MidTerm Exam Review for Problem 4-37 in Electrical Engineering, Slides of Computational Methods

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MidTerm Exam

Review

Problem 4-

• Possible

Confusion in Text

Book

• All R’s Should be

in kΩ

• Plot at right shows

the large currents

Generated by Not

Using kΩ

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

P4-37 Electrical Engineering

• Analyze this Circuit

using Methods from

ENGR

 Specifically use

• Kirchoff’s Voltage Law

(KVL)

• Ohm’s Law (V = IR)
• Kirchoff’s Current Law

(KCL)

 These Laws yield

Eqns for the currents 235

1 2 4

5 5 3 3 2

4 4 2 2 5 5

1 1 1 4 4

0

0

0

i i i

i i i

R i R i v

R i R i R i

v R i R i

= +

= +

− + + =

− + + =

− + + =

P4-37 Applied Math

• The 5 eqns for the 5 currents can be cast into

Matrix form:

































= −

































































−

−

−

−

0

0

0

0 1 1 0 1

1 1 0 1 0

0 0 0

0 0

0 0 0

2

1

5

4

3

2

1

3 5

2 4 5

1 4

v

v

i

i

i

i

i

R R

R R R

R R

A C^ =^ V

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

4-37(b)  v min

(R3), v max

(R3)

• Now Allow R3 to vary: 150 kΩ 250 kΩ

 Use solution to part (a) as basis

• Vary R3 with For-Loop, then

Chk v2 as in part (a)

150 k  250 k

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

Bruce Mayer, PE

Licensed Electrical & Mechanical Engineer

[email protected]

Engr/Math/Physics 25

Appendix

( ) 2 7 9 6

3 2 f x = x − x + x −

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 %

Prob4_37_KVL_KCL_Calc.m - 2

% HIGH Loop % Initialize Vars v2 = 300; 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 Vector V V = [v1; 0; -v2; 0; 0]; % find soltion vector for currents, C C = A\V; % Collect last conforming set v2_hi = v2; i1_hi = C(1); i2_hi = C(2); i3_hi = C(3); i4_hi = C(4); i5_hi = C(5); %increment v2 by 10 mV UP v2 = v2 + 0.01; end

%display "hi" vars v2_hi display('currents in mA') i1_high = 1000i1_hi i2_high = 1000i2_hi i3_high = 1000i3_hi i4_high = 1000i4_hi i5_high = 1000*i5_hi