LEARNING EXAMPLE

CONVERT THIS Y INTO A DELTA?

SHOULD KEEP THESE TWO NODES!

IF WE CONVERT TO Y INTO A DELTA THERE ARE SERIES PARALLEL REDUCTIONS!

++=

RRRRRRRR

b

accbba1

3*12 *12 3612k k k

kΩ Ω

= = ΩΩ

++

++=

RRRRRRR

RRRRRRR

R

c

accbba

b

2

12kΩ

36k12k

THE RESULTINGCIRCUIT IS A CURRENT DIVIDER

Δ−

++=

YR

RRRRRRRa

accbba3 4mA 36k

36k12k OV

+

−

CIRCUIT AFTER PARALLEL RESISTORREDUCTION

4mA 36k

36 ||12 9k k k=

+

REDUCTION

OI36 8k

OV−

9k 36 8436 18 3O

kI mA mAk k

= × =+

89 9 243O OV k I k mA V= Ω× = Ω× = NOTICE THAT BY KEEPING

THE FRACTION WE PRESERVEFULL NUMERICAL ACCURACY

NODAL AND LOOP ANALYSIS TECHNIQUES

NODAL ANALYSISLOOP ANALYSIS

Develop systematic techniques to determine all the voltagesp y q gand currents in a circuit

NODE ANALYSIS

• One of the systematic ways to determine every voltage and y gcurrent 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 nodereference node

A PROBLEM SOLVED BEFORE USING SERIES/PARALLEL RESISTOR COMBINATIONS

COMPUTE ALL THE VOLTAGES AND CURRENTS IN THIS CIRCUITCOMPUTE ALL THE VOLTAGES AND CURRENTS IN THIS CIRCUIT

k12kk 12||4

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

V

k63I

kVI a62 = :SOHM' 0321 =−− III :KCL

3*3 IkVb = :SOHM' …OTHER OPTIONS...

12

3

4

34

*4124

12

IkV

II

b =+

=

345 0III =−+ :KCL

kk 6||6FIRST REDUCE TO A SINGLE LOOP CIRCUIT

5*3 IkVC = :SOHM'

V12kVI

1212

1 =)12(

933+

=aV

DEFINING THE REFERENCE NODE IS VITAL

−+ 12V

+V4

−V2

− +

SMEANINGLESIS4V VSTATEMENT THE 1 =

UNTIL THE REFERENCE POINT IS DEFINEDUNTIL THE REFERENCE POINT IS DEFINED

BY CONVENTION THE GROUND SYMBOLSPECIFIES THE REFERENCE POINT.

ALL NODE VOLTAGES ARE MEASURED WITHRESPECT TO THAT REFERENCE POINT

THE STRATEGY FOR NODE ANALYSIS1. IDENTIFY ALL NODES AND SELECT

A REFERENCE NODESV aV bV cV

2. IDENTIFY KNOWN NODE VOLTAGES

3. AT EACH NODE WITH UNKNOWN VOLTAGE WRITE A KCL EQUATION

( 0)(e.g.,SUM OF CURRENT LEAVING =0)

0:@ 321 =++− IIIVa

4. REPLACE CURRENTS IN TERMS OF NODE VOLTAGES

VVVVV

REFERENCE

0369

=−

++−

kVV

kV

kVV baasa AND GET ALGEBRAIC EQUATIONS IN

THE NODE VOLTAGES ...

0:@ 543 =++− IIIVb

0:@ =+− IIV

0943

=−

++−

kVV

kV

kVV cbbab

SHORTCUT:SHORTCUT: SKIP WRITING THESE EQUATIONS...

AND PRACTICE WRITING 0:@ 65 =+− IIVc

039

=+−

kV

kVV cbc

THESE DIRECTLY

CIRCUITS WITH ONLY INDEPENDENT SOURCES

@ NODE 1

02

21

1

1 =−

++−R

vvRviA SRESISTANCE USING

@ NODE 2@ NODE 2

THE MODEL FOR THE CIRCUIT IS A SYSTEMOF ALGEBRAIC EQUATIONS

EXAMPLE

WRITE THE KCL EQUATIONS

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

REPEAT THE PROCESS AT NODE 2REPEAT THE PROCESS AT NODE 2

03

12

4

122 =

−+

−+−

Rvv

Rvvi

LEARNING EXTENSION: FIND NODE VOLTAGES V1 and V2

REARRANGE AND MULTIPLY BY 10k

][402 21 =− VVV eqs.addand2/*02

][402

21

21

=+ VVVVV eqs.addand 2/

VVVV 16805 11 =⇒=

04:@ 2111 =

−+−

VVmAVV

NODE EQUATIONSVVVV 8

2 21

2 −=⇒−=

010

410

:@ 1 +k

mAk

V

010

210

:@ 2122 =++

−k

VIkVVV O

CONTROLLING VARIABLE (IN TERMS ON NODEVOLTAGES)

kVIO 10

1=k10

REPLACE

010

410

211 =−

+−kVVmA

kV

01010

210

10102112 =++

−k

Vk

VkVV

kk

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

CIRCUITS WITH INDEPENDENT VOLTAGE SOURCES Find V1, V2 and V3

principle one needs 3 equations...

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

O l C i …Only one KCL is necessary

012126

12322 =−

+−

+kVV

kVV

kV

Q O SSO GHi t E h lt

][6][12

3

1

VVVV

−==

THESE ARE THE REMAININGTWO NODE EQUATIONS

0)()(2 12322 VVVVV =−+−+EQUATIONSTHESOLVINGHint: Each voltage source

connected to the referencenode saves one node equation

][5.1][64 22 VVVV =⇒=

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

THE SUPERNODE TECHNIQUE

SUPERNODE

SI

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

Conventional node analysis requires all currents at a node

@V 1 06 1 =++ IVmA

parallel, inside a surface.

Apply KCL to the surface!!!

046 21 =+++− mAVVmA@V_1 06

6 =++− SIk

mA

@V_2 012

4 2 =++−k

VmAIS

04126

6 +++ mAkk

mA

The source current is interiorto the surface and is not required

2 eqs, 3 unknownsThe current through the source is notrelated to the voltage of the source

h l i dd i

We STILL need one more equation

][621 VVV =−Math solution: add one equation

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

Equations The

ALGEBRAIC DETAILS

][6

046126

21

VVV

mAmAk

Vk

V=+−+

(2)

(1)

][621 VVV =− (2)

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

][6][242

21

21

VVVVVV

=−=+

][21

][10][303 11 VVVV =⇒=2Veliminateto equations Add2.

][][ 11

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

][][612 VVVV

SUPERNODE

LEARNING EXTENSION

VV 6 SOURCES CONNECTED TO THE

VVVV4

6

4

1

−== SOURCES CONNECTED TO THE

REFERENCE

CONSTRAINT EQUATION VVV 1223 =−

KCL @ SUPERNODE

02

)4(212

6 3322 =−−

+++−

kV

kV

kV

kV

k2/*

VVV 223 32 =+VVV 1232 =+− addand 3/*VV 385 3 =

V

OIV FORNEEDED NOT IS 2

mAk

VIO 8.32

3 == LAW SOHM'

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

WE MUST ADD ONE EQUATION FOR EACHWE MUST ADD ONE EQUATION FOR EACHCONTROLLING VARIABLE

SUPER NODE WITH DEPENDENT SOURCE Find Io

VOLTAGE SOURCE CONNECTED TO REFERENCE

VV 63 =

SUPERNODE CONSTRAINT xVVV 221 =−

KCL AT SUPERNODE

CONTROLLING VARIABLE IN TERMS OF NODES

2VVx = 21 3VV =⇒

k12/*

062)6(2 2211 =−+++− VVVV1833 VV 184V1833 21 =+ VV 184 1 =⇒ V