lect4eee 2021 nodal analysis dr. holbert january 28, 2008
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Lect4 EEE 202 1
Nodal Analysis
Dr. Holbert
January 28, 2008
Lect4 EEE 202 2
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 3
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 4
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 5
1. Reference Node
The reference node is called the ground node where V = 0
+
–
V 500
500
1k
500
500I1 I2
Lect4 EEE 202 6
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 7
2. Node Voltages
V1, V2, and V3 are unknowns for which we solve using KCL
500
500
1k
500
500I1 I2
1 2 3
V1 V2 V3
Lect4 EEE 202 8
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 9
Currents and Node Voltages
500
V1500V1 V2
50021 VV
5001V
Lect4 EEE 202 10
3. KCL at Node 1
500
500I1
V1 V2
500500
1211
VVVI
Lect4 EEE 202 11
3. KCL at Node 2
500
1k
500 V2 V3V1
0500k1500
32212
VVVVV
Lect4 EEE 202 12
3. KCL at Node 3
2323
500500I
VVV
500
500
I2
V2 V3
Lect4 EEE 202 13
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 14
+
–
V 500
500
1k
500
500I1 I2
4. Summing Circuit Solution
Solution: V = 167I1 + 167I2
Lect4 EEE 202 15
A Linear Large Signal Equivalent to a Transistor
5V100Ib
+
–
Vo
50
Ib
2k1k+–
+ –
0.7V
Lect4 EEE 202 16
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 17
Linear Large Signal Equivalent
5V 100Ib
+
–
Vo
50
Ib
2k
1k
0.7V
12 3 4
V1V2 V3 V4
+–
+ –
Lect4 EEE 202 18
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 19
KCL @ Node 4
k2100
50443 V
IVV
b
100Ib
+
–
Vo
50
Ib
2k
1k+–
0.7V
12 3 4
V1 V2 V3 V4
5V
+ –
Lect4 EEE 202 20
The Dependent Source
• We must express Ib in terms of the node voltages:
• Equation from Node 4 becomes
k1
21 VVIb
0k2k1
10050
42143
VVVVV
Lect4 EEE 202 21
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 22
100Ib
+
–
Vo
50Ib
2k
1k
0.7V
14
V1 V2 V3 V4
+–
+ –
050k1
4312
VVVV
KCL at Supernode
Lect4 EEE 202 23
Class Examples
• Drill Problems P2-8, P2-9, P2-10, P2-11