CSE 245: Computer Aided Circuit Simulation and Verification Instructor: Prof. Chung-Kuan Cheng
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Transcript CSE 245: Computer Aided Circuit Simulation and Verification Instructor: Prof. Chung-Kuan Cheng
CSE 245: Computer Aided Circuit
Simulation and Verification
Winter 2003
Lecture 1: Formulation
Instructor:
Prof. Chung-Kuan Cheng
Agenda
RCL Network
Sparse Tableau Analysis
Modified Nodal Analysis
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History of SPICE
SPICE -- Simulation Program with
Integrated Circuit Emphasis
1969, CANCER developed by Laurence
Nagel on Prof. Ron Roher’s class
1970~1972, CANCER program
May 1972, SPICE-I release
July ’75, SPICE 2A, …, 2G
Aug 1982, SPICE 3 (in C language)
No new progress on software package
since then
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RCL circuit
R
L
C
Vs
1 v2 0
c 0 v2 0
0 l i 1 r i v
1
1 s
1
0
v
v2 0
2
c
vs
i 1
r i1
1
l
l
l
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RCL circuit (II)
General Circuit Equation
Y AY BU
Consider homogeneous form first
Y AY
Y e Y0
At
2 2
and
k k
At A t
At
e I
...
...
1!
2!
k!
At
Q: How to Compute Ak ?
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Solving RCL Equation
Assume A has non-degenerate eigenvalues
1 , 2 ,..., k and corresponding linearly
independent eigenvectors 1 , 2 ,..., k , then A
can be decomposed as
1
A
1 0
0
where
2
0
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0
and , ,...,
1
2
k
k
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Solving RCL Equation (II)
What’s the implication then?
1
A
1 t
e e
At
1 2
1
A A
2
2
where e t
To compute the eigenvalues:
e 1
0
0
2
k
e
0
e 2
0
det( I A) n cn 1n 1 ... c0
( p0 )(2 p1 p2 )...(...) 0
real
eigenvalue
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Conjugative
Complex
eigenvalue
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Solving RCL Equation (III)
0
In the previous example v2 e At X e 1/ l
0
i
1
1/ c
t
r / l
v2 (0)
i (0)
1
0 1
1
1
Let c=r=l=1, we have A
1 1
0
1 3 j
1 3 j
1
j
2
2
where
3 1 3 j
1 3 j
1
2
2
e
0
At
1
e
hence
0 e
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1
1 3 j
2
1
0
1
1 3 j
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Solving RCL Equation (IV)
What if matrix A has degenerated
eigenvalues? Jordan decomposition !
1
A J
J is in the Jordan Canonical form
And still
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e At 1e Jt
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Jordan Decomposition
1
J
0
t
1
0
1
e
e Jt
t
0 1 0
0
tet
t
e
similarly
1 0
J 0 1
0 0
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t
e
1 0 0 1 0
e Jt 0 1 0 0 1 t 0
0
0 0 1 0 0
10
te
t
e t
0
t 2 t
e
2!
tet
e t
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Agenda
RCL Network
Sparse Tableau Analysis
Modified Nodal Analysis
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Equation Formulation
KCL
Converge of node current
KVL
Closure of loop voltage
Brach equations
I, R relations
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Types of elements
Resistor
Capacitor
dQ Q(v) dv
dv
i
c (v )
dt
v dt
dt
Inductor
d (i ) di
v
dt
i
dt
l (i )
di
dt
L is even dependent on frequency due to skin
effect, etc…
Controlled Sources
VCVS, VCCS, CCVS, CCCS
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Cut-set analysis
1
3
5
1. Construct a spanning tree
2. Take as much capacitor
branches as tree branches as
possible
2
4
6
3. Derive the fundamental cut-set,
in which each cut truncates exactly
one tree branch
4. Write KCL equations for each
cut
5. Write KVL equations for each
tree link
6. Write the constitution equation
for each branch
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KCL Formulation
Ai 0
c12
c24
c34
c45
c46
i12 i13 i24
1 1 0
0 1 1
0 1 0
0 0 0
0 0 0
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i34
0
0
1
0
0
i35 i45
0 0 0
0 0 0
1 0 0
1 1 0
0 0 1
1
3
5
2
4
6
i12
i46 i56 i13
0
0 i24
0
0 i34
0
0 i35
0
1 i45
0
1
i
46
i56
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#nodes-1 lines
#braches columns
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KCL Formulation (II)
Permute the columns to achieve a
systematic form
1
~
I|Ai 0
i12 i24 i34 i45 i46 i13 i23
c12
c24
c34
c45
c46
1
0
0
0
0
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0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0 1
0 1
0 1
0 0
1 0
0
0
1
1
0
i12
2
i56
i24 0
0 i34
0
0 i45
0
0 i46
0
1 i13
0
1 i23
i56
16
3
5
4
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KVL Formulation
AT e V
1
1
0
0
0
0
0
0
0 0 0
1 1 0
1 0 0
0 1 0
0 1 1
0 0 1
0 0 0
0 0 1
0
v12
v
0
13
v12
0 v24
v24
0 v34
v34
0 v35
v45
v
0
v46 45
v
1
46
1
v56
Remove the
equations for
tree braches
and systemize
1
3
5
2
4
6
v12
v
24
v34
l13 1 1 1 0 0 1 0 0
v
l35 0 0 1 1 0 0 1 0 45 0
v46
l56 0 0 0 1 1 0 0 1
v13
v
35
v56
BV 0
~
B B
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I
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Cut & Loop relation
A e V
BV 0
T
In the previous example
1
1
~
A 1
0
0
BA T 0
~T
~
[ B I ][ IA] 0
~T
B A 0
~T
~
B A
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0 0
0 0
1 0
1 1
0 1
1 1 1 0 0
~
B 0 0 1 1 0
0 0 0 1 1
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Sparse Tableau Analysis (STA)
n=#nodes, b=#branches
b
(n-1) KCL
b
KVL
b branch
relations
A
0
K i
b
0
I
Kv
n-1
0 i
T
A v
0 e
0
b 0
n-1
S
b
Due to independent
sources
Totally 2b+n-1 variables, 2b+n-1 equations
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STA (II)
Advantages
Covers any circuit
Easy to assemble
Very sparse
Ki, Kv, I each has exactly b non-zeros. A
and AT each has at most 2b non-zeros.
Disadvantages
Sophisticated data structures &
programming techniques
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Agenda
RCL Network
Sparse Tableau Analysis
Modified Nodal Analysis
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Nodal Analysis
Derivation
From STA:
(3) x Ki-1
(4) x A
Using (a)
e
Ai 0 (1)
v AT e (2)
Kii Kvv S
(3)
1
1
i K i K v v K i S (4)
1
1
Ai AK i K v v AK i S
1
T
1
AK i K v A e AK i S
(5)
(6)
Tree trunk voltages
Substitute
with node voltages (to a given reference),
we get the nodal analysis equations.
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Nodal Analysis (II)
1
AK K v A e AKi S
1
i
T
1 1
0 1
A 0 1
0 0
0 0
y12
K i1
y13
y24
y34
y35
y 45
Kv I
y 46
y56
y12 y13
y
13
1
T
AKi K v A e y13
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y13
y13 y24
y13
23
y13
y13
y13 y34 y35
y35
0
1
0
0
0
1
3
5
2
4
6
0
0
1
0
0
0
0
1
1
0
y35
y35 y45 y56
y56
0
0
0
1
0
0 0
0 0
0 0
0 1
1 1
v12
v
24
v34
y56 v45
y46 y56 v46
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Modified Nodal Analysis
General Form
Independent
current source
Node Conductance
matrix
KCL
Yn
Y AT
2 2
A2 Vn J 2
Z 2 I 2 W2
Due to non-conductive
elements
Independent
voltage source
Yn can be easily derived
Add extra rows/columns for each non-conductive elements
using templates
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1
3
2
4
5
MNA (II)
Fill Yn matrix according to incidence matrix
n1
A
n2
n3
n4
n5
n6
i12 i13 i24 i34 i35 i45 i46 i56
1 1 0 0 0 0 0 0
1 0 1 0 0 0 0 0
0 1 0 1 1 0 0 0
0
0
0
1
0
1
1
0
0 0 0 0 1 1 0 1
0 0 0 0 0 0 1 1
6
Yn AYe A
T
Choose n6 as reference node
y12 y13
y
12
Yn y13
0
0
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y12
y12 y 24
0
y 24
0
y13
0
y13 y34 y35
y34
y35
25
0
y 24
y34
y 24 y34 y 45 y 46
y 45
0
0
y35
y 45
y35 y 45 y56
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MNA Templates
Add to the right-hand
side of the equation
j
Independent
current source
j i j
j' ij
ij
j'
j
Independent
voltage source
vj
ij
vg
j
j'
m 1 1
v j'
ij
1
1
1
v g
j'
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MNA Templates (II)
vj
CCVS
j
ij
CCCS
k'
j
k
j'
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vk
vk '
ij
ik
1
j'
1
k
1
k'
m 1 1 1
m 2
1 1 r
ri j
j'
ij
j
k
ik
v j'
1
N N N c N c i j
N
N
N c
i j
N c
m 1
k'
27
1 1
1
1
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MNA Templates (III)
VCVS
j
v j v j ' vk vk ' ik
1
j
j'
1
k
k'
m 1 1 1
k
+
vj
ik
v j
-
VCCS
j'
k
j
k
vj
+
vj
k
k'
g mv j
-
j'
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gm
g
m
v j'
gm
g m
k'
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MNA Templates (IV)
Mutual inductance
j
i1
M
v j v j ' vk vk '
i2
j
1
1
j '
k
'
k
m 1 1 1
jL1
m 2
1 1 jM
k
+
+
vj
vk
-
L1
L2
j'
k
-
'
Operational Amplifier
j
j'
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i
i1
vj
v j'
m 1 1 1
j
j'
k
k'
k
k'
29
vk
vk '
i2
1
1
jM
jL2
i
1
1
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MNA Example
4
C4
G2
n1
G3
n3
n2
C5
vg
n4
+
n1
v7 = v6
v6
n2
2
1
_
n3
3
5
ig
n0
6
n4
7 i vcvs
n0
Circuit Topology
MNA
Equations
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G2
G
2
0
0
1
0
G2
0
0
G2 G3 jC4
G3
jC 4
G3
G3 jC5
0
jC4
0
jC 4
0
0
0
0
1
30
1 0 vn1 0
0 0 vn 2 0
0 0 v n 3 0
0 1 vn 4 0
0 0 i g v g
0 0 ivcvs 0
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MNA Summary
Advantages
Covers any circuits
Can be assembled directly from input
data. Matrix form is close to Yn
Disadvantages
We may have zeros on the main diagonal.
Principle minors could be singular
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