Download ENGR-25_Lec-16A_MTE_Review_Prob4-37.ppt

Engr/Math/Physics 25
MidTerm Exam
Review
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering/Math/Physics 25: Computational Methods
1
Bruce Mayer, PE
[email protected] • ENGR-25_Plot_Model-2.ppt
Problem 4-37
 Possible
Confusion in
Text Book
i1
i2
i3
i4
i5
800
600
Resitor Current(mA)
 All R’s Should
be in kΩ
Resistor Network currents
1000
 Plot at right
shows the large
currents
Generated by
Not Using kΩ
400
200
In NO case are ALL
Currents  1mA
0
I5  0 In all Cases
-200
-400
0
20
40
60
80
100
120
Supply-2 Potential (V)
140
Prob4_31_KVL_KCL_Plot.m
Engineering/Math/Physics 25: Computational Methods
2
Bruce Mayer, PE
[email protected] • ENGR-25_Plot_Model-2.ppt
160
180
200
Prob 4-37
 v1 = 100 V
 v2 Variable
 All Resistances kΩ
 Max Resistor
Current = 1 mA
Engineering/Math/Physics 25: Computational Methods
3
Bruce Mayer, PE
[email protected] • ENGR-25_Plot_Model-2.ppt
P4-37 Electrical Engineering
 Analyze this Circuit
using Methods from
ENGR43
 Specifically use
• Kirchoff’s Voltage Law
(KVL)
• Ohm’s Law (V = IR)
• Kirchoff’s Current Law
(KCL)
 These Laws yield
Eqns for the currents
Engineering/Math/Physics 25: Computational Methods
4
 v1  R1i1  R4i4  0
 R4i4  R2i2  R5i5  0
 R5i5  R3i3  v2  0
i1  i2  i4
i2  i3  i5
Bruce Mayer, PE
[email protected] • ENGR-25_Plot_Model-2.ppt
P4-37 Applied Math
 The 5 eqns for the 5 currents can be
cast into Matrix form:
R4
0   i1   v1 
 R1 0 0


0 R



i
0

R
R
0
2
2
4
5

  

 0 0 R3
0
 R5  i3    v2 

  

1
0  i4   0 
 1 1 0
 0  1 1
0
1  i5   0 
A
Engineering/Math/Physics 25: Computational Methods
5
C V
Bruce Mayer, PE
[email protected] • ENGR-25_Plot_Model-2.ppt
4-37 MATLAB Numerical Processor
 Recall:
A  Coefficien t Matrix
V  Constraint Vector
C  Solution Vector
 Use MATLAB’s LEFT Division to Find
the solution vector C
 PLOT Results to Analyze Circuit
BEHAVIOR
• Remember – When in Doubt PLOT
Engineering/Math/Physics 25: Computational Methods
6
Bruce Mayer, PE
[email protected] • ENGR-25_Plot_Model-2.ppt
Plot v2 over 1-400V
Resistor Network currents
1.5
Green Zone
Prob4_31_KVL_KCL_Calc.m
Resitor Current(mA)
1
0.5
0
-0.5
i1
i2
i3
i4
i5
-1
-1.5
0
50
100
150
200
250
Supply-2 Potential (V)
Engineering/Math/Physics 25: Computational Methods
7
300
350
400
Bruce Mayer, PE
[email protected] • ENGR-25_Plot_Model-2.ppt
4-37(b)  v2min(R3), v2max(R3)
 Now Allow R3 to vary: 150 kΩ 250 kΩ
150 k  250 k
 Use solution to part (a) as basis
• Vary R3 with For-Loop, then
Chk v2 as in part (a)
Engineering/Math/Physics 25: Computational Methods
8
Bruce Mayer, PE
[email protected] • ENGR-25_Plot_Model-2.ppt
Plot v2 Max & Min as f(R3)
 Use Solution from part(a) in FOR loop
that Varies R3 to produce Plot
Problem 4-31(b): Allowable v2 as a function of R3
400
350
Part (a) Case
Allowable v2 (V)
300
250
v2,max
v2,min
200
150
100
Part (a) Case
50
0
150
160
170
180
190
Engineering/Math/Physics 25: Computational Methods
9
200
R3 (k)
210
220
230
240
Bruce Mayer, PE
[email protected] • ENGR-25_Plot_Model-2.ppt
250
All Done for Today
APNext™ 2X-Inj Dep & Etch Profiles
5.5
This Space
For
Rent
Dep or Etch Depth on Wafer or Seal Plt (a. u.)
5.0
Static Print Assumptions
• Outer Etch Static Width = 16 mm @ ±52 mm
• Inner Etch Static Width = 28 mm @
• Dep Static Width = 28mm @ ±60 mm
4.5
4.0
3.5
3.0
2.5
~18 mm
2.0
1.5
1.0
Etch
0.5
0.0
-250
-200
-150
-100
-50
Dep
0
50
100
Distance from Wafer CenterLIne (mm)
Engineering/Math/Physics 25: Computational Methods
10
Bruce Mayer, PE
[email protected] • ENGR-25_Plot_Model-2.ppt
150
200
250
file = 2Xvs3X.xls
Engr/Math/Physics 25
Appendix
f x   2 x  7 x  9 x  6
3
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering/Math/Physics 25: Computational Methods
11
Bruce Mayer, PE
[email protected] • ENGR-25_Plot_Model-2.ppt
Prob4_37_KVL_KCL_plot.m - 1
% Bruce Mayer, PE * 08Jul05
% ENGR25 * Problem 4-31
% 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];
%
Engineering/Math/Physics 25: Computational Methods
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% 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,1000*i1,vplot,1000*i2,
vplot,1000*i3, vplot,1000*i4, vplot,1000*i5
),...
ylabel('Resitor
Current(mA)'),xlabel('Supply-2 Potential
(V)'),...
title('Resistor Network currents'),
grid, legend('i1', 'i2', 'i3', 'i4', 'i5')
Bruce Mayer, PE
[email protected] • ENGR-25_Plot_Model-2.ppt
Prob4_31_KVL_KCL_Calc.m - 1
% Bruce Mayer, PE * 08Jul05
% ENGR25 * Problem 4-31
% 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
Engineering/Math/Physics 25: Computational Methods
13
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 = 1000*i1_lo
i2_low = 1000*i2_lo
i3_low = 1000*i3_lo
i4_low = 1000*i4_lo
i5_low = 1000*i5_lo
%
Bruce Mayer, PE
[email protected] • ENGR-25_Plot_Model-2.ppt
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
Engineering/Math/Physics 25: Computational Methods
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%display "hi" vars
v2_hi
display('currents in mA')
i1_high = 1000*i1_hi
i2_high = 1000*i2_hi
i3_high = 1000*i3_hi
i4_high = 1000*i4_hi
i5_high = 1000*i5_hi
Bruce Mayer, PE
[email protected] • ENGR-25_Plot_Model-2.ppt