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Transcript
EE 616
Computer Aided Analysis of Electronic Networks
Lecture 2
Instructor: Dr. J. A. Starzyk, Professor
School of EECS
Ohio University
Athens, OH, 45701
1
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Review and Outline


2
Review of the previous lecture
-- Class organization
-- CAD overview
Outline of this lecture
* Review of network scaling
* Review of Thevenin/Norton Analysis
* Formulation of Circuit Equations
-- KCL, KVL, branch equations
-- Sparse Tableau Analysis (STA)
-- Nodal analysis
-- Modified nodal analysis
EE 616
Network scaling
3
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Network scaling (cont’d)
4
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Network scaling (cont’d)
5
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Review of the Thevenin/Norton Analysis
ZTh
Voc
+
–
Thevenin equivalent
circuit
Isc
ZTh
Norton equivalent
circuit
Note: attention to the voltage and current direction
6
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Review of the Thevenin/Norton Analysis
1. Pick a good breaking point in the circuit (cannot split a dependent
source and its control variable).
2.Replace the load by either an open circuit and calculate the voltage E
across the terminals A-A’, or a short circuit A-A’ and calculate the
current J flowing into the short circuit. E will be the value of the
source of the Thevenin equivalent and J that of the Norton
equivalent.
3. To obtain the equivalent source resistance, short-circuit all
independent voltage sources and open-circuit all independent current
sources. Transducers in the network are left unchanged. Apply a unit
voltage source (or a unit current source) at the terminals A-A’ and
calculate the current I supplied by the voltage source (voltage V
across the current source). The Rs = 1/I (Rs = V).
7
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Modeling
8
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Formulation of circuit equations (cont’d)
9
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Ideal two-terminal elements
10
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Ideal two-terminal elements
Topological equations
11
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KVL and KCL

Determined by the topology of the circuit

Kirchhoff’s Current Law (KCL): The algebraic
sum of all the currents leaving any circuit node is zero.

Kirchhoff’s Voltage Law (KVL): Every circuit
node has a unique voltage with respect to the reference node.
The voltage across a branch eb is equal to the difference
between the positive and negative referenced voltages of the
nodes on which it is incident
12
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Formulation of circuit equations (cont’d)

Unknowns
–
–
–

(i)
(e)
(v)
Equations
–
–
–
13
B branch currents
N node voltages
B branch voltages
KCL: N equations
KVL: B equations
Branch equations: B equations
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Branch equations

Determined by the mathematical model of the
electrical behavior of a component
–

14
Example: V=R·I
In most of circuit simulators this mathematical
model is expressed in terms of ideal elements
EE 616
Matrix form of KVL and KCL
B equations
N equations
15
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Branch equation
 1
 R
 1
 0
 0


 0

 0
16
0
0
0
0  G2
1
0 
R3
0
0
0
0
0
0

Kvv + i = is
1
R4
0

0
v
i
0
  1   1   
0 v
i2   0 
2


0 v3   i3    0 
 v4  i4   0 
0 v  i  i 
  5   5   s5 
0
B equations
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Node branch incidence matrix
branches
1 2 3
n
o 1
d 2
e
s i
j
B
(+1, -1, 0)
N
{
Aij =
17
+1 if node i is terminal + of branch j
-1 if node i is terminal - of branch j
0 if node i is not connected to branch j
PROPERTIES
•A is unimodular
•2 nonzero entries in
each column
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Equation Assembly for Linear Circuits
–
Sparse Table Analysis (STA)

–
Modified Nodal Analysis (MNA)

18
Brayton, Gustavson, Hachtel
McCalla, Nagel, Roher, Ruehli, Ho
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Sparse Tableau Analysis (STA)
19
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Advantages and problems of STA
20
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Nodal analysis
1.
2.
3.
Write KCL
A·i=0
(N equations, B unknowns)
Use branch equations to relate branch currents to
branch voltages
i=Yv
(B equations, B unknowns)
Use KVL to relate branch voltages to node
voltages
v=ATe
(B equations, N unknowns)
Yne=ins
21
Nodal Matrix
N equations
N unknowns
N = # nodes
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Nodal analysis
22
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Nodal analysis – Resistor “Stamp”
Spice input format: Rk
N+
N+
Rk
N-
i
 1
N+ 
R
 k
 1
 Rk
N-
Rkvalue
N1

Rk
1
Rk





1
 iothers  R eN   eN     is
k
KCL at node N+
1
eN   eN     is
Rk
KCL at node N-
 iothers 
23
N+ N-
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Nodal analysis – VCCS “Stamp”
Spice input format: Gk
NC+
NC-
-
i
i
others
others
24
N+
+
vc
N+ N- NC+ NC-
Gkvalue
NC+
N+  G
k
 G
k
N- 
Gkvc
NC-
 Gk 
Gk 
N-
 Gk eNC   eNC     is
 Gk eNC   eNC     is
KCL at node N+
KCL at node NEE 616
Nodal analysis- independent current sources “stamp”
25
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Nodal analysis- by inspection
Rules (page 36):
1.
The diagonal entries of Y are positive and
y jj  admittances connected to node j
2. The off-diagonal entries of Y are negative and are given by
y jk   admittances connected between nodes j and k
3. The jth entry of the right-hand-side vector J is
J j   currents from independent sources entering node j
26
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Example of nodal analysis by inspection
Exercise
Formulate nodal equations by inspection
27
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Example of nodal analysis by inspection
28
EE 616
Example of nodal analysis by inspection
Exercise
Formulate nodal equations by inspection
29
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Example of nodal analysis by inspection
Exercise
30
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Nodal analysis (cont’d)
31
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Modified Nodal Analysis (MNA)
32
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Modified Nodal Analysis (2)
33
1
1
  G2 
R3
 R1
1



R3


0



0


0

E7


1 
  G2  
R3 

1
1

R3 R4
0
0
0
0
1
1
R8
1

R8
1
 E7
0
0
0
1
R8
1
R8
0

1

0
e   0 
 1
 1 0  e   i 
 2   s5 
 e   0 
3
1 0    

 e4   0 
  

0 1  i6   ES 6
  i7   0 
0 0

0 0
0
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Modified Nodal Analysis (3)
34
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General rules for MNA
35
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Example 4.4.1(p.143)
36
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Advantages and problems of MNA
37
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Analysis of networks with VVT’s & Op Amps
38
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Example 4.5.2 (p.145)
39
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Example 4.5.5 (p. 148)
40
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Example 4.5.5 (cont’d)
41
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