EE 205
Coifman
Node voltages:
vA
vB
vC
Notation
+
vA
+
vB
+
vC
-
-
Interpretation
6-1
EE 205
Coifman
If the Kth two-terminal element is connected between
nodes X and Y, then the element voltage can be
expressed in terms of the two node voltages as
vk=vx-vy
where X is the node connected to the positive reference
for element voltage vk
vA
vB
+
v1
-
Case A
vA
+
vB
v1
Case B
6-2
-
EE 205
Coifman
Node Voltage Method
1) select a reference node for ground, assign a node
voltage to all other nodes and a current through every
element.
2) write node equations via KCL
3) element equations (using conductance)
4) combine steps 2 & 3 to eliminate (non-source) element
currents. In the process, reorder node equations to put
all dependent values on LHS and independent values
on RHS
5) solve for node voltages using linear algebra (you should
know how to do this, see appendix B for a refresher)
6-3
EE 205
Coifman
Did somebody say example?
vA i 2
i0
iS
R2
vB
i1
i3
R1
R3
6-4
EE 205
Coifman
Making things quicker...
Replace steps 2-4 with modified step 2:
vX•(sum of all conductances entering node X) Σ(conductance i entering node X)•vi + other currents = 0
where vi is the node voltage on the opposite side of the i-th
resistor entering node X
6-5
EE 205
Coifman
Oh what I’d give for an example...
vA
2 kΩ
vB
+
i S1
1 kΩ
i S2
500 Ω
vO
-
6-6
EE 205
Coifman
But what about voltage sources?
method 1: convert to current source
method 2: set ground at one end of voltage source. Now
vA≡vS
may not be able to do this for all voltage sources in a circuit
method 3: Super node: apply KCL to region and note that
vA-vB=vS
vA
vA
Rest
RS
vC
of the
+
-
vS
Rest
vS
RS
of the
RS
circuit
circuit
vB
vB
Method 1
Supernode
vA
vA
Rest
+
-
Rest
+
-
of the
vS
vB
of the
vS
circuit
circuit
vB
Method 2
Method 3
6-7
EE 205
Coifman
R1
vS1 +
-
R2
+
vO
R3
-
6-8
+
-
vS2
EE 205
Coifman
+
vS1
vA
i2
i3
vB
R2
i1
R1
R3
+
-
vS2
i4
R4
D
6-9
vC
+
vO
-