CIRCUITS AND ELECTRONICS. Basic Circuit Analysis Method (KVL and KCL method)

(1)

6.002 CIRCUITS

AND

ELECTRONICS

Basic Circuit Analysis Method

(KVL and KCL method)

(2)

0 =

∂

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

(3)

LMD allows us to create the

lumped circuit abstraction

Lumped circuit element

+

-v

i

power consumed by element =

vi

(4)

KVL:

0 loop

KCL:

=

∑

j

ν

j

0 =

∑

j j

i

Review

Maxwell’s equations simplify to

algebraic KVL and KCL under LMD!

(5)

KVL

0 =

+

_ab

_bc

ca

v

0 =

+

_da

_ba

ca

i

KCL

DEMO

1 R

2 R

4 R

5 R

3 R

a

b

_d

c

+

–

Review

(6)

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

(7)

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

(8)

KVL, KCL Example

The Demo Circuit

+

–

1 R

2 R

4 R

5 R

3 R

a

b

_d

c

0

0 =

V

ν

+

–

1 ν

+

–

5 ν

+

–

3 ν

+ –

2 ν

+

–

4 ν

+

–

(9)

Associated variables discipline

ν

i

+

-Element e

Then power consumed

by element e

=

ν

i

is positive

Current is taken to be positive going

into the positive voltage terminal

(10)

KVL, KCL Example

The Demo Circuit

+

–

1 R

2 R

4 R

5 R

3 R

a

b

_d

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

(11)

Analyze

12 unknowns

5

0

5

0 ν

,

ι

ν

…

1. Element relationships

3. KVL for loops

0

0 V

v

=

1

1 i

R

v

=

2

2 i

R

v

=

3

3 i

R

v

=

4

4 i

R

v

=

5

5 i

R

v

=

given

2. KCL at the nodes

redundant

0

4

3

1 +

v

−

v

=

v

0

2

1

0 +

+

=

−

v

0

2

5

3 +

v

−

v

=

v

0

5

4

0 +

+

=

−

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

0 −

−

=

−

i

a:

b:

d:

e:

6 equations

3 independent

equations

3 independent

equations

12 un

know

ns

12 equ

ati

ons

/

( )

v

,

i

L1:

L2:

L3:

L4:

(12)

Other Analysis Methods

Method 2— Apply element combination rules

B

C

D

⇔

R

1 +

R

2 +

+

R

N

⇔

1 G

G

₂

G

_N

G

₁

+

G

₂

+

G

_N

i

R

G

=

1 ⇔

+ –

1 V

V

₂

V

₁

+

V

₂

⇔

J

I

J

I

+

I

A

R

1 R

2 R

3 …

R

N

(13)

Other Analysis Methods

Method 2— Apply element combination rules

V

I

3

2

3

2 R

R

+

V

I

3

2

3

2

1 R

R

+

=

+

–

V

?

=

I

1 R

3 R

2 R

+

–

+

–

_R

Example

1 R

V

=

(14)

1.

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

(15)

Example: Old Faithful

plus current source

0 V

1 R

2 R

4 R

5 R

3 R

J

I

1

0 V

+

–

e

1

2 e

Step 1

_{Step 2}

(16)

Example: Old Faithful

plus current source

0 )

(

)

(

)

(

e

₁

−

V

₀

G

₁

+

e

₁

−

e

₂

G

₃

+

e

₁

G

₂

=

KCL at

e

1

0 )

(

)

(

)

(

e

−

e

G

+

e

−

V

G

+

e

G

−

I

=

KCL at

e

₂

for

convenience,

write

i

R

G

=

1

0 V

1 R

2 R

4 R

5 R

3 R

J

1 e

1 I

0 V

+

–

2 e

(17)

Example: Old Faithful

plus current source

0 )

(

)

(

)

(

e

₁

−

V

₀

G

₁

+

e

₁

−

e

₂

G

₃

+

e

₁

G

₂

=

KCL at

e

₁

0 )

(

)

(

)

(

e

₂

−

e

₁

G

₃

+

e

₂

−

V

₀

G

₄

+

e

₂

G

₅

−

I

₁

=

KCL at

l

₂

move constant terms to RHS & collect unknowns

)

(

)

(

)

(

₁

₂

₃

₂

₃

₀

₁

1 G

G

e

G

V

G

e

+

−

=

1

4

0

5

4

3

2

3

1 (

G

)

e

(

G

)

V

(

G

)

I

e

−

+

=

+

i

R

G

=

1 2 equations, 2 unknowns

Solve for e’s

0 V

1 R

2 R

4 R

5 R

3 R

J

1 e

1 I

0 V

+

–

2 e

(18)

In matrix form:

⎥

⎦

⎤

⎢

⎣

⎡

+

=

⎥

⎦

⎤

⎢

⎣

⎡

⎥

⎦

⎤

⎢

⎣

⎡

+

−

+

1 0 4 0 1 2 1 5 4 3 3 3 3 2 1

I

V

G

V

G

e

G

conductivity

matrix

unknown

node

voltages

sources

(

)(

)

2 3 5 4 3 3 2 1 1 0 4 0 1 3 2 1 3 3 5 4 3 2 1

G

I

V

G

V

G

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

+

=

( )(

) (

)(

)

2 1 0 4 3 2 1 0 1 3 2

G

I

V

G

V

G

e

+

=

(19)

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

(

)(

)

(

) (

)

2

3 G

5 G

4 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

9 .

3

1

2 .

8

1

3 G

2 G

1 G

+

=

+

=

1

2 .

8

1

9 .

3

1

5 .

1

1 G

G

₃

+

₄

+

₅

=

+

=

0

2

2 V

5 .

1

9 .

3

1

5 .

1

2 .

8

1 e

−

×

+

×

=

0

2

0 .

6 V