Basic Circuit Analysis Method

6.002

CIRCUITS AND ELECTRONICS

Basic Circuit Analysis Method (KVL and KCL method)

6.002 Fall 2000

Lecture 2

1

Review Lumped Matter Discipline LMD:

Constraints we impose on ourselves to simplify our analysis

∂φ B =0 ∂t ∂q =0 ∂t

Outside elements Inside elements wires resistors sources

Allows us to create the lumped circuit abstraction

6.002 Fall 2000

Lecture 2

2

Review

LMD allows us to create the lumped circuit abstraction i

+

v

Lumped circuit element

power consumed by element = vi

6.002 Fall 2000

Lecture 2

3

Review Review Maxwell’s equations simplify to algebraic KVL and KCL under LMD! KVL:

∑ jν j = 0 loop

KCL:

∑jij = 0 node

6.002 Fall 2000

Lecture 2

4

Review a R1

+ –

b

R4

R3

R2

d R5

c

DEMO

6.002 Fall 2000

vca + vab + vbc = 0

KVL

ica + ida + iba = 0

KCL

Lecture 2

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Method 1: Basic KVL, KCL method of Circuit analysis Goal: Find all element v’s and i’s 1. write element v-i relationships (from lumped circuit abstraction) 2. write KCL for all nodes 3. write KVL for all loops

lots of unknowns lots of equations lots of fun solve

6.002 Fall 2000

Lecture 2

6

Method 1: Basic KVL, KCL method of Circuit analysis

Element Relationships For R,

V = IR

For voltage source, V = V0

R +–

V0 For current source, I = I 0 J Io 3 lumped circuit elements

6.002 Fall 2000

Lecture 2

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KVL, KCL Example a +

ν1 +

ν 0 = V0 –

–

+ –

+

ν2 –

+

R1

R3

b

+ν 3 – R2

ν4 –

R4

d +

ν5 –

R5

c The Demo Circuit

6.002 Fall 2000

Lecture 2

8

Associated variables discipline i

+ ν

Element e

Current is taken to be positive going into the positive voltage terminal

Then power consumed by element e

6.002 Fall 2000

Lecture 2

= νi is positive

9

KVL, KCL Example a +

+

ν 0 = V0 –

+ –

ν1

i0

L1

–

+

ν2 –

i4 i1 L 2 + R1 ν 4 R4 – R3 b i3 d +ν 3 – i2 i5 + R2 ν 5 R5 L3 –

c The Demo Circuit

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Lecture 2

L4

10

Analyze ν 0 …ν 5 ,ι0 …ι5 1. Element relationships (v, i ) given v3 = i3 R3 v0 = V0 v4 = i4 R4 v1 = i1 R1 v5 = i5 R5 v2 = i2 R2

12 unknowns 6 equations

2. KCL at the nodes a: i0 + i1 + i4 = 0 3 independent b: i2 + i3 − i1 = 0 equations d: i5 − i3 − i4 = 0 e: − i0 − i2 − i5 = 0 redundant 3. KVL for loops L1: − v0 + v1 + v2 = 0 3 independent equations L2: v1 + v3 − v4 = 0 L3: v3 + v5 − v2 = 0 s L4: − v0 + v4 + v5 = 0 redundant n o i t ns a w u o n k eq n u 1 2 12

/

ugh @#! 6.002 Fall 2000

Lecture 2

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Other Analysis Methods Method 2— Apply element combination rules

A B

C

D

R1

R2 R3

G1

G2

V1

V2

+–

+–

GN

⇔

⇔

⇔

⇔

R1 + R2 +

G1 + G2

1 Gi = Ri

+ RN

+ GN

V1 + V2 +–

J

I2

J

J

I1

…

RN

I1 + I 2

Surprisingly, these rules (along with superposition, which you will learn about later) can solve the circuit on page 8

6.002 Fall 2000

Lecture 2

12

Other Analysis Methods Method 2— Apply element combination rules

I =?

Example

R1

V + –

R3

R2

I

I V + –

R1

R2 R3 R2 + R3

V + –

R = R1 +

R R2 R3 R2 + R3

V I= R 6.002 Fall 2000

Lecture 2

13

Method 3—Node analysis Particular application of KVL, KCL method 1. Select reference node ( ground) from which voltages are measured. 2. Label voltages of remaining nodes with respect to ground. These are the primary unknowns. 3. Write KCL for all but the ground node, substituting device laws and KVL. 4. Solve for node voltages. 5. Back solve for branch voltages and currents (i.e., the secondary unknowns)

6.002 Fall 2000

Lecture 2

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Example: Old Faithful plus current source

V0

Step 1

6.002 Fall 2000

e2

R2

R5

J

+ V e1 – 0

R4

R1 R 3

I1

Step 2

Lecture 2

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Example: Old Faithful plus current source

V0

R2

R4 e2

R5

J

+ V e1 – 0

R1 R 3

for I1 convenience, write 1 Gi = Ri

KCL at e1 (e1 − V0 )G1 + (e1 − e2 )G3 + (e1 )G2 = 0

KCL at e2 (e2 − e1 )G3 + (e2 − V0 )G4 + (e2 )G5 − I1 = 0 Step 3

6.002 Fall 2000

Lecture 2

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Example: Old Faithful plus current source

V0 e2

R2

R5

J

+ V e1 – 0

R4

R1 R 3

I1

Gi =

KCL at e1 (e1 − V0 )G1 + (e1 − e2 )G3 + (e1 )G2 = 0

1 Ri

KCL at l2 (e2 − e1 )G3 + (e2 − V0 )G4 + (e2 )G5 − I1 = 0 move constant terms to RHS & collect unknowns

e1 (G1 + G2 + G3 ) + e2 (−G3 ) = V0 (G1 ) e1 (−G3 ) + e2 (G3 + G4 + G5 ) = V0 (G4 ) + I1 2 equations, 2 unknowns (compare units) 6.002 Fall 2000

Lecture 2

Solve for e’s Step 4 17

In matrix form: G1 + G2 + G3  − G3 

− G3   e1   G1V0  = G V + I  G3 + G4 + G5  e2   4 0 1

conductivity matrix

sources

unknown node voltages

Solve G3 G3 + G4 + G5   G1V0  G3 G1 + G2 + G3  G4V0 + I1   e1   e  = (G1 + G2 + G3 )(G3 + G4 + G5 ) − G3 2  2

(

)(

) ( )(

)

G +G +G G V + G G V + I 3 4 5 1 0 3 4 0 1 e = 1 G G +G G +G G +G G +G G +G G +G 2 +G G +G G 1 3 1 4 1 5 2 3 2 4 2 5 3 3 4 3 5 e2 =

(G3 )(G1V0 ) + (G1 + G2 + G3 )(G4V0 + I 1 ) 2

G1G3 + G1G4 + G1G5 + G2G3 + G2G4 + G2 G5 + G3 + G3G4 + G3G5

(same denominator)

Notice: linear in V0 , I1 , no negatives in denominator 6.002 Fall 2000

Lecture 2

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Solve, given G1  1 =  G5  8.2 K

G2  1 = G4  3.9 K

1 G3 = 1.5 K

I1 = 0

(

)(

)

G G V + G +G +G G V + I e = 3 10 1 2 3 40 1 2 G + G + G + G + G + G −G 2 1 2 3 3 4 5 3 1 1 1 G +G +G = + + =1 1 2 3 8.2 3.9 1.5

(

G3 + G4 + G5 =

)(

)

1 1 1 + + =1 1.5 3.9 8.2

1 1 1 × + 1× 3.9 V e2 = 8.2 1.5 0 1 1− 2 1.5

Check out the DEMO

e2 = 0.6V0

If V0 = 3V , then e2 = 1.8V0 6.002 Fall 2000

Lecture 2

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