Circuits & Electronics P2

6.002 CIRCUITS ANDELECTRONICS Basic Circuit Analysis Method KVL and KCL method... LMD allows us to create the lumped circuit abstraction Lumped circuit element + -v i power consumed by

6.002 CIRCUITS AND

ELECTRONICS

Basic Circuit Analysis Method

(KVL and KCL method)

=

∂

∂

t

B

φ

0

=

∂

∂

t q

Outside elements Inside elements

Allows us to create the lumped circuit

abstraction

wires resistors sources

Review

Lumped Matter Discipline LMD:

Constraints we impose on ourselves to simplify our analysis

LMD allows us to create the

lumped circuit abstraction

Lumped circuit element +

-v

i

power consumed by element = vi

Review

loop

KCL:

node

0

=

∑jν j

0

=

∑j j i

Review Review

Maxwell’s equations simplify to

algebraic KVL and KCL under LMD!

KVL 0

= +

+ ab bc

v

0

= +

+i i

DEMO

1

R

2

R

4

R

5

R

3

R

a

c

+

–

Review

Method 1: Basic KVL, KCL method of

Circuit analysis

Goal: Find all element v’s and i’s

write element v-i relationships (from lumped circuit abstraction) write KCL for all nodes

write KVL for all loops

1

2

3

lots of unknowns lots of equations lots of fun

solve

Method 1: Basic KVL, KCL method of

Circuit analysis

For R,

For voltage source,

For current source,

Element Relationships

IR

V =

0

V

V =

0

I

I =

3 lumped circuit elements

R

0

V

o

I

+ –

J

KVL, KCL Example

The Demo Circuit

+ –

1

R

2

R

4

R

5

R

3

R

a

c

0

0 =V

ν+

–

1

ν+ –

5

ν+ –

3

ν + –

2

ν+ –

4

ν+ –

Associated variables discipline

ν

i

+

-Element e

Then power consumed

by element e =νi is positive Current is taken to be positive going

into the positive voltage terminal

KVL, KCL Example

The Demo Circuit

+ –

1

R

2

R

4

R

5

R

3

R

a

c

0

0 =V

ν+

–

1

ν+ –

5

ν+ –

3

ν + –

1

L

2

L

4

L

3

L

2

ν+ –

4

ν+ –

2

i

1

i 0

i

5

i

3

i

4

i

12 unknowns 5

0 5

0 ν , ι ι

1 Element relationships

3 KVL for loops

0

v =

1 1

1 i R

v =

2 2

2 i R

v =

3 3

3 i R

v =

4 4

4 i R

v =

5 5

5 i R

v = given

2 KCL at the nodes

redundant

0

4 3

1 + v − v =

v

0

2 1

−v v v

0

2 5

3 + v −v =

v

0

5 4

− v v v redundant

0

4 1

0 + i +i =

i

0

1 3

2 + i −i =

i

0

4 3

5 −i −i =

i

0

5 2

−i i i

a:

b:

d:

e:

6 equations

3 independent equations

3 independent equations

ns

12 equ

ations

/

( )v, i

L1:

L2:

L3:

L4:

Other Analysis Methods Method 2— Apply element combination rules

B

C

D

⇔ R1 + R2 + + R N

⇔

1

G G2 G N G1 + G2 +G N

i

i

R

G = 1

⇔

1

V V2 V1 +V2

⇔

1

I I2 J I1 + I2

A R1 R2 R3 … R N

Other Analysis Methods Method 2— Apply element combination rules

V

I

3 2

3 2

R R

R

R

+

V

I

3 2

3

2 1

R R

R

R R

R

+

+

=

+ –

V

?

=

I

1

R

3

R

2

R

+ –

+

Example

1

R

V

2

3

4

5

Select reference node ( ground)

from which voltages are measured

Label voltages of remaining nodes

with respect to ground

These are the primary unknowns

Write KCL for all but the ground

node, substituting device laws and

KVL

Solve for node voltages

Back solve for branch voltages and

Particular application of KVL, KCL method Method 3—Node analysis

Example: Old Faithful

plus current source

0

V

1

R

2

R

4

R

5

R

3

R

J I1

0

V

+

Step 1 Step 2

Example: Old Faithful

plus current source

0 )

( )

( )

(e1 −V0 G1 + e1 −e2 G3 + e1 G2 = KCL at e1

0 )

( )

( )

(e2 − e1 G3 + e2 −V0 G4 + e2 G5 − I1 = KCL at e 2

for convenience, write

i

i

R

G = 1

0

V

1

R

2

R

4

R

5

R

3

R

1

e

1

I

0

V

+

–

2

e

Example: Old Faithful

plus current source

0 )

( )

( )

(e1 −V0 G1 + e1 −e2 G3 + e1 G2 = KCL at e1

0 )

( )

( )

(e2 − e1 G3 + e2 −V0 G4 + e2 G5 − I1 = KCL at l2

move constant terms to RHS & collect unknowns

) ( )

( )

) (

) (

)

i

i

R

G = 1

0

V

1

R

2

R

4

R

5

R

3

R

1

e

1

I

0

V

+

–

2

e

In matrix form:













+

=

























+ +

−

− +

+

1 0 4

0 1 2

1 5

4 3 3

3 3

2 1

I V G

V G e

e G

G G G

G G

G G

conductivity

matrix unknownnode

voltages

sources

3 5

4 3 3 2 1

1 0 4

0 1 3

2 1 3

3 5

4 3

2

1

G G

G G G G G

I V G

V G G

G G G

G G

G G e

e

− +

+ +

+













+













+ +

+ +

=













Solve

( )( ) ( )( )

5

G 3

G 4

G 3 G

2 3

G 5

G 2

G 4

G 2

G 3

G 2

G 5

G 1

G 4

G 1

G 3

G 1 G

1

I 0

V 4

G 3

G 0

V 1

G 5

G 4

G 3

G 1

e

+ +

+ +

+ +

+ +

+ +

+ +

=

5 3 4 3

2 3 5 2 4 2 3 2 5 1 4 1 3 1

1 0 4 3 2 1 0

1 3 2

G G G G G G G G G G G G G G G G G

I V G G G G V

G G e

+ +

+ +

+ +

+ +

+ +

+ +

=

Solve, given

K 2 8

1 G

G

5

1 =







K 9 3

1 G

G

4

2 =







K 5 1

1

G 3 =

0

1 =

I

3

G 5

G 4

G 3

G 3

G 2

G 1 G

1

I 0

V 4

G 3

G 2

G 1

G 0

V 1

G 3

G 2

e

− +

+ +

+ +

+ +

+

+

=

1 5 1

1 9

3

1 2

8

1 3

G 2

G 1

1 2 8

1 9

3

1 5

1

1 G

G

G 3 + 4 + 5 = + + =

0 2

5 1

1 1

9 3

1 1

5 1

1 2

8

1 e

−

× +

×

=