[email protected] engr-43_lec-03-1b_nodal_analysis.ppt 1 bruce mayer, pe engineering-43:...
TRANSCRIPT
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt1
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Bruce Mayer, PERegistered Electrical & Mechanical Engineer
Engineering 43
Chp 3.1bChp 3.1bNodal Nodal
AnalysisAnalysis
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt2
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Ckts with Voltage SourcesCkts with Voltage Sources Need Only
ONE KCL Eqn
012126
12322
k
VV
k
VV
k
V
The Remaining Eqns From the Indep Srcs
][6
][12
3
1
VV
VV
Solving The Eqns
][5.1][64
0)()(2
22
12322
VVVV
VVVVV
3 Nodes Plus the Reference. In Principle Need 3 Equations...• But two nodes are connected
to GND through voltage sources. Hence those node voltages are KNOWN
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt3
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ExampleExample
+-
2SI
3SI 1SV
1SI 1R2R
3R
4R
1V2V 3V
4V
OV
R1 = 1k; R2 = 2k, R3 = 1k, R4 = 2kIs1 =2mA, Is2 = 4mA, Is3 = 4mA,Vs1 = 12 V
210 VVV Need Only V1 and V2
to Find Vo
Known Node Potential
][12:@ 133 VVVV S
Now KCL at Node 1
021
][2
0:@
121
4
1
1
2111
k
V
k
VVmA
R
V
R
VVIV S
Find Vo
To Start• Identify & Label All Nodes
• Write Node Equations
• Examine Ckt to Determine Best Solution Strategy
Notice
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt4
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example cont.Example cont.
+-
2SI
3SI 1SV
1SI 1R2R
3R
4R
1V2V 3V
4V
OV
R1 = 1k; R2 = 2k, R3 = 1k, R4 = 2kIs1 =2mA, Is2 = 4mA, Is3 = 4mA,Vs = 12 V
021
12
1][4
0:@
42212
2
42
3
32
1
1232
k
VV
k
V
k
VVmA
R
VV
R
VV
R
VVIV S
At Node 4
02
][4][2
0:@
24
2
24214
k
VVmAmA
R
VVIIV SS
To Solve the System of Equations Use LCD-multiplication and Gaussian Elimination
At Node 2
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt5
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example cont.Example cont. The LCDs
021
][2 121
k
V
k
VVmA
021
12
1][4 42212
k
VV
k
V
k
VVmA
02
][4][2 24
k
VVmAmA
][423 21 VVV *2k
*2k]3252 421 VVVV
*2k][442 VVV
(1)
(2)
(3)
Now Add Eqns (2) & (3) To Eliminate V4
][182][3642 2121 VVVVVV (4)
Now Add Eqns (4) & (1) To Eliminate V2
][11][222 11 VVVV
][5.14][182][11 22 VVVVV
][5.3][5.14][11210 VVVVVV
BackSub into (4) To Find V2
Find Vo by Difference Eqn
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt6
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
SuperNode TechniqueSuperNode Technique Consider This
Example Conventional Node
Analysis Requires All Currents At A Node
2 eqns, 3 unknowns...Not Good • Recall: The Current thru the
Vsrc is NOT related to the Potential Across it
But Have Ckt V-Src Reln
][621 VVV More Efficient solution:
• Enclose The Source, And All Elements In Parallel, Inside A Surface. – Call That a SuperNode
SUPERNODE
SI
06
6@ 11 SIk
VmAV
012
4@ 22
k
VmAIV S
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt7
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Supernode cont.Supernode cont. Apply KCL to the
Surface
Now Have 2 Equations in 2 Unknowns
Then The Ckt Solution Using LCD Technique• See Next Slide
SUPERNODE
SI
04126
6 21 mAk
V
k
VmA
][621 VVV
• The Source Current Is interior To The Surface And Is NOT Required
Still Need 1 More Equation – Look INSIDE the Surface to Relate V1 & V2
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt8
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Now Apply Gaussian ElimNow Apply Gaussian Elim
The Equations
][6 (2)
046126
(1)
21
21
VVV
mAmAk
V
k
V
Mult Eqn-1 by LCD (12 kΩ)
][6
][242
21
21
VVV
VVV
Add Eqns to Elim V2
][10][303 11 VVVV
][4][612 VVVV
Use The V-Source Rln Eqn to Find V2
SUPERNODE
SI
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt9
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
1V 2V
1sI
2sI
1R 2R
3R
SV
][6],[10],[20
4,10
21
321
mAImAIVV
kRkRR
ssS
Find the node voltagesAnd the power suppliedBy the voltage source
2012 VV
0101010
21 mAk
V
k
V ][10010* 21 VVVk
][2021 VVV
][40100
][1202 :adding
21
2
VVV
VV
To compute the power supplied by the voltage source We must know the current through it: @ node-1
VI
k
VVmA
k
VIV 4
610
211 mA5
BASED ON PASSIVE SIGN CONVENTION THEPOWER IS ABSORBED BY THE SOURCE!!
mWmAVP 100][5][20
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt10
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Illustration using ConductancesIllustration using Conductances Write the Node Equations
• KCL At v1
At The SuperNode Have V-Constraint• v2 − v3 = vA
KCL Leaving Supernode
Now Have 3 Eqnsin 3 Unknowns• Solve Using Normal Techniques
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt11
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ExampleExample Find Io
Known Node Voltages
Now KCL at SuperNode
SUPERNODE
123V
VVVV 12,6 42 The SuperNode
V-Constraint
VVV 1231
Or
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt12
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Student Exercise Student Exercise
Lets Turn on the Lights for 5-7 min Students are invited to Analyze the
following Ckt• Hint: Use SuperNode
Determine the OutPut Current, IO
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt13
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical ExampleNumerical Example Find Io Using
Nodal Analysis Known Voltages for Sources Connected
to GND
Now KCL at SuperNode
VVVV 4,6 41
The Constraint Eqn
SUPERNODE
VVV 1223
k
k
V
k
V
k
V
k
V20
2
)4(
212
6 3322
Now Notice That V2 is NOT Needed to Find Io
• 2 Eqns in 2 Unknowns
VVVV
VVV
VVV
6.7385
------------------
eqns add and312
223
33
32
32
By Ohm’s LawmA
k
V
k
VIO 8.3
2
6.7
2 3
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt14
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Complex SuperNodeComplex SuperNode Write the Node Eqns Set UP
• Identify all nodes
• Select a reference
• Label All nodes
Nodes Connected To Reference Through A Voltage Source
Eqn Bookkeeping:• KCL@ V3
• KCL@ SuperNode,
• 2 Constraint Equations
• One Known Node
+-
+ -
+-1R
2R
3R
4R
5R
6R
7R
supernode
1V
2V 3
V
4V
5V
Voltage Sources In Between Nodes And Possible Supernodes
• Choose to Connect V2 & V4
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt15
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Complex SuperNode cont.Complex SuperNode cont. Now KCL at Node-3
+-
+ -
+-1R
2R
3R
4R
5R
6R
7R
supernode
1V
2V 3
V
4V
5V
Constraints Due to Voltage Sources
07
3
5
43
4
23
R
V
R
VV
R
VV
Now KCL at Supernode• Take Care Not to Omit
Any Currents
04
32
5
34
6
4
3
5
2
15
1
12
R
VV
R
VV
R
V
R
V
R
VV
R
VV
11 SVV 252 SVVV 345 SVVV
Vs1
Vs2 Vs3
5 Equations 5 Unknowns → Have to Sweat Details
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt16
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Dependent SourcesDependent Sources
Circuits With Dependent Sources Present No Significant Additional Complexity
The Dependent Sources Are Treated As Regular Sources
As With Dependent CURRENT Sources Must Add One Equation For Each Controlling Variable
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt17
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical Example – Dep INumerical Example – Dep Isrcsrc
Find Io by Nodal Analysis
Notice V-Source Connected to the Reference Node
KCL At Node-2 Sub Ix into KCL Eqn
Mult By 6 kΩ LCD
VV 31
Then Io
0263
212
xIk
V
k
VV
Controlling Variable In Terms of Node Potential
k
VI x 6
2
06
263
2212
k
V
k
V
k
VV
VVVV 602 212
mAk
VVIO 1
321
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt18
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Dep V-Source ExampleDep V-Source Example Find Io by Nodal Analysis
Notice V-Source Connected to the Reference Node
SuperNode Constraint
KCL at SuperNode
Mult By 12 kΩ LCD
VV 63
xVVV 221
212 3VVVVx 062)6(2 2211 VVVV
Controlling Variable in Terms of Node Voltage
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt19
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Dep V-Source Example contDep V-Source Example cont Simplify the LCD Eqn
By Ohm’s Law
mAk
V
k
VIo 8
3
24
9
121
VV
VV
VV
VVV
5.4
184
3 and
1833
1
1
12
21
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt20
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Current Controlled V-SourceCurrent Controlled V-Source Find Io
Supernode Constraint
Controlling Variable in Terms of Node Voltage
xkIVV 212
k
VI x 2
1
121 22 VVkIV x KCL at SuperNode
02
22
4 21 k
VmA
k
VmA
Multiply by LCD of 2 kΩ
][421 VVV 02 21 VV Recall
Then VV 83 2
So Finally
mAk
VIO 3
4
22
[email protected] • ENGR-43_Lec-03-1b_Nodal_Analysis.ppt21
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
WhiteBoard WorkWhiteBoard Work
Let’s Work This Problem
Find the OutPut Voltage, VO
1K1K 1K
12V
VO
+
-
1K
IO2IX IX