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.

Question

Anil Kumar 2018-10-18

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/424771-how-to-solve-non-linear-electronutrility-condition-in-space-region-region-in-semiconductor-using-vp

评论： Star Strider 2018-10-19

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)

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Anil Kumar 2018-10-18

I have used vpasolve but getting error [empty syms ]! Bit new to matlab. anyhelp will be appreciated

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Answer 1

Star Strider 2018-10-18

2
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/424771-how-to-solve-non-linear-electronutrility-condition-in-space-region-region-in-semiconductor-using-vp#answer_342116

在 MATLAB Online 中打开

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

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Star Strider 2018-10-18

As always, my pleasure!

Anil Kumar 2018-10-19

In the same fashion I tried to solve the same equation in other form(more relevant for my calculation and theory. BUt I am getting the error as"Not enough input arguments" and

Error in @(E0)(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)])

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Answer 2

Anil Kumar 2018-10-19

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/424771-how-to-solve-non-linear-electronutrility-condition-in-space-region-region-in-semiconductor-using-vp#answer_342250

在 MATLAB Online 中打开

if true
  % code
end
r=80*10^-9;
T=500;
N=10^11;
K=1.3807*10^-23;
es=12*8.85*10^-12;
V=4/3*pi*r^3;
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;
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;
fcn =@(E0)(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)])
[E0s,fval]=fsolve(fcn, 2)

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Star Strider 2018-10-19

在 MATLAB Online 中打开

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

在 MATLAB Online 中打开

And

...exp*(E0...

only makes sense if you have a variable called "exp" somewhere in your code (which is not the case).

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Answer 3

Anil Kumar 2018-10-19

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/424771-how-to-solve-non-linear-electronutrility-condition-in-space-region-region-in-semiconductor-using-vp#answer_342268

在 MATLAB Online 中打开

if true
  % code
end
r=80*10^-9;
T=500;
N=10^11;
K=1.3807*10^-23;
es=12*8.85*10^-12;
V=4/3*pi*r^3;
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;
Ld=sqrt(es*(K*T)/(e^2*Nd))
es=12*8.85*10^-12;
fun1= Nd*exp(E0+1/6*(r/Ld)^2);
e=1.6*10^-19;
Ld=sqrt(es*(K*T)/(e^2*Nd));
%E=E0+1/6*(r/Ld)^2;
%fcn =@(E0)(Nd*integral(@(V)fun1(E0),0,V)+ 4*pi*r^2*(Nt/1+2*exp(((y)/(K*T))+E0+1/6*(r/Ld)^2)))
%[E0s,fval]=fsolve(fcn, 2)
eqn=(Nd*integral(@(V)fun1(E0),0,V)+ 4*pi*r^2*(Nt/1+2*exp(((y)/(K*T))+E0+1/6*(r/Ld)^2)))
result=fsolve(@(E0)eqn(E0),[0,50])
% fcn =@(E0)(Nd*integral(@(V)Nd*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)))
% [E0s,fval]=fsolve(fcn, 2)

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Star Strider 2018-10-19

在 MATLAB Online 中打开

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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