5 Density of States - nanoHUB5_Density_of_StatesV2.pdf · 2015. 9. 29. · DOS: k-space vs. energy...
Transcript of 5 Density of States - nanoHUB5_Density_of_StatesV2.pdf · 2015. 9. 29. · DOS: k-space vs. energy...
Lundstrom ECE-656 F15
ECE 656: Electrothermal Transport in Semiconductors Fall 2015
Density of States
Professor Mark Lundstrom
Electrical and Computer Engineering Purdue University, West Lafayette, IN USA
Revised: 9/29/15
density-of-states in k-space
2
Nk =2 ×
L2π
⎛⎝⎜
⎞⎠⎟=
Lπ
Nk =2 ×
A4π 2
⎛⎝⎜
⎞⎠⎟=
A2π 2
Nk =2 ×
Ω8π 2
⎛⎝⎜
⎞⎠⎟=
Ω4π 3
1D:
2D:
3D:
dk
x ydk dk
x y zdk dk dk
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DOS: k-space vs. energy space
3 Lundstrom ECE-656 F15
xdk
dE
E
k
!2kx2
2m*
2 xLπStates are uniformly
distributed in k-space,
but non-uniformly distributed in energy space. Depends on E(k) (e.g. different for parabolic bands and linear bands)
( ) ( )D E dE N k dk=
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conduction band non-parabolicity
E k( )
k1ε
2k 2
2m*
′E 1+α ′E( ) =
2k 2
2m*(0)
′E = E − ε1
effect on DOS
5 Lundstrom ECE-656 F15
E
k
dk2 Lπ
E k( )
dEHow many states in this range of energies?
effect on DOS
6
E
k
dk2 Lπ
E k( )
dEHow many states in this range of energies?
d ′k
′E k( )
effect on DOS
7
E
k
2 Lπ
E k( )
Nonparabolicity increases the DOS (E).
′E k( )parabolic
nonparabolic
Example: 1D
8 Lundstrom ECE-656 F15
E
k
dk2 Lπ
E k( )
Nkdk = dk
2π L( ) × 2dE
D1D E( )dE =
NkdkL
D1D E( ) #
J-m
1D DOS
9 Lundstrom ECE-656 F15
Nkdk = dk
2π L( ) × 2
D1D E( )dE =
NkdkL
D1D E( ) #
J-m
E = !
2k 2
2m*
D1D E( )dE = 1
πdk
dE = !
2kdkm*
k = 2m*E
!
dk = m*dE
!2k
D1D E( )dE = 1
πdk
D1D E( )dE = 1
π!m*
2EdE
1D DOS
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D1D E( )dE = 2
π!m*
2EdE
2 Lπ
dk
E
k
dk
dE
Multiply by 2 to account for the negative k-states.
1D DOS
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D1D E( )dE = 2
π!m*
2EdE = 2
π!υdE
2 Lπ
dk
E
k
dk
dE
υ = !k
m* = 2E m*
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alternative expression for DOS
D1D (E)dE = 1
LΔE ,Ek
k∑
D2 D (E)dE = 1
AΔE ,Ek
k∑
D3D (E) = 1
ΩΔE ,Ek
k∑
like a “Kronecker delta” one if: otherwise zero
E − dE 2 < Ek < E + dE 2
13
example: 2D DOS for parabolic energy bands
D2 D (E1) = 1
AΔ ′E ,Ek!
k∑
D2 D (E1) = 1A
δ E1,Ekk∑ → 1
AgV
A
2π( )2 × 2 2π k dkδ E1 − E k( )( )0
∞
∫
D2 D (E1) = gV
1π× k dkδ E1 − E k( )( )
0
∞
∫
E k( ) =
2k 2
2m* dE =
2 2kdk2m*
kdk = m*
2 dE
D2 D (E1) = gV
m*
π2 × dEδ E1 − E k( )( )0
∞
∫
14
example: 2D DOS for parabolic energy bands
D2 D (E1) = 1
AΔE1,Ek!
k∑ = gV
m*
π"2 ✓
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parabolic bands: 1D, 2D, and 3D
15
D2D E( ) = gV
m*
π2 Θ E − ε1( )
D3D E( ) = gV
m* 2m* E − EC( )π 23 Θ E − EC( )
D1D E( ) = 1
π2m*
E − ε1( )Θ E − ε1( )
D3D
E
D2D
E
D1D
E
E k( ) = EC + 2k 2 2m*( ) 15
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graphene (2D)
16
ky
( )E k
f1 E( )
“neutral point”
(“Dirac point”)
FE
kx
E k( ) = ±υF k = ±υF kx
2 + ky2
D2D
E
D2 D E( ) = 2 E
π2υF2
(parabolic)
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DOS for bulk Si
–5 –4 –3 –2 –1 0 1 2 3 4 5 60
2
4
6
ENERGY (eV)
DO
S (
1022
cm
–1 e
V–1
)
The DOS is calculated with nonlocal empirical pseudopotentials including the spin-orbit interaction. (Courtesy Massimo Fischetti, August, 2011.)
D3D (E1) = 1
ΩΔE1,E!k!
k∑
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DOS for a Si quantum well
E = εi +
2k||2
2mi*
E kx( )
k||
ε2ε1
2k||2
2mi*
g2D
i (E) = gVmi*
π2
g2D E( )
Eε2ε1
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DOS for a Si quantum well
g2D E( )
Eε2ε1
g2D
i (E) = gVmi*
π2
sp3s*d5 TB calculation by Yang Liu, Purdue University, 2007
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DOS for a Si quantum well
sp3s*d5 TB calculation by Yang Liu, Purdue University, 2007
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summary
1) DOS in energy depends on dimension and on the dispersion.
2) The DOS becomes complicated at high energies.