Need help with correcting the code below
11 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Hello,
I am trying to plot this curve but failing to do so. I want to plot Vth as a function of shiM (3.5:0.25:5.5)
But maybe I am not using my for loop correctly. Could anyone please have a look? I really need help. Here is the code:
clc;
mo= 9.11e-31;
mn= 1.1*mo
mh=0.5*mo
k=1.3e-23;
T=300;
Eg=1.166*1.6e-19 %%Eg=1.166ev
h=6.63e-34
q= 1.6e-19;
alpha= 0.000473*q
B=636;
Na= 1e21; %%unit in meter cube,,, we have to change it for (b)
X=4.1*q %%%electron affinity value in eV CHECK
tox= 0.6e-9; %%in meter .. i took tox=0.6nm
eps0 = 8.85e-12;
Kox = 3.95; %%%oxide dielectric constant
Ks = 11.7;%%%semiconductor dielectric constant
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Nv= 2*((2*3.142*mh*k*T)/h^2)^1.5
Nc= 2*((2*3.142*mn*k*T)/h^2)^1.5
ni=sqrt(Nc*Nv)*exp((-Eg)/(2*k*T))
Egt= Eg- ((alpha*(T^2))/(B+T))
shiF= k*T* log(Na/ni) %%%joules
shiS= abs(X+ (Eg/2)+ shiF) %%joule
Cox= (Kox * eps0)/tox
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
i=1;
for shiM=3.5*q:0.25*q:5.5*q %%%assumed
shiMS(i)= abs(shiM - shiS) %%joule
%%now calculatetion of Vth
Vth= abs(shiM - shiS)+ (2*shiF) + sqrt(4*q*Ks*eps0*Na*shiF)/Cox %%unit in volts
i=i+1;
end
shiM=3.5*q:0.25*q:5.5*q
plot(shiM, Vth, '-*r')
title('Threshold voltage vs given parameter curve'), xlabel('phiM'), ylabel('Threshold voltage')
0 commentaires
Réponse acceptée
Walter Roberson
le 8 Oct 2017
Change
Vth= abs(shiM - shiS)+ (2*shiF) + sqrt(4*q*Ks*eps0*Na*shiF)/Cox %%unit in volts
to
Vth(i) = abs(shiM - shiS)+ (2*shiF) + sqrt(4*q*Ks*eps0*Na*shiF)/Cox %%unit in volts
0 commentaires
Plus de réponses (1)
Rizbi
le 8 Oct 2017
2 commentaires
Walter Roberson
le 8 Oct 2017
Yes, that is what your formula leads to. For the K'th entry, your formula gives approximately
abs(4.0e-20*K - 2.851588058088316e-19) + 0.000000000001044453337529677
Voir également
Catégories
En savoir plus sur Semiconductors and Converters dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!