ECE 546
Lecture ‐16
MNA and SPICE
Spring 2016
Jose E. Schutt-Aine
Electrical & Computer Engineering
University of Illinois
jesa@illinois.edu
ECE 546 – Jose Schutt‐Aine
1
Nodal Analysis
The Node Voltage method consists in determining
potential differences between nodes and ground
(reference) using KCL
For Node 1:
V1  Vm V1 V1  V2


0
ZA
ZB
ZC
ECE 546 – Jose Schutt‐Aine
2
Nodal Analysis
For Node 2:
V2  V1 V2 V2  Vn


0
ZC
ZD
ZE
Rearranging the terms gives:
1
1 
 1
 1
 Z  Z  Z V1   Z
B
C 
 A
 C
 1

 ZC
Defining:

 1 
V2   Z Vm
 A

1
1 

 1
 1 
V1   Z  Z  Z V2    Z Vn
D
E 
 E

 C
GA 
1
1
1
1
1
, GB 
, GC 
, GD 
, GE 
ZA
ZB
ZC
ZD
ZE
ECE 546 – Jose Schutt‐Aine
3
Nodal Analysis
Rearranging the terms gives:
 GA  GB  GC 

GC

 V1   GAVm 
GC
   
 GC  GD  GE  V2   GEVm 
The system can be solved to yield V1 and V2.
ECE 546 – Jose Schutt‐Aine
4
Nodal Analysis
For Node 1:
For Node 2:
V1  50 V1  j1045 V1  V2


0
10
j5
2  j2
V2  V1
V2
V2

 0
2  j2 3  j4 5
ECE 546 – Jose Schutt‐Aine
5
Nodal Analysis
Rearranging the terms gives:
1 
50 1045
1 1
 1 


V

V


 10 j 5 2  j 2  1  2  j 2  2
j5
10




1
1
 1 
 1

V1  

 V2  0

 2  j2 
 2  j2 3  j4 5 
ECE 546 – Jose Schutt‐Aine
6
Nodal Analysis
 1 1 1 

1
 500 





 5 j 2 4 
 V
 5 


4




1







1
1   V2   5090 
1
1

 
  

 5 
4
 4  j 2 2 

[Y]
[v]=[i]
ECE 546 – Jose Schutt‐Aine
7
Nodal Analysis - Solution
10
0.25
j 25 0.75  j 0.5
13.556.3
V1 

 24.772.25 V
0.45  j 0.5
0.25
0.546  15.95
0.25
0.75  j 0.5
0.45  j 0.5 10
0.25
j 25
18.3537.8
V1 

 33.653.75 V
0.45  j 0.5
0.25
0.546  15.95
0.25
0.75  j 0.5
ECE 546 – Jose Schutt‐Aine
8
Nodal Formulation
V1  V2
0
Node 1: 3 
5
V2  V1 V2 V2  V3
 
0
Node 2:
5
10
5
V3  V2 V3
 0
Node 3:
5
10
ECE 546 – Jose Schutt‐Aine
9
Nodal Solution
Arrange in matrix form: [G][V]=[I]
0  V1   3
 0.2 0.2
 0.2 0.5 0.2  V   0 

 2  
0.2 0.3  V3  0 
 0
Use Gaussian elimination to form an upper triangular matrix
0  V1  3
0.2 0.2
 0
0.3 0.2  V2   3

   
0
0.25  V3  3
 0
Solve for V1, V2 and V3 using backward substitution
This can always be solved no matter how large the matrix is
ECE 546 – Jose Schutt‐Aine
10
Why SPICE ?
 Established platform
 Powerful engine
 Source code available for free
 Extensive libraries of devices
 New device installation procedure easy
ECE 546 – Jose Schutt‐Aine
11
SPICE
From
Netlist
Parser
Device
Stamp
To
Solver
I=YV
SPICE Directory Structure
spice3f4
conf
examples
helpdir
scripts
lib
man
man1
man3
ckt
cp
asrc
bjt bsim1 bsim2 cap
dev
fte
hlp
patches
notes
man5
inp
mfb
src
tmp
util
bin include lib unsuppo
mfbpc
misc
ni
doc
lib skeleton
sparse mac
cccs ccvs csw dio disto ind isrc jfet ltra mes
mos1 mos2 mos3 mos6 res
sw
tra urc vccs vcvs vsrc
ECE 546 – Jose Schutt‐Aine
12
MOS SPICE Parameters
Symbol
Description
Value
Units
Ldrawn
Device length (drawn)
0.35
m
Leff
Device length (effective)
0.25
m
tox
Gate oxide thickness
70
A
Na
Density of acceptor ions in NFET channel
1.0  1017
cm-3
Nd
Density of donor ions in PFET channel
2.5  1017
cm-3
VTn
NFET threshold voltage
0.5
V
VTp
PFET threshold voltage
-0.5
V

Channel modulation parameter
0.1
V-1

Body effect parameter
0.3
V1/2
Vsat
Saturation velocity
1.7  105
m/s
n
Electron mobility
400
cm2/Vs
p
Hole mobility
100
cm2/Vs
kn
NFET process transconductance
200
A/V2
kp
PFET process transconductance
50
A/V2
Cox
Gate oxide capacitance per unit area
5
fF/m2
CGSO,CGDO
Gate source and drain overlap capacitance
0.1
fF/m
CJ
Junction capacitance
0.5
fF/m2
CJSW
Junction sidewall capacitance
0.2
fF/m
Rpoly
Gate sheet resistance
4
/square
Rdiff
Source and drain sheet resistance
4
/square
ECE 546 – Jose Schutt‐Aine
13
Problems
• Nonlinear Devices
Diodes, transistors cannot be simulated in the frequency domain
Capacitors and inductors are best described in the frequency domain
Use time‐domain representation for reactive elements (capacitors and inductors)
• Circuit Size
Matrix size becomes prohibitively large
ECE 546 – Jose Schutt‐Aine
14
Motivations
Zo
Zo
C
- Loads are nonlinear
- Need to model reactive elements in the time domain
- Generalize to nonlinear reactive elements
ECE 546 – Jose Schutt‐Aine
15
Time-Domain Model for Linear Capacitor
For linear capacitor C with voltage v and current i which must satisfy
dv
i  C
dt
Using the backward Euler scheme, we discretize time
and voltage variables and obtain at time t = nh
vn1  vn  hvn' 1
ECE 546 – Jose Schutt‐Aine
16
Time-Domain Model for Linear Capacitor
After substitution, we obtain
v 'n1 
in1
C
so that
in1
vn1  vn  h
C
The solution for the current at tn+1 is, therefore,
C
C
in1  vn1 - vn
h
h
ECE 546 – Jose Schutt‐Aine
17
17
Time-Domain Model for Linear Capacitor
i
C
v
Backward Euler companion model at t=nh
Trapezoidal companion model at t=nh
ECE 546 – Jose Schutt‐Aine
18
Time-Domain Model for Linear Capacitor
Step response comparisons
0.4
Vo (volts)
0.3
Exact
Backward Euler
Trapezoidal
0.2
0.1
0.0
0
10
20
30
40
50
60
Time (ns)
ECE 546 – Jose Schutt‐Aine
19
Time-Domain Model for Linear Inductor
di
vL
dt
Backward Euler : in1  in  hin1
in1 
vn1
L
L
L
vn1  in1  in
h
h
ECE 546 – Jose Schutt‐Aine
20
Time-Domain Model for Linear Inductor
If trapezoidal method is applied
in1  in 
h
in1  in 
2
2L
 2L

 vn 
vn1 
in1  
h
 h

ECE 546 – Jose Schutt‐Aine
21
The Diode
+
I
V
•
Diode Properties
– Two-terminal device that conducts current freely in one direction but blocks
current flow in the opposite direction.
– The two electrodes are the anode which must be connected to a positive voltage
with respect to the other terminal, the cathode in order for current to flow.
ECE 546 – Jose Schutt‐Aine
22
Diode Circuits
V V
out
D


I D  I S eVD / VT  1
VS  RI D  VD  RI D (VD )  VD
Nonlinear transcendental system  Use graphical method
ID
Diode characteristics
Vs/R
Load line
(external characteristics)
Vout
VS
VD
Solution is found at itersection of load line characteristics
and diode characteristics
ECE 546 – Jose Schutt‐Aine
23
BJT Ebers-Moll Model
C
NPN Transistor
B
E
I 
iE   S  evBE / VT  1  I S evBC / VT  1
 F 




I 
iC  I S evBE / VT  1   S  evBC / VT  1
 R 




I 
I 
iB   S  evBE / VT  1   S  evBC / VT  1
 F 
 R 

F 
F
1F


R 

R
1R
Describes BJT operation in all of its possible modes
ECE 546 – Jose Schutt‐Aine
24
Newton Raphson Method
Problem: Wish to solve for f(x)=0
Use fixed point iteration method:
Define F ( x)  x  K ( x) f ( x)
 : xk 1  F ( x k )  x k  K ( x k ) f ( x k )
With Newton Raphson:
 df 
1

K ( x)  [ f ( x)]   
 dx 
1
therefore,  : x k 1  xk  [ f ( xk )]1 f ( xk )
ECE 546 – Jose Schutt‐Aine
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Newton Raphson Method
(Graphical Interpretation)
ECE 546 – Jose Schutt‐Aine
26
Newton Raphson Algorithm
N  R : xk 1  xk  Ak f  xk 
1
Ak xk 1  Ak xk  f ( xk )  S k .
xk+1 is the solution of a linear system of equations.
Ak x  Sk  LU fact
Forward and backward substitution.
Ak is the nodal matrix for Nk
Sk is the rhs source vector for Nk.
ECE 546 – Jose Schutt‐Aine
27
Newton Raphson Algorithm
current controlled


0. k  0, gives
V0
,
i0
1. Find Vk , ik compute companion mod els.
Gk , I k , Rk , Ek
 
voltage controlled
Vc
Cc
2. Obtain Ak , Sk .
3. Solve Ak xk  S k .
4. xk 1  Solution
5. Check for convergence xk 1  xk   .
If they converge, then stop.
6. k  1  k , and go to step 1.
ECE 546 – Jose Schutt‐Aine
28
Newton Raphson - Diode
It is obvious from the circuit that the solution must satisfy
f(V) = 0
We also have
1 I s V / Vt
f '(V )   e
R Vt
The Newton method relates the solution at the (k+1)th
step to the solution at the kth step by
f (Vk )
 Vk
Vk 1  f '(Vk )
Vk  E
 I s  eVk / Vt 1
Vk 1  Vk - R
1 I s Vk / Vt
 e
R Vt
ECE 546 – Jose Schutt‐Aine
29
Newton Raphson
After manipulation we obtain
E
1

  g k  Vk 1  - J k
R
R

I s Vk / Vt
gk 
e
Vt
J k  I s ( eVk / Vt -1) - Vk g k
ECE 546 – Jose Schutt‐Aine
30
Diode Circuit – Iterative Method
Newton-Raphson Method
Use:
x k 1  x k   f '( x k ) 
1
x ( k 1)  x ( k )   f '( x ( k ) ) 
f (VD ) 
1
f ( xk )
Vout  VD
f ( x(k ) )
VD  VS
 I S eVD / VT  1  0
R
1 I
f '(VD )   S eVD / VT
R VT


VD( k )  VS
( k 1)
D
V
V
(k )
D

R

V ( k ) / VT
 IS e
D

1
1 I S VD( k ) / VT
 e
R VT
VD( k )
Where is the value of VD at the kth iteration
Procedure is repeated until convergence to final (true) value of VD which is the
solution. Rate of convergence is quadratic.
ECE 546 – Jose Schutt‐Aine
31
Newton Raphson for Diode
Newton-Raphson representation of diode circuit at kth iteration
gk 
I s Vk / Vt
e
Vt
J k  I s ( eVk / Vt -1) - Vk g k
ECE 546 – Jose Schutt‐Aine
32
Current Controlled
Companion
dh(i )
Rk 
di i ik
Ek  h(ik )  Rk ik
ECE 546 – Jose Schutt‐Aine
33
General Network
Let x = vector variables in the network to be solved for. Let f(x) =
0 be the network equations. Let xk be the present iterate, and
define
Ak  f ( xk )  Jacobian of f at x  xk
Let Nk be the linear network where each non-linear resistor is
replaced by its companion model computed from xk.
I j  g j (V j )
ECE 546 – Jose Schutt‐Aine
34
General Network
V jk  Pj k  Pj k
Companion
model
I k  g Vk   GkVk
dg (V )
Gk 
dV V Vk
ECE 546 – Jose Schutt‐Aine
35
Nonlinear Reactive Elements
in+1
in+1
+
V
-
+
+
q(V)
h
C(v)  Vn+1
Jn  Vn+1 gk
Jk
Jn
-
-
dq
q  f (v ) , i 
dt
dq
qn1  qn  h
dt t tn1
qn1 qn
f (vn1 )
or , in1 

 in1 (vn1 ) 
h
h
h
ECE 546 – Jose Schutt‐Aine
36
General Element
ECE 546 – Jose Schutt‐Aine
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