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ECE 3144 Lecture 12
Dr. Rose Q. Hu
Electrical and Computer Engineering Department
Mississippi State University
1
Review for Chapter 1
• Charge (q), current (i), and their relationship
dq (t )
i (t ) 
dt
t
t

0
q(t )   i( )d   i( )d  q(0)
t2
•
Q   i(t )dt
t1
Voltage (v), work/energy (w), power (p), and their relationships.
dw dw dq
p

 v *i
dt dq dt
dw
v 
dq
t2
W   p(t )dt
t1
• Steady state voltage V (voltage is constant), steady state current I
(current is constant) and steady state power P (power is constant).
P V *I
W  P * t
2
Review for Chapter 1
• Passive sign convention
– The positive reference of voltage v(t) is at the same terminal the the
current variable i(t) is entering.
– How to determine absorbing or supplying energy?
• Active elements and passive elements
– Active elements
 Independent sources
 Independent voltage sources and independent current sources
 Dependent sources
 Voltage controlled voltage source
 Current controlled voltage source
 Voltage controlled current source
 Current controlled current source
– Passive element: resistor
3
Review for Chapter 2
Resistor, Resistance, Resistivity: R = *L/A
R=1/G
Ohms’law v(t) = i(t)*R: only linear resistors satisfy Ohm’s law.
Power p(t) = v(t)*i(t)= v(t)2/R = i(t)2R
Kirchoff’s current law (KCL): The algebraic sum of the currents
leaving (entering) a node is zero.
• Kirchoff’s voltage law: The algebraic sum of the voltages around any
loop path is zero.
• Single loop circuit: resistors in series => voltage divider
•
•
•
•
N
Rs   R j ,
j 1
Ri
vi 
vs
N
 Rj
j 1
,
vi
Ri

vj
Rj
• Voltage sources in series
n
v(t )   vi (t )
i 1
• Single node circuit: resistors in parallel =>current divider
N
Gn  i
im Gm R n
in 
G
  Gn


parallel n1
N
,
,
in Gn Rm
 Gn
n1
4
Review for Chapter 2
• Current sources in parallel
n
i(t )   ik (t )
k 1
• Circuits containing a single source and a series-parallel
interconnection of resistors
• How to perform series-parallel combinations between two
terminals.
• Wye-delta or Delta-to-Wye transformations
• Circuits containing dependent sources: not covered in the
exam 1.
5
Reminder from Lecture 11
•Nodal analysis case 1: with independent current sources and resistors only
GV = I. G matrix is symmetric.
In general, KCL is applied to node j with node voltage vj. The coefficient of vj,
which is the element gjj of G matrix, is the sum of all the conductances connected to
node j. The coefficient of any other node voltage, say i (ij), is the negative of the sum
of the conductances connected directly between node j and node i. The right hand side
of the equation is equal to the sum of the currents entering the node j via independent
current resources.
•Nodal analysis case 2: with dependent current sources
GV = I
However G matrix may not be symmetric anymore
Treat the dependent sources as independent sources first.
Derive the KCL equations at each node.
Replace the dependent sources by the control formula given.
Solve the equations for nodal voltages.
•Nodal analysis case 3: with independent voltage sources
Scenario 1:The independent voltage source is connected to the reference node.
Any time an independent voltage source is connected between an reference node
and a nonreference node, the voltage for the nonreference node is known.
Scenario 2: will be introduced in this lecture
6
Reminder: nodal analysis with independent voltage sources scenario 1
-independent voltage sources are connected to the reference node
•Any time an independent voltage source is connected between a reference node and a
nonreference node, the voltage for the nonreference node is known.
•So the presence of voltages sources in this case will make the voltage equations simpler.
Say the given network has N nodes. If k independent voltages sources are connected
between a reference node and a nonreference node, then the number of linearly independent
equations needed to solve the node voltages is reduced from (N-1) to (N-1-k).
9k
-
12k
V3
6k
+
+ 12 V
V2
-
12k
V1
6V
•For the circuit given, immediately we know
V1 = 12 V and V2 = -6V.
•The only unknown voltage variable is V3.
• Applying KCL at node 2
V1  V2 V3  V2 V2


0
12k
12 K
6K
12  V2  6  V2 V2


0
12k
12K
6K
V2 =1.5 V
7
Scenario 2: independent voltage source are connected between the
nonreference nodes
V1
6V
+
-
V2
12k
6 mA
6k
•If we follow the nodal analysis brute force
manner, we will have a problem: the current
4 mA on the branch where independent voltage
source is can be not calculated, which means
KCL equations at node 1 and node 2 can not
be established.
•A method called supernode technique is introduced.
First circle the voltage source and the two connecting nonreference nodes to form
a supernode.
Write the equations that defines the voltage relationships between the two nonreference
nodes as a result of the presence of the voltage source.
(1)
V1 – V2 = 6
Write the KCL equation for the super node.
V
V
(2)
 6mA  1  2  4mA  0
6k 12k
Then we have two equations (1)& (2) and two unknowns => V1 = 10 V, V2 = 4 V
8
Example 1: circuits with independent voltage sources
connected between nonreference nodes
0
+
2 k
2 k
Vo
2 mA
1
Problem 3.28: Using nodal analysis to find Vo for
the given network

4 k
12 V
2 k
V1  V 2  12
At node 0:
Vo  V1
V
 o  2 mA  0
2 k
2 k
2Vo  V1  4
2
At super node:
(1)

(2)
V1  Vo
V
V
 1  2  2 mA  0
2 k 4 k 2 k
2Vo  3V1  2V2  8
(3)
Put equations (1), (2) and (3) in matrix form:
0
2

 2
1 1 Vo  12 
1 0  V1    4 
3
2  V2   8

V0   4.5 
 V    5.0 
 1 

V 2   7.0
9

Nodal analysis case 4: circuits containing dependent
voltages sources
• Circuits containing dependent voltages sources are treated the same
way as the independent voltage sources except that the voltage
controlling equations for the dependent voltage sources should be used
instead of the given independent voltage source values.
– Scenario 1: if a dependent voltage source is connected between a
reference node and a nonreference node, the controlling equation for
voltage at the nonreference node is immediately known.
– Scenario 2: a dependent voltage source is connected between a reference
node and a nonreference node
 First circle the dependent voltage source and the two connecting
nonreference nodes to form the supernode.
 Write the equation that defines the voltage relationship between the
nonreference nodes as a result of the presence of the dependent
voltage source.
 Write the KCL equation for the supernode and the rest normal nodes.
 Write the controlling equation for the dependent voltage source
10
Case 4 example : nodal analysis with dependent voltage sources
connected between the reference node and a nonreference node
1
0
Ix
+
10 k
4000 Ix
10 k
Problem 3.37: Find V0 in the circuit given
Vo
10 k
4 mA

The dependent voltage source is connected to the reference node

V1  4000I x  0
At node 1:
V1  4000I x
At node 0:
Vo  V1
Vo
 4 mA 
 0
10 k
10 k
(1)

2Vo  V1  40
(2)
Write the controlling equation for the dependent voltage source
Ix 
Vo  V1
10 k

Vo  V1  10000I x  0
(3)
Put equations (1),(2) and (3) in matrix form
4000  Vo   0 
0 1
 2 1
 V    40
0

 1  
1 1 10000  I x   0 

Vo   15V 
V     10V 
 1 

 I x  2.5mA
11
Case 4 example: nodal analysis with dependent voltage
sources connected between two nonreference nodes
2kIx
Problem 3.35: Find Vo in the circuit network given.
1 k
2
1
1 k
0
+
1 k
12 V
1 k
Ix
Vo

The dependent voltage source connected to two
nonreference nodes 0 and 2. Thus node 0 and
node 2 form a supernode.

Vo  V2  2kI x
Vo  V2  2kI x  0
(1)
A voltage source is connected between the reference node and node 1.
V1  12 V
Apply KCL to the supernode:
V  V1
Vo
V2  V1
V2

 o

0
1 k
1 k
1 k
1 k

Vo  V1  V2  0
(2)
The controlling function for the voltage source is:
Ix 
V2
1 k

V2  1000 I x  0
(3)
12
Example 3: cont’d
Put the three equations (1), (2) and (3) into matrix format:
1
0

1

0

0
1
1
1
0
1
0
1
2000  Vo   0 
V  12 
0 
 1   
0  V2   0 
   
1000  
Ix 
 0
Vo   9.0V 
V  12.0V 
 1  

V2   3.0V 
  

I
3
mA
 x 

13
Homework for lecture 12
• Problem 3.31, 3.32, 3.34, 3.36, 3.39
• Due Feb 11
14