Warning: Matrix is singular, close to singular or badly scaled. Results may be inaccurate. RCOND = NaN.

I'm trying to write a function that will solve for the exact DC solution of my circuit through Newton's Method. However I get this error "Warning: Matrix is singular, close to singular or badly scaled. Results may be inaccurate. RCOND = NaN." Was hoping anyone could help me rectify this. My output should be a solution vector and the number of iterations needed to for convergence. Below is my function and main code.

%%My Function 
  function [x, iter] = bjt(x0,G)
F = zeros(5,1) ;
J = zeros(5,5) ; 
VT = 25.3*exp(-3); 
BF = 100; 
BR = 1; 
Is = 1*exp(-16); 
i = 1; 
x = x0; 
while (norm(F,inf)>(1*exp(-9)))|(i==1)

    F = G*x; 
    F(5) = F(5)-12 ; 
    J = G; 
    ID1 = (Is/BR)*((exp(x(1)-x(2))/VT)-1);

    F(1) = F(1)+ID1;
    F(2) = F(2)-ID1; 

    J(1,1) = J(1,1) + (Is/(BR*VT))*(exp(x(1)-x(2))/VT);
    J(1,2) = J(1,2) - (Is/(BR*VT))*(exp(x(1)-x(2))/VT);
    J(2,1) = J(2,1) - (Is/(BR*VT))*(exp(x(1)-x(2))/VT);
    J(2,2) = J(2,2) + (Is/(BR*VT))*(exp(x(1)-x(2))/VT);

    ID2 = (Is/BF)*((exp(x(1)-x(3))/VT)-1);

    F(1) = F(1)+ID2;
    F(3) = F(3)-ID2;

    J(1,1) = J(1,1) + (Is/(BF*VT))*(exp(x(1)-x(3))/VT);
    J(1,3) = J(1,3) - (Is/(BF*VT))*(exp(x(1)-x(3))/VT);
    J(3,1) = J(3,1) - (Is/(BF*VT))*(exp(x(1)-x(3))/VT);
    J(3,3) = J(3,2) + (Is/(BF*VT))*(exp(x(1)-x(3))/VT);

    I0 = Is*(exp((x(1)-x(3))/VT) - exp((x(1)-x(2))/VT));

    F(2) = F(2) + I0; 
    F(3) = F(3) - I0;

    y = (-J)\F;  
    x = x+y; 
    i = i+1; 
end 
iter = i-1;  
%%MAIN CODE 
Vcc = 12; 
r1 = 8000;
r2 = 4000;
re = 3300;
rc = 6000; 
G = [((1/r1)+(1/r2)), (1/r2), 0, (-1/r1), 0;...
    0, (1/rc), 0, (-1/rc), 0;...
    0, 0, (1/re), 0, 0;... 
    0, (-1/rc), 0, (1/rc), 1;... 
    0, 0, 0, 1, 0]; 
x0 = [4; 6; 3.3; 12; -0.002]; %%initial guess
[x, iter] = bjt(x0,G);  
end

Answer 1

[Walter Roberson]

The 3 x 3 J you are producing has rows that are not linearly independent.

You have a number of terms of the form

(Is/(BR*VT))*(exp(x(1)-x(2))/VT)

If your x(1) or x(2) grew to be large and positive then the exp(x(1)-x(2)) could grow to infinity, which would lead to evaluation problems. If they grew to be large and negative then exp(x(1)-x(2)) could shrink to 0, probably leading to duplicate rows or columns.