Review Two - Computational Methods - Lecture Slides, Slides of Computational Methods

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MidTerm Exam

Review

Problem 4-

• Error in Text Book
• All R’s Should be

in kΩ

• Plot at right shows

the large currents

Generated by this

Error

0 20 40 60 80 100 120 140 160 180 200

0

200

400

600

800

1000

Resitor Current(mA)

Supply-2 Potential (V)

Resistor Network currents

i i i i i

Prob4_31_KVL_KCL_Plot.m

In NO case are ALL

Currents ≤ 1mA

I 5 ≅ 0 In all Cases

Plot v2 over 1-400V

Resitor Current(mA)

Supply-2 Potential (V)

Resistor Network currents

i

i

i

i

i

Green Zone

Prob4_31_KVL_KCL_Calc.m

All Done for Today

This Space

For

Rent

APNext™ 2X-Inj Dep & Etch Profiles

-250 -200 -150 -100 -50 0 50 100 150 200 250 Distance from Wafer CenterLIne (mm)

Dep or Etch Depth on Wafer or Seal Plt (a. u.) Etch Dep

file = 2Xvs3X.xls

Static Print Assumptions

• Outer Etch Static Width = 16 mm @ ±52 mm
• Inner Etch Static Width = 28 mm @
• Dep Static Width = 28mm @ ±60 mm

~18 mm

Prob4_31_KVL_KCL_plot.m - 1

% Bruce Mayer, PE * 08Jul % ENGR25 * Problem 4- % file = Prob4_31_KCL_KVL.m % % INPUT SECTION %R1 = 5; R2= 100; R3 = 200; R4 = 150; % SingleOhm case R5 = 250e3; R1 = 5e3; R2= 100e3; R3 = 200e3; R4 = 150e3; % kOhm case % Coeff Matrix A v1 = 100 A = [R1 0 0 R4 0; 0 R2 0 -R4 R5; 0 0 R3 0 - R5;... -1 1 0 1 0; 0 -1 1 0 1]; %

% Make Loop with v2 as counter in units of Volts for v2 =1:400 % units of volts %Constraint Vector V V = [v1; 0; -v2; 0; 0]; % find soltion vector for currents, C C = A\V; % Build plotting vectors for current vplot(v2) = v2; i1(v2) = C(1); i2(v2) = C(2); i3(v2) = C(3); i4(v2) = C(4); i5(v2) = C(5); end % PLOT SECTION plot(vplot,1000i1,vplot,1000i2, vplot,1000i3, vplot,1000i4, vplot,1000*i ),... ylabel('Resitor Current(mA)'),xlabel('Supply-2 Potential (V)'),... title('Resistor Network currents'), grid, legend('i1', 'i2', 'i3', 'i4', 'i5')

Prob4_31_KVL_KCL_Calc.m - 1

% Bruce Mayer, PE * 08Jul % ENGR25 * Problem 4- % file = Prob4_31_KCL_KVL.m % % INPUT SECTION %R1 = 5; R2= 100; R3 = 200; R4 = 150; % SingleOhm case R5 = 250e3; R1 = 5e3; R2= 100e3; R3 = 200e3; R4 = 150e3; % kOhm case % Coeff Matrix A v1 = 100; % in Volts A = [R1 0 0 R4 0; 0 R2 0 -R4 R5; 0 0 R3 0 - R5;... -1 1 0 1 0; 0 -1 1 0 1]; % % LOW Loop % Initialize Vars v2 = 40; C = [0;0;0;0;0]; % use element-by-element logic test on while % Must account for NEGATIVE Currents

while abs(C) < 0.001[1;1;1;1;1] % Constraint Col Vector V V = [v1; 0; -v2; 0; 0]; % find solution vector for currents, C C = A\V; % Collect last conforming Value-set v2_lo = v2; i1_lo = C(1); i2_lo = C(2); i3_lo = C(3); i4_lo = C(4); i5_lo = C(5); %increment v2 by 10 mV DOWN v2 = v2 - 0.01; end %display "lo" vars v2_lo display('currents in mA') i1_low = 1000i1_lo i2_low = 1000i2_lo i3_low = 1000i3_lo i4_low = 1000i4_lo i5_low = 1000i5_lo %