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abdalmoed-alaiashy
about 7 years ago
Physics of semiconductor devices
schottky diode numerical simulation
i need help to solve this problem
%% Counters and Constants
k=0;
m=0;
Ht=4E-13;
%% Input Parameters
N=19;
D1=-2.0;
IA=50;
VS=zeros(1,19);
U=zeros(1,19);
H=2.5E-5;
eps=9.47E-13;
q=1.602E-19;
pi=3.1415;
T=300;
k=1.38E-23;
Vt=(k*T)/q;
%% Boundary Conditions
U1(1)=0.5;
U10=0.5;
U1(19)=2.5;
%% Initial Electron Density
An1(1) = 6E14;
An(19) = 6E14;
An1(19) = 6E14;
And = 6E14;
for i = 1:19;
An(i)=2E15;
end
%% Elements of Matrix A
A=zeros(18,18);
for i=1:18;
j=i;
A(i,j)=-2;
end
for i=1:18;
j=i+1;
A(i,j)=1;
A(j,i)=1;
end
Ainv=inv(A);
for k=1:145
%% Elements of Matrix C
for j=1:19;
C(j)=((An(j)-And)/eps)\*H\*H\*q;
end
C(1)=C(1)-0.5;
C(19)=C(19)-2.5;
for i=1:19
U1(i)=U1(i)+Ainv(i,j)\*C(j);
end
%% Electric Field
E10=-(3.0*U10+4*U1(1)-U1(2))/(2*H);
E1(1)=-(U1(2)-0.5)/(2*H);
for j=2:18;
E1(j)=-(U1(j+1)-U1(j-1))/(2\*H);
end
E1(19)=-(3*U1(19)-4*U1(18)+U1(17))/(2*H);
C1=sqrt(q/(4*pi*eps));
C2=q*C1;
D=abs(E10);
%% Depletion Width
Vb1=1;
Vab=1.5;
Vb=abs(Vb1+Vab);
W=sqrt((2*eps*Vb)/(q*And));
Deltab=C1*sqrt(D);
%% Input Velocity
E0=4E+3;
S0=8000;
Vsa=0.85E+7;
Vn0=(S0*abs(E10)+Vsa*(E10/E0)^4)/(1+(E10/E0)^4);
for i=1:19;
Vn(i)=(S0\*abs(E1(i))+Vsa\*(E1(i)/E0)^4)/(1+(E1(i)/E0)^4);
end
%% Mobility
Sc0=Vn0/abs(E10);
for i=1:19;
S(i)=Vn(i)/abs(E1(i));
end
%% Total Velocity (Drift + Diffusion)
Vn10=Vn0-0.5*Sc0*Vt*((11*An(1)+18*An(1)-9*An(2)+2*An(3))/(61*An(1)*H));
Vn1(1)=Vn(1)-0.5*S(1)*Vt*((An(2)-And)/(H*An(1)));
for j=2:18
Vn1(j)=Vn(j)-S(j)\*Vt\*((An(j+1)-An(j-1))/(2\*An(j)\*H));
end
Vn1(19)=Vn(19)-S(19)+Vt*(3*An(19)-4*An(18)+An(17))/(2*An(19)*Vt*H);
%% Currnt Density Test
Dif=(fjmax(C2,D)-An(1)*Vn10)*(Ht/H);
%% update Electron density
An10=An(1)+(fjmax(C,D)-An(1)*Vn10)*(Ht/H);
% An1(1)=An(1)+(fjmax(C,D)-An(1)*Vn1(1))*(Ht/H);
for i=2:19;
An1(i)=An(i)-(An(i)\*Vn1(i)-An(i-1)\*Vn1(i-1))\*(Ht/H);
end
%% Displacement Current
Disp(1)=((U1(2)-U10)-(U(2)-U1(1)))*(1/(H*Ht))*(eps/q);
for i=2:18
Disp(i)=((U1(i+1)-U1(i-1))-(U(i+1)-U(i-1)))\*(1/(H\*Ht))\*(eps/q);
end
Disp(19)=Disp(18);
%% PARTICAL CURRENT
Vs=zeros(50);
for i=1:19
Anvn(i)=0.5\*(An1(i)+An(i))\*(Vn1(i)+Vs(i));
end
%% TOTAL CURRENT
for i=1:18
Ajt(i)=-0.5\*Anvn(i)+0.5\*Disp(i);
end
Ajt(19)=Ajt(18);
for I=1:19
Ajc(i)=-q\*An1(i)\*Vn1(i);
end
Ajc(19)=Ajc(18);
An0=An10;
for i=1:19;
An(i)=An1(i)
end
for i=1:N
Vs(i)=Vn1(i)
end
Vs(19)=Vn1(19);
for i=1:N
U(i)=U1(i)
end
U(19)=U1(19);
end
0
Spike69fei
almost 3 years ago
Physics of semiconductor devices
Determine the number of atoms per unit cell in a diamond lattice.
about solid state physics
1
Solved
abdalmoed-alaiashy
about 7 years ago
Physics of semiconductor devices
schottky diode numerical simulation
i need help to solve this problem
%% Counters and Constants
k=0;
m=0;
Ht=4E-13;
%% Input Parameters
N=19;
D1=-2.0;
IA=50;
VS=zeros(1,19);
U=zeros(1,19);
H=2.5E-5;
eps=9.47E-13;
q=1.602E-19;
pi=3.1415;
T=300;
k=1.38E-23;
Vt=(k*T)/q;
%% Boundary Conditions
U1(1)=0.5;
U10=0.5;
U1(19)=2.5;
%% Initial Electron Density
An1(1) = 6E14;
An(19) = 6E14;
An1(19) = 6E14;
And = 6E14;
for i = 1:19;
An(i)=2E15;
end
%% Elements of Matrix A
A=zeros(18,18);
for i=1:18;
j=i;
A(i,j)=-2;
end
for i=1:18;
j=i+1;
A(i,j)=1;
A(j,i)=1;
end
Ainv=inv(A);
for k=1:145
%% Elements of Matrix C
for j=1:19;
C(j)=((An(j)-And)/eps)\*H\*H\*q;
end
C(1)=C(1)-0.5;
C(19)=C(19)-2.5;
for i=1:19
U1(i)=U1(i)+Ainv(i,j)\*C(j);
end
%% Electric Field
E10=-(3.0*U10+4*U1(1)-U1(2))/(2*H);
E1(1)=-(U1(2)-0.5)/(2*H);
for j=2:18;
E1(j)=-(U1(j+1)-U1(j-1))/(2\*H);
end
E1(19)=-(3*U1(19)-4*U1(18)+U1(17))/(2*H);
C1=sqrt(q/(4*pi*eps));
C2=q*C1;
D=abs(E10);
%% Depletion Width
Vb1=1;
Vab=1.5;
Vb=abs(Vb1+Vab);
W=sqrt((2*eps*Vb)/(q*And));
Deltab=C1*sqrt(D);
%% Input Velocity
E0=4E+3;
S0=8000;
Vsa=0.85E+7;
Vn0=(S0*abs(E10)+Vsa*(E10/E0)^4)/(1+(E10/E0)^4);
for i=1:19;
Vn(i)=(S0\*abs(E1(i))+Vsa\*(E1(i)/E0)^4)/(1+(E1(i)/E0)^4);
end
%% Mobility
Sc0=Vn0/abs(E10);
for i=1:19;
S(i)=Vn(i)/abs(E1(i));
end
%% Total Velocity (Drift + Diffusion)
Vn10=Vn0-0.5*Sc0*Vt*((11*An(1)+18*An(1)-9*An(2)+2*An(3))/(61*An(1)*H));
Vn1(1)=Vn(1)-0.5*S(1)*Vt*((An(2)-And)/(H*An(1)));
for j=2:18
Vn1(j)=Vn(j)-S(j)\*Vt\*((An(j+1)-An(j-1))/(2\*An(j)\*H));
end
Vn1(19)=Vn(19)-S(19)+Vt*(3*An(19)-4*An(18)+An(17))/(2*An(19)*Vt*H);
%% Currnt Density Test
Dif=(fjmax(C2,D)-An(1)*Vn10)*(Ht/H);
%% update Electron density
An10=An(1)+(fjmax(C,D)-An(1)*Vn10)*(Ht/H);
% An1(1)=An(1)+(fjmax(C,D)-An(1)*Vn1(1))*(Ht/H);
for i=2:19;
An1(i)=An(i)-(An(i)\*Vn1(i)-An(i-1)\*Vn1(i-1))\*(Ht/H);
end
%% Displacement Current
Disp(1)=((U1(2)-U10)-(U(2)-U1(1)))*(1/(H*Ht))*(eps/q);
for i=2:18
Disp(i)=((U1(i+1)-U1(i-1))-(U(i+1)-U(i-1)))\*(1/(H\*Ht))\*(eps/q);
end
Disp(19)=Disp(18);
%% PARTICAL CURRENT
Vs=zeros(50);
for i=1:19
Anvn(i)=0.5\*(An1(i)+An(i))\*(Vn1(i)+Vs(i));
end
%% TOTAL CURRENT
for i=1:18
Ajt(i)=-0.5\*Anvn(i)+0.5\*Disp(i);
end
Ajt(19)=Ajt(18);
for I=1:19
Ajc(i)=-q\*An1(i)\*Vn1(i);
end
Ajc(19)=Ajc(18);
An0=An10;
for i=1:19;
An(i)=An1(i)
end
for i=1:N
Vs(i)=Vn1(i)
end
Vs(19)=Vn1(19);
for i=1:N
U(i)=U1(i)
end
U(19)=U1(19);
end
0
Spike69fei
almost 3 years ago
Physics of semiconductor devices
Determine the number of atoms per unit cell in a diamond lattice.
about solid state physics
1
Solved
abdalmoed-alaiashy
about 7 years ago
Physics of semiconductor devices
schottky diode numerical simulation
i need help to solve this problem
%% Counters and Constants
k=0;
m=0;
Ht=4E-13;
%% Input Parameters
N=19;
D1=-2.0;
IA=50;
VS=zeros(1,19);
U=zeros(1,19);
H=2.5E-5;
eps=9.47E-13;
q=1.602E-19;
pi=3.1415;
T=300;
k=1.38E-23;
Vt=(k*T)/q;
%% Boundary Conditions
U1(1)=0.5;
U10=0.5;
U1(19)=2.5;
%% Initial Electron Density
An1(1) = 6E14;
An(19) = 6E14;
An1(19) = 6E14;
And = 6E14;
for i = 1:19;
An(i)=2E15;
end
%% Elements of Matrix A
A=zeros(18,18);
for i=1:18;
j=i;
A(i,j)=-2;
end
for i=1:18;
j=i+1;
A(i,j)=1;
A(j,i)=1;
end
Ainv=inv(A);
for k=1:145
%% Elements of Matrix C
for j=1:19;
C(j)=((An(j)-And)/eps)\*H\*H\*q;
end
C(1)=C(1)-0.5;
C(19)=C(19)-2.5;
for i=1:19
U1(i)=U1(i)+Ainv(i,j)\*C(j);
end
%% Electric Field
E10=-(3.0*U10+4*U1(1)-U1(2))/(2*H);
E1(1)=-(U1(2)-0.5)/(2*H);
for j=2:18;
E1(j)=-(U1(j+1)-U1(j-1))/(2\*H);
end
E1(19)=-(3*U1(19)-4*U1(18)+U1(17))/(2*H);
C1=sqrt(q/(4*pi*eps));
C2=q*C1;
D=abs(E10);
%% Depletion Width
Vb1=1;
Vab=1.5;
Vb=abs(Vb1+Vab);
W=sqrt((2*eps*Vb)/(q*And));
Deltab=C1*sqrt(D);
%% Input Velocity
E0=4E+3;
S0=8000;
Vsa=0.85E+7;
Vn0=(S0*abs(E10)+Vsa*(E10/E0)^4)/(1+(E10/E0)^4);
for i=1:19;
Vn(i)=(S0\*abs(E1(i))+Vsa\*(E1(i)/E0)^4)/(1+(E1(i)/E0)^4);
end
%% Mobility
Sc0=Vn0/abs(E10);
for i=1:19;
S(i)=Vn(i)/abs(E1(i));
end
%% Total Velocity (Drift + Diffusion)
Vn10=Vn0-0.5*Sc0*Vt*((11*An(1)+18*An(1)-9*An(2)+2*An(3))/(61*An(1)*H));
Vn1(1)=Vn(1)-0.5*S(1)*Vt*((An(2)-And)/(H*An(1)));
for j=2:18
Vn1(j)=Vn(j)-S(j)\*Vt\*((An(j+1)-An(j-1))/(2\*An(j)\*H));
end
Vn1(19)=Vn(19)-S(19)+Vt*(3*An(19)-4*An(18)+An(17))/(2*An(19)*Vt*H);
%% Currnt Density Test
Dif=(fjmax(C2,D)-An(1)*Vn10)*(Ht/H);
%% update Electron density
An10=An(1)+(fjmax(C,D)-An(1)*Vn10)*(Ht/H);
% An1(1)=An(1)+(fjmax(C,D)-An(1)*Vn1(1))*(Ht/H);
for i=2:19;
An1(i)=An(i)-(An(i)\*Vn1(i)-An(i-1)\*Vn1(i-1))\*(Ht/H);
end
%% Displacement Current
Disp(1)=((U1(2)-U10)-(U(2)-U1(1)))*(1/(H*Ht))*(eps/q);
for i=2:18
Disp(i)=((U1(i+1)-U1(i-1))-(U(i+1)-U(i-1)))\*(1/(H\*Ht))\*(eps/q);
end
Disp(19)=Disp(18);
%% PARTICAL CURRENT
Vs=zeros(50);
for i=1:19
Anvn(i)=0.5\*(An1(i)+An(i))\*(Vn1(i)+Vs(i));
end
%% TOTAL CURRENT
for i=1:18
Ajt(i)=-0.5\*Anvn(i)+0.5\*Disp(i);
end
Ajt(19)=Ajt(18);
for I=1:19
Ajc(i)=-q\*An1(i)\*Vn1(i);
end
Ajc(19)=Ajc(18);
An0=An10;
for i=1:19;
An(i)=An1(i)
end
for i=1:N
Vs(i)=Vn1(i)
end
Vs(19)=Vn1(19);
for i=1:N
U(i)=U1(i)
end
U(19)=U1(19);
end
0
Physics of semiconductor devices