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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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)
采纳的回答
Star Strider
2018-10-18
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 个评论
更多回答(2 个)
Anil Kumar
2018-10-19
5 个评论
Star Strider
2018-10-19
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
2018-10-19
And
...exp*(E0...
only makes sense if you have a variable called "exp" somewhere in your code (which is not the case).
Anil Kumar
2018-10-19
1 个评论
Star Strider
2018-10-19
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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