Can anyone provide me the MATLAB code for Kronig-Penney model to draw band structure for 1-D periodic potential well structure?

2 commentaires

Adam Danz
Adam Danz le 9 Fév 2021
You'll likely have more success searching for it in an internet search engine.
Have you got the code?

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Réponses (2)

RAVI SHANKAR KUMAR
RAVI SHANKAR KUMAR le 30 Août 2023
Modifié(e) : Walter Roberson le 27 Sep 2023

2 votes

close all;
set(0,'defaultlinelinewidth',1.5)
%Constants
h = 6.626e-34;
c = 2.998e8;
h_cut = 1.055e-34;
m0 = 9.109e-31;
e_const = 1.602e-19;
%User inputs
U_eV=1;
a=3e-10;
b=4e-10;
%Derived values
U=U_eV*e_const;
ap0 = sqrt(2*m0*U/(h_cut^2));
f=@(g) (1-2*g)./(2*sqrt(g.*(g-1))).*sin(a*ap0*sqrt(g))...
.*sin(b*ap0*sqrt(g-1))...
+cos(a*ap0*sqrt(g)).*cos(b*ap0*sqrt(g-1));
g=linspace(.1, 10, 1e5);
fg=f(g);
g(isnan(g))=1;
plot(g,fg)
hold on
plot([g(1) g(end)], [1, 1], 'r--')
plot([g(1) g(end)], [-1, -1], 'g--')
ylim([min(fg)-.5, 3])
xlabel('\zeta (=E/U_o) \rightarrow');
ylabel('f(\zeta) (=RHS) \rightarrow');
title(['Plot of the RHS of the equation f(\zeta) vs. \zeta; '...
'for U=' num2str(U_eV), ' eV; a=' num2str(a*1e10) char(197)...
' and b=' num2str(b*1e10) char(197)])
grid on
flg=abs(fg)<=1;
figure
h1=gca;
hold on
xlabel('Crystal momentum, k(radian/meter) \rightarrow');
ylabel('Energy, E (eV) \rightarrow');
title(['Reduced zone representation of the E-k relationship '...
'for U=' num2str(U_eV), ' eV; a=' num2str(a*1e10) char(197)...
' and b=' num2str(b*1e10) char(197)])
xticks([-pi -pi/2 0 pi/2 pi]/(a+b));
xticklabels({'-\pi/(a+b)','-\pi/(2(a+b))','0','\pi/(2(a+b))','\pi/(a+b)'})
grid on
figure
h2=gca;
hold on
xlabel('Crystal momentum, k(radian/meter) \rightarrow');
ylabel('Energy, E (eV) \rightarrow');
title(['Extended zone representation of the E-k relationship '...
'for U=' num2str(U_eV), ' eV; a=' num2str(a*1e10) char(197)...
' and b=' num2str(b*1e10) char(197)])
xticks([-6*pi -5*pi -4*pi -3*pi -2*pi -pi 0 pi 2*pi 3*pi 4*pi 5*pi 6*pi]/(a+b))
xticklabels({'-6\pi/(a+b)','-5\pi/(a+b)','-4\pi/(a+b)' ...
'-3\pi/(a+b)','-2\pi/(a+b)','-\pi/(a+b)','0','\pi/(a+b)',...
'2\pi/(a+b)','3\pi/(a+b)'...
'4\pi/(a+b)','5\pi/(a+b)','6\pi/(a+b)'})
xtickangle(45)
grid on
prd=pi/(a+b);
plst=1;
k=1;
while ~isempty(flg) && k<6
pos=find(flg);
if isempty(pos)
break
end
pfst=plst+pos(1)-1;
flg=flg(pos(1):end);
pos=find(~flg);
if isempty(pos)
break
end
plst=pfst+pos(1)-1;
flg=flg(pos(1):end);
kv=acos(fg(pfst:plst-1))/(a+b);
ev=g(pfst:plst-1)*U_eV;
if mod(k,2)
plot(h1,[-fliplr(kv), kv], [fliplr(ev),ev], 'b');
if k==1
plot(h2,[-fliplr(kv), kv], [fliplr(ev),ev], 'b');
else
plot(h2,kv+prd*(k-1), ev, 'b');
plot(h2,-fliplr(kv)-prd*(k-1), fliplr(ev), 'b');
end
else
plot(h1, [kv, -fliplr(kv)], [ev,fliplr(ev)], 'b');
plot(h2,kv-prd*k, ev, 'b');
plot(h2,-fliplr(kv)+prd*k, fliplr(ev), 'b')
end
k=k+1;
end
Krishna Kumar
Krishna Kumar le 26 Sep 2023
Modifié(e) : Walter Roberson le 27 Sep 2023

0 votes

close all;
set(0,'defaultlinelinewidth',1.5)
%Constants
h = 6.626e-34;
c = 2.998e8;
h_cut = 1.055e-34;
m0 = 9.109e-31;
e_const = 1.602e-19;
%User inputs
U_eV=1;
a=3e-10;
b=4e-10;
%Derived values
U=U_eV*e_const;
ap0 = sqrt(2*m0*U/(h_cut^2));
f=@(g) (1-2*g)./(2*sqrt(g.*(g-1))).*sin(a*ap0*sqrt(g))...
.*sin(b*ap0*sqrt(g-1))...
+cos(a*ap0*sqrt(g)).*cos(b*ap0*sqrt(g-1));
g=linspace(.1, 10, 1e5);
fg=f(g);
g(isnan(g))=1;
plot(g,fg)
hold on
plot([g(1) g(end)], [1, 1], 'r--')
plot([g(1) g(end)], [-1, -1], 'g--')
ylim([min(fg)-.5, 3])
xlabel('\zeta (=E/U_o) \rightarrow');
ylabel('f(\zeta) (=RHS) \rightarrow');
title(['Plot of the RHS of the equation f(\zeta) vs. \zeta; '...
'for U=' num2str(U_eV), ' eV; a=' num2str(a*1e10) char(197)...
' and b=' num2str(b*1e10) char(197)])
grid on
flg=abs(fg)<=1;
figure
h1=gca;
hold on
xlabel('Crystal momentum, k(radian/meter) \rightarrow');
ylabel('Energy, E (eV) \rightarrow');
title(['Reduced zone representation of the E-k relationship '...
'for U=' num2str(U_eV), ' eV; a=' num2str(a*1e10) char(197)...
' and b=' num2str(b*1e10) char(197)])
xticks([-pi -pi/2 0 pi/2 pi]/(a+b));
xticklabels({'-\pi/(a+b)','-\pi/(2(a+b))','0','\pi/(2(a+b))','\pi/(a+b)'})
grid on
figure
h2=gca;
hold on
xlabel('Crystal momentum, k(radian/meter) \rightarrow');
ylabel('Energy, E (eV) \rightarrow');
title(['Extended zone representation of the E-k relationship '...
'for U=' num2str(U_eV), ' eV; a=' num2str(a*1e10) char(197)...
' and b=' num2str(b*1e10) char(197)])
xticks([-6*pi -5*pi -4*pi -3*pi -2*pi -pi 0 pi 2*pi 3*pi 4*pi 5*pi 6*pi]/(a+b))
xticklabels({'-6\pi/(a+b)','-5\pi/(a+b)','-4\pi/(a+b)' ...
'-3\pi/(a+b)','-2\pi/(a+b)','-\pi/(a+b)','0','\pi/(a+b)',...
'2\pi/(a+b)','3\pi/(a+b)'...
'4\pi/(a+b)','5\pi/(a+b)','6\pi/(a+b)'})
xtickangle(45)
grid on
prd=pi/(a+b);
plst=1;
k=1;
while ~isempty(flg) && k<6
pos=find(flg);
if isempty(pos)
break
end
pfst=plst+pos(1)-1;
flg=flg(pos(1):end);
pos=find(~flg);
if isempty(pos)
break
end
plst=pfst+pos(1)-1;
flg=flg(pos(1):end);
kv=acos(fg(pfst:plst-1))/(a+b);
ev=g(pfst:plst-1)*U_eV;
if mod(k,2)
plot(h1,[-fliplr(kv), kv], [fliplr(ev),ev], 'b');
if k==1
plot(h2,[-fliplr(kv), kv], [fliplr(ev),ev], 'b');
else
plot(h2,kv+prd*(k-1), ev, 'b');
plot(h2,-fliplr(kv)-prd*(k-1), fliplr(ev), 'b');
end
else
plot(h1, [kv, -fliplr(kv)], [ev,fliplr(ev)], 'b');
plot(h2,kv-prd*k, ev, 'b');
plot(h2,-fliplr(kv)+prd*k, fliplr(ev), 'b')
end
k=k+1;
end

1 commentaire

Walter Roberson
Walter Roberson le 27 Sep 2023
What is the difference between this and the Answer given by @RAVI SHANKAR KUMAR ?

Connectez-vous pour commenter.

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