EMLAB

1

Chapter 3. Nodal and loop analysis techniques

2014. 9. 12.

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2Contents

1. Nodal analysis

2. Loop analysis

3. Application examples

4. Design examples

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33.1 Nodal analysis

• Unknowns are node voltages.

• The node voltages are defined with respect to a common point in the circuit which is called ground.

• Ground node is the one to which the largest number of branches are connected.

• Ground is at zero potential, and it sometimes represents the chassis or ground line in a practical circuit.

• Once ground node in an N-node circuit has been selected, our task is reduced to identifying the remaining non-ground nodes and writing one KCL equation at each of them.

unknowns

0 [V]

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4Example 3.1

Suppose that the network in the figure has the following parameters: IA = 1mA, R1 = 12kΩ, R2= 6k Ω, IB=4mA, R3= 6k Ω. Let us determine all node voltages and branch currents.

Node 1: 021 iiiA

Node 2: 032 iii B

00

2

21

1

1

RR

iA

00

3

2

2

12

Ri

R B

AiRRR

2

21

21

111 BiRRR

2

322

1 11

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5

m

m

V

V

kk

kk4

1

3

1

6

16

1

4

1

2

1

48

12

42

23

2

1

V

V

15

6

366

2412

2

1

12

3

32

24

2

1

48

12

32

24

8

1

2

1

V

V

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6Example 3.2Let us apply what we have just learned to write the equations for the network in the figure by inspection. (IA = 4mA, IB=2mA, R1 = R2= 2kΩ, R3= R4= 4kΩ, R5= 1kΩ)

00

1

31

2

1

Ri

R A

00

4

32

3

2

R

iR

i BA

00

1

13

5

3

4

23

RRR

υ1 = -4.3636 Vυ2 = 3.6364 Vυ3 = -0.7273 V

011111

1110

10

11

3

2

1

54141

443

121

BA

A

ii

i

RRRRR

RRR

RRR

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7Example 3.4

Let us determine the set of linearly independent equations that when solved will yield the node voltages in the network in Fig. 3.10. Then given the following com-ponent values, we will compute the node voltages using MATLAB:R1 = 1 kΩ, R2 = R3 = 2 kΩ, R4 = 4 kΩ, iA = 2 mA, iB = 4 mA, and α= 2.

00

1

21

3

1

AiRR

02

32

1

12

Ri

R xA

04

3

2

23

BiRR

)( 32 x0)(

2

3232

1

12

Ri

R A

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8

00

1

21

3

1

AiRR

04

3

2

23

BiRR

B

A

A

i

i

i

RRR

RRRR

RRR

3

2

1

422

2211

231

1110

1111

0111

0)(2

3232

1

12

Ri

R A

][9940.15

][9910.15

][9940.11

3

2

1

V

V

V

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9Example 3.6

Suppose we wish to find the currents in the two resistors in the circuit of the figure.

I

6,0412

0

06

06

212

1

kI

Ik

m6,04

12

0

6

06 21

21

kk

m

Super node

][4],[10 21 VV

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103.2 Loop analysis

• In contrast to nodal analysis, a loop analysis uses KVL to determine a set of loop currents in the circuit.

• There are exactly (E-V+1) linearly independent KVL equations for any net-work, where E is the number of branches in the circuit and V is the number of nodes(vertices). (Euler’s theorem for planar graphs)

• For each loop path, KCL is automatically satisfied.

D

DC

C

DB

CA

B

BA

A

ii

iii

ii

iii

iii

ii

iii

ii

8

7

6

5

4

3

2

1

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11Euler’s theorem for planar graph

v : number of verticese : number of edges (branches).f : number of facets

• for graphs with vertices and edges : v - e = 1• for planar graphs with vertices, edges, and facets : v – e + f = 1• for solid with vertices, edges, and facets : v – e + f = 2

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12Example 3.12

033)(6

0)(6612

212

211

IkIIk

IIkIk

Consider the network in the figure. We wish to find the current Io.

396

12612

21

21

IkIk

IkIk

][75.0

][5.0],[25.1

210

21

mAIII

mAImAI

033)(612

06)(612

221

121

IkIIk

IkIIk

Alternately, loop paths could be chosen as follows, which yields the same solution.

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13Example 3.13

012)(6)(3

0)(369

0)(646

31323

322

311

IkIIkIIk

IIkIk

IIkIk

0

6

6

2136

3120

6010

3

2

1

I

I

I

kkk

kk

kk

][1261.0

][4685.0

][6757.0

3

2

1

VI

VI

VI

Let us write the mesh equations by inspection for the network

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14Example 3.16Let us find Io.

][2,4

0)(116

02)(2

132

133

212

mAImII

IIkVIk

IkVIIk

x

x

0)(1)(2

)(216

][4],[2

13132

323

21

IIkIIIk

IIkIk

mAImAI

][3

23 mAI mII

IkIk

4

02412

32

32

][3

10],[

3

223 mAImAI

][3

42 221 mAImIIIo

][3

4321 mAIIIIo

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15Example 3.18

Let us find V0 in the circuit in the figure, which contains a voltage-controlled current source

)(4

06)(23

][2,2000

12

313

21

IIkV

IkIIk

mAIV

I

x

x

114238

][482

3

11

mkIk

mAIIk

][25.84

33],[

8

1103 VVmAI

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16Example 3.19

The network in the figure contains both a current-controlled voltage source and a voltage controlled current source. Determine the loop currents.

241321

2434

4313

),(2,2

],[4

012)(1)(1

0)(1)(21

IIIIIkVk

VImAI

IIkIIk

IIkIIkIk

xxx

x

0

12211

8231

321

423

432

III

IkIkIk

IkIkIk

][1],[2],[6 432 mAImAImAI

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17Some Matlab command

754

132A A= [2,3,1; 4, 5, 7]

97531A A= [1:2:9]

TAA A= A’

)(inv AA= A-1

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18

01

11

1 axaxaxay nn

nn

);x,a(polyvaly

];a,a,,a,a[a

1.e3];:2:[1x

011nn

01

11

1

01

11

1

bxbxbxb

axaxaxaH

nn

nn

mm

mm

./den;numH

x);polyval(b,den

x);polyval(a,num

];b,b,,b,[bb

];a,a,,a,[aa

1e3];:2:1[x

011nn

011mm