ECE495N-F08-HW_1_solution

# ECE495N-F08-HW_1_solution - As you can notice from the...

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HW_1 Problem # 1 (a) Voc=0 [V], Isc=0 [A] (b) Voc=-0.067 [V], Isc=4.2*e-10 [A], external current flows from Source to Drain. (c) Voc=0.067 [V], Isc=-4.2*e-10 [A], external current flows from Drain to Source. ----------------------- clear all %Parameters hbar=1.06e-34;q=1.6e-19; kT1=0.025;kT2=0.03;D=0.1;g=10e-3; kT=(kT1+kT2)/2;Ef=0;dE=1e-3; g1=g;g2=g; E=[0.0:dE:0.5]; %Bias ii=1;dV=1e-4; for V=-4e-2:dV:4e-2 mu1=Ef+V/2;mu2=Ef-V/2; f1=1./(1+exp((E-mu1)./kT1)); f2=1./(1+exp((E-mu2)./kT2)); f=(g1*f1+g2*f2)./(g1+g2); curr(ii)=(q*q*D*dE*g/2/hbar)*sum(f1-f2); volt(ii)=V;ii=ii+1; end hold on h=plot(volt,curr, 'b-.' ); set(h, 'linewidth' ,[3.0]) set(gca, 'Fontsize' ,[24]) xlabel( ' Voltage (V) ---> ' );

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ylabel( ' Current (A) ---> ' ); grid on ----------------------- Problem # 2
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Unformatted text preview: As you can notice from the above figure, there is a slope change of I-V characteristics. It has a symmetric I-V curve, which is a result of electrostatic of the channel. ----------------------- clear all %Parameters hbar=1.06e-34;q=1.6e-19; kT1=0.025;kT2=0.025;D=0.1;g=10e-3; kT=(kT1+kT2)/2;Ef=0.1;dE=1e-3; g1=g;g2=g; E=[0.0:dE:0.5]; %Bias ii=1;dV=1e-4; for V=-4e-1:dV:4e-1 mu1=Ef+V/2;mu2=Ef-V/2; f1=1./(1+exp((E-mu1)./kT1)); f2=1./(1+exp((E-mu2)./kT2)); f=(g1*f1+g2*f2)./(g1+g2); curr(ii)=(q*q*D*dE*g/2/hbar)*sum(f1-f2); volt(ii)=V;ii=ii+1; end hold on h=plot(volt,curr, 'y-.' ); set(h, 'linewidth' ,[3.0]) set(gca, 'Fontsize' ,[24]) xlabel( ' Voltage (V) ---> ' ); ylabel( ' Current (A) ---> ' ); grid on-----------------------...
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## This note was uploaded on 12/30/2010 for the course ECE 495N taught by Professor S.datta during the Spring '08 term at Indiana University-Purdue University Fort Wayne.

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ECE495N-F08-HW_1_solution - As you can notice from the...

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