Handouts 6002 L2 Oei12 Gaps

7/28/2019 Handouts 6002 L2 Oei12 Gaps

1/27

Basic Circuit Analysis Methods

(KVL and KCL method, Node method)

6.002xCIRCUITS ANDELECTRONICS


2/27

Remember, our EECS playgroun

Review

Observe the lumpedmatter discipline LMD


3/27

Lumped circuit element+

-

power consumed by element =

Review

i

v

vi


4/27

LMD allows us to create thelumped circuit abstraction

Review

+!! R1

R3R2

V


5/27

KVL:

For all loops

KCL:

For all nodes

0=j j

0=j ji

ReviewReview

Maxwells equations simplify toalgebraic KVL and KCL under LMD!

dlE =


6/27

DE

+!!

Review

R1 R4R3

R2 R5

d

c

b

a

V0


7/27

Lets Begin by Building aToolchest ofAnalysis Techniques

+!!

R1 R4

R3

R2 R5

Analyzingcircuit me

Find allelemenand is

V0


8/27

Method 1: Basic KVL, KCL method ofCircuit analysis

Goal: Find all element vs and is

1. write element v-i relationships(from lumped circuit abstraction)

2. write KCL for all nodes

3. write KVL for all loopslots of unknownslots of equationslots of fun

solve


9/27

!

i+

-

Element e

Current is taken to bepositive going into thepositive voltage terminal

Method 1: Basic KVL, KCL method of Circuit analy

Then powerconsumedby element e

is = vi

This

convecalledAssocvaria

discip

Goal: Find all element vs and isLabeling element vs and is


10/27

Method 1: Basic KVL, KCL method ofCircuit analysis

For R

For voltage source

For current source

You will need this for step 1: Element Relations


11/27

Lets Apply KVL, KCL Method to this

The Demo Circuit

+!!

R1 R4

R3

R2 R5

V0

Goal: Find all element vs and is


12/27

KVL, KCL Example

a

d

c

+!!

Note the use of associated var

R4

R5R2

R1

R3bV0

Label all vs and is

0

12 u

a


13/27

Step 1 of KVL, KCL Method12 unknowns

1. Element relationshipsiv,

c

1

+

2

+

0i

+!!V00+

a

5050, ii

L1

a


14/27

Step 2 of KVL, KCL Method

12 unknowns

2. KCL at the nodes

(use conventiosum currents the node)

5050, ii

c

1

+

2

+

0i

+!!V00+

a

L1

a


15/27

Step 3 of KVL, KCL Method12 unknowns

3. KVL for loops

(use conventyou go arounassign first sign to each

5050, ii

c

1

+

2

+

0i

+!!V00+

a

L1

KVL KCL M th d


16/27

KVL, KCL Method1. Element v, i relationships

v0 = V0v1 = i1R1

v2 = i2R2

v3 = i3R3v4 = i4R4

v5 = i5R5

2. KCL at the nodes

redundant

0410=++ iii

0132=+ iii

0435=

iii

0520= iii

a:

b:

d:

c:

3. KVL for loops

431=+ vvv

210=++ vvv

253

=+ vvv

540=++ vvv

L1:

L2:

L3:

L4:

Method 3 the node method will be much better!


17/27

Other Analysis MethodsMethod 2 Apply element combination rules

A R1 R2 R3 RN

B G2G1 GN


18/27

Method 2 Apply element combination ru


19/27

Method 2 Apply element combination ru

+!!

Example

R1

R3R2

V

Method 3 Node analysis


20/27

1.

2.

3.

4.

5.

Select reference node ( ground) fromwhich voltages are measured.

Label voltages of remaining nodes withrespect to ground. These are theprimary unknowns.

Write KCL for all but the ground node,substituting device laws and KVL.

Solve for node voltages.

Back solve for branch voltages andcurrents (i.e., the secondary unknowns).

ParticularapplicatioKVL, KCLmethod


6.002xworkhorse



21/27

Example: Old Faithful, plus current source

+!!

R1 R4R3

R2R5

V0


1. Select rground n

2. Label nowith resground.

g

I1

St 3 f N d M th dV0


22/27

Step 3 of Node Method

For convenience, writei

iR

G1

=

To avoid mistakes, use convention

E.g., always sum the currents leaving a node

+!! e1

R

V0

V0

3. Writnodes

devic


V0


23/27


Move constant terms to RHS & collect unknowns

2 equations, 2 unknowns Solve for es

(compare units)

0)()()( 21321101 =++ GeGeeGVe

0)()()( 152402312 =++ IGeGVeGee

KCL at e1

KCL at e2

4. Solvvolt

+!! e1

R

V0

0


V0


24/27


5. Bacbran

curr

+!! e1

R

V0

0

e1 e2Once you have solved for and ,easy to find branch vs and is

For example:

i+

v1

Revisit Step 4 of Node Method for Cultural Intere


25/27

In matrix form:

=

++

++

04

1

2

1

5433

3321

VG

VG

e

e

GGGG

GGGG

conductivitymatrix

unknownnode

voltages

Revisit Step 4 of Node Method for Cultural Intere

)()()( 10323211 GVGeGGGe =+++

140543231 )()()( IGVGGGeGe +=+++

4. Solve for no

Step 4 of Node Method 4 S l f


26/27

+

=

++

++

104

01

2

1

5433

3321

IVG

VG

e

e

GGGG

GGGG

( )( ) 23543321

104

01

3213

3543

2

1

GGGGGGG

IVG

VG

GGGG

GGGG

e

e

++++

+

++

++

=

Solve

5G

3G

4G

3G

2

3G

5G

2G

4G

2G

3G

2G

5G1

G4

G1

G3

G1

G

1I0V4G3G0V1G5G4G3G

1e

++++++++

++++

=

( )( ) ( )( )

5343

2

3524232514131

104321013

2

GGGGGGGGGGGGGGGGG

IVGGGGVGGe

++++++++

++++

=

(same denominator)

Nin

nein th

Step 4 of Node Method 4. Solve for no

Step 4 of Node Method V0


27/27

E.g., solve for , given

K2.8

1

G

G

5

1=

K9.3

1

G

G

4

2=

K5.1

1G

3= 0

1=I

026.0 Ve =

If , thenVV 30= 02

8.1 Ve =


+!! e1

R

V0

e2

Handouts 6002 L2 Oei12 Gaps

Documents

Transcript of Handouts 6002 L2 Oei12 Gaps