The Flow of PMOS’s Mobility (Part2)
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Student 光光光 ： Date ： 2010/04/15
description
The Flow of PMOS’s Mobility (Part2). Student ：光心君 Date ： 2010/04/15. 推導 Poisson equation 為矩陣表示. Proof: 經整理變矩陣程式. V 3. V 1. 推廣. V 2. 推廣. A. V. Rho. V0=A\Rho initial guess(V0=A -1 R). Poisson equation 的 Boundary Condition. Code: - PowerPoint PPT Presentation
Transcript of The Flow of PMOS’s Mobility (Part2)
Student ：光心君Date ： 2010/04/15
Proof:
經整理變矩陣程式
21 2 1 1
2 2
21 3 22 2
2 2
23 2 3 3
2 2
2
2
2
si
si
si
d V V V
dx x
V V Vd V
dx x
d V V V
dx x
2
[ ] [ ] [ ]1 12
d VA V RhoN N N Ndx si
V1 V2
V3
2 21 1 12
2 2 2 2
d V V V V V Vn n n n n n
dx x x x
推廣
12 2
1
222 2 2
33
2 2
2 10
1 2 1
1 20
x x V
Vx x x
V
x x
2 2
1
2 2 22
2 2 2
1
2 2
2 10 0
1 2 10 0
1 2 10
0 0
1 20 0
N N
N N
x xV
Vx x x
x x x
V
x x
推廣
A V Rho
V0=A\Rho initial guess(V0=A-1R)
Code: A = zeros(N,N); % Matrix for 2nd differential
operator A(1,1)=1/dx0^2; %boundary condition
Vsurface=Vs A(N,N)=1/dx(N-1)^2; Explain:Initial Condition V(1)=Vs , V(N)=01.V(1)=Vs 2.V(N)=0
VsVAV
dx
VsRhoand
dxA
1
20
20
)1( 1
)1,1(
0
0)( 1
),(2
)1(
N
N
VAV
NRhoanddx
NNA
for j=2:N-1 avgdx=(dx(j-1)+dx(j))/2; avgdx=Δx A(j,j-1) = 1/dx(j-1)/avgdx; A(j,j-1) = 1/ Δx2
A(j,j) = -(1/avgdx)*(1/dx(j-1)+1/dx(j)); A(j,j)=-2/ Δx2
A(j,j+1) = 1/dx(j)/avgdx; A(j,j+1) = 1/ Δx2
end; 2 2
1
2 2 22
2 2 2
1
2 2
2 10 0
1 2 10 0
1 2 10
0 0
1 20 0
N N
N N
x xV
Vx x x
x x x
V
x x
J=2
J=N-1
%****************CALCULATED PARAMETERS****************
A(1,1)=1/dx0^2; %boundary condition Vsurface=Vs
A(N,N)=1/dx(N-1)^2;%**************POISSON EQUATION
SETUP***************************** Rho(1)=Vs/dx0^2; %bondary condition on the
surface Rho(N)=0; %bondary condition V(N)=0
2
1 2102 2 2
22
2 2 2
1
1
2
10 0 0
1 2 10 0
1 2 10
0 001
0 0 0
s
NN N
N N
x VRhoV
dxVx x x Rho
x x xRho
V
x
%**************************scale set up *****************************
xscale = linspace(xstart,xend,N).'; % New scale cmdx0=(xend-xstart)/real(N);dx= dx0/au; % Mesh separation in a.u.
au = 0.5262E-8; % atomic unit in cm( 波耳氫原子半徑 )
dd=1/2/(dx^2); % (a.u.)^-2
Xscale xstart~xend 中有 N 個元素的行向量
From subprogram: shhole01R.m
2
'( )
'
.1
2
end startx xx
real N
xx
a u
ddx
找由1 至 N 所對應的 potential值以帶入薛丁格方程式
%**************potential set up*****************************
V=zeros(N,1);% Potential in Hr (Hr = 27.212; % 1 Hartree in eV)for j=2:(N-1) V(j) = interp1(xscaleI,VI,xscale(j))/Hr;endV(1)=20; %boundary conditionV(N)=20; %boundary condition
Interp1做一維的內插法 Hr Hartree energy ,the atomic unit of energy.
Time independent equation:
其中 H= , 且
2 2
2
( )( ) ( ) ( )
2
d xV x x E x
m dx
n n nE 2 2
2
( )
2
d xV
m dx
1 1N N N NE
pf
21 2 1
2 2
21 3 22
2 2
23 2 3
2 2
2
2
2
d
dx x
d
dx x
d
dx x
22 1
1 12
21 3 2
2 22
22 3
3 32
2( )
2
2( )
2
2( )
2
Vm x
Vm x
Vm x
21
2
2( )
2N N
N NVm x
推廣
0 10, 0N
xstart xend
V=∞Ψ0=0
V=∞ΨN+1=0
矩陣形式
12 2
12
2 22 2 2
3
32 2
2 10
1 2 1
21 2
0
Vx x
Vm x x x
Vx x
2 2
12 2
12 2 2
2 22 2 2
2 2 3
32 2
02
2 2
02
Vm x m x
Vm x m x m x
Vm x m x
推廣 1 1
2 2
3
1
0 0
0 0
0
0 0
0 0 0 N NN N N
A B
B A B
B A B
A
2
2
2
22
j jA Vm x
Bm x
%******** Schrodinger Equation ***************H = zeros(N,N);% light hole for j=2:(N-1) H(j,j) = V(j)+2*dd/m1; end H(N,N)=V(N)+2*dd/m1; H(1,1)=V(1)+2*dd/m1; for j=2:N H(j-1,j) = -dd/m1; H(j,j-1) = -dd/m1; end
21
1
2B
x m
21
12
2j jA Vx m
解波函數 Ψ 與 E [Y,D]=eig(H); % Eigen vectors(Y) and Eigen
values(D) [lambda1,key1] =sort(diag(D));
%sort: 以行為單位 , 將每一行的向量由小到大排列 Y1 = Y(:,key1); 取 key1行的一整列的元素
E1=lambda1*Hr+Ev1;
Y=eigenvector : [Ψ]N ×1(Ψ 以行向量的方式儲存在矩陣 Y 裡 )
D=eigenvalue : (Ei存放在矩陣 D 的對角元素 )
1 1N N N NE
1 0 0
0 0
0 0 N
E
E
藉由 Ψ 去計算 n(x) %**************** Calculating hole densities ****************** for j=1:N hole density /cm2
p1(j)=Do1*k*T*log(1+exp((Ef-E1(j))/k/T)); %hole den in heavy p2(j)=Do2*k*T*log(1+exp((Ef-E2(j))/k/T)); %hole den in light endDo1 = md1*m0/3.1415/(hb)^2/6.24146E11; %density of state (#/eV/cm2)Do2 = md2*m0/3.1415/(hb)^2/6.24146E11; %density of state (#/eV/cm2) for j=1:N jj=1:Nhole density at each valley /cm3
YY1(j,jj)=(Y1(j,jj))^2*p1(jj)/dx0; %hole den. (heavy band in #/cc)
YY2(j,jj)=(Y2(j,jj))^2*p2(jj)/dx0; %hole den. (light band in #/cc)
02 2( ) ( ) ( )D DE Ep x D E f E dE
0
2
3 2( ) ( ) ( )D i DE Ep x D E f E dE
*
2 2( )D
mD E
求 R: hole density /cm3R(j) = R(j)+(YY1(j,jj)+YY2(j,jj));
if xscale(j)>=xstart & xscale(j)<=xend; Nep(j) = interp1(xscaleO,R0,xscale(j));
else Nep(j)=+Nep0*exp(-beta*V0(j));
Nen=+ni^2./Nep( 只考慮 classical case)
( )exp[ ]c f
B
E Ep Nc
k T
0
2
3 2( ) ( ) ( )D i DE Ep x D E f E dE
xstart=0.0;
xend=fregion;
