Systematic Circuit Analysis (T&R Chap 3)carmenere.ucsd.edu/.../mae140/f09/lectures/3analysis.pdf ·...

MAE140 Linear Circuits 40

Systematic Circuit Analysis (T&R Chap 3)

Node-voltage analysisUsing the voltages of the each node relative to a ground

node, write down a set of consistent linear equations forthese voltages

Solve this set of equations using, say, Cramer’s Rule

Mesh current analysisUsing the loop currents in the circuit, write down a set of

consistent linear equations for these variables. Solve.

This introduces us to procedures for systematicallydescribing circuit variables and solving for them


Nodal Analysis

Node voltagesPick one node as the ground nodeLabel all other nodes and assign voltages vA, vB, …, vN

and currents with each branch i1, …, iMRecognize that the voltage across a branch

is the difference between the end nodevoltagesThus v3=vB-vC with the directionas indicated

Write down the KCL relations at each nodeWrite down the branch i-v relations to express branch

currents in terms of node voltagesAccommodate current sourcesObtain a set of linear equations for the node voltages

v5

C

B

A

vC

vB

vA

v4

v3

v2

v1+

+

+

++

--

-

- -


Nodal Analysis – Ex 3-1 (T&R, 5th ed, p.72)

Apply KCL

Write the element/branch eqns

Substitute to get node voltage equations

Solve for vA, vB, vC then i0, i1, i2, i3, i4, i5

vCvB

vA

R4

R2R1

R3

iS2

iS1 i4i3

i2

i1i0

Reference node

i5


Nodal Analysis – Ex 3-1 (T&R, 5th ed, p.72)

Apply KCLNode A: i0-i1-i2=0Node B:i1-i3+i5=0Node C: i2-i4-i5=0

Write the element/branch eqnsi0=iS1 i3=G3vB

i1=G1(vA-vB) i4=G4vC

i2=G2(vA-vC) i5=iS2

Substitute to get node voltage equationsNode A: (G1+G2)vA-G1vB-G2vC=iS1

Node B: -G1vA+(G1+G3)vB=is2Node C: -G2vA+(G2+G4)vC=-iS2

Solve for vA, vB, vC then i0, i1, i2, i3, i4, i5

vCvB

vA

R4

R2R1

R3

iS2

iS1 i4i3

i2

i1i0

Reference node

i5

!!

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#

$$

%

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#

$$

%

&

!!

"

#

$$

%

&

'

=

+'

+'

''+

2

2

1

4202

0311

2121

Si

Si

Si

Cv

Bv

Av

GGG

GGG

GGGG


Systematic Nodal Analysis

Writing node equations by inspectionNote that the matrix equation looks

just like Gv=i for matrix G and vector v and iG is symmetric (and non-negative definite)

Diagonal (i,i) elements: sum of all conductances connectedto node i

Off-diagonal (i,j) elements: -conductance between nodes iand j

Right-hand side: current sources entering node iThere is no equation for the ground node – the column sums

give the conductance to ground

!!

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!!

"

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$$

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=

+'

+'

''+

2

2

1

4202

0311

2121

Si

Si

Si

Cv

Bv

Av

GGG

GGG

GGGG

vCvB

vA

R4

R2R1

R3

iS2

iS1 i4i3

i2

i1i0

Reference node

i5


Nodal Analysis Ex. 3-2 (T&R, 5th ed, p.74)

Node A:Conductances

Source currents entering

Node B:Conductances


Node C:Conductances


iS

vBvA vC

R/2

2R2R

R/2 R


Nodal Analysis Ex. 3-2 (T&R, 5th ed, p.74)

Node A:Conductances

G/2B+2GC=2.5GSource currents entering= iS

Node B:Conductances

G/2A+G/2C+2Gground=3G Source currents entering= 0

Node C:Conductances

2GA+G/2B+Gground=3.5G Source currents entering= 0

iS

vBvA vC

R/2

2R2R

R/2 R

!!!

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=!!!

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#

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!!!

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''

''

''

0

0

5.35.02

5.035.0

25.05.2 S

C

B

A i

v

v

v

GGG

GGG

GGG


Nodal Analysis – some points to watch

1. The formulation given is based on KCL with thesum of currents leaving the node0=itotal=GAtoB(vA-vB) +GAtoC(vA-vC)+…+ GAtoGroundvA+ileavingA

This yields0=(GAtoB+…+GAtoGround)vA-GAtoBvB-GAtoCvC…-ienteringA

(GAtoB+…+GAtoGround)vA-GAtoBvB-GAtoCvC…=ienteringA

2. If in doubt about the sign of the current source,go back to this basic KCL formulation

3. This formulation works for independent currentsourcesFor dependent current sources (introduced later) use your

wits


Nodal Analysis Ex. 3-2 (T&R, 5th ed, p. 72)

+

_

vA vB

500Ω1KΩ

2KΩ

iS1 iS2vO



Solve this using standard linear equation solversCramer’s ruleGaussian eliminationMatlab

+

_

vA vB

500Ω1KΩ

2KΩ

iS1 iS2vO

!"

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&

'=!

"

#$%

&!!"

#$$%

&

(('

('(''

''

2

1

33

33

105.2105.0

105.0105.1

S

S

B

A

i

i

v

v


Nodal Analysis with Voltage Sources

Current through voltage source is not computablefrom voltage across it. We need some tricks!They actually help us simplify things

Method 1 – source transformation

+_

Restof thecircuitvS

RS

vA

vB

Restof thecircuit

vA

vB

RSSR

Sv!

Then use standard nodal analysis – one less node!


Nodal Analysis with Voltage Sources 2

Method 2 – grounding one node

Restof thecircuit

vA

vB

+_ vS

This removes the vB variable – simpler analysisBut can be done once per circuit


Nodal Analysis with Voltage Sources 3

Method 3Create a supernode

Act as if A and B were one nodeKCL still works for this node

Sum of currents enteringsupernode box is 0

Write KCL at all N-3 other nodesN-2 nodes less Ground node

using vA and vB as usualWrite one supernode KCLAdd the constraint vA-vB=vS

These three methods allow us to deal with all cases

Restof thecircuit

vA

vB

+_ vS

supernode



This is method 1 – transform the voltage sourcesApplicable since voltage sources appear in series with Resist

Now use nodal analysis with one node, A

+_ +__+

v0vS1 vS2R3

R2R1

1

1

R

Sv

2

2

R

SvR1 R2 R3

_

+

v0

vA

!



This is method 1 – transform the voltage sourcesApplicable since voltage sources appear in series with Resist

Now use nodal analysis with one node, A

+_ +__+

v0vS1 vS2R3

R2R1

1

1

R

Sv

2

2

R

SvR1 R2 R3

_

+

v0

vA

!

( )

321

2211

2211321

GGG

vGvGv

vGvGvGGG

SSA

SSA

++

+=

+=++



vS

vBvA vC

R/2

2R2R

R/2 R+_

Riniin

What is the circuit input resistance viewed through vS?



Rewrite in terms of vS, vB, vCThis is method 2

Solve

vS

vBvA vC

R/2

2R2R

R/2 R+_

Riniin

What is the circuit input resistance viewed through vS?

05.35.02

05.035.0

=+!!

=!+!

=

CvBGvAGv

CGvBGvAGv

SvAv

SGvCGvBGvSGvCGvBGv

25.35.0

5.05.03=+−

=−

RR

inR

R

Sv

R

CvSv

R

BvSv

ini

Sv

CvSv

Bv

872.075.11

25.10

25.10

75.11

2/2

25.10

25.6,

25.10

75.2

==

=!

+!

=

==



Method 3 – supernodes

vBvA vC

vS1

R3R2

_

_

+

+

vS2 R4R1

reference

i4

i3i2

i1+

-vO

supernode



Method 3 – supernodesKCL for supernode:Or, using element equations

Now use

Other constituent relation

vBvA vC

vS1

R3R2

_

_

+

+

vS2 R4R1

reference

i4

i3i2

i1+

-vO

supernode

04321 =+++ iiii

04)(3)(21 =+!+!+ CvGBvCvGBvAvGAvG

2SvBv =

2)32()43()31( SvGGCvGGAvGG +=+++

1SvCvAv =!

Two equations in two unknowns


Summary of Nodal Analysis

1. Simplify the cct by combining elements in series or parallel

2. Select as reference node the one with most voltage sourcesconnected

3. Label node voltages and supernode voltages – do not labelthe nodes directly connected to the reference

4. Use KCL to write node equations. Express element currentsin terms of node voltages and ICSs

5. Write expressions relating node voltages and IVSs

6. Substitute from Step 5 into equations from Step 4Write the equations in standard form

7. Solve using Cramer, Gaussian elimination or matlab


Solving sets of linear equations

Cramer’s Rule Thomas & Rosa Appendix B pp. A-2 to A-11

!!!

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'=!!!

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!!!

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6

10

4

833

275

325

3

2

1

x

x

x

( ) ( ) ( )

5075125250

)3(7)2()2(3)3()3(8)2(5)2()3(875

27

32)3(

83

32)5(

83

275

833

275

325

=!!=

!"!!"!!!"!!"!+!"!!"=

!

!!!+

!

!!!!

!

!=

!!

!!

!!

=#

( ) ( ) ( )

100150250200

)3(7)2()2(6)3()3(8)2(10)2()3(874

27

326

83

32)10(

83

274

836

2710

324

1

=+!=

!"!!"!+!"!!"!+!"!!"=

!

!!+

!

!!!!

!

!=

!

!!

!!

=#

250

10011 ==

!

!=x


Solving sets of linear equations (contd)

Notes:This Cramer is not as much fun as Cosmo Kramer in SeinfeldI do not know of any tricks for symmetric matrices

24114250340

210

34)3(

86

34)5(

86

2105

863

2105

345

2

=++!=

!!

!!+

!!!

!!=

!

!!!

!

="

8424060

107

42)3(

63

42)5(

63

1075

633

1075

425

3

=+!=

!

!!+

!

!!!

!

!=

!!

!!

!

="

48.050

2422 ==

!

!=x

68.150

8433 ==

!

!=x


Solving Linear Equations: Gaussian elimination

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&'=

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6

10

4

3

2

1

833

275

325

x

x

x

!!!

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''

''

6

10

4

833

275

325

!!!!

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5

42

6

4

5

31

5

210

550

325

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'

''

42

6

4

31210

550

325

!!!!

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''

5

84

6

4

1000

550

325

!!!!

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#

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''

''

50

84

6

4

100

550

325

!!!!

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#

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%

&''

5084

5024

4

100

010

325

!!!!

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5084

5024

50100

100

010

001

Augmentmatrix

Rowoperationsonly

row2+row1row3+row1×3/5 row3×5 row3+row2×21/5

row3÷10 (row2+row3×5) ÷5 (row1+row2 ×2 + row3×3) ÷5


Solving Linear Equations - matlabA=[5 –2 –3; -5 7 –2; -3 –3 8]

A = 5 –2 –3 -5 7 –2 -3 –3 8

B=[4;-10;6]

B = 4 -10 6

inv(A)

ans = 1.0000 0.5000 0.5000 0.9200 0.6200 0.5000 0.7200 0.4200 0.5000

inv(A)*B

ans = 2.0000 0.4800 1.6800

A\B

ans = 2.0000 0.4800 1.6800


Mesh Current AnalysisDual of Nodal Voltage Analysis with KCL

Mesh Current Analysis with KVLMesh = loop enclosing no elements

Restricted to Planar Ccts – no crossovers (unless you are reallyclever)

Key Idea: If element K is contained in both mesh i and mesh j thenits current is ik=ii-ij where we have taken the reference directionsas appropriate

Same old tricks you already know

++ +++ + +- --

-

-

-

-

R1 R2

R3vS1 vS2v4v3

v2v1

v0

iA iB

Mesh A: -v0+v1+v3=0Mesh B: -v3+v2+v4=0

(R1+R3)iA-R3iB=vS1-R3iA+(R2+R3)iB=-vS2

v1=R1iA v0=vS1v2=R2iB v4=vS2v3=R3(iA-iB)

!"

#$%

&!"

#$%

&!"

#$%

&'

=+'

'+

2

1

323

331

Sv

Sv

Bi

Ai

RRR

RRR


Mesh Analysis by inspectionMatrix of Resistances R

Diagonal ii elements: sum of resistances around loopOff-diagonal ij elements: - resistance shared by loops i and j

Vector of currents iAs defined by you on your mesh diagram

Voltage source vector vSSum of voltage sources assisting the current in your mesh

If this is hard to fathom, go back to the basic KVL to sortthese directions out

SvRi =

-

+

+

-vS1

vS2R1

R3R2R4

iA iB

iC

+

-v0


Mesh Analysis by inspectionMatrix of Resistances R

Diagonal ii elements: sum of resistances around loopOff-diagonal ij elements: - resistance shared by loops i and j

Vector of currents iAs defined by you on your mesh diagram

Voltage source vector vSSum of voltage sources assisting the current in your mesh

If this is hard to fathom, go back to the basic KVL to sortthese directions out

SvRi =

-

+

+

-vS1

vS2R1

R3R2R4

iA iB

iC

+

-v0 !

!!

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#

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%

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'

'

=!!!

"

#

$$$

%

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!!!

"

#

$$$

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+''

'+

'+

1

2

2

3232

343

221

0

0

S

S

S

C

B

A

v

v

v

i

i

i

RRRR

RRR

RRR


Mesh Equations with Current Sources

Duals of tricks for nodal analysis with voltage sources1. Source transformation to equivalentT&R, 5th ed, Example 3-8 p. 91

+-5V

2KΩ

3KΩ

1KΩ

5KΩ

4KΩ2mA

iO

+-

2KΩ

3KΩ

1KΩ

5KΩ

iA iB

iO

+-

4KΩ

8V5V

!"

#$%

&

'=!

"

#$%

&!"

#$%

&

'

'

8

5

110002000

20006000

B

A

i

i iA=0.6290 mAiB=-0.6129 mAiO=iA-iB=1.2419 mA


Mesh Analysis with ICSs – method 2

Current source belongs to a single mesh

+-5V 2KΩ

3KΩ

1KΩ

5KΩ

4KΩ 2mA

iO

iA iB iC

mA2

04000110002000

520006000

!=

=!+!

=!

C

CBA

BA

i

iii

ii

Same equations!Same solution

Same example


Mesh Analysis with ICSs – Method 3

Supermeshes – easier than supernodesCurrent source in more than one mesh and/or not in parallel

with a resistance1. Create a supermesh by eliminating the whole branch

involved2. Resolve the individual currents last

R4

iS1iS2

R3

R2R1

vO

+

-iA iC

iB

Supermesh

Excluded branch

0)(342)(1 =!+++! AiCiRCiRBiRAiBiR

1SiAi =

2SiCiBi =!


Summary of Mesh Analysis

1. Check if cct is planar or transformable to planar

2. Identify meshes, mesh currents & supermeshes

3. Simplify the cct where possible by combiningelements in series or parallel

4. Write KVL for each mesh

5. Include expressions for ICSs

6. Solve for the mesh currents


Linearity & Superposition

Linear cct – modeled by linear elements andindependent sourcesLinear functionsHomogeneity: f(Kx)=Kf(x)Additivity: f(x+y)=f(x)+f(y)

Superposition –follows from linearity/additivityLinear cct response to multiple sources is the sum of the

responses to each source1. “Turn off” all independent sources except one and

compute cct variables2. Repeat for each independent source in turn3. Total value of all cct variables is the sum of the values

from all the individual sources


Superposition

Turning off sourcesVoltage source

Turned off when v=0 for all ia short circuit

Current sourceTurned off when i=0 for all v

an open circuit

We have already used this inThévenin and Norton equiv

+_vS

i+

v

-

i

v

i+

v

-

i

v

iS

i+

v

-

i

v

i

v

i+

v

-


Where are we now?

Finished resistive ccts with ICS and IVSTwo analysis techniques – nodal voltage and mesh current

Preference depends on simplicity of the case at handThe aim has been to develop general techniques for access

to analytical tools like matlab

Where to now?Active ccts with resistive elements – transistors, op-amps

Life starts to get interesting – design introducedCapacitance and inductance – dynamic ccts

Frequency response – s-domain analysisFilters

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