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.

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)

[Anil Kumar]

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

Answer 1

[Star Strider]

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

[Anil Kumar]

Thank You sir for your time and help! I will check if it works for me. Thanks again.

[madhan ravi]

+1 Perfect analysis

[Star Strider]

As always, my pleasure!

[Anil Kumar]

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)])

Answer 2

[Anil Kumar]

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)

[Torsten]

fsolve doesn't work with symbolic integration. Use "integral" instead of "int".

[Star Strider]

@Torsten — Thank you.

[Anil Kumar]

Even after writing integral instead of int, the same error is happening.

[Star Strider]

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]

And

...exp*(E0...

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

Answer 3

[Anil Kumar]

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)

[Star Strider]

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.