Diode i-v Curve Graph

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Fulden Ece Ugur
Fulden Ece Ugur el 22 de Nov. de 2020
Respondida: Md. Zamil Hasan Shovon el 13 de En. de 2023
Hello,
I need help to graph that diode i-v curve. my code is here;
clc
clear all
n = 1.65; %Ideality factor
Is = 220*10^-12; % diode reverse saturated current
q = 1.602*10^-19; % electron charge
K = 1.38*10^-23; %Boltzmann constant
T = 300; % Absolute temperature
Vd = ((n.*K.*T)./q).*log((Is./Id)+1);
Id = Is*(exp((q*Vd)./(1.65*K*T))-1);
plot(Vd,Id)
hold on
ylabel('Id')
xlabel('Vd')
Id and Vd values are depending each other so how can graph that depending values?

Respuestas (4)

Mohamad
Mohamad el 22 de Nov. de 2020
Try this , change voltage vaues as you need .
clc
clear all
n = 1.65; %Ideality factor
Is = 220*10^-12; % diode reverse saturated current
q = 1.602*10^-19; % electron charge
K = 1.38*10^-23; %Boltzmann constant
T = 300; % Absolute temperature
fs=1000;
dv=1/fs;
v0=-0.3 ; % change as you want
vend=1.1 ; % change as you want
Vd=v0:dv:vend;
Id = Is*(exp((q*Vd)./(1.65*K*T))-1);
plot(1000*Vd,Id) ; grid ; xlabel(' Diode Voltage in mV ' ) ; ylabel(' Diode Current in Amp. ')
  3 comentarios
Mohamad
Mohamad el 23 de Nov. de 2020
Ok , thanks , please accept the answer for the question
Ashish Kulkarni
Ashish Kulkarni el 6 de Sept. de 2021
How will you plot the reverse breakdown voltage here? For which diode are these paramaters specified.
I tried upto 100 V, but it is giving only a straight line.

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Sumit  Debnath
Sumit Debnath el 13 de Nov. de 2021
n = 1.65; %Ideality factor
Is = 220*10^-12; % diode reverse saturated current
q = 1.602*10^-19; % electron charge
K = 1.38*10^-23; %Boltzmann constant
T = 300; % Absolute temperature
Vd = ((n.*K.*T)./q).*log((Is./Id)+1);
Id = Is*(exp((q*Vd)./(1.65*K*T))-1);
plot(Vd,Id)
hold on
ylabel('Id')
xlabel('Vd')

Khuzaim
Khuzaim el 17 de Nov. de 2022
Use “exponential model with graphical analysis” to
determine VD1, VD2, VD3, ID1, ID2, ID3. Assume that the diode has a current of 0.5 mA at a
voltage of 0.7 V.

Md. Zamil Hasan Shovon
Md. Zamil Hasan Shovon el 13 de En. de 2023
I = Iph - Io [exp(V/n*Vt) – 1]

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