Circuits/Nodes
ECSE 210: Circuit Analysis
Lecture #2: Nodal Analysis
Circuit Elements
+
v
Circuit Variables
i4
-
i2
i3
i
+
v
-
v = iR
i5
+
i
v
-
i
+
-
E
+
v
-
v=E
i
i1
i6
i7
,
i=-I
• 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?
Nodal Equations (KCL at each node)
Nodal Voltages
g3
i4
i2
i3
(b)
g1
va
(d)
(a)
vb
g2
vd
vc
(c)
i5
i7
i6
i1
g4
g5
,
g6
(e)
(a)
(b)
(c)
(d)
i1 + i2 = 0
- i2 + i3 = 0
-i3 + i4 +i5 = 0
-i4 – i5 + i6 +i7 = 0
KCL equations
¦i
0
out
5 nodes Æ 4 equations
Redundant!
Linear combination of first 4 eqs.
(e) -i1 – i6 – i7 = 0
• 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.
Voltage Across Elements
Kirchoff’s Voltage Law (KVL)
g3
g
va
+
v
-
g1
va
vb
vb
g2
vd
vc
i
,
Loop #1
g4
g5
g6
v = va - vb
i = (va – vb) g
Note: The conductance g = 1/R
Therefore, V=IR or I=gV are equivalent forms of Ohm’s Law.
Check KVL
(va – vb) + (vb – vc) + (vc – vd) + (vd – 0) + (0 – va) = 0
Æ Using “node voltages” means that KVL will always be satisfied.
Nodal Equations
Nodal Equations
g3
g1
va
vb
g2
(a)
(b)
(c)
(d)
g1
va
vd
vc
,
g3
g4
g5
g6
(va – vb)g1 – I = 0 Æ (va – vb)g1 = I
(vb – va)g1 + (vb – vc)g2 = 0
(vc – vb)g2 + (vc – vd)g4 + (vc – vd)g3 = 0
(vd – vc)g4 + (vd – vc)g3 + (vd – 0)g5 + (vd – 0)g6 =0
4 equations in 4 unknowns Æ solve using linear algebra
Note: 5 nodes including ground Æ (5 -1) equations
,
Æ See appendix A of textbook for a review of linear algebra
g2
vd
vc
i1
g4
g6
g5
Now we have solved for all voltages.
i1= (va - vb)g1
Æ Similarly we can find all currents.
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.
vb
Nodal Analysis – Example
g3
va
g2
g5
vb
vc
vc
g1
,a
g4
,b
(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
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
Nodal Analysis – Example
ª g1 g 2 g 3
« g
2
«
«¬
g3
g2
g 2 g 4 g5
g5
g 3 º ªv a º
g 5 »» ««vb »»
g 3 g 5 »¼ «¬ vc »¼
ª Ia º
« 0 »
«
»
«¬ I b »¼
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)
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
ª g1 g 2 g 3
« g
2
«
«¬
g3
g2
g 2 g 4 g5
g5
g 3 º ªva º
g 5 »» ««vb »»
g 3 g 5 »¼ «¬ vc »¼
ª Ia º
« 0 »
«
»
«¬ I b »¼
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.
Nodal Analysis by Inspection
Nodal Analysis by Inspection
g3
v1
g1
g2
g5
v2
col j
row i
g
-g
row j
-g
g
Node j
v3
g4
,a
col i
g
Node i
,b
col i
g
Node i
row i
ª g1 g 2 g 3
« g
2
«
«¬
g3
g2
g 2 g 4 g5
g5
g 3 º ª v1 º
g 5 »» ««v2 »»
g 3 g 5 »¼ «¬v3 »¼
Nodal Analysis by Inspection
Nodal Analysis by Inspection
RHS vector
Node i
,a
g
ª Ia º
« 0 »
«
»
«¬ I b »¼
row i
,a
row j
,a
Node j
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).
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.
Dependent Current Source
Nodal Analysis by Inspection: Example
g1
v2
,a
v1
g2
,b
g3
g4
g6
g5
v4 ,c
,a
v5
,a 3
,c 4
,c
2
3
4
5
1
g1 + g2
-g1
0
0
0
v1
2
-g1
g1+ g4
0
-g4
0
v2
3
0
0
g3+ g6
4
0
-g4
0
5
0
0
-g6
-g6
v3
g4 + g5 0
v4
g6
v5
0
0
1
,b 2
=
v3
Æ Check to see that each row
represents the KCL equation
at the corresponding node.
5
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=0
KCL at node 3: (v3-v2)g5+(v3-v1)g3+(v1-v2)gm=0
ª g1 g 2 g 3
« g
2
«
«¬ g 3 g m
g3
1
g2
g 2 g 4 g5
g5 g m
g 3 º ª v1 º
g 5 »» ««v2 »»
g 3 g 5 »¼ «¬v3 »¼
ªI a º
«0»
« »
«¬ 0 »¼
Dependent sources can destroy symmetry.
g2
v1
+
g1
,a
g5
v2
v3
vs g4
gmvs Given
gm(v1-v2)
KCL at node 1: v1g1+(v1-v3)g3+(v1-v2)g2=Ia
KCL at node 2: v2g4+(v2-v1)g2+(v2-v3)g5=0
KCL at node 3: (v3-v2)g5+(v3-v1)g3+(v1-v2)gm=0