How to Solve Non Linear Electronutrility condition in Space region region in semiconductor? Using vpasolve it is showing [empty syms] error while a theortical solution exist.
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Anil Kumar
el 18 de Oct. de 2018
Comentada: Star Strider
el 19 de Oct. de 2018
if true
% code
end
r=40*10^-9;
T=500;
N=10^13;
K=1.3807*10^-23;
es=12*8.85*10^-12;
Nd=10^17;
Eg=3.6*1.6*10^-19;
e=1.6*10^-19;
x=0.21*1.6*10^-19; %(Ecb-Ef)=x
y=1*1.6*10^-19; %(Ecs-Eas)=y
%Eg=3.6*1.6*10^-19;
Nc=2.4154*10^24;
Nv=1.7959*10^25;
Ni=sqrt((Nc*Nv)*exp(-Eg/(K*T)))
Nb=Nc*exp(-(x/(K*T)))
Pb=Nv*exp(-(3.6*e/(K*T)))
ub=log(Nb/Ni)
Ld=sqrt(es*(K*T)/(e^2*Nd))
es=12*8.85*10^-12;
e=1.6*10^-19;
Ld=sqrt(es*(K*T)/(e^2*Nd));
%E=E0+1/6*(r/Ld)^2
%Qsc=sqrt(2)*(Nb+Pb)*e*Ld*sqrt(cosh(ub+E0+1/6*(r/Ld)^2/cosh(ub))-(E0+1/6*(r/Ld)^2)*tanh(ub)-1);
%syms r
%Nd*int(1-exp(E0+1/6*(r/Ld)^2),0,2.64*10^-22);
%E=[-50,50]
%F= Nd*int(1-exp*(E0+1/6*(r/Ld)^2),0,V)+ 4*pi*r^2*[Nt/1+2*exp(((y)/(K*T))+E0+1/6*(r/Ld)^2)]
%fsolve(@myfun,E)
%sqrt(2)*(Nb+Pb)*e*Ld*sqrt(cosh(ub+(E0+1/6*(r/Ld)^2/cosh(ub)))-(E0+1/6*(r/Ld)^2)*tanh(ub)-1)+ 4*pi*r^2*(N/1+2*exp(((-1.21*1.6*10^-19)/(K*T))+(E0+1/6*(r/Ld)^2)))=0
syms E0
vpasolve(sqrt(2)*(Nb+Pb)*e*Ld*sqrt(cosh(ub+(E0+1/6*(r/Ld)^2/cosh(ub)))-(E0+1/6*(r/Ld)^2)*tanh(ub)-1)+ 4*pi*r^2*(N/1+2*exp(((-1.21*1.6*10^-19)/(K*T))+(E0+1/6*(r/Ld)^2)))==0,E0)
Respuesta aceptada
Star Strider
el 18 de Oct. de 2018
There appears to be a minimum, but not a root.
To illustrate —
fcn = @(E0) (sqrt(2)*(Nb+Pb)*e*Ld*sqrt(cosh(ub+(E0+1/6*(r/Ld)^2/cosh(ub)))-(E0+1/6*(r/Ld)^2)*tanh(ub)-1)+ 4*pi*r^2*(N/1+2*exp(((-1.21*1.6*10^-19)/(K*T))+(E0+1/6*(r/Ld)^2))))
[E0s, fval] = fsolve(fcn, 1)
E0s =
-34.966552734375
fval =
0.624442049383109
With a complex initial estimate:
[E0s, fval] = fsolve(fcn, 1+1i)
E0s =
-31.6877692741181 + 3.04633319691543i
fval =
0.201062230775014 - 0.0783028987275935i
4 comentarios
Más respuestas (2)
Anil Kumar
el 19 de Oct. de 2018
5 comentarios
Star Strider
el 19 de Oct. de 2018
In this call:
... int(1-exp*(E0+1/6*(r/Ld)^2),0,V) ...
the int function needs to know what the variable of integration is. That is the second argument (after the function), with the limits of integration being the third and fourth arguments.
We need to see your integral call.
Torsten
el 19 de Oct. de 2018
And
...exp*(E0...
only makes sense if you have a variable called "exp" somewhere in your code (which is not the case).
Anil Kumar
el 19 de Oct. de 2018
1 comentario
Star Strider
el 19 de Oct. de 2018
Several problems, incluiding a reference to non-existent ‘Nt’.
Try this:
... CODE ...
fun1 = @(E0) Nd*exp(E0+1/6*(r/Ld)^2);
e=1.6E-19;
Ld=sqrt(es*(K*T)/(e^2*Nd));
eqn = @(E0)(Nd*integral(@(V)fun1(E0),0,V, 'ArrayValued',1)+ 4*pi*r^2*(Nd/1+2*exp(((y)/(K*T))+E0+1/6*(r/Ld)^2)))
result=fsolve(@(E0)eqn(E0),[0,50])
With those changes, it runs. You must determine if it gives reasonable results.
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