Transcript Networks I for M.E. ECE 09.201

Networks I for M.E.
ECE 09.201 - 2
James K. Beard, Ph.D.
Rowan Hall 238A
[email protected]
http://rowan.jkbeard.com
September 5, 2006
Passive Sign Convention



Always mark one terminal with
a + sign
Voltage is positive measured
from the other terminal
Current is positive going into
the terminal marked with the +
sign
September 5, 2006
Networks I for M.E.


iR
Slide 2
Source Sign Convention



Voltage sources marked
with + and – signs inside
a circle or diamond
Current sources marked
with an arrow inside a
square, circle or diamond
Positive






Current out of a voltage
source
Voltage out of a current
source
September 5, 2006
Networks I for M.E.
Slide 3
General Methodology

Write the loop equations










Pick small or simple loops
Make all of them one direction – clockwise or counterclockwise
Make sure that every trace is covered once
Use node voltage equations
Leverage supernodes – treat as single nodes
Write the node equations -- Currents are positive into each node
Write the Ohm’s Law equations
Rearrange the equations to allow vector-matrix notation
Check your work
Use computer to solve the matrix equation



TI-89 for smaller problems
Matlab or Mathcad for matrices larger than 4 X 4 or so
Note that free linear algebra packages are available for most HLLs
September 5, 2006
Networks I for M.E.
Slide 4
Approach Used Here for 4.4-7




Use voltage node notation
Leverage supernodes
Write Ohm’s Law equations first
Knowns are




Resistances
Source voltages and currents
Controlled source multipliers
Unknowns are



Node voltages
Currents through voltage sources
Voltages across current sources
September 5, 2006
Networks I for M.E.
Slide 5
Technique for Loop Equations





Add voltage drops around the loop
The sum of voltage drops around a closed loop must be
zero
Voltage drop is voltage at the present node minus
voltage on the other terminal of the resistor or source
Drop is positive through a resistor when the loop goes
into the “+” terminal and out the “-” terminal and its
current is positive
Look at the equation with and without Ohm’s Law


Supernodes will identify themselves
Node voltages instead of currents often gives simpler equations
September 5, 2006
Networks I for M.E.
Slide 6
Voltage Drop
Direction of loop
VIN
VOUT
VDROP  VIN  VOUT
September 5, 2006
Networks I for M.E.
Slide 7
Technique for Node Equations



Current into a node through each resistor is the
voltage on the other side of each resistor divided
by the resistance
Subtract the voltage at that node times the sum
of the reciprocals of all the resistors connected
to the node
Add currents through sources
 Current
sources connected to the node
 Currents through voltage sources connected to the
node

Use Ohm’s Law to pose the currents through
resistors in terms of the node voltages
September 5, 2006
Networks I for M.E.
Slide 8
Matrix Notation Is
A way of organizing several linear
equations.
 Nothing is changed from the original
equations.
 TI-89, Matlab, and Mathcad can solve the
problem for you from there.

September 5, 2006
Networks I for M.E.
Slide 9
Problem 4.4-7
iA


1
ivv
RA


iiv
 2
VS  10 v

2
RB
 
Kiv  iB
3
Kvv VA

iB

 3
RD

iD 1

 V
S
iC
K iv  8 
K vv  3
R A  2
RB  8
RD  4 
Waaaay too messy for a quiz!
September 5, 2006
Networks I for M.E.
Slide 10
Ohm’s Law
V1  VS
iA 
RA
V3  VS
iB 
RB
V3
iD 
RD
Use to make node voltages the unknowns
September 5, 2006
Networks I for M.E.
Slide 11
Loop Equations
Loop 1:  0   V    V
S
2
 V3   V3  0   0
V2  VS (Supernode with ground)
Loop 2: K IV  iB  V3   VS      VS   V1   0
K IV
 V3  VS   V3  V1  0
RB
Loop 3:
 0  V3   K IV  iB  KVV VA  0
K IV
V3 
 V3  VS   KVV  V1  VS   0
RB
September 5, 2006
Networks I for M.E.
Slide 12
Node Equations
V1  VS
Node 1: ivv  iiv 
0
RA
V1  VS V3  VS
Node 2: 

 iC  0
RA
RB
V3 V3  VS
Node 3: iiv 

0
RD
RB
September 5, 2006
Networks I for M.E.
Slide 13
Rearranging into Matrix Format
V1
Loop 2
Loop 3
Node 1
Node 2
Node 3

 1


 K vv

 1

 RA
 1

 RA

 0

September 5, 2006
V3
iiv
K iv
1
RB
0
K iv
1 
RB
0
0
1
1
RB
0
1
1


RB RD
0
ivv
iC
K iv




V
S



R
B
0 0


  K iv



 K vv  Vs 
 V1   
0 0      RB


V

  3 
VS

  ivv   
1 0   
RA


  iiv   

  iC     1  1  VS 
0 -1
  RA RB  



VS



1 0 


R

B


Networks I for M.E.
Slide 14
Solution
VA  5v
VB  2.5v
VD  12.5v
iA  2.5a
iB  0.3125a iD  3.125a
V1  15v
V2  10v
V3  12.5v
ivv  5.9375a iiv  3.4375a iC  2.8125a
September 5, 2006
Networks I for M.E.
Slide 15
Problem 4.4-7 Solution
13
2 16
a
10v
2 12 a
2
15v
15
5 16
a

15v
September 5, 2006
5
16
8
 2 12 v
a
12 12 v
3 167 a
4
Networks I for M.E.

10v

3 18 a
Slide 16
Intuitive Approach

From Loop 2 plus Loop 3:
KVV VA  VA  VS

From Loop 2
 Kiv 
VA   1 
 VB
RB 

 The rest is found from Ohm’s Law
Good for many homework problems, but not for most real-world problems
September 5, 2006
Networks I for M.E.
Slide 17
Homework Problem

Put problem 4.4-7 in Matlab and
solve it with the matrix method.
 Reproduce
the results given here.
 Save your code, print it out, and turn it
in with the homework.



Use your program to solve problem
4.4-7 with the parameters to the
right
Why can’t Kvv be 1.0?
What is special about the circuit that
allows simple intuitive solutions?
September 5, 2006
Networks I for M.E.
VS  15 v
K iv  10 
K vv  4
RA  10
RB  20
RD  5
Slide 18