How to write the equation in matlab I= A A* T² exp (qob/kT) exp (q(V - Rs I /kT)) (exp (-1/n) ? Why is the curve not shown in the drawing?

clear; close all; clc; n=5.2; %Ideality factor A =8.16; %Richardson constant q = 1.602e-19; % electron charge K = 1.380e-23; %Boltzmann constant T =140; % Absolute temperature phi_b = 0.43; %Barrier Height V =(0:0.1:2); Rs=7.74e9; %Resistance a=2.8e-5; %the effective area I1=1.5; I =A*a* (T.* T).* exp((-q.* phi_b)/(K* T).*exp((q.* V) ./ ( K* T))) .*exp((-q .*I1.*Rs) ./ (K* T).*(-1/n)); %corrected the paranthesis plot(V,I)

[Dyuman Joshi]

Are you sure the original formula is correct?

What are you trying to find with the formula? Looks like something related to Semi-conductor.

n = 5.2; 
%Ideality factor
A = 8.16; 
%Richardson constant
q = 1.602e-19;
% electron charge 
K = 1.380e-23;
%Boltzmann constant
T = 140;
%Absolute temperature
phi_b = 0.43;
%Barrier Height
V = (0:0.1:2);
Rs = 7.74e9; 
%Resistance
a = 2.8e-5; 
%the effective area 
I1 = 1.5; 

I=A*a*T*T*exp(-q*phi_b/(K*T))*exp(-1/n)*exp(q*(V-Rs*I1/(K*T)));
%corrected the paranthesis
plot(V,I)

[Han]

I am trying to find the relationship between current and voltage and it is related to Schottky diodes

[Dyuman Joshi]

"Why is the curve not shown in the drawing?"

There is a curve as you can on the plot as you can see above, but the value might not be what you expected it to be.

"I am trying to find the relationship between current and voltage and it is related to Schottky diodes"

Alright, however, you did not answer my question - Is the original formula you posted, as the problem title, correct?

I was not able to find the exact equation on the internet, though the closest I got was equation 4.6 on this link

https://in.ncu.edu.tw/ncume_ee/SchottkyDiode.htm

[Han]

Yes, the formula is correct

[Dyuman Joshi]

If the formula is correct, then so is the output and the plot from the code in my comment above.

The formula in above code is according to the formula in the title.

Answer 1

[Torsten]

Maybe you mean

I = A*a*T^2*exp(q*phi_b/(K*T))*exp(q*(V-I1*Rs/(K*T)))*exp(-1/n) 

?

Answer 2

[John D'Errico]

n=5.2;
%Ideality factor
A =8.16;
%Richardson constant
q = 1.602e-19;
% electron charge
K = 1.380e-23;
%Boltzmann constant
T =140;
% Absolute temperature
phi_b = 0.43;
%Barrier Height
V =(0:0.1:2);
Rs=7.74e9;
%Resistance
a=2.8e-5;
%the effective area
I1=1.5;
I =A*a* (T.* T).* exp((-q.* phi_b)/(K* T).*exp((q.* V) ./ ( K* T))) .*exp((-q .*I1.*Rs) ./ (K* T).*(-1/n));
%corrected the paranthesis

I

I = 1×21

Inf NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN

Before you plot it, why not look at the numbers you generated?

Remember that MATLAB uses DOUBLE PRECISION floating point arithmetic.

I would strongly suggest you compute the log of that expression. No, don't compute the variable I, then try to take the log. Take the log, by adding the logs of each individual term. Don''t even try to compute those exponentials. Remember that log(exp(x))=x.