NodeAnalysis7Ed

NODAL AND LOOP ANALYSIS TECHNIQUES

LEARNING GOALS

CIRCUITS WITH OPERATIONAL AMPLIFIERS

Develop systematic techniques to determine all the voltagesand currents in a circuit

NODAL ANALYSISLOOP ANALYSIS

Op-amps are very important devices, widely available,that permit the design of very useful circuit…

and they can be modeled by circuits with dependent sources

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 WITHA RECAP OF A PROBLEM SOLVED BEFORE USING SERIES/PARALLEL RESISTOR COMBINATIONS

COMPUTE ALL THE VOLTAGES AND CURRENTS IN THIS CIRCUIT

SECOND: “BACKTRACK” USING KVL, KCL OHM’S

k

VI a

62 :SOHM' 13 2KCL: I I I

4

34

*4124

12

IkV

II

b

5

345

*3

0

IkV

III

C

:SOHM'

:KCL

k12kk 12||4

k6

kk 6||6

k

VI

12

121

)12(93

3

aV

3I

FIRST REDUCE TO A SINGLE LOOP CIRCUIT

THE NODE ANALYSIS PERSPECTIVETHERE ARE FIVE NODES. IF ONE NODE IS SELECTED AS REFERENCE THEN THERE AREFOUR VOLTAGES WITH RESPECTTO THE REFERENCE NODE

THEOREM: IF ALL NODE VOLTAGES WITHRESPECT TO A COMMON REFERENCE NODEARE KNOWN THEN ONE CAN DETERMINE ANY OTHER ELECTRICAL VARIABLE FORTHE CIRCUIT

aS

aS

VVV

VVV

1

10

ba

ba

VVV

VVV

3

3 0

______caV

QUESTIONDRILL

REFERENCE

aV

bV

cV

KVL KVL

WHAT IS THE PATTERN???

KVL

05 bc VVV

ONCE THE VOLTAGES ARE KNOWN THE CURRENTS CANBE COMPUTED USING OHM’SLAW

A GENERAL VIEW

Rv

NmR vvv

5 cbV V V

THE REFERENCE DIRECTION FOR CURRENTS IS IRRELEVANT

'i

Rv

USING THE LEFT-RIGHT REFERENCE DIRECTIONTHE VOLTAGE DROP ACROSS THE RESISTOR MUSTHAVE THE POLARITY SHOWN

R

vvi Nm LAW SOHM'

IF THE CURRENT REFERENCE DIRECTION IS REVERSED ...

'Rv

THE PASSIVE SIGN CONVENTION WILL ASSIGNTHE REVERSE REFERENCE POLARITY TO THE VOLTAGE ACROSS THE RESISTOR

R

vvi mN 'LAW SOHM'

'ii PASSIVE SIGN CONVENTION RULES!

DEFINING THE REFERENCE NODE IS VITAL

SMEANINGLES IS4V VSTATEMENT THE 1 UNTIL THE REFERENCE POINT IS DEFINED

BY CONVENTION THE GROUND SYMBOLSPECIFIES THE REFERENCE POINT.

V4

ALL NODE VOLTAGES ARE MEASURED WITHRESPECT TO THAT REFERENCE POINT

V2

_____?12 V

12V

THE STRATEGY FOR NODE ANALYSIS 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)

0:@321 IIIV

a

4. REPLACE CURRENTS IN TERMS OF NODE VOLTAGES

0369

k

VV

k

V

k

VV baasa AND GET ALGEBRAIC EQUATIONS IN THE NODE VOLTAGES ...

REFERENCE

SV

aV

bV

cV

0:@543 IIIV

b

0:@65 IIV

c

0943

k

VV

k

V

k

VVcbbab

039

k

V

k

VVcbc

SHORTCUT:SHORTCUT: SKIP WRITING THESE EQUATIONS...

AND PRACTICE WRITING

THESE DIRECTLY

a b c

d

aV

bV

cV

dV

1R 3R

2R

WHEN WRITING A NODE EQUATION...AT EACH NODE ONE CAN CHOSE ARBITRARYDIRECTIONS FOR THE CURRENTS

AND SELECT ANY FORM OF KCL.WHEN THE CURRENTS ARE REPLACED IN TERMSOF THE NODE VOLTAGES THE NODE EQUATIONSTHAT RESULT ARE THE SAME OR EQUIVALENT

00321

321

R

VV

R

VV

R

VVIII cbdbba

0LEAVING CURRENTS

00321

321

R

VV

R

VV

R

VVIII cbdbba

0 NODE INTO CURRENTS

2I3I1I

a b c

d

aV

bV

cV

dV

1R 3R

2R '2I

'3I'

1I

00321

'3

'2

'1

R

VV

R

VV

R

VVIII bcdbab

0LEAVING CURRENTS

00321

'3

'2

'1

R

VV

R

VV

R

VVIII bcdbab

0 NODE INTO CURRENTS

WHEN WRITING THE NODE EQUATIONSWRITE THE EQUATION DIRECTLY IN TERMSOF THE NODE VOLTAGES.BY DEFAULT USE KCL IN THE FORMSUM-OF-CURRENTS-LEAVING = 0

THE REFERENCE DIRECTION FOR THECURRENTS DOES NOT AFFECT THE NODEEQUATION

CIRCUITS WITH ONLY INDEPENDENT SOURCES

HINT:HINT: THE FORMAL MANIPULATION OF EQUATIONS MAY BE SIMPLER IF ONE USES CONDUCTANCES INSTEAD OF RESISTANCES.

@ NODE 1

02

21

1

1

R

vv

R

viA SRESISTANCE USING

0)( 21211 vvGvGiA ESCONDUCTANC WITH

REORDERING TERMS

@ NODE 2

REORDERING TERMS

THE MODEL FOR THE CIRCUIT IS A SYSTEMOF ALGEBRAIC EQUATIONS

THE MANIPULATION OF SYSTEMS OF ALGEBRAICEQUATIONS CAN BE EFFICIENTLY DONEUSING MATRIX ANALYSIS

LEARNING BY DOING

@ NODE 1 WE VISUALIZE THE CURRENTSLEAVING AND WRITE THE KCL EQUATION

REPEAT THE PROCESS AT NODE 2

03

12

4

122

R

vv

R

vvi

OR VISUALIZE CURRENTS GOING INTO NODE

WRITE THE KCL EQUATIONS

mA15

A

B

C

k8 k8k2 k2

SELECT ASREFERENCE

AV

BV

MARK THE NODES (TO INSURE THAT NONE IS MISSING)

ANOTHER EXAMPLE OF WRITING KCL

WRITE KCL AT EACH NODE IN TERMS OFNODE VOLTAGES 015

82@ mA

k

V

k

VA AA

01528

@ mAk

V

k

VB BB

A MODEL IS SOLVED BY MANIPULATION OFEQUATIONS AND USING MATRIX ANALYSIS

THE NODE EQUATIONS

kRRkR

mAimAi BA

6,12

4,12

321

THE MODEL

REPLACE VALUES AND SWITCH NOTATIONTO UPPER CASE

NUMERICAL MODEL

USE GAUSSIAN ELIMINATION

ALTERNATIVE MANIPULATION

k12/*

k6/*

242

1223

21

21

VV

VV

RIGHT HANDSIDE IS VOLTS.COEFFS ARENUMBERS

][122 1 VV ADD EQS

equations) add (and3/*][604 2 VV

LEARNING EXAMPLE

SOLUTION USING MATRIX ALGEBRA

PLACE IN MATRIX FORM

USE MATRIX ANALYSIS TO SHOW SOLUTION

PERFORM THE MATRIX MANIPULATIONS

||

)(1

A

AAdjA

FOR THE ADJOINT REPLACEEACH ELEMENT BY ITSCOFACTOR

AND DO THE MATRIX ALGEBRA ...

kkkV

6

104

3

10118

332

1

SAMPLE

AN EXAMPLE OF NODE ANALYSIS

1@v

2@ v

3@ v

Rearranging terms ...

2&1 BETWEEN ESCONDUCTANC

3&1 BETWEEN ESCONDUCTANC

3&2 BETWEEN ESCONDUCTANC

NODE TO CONNECTED ESCONDUCTANC

COULD WRITE EQUATIONS BY INSPECTION

WRITING EQUATIONS “BY INSPECTION”

FOR CIRCUITS WITH ONLY INDEPENDENTSOURCES THE MATRIX IS ALWAYS SYMMETRIC

THE DIAGONAL ELEMENTS ARE POSITIVE

THE OFF-DIAGONAL ELEMENTS ARE NEGATIVE

Conductances connected to node 1

Conductances between 1 and 2

Conductances between 1 and 3

Conductances between 2 and 3

VALID ONLY FOR CIRCUITSWITHOUT DEPENDENTSOURCES

k

VV

k

VmAV

1264:@ 211

1

USING KCL

0126

2:@ 1222

k

VV

k

VmAV

BY “INSPECTION”

mAVk

Vkk

412

1

12

1

6

121

mAVkkk

212

1

6

1

12

12

LEARNING EXTENSION

mA6

1I

062:@ 11 mAmAIVKCL

2I

3I

036

6:@ 222 k

V

k

VmAV

mAI

mAI

VV

4

2

12

3

2

2

CURRENTS COULD BE COMPUTED DIRECTLYUSING CURRENT DIVIDER!!

mAmAkk

kI

mAmAkk

kI

4)6(63

6

2)6(63

3

3

2

LEARNING EXTENSION

IN MOST CASES THEREARE SEVERAL DIFFERENTWAYS OF SOLVING APROBLEM

NODE EQS. BY INSPECTION

mAVVk

6202

121

mAVkk

V 63

1

6

10 21

k

VI

k

VI

k

VI

3622

32

21

1

CIRCUITS WITH DEPENDENT SOURCES CANNOTBE MODELED BY INSPECTION. THE SYMMETRYIS LOST.

A PROCEDURE FOR MODELINGA PROCEDURE FOR MODELING•WRITE THE NODE EQUATIONS USING DEPENDENT SOURCES AS REGULAR SOURCES.•FOR EACH DEPENDENT SOURCE WE ADD ONE EQUATION EXPRESSING THE CONTROLLING VARIABLE IN TERMS OF THE NODE VOLTAGES

02

21

1

1

R

vv

R

vio

02

12

3

2

R

vv

R

viA

MODEL FOR CONTROLLING VARIABLE

3

2

R

vio

NUMERICAL EXAMPLE

mAvkk

vk

vkk

vkk

23

1

12

1

6

1

06

1

3

2

6

1

12

1

21

21

0111

223

121

v

RRv

RR

REPLACE AND REARRANGE

AivRR

vR

2

321

2

111

k4/*

k6/*

][123

02

21

21

VVV

VV

ADDING THE EQUATIONS ][125 2 VV

VV5

241

CIRCUITS WITH DEPENDENT SOURCES LEARNING EXAMPLE

LEARNING EXAMPLE: CIRCUIT WITH VOLTAGE-CONTROLLED CURRENT

WRITE NODE EQUATIONS. TREAT DEPENDENTSOURCE AS REGULAR SOURCE

EXPRESS CONTROLLING VARIABLE IN TERMS OFNODE VOLTAGES

FOUR EQUATIONS IN OUR UNKNOWNS. SOLVEUSING FAVORITE TECHNIQUE

OR USE MATRIX ALGEBRA

REPLACE AND REARRANGE

CONTINUE WITH GAUSSIAN ELIMINATION...

USING MATLAB TO SOLVE THE NODE EQUATIONS

]/[2

,4,2,4

,2,1

4

321

VA

mAimAikR

kRRkR

BA

» R1=1000;R2=2000;R3=2000;R4=4000; %resistances in Ohm» iA=0.002;iB=0.004; %sources in Amps» alpha=2; %gain of dependent source

DEFINE THE COMPONENTS OF THE CIRCUIT

DEFINE THE MATRIX G » G=[(1/R1+1/R2), -1/R1, 0; %first row of the matrix-1/R1, (1/R1+alpha+1/R2), -(alpha+1/R2); %second row0, -1/R2, (1/R2+1/R4)], %third row. End in comma to have the echo

G =

0.0015 -0.0010 0 -0.0010 2.0015 -2.0005 0 -0.0005 0.0008

Entries in a row areseparated by commas (or plain spaces).Rows are separated bysemi colon

DEFINE RIGHT HAND SIDE VECTOR » I=[iA;-iA;iB]; %end in ";" to skip echo

SOLVE LINEAR EQUATION» V=G\I % end with carriage return and get the echo

V = 11.9940 15.9910 15.9940

LEARNING EXTENSION: FIND NODE VOLTAGES

010

410

:@ 2111

k

VVmA

k

VV

NODE EQUATIONS

010

210

:@ 2122

k

VI

k

VVV O

CONTROLLING VARIABLE (IN TERMS ON NODEVOLTAGES)

k

VIO 10

1

REPLACE

01010

210

010

410

2112

211

k

V

k

V

k

VVk

VVmA

k

V

REARRANGE AND MULTIPLY BY 10k

02

][402

21

21

VV

VVV eqs. add and 2/*

VVVV 16805 11

VVV

V 82 21

2

O V VOLTAGETHE FIND

NODE EQUATIONS

063

2 k

V

k

VmA xx

NOTICE REPLACEMENT OF DEPENDENT SOURCE IN TERMS OF NODE VOLTAGE

012126

k

Vk

V

k

V OOx

k6/*

k12/*

][4022

][4][123

VVVV

VVVV

OxO

xx

LEARNING EXTENSION

3 nodes plus the reference. In principle one needs 3 equations...

…but two nodes are connected tothe reference through voltage sources. Hence those node voltages are known!!!

…Only one KCL is necessary

012126

12322

k

VV

k

VV

k

V

][5.1][64

0)()(2

22

12322

VVVV

VVVVV

EQUATIONS THE SOLVING

Hint: Each voltage sourceconnected to the referencenode saves one node equation

One more example ….

CIRCUITS WITH INDEPENDENT VOLTAGE SOURCES

][6

][12

3

1

VV

VV

THESE ARE THE REMAININGTWO NODE EQUATIONS

+-

2SI

3SI 1SV

1SI 1R2R

3R

4R

Problem 3.67 (6th Ed) Find V_0 R1 = 1k; R2 = 2k, R3 = 1k, R4 = 2kIs1 =2mA, Is2 = 4mA, Is3 = 4mA,Vs = 12 V

IDENTIFY AND LABEL ALL NODES

WRITE THE NODE EQUATIONS

][12:@ 33 VVVV VS VOLTAGENODE KNOWN

021

][2

0:@

121

4

1

1

2111

k

V

k

VVmA

R

V

R

VVIV S

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

02

][4][2

0:@

24

2

24214

k

VVmAmA

R

VVIIV SS

NOW WE LOOK WHAT IS BEING ASKED TO DECIDE THE SOLUTIONSTRATEGY.

210 VVV

1V2V 3V

4V

OV

FOR NEEDED AREONLY OVVV 21,

021

][2 121

k

V

k

VVmA

021

12

1][4 42212

k

VV

k

V

k

VVmA

02

][4][2 24

k

VVmAmA

TO SOLVE BY HAND ELIMINATE DENOMINATORS

][423 21 VVV */2k

*/2k][3252 421 VVVV

*/2k][442 VVV

Add 2+3 ][3642 21 VVV

][10][404 11 VVVV

][14][564 22 VVVV

FINALLY!! ][4210 VVVV

4

32

4

110

152

023

3

2

1

V

V

V

ALTERNATIVE: USE LINEAR ALGEBRA

(1)

(2)

(3)

So. What happens when sources are connected between two non reference nodes?

][423 21 VVV add and 2/*

We will use this example to introduce the concept of a SUPERNODE

Conventional node analysis requires all currents at a node

@V_1 06

6 1 SIk

VmA

@V_2 012

4 2 k

VmAIS

2 eqs, 3 unknowns...Panic!! The current through the source is notrelated to the voltage of the source

Math solution: add one equation

][621 VVV

Efficient solutionEfficient solution: enclose the source, and all elements in parallel, inside a surface.

Apply KCL to the surface!!!

04126

6 21 mAk

V

k

VmA

The source current is interiorto the surface and is not required

We STILL need one more equation

][621 VVV Only 2 eqs in two unknowns!!!

SUPERNODE

THE SUPERNODE TECHNIQUE

SI

k 12 / *

][6

046126

21

21

VVV

mAmAk

V

k

V

(2)

(1)

Equations The

ALGEBRAIC DETAILS

...by Multiply Eq(1). in rsdenominato Eliminate 1.

Solution

][6

][242

21

21

VVV

VVV

][10][303 11 VVVV 2 Veliminate to equations Add2.

][4][612 VVVV 2 Vcompute to Eq(2) Use 3.

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

][60

21

2

VVV

VV

:adding

TO COMPUTE THE POWER SUPPLIED BY VOLTAGE SOURCEWE MUST KNOW THE CURRENT THROUGH IT

VI

k

VVmA

k

VIV 10

610

211 mA8

BASED ON PASSIVE SIGN CONVENTION THEPOWER IS RECEIVED BY THE SOURCE!!

mWmAVP 160][8][20

WRITE THE NODE EQUATIONS

1@v

:supernode) (leavingKCL

:CONSTRAINT

SUPERNODE @

Avvv

32

THREE EQUATIONS IN THREE UNKNOWNS

LEARNING EXAMPLE

OI FIND

SUPERNODE

1231 VVCONSTRAINT SUPERNODE

123V

KCL @ SUPERNODE

VVVV 12,642 KNOWN NODE VOLTAGES

LEARNING EXAMPLE

VV

VV

4

6

4

1

SOURCES CONNECTED TO THE

REFERENCE

SUPERNODE

CONSTRAINT EQUATION VVV 1223

KCL @ SUPERNODE

02

)4(

212

63322

k

V

k

V

k

V

k

Vk2/*

VVV 22332VVV 12

32 add and 3/*

VV 385 3

mAk

VIO

8.32

3 LAW SOHM'

OIV FOR NEEDED NOT IS

2

LEARNING EXTENSION

+-

+ -

+-1R

2R

3R

4R

5R

6R

7R

Supernodes can be more complex

Identify all nodes, select areference and label nodes

Nodes connected to reference througha voltage source

Voltage sources in between nodes and possible supernodes

EQUATION BOOKKEEPING:KCL@ V_3, KCL@ supernode, 2 constraints equationsand one known node

KCL@V_3 07

3

5

43

4

23

R

V

R

VV

R

VV

KCL @SUPERNODE (Careful not to omit any current)

04

32

5

34

6

4

3

5

2

15

1

12

R

VV

R

VV

R

V

R

V

R

VV

R

VV

CONSTRAINTS DUE TO VOLTAGE SOURCES

11 SVV

252 SVVV

345 SVVV

5 EQUATIONS IN FIVE UNKNOWNS.

supernode

1V

2V 3

V

4V

5V

WRITE THE NODE EQUATIONS

CIRCUITS WITH DEPENDENT SOURCESPRESENT NO SIGNIFICANT ADDITIONAL COMPLEXITY. THE DEPENDENT SOURCESARE TREATED AS REGULAR SOURCES

WE MUST ADD ONE EQUATION FOR EACHCONTROLLING VARIABLE

VOLTAGE SOURCE CONNECTED TO REFERENCE

VV 31

0263

: 212

xI

k

V

k

VV VKCL@ 2

CONTROLLING VARIABLE IN TERMS OF NODE VOLTAGES k

VIx 6

2

REPLACE

06

263

2212

k

V

k

V

k

VVk6/*

VVVV 602212

mAk

VVIO

13

21

OI FIND

LEARNING EXAMPLE

SUPER NODE WITH DEPENDENT SOURCE

VOLTAGE SOURCE CONNECTED TO REFERENCE

VV 63

SUPERNODE CONSTRAINT xVVV 2

21

KCL AT SUPERNODE

k12/*

062)6(22211

VVVV

CONTROLLING VARIABLE IN TERMS OF NODES

2VV

x

213VV

183321 VV 184

1 V

CURRENT CONTROLLED VOLTAGE SOURCE

CONSTRAINT DUE TO SOURCE xkIVV 2

12

KCL AT SUPERNODE 02

22

4 21 k

VmA

k

VmA

CONTROLLING VARIABLE IN TERMS OF NODES

k

VIx 2

1121 22 VVkIV x

][421

VVV 02

21 VV

add and 2/*

][832

VV

mAk

VIO 3

4

22

k2 k3

k6k2xaI1000

xISV

An example with dependent sources

2 nodes are connected to the reference through voltage sources

EXPRESS CONTROLLING VARIABLE IN TERMS OF NODE VOLTAGES

k

VI XX 2

‘a’ has units of [Volt/Amp]

22XaV

V

What happens when a=8?

1V X

V 2V

X

S

aIV

VV

10002

1

0322

2 k

vV

k

V

k

VV XXSX

KCL @ Vx

k

aVkV X

2

*12

REPLACE Ix IN V2

SX

XXXSX

VVa

aVVVVV

3)8(

0)2/(23)(3

REPLACE V2 IN KCL

IDENTIFY AND LABEL NODES