Density of States

ECE 656: Electronic Transport in Semiconductors Fall 2017

Density of States

Mark Lundstrom

Electrical and Computer Engineering Purdue University, West Lafayette, IN USA

Lundstrom8/29/17 ECE-656 F17 Outline

1) Counting states 2) DOS in k-space vs. DOS in E-space 3) Examples 4) Realistic DOS in semiconductors 5) DOS for phonons

Lundstrom ECE-656 F17 2 Energy levels in Si

Si atom (At. no. 14)

4S0

3P2 4 valence electrons 3S2 8 valence states energy 2P6 core 2 2S levels 1S2 3 Lundstrom ECE-656 F17 States in a Si crystal

• Only the valence states are of interest to us.

N ≈ 5 × 1022 cm-3 atoms • 3s and 3p orbitals hybridize and produce bonding and anti-bonding states.

• The interaction of the electron wavefunctions in the crystal broadens the discrete energy levels of the isolated Si atoms into energy bands.

4 5.43 A Lundstrom ECE-656 F17 Energy levels à energy bands

Si atom (At. no. 14) Si crystal

Etop C D E C ( ) 3P2 conduction band E bot • • • • • • • • • s-like 2 C energy 3S forbidden gap Etop V • • • • • • • • • p-like valence band D(E) tells us how D E the states are V ( ) E bot distributed with a V band. Lundstrom ECE-656 F17 5 States in a finite volume of semiconductor

0, L ,−L ( y z ) Finite volume, Ω 0, L ,0 (part of an infinite volume) ( y )

Finite number of states Ω = LLLxyz

0,0,0 L ,0,0 Periodic boundary conditions: ( ) ( x ) yˆ ψ 0,0,0 =ψ L ,0,0 ( ) ( x ) xˆ

zˆ Lundstrom ECE-656 F17 6 x-direction dk

x 0 2π Lx

ikx x ψ (xuxe) = k () dkx # of states= × 2 = Ndkk (2π Lx ) ikxx L ψψ(01) = (Lex ) → = L Nk=x = density of states in -space k π kLxx==21,2,3,...π j j

2π 2π j kj= L = Na k j = N x x x = max Lx a N 2π 2π “Brillouin zone” 0 < k < kmax = 7 a a Outline

1) Counting states 2) DOS in k-space vs. DOS in E-space 3) Examples 4) Realistic DOS in semiconductors 5) DOS for phonons

Lundstrom ECE-656 F17 8 Density-of-states in k-space

1D: ⎛ L ⎞ L N =2 × = dk k ⎜ 2 ⎟ ⎝ π ⎠ π 2D: ⎛ A ⎞ A N =2 × = dk dk k ⎜ 4 2 ⎟ 2 2 xy ⎝ π ⎠ π 3D: ⎛ Ω ⎞ Ω N =2 × = dk dk dk k ⎜ 8 2 ⎟ 4 3 xyz ⎝ π ⎠ π

Lundstrom ECE-656 F17 9 DOS: k-space vs. energy space

2 2 ! kx 2m* E but non-uniformly distributed in energy space.

dE Depends on E(k) (e.g. different for parabolic bands and linear bands) dkx k

2π Lx States are uniformly distributed in k-space, DEdE()= Nkdk ()

10 Lundstrom ECE-656 F17 Effect of E(k) on the DOS

How does non-parabolicity affect DOS(E)?

2k 2 E k * “Kane bands” ( ) 2m 2k 2 E′ 1+α E′ = ( ) 2m*(0)

E E ′ = − ε1

ε1 k

Lundstrom ECE-656 F17 11 Effect on DOS

E k E′ k E ( ) ( )

Nonparaboliciy increases the dE Kane band DOS (E).

2 L π dk dk′ 12 Outline

1) Counting states 2) DOS in k-space vs. DOS in E-space 3) Examples 4) Realistic DOS in semiconductors 5) DOS for phonons

Lundstrom ECE-656 F17 13 DOS(E) for 1D nanowire

ˆ L z yˆ

xˆ

Find DOS(E) per unit energy, per unit length, a single subband assuming parabolic energy bands.

2 2 ! kx E = εn + * 2m

Lundstrom ECE-656 F17 14 Example: 1D (single subband )

E k E ( )

N dk D E dE k dE 1D ( ) = dk L N dk = × 2 k (2π L) # D E 1D ( ) J-m L N k dk = dk π ε n k

2π L dk 15 Lundstrom ECE-656 F17 1D DOS

N dk D E dE k 1D ( ) = L 1 D E dE dk 1D ( ) = π L N k dk = dk 2 * π ! kdk m dE dE = * dk = 2 m ! k !2k 2 E = ε + 2m* E n * ( − ε n ) 2m k = ! 1 m* D E dE = dE 1D ( ) π! 2 E − ε ( n ) 16 Lundstrom ECE-656 F17 Don’t forget to multiply by 2

dk dk k 2π L k = 0 Multiply by 2 to account for the 2 m* D E dE dE (parabolic 1D ( ) = negative k- π! 2 E − ε energy bands) ( n ) states. 17 Lundstrom ECE-656 F17 Multiple subbands

2 2 1 2m* ! k D n (E) = i E 1D = εi + * E 2m π! ( − εn ) i E(k) D E D1 (E) D2 (E) 1D ( ) 1D 1D

!2k 2 * 2mi ε2

ε1

k ε1 ε2 E

18 Lundstrom ECE-656 F17 In terms of velocity

2 m* E D E dE = dE 1D ( ) π! 2 E − ε ( n ) dE !k υ = = 2 E − ε m* m* ( n ) dk dk k 2π L Exercise: Show that 2 the final expression is D E dE dE 1D ( ) = independent of π!υ bandstructure. 19 Lundstrom ECE-656 F17 Example 2: DOS(E) for 2D electrons

zˆ

yˆ A xˆ t

(large) normalization area

Find DOS(E) per unit energy, per unit area, for a single subband assuming parabolic energy bands.

2 2 ! k|| E = εn + * 2m 20 Lundstrom ECE-656 F17 Example 2: DOS(E) for 2D electrons

N(k) 2 D(E)dE = dk ky A 2k 2 2 kdk ! π E = * ⎛ A ⎞ A 2m N =2 × = k ⎜ 4 2 ⎟ 2 2 ⎝ π ⎠ π k 1 D(E)dE 2 kdk = 2 π k 2π x

Exercise: Show that:

m* 4 2 D2 D (E) = 2 π π! Area of each state in k-space: A 21 Lundstrom ECE-656 F17 Valley degeneracy z

y m* 2m* E − E D E ( C ) E E 3D ( ) = 2 3 Θ( − C ) π ! x m* 2m* E − E g D D ( C ) E E ⇒ V 2 3 Θ( − C ) π ! 1/3 m* ≡ m m2 D ( ℓ t )

Conduction band of Si:

6 equivalent valleys: gV = 6 (bulk) 22 Parabolic bands: 1D, 2D, and 3D

* 1 2m D1D D E = g Θ E − ε 1D ( ) V π! E − ε ( 1 ) ( 1 ) E ε D2D 1 m* D E g E 2D ( ) = V 2 Θ( − ε1 ) π! E ε D3D 1 m* 2m* E − E D E g ( C ) E E 3D ( ) = V 2 3 Θ( − C ) π  E E 2 2 * C 23 E k = E +  k 2m ( ( ) C ) 23 Graphene (2D)

Ek( ) Exercise: Show that for graphene, the 2D DOS is:

E 2 E g = 2 D E g Vf E 2 D ( ) = V 2 2 = 2 2 1 ( ) π υ π υ k F F y D2D k neutral point x (Dirac point) E

E k = ±υ k = ±υ k 2 + k 2 ( ) F F x y 24 24 Lundstrom ECE-656 F17 Outline

1) Counting states 2) DOS in k-space vs. DOS in E-space 3) Examples 4) Realistic DOS in semiconductors 5) DOS for phonons

Lundstrom ECE-656 F17 25 DOS for bulk Si

6 ) –1

eV 4 –1 cm 22

2 DOS (10

0 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 ENERGY (eV)

D(E) ∝ E − EC

The DOS is calculated with nonlocal empirical pseudopotentials including the spin-orbit interaction. (Courtesy Massimo Fischetti, August, 2011.) 26 Lundstrom ECE-656 F17 DOS for a Si quantum well

!2k 2 m* E = ε + || D j (E) = g n j 2m* 2 D Vj π!2 n primed E(k|| ) gV = 4

D2 D (E) unprimed

ε 2 2 2  k|| 2m* gV = 2 i ε1

ε1 ε2 k|| E

27 Lundstrom ECE-656 F17 DOS for a Si quantum well

D (E) 2 D m* D j (E) g j 2 D = Vj 2 π!

ε ε1 2 E

sp3s*d5 TB calculation by Yang Liu, Purdue University, 2007 Lundstrom ECE-656 F17 28 DOS for a Si quantum well

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

Lundstrom ECE-656 F17 29 Outline

1) Counting states 2) DOS in k-space vs. DOS in E-space 3) Examples 4) Realistic DOS in semiconductors 5) DOS for phonons

Lundstrom ECE-656 F17 30 Debye model

Exercise: Show that the DOS Debye for 3D phonons assuming a linear dispersion is

2 ω = υ q BW 3 !ω −1 S D ( ) J-m3 ph (!ω ) = 3 ( ) q 2π 2 !υ ( S ) −π a π a

Lundstrom ECE-656 F17 31 Realistic phonon DOS (Si)

#1024 15 Optical )

-1 phonons eV -3 10 D !ω ∝ E 2 ph ( )

5 Phonon DOS (cm 0 0 10 20 30 40 50 60 Near room Energy (meV) temperature and BW only a few kT! above, all states are occupied. (Figure provided by J. Maassen, Dalhousie Univ., CA, 2017) 32 Summary

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

2) The DOS becomes complicated at high energies.

3) The phonon DOS is generally complicated over the relevant energy range.

33 Lundstrom ECE-656 F17