Diode i-v Curve Graph

102 次查看(过去 30 天)
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?

采纳的回答

Mohamad
Mohamad 2020-11-22
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 个评论
Mohamad
Mohamad 2020-11-23
Ok , thanks , please accept the answer for the question
Ashish Kulkarni
Ashish Kulkarni 2021-9-6
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.

请先登录,再进行评论。

更多回答(3 个)

Sumit  Debnath
Sumit Debnath 2021-11-13
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 2022-11-17
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
I = Iph - Io [exp(V/n*Vt) – 1]

类别

Help CenterFile Exchange 中查找有关 MATLAB 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by