Lecture 31: Balance Equation Approach: II

ECE-656: Fall 2011

Lecture 31:

Balance Equation Approach: II

Mark Lundstrom Purdue University

West Lafayette, IN USA

1 11/11/11

2

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Lundstrom ECE-656 F11

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(Reference: Chapter 5, Lundstrom, FCT)

1) Review of L30 2) Energy balance equation 3) Energy flux balance equation 4) Terminating the hierarchy 5) Summary

Lundstrom ECE-656 F11 3

moment of f

Physical quantities are moments of

e.g.

Can we bypass solving the BTE and solve directly for the physical quantities of interest?

4

the continuity equation

A familiar balance equation:

in-flow

out-flow

∂p ∂t

in-flow -out-flow = - divergence of flux

recombination generation

5

balance equations for physical quantities

nφ (x,t) = 1

Ld φ( p) f x, p,t( )p∑

∂p∂t

= −∇ •J p

q+ Gp − Rp →

∂nφ

∂t= −∇ •

Fφ + Gφ − Rφ

electron continuity equation

current balance equation

kinetic energy balance equation


putting it all together

∂f∂t

+υx

∂f∂x

−qE x

∂f∂kx

= C f⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪p∑

Gφ = −qE x

1Ld

∂φ∂px

fp∑

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

Fφx ≡

1Ld φ p( )υx f x, p,t( )

p∑

nφ (x,t) = 1

Ld φ( p) f x, p,t( )p∑


0th moment of the BTE

∂nφ

∂t= −

dFφx

dx+ Gφ − Rφ →

∂n∂t

= −d Jnx (−q)⎡⎣ ⎤⎦

dx

In steady-state, the current is constant because we have assumed that there is no generation-recombination of electrons.

∂n∂t

=1q

dJnx

dx


1st moment of the BTE

Rφ =

Px

τm Gφ = −q( )nE x

nφ (x,t) = Px Fφ =Wxx = n

pxυx

2

∂nφ

∂t= −

dFφx

dx+ Gφ − Rφ →

∂Px x,t( )∂t

= −d 2Wxx( )

dx− nqE x −

Px

τm

“momentum balance equation”

Px = n px


current equation

Px = n px = nm* υx

∂Px x,t( )∂t

= −d 2Wxx( )

dx− nqE x −

Px

τm

Jnx = (−q)n υx

Jnx =

(−q)m* Px

Jnx + τm

∂Jnx x,t( )∂t

= nqµnE x + 2µn

dWxx

dx


drift-diffusion equation

τm

∂Jnx x,t( )∂t

<< Jnx

assume:

q τm

m* = µn

f <<

1τm

= 2 THz

Jnx + τm

∂Jnx x,t( )∂t

= nqµnE x + 2µn

dWxx

dx


DD equation with temperature gradients

Wxx = n 1

2m*υx

2 ≈ nkBTe

2 τm

∂Jnx x,t( )∂t

<< Ix

assume:

Jnx = nqµnE x + 2µn

dWxx

dx

2µn

dWxx

dx= 2µn

ddx

nkBTe

2⎛

⎝⎜⎞

⎠⎟= µn

ddx

nkBTe( )

n = NCe Fn −EC( ) kBTe =

14

2m*kBTe

π2

⎛

⎝⎜

⎞

⎠⎟

3/ 2

e Fn −EC( ) kBTe = n x,Te( )

Te ≠ constant

12

DD equation with temperature gradients (ii)


dWxx

dx

2µn

dWxx

dx= µn

ddx

nkBT( ) = µn kBT ∂n∂x

+ nkB

dTe

dx+ kBTe

∂n∂Te

dTe

dx⎡

⎣⎢

⎤

⎦⎥

2µn

dWxx

dx= kBTeµn

∂n∂x

+ nkBµn

32−

Fn − EC( )kBTe

⎡

⎣⎢⎢

⎤

⎦⎥⎥

dTe

dx


n =

14

2m*kBTe

π2

⎛

⎝⎜

⎞

⎠⎟

3/ 2

e Fn −EC( ) kBTe

13

final current equation

Jnx = nqµnE x + qDn

dndx

− ST

dTe

dx

µn =

q τm

m*

Dn =

kBTe

qµn

ST = nLqµn

kB

−q( )32−ηF

⎡

⎣⎢

⎤

⎦⎥

(Soret coefficient)


Note error in eqn. (5.101), p. 236 of Lundstrom, FCT.


Seebeck coefficient assume dn/dx = 0, and solve for Ex:

E x =

Jnx

nqµn

+ST

nqµn

dTe

dx

E x = ρn Jnx + Sn

dTe

dx

Jnx = nqµnE x + qDn

dndx

− ST

dTe

dx

Sn =

kB

−q( )32−ηF

⎡

⎣⎢

⎤

⎦⎥

recall: Lecture 12

S3D =

kB

−q( ) r + 2( ) −ηF⎡⎣ ⎤⎦


recap: moments of the BTE

∂f∂t

+υx

∂f∂x

−qE x

∂f∂kx

= C f →∂nφ

∂t= −∇ •

Fφ + Gφ − Rφ

Now we need an equation for Wxx ….

∂n∂t

=dJnx

dx1) 0th moment:

2) 1st moment:

Jnx + τm

∂Jnx x,t( )∂t

= nqµnE x + 2µn

dWxx

dx

φ px( ) = −q( ) px

1

m*


terminating the hierarchy

Wxx =

1Ω

pxυx

2f x, p,t( )

p∑ = nuxx

uxx ≈

kBTe

2

But what is Te? Answer: Near equilibrium: Te≈ TL

Wxx = n

kBTe

2

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18

2nd moment of the BTE

nφ (x,t) = 1

Ωφ( p) f x, p,t( )

p∑

φ( p) = E p( )

W x,t( ) = 1

ΩE p( ) f x, p,t( )

p∑

nφ =W =

1Ω

E p( )p∑ f = n E = nu


Wxx =

1Ω

pxυx

2f x, px ,t( )

p∑



φ( p) = E p( )

Fφx ≡

1Ω

φ p( )υx f x, p,t( )p∑ =

1L

E p( )υx f x, p,t( )p∑ = FWx

Gφ = −qE x

1Ω

∂φ∂px

fp∑

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= −qE x

1Ω

∂E p( )∂px

fp∑

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= JnxE x

Rφ ≡nφ − nφ

0

τφ

=W −W0

τ E



Fφ = FWx

Gφ = JnxE x

∂nφ

∂t= −

dFφx

dx+ Gφ − Rφ ⇒

∂W x,t( )∂t

= −dFWx

dx+ JnxE x −

W −W0( )τ E


recap

∂W x,t( )∂t

= −dFWx

dx+ JnxE x −

W −W0( )τ E

∂n x,t( )∂t

= −d Jnx (−q)⎡⎣ ⎤⎦

dx

τm

∂Jnx x,t( )∂t

+ Jnx = nqµnE x + 2µn

dWxx

dx

0th moment of BTE

1st moment of BTE

2nd moment of BTE

Now we need an equation for FW !

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23

energy flux balance equation

Fφ ≡

1Ω

E( p)υx( )υx f x, p,t( )p∑ ≡ n Eυx

2

nφ (x,t) = 1

Ωφ( p) f x, p,t( )

p∑ = FWx

φ( p) = E p( )υx

Fφ = nL Eυx

2 =2

m* nL E2 = X (parabolic energy bands)


Gφ = −qE x

1L

∂φ∂px

fpx∑

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= −

3qm*E xW

Rφ ≡nφ − nφ

0

τφ

=FW

τ FW


energy flux balance equation

∂nφ

∂t= −

dFφx

dx+ Gφ − Rφ →

∂FWx x,t( )∂t

= −dXdx

−3qm* WE x −

FWx

τ FW

Now we need an equation for X !

Fφ ≡

1Ω

E( p)υx( )υx f x, p,t( )p∑ ≡ n Eυx

2 = X

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the four balance equations

∂FW x,t( )∂t

= −dXdx

−3qm* WE x −

FWx

τ FW

∂W x,t( )∂t

= −dFWx

dx+ JnxE x −

W −W0( )τ E

∂n x,t( )∂t

= −d Jnx (−q)⎡⎣ ⎤⎦

dx

τm

∂Jnx x,t( )∂t


dWxx

dx

0th moment of BTE

1st moment of BTE

2nd moment of BTE

3rd moment of BTE

27

terminating at the 1st moment

∂n∂t

= −d Jnx (−q)⎡⎣ ⎤⎦

dx0th moment of BTE

1st moment of BTE

Wxx ≈

W3≈ n

u0

3

Wxx ≈

kBTL

2assume: τm ∂Jnx x,t( ) ∂t << Jnx

Jnx = nqµnE x + kBTLµn

dndx

= nqµnE x + qDn

dndx


τm

∂Jnx x,t( )∂t


dWxx

dx


terminating at the second moment

∂n∂t

= −d Jnx (−q)⎡⎣ ⎤⎦

dx

τm

∂Jnx x,t( )∂t


dWxx

dx

0th moment of BTE

1st moment of BTE

2nd moment of BTE

We need to approximate FW in terms of the 3 unknowns

∂W x,t( )∂t

= −dFWx

dx+ JnxE x −

W −W0( )τ E


approximating the third moment

∂FWx

∂t= −

dXdx

−3qm* WE x −

FWx

τ FW

τ FW

∂FWx

∂t+ FW = −3

q τ FW

m* WE x − τ FW

dXdx

τ FW

∂FW x,t( ) ∂t << FWx

X ≈

1Ω

E( p)υx2( ) fS x, p,t( )

p∑ =

3kBTe

m* W


the energy flux

X ≈

3kBTe

m* W τ FW

∂FWx

∂t+ FW = −3

q τ FW

m* WE x − τ FW

dXdx

FWx = −3µEWE x − 3µE

d W kBTe / q( )dx

µE =

q τ FW

m* W ≈ n

kBTe

2


four balance equations

∂n∂t

= −d Jnx (−q)⎡⎣ ⎤⎦

dx electron continuity equation


dWxx

dxcurrent equation

∂W x,t( )∂t

= −dFWx

dx+ JnxE x −

W −W0( )τ E

FWx = −3µEWE x − 3µE

d W kBTe / q( )dx

energy-balance equation

energy-flux equation

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summary

1) The first moment of the BTE gives the current continuity equation.

2) The second moment gives the current density.

3) The third moment gives the kinetic energy.

4) Each moment involves the next moment – the hierarchy must be terminated.


unfinished business

1)  Can we define temperature more clearly?


questions?


Lecture 31: Balance Equation Approach: II

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Transcript of Lecture 31: Balance Equation Approach: II