NODE ANALYSIS

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Transcript NODE ANALYSIS

NODAL AND LOOP ANALYSIS TECHNIQUES
LEARNING GOALS
NODAL ANALYSIS
LOOP ANALYSIS
Develop systematic techniques to determine all the voltages
and currents in a circuit
NODE ANALYSIS
• One of the systematic ways to
determine every voltage and
current in a circuit
The variables used to describe the circuit will be “Node Voltages”
-- The voltages of each node with respect to a pre-selected
reference node
LEARNING EXAMPLE
V1 V1  V2
@ V1 :  4mA  
0
6k 12k
@ V2 : 2mA 
V2 V2  V1

0
6k
12k
BY “INSPECTION”
1 
1
 1

V

V2  4mA

 1
12k
 6k 12k 
1
1 
1
 V1   
V2  2mA
12k
 6k 12k 
USING KCL
LEARNING EXAMPLE
SUPERNODE
V1  6V
V4  4V
SOURCES CONNECTED TO THE
REFERENCE
CONSTRAINT EQUATION
V3  V2  12V
KCL @ SUPERNODE
V2  6 V2 V3 V3  (4)
 

 0 * / 2k
2k
1k 2k
2k
V2 IS NOT NEEDED FOR I O 3V2  2V3  2V
 V2  V3  12V * / 3 and add
5V3  38V
V
OHM'S LAW I O  3  3.8mA
2k
CIRCUITS WITH DEPENDENT SOURCES
PRESENT NO SIGNIFICANT ADDITIONAL
COMPLEXITY. THE DEPENDENT SOURCES
ARE TREATED AS REGULAR SOURCES
WE MUST ADD ONE EQUATION FOR EACH
CONTROLLING VARIABLE
LEARNING EXAMPLE
FIND IO
VOLTAGE SOURCE CONNECTED TO REFERENCE
V1  3V
V V V
[email protected] V2 : 2 1  2  2 I x  0
3k
6k
CONTROLLING VARIABLE IN
TERMS OF NODE VOLTAGES
Ix 
V2  V1 V2
V

 2 2  0 * / 6k
3k
6k
6k
V2  2V1  0  V2  6V
IO 
V1  V2
 1mA
3k
REPLACE
V2
6k
SUPER NODE WITH DEPENDENT SOURCE
VOLTAGE SOURCE CONNECTED TO REFERENCE
V3  6V
SUPERNODE CONSTRAINT
V1  V2  2Vx
CONTROLLING VARIABLE IN TERMS OF NODES
KCL AT SUPERNODE
Vx  V2  V1  3V2
* /12k
2(V1  6)  V1  2V2  V2  6  0
3V1  3V2  18  4V1  18
LEARNING EXAMPLE
FIND THE VOLTAGE Vo
@V4 : V4  4V
AT SUPER NODE
V1  V2  2VX
V2 V2  V3 V1  V3 V1  4V

2
mA




0
1k 
1k
1k
1k
1k
1k  @V :  2mA  V  V  V  V  0
3
3
2
1k
CONTROLLING VARIABLE
3
1
1k
VX  V2
SOLVE EQUATIONS NOW
V1  3VX
2V1  2VX  2V3  6V
V1  VX  2V3  2V
VARIABLE OF INTEREST
VO  V1  V3
LEARNING EXAMPLE
Find the current Io
@ V2 : V2  12V
@ V3 : V3  2VX
@ super node:
V4  V1  6V (constraint eq.)
FIND NODES – AND SUPER NODES
V  V3 V4  V5 V4
V1  V2 V1  V3

 2I X  4


1k
1k
1k
1k
1k
0
@V5 :  2 I X 
V5  V4 V5

0
1k
1k
CONTROLLING VARIABLES
VX  V1  V2
7 eqs in 7 variables
IX 
V4
1k
VARIABLE OF INTEREST
IO 
V5
1k