Transcript Chapter 3 - Nodal Analysis(PowerPoint Format)

Chapter 3 – Nodal Analysis
Read pages 65 - 80
Nodal Analysis: Nodal analysis is a systematic application of
KCL that generates a system of equations which can be solved
to find voltage at each node in a circuit. (We sum currents at
each node to find the node voltages.)
Homework:
•online HW, Nodal #1 and Nodal #2
•3FE-1 and 3FE-3
•Due 9/24/01
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Nodal Analysis Steps:
1. Label all nodes in the circuit,
2. Select one node as the reference node (also called common).
The voltage at every other other node in the circuit is
measured with respect to the reference node.
3. Write a KCL equation (i = 0) at each node.
4. Solve the resulting set of equations for the node voltages.
Branches connected to a node will have one of three types of
elements:
current sources (independent or dependent)
resistors
voltage sources (independent or dependent)
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Nodal Analysis – Branches With Curent Sources
Since we are applying KCL, current sources (either independent
or dependent) connected to a node provide terms for our KCL
equation that we can write down by inspection.
IS = IR1 + IR2 + IR3
The next step is to write each resistive current in terms of the
node voltages.
If a current source is dependent, we must also write the
dependent current in terms of the node voltages.
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Nodal Analysis – Resistive Branches
Consider a single resistor connected
between two arbitrary nodes:
A
+ VAB R
B
IAB
0V
By KVL, the voltage drop from node-A to node-B is the
difference between the voltage at node-A (VA0 = VA) and the
voltage at node-B (VB0 = VB) .
VAB VA  VB


R
R
The current leaving node-A going toward
node-B, IAB, is:
I AB
The current leaving node-B going toward
node-A is:
I BA 
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VBA VB  VA

R
R
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Example 1
If we apply the previous techniques to the resistors connected to
node-X in the following circuit and apply KCL at node-X, we get
the following equation. Note that the equation should have five
terms since there are five branches connected to node-X and each
branch will have a corresponding current
B
A
I2
X
R1
R2
0
C
R3
I1
VX  VA VX  VC VX  VE


 I 2  I1  0
R1
R2
R3
E
VX  VA VX  VC VX  VE


 I1  I 2  0
R1
R2
R3
D
currents leaving node-X
resistive branches
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currents entering node-X
current Sources
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Example 2
Use nodal analysis to find Ix.
12 k
Ix
4 mA
6 k
Step 1, Label nodes:
4 mA
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V1
2 mA
6 k
12 k
Ix
6 k
ENGR201 Nodal Analysis
V2
6 k
2 mA
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Example 2 - continued
Use nodal analysis to find Ix.
Step 2: Write KCL equations at each node (except reference node):
V1
4 mA
Ix
6 k
6 k
V1  V2 V1  0

 4mA
12k  6k 
V2  V1 V2  0

 2mA
12k  6k 
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V2
12 k

2 mA
 1 1
1
V1     V2    4V
12 6 
12 
1
 1 1
V1    V2     2V
12 
12 6 
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Example 2 - continued
V1
12 k
Ix
4 mA
Use nodal analysis to find Ix.
6 k
In matrix form:
V2
6 k
2 mA
 1 1  1 
 12  6   12  V   4 


 1    V
 1  1 1   V2   2 
  12  12  6  



Solving these equations (shown on the following slide) yields:
V1 = -15 V and V2 = 3 V.
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Example 2 - continued
Use nodal analysis to find Ix.
V1 = -15 V 12 k V2 = 3 V
4 mA
Ix
6 k
6 k
2 mA
In terms of the node voltages:
Ix = (V1 - V2)/12k = (-15 – 3)/12 k = -18v/ 12k
Ix = -1.5mA
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TI-86 Solution
 1 1
1
V1     V2    4V
12 6 
12 
1
 1 1
V1    V2     2V
12 
12 6 
3
1
2
4
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TI-86 Solution
 1 1
1
V1     V2    4V
12 6 
12 
1
 1 1
V1    V2     2V
12 
12 6 
5
7
6
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Dependent Sources
• Circuits containing dependent sources generally introduce
another unknown - the parameter (voltage or current) that
controls the dependent source.
• This requires that the additional unknown be eliminated by
writing an equation that expresses the controlling
parameter in terms of the node voltages.
• The resulting equations, with the additional unknown
eliminated, are solved in a conventional manner.
• The following example illustrates.
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Dependent Source Example
V1
Io
V2
The nodal equations are:
R2
R1
Node #1:
R3
Io
IS
V1 V1  V2

  Io  0
R1
R2
V2 V2  V1

 IS
Node #2:
R3
R2
• There are three unknowns in the equations, V1, V2 and Io.
• Another equation is needed that relates Io to V1 and/or V2.
The additional equation can be formed by noting the Io is the
current through R3, and by Ohm’s law Io = V2/R3. This relation
can be used to form a system of three- equations or to eliminate Io
from the first equation, leaving a two-by-two system to solve.
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Example 3 Use nodal analysis to find node voltages V1 and V2.
V1
V2
2Io
Io
The node equations are:
V1
V V
 1 2  4ma
10k  10k 
V 2 V2  V1

 2Io  0
10k  10k 
The “extra” unknown, Io, can be
expressed as:
V
Io  1
10k
The equations become:
V1
V  V2
 1
 4ma
10k 10k
V2
V  V1
 V 
 2
 2 1   0 2V  V  40 volts

1
2
10k 10k
 10k 
  V1  16v and V2  8v
V1  2V2  0

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