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ECE-656: Fall 2011
Lecture 31:
Balance Equation Approach: II
Mark Lundstrom Purdue University
West Lafayette, IN USA
1 11/11/11
2
outline
Lundstrom ECE-656 F11
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 United States License. http://creativecommons.org/licenses/by-nc-sa/3.0/us/
(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
Lundstrom ECE-656 F11 6
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∑
Lundstrom ECE-656 F11 7
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
Lundstrom ECE-656 F11 8
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
Lundstrom ECE-656 F11 9
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
Lundstrom ECE-656 F11 10
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
Lundstrom ECE-656 F11 11
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)
Jnx = nqµnE x + 2µn
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
Lundstrom ECE-656 F11
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)
Lundstrom ECE-656 F11
Note error in eqn. (5.101), p. 236 of Lundstrom, FCT.
Lundstrom ECE-656 F11 14
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⎡⎣ ⎤⎦
Lundstrom ECE-656 F11 15
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*
Lundstrom ECE-656 F11 16
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
17
outline
Lundstrom ECE-656 F11
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 United States License. http://creativecommons.org/licenses/by-nc-sa/3.0/us/
(Reference: Chapter 5, Lundstrom, FCT)
1) Review of L30 2) Energy balance equation 3) Energy flux balance equation 4) Terminating the hierarchy 5) Summary
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
Lundstrom ECE-656 F11
Wxx =
1Ω
pxυx
2f x, px ,t( )
p∑
Lundstrom ECE-656 F11 19
2nd moment of the BTE
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
Lundstrom ECE-656 F11 20
2nd moment of the BTE
Fφ = FWx
Gφ = JnxE x
∂nφ
∂t= −
dFφx
dx+ Gφ − Rφ ⇒
∂W x,t( )∂t
= −dFWx
dx+ JnxE x −
W −W0( )τ E
Lundstrom ECE-656 F11 21
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 !
22
outline
Lundstrom ECE-656 F11
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 United States License. http://creativecommons.org/licenses/by-nc-sa/3.0/us/
(Reference: Chapter 5, Lundstrom, FCT)
1) Review of L30 2) Energy balance equation 3) Energy flux balance equation 4) Terminating the hierarchy 5) Summary
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)
Lundstrom ECE-656 F11
Gφ = −qE x
1L
∂φ∂px
fpx∑
⎧⎨⎪
⎩⎪
⎫⎬⎪
⎭⎪= −
3qm*E xW
Rφ ≡nφ − nφ
0
τφ
=FW
τ FW
Lundstrom ECE-656 F11 24
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
25
outline
Lundstrom ECE-656 F11
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 United States License. http://creativecommons.org/licenses/by-nc-sa/3.0/us/
(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 26
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
+ Jnx = nqµnE x + 2µn
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
Lundstrom ECE-656 F11
τm
∂Jnx x,t( )∂t
+ Jnx = nqµnE x + 2µn
dWxx
dx
Lundstrom ECE-656 F11 28
terminating at the second moment
∂n∂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
We need to approximate FW in terms of the 3 unknowns
∂W x,t( )∂t
= −dFWx
dx+ JnxE x −
W −W0( )τ E
Lundstrom ECE-656 F11 29
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
Lundstrom ECE-656 F11 30
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
Lundstrom ECE-656 F11 31
four balance equations
∂n∂t
= −d Jnx (−q)⎡⎣ ⎤⎦
dx electron continuity equation
Jnx = nqµnE x + 2µn
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
32
outline
Lundstrom ECE-656 F11
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 United States License. http://creativecommons.org/licenses/by-nc-sa/3.0/us/
(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 33
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.
Lundstrom ECE-656 F11 34
unfinished business
1) Can we define temperature more clearly?
Lundstrom ECE-656 F11 35
questions?
1) Review of L30 2) Energy balance equation 3) Energy flux balance equation 4) Terminating the hierarchy 5) Summary