ECE 453 Assignment 6 Solutions[1]

ECE 453 Assignment - Np = 25 t0 = 1 E =-0.5:0.01:4.5 U = 1 H = 2*t0*eye(Np t0*diag(ones(1,Np-1,1 t0*diag(ones(1,Np-1-1%H(13,13 = H(13,13 U Sig1 =

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Problem 1 a
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1 b
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Problem 2 clear all ; clc; q = 1.6e-19; h = 6.64e-34; m = 9.1e-31; z = 1i*1e-6; eps1 = 50; eps2 = -50; t = 100; gam = 50; E = -200:1:200; H = [eps1 t;t eps2]; Sig1 = -1i*[gam/2 0;0 0]; Sig2 = -1i*[0 0;0 gam/2]; Gam1 = 1i*(Sig1-Sig1'); Gam2 = 1i*(Sig2-Sig2'); T = zeros(length(E),1); for ii = 1:length(E) G = inv((E(ii)+z)*eye(2)-H-Sig1-Sig2); T(ii) = trace(Gam1*G*Gam2*G'); end plot(E,(q*q/h/1000)*T); The peaks are close to the eigen values of H, but not exactly the same. This is due to the broadening introduced by the contacts. Do it yourself: change the value of gam in the code and see how the plot changes. -200 -100 0 100 200 0 0.5 1 1.5 2 2.5 3 3.5 x 10 -8 E(meV)-> I(E) (A/meV)-> I(E) v E
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Problem 3 clear all ; clc; q = 1.6e-19; h = 6.64e-34; m = 9.1e-31; z = 1i*1e-6;
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Unformatted text preview: Np = 25; t0 = 1; E = -0.5:0.01:4.5; U = 1; H = 2*t0*eye(Np)- t0*diag(ones(1,Np-1),1) - t0*diag(ones(1,Np-1),-1); %H(13,13) = H(13,13)+U; Sig1 = zeros(Np); Sig2 = zeros(Np); T = zeros(length(E),1); Tan = zeros(length(E),1); for ii = 1:length(E) ka = acos(1-(E(ii)+z)/2/t0); sig = -t0*exp(1i*ka); Sig1(1,1) = sig; Sig2(Np,Np) = sig; Gam1 = 1i*(Sig1-Sig1'); Gam2 = 1i*(Sig2-Sig2'); G = inv((E(ii)+z)*eye(Np)-H-Sig1-Sig2); T(ii) = trace(Gam1*G*Gam2*G'); %Tan(ii) = 1/(1+(U/(2*t0*sin(ka)))^2); end plot(E,T); %plot(E,T,E,Tan); 3a 1 2 3 4 0.2 0.4 0.6 0.8 1 E(eV)-> T(E)-> T(E) v E 3b i & ii) 0.2 0.4 0.6 0.8 1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 E(eV)-> T(E)-> T(E) v E Numerical Analytical...
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This note was uploaded on 11/29/2010 for the course ECE 453 taught by Professor Supriyodatta during the Spring '10 term at Purdue University-West Lafayette.

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ECE 453 Assignment - Np = 25 t0 = 1 E =-0.5:0.01:4.5 U = 1 H = 2*t0*eye(Np t0*diag(ones(1,Np-1,1 t0*diag(ones(1,Np-1-1%H(13,13 = H(13,13 U Sig1 =

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