Diode i-v Curve Graph
29 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Fulden Ece Ugur
am 22 Nov. 2020
Beantwortet: Md. Zamil Hasan Shovon
am 13 Jan. 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?
0 Kommentare
Akzeptierte Antwort
Mohamad
am 22 Nov. 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 Kommentare
Ashish Kulkarni
am 6 Sep. 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.
Weitere Antworten (3)
Sumit Debnath
am 13 Nov. 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')
0 Kommentare
Khuzaim
am 17 Nov. 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.

0 Kommentare
Siehe auch
Kategorien
Mehr zu Power Converters finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!