Transition Preparing to Move from Part C (EI) to Part B (ECSE) ECSE 672 Fall 2014.
ECSE 210: Circuit Analysis
Transcript of ECSE 210: Circuit Analysis
![Page 1: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/1.jpg)
ECSE 210: Circuit AnalysisECSE 210: Circuit Analysis
Lecture #2: Nodal Analysis
![Page 2: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/2.jpg)
Circuits/Nodes
![Page 3: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/3.jpg)
Circuit Elements
+ -
i
v
v = iR
i-v
+
-v
+ i+-
E
v = E
-v
+
Ι
i
i = - I
![Page 4: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/4.jpg)
Circuit Variables
i2
i1
i3
i4
i5 i6i7
• Why did we chose the above current directions?• Do we need to add voltage variables across elements?• If we do add voltages… in what orientation?
![Page 5: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/5.jpg)
Nodal Equations (KCL at each node)
i2
i1
i3
i4
i5 i6i7
(a)
(b) (c)
(d)
(e)
(a) i1 + i2 = 0 (b) - i2 + i3 = 0(c) -i3 + i4 +i5 = 0(d) -i4 – i5 + i6 +i7 = 0
(e) -i1 – i6 – i7 = 0 Redundant!Linear combination of first 4 eqs.
KCL equations 0=∑out
i
5 nodes 4 equations
![Page 6: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/6.jpg)
Nodal Voltages
Ι
g1 g2
g3
g4
g5 g6
va vb
vc
vd
• Choose reference node (ground).• The voltage at the reference node is defined to be zero.• Nodal voltages are defined with respect to the reference node.• For example, vb is the potential difference between point b and ground.
![Page 7: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/7.jpg)
Voltage Across Elements
gva vb
+ -v i
v = va - vb
i = (va – vb) g
Note: The conductance g = 1/RTherefore, V=IR or I=gV are equivalent forms of Ohm’s Law.
![Page 8: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/8.jpg)
Kirchoff’s Voltage Law (KVL)
Ι
g1 g2
g3
g4
g5 g6
va vb
vc
vd
Loop #1
Check KVL
(va – vb) + (vb – vc) + (vc – vd) + (vd – 0) + (0 – va) = 0
Using “node voltages” means that KVL will always be satisfied.
![Page 9: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/9.jpg)
Nodal Equationsg3
Ι
g1 g2
g4
g5 g6
va vb
vc
vd
(a) (va – vb)g1 – I = 0 (va – vb)g1 = I(b) (vb – va)g1 + (vb – vc)g2 = 0(c) (vc – vb)g2 + (vc – vd)g4 + (vc – vd)g3 = 0(d) (vd – vc)g4 + (vd – vc)g3 + (vd – 0)g5 + (vd – 0)g6 =0
4 equations in 4 unknowns solve using linear algebraNote: 5 nodes including ground (5 -1) equations
![Page 10: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/10.jpg)
Nodal Equationsg3
Ι
g1 g2
g4
g5 g6
va vb
vc
vd
i1= (va - vb)g1
Now we have solved for all voltages.
i1
Similarly we can find all currents.
![Page 11: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/11.jpg)
Nodal Analysis – Basic Steps
(1) Define all node voltage variables. Select the reference node to be used as ground.
(2) Arbitrarily define circuit branch currents. (3) Write KCL for each node in terms of the circuit
branch currents.(4) Use Ohm’s law to express branch currents in
terms of node voltage variables according to the passive sign convention.
(5) Solve the simultaneous algebraic equations for the unknown node voltages – any way you like.
(6) Determine the required characteristics from the circuit elements and node voltages.
See appendix A of textbook for a review of linear algebra
![Page 12: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/12.jpg)
Nodal Analysis – Example
Ιbg1
g2
g4
g5
g3
va vb
vc
vc
Ιa
(a)g1va + g3(va – vc) + g2(va – vb) = Ia(b)g2(vb-va) + g4vb + g5(vb-vc) = 0(c) g5(vc-vb) + g3(vc-va) = -Ib
![Page 13: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/13.jpg)
Nodal Analysis – Example
(a)g1va + g3(va – vc) + g2(va – vb) = Ia(b)g2(vb-va) + g4vb + g5(vb-vc) = 0(c) g5(vc-vb) + g3(vc-va) = -Ib
(a) (g1 + g3 + g2)va – g2vb –g3vc = Ia(b) -g2 va + (g2+g4+g5)vb – g5vc = 0(c) -g3va –g5vb +(g3+g5)vc = -Ib
![Page 14: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/14.jpg)
Nodal Analysis – Example
(a) (g1 + g3 + g2)va – g2vb –g3vc = Ia(b) -g2 va + (g2+g4+g5)vb – g5vc = 0(c) -g3va –g5vb +(g3+g5)vc = -Ib
−=
+−−−++−−−++
b
a
c
b
a
I
I
vvv
gggggggggggggg
0
5353
55422
32321
![Page 15: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/15.jpg)
Nodal Analysis – Example
−=
+−−−++−−−++
b
a
c
b
a
I
I
vvv
gggggggggggggg
0
5353
55422
32321
1. Each row represents a KCL equation.2. Matrix is symmetric – This is not a coincidence!3. All circuits containing resistors and current sources
will have symmetric matrices.4. Can use matrix methods to find the solution
(see Appendix A of textbook)
![Page 16: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/16.jpg)
Summary
1. We have studied nodal analysis for circuits containing resistors and current sources.
2. Nodal analysis is based on applying KCL for each node in the circuit, except the reference node.
3. For a circuit containing N nodes (including ground) we therefore have N-1 equations. Each equation represents KCL at a different node.
4. Node voltages are defined with respect to the reference node (ground).
5. In this lecture, we always summed the currents leaving a node when applying KCL. Summing the currents entering a node is also valid but it is better stick to one approach.
![Page 17: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/17.jpg)
Nodal Analysis by Inspection
Ιbg1
g2
g4
g5
g3
v1 v2 v3
Ιa
−=
+−−−++−−−++
b
a
I
I
vvv
gggggggggggggg
0
3
2
1
5353
55422
32321
![Page 18: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/18.jpg)
Nodal Analysis by Inspection
gNode i Node j
Node ig
col j
g -g
g-grow j
row i
col i
grow i
col i
![Page 19: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/19.jpg)
Nodal Analysis by Inspection
Node i
Node j
Ιa
Ιa
−Ιarow j
row i
RHS vector
1. Be careful about the signs.2. If one of the terminals is connected to
ground, then there is no corresponding KCL equation and that node simply does not appear in the equations (we get only one entry in the RHS vector).
![Page 20: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/20.jpg)
Nodal Analysis by Inspection
1. Designate a reference node.2. Number the remaining nodes.3. Size of the matrix (number of eqs.) is equal to
the number of nodes (not including the reference node).
4. Add the contribution of each element to the matrix equations.
5. Resistors contribute to the LHS equations or to the matrix itself.
6. Current sources contribute to the right hand side (RHS) vector.
![Page 21: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/21.jpg)
Nodal Analysis by Inspection: Example
v3Ιb
g6
Ιa
Ιc
g1
g2
g3
g5g4
v1
v2
v4v5
g1 -g1
-g1 g1
+ g2
g3
+ g4 -g4
-g4 g4 + g5
+ g6 -g6
-g6 g6
1 2 3 4 5
1
2
3
4
5
v1
v2
v3
v4
v5
0 0 0
0 0
0 0 0
0 0
000
0
=
Ιa
−Ιa−ΙcΙc
−Ιb
1
2
3
4
5
Check to see that each row represents the KCL equation at the corresponding node.
![Page 22: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/22.jpg)
Dependent Current Source
g1
g2
g4
g5v1 v2 v3
Ιa
g3
gmvs Given
KCL at node 1: v1g1+(v1-v3)g3+(v1-v2)g2=Ia
KCL at node 2: v2g4+(v2-v1)g2+(v2-v3)g5=0KCL at node 3: (v3-v2)g5+(v3-v1)g3+(v1-v2)gm=0
+ -vs
gm(v1-v2)
![Page 23: ECSE 210: Circuit Analysis](https://reader030.fdocuments.in/reader030/viewer/2022032608/6237a389f36da7430120cbac/html5/thumbnails/23.jpg)
Dependent Current Source
KCL at node 1: v1g1+(v1-v3)g3+(v1-v2)g2=Ia
KCL at node 2: v2g4+(v2-v1)g2+(v2-v3)g5=0KCL at node 3: (v3-v2)g5+(v3-v1)g3+(v1-v2)gm=0
=
+−−+−−++−−−++
00
3
2
1
5353
55422
32321 a
mm
I
vvv
gggggggggggggggg
Dependent sources can destroy symmetry.