Transcript Nodal Analysis - Keith E. Holbert

Nodal Analysis
Dr. Holbert
January 28, 2008
Lect4
EEE 202
1
Node and Loop Analysis
• Node analysis and loop analysis are both
circuit analysis methods which are
systematic and apply to most circuits
• Analysis of circuits using node or loop
analysis requires solutions of systems of
linear equations
• These equations can usually be written by
inspection of the circuit
Lect4
EEE 202
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Steps of Nodal Analysis
1. Choose a reference (ground) node.
2. Assign node voltages to the other nodes.
3. Apply KCL to each node other than the
reference node; express currents in terms
of node voltages.
4. Solve the resulting system of linear
equations for the nodal voltages.
Lect4
EEE 202
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Example: A Summing Circuit
• The output voltage V of this circuit is
proportional to the sum of the two input
currents I1 and I2
• This circuit could be useful in audio
applications or in instrumentation
• The output of this circuit would probably
be connected to an amplifier
Lect4
EEE 202
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1. Reference Node
500W
500W
+
I1
500W
V
1kW
500W
I2
–
The reference node is called the ground node
where V = 0
Lect4
EEE 202
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Steps of Nodal Analysis
1. Choose a reference (ground) node.
2. Assign node voltages to the other
nodes.
3. Apply KCL to each node other than the
reference node; express currents in terms
of node voltages.
4. Solve the resulting system of linear
equations for the nodal voltages.
Lect4
EEE 202
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2. Node Voltages
V1
500W V 500W
2
1
I1
2
500W
V3
3
1kW
500W
I2
V1, V2, and V3 are unknowns for which we
solve using KCL
Lect4
EEE 202
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Steps of Nodal Analysis
1. Choose a reference (ground) node.
2. Assign node voltages to the other nodes.
3. Apply KCL to each node other than the
reference node; express currents in
terms of node voltages.
4. Solve the resulting system of linear
equations for the nodal voltages.
Lect4
EEE 202
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Currents and Node Voltages
V1
500W
V2
V1
500W
V1  V2
500 W
Lect4
V1
500 W
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3. KCL at Node 1
V1
I1
Lect4
500W
V2
V1  V2
V1
I1 

500 W 500 W
500W
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3. KCL at Node 2
V1
500W
V2 500W
V3
1kW
V2  V1 V2 V2  V3


0
500 W 1kW 500 W
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EEE 202
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3. KCL at Node 3
V2 500W
V3
500W
Lect4
I2
EEE 202
V3  V2
V3

 I2
500 W 500 W
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Steps of Nodal Analysis
1. Choose a reference (ground) node.
2. Assign node voltages to the other nodes.
3. Apply KCL to each node other than the
reference node; express currents in terms
of node voltages.
4. Solve the resulting system of linear
equations for the nodal voltages.
Lect4
EEE 202
13
4. Summing Circuit Solution
500W
500W
+
I1
500W
V
1kW
500W
I2
–
Solution: V = 167I1 + 167I2
Lect4
EEE 202
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A Linear Large Signal Equivalent
to a Transistor
0.7V
Ib
5V
+
–
1kW
+ –
50W
100Ib
2kW
+
Vo
–
Lect4
EEE 202
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Steps of Nodal Analysis
1. Choose a reference (ground) node.
2. Assign node voltages to the other nodes.
3. Apply KCL to each node other than the
reference node; express currents in terms
of node voltages.
4. Solve the resulting system of linear
equations for the nodal voltages.
Lect4
EEE 202
16
Linear Large Signal Equivalent
0.7V
1
5V
Ib V2
V1
+
–
1kW
2
+ –
V3
V4
3 50W
4
Vo
100Ib
2kW
Lect4
EEE 202
+
–
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Steps of Nodal Analysis
1. Choose a reference (ground) node.
2. Assign node voltages to the other nodes.
3. Apply KCL to each node other than the
reference node; express currents in terms
of node voltages.
4. Solve the resulting system of linear
equations for the nodal voltages.
Lect4
EEE 202
18
KCL @ Node 4
0.7V
1
5V
Ib V2
V1
+
–
1kW
2
+ –
V3
V4
3 50W
100Ib
4
+
Vo
2kW
–
V3  V4
V4
 100 I b 
50W
2 kW
Lect4
EEE 202
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The Dependent Source
• We must express Ib in terms of the node
voltages:
V1  V2
Ib 
1kW
• Equation from Node 4 becomes
V3  V4
V1  V2
V4
 100

0
50W
1 kW 2kW
Lect4
EEE 202
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How to Proceed?
• The 0.7-V voltage supply makes it
impossible to apply KCL to nodes 2 and 3,
since we don’t know what current is
passing through the supply
• We do know that
V2 – V3 = 0.7 V
• The above is a needed constraint equation
Lect4
EEE 202
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KCL at
Supernode
V2  V1 V3  V4

0
1kW
50W
0.7V
1
V2
V1
+
–
1kW
Ib
+ –
V3
50W
100Ib
V4
4
Vo
2kW
Lect4
EEE 202
+
–
22
Class Examples
• Drill Problems P2-8, P2-9, P2-10, P2-11
Lect4
EEE 202
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