass
4) Conclusions
Alam ECE‐606 S09 15
Measurement of Effective Mass
ν0=24 GHz B field variable …
(fixed)
Iin Iout
kz
qBcon
m* =
2π v0
ky kx
Iout - Iin
Bcon B 16
Motion in Real Space and Phase Space
Energy=constant.
x Liquid He temperature …
ky
(kx,0)
((0,k
, y)
y
kx
(-kx,0) (0,-ky)
kz
ky kx
qB0
Derive the Cyclotron Formula m* =
2π v0
For an particle in (x‐y) plane with B‐field in z‐direction,
the Lorentz force is
the Lorentz force is …
B
m* υ 2
= qυ × Bz = qυ Bz
r0
qB0 r0
υ=
m*
2π r0 2π m*
τ= =
υ qB0
1 qB0
ν0 ≡ =
τ 2π m*
Alam ECE‐606 S09 18
Effective mass in Ge
4 angles between B field and the ellipsoids …
Recall the HW1
Alam ECE‐606 S09 19
Derivation for the Cyclotron Formula
1 cos 2 θ sin 2 θ
Show that 2
= 2
+ Given three mc and three θ,
mc mt ml mt we will Find mt, and ml
The Lorentz force on electrons in a B‐field
dυ
F = qυ × B = [ M ]
dt
In other words, B
dυ x
Fx = q (υ y Bz − υ z By ) = m*
t
dt
dυ
Fy = q (υ z Bx − υ x Bz ) = mt* y
dt
* dυ z
Fz = q (υ x By − υ y Bx ) = ml
dt
Alam ECE‐606 S09 20
Continued … kz
Let (B) make an angle (θ) with longitudinal axis of the
ellipsoid (ellipsoids oriented along kz)
B0
Bx = B0 cos (θ ) , By = 0 , Bz = B0 sin (θ ) , ky
kx
Differentiate (vy) and use other equations to find …
d 2υ y
+ υ yω 2 = 0 with ω 2 ≡ ⎡⎣ωt wl sin 2 θ + ωt 2 cos 2 θ ⎤⎦
dt 2
q 0
qB qB
q 0 qB
q 0
ω0 ≡ * ωt ≡ * ωl ≡ *
mc mt ml
1 sin 2 θ cos 2 θ
so that … = +
(m )
* 2
c
ml mt mt 2
Alam ECE‐606 S09 21
Measurement of Effective Mass
B=[0.61, 0.61, 0.5]
[110]
qB1
Three peaks B
Three peaks B1, B B2, B
B3 mc =
1 cos 2 θ sin 2 θ 2π v0
2
= 2
+ Three masses mc1,mc2,mc3
mc mt ml mt Three unique angles: 7, 65, 73
Known θ and mc allows calculation of mt and ml.
Alam ECE‐606 S09 22
Valence Band Effective Mass
Alam ECE‐606 S09 23
Conclusions
1) Measurement of Effective mass and band gaps define
the energy‐band of a material.
2) Only a fraction of the available states are occupied. The
number of available states change with energy. DOS
captures this variation.
3)) DOS is an important and useful characteristic of a
p
material that should be understood carefully.
Alam ECE‐606 S09 24