PRINCIPLES OF CIRCUIT SIMULATION
Lecture 7. Element Stamping Guoy ong Shi , PhD
[email protected] School of Microele ctron ics Shanghai Ji ao Tong Universit y Sprin g 2010
2010-9-27
1 Slide
Outline • Basi c Con cepts – KVL/KCL – Cir cui t Ele ment E quation s
• Sparse Ta bu lar A nalys is (STA) • Nodal Analysi s • Mod if ied N od al An aly si s (MNA) • Part 1: Static Eleme nt Stampi ng
2010-9-27
Lecture 7
slide 2
Formulation of Circuit Equations • Kirchoff Curre nt La w (KCL) • Kirchof f Volta ge Law (KVL) • Circ uit Element Equa ti ons
2010-9-27
Lecture 7
slide 3
Basic Concepts Node num
Circuit Element
R3
1
G2v3
2
v3 +-
IS5
R1
R4 0
Branch
Reference/datum node
2010-9-27
Lecture 7
slide 4
Basic Physical Quantities Every circuit element (1 or 2 ports) is characterized by (i,v)
Node/Terminal Voltage e
KCL for nodes KVL for loops
R3
1
G2v3
equations
2
v3 +-
R1
IS5 R4 0 Reference node: e = 0
Branch current i Branch voltage v
2010-9-27
Lecture 7
slide 5
Circuit Element Equations •
Mathema ti cal models of c ir cui t co mpo nents a re expr ess ed in terms o f i deal ele ments: – Inductors – Capacitors – Resistors – Curr ent Sour ces – Voltage S ou rc es – Two Ports – ……….
•
Physical qua ntities – curre nt, volta ge
•
Some ti mes, we ne ed t o u se quantit ies: c harge ( non li near capacit or), flux (nonl inea r induc tor)
2010-9-27
Lecture 7
slide 6
Reference Directions Two-terminal
+ V
Two-port
i + V1 -
-
i1
i2
i1
i2
+ V2 -
• i and v are branch currents and voltages, respectively • (Default) For each branch, current is directed from higher potential to lower potential
2010-9-27
Lecture 7
slide 7
Resistor Resistors
Symbol
Linear
i
+
V
Nonlinear
V
(1/R) =i v
Current controlled
R i=v
i
+
2010-9-27
Voltage controlled
(v) = ii
v= (i)
-
Lecture 7
slide 8
Capacitor Capacitor
Symbol
Voltagecontrolled
Linear
i
v C= q i = dq / dt Time-invariant C:
+
V
Nonlinear
i
i = C dv /dt (v) q=q iTime-invariant = dq / dt C:
+
2010-9-27
V
-
Lecture 7
i = C(v) dv/dt
slide 9
Two-Port Elements Controlled Sources
Symbol
linear
Nonlinear
v k = Ek v c
v k = v k (v c )
ic = 0
ic = 0
Ik
i k = Fk i c
i k = i k (i c )
+ Vk
vc = 0
vc = 0
VCVS
+
Ic
Vc CCCS
Ek
Ic + Vc -
2010-9-27
Ik +
Fk
+ -
Vk -
-
Lecture 7
slide 10
Topological Equations KCL (branch c urrents) Current l eavi ng a node is " +"
1
1
i1
i2 i3 4
2
i1-i2-i3 = 0
3
KVL (nod al vol tage s) Voltage dropp ing is " +"
e1
e2
1
2
+
v1
v 1 + e2 – e1 = 0
-
GND 2010-9-27
Lecture 7
slide 11
Matrix Forms 1
i1 + v1 -
i2 G2v3
2
+
1 1 0 0
2
v3 -
+
e=0
i
i4 R4
1
0
1 1
+
IS5
v5 0
Nodal vol tage
branch vo lt age vector
i1 0 1
i2 i3 i4
0 0
i5
KCL: A i = 0
v1
1
0
v2
1
0
v3
1
1
v4 v5
0 0
1 1
KVL
Telle gen’s equa tion i Tv = 0 (conservation of ene rgy ) 2010-9-27
i5
v4 -
R1
A 1
R3
i3
Lecture 7
0 e1 e2
0 0 0 0
v – A Te = 0 slide 12
Incidence Matrix A branch 1 2 3….. j… e d o n
1 2 3 . i :
A ij =
(+1,-1,0)
+1; if node i is + terminal of branch j -1; if node i is – termina l of branch j 0; if node i is not connecte d to the branch j
A
i1 i2 + Each branch i s d irecte
Properties • • •
A is unimodular (all minors equal to 1, -1, or 0) Only 2 non zero entri es in any c olu mn Sum of all ro ws o f A is a z ero v ecto r. Thus , pick a node a s t he re fere nce (ground ) node
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Lecture 7
slide 13
Equation Assembly •
How do es a com pu ter a ss emb le equ ati on s fro m the cir cui t de scr ipt ion ( netli st) ?
•
Two s ys temati c me tho ds : 1. Sparse Table au A nalysis (STA) Used b y earl y ASTAP si mu lator (IBM) 2. Mod if ied Nod al An aly si s (MNA) Used b y SPICE sim ul ato rs
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Lecture 7
slide 14
Sparse Tableau Analysis (STA) Pro pos ed by (Br ayt on, G us tavson , Hach tel 1969-71 ) – Wri te KCL : Ai = 0T n equations (one for each node) – Wri te KVL : v – A e = 0 b equations (one for each branch) – Wri te Cir cui t Ele ment (Branch) E quatio ns :
K ii + K vv = S
b equations sources
current
voltage
controlled
controlled
2010-9-27
Lecture 7
slide 15
Sparse Tableau Analysis Put all (n + 2b) equations together:
A
0
0
l
K
K
i
v
0 AT
i
0
v
0
0
e
S
n + 2b unknowns
sparse tableau
n = #nodes b = #branches
2010-9-27
Lecture 7
slide 16
Advantages of STA • • •
STA can be a ppl ied to any (li neari zed) ci rc uit STA equa ti ons can be ass embl ed di rectly fro m netlis t STA c oeffi cient matrix is very sparse 2b
b
2b
⎛A 0 ⎜ ⎜0 I ⎜ K iK v ⎝ b
0 ⎞ ⎛ i⎞ ⎛ ⎞ 0 ⎟ − AT ⎟ ⎜v⎟ =⎜ ⎟ 0
⎜ ⎟ ⎜⎟ e0 S⎟⎠ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠
(2b+2b+b+b+b) non zero sp ars it y is
7b (n 2b )2
b
Caution: Sophi sti cated pr ogramming techni ques a nd d ata struct ures are re qui red for achi evi ng t he ti me and m emor y effic iency 2010-9-27
Lecture 7
slide 17
Modified Nodal Analysis (MNA) • A more compact formulation • In MNA, every eleme nt is in conductance form! • We’l l revi ew t he ste ps ho w MNA i s do ne. •
Intr od uc ed by McCall a, Nagel, Rohr er, Ruehl i, Ho (1975)
2010-9-27
Lecture 7
slide 18
Nodal Analysis i1 + v1
1
R3
i3
i2 G2v3
+
v3 -
R1
-
Step 1: Write KCL:
2
+ v4 e=0
i5 i4 R4
+ v5
IS5
0
i1 + i2 + i3 = 0 (node 1) -i3 + i4 – i5 = 0 (node 2)
Step 2: Subs tit ute bra nch equations t o rewrite KC L in br anch voltages:
1 v 1 G2*v 3 1 v 3 0 R1 R3 1 v 3 1 v 4 IS5 R3 R4 2010-9-27
Lecture 7
(1) (2) slide 19
Nodal Analysis Step 3 : Substit ute bra nch volt ages b y n odal vol tages (using KVL) :
1 1 R1e1 G2(e1 e2) R3 (e1 e2)0 1 1 (e1- e2) e2 IS5 R3 R4
Put i n matrix form
1 ⎡1 + G + − − G2 2 ⎢R R3 ⎢ 1
⎢⎢ ⎣
− 1 R3
(1) (2)
1⎤
R 3 ⎥ ⎛ e1 ⎞ ⎛ 0 ⎞ ⎥⎜ ⎟ = ⎜ ⎟ 1 + 1 ⎥ ⎝e2 ⎠ ⎝ I S 5 ⎠ R 4 R 3 ⎥⎦
Yn e = S 2010-9-27
Lecture 7
slide 20
Regularity in MNA Matrix R3 1
i1 + v1 -
i3
i2 G2v3
i5
2
+
v3 -
R1
+ i4 v4 R4 -
e=0
+ v5 -
•
Each element contr ibutes (in condu ctance form ) only to t he entries with row- column positions corr espond ing to the node numbers.
•
Such a regul ar f orm a is called a “ stamp ”
IS5
0
Stamping 2
1
⎡1 1 2
+ G+
2 ⎢R1 1 ⎢ − ⎢ R3 ⎣
1
− −G 2 R3
1⎤
R3 ⎥ 1 1 ⎥ + R 4 R 3 ⎥⎦ Coefficient matrix
2010-9-27
Lecture 7
slide 21
Resistor Stamp SPICE Netl is t Form at (R) Rk N+ N- value_of_Rk N+
N-
N+ N+
1
Rk
Rk
N-
− 1 Rk
N2010-9-27
Lecture 7
−
1
Rk
1
Rk
slide 22
VCCS Stamp SPICE Netl is t For mat (VCCS) Gk N+ N- NC+
NC-
value_of_Gk
Similar to a resistor; but note tha t the row/ col indices are different. Nc+
Vc+
Vc-
+ Vc -
N+
N+
Gk
N-
N-
−G k
2010-9-27
Nc-
Lecture 7
−G k
Gk
slide 23
Current Source Stamp SPICE Netli st For mat ( Curr ent Sou rc e) ISK N + N- valu e_of _I k N+
Note the sig ns in this case!
⎛ ⎜ −I ⎜ k ⎜ ⎜ ⎜ +I k ⎝⎜
⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠⎟
+ N+
Ik N-
N-
Rig ht -Hand Side (RH S) 2010-9-27
Lecture 7
slide 24
Relation between STA and NA K i−1
⎛ Ki K− v ⎜0 I ⎜⎝ A 0
−A
⎛ I K−K i−1 ⎜ I ⎜0 ⎜A 0 ⎝
i0⎞⎛S⎞ ⎛ ⎞
T ⎟ ⎜⎟ − A⎟⎜ v = 0 0⎟⎜ 0 ⎠⎝⎟⎠ ⎜⎝ ⎟⎠e
v
⎛ I K−K i−1 v ⎜ I ⎜0 ⎜ 0 AK i−1Kv ⎝ 2010-9-27
−1 ⎞ i0⎛ ⎞ K⎛S ⎞ i ⎜ ⎟ T ⎟⎜ ⎟ A v 0 − = ⎟ ⎜⎟ ⎜ ⎟ ⎟ 0⎜ ⎟ ⎜⎝ e ⎟⎠ 0 ⎠ ⎝⎠ −1 ⎞⎛KS⎞ i ⎟⎜ ⎟ 0 ⎟⎜ v ⎟ = ⎟ ⎜⎝ e ⎟⎠ ⎜ − AK i−1S ⎟ 0 ⎠ ⎝ ⎠
⎞ i ⎛0 ⎟ ⎜T ⎟−⎜A
Lecture 7
slide 25
Relation between STA and NA − AK i−1Kv
⎛ I K−K i−1 ⎜ I ⎜0
v
⎞ i⎛0 ⎟ ⎜T ⎟−⎜A
⎜⎝ 0 AK i−1Kv ⎠⎟ ⎝⎜0
−1 ⎞⎛KS⎞ i ⎟⎜ ⎟ 0 ⎟⎜ v ⎟ = ⎠⎟⎝⎜ e ⎠⎟ − AK i−1S
Tabl eau Matri x
⎛ I KK− i−1 ⎜ I ⎜0 ⎜⎝ 0
0
v
−1 ⎞0i⎛ ⎞ KS ⎛ ⎞ i ⎟⎜ ⎟ − ⎟A⎜T ⎟ ⎜⎜ v ⎟⎟ = 0 AK i−1⎠⎟K⎝⎜ v A⎠⎟T ⎝⎜ e ⎟⎠ − AK i−1S
MNA Yn
After solving e, we get v, th en get i . 2010-9-27
Lecture 7
I s
Y n e = Is slide 26
Nodal Analysis -- Advantages & Problem • Advantages: – Circ uit equations can b e assemble d b y s tampin g – Yn is sparse ( but not as s parse a s STA) and small ( nxn), sm all er t han STA ( n + 2b*n + 2b ) – Yn has non-z ero d iagonal e ntr ies a nd i s of ten diagonally dominant
• Problem: Noda l A nalys is c annot handle the follow ing – Flo ati ng i ndepe ndent vol tage so urc e (not c onn ect ed to ground) – – – –
VCVS CCCS (VCCSok!) CCVS
2010-9-27
(E-ELEMENT) (F-ELEMENT) (G-ELEMENT) (H-ELEMENT) Lecture 7
slide 27
Modified Nodal Analysis (MNA)
R3
1
2 +
R1
ES6
G2v3
- +
3
v3 R4
+
R8
IS5 - +
0
2010-9-27
Lecture 7
E7v3
4
slide 28
Modified Nodal Analysis (MNA) R3
1 i1
i2 R1
i3
2
i6 i4
+ v3 -
G2v3
R4
0
ES6
3
- +
i
i5
R8
+ IS5 + E7v3
i7 4
Step 1: Wri te KCL i1 + i2 + i3 = 0 -i3 + i4 – i5 – i6 = 0 i6 + i 8 = 0
(1) (2)
i7 – i8 = 0
2010-9-27
(3) (4)
Lecture 7
slide 29
Modified Nodal Analysis (MNA) Step 2: Subs ti tut e branch c urr ents by br anch v olt ages
1 v 12G v 3 1 v 3 0 R1 R3 1 v 3 1 vi4 R3 R4
(2)
i 6 R18v 8 0
(3)
1 v8 0 R8
(4)
i7 2010-9-27
6 IS5
(1)
Lecture 7
slide 30
Modified Nodal Analysis (MNA) Step 3 : Write down unu sed branch equation s
v v
7
6
ES
(4)
0
(5)
6
Ev
7 3
R3
1
2
i6
i2 R1
+ v3 -
i4 R4
G2v3
0 2010-9-27
3
- +
i3 i1
ES6
Lecture 7
i5
i8 R8
+ IS5 + E7v3
i7 4 slide 31
Modified Nodal Analysis (MNA) Step 4: Substi tut e branch v ol tage s by n odal vol tage s
1
1 e1 G2(e1 e2) (e1 e2) R1 R3 1 1 e2 i 6 IS5 (e1 e2) R3 R4 1 i6 (e3 e 4) 0 R8 1 i7 (e3 e 4) 0 R8 (e3 e2) e 4 E 7(ee 1 2010-9-27
2)
0
(1) (2) (3) (4)
ES6
(5)
0
(6) Lecture 7
slide 32
Modified Nodal Analysis (MNA) branch branch
1 node-1
G2
R1
node-2
R3
node-3
0
branch-6 branch-7
G2
0
0
0
1
R3 R 4 0
0
1
E7
E7 node-2
⎛Yn ⎜C ⎝
0 1
0 1
R8 1
R8 1
R8 1 0
1 0 1
0
0
1
R8 0
0
0
1
0
0
node-3 node-4
2
0
R3 1
00
node-1
2010-9-27
1
R3 1
node-4
1
ES6
3
I6
e1
0
e2 e3
IS5 0
e4
0
i6
ES6
branch-6
i7
0
branch-7
nod e voltage s
B ⎞ ⎛e ⎞ = RHS ⎟ ⎜ ⎟ 0⎠ ⎝ ⎠ i some bra nch currents Lecture 7
slide 33
Voltage Source Stamp SPICE Netl is t For mat (Floatin g vol tage sou rc e) VK
N+
N-
value_of_Vk
current introduced!
N+ NN+ Nbranch k
i
0
0
k 1
0
0
-1
1
-1
0
RHS
0 0 Vk
i
N+ k
+
Vk
N-
2010-9-27
Lecture 7
slide 34
CCCS Stamp SPICE Netl is t For mat (CCCS) FK N+ N- Vname Vname NC+ NC- value
N+ NNC+ NC-
br Vc
N + N- NC + NC - i c Fk -Fk 1 -1 1 -1
value_of_FK
RHS
0 0 0 0
Vc
ik
NC+
ic
N+
Vc
NC-
k
ic
N-
* If ‘Vna me’ is u sed a s a CC for mul tip le tim es, it is st amped only once though! 2010-9-27
Lecture 7
slide 35
CCVS Stamp SPICE Netli st For mat (CCVS) HK
N+ N- Vname
value_of_HK
Vname NC+ NC- value
N+ N- NC + NC- i k N+ 1 N-1
1 -1
1
-1
RHS
NC+
N+ i
0 0 1 -1
NC+ NC-
br-k br-c
ic
-HK
ic
+ H -
Vc
0 0
Vc
Lecture 7
N-
NC-
* If ‘Vna me’ is u sed a s a CC for mult iple times, it is stamped once though! 2010-9-27
k
only slide 36
VCVS Stamp SPICE Netli st For mat (VCVS) EK N+ N- NC+ NC-
value_of_EK
N + N- NC + NC- i k N+ 1 N-1 NC + NCbr k 1 -1 -Ek Ek
NC+ i
Lecture 7
k
+ Ek -
Vj
NC-
2010-9-27
N+
N-
slide 37
General Rules for MNA • A branch current is introduced as an add it io nal v ari abl e fo r a vol tage sour ce or an
inductor • For c ur rent s our ces, resi st ors , co ndu ct anc e and capaci tor s, the br anc h c ur rent is introduc ed only if – Any circuit element depends on that branch curre nt; or – The branch cu rr ent i s r equested a s an o utp ut.
2010-9-27
Lecture 7
slide 38
Modified Nodal Analysis (MNA) Advantages of MNA • MNA c an b e app li ed t o any ci rc ui t • MNA equa ti ons can b e ass embl ed “ di rect ly ” fr om a ci rc ui t de sc ri pti on ( e.g. netl is t)
Problem • Som eti mes zero s app ear on th e main di ago nal; causi ng so me pri nci ple mino rs t o be sin gula r (num eri cal i nst abil it y.) 2010-9-27
Lecture 7
slide 39
Summary • KVL/ KCL + Cir cu it Eleme nt Equ ati on s • Equ ati on s fo rm ul ati on : STA and M NA • MNA w as i mpl emented in m os t s im ulators (SPICE) • Eleme nt s tamps • A key observation: – Circuit matrix s truct ure will not c hange ! (exploited by SPICE fo r s peedu p – sym boli c facto riza tion )
2010-9-27
Lecture 7
slide 40
Assignment 3 • Impl ement a ne tlis t parser tha t rea ds a simpl e netli st with the follow ing elements – R – Vsou rc e, Isou rc e – VCVS, CCCS, VCCS, CCVS
Pri nt t he st amp s and t he RHS wi th r ow and co lum n indices.
2010-9-27
Lecture 7
slide 41
PRINCIPLES OF CIRCUIT SIMULATION
Part 2. Dynamic Element Stamping
2010-9-27
Slide 42
Outline • Discre tiz ation Formula s for d/dt • Eleme nt Stamps fo r Li near Capaci to rs and Inductors
2010-9-27
Lecture 7
slide 43
Circuit with dynamic element RC
dVc
+ Vc = Vs , Vc (0) = 0
R Vs
dt
+ -
+ Vc
C -
Analytical solution t ⎛ − Vc (t ) = Vs ⎜1 − e ⎜ ⎝
τ
⎞ ⎟, ⎟
Vc (t ) τ
= RC
• How to so lve it numerica lly ? 2010-9-27
Lecture 7
Vs
t
slide 44
Numerical Solution dV + V = Vs , V (0) = 0 dt
Assuming
= RC = 1
• Replace the derivative by diffe rence V (t +h ) −V(t) h
+ V (t ) = Vs
h = tim e step ( small)
There re many ays to Thereaare ma nywways to do do ddiscretization isc retiza tion.. V (t +h =) V t+( h) V− [Vst 2010-9-27
( )] Lecture 7
Becomes itera ti on slide 45
Forward Euler (FE) tangent at t n-1
y(t ) h
.
y(t) =
dy(t) = f(y(t)) dt
tn−1
tn curr ent time
i y t( )n y−(t ) n −1 ≈ y t( n −)1 f=y( t( ))n −1 h
y t( )n (y=t ) 2010-9-27
1f n −h
(y t )i ( +
n −1
Lecture 7
)
Direct it eration ; (no linear / no nlin ear so lv e involved) slide 46
Backward Euler (BE) tangent at t n .
y(t) =
dy(t) = f(y(t)) dt
y(t ) h
tn−1
i y t( )n y−(t ) n −1 ≈ y t( n) =f y( t( ))n h
y t( )n( y=t )
1f n −h
+y(t) i (
tn curr ent time
n
)
y(tn) appears o n bo th s ides of t he equ ation; n eed to so lve y(tn) by
iterations if f( ·) is n onl inear . 2010-9-27
Lecture 7
slide 47
FE and BE For ward Euler:
y t( )n (y=t )
1f n −h
+(y t )i (
n −1
Backward Eul er:
y t( )n( y=t )
1f n −h
+y(t) i (
n
)
)
• Although B.E. requires more computation, it is pr eferr ed in pr act ic e because it is mo re num eri call y sta bl e. • Will dis cus s on t his l ater.
2010-9-27
Lecture 7
slide 48
Trapezoidal Rule (TR) dy (t ) yt () = fyt=( ()) dt i
tange nts at bot h t n-1 and t n
y(t )
y t( )n y−(t ) n −1 ≈ 1 ⎡y. t( ) y+(t . ) ⎤ ⎢ n n −1 ⎥ h 2⎣ ⎦ sl ope of seca nt
tn−1
tn
averaged tangent
y t( )n y−(t ) n −1 1 ≈ [f y( t( )) n f +y( t( h 2
yt( n) y=nt (
h
) −1 f + y(th( [))nf yt( ( n 2
))n −1
)) +
] −1
]
Again, requires solving y(tn) from a set of non linea r equa tion s 2010-9-27
Lecture 7
slide 49
FE, BE and TR .
y(t) =
dy(t) = f(y(t)) dt
For ward Euler:
y t( )n (y=t )
1f n −h
(y t )i ( +
n −1
Back ward Euler:
y t( )n( y=t )
1f n −h
+y(t) i (
n
Trapez oi dal Rul e: yt( n) y=nt (
h
) −1 f + y(t ( [))nf yt( ( n 2
)
)
)) +
−1
• Trapezoi dal r ule has even b ett er n umerica l pr op ert y th an B.E. 2010-9-27
Lecture 7
slide 50
Interpretation of Trapezoidal Rule Trapezoida l r ule come s f rom numerica l i ntegra tion – com puti ng the a rea unde r a curv e
i
area und er t he curv e
x t( ) =g (t )
tn
xt ( )n xt=( )
(d )∫ n −1g +
τ
h
τ
t n −1
xt( )n x≈t(n )
−1
h (nt ) n⎡⎣g(t ) g+ 2
tn−1
+
−1
Lecture 7
tn
⎤⎦ appr oxi mate d by a trape zoid al
i x t( )n x−(t ) n −1 1⎡i ⎤ ≈ ⎢x t( )n + x(t ) n −1 ⎥ 2⎣ h ⎦
2010-9-27
g(t )
Trapezoidal Rul e slide 51
Stamps of Dynamic Elements • Stamps for dyna mic elements can be deriv ed from a discre tiz ation method: – Capaci to r (C) – Induc tor (L) – Oth ers ...
2010-9-27
Lecture 7
slide 52
Capacitor Stamp dv (t ) i (t ) = C dt
C
i N+
N
+ v(t ) −
discre tiz ed by B. E.
it()
C C = v()t [ (v−t h− ) = (v) t −] ( vt− )h h h
C h MNA Stamp
NA Stamp
N+
N−
N+
N−
RHS
C h C − h
−C h
C v(t − h) h C − v(t − h) h
2010-9-27
C h
trea ted a s so urc e
N+ N−
N+ N− branch C Lecture 7
C v t() ( i−t) h
C h
−
C h
C v(=t h ) h
i
RHS
1 −1
0 0
−1
C v(t − h h
−
slide 53
Inductor Stamp
di
N+
i t ( i)t−h ( − )
vt( )L= dtL ≈ L L vt( ) − it ( =−) it −h ( h h
L
i
discre tiz ed by B. E.
+
h
−
i
RHS
N+
1
0
N−
−1
0
N+ N−
)
v(t )
N−
branch L
1 − 1 − L − L i(t − h) h h
(must i ntro duce curr ent) MNA Stamp 2010-9-27
Lecture 7
slide 54
Stamps for C & L NA Sta mp for C
N+
N−
MNA Stamp f or C
N+
N−
RHS
C h C − h
−C h
C v(t − h) h C − v(t − h) h
C h
N+ N−
N+ N−
C h
−
C h
i
RHS
1
0
−1
0
−1
C v(t − h) h
MNA Stamp f or L
i
RHS
N+
1
0
N−
−1
0
N+ N−
1 −1 − 2010-9-27
L h
Note that: Stamps for C or L depend on the dis cr eti zati on method used!
L h
− i(t − h) Lecture 7
slide 55
A Circuit Example R
1
Vin
L
iV
C i ct( ) = v [t vC ( t) − C ( − 1)] h C C = [v 2t(−0v) t] [=−( ) 2 v−t − 0 −]v (t h h
2
iL
iC
C
0
v L (t ) = L
di L (t ) dt
L v1t( )v− t 2 (= ) i t − [i Lt(− ) h
L
(
1)]
0
1
0
C h
0
1
0
1 R
2 −
C h
branch L branch V
2010-9-27
in
1
−
−
C h
−
1 R
C ⎞ ⎛ 1 ⎜R + h ⎟ ⎝ ⎠
R
3
0
1
4
-1
1
Lecture 7
2
-1
iL
iV
3
4
0
-1
1
1
-1 L − h
1)
(
RHS −
C v C (t − h ) h
0 C h
v C (t − h ) L h
− i L (t − h )
Vin slide 56
Assignment 4 • Deri ve th e st amp s f or C and L u si ng th e Trapezoi dal Rul e (bo th in MNA).
2010-9-27
Lecture 7
slide 57
Classical Papers 1.
G.D. Hacht el, R.K. Bra yto n and F.G. Gust avso n, “ The sparse table au appr oach to networ k analy si s and desig n ,” IEEE Trans. Circuit Theory , vol.CT-18, Jan. 1971, pp. 101-119.
2.
C.W. Ho, A.E. Ruehl i and P.A. Br enn an, “ The mod if ied nodal appro ach to n etwor k ana lys is ” , IEEE Trans. Circ ui ts and Systems , CAS-22, June 1975, pp. 504-509
2010-9-27
Lecture 7
slide 58
