NODE ANALYSIS
• One of the systematic ways to
determine every voltage and
current in a circuit
The variables used to describe the circuit will be “Node Voltages”
-- The voltages of each node with respect to a pre-selected
reference node
IT IS INSTRUCTIVE TO START THE PRESENTATION WITH
A RECAP OF A PROBLEM SOLVED BEFORE USING SERIES/
PARALLEL RESISTOR COMBINATIONS
COMPUTE ALL THE VOLTAGES AND CURRENTS IN THIS CIRCUIT
4k || 12k 12k
SECOND: “BACKTRACK” USING KVL, KCL OHM’S
6k
I3
OHM' S : I 2 =
Va
6k
KCL : I1 − I 2 − I 3 = 0
OHM' S : Vb = 3k * I 3
…OTHER OPTIONS...
6k || 6k
FIRST REDUCE TO A SINGLE LOOP CIRCUIT
KCL : I 5 + I 4 − I 3 = 0
OHM' S : VC = 3k * I 5
I1 =
12
I3
4 + 12
Vb = 4k * I 4
I4 =
12V
12k
Va =
3
(12)
3+9
THE NODE ANALYSIS PERSPECTIVE
VS
+ V1 − V
a
KVL
+ V3 − V
b
KVL
THERE ARE FIVE NODES.
IF ONE NODE IS SELECTED AS
REFERENCE THEN THERE ARE
FOUR VOLTAGES WITH RESPECT
TO THE REFERENCE NODE
+ V5 − V
c
KVL
− Vc − V5 + Vb = 0
− VS + V1 + Va = 0 − Va + V3 + Vb = 0
V1 = VS − Va
V3 = Va − Vb
V5 = Vb − Vc
REFERENCE
WHAT IS THE PATTERN???
THEOREM: IF ALL NODE VOLTAGES WITH
RESPECT TO A COMMON REFERENCE NODE
ARE KNOWN THEN ONE CAN DETERMINE
ANY OTHER ELECTRICAL VARIABLE FOR
THE CIRCUIT
ONCE THE VOLTAGES ARE
KNOWN THE CURRENTS CAN
BE COMPUTED USING OHM’S
LAW
v R = vm − v N
+ vR −
DRILL QUESTION
V ca = ______
A GENERAL VIEW
THE REFERENCE DIRECTION FOR CURRENTS IS IRRELEVANT
+ vR
−
v R'
USING THE LEFT-RIGHT REFERENCE DIRECTION
THE VOLTAGE DROP ACROSS THE RESISTOR MUST
HAVE THE POLARITY SHOWN
v − vN
OHM' S LAW i = m
R
' PASSIVE SIGN CONVENTION RULES!
i =−i
−
+
i'
IF THE CURRENT REFERENCE DIRECTION IS
REVERSED ...
THE PASSIVE SIGN CONVENTION WILL ASSIGN
THE REVERSE REFERENCE POLARITY TO THE
VOLTAGE ACROSS THE RESISTOR
OHM' S LAW i ' =
v N − vm
R
DEFINING THE REFERENCE NODE IS VITAL
+ V12 −
+
4V
−
−
2V
+
THE STATEMENT V1 = 4V IS MEANINGLES
S
UNTIL THE REFERENCE POINT IS DEFINED
BY CONVENTION THE GROUND SYMBOL
SPECIFIES THE REFERENCE POINT.
ALL NODE VOLTAGES ARE MEASURED WITH
RESPECT TO THAT REFERENCE POINT
V12 = _____?
THE STRATEGY FOR NODE ANALYSIS
VS
Va
Vb
Vc
1. IDENTIFY ALL NODES AND SELECT
A REFERENCE NODE
2. IDENTIFY KNOWN NODE VOLTAGES
3. AT EACH NODE WITH UNKNOWN
VOLTAGE WRITE A KCL EQUATION
(e.g.,SUM OF CURRENT LEAVING =0)
@Va : − I1 + I 2 + I 3 = 0
Va − Vs Va Va − Vb
+
+
=0
9k
6k
3k
@Vb : − I 3 + I 4 + I 5 = 0
REFERENCE
4. REPLACE CURRENTS IN TERMS OF
NODE VOLTAGES
AND GET ALGEBRAIC EQUATIONS IN
THE NODE VOLTAGES ...
Vb − Va Vb Vb − Vc
+
+
=0
3k
4k
9k
SHORTCUT: SKIP WRITING
THESE EQUATIONS...
@Vc : − I 5 + I 6 = 0
Vc − Vb Vc
+
=0
9k
3k
AND PRACTICE WRITING
THESE DIRECTLY
WHEN WRITING A NODE EQUATION...
AT EACH NODE ONE CAN CHOSE ARBITRARY
DIRECTIONS FOR THE CURRENTS
a
Va
Vb
R1
b
R3
I 1'
c
d
CURRENTS LEAVING = 0
Va − Vb Vb − Vd Vb − Vc
+
+
=0
R1
R2
R3
CURRENTS INTO NODE = 0
I1 − I 2 − I 3 = 0 ⇒
R3
I 3'
R2
c
Vc
I 2'
d
I2
AND SELECT ANY FORM OF KCL.
WHEN THE CURRENTS ARE REPLACED IN TERMS
OF THE NODE VOLTAGES THE NODE EQUATIONS
THAT RESULT ARE THE SAME OR EQUIVALENT
− I1 + I 2 + I 3 = 0 ⇒ −
b
Vd
I3
R2
Vd
∑
Va
R1
Vc
I1
∑
a
Vb
Va − Vb Vb − Vd Vb − Vc
−
−
=0
R1
R2
R3
∑
CURRENTS LEAVING = 0
I1' + I 2' − I 3' = 0 ⇒
∑
Vb − Va Vb − Vd Vc − Vb
+
−
=0
R1
R2
R3
CURRENTS INTO NODE = 0
− I1' − I 2' + I 3' = 0 ⇒ −
Vb − Va Vb − Vd Vc − Vb
−
+
=0
R1
R2
R3
WHEN WRITING THE NODE EQUATIONS
WRITE THE EQUATION DIRECTLY IN TERMS
OF THE NODE VOLTAGES.
BY DEFAULT USE KCL IN THE FORM
SUM-OF-CURRENTS-LEAVING = 0
THE REFERENCE DIRECTION FOR THE
CURRENTS DOES NOT AFFECT THE NODE
EQUATION
CIRCUITS WITH ONLY INDEPENDENT SOURCES
HINT: THE FORMAL MANIPULATION OF
EQUATIONS MAY BE SIMPLER IF ONE
USES CONDUCTANCES INSTEAD OF
RESISTANCES.
@ NODE 1
USING RESISTANCES − i A +
v1 v1 − v2
+
=0
R1
R2
WITH CONDUCTANCES − i A + G1v1 + G2 (v1 − v 2 ) = 0
REORDERING TERMS
@ NODE 2
REORDERING TERMS
THE MODEL FOR THE CIRCUIT IS A SYSTEM
OF ALGEBRAIC EQUATIONS
EXAMPLE
WRITE THE KCL EQUATIONS
@ NODE 1 WE VISUALIZE THE CURRENTS
LEAVING AND WRITE THE KCL EQUATION
REPEAT THE PROCESS AT NODE 2
− i2 +
v2 − v1 v2 − v1
+
=0
R4
R3
OR VISUALIZE CURRENTS GOING INTO NODE
ANOTHER EXAMPLE OF WRITING NODE EQUATIONS
V
BB
MARK THE NODES
(TO INSURE THAT
NONE IS MISSING)
15mA
A
VA
8k
2k
8k
2k
C
WRITE KCL AT EACH NODE IN TERMS OF
NODE VOLTAGES
SELECT AS
REFERENCE
VA VA
+
+ 15mA = 0
2 k 8k
V V
@ B B + B − 15mA = 0
8k 2 k
@A
LEARNING EXTENSION
V1 V1 − V2 USING
+
6k
12k
V V −V
@V2 : 2mA + 2 + 2 1 = 0
6k
12k
@V1 : − 4mA +
BY “INSPECTION”
1 ⎞
1
⎛ 1
V
−
V2 = 4mA
+
⎟ 1
⎜
12k
⎝ 6k 12k ⎠
1 ⎛ 1
1 ⎞
−
+⎜ +
⎟V2 = −2mA
12k ⎝ 6k 12k ⎠
KCL
LEARNING EXTENSION
6mA
I3
I1
I2
Node analysis
V
@ V1 : 1 + 2mA + 6mA = 0 ⇒ V1 = −16V
2k
V V
@V : − 6mA + 2 + 2 = 0 ⇒ V2 = 12V
2
6k
IN MOST CASES THERE
ARE SEVERAL DIFFERENT
WAYS OF SOLVING A
PROBLEM
NODE EQS. BY INSPECTION
1
V1 + (0 )V2 = −(2 + 6 )mA
2k
(0)V1 + ⎛⎜ 1 + 1 ⎟⎞V2 = 6mA
⎝ 6k 3k ⎠
3k
I 1 = −8mA
3k
I2 =
(6mA) = 2mA
3k + 6k
6k
I3 =
(6mA) = 4mA
3k + 6k
CURRENTS COULD BE COMPUTED DIRECTLY
USING KCL AND CURRENT DIVIDER!!
Once node voltages are known
I1 =
V1
2k
I2 =
V2
6k
I3 =
V2
3k
CIRCUITS WITH INDEPENDENT VOLTAGE SOURCES
3 nodes plus the reference. In
principle one needs 3 equations...
…but two nodes are connected to
the reference through voltage
sources. Hence those node
voltages are known!!!
…Only one KCL is necessary
Hint: Each voltage source
connected to the reference
node saves one node equation
V2 V2 − V3 V2 − V1
+
+
=0
6k
12k
12k
V1 = 12[V ] THESE ARE THE REMAINING
V3 = −6[V ]
TWO NODE EQUATIONS
SOLVING THE EQUATIONS
2V2 + (V2 − V3 ) + (V2 − V1 ) = 0
4V2 = 6[V ] ⇒ V2 = 1.5[V ]
THE SUPERNODE TECHNIQUE
We will use this example to introduce the concept of a SUPERNODE
SUPERNODE
IS
Conventional node analysis
requires all currents at a node
@V_1
@V_2
V
− 6mA + 1 + I S = 0
6k
V2
− I S + 4mA +
=0
12k
Efficient solution: enclose the
source, and all elements in
parallel, inside a surface.
Apply KCL to the surface!!!
− 6mA +
V1 V2
+
+ 4mA = 0
6k 12k
The source current is interior
to the surface and is not required
We STILL need one more equation
2 eqs, 3 unknowns...Panic!!
The current through the source is not
related to the voltage of the source
1
2
V −V = 6[V]
Math solution: add one equation
V1 −V2 = 6[V]
Only 2 eqs in two unknowns!!!
ALGEBRAIC DETAILS
The Equations
V1 V2
(1)
+
− 6mA + 4mA = 0
6k 12k
(2) V1 − V2 = 6[V ]
Solution
1. Eliminate denominato rs in Eq(1). Multiply by ...
2V1 + V2 = 24[V ]
V1 − V2 = 6[V ]
2. Add equations to eliminate V2
3V1 = 30[V ] ⇒ V1 = 10[V ]
3. Use Eq(2) to compute V2
V2 = V1 − 6[V ] = 4[V ]
SUPERNODE
V1 = 6V
V4 = −4V
SOURCES CONNECTED TO THE
REFERENCE
CONSTRAINT EQUATION
V3 − V2 = 12V
KCL @ SUPERNODE
V2 − 6 V2 V3 V3 − (−4)
+ +
+
= 0 * / 2k
2k
1k 2k
2k
V2 IS NOT NEEDED FOR I O 3V2 + 2V3 = 2V
− V2 + V3 = 12V * / 3 and add
5V3 = 38V
V
OHM' S LAW I O = 3 = 3.8mA
2k