ESE 271 / Spring 2013 / Lecture 6
Last time: Proportionality and Superposition.
Example we considered last time:
Example we considered last time:
For linear circuits
For any
Proportionality: Scaling of all independent sources
Scaling of all independent sources scales all circuit variables.
Superposition:
Response of circuit to several sources is equal to sum of responses to each individual source. 1
ESE 271 / Spring 2013 / Lecture 6
Last time: Nodal Analysis.
Idea:
If known
If known Then:
Reference node.
All voltages will be calculated with respect to this reference.
2
ESE 271 / Spring 2013 / Lecture 6
Last time: Example of Nodal Analysis
Node 2
Node 3
Node 1
Let’s choose Node 4 as a reference node hence V4 = 0.
reference node, hence V
=0
Node 4
3
ESE 271 / Spring 2013 / Lecture 6
Last time: Nodal Analysis ‐ Equation pattern
Node 2
Node 3
Node 1
Let’s choose Node 4 as a reference node hence V4 = 0.
reference node, hence V
=0
Node 4
Currents from current sources that enter the h
h
node
Sum of conductances of the elements connected
the elements connected to node multiplied by nodal voltage
Other nodal voltages multiplied by MINUS conductances of the
by MINUS conductances of the elements connecting other nodes to the equation node.
4
ESE 271 / Spring 2013 / Lecture 6
Last time: Nodal analysis of circuits containing voltage sources
Can not find current even if nodal voltages are known. What to do?
Example we have considered last time
Supernode A contains Node 1 and Reference Node, hence: Supernode B contains Node 4 and Supernode
B contains Node 4 and
Node 5, hence: Summary: In this circuit there are five non‐
reference nodes and two voltage sources, hence only three unknown nodal voltages:
nodal voltages:
5
ESE 271 / Spring 2013 / Lecture 6
Example 2
6
ESE 271 / Spring 2013 / Lecture 6
Example 3 – with dependent current source
Reference node
Unknown nodal voltages
7
ESE 271 / Spring 2013 / Lecture 6
Example 4 – with dependent voltage source
Reference node
Unknown nodal voltages:
But
Only
hence
are unknown
Also
KCL at node 4:
KCL at supernode:
8
ESE 271 / Spring 2013 / Lecture 6
Mesh analysis
In its basic form it is suitable for planar circuits only.
In
its basic form it is suitable for planar circuits only
Hence not as universal as nodal analysis but more convenient in certain cases.
EXAMPLE – circuit containing four meshes (loops with no loops inside)
ACTION ITEMS
1. Assign current to each mesh.
2. Write KVL for each mesh.
3. Solve system of linear equations for mesh currents.
4. Find actual currents and voltages.
Mesh 1:
Mesh 2:
Mesh 3:
Mesh 4:
9
ESE 271 / Spring 2013 / Lecture 6
Mesh analysis: Example 1.
1. Assign current to each mesh.
2. Write KVL for each mesh.
3. Solve for mesh currents:
4. Find actual currents and voltages.
10
ESE 271 / Spring 2013 / Lecture 6
Mesh analysis: Example 2.
11
ESE 271 / Spring 2013 / Lecture 6
Mesh analysis of circuits containing current sources.
Voltage drop across current source is not defined.
Voltage
drop across current source is not defined
However, presence of current source reduces number of equations by one.
EXAMPLE – circuit containing four meshes and two current sources
Four meshes, hence, four mesh currents:
BUT:
Only two mesh currents remain unknown.
Hence, only two equations are needed.
Write KVL for supermeshes.
f
p
12
ESE 271 / Spring 2013 / Lecture 6
Example 1.
Four meshes, hence, four mesh currents:
BUT:
Hence only two unknowns:
and
Observe:
KVL for mesh 2:
KVL for supermesh 1+3+4:
13
ESE 271 / Spring 2013 / Lecture 6
Example 1 ‐ cont.
14
ESE 271 / Spring 2013 / Lecture 6
Example 2.
Two mesh currents:
BUT:
15
ESE 271 / Spring 2013 / Lecture 6
Nodal analysis of circuits with OpAmps.
Example 1
Example 1.
virtual short
it l h t
KCL at Node 2:
No current into OpAmp
KCL at Node 3:
16
ESE 271 / Spring 2013 / Lecture 6
Nodal analysis of circuits with OpAmps.
Example 2
Example 2.
17
ESE 271 / Spring 2013 / Lecture 6
Example 3.
18