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

Lundstrom ECE-656 F15

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=

Lundstrom ECE-656 F15 4

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

10 Lundstrom ECE-656 F15

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

11 Lundstrom ECE-656 F15

D1D E( )dE = 2

π!m*

2EdE = 2

π!υdE

2 Lπ

dk

E

k

dk

dE

υ = !k

m* = 2E m*

Lundstrom ECE-656 F15 12

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 ✓

Lundstrom ECE-656 F15

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

Lundstrom ECE-656 F15 16

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)

Lundstrom ECE-656 F15 17

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∑

Lundstrom ECE-656 F15 18

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

Lundstrom ECE-656 F15 19

DOS for a Si quantum well

g2D E( )

Eε2ε1

g2D

i (E) = gVmi*

π2

sp3s*d5 TB calculation by Yang Liu, Purdue University, 2007

Lundstrom ECE-656 F15 20

DOS for a Si quantum well

sp3s*d5 TB calculation by Yang Liu, Purdue University, 2007

Lundstrom ECE-656 F15 21

summary

1)  DOS in energy depends on dimension and on the dispersion.

2)  The DOS becomes complicated at high energies.