EECS 215 F07 Problem Set 5 solutions

EECS 215 F07 Problem Set 5 solutions - 5 4.5 4 3.5 3 ID...

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 x 10 -3 X: 0.7308 Y: 0.001611 V D (Volts) I D (Amps) Problem 2, part a Diode Current Resistor Current %Code for Problem #2, Part a VD = linspace(0,.78,1000); ID_diode = 10.^(-15).*(exp(VD./0.026)-1); ID_res = (4-VD)./2000; hold on ; plot(VD, ID_diode); plot(VD, ID_res); xlabel( 'V_D (Volts)' ); ylabel( 'I_D (Amps)' ); title( 'Problem 2, part a' );
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0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x 10 -3 V DS (V) I D (A) EECS 215, Fall 2007, Problem 5, Part A
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%EECS 215 Problem 5, Homework 5, Part A (October 10, 2007) hold on ; VT = 1.1; %V k = 0.25e-3; %(A/V^2) VDS = linspace(0,5,500); %for VGS = 0, 0.5, and 1 V, VGS < VT so ID = 0. ID = 0.*VDS; plot(VDS,ID, 'LineWidth' ,2); ID = 0.*VDS; plot(VDS,ID, 'LineWidth' ,2); ID = 0.*VDS; plot(VDS,ID, 'LineWidth' ,2); %The above three plots overlap so they are redundant, but I'm just doing it %to be thorough. %for VGS > VT . .. for n = 0:7 VGS = 1.5+n.*0.5; VDSsat = VGS-VT; %Saturation Voltage %%%%%%%%%%%%%%% %Linear %%%%%%%%%%%%%%% VDS = linspace(0,VDSsat,VDSsat.*100);
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EECS 215 F07 Problem Set 5 solutions - 5 4.5 4 3.5 3 ID...

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