bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

[email protected] • ENGR-43_Lec-02b_SuperNM_Thevenin-Norton.pptx1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

Super Node/Mesh

Thevenin/Norton



ReCall Node (KCL) Analysis Need Only

ONE KCL Eqn0

1212612322

k

VV

k

VV

k

V

The Remaining Eqns From the Indep Srcs

][6

][12

3

1

VV

VV

Solving The Eqns

][5.1][64

0)()(2

22

12322

VVVV

VVVVV

3 Nodes Plus the Reference. In Principle Need 3 Equations...• But two nodes are connected

to GND through voltage sources. Hence those node voltages are KNOWN



SuperNode Technique Consider This

Example Conventional Node

Analysis Requires All Currents at a Node

2 eqns, 3 unknowns...Not Good (IS unknown) • Recall: The Current thru the

Vsrc is NOT related to the Potential Across it

But Have Ckt V-Src Reln

][621 VVV More Efficient solution:

• Enclose The Source, and All Elements In parallel, Inside a Surface. – Call That a SuperNode

SUPERNODE

SI

06

6@ 11 SIk

VmAV

012

4@ 22

k

VmAIV S



Supernode cont. Apply KCL to the

Surface

Now Have 2 Equations in 2 Unknowns

Then The Ckt Solution Using LCD Technique• See Next Slide

SUPERNODE

SI

04126

6 21 mAk

V

k

VmA

][621 VVV

• The Source Current Is interior to the Surface and is NOT Required

Still Need 1 More Equation – Look INSIDE the Surface to Relate V1 & V2



Now Apply Gaussian Elim

The Equations

][6 (2)

046126

(1)

21

21

VVV

mAmAk

V

k

V

Mult Eqn-1 by LCD (12 kΩ)

][6

][242

21

21

VVV

VVV

Add Eqns to Elim V2

][10][303 11 VVVV

][4][612 VVVV

Use The V-Source Rln Eqn to Find V2

SUPERNODE

SI



1V 2V

1sI

2sI

1R 2R

3R

SV

][6],[10],[20

4,10

21

321

mAImAIVV

kRkRR

ssS

Find the node voltagesAnd the power suppliedBy the voltage source

2012 VV

0101010

21 mAk

V

k

V ][100 10* 21 VVVk

][2021 VVV

][40100

][1202 :adding

21

2

VVV

VV

To compute the power supplied by the voltage source We must know the current through it: @ node-1

VI

k

VVmA

k

VIV 4

610

211 mA5

BASED ON PASSIVE SIGN CONVENTION THEPOWER IS ABSORBED BY THE SOURCE!!

mWmAVP 100][5][20



Illustration using Conductances Write the Node Equations

• KCL At v1

At The SuperNode Have V-Constraint• v2 − v3 = vA

KCL Leaving Supernode

Now Have 3 Eqnsin 3 Unknowns• Solve Using Normal Techniques



Example Find Io

Known Node Voltages

Now use KCL at SuperNode to Find V3

VVVV 12 ,6 42 The SuperNode

V-Constraint

VVVVVV 12or12 3131

Mult by 2 kΩLCD, collectTerms to Find:

SUPERNODE

123V



Numerical Example Find Io Using

Nodal Analysis Known Voltages for

Sources Connected to GND

Now KCL at SuperNode

VVVV 4,6 41

The Constraint Eqn

SUPERNODE

VVV 1223

k

k

V

k

V

k

V

k

V20

2

)4(

212

6 3322

Now Notice That V2 is NOT Needed to Find Io

• 2 Eqns in 2 Unknowns

VVVV

VVV

VVV

6.7385

------------------

eqns add and312

223

33

32

32

By Ohm’s LawmA

k

V

k

VIO 8.3

2

6.7

2 3



Dependent Sources

Circuits With Dependent Sources Present No Significant Additional Complexity

The Dependent Sources Are Treated As Regular Sources

As With Dependent CURRENT Sources Must Add One Equation For Each Controlling Variable



Numerical Example – Dep Isrc

Find Io by Nodal Analysis

Notice V-Source Connected to the Reference Node

KCL At Node-2 Sub Ix into KCL Eqn

Mult By 6 kΩ LCD

VV 31

Then Io

0263

212

xIk

V

k

VV

Controlling Variable In Terms of Node Potential k

VI x 6

2

06

263

2212

k

V

k

V

k

VV

VVVV 602 212

mAk

VVIO 1

321



Current Controlled V-Source Find Io

Supernode Constraint

Controlling Variable in Terms of Node Voltage

xkIVV 212

k

VI x 2

1

121 22 VVkIV x KCL at SuperNode

02

22

4 21 k

VmA

k

VmA

Multiply by LCD of 2 kΩ

][421 VVV 02 21 VV Recall

Then 38 83 22 VVVV

So Finally

mAk

VIO 3

4

22



ReCall MESH Analysis Use Mesh Analysis

So Vo

1I

2I

mAI 4 :1Mesh 1

Sub for I1 to Find I2

0124 2Mesh 21 kIkI

mAmAI

I3

4

12

16

12

4 12

][8333.16

6 2

VmAkV

IkV

O

O



SuperMeshes1. Create Mesh Currents

2. Write Constraint Equation Due To Mesh Currents SHARING Current Sources

SUPERMESH

mAII 432 3. Write Equations For

Remaining Meshes

mAI 21

4. Define A Supermesh By (Mentally) Removing The Shared Current



SuperMeshes cont.5. Write KVL For The SuperMesh as we do NOT know

the Voltage Across the 4 mA Current-Source

We Now have 3 Eqns in 3 Unknowns and the Math Model is Complete • Solve for I1, I2, I3 using

standard techniques

SUPERMESH

0)(1)(2216 131223 IIkIIkkIkI

KVL



SuperNODE vs SuperMESH

Use superNODE to AVOID a V-source current, IVs, in KCL Eqns

12V

2K

1K2K

2m A

IO

4m A

IO

Vx

Use superMESH to AVOID an I-source ΔVIs in KVL Eqns



Shared Isrc – General Loop Approach Strategy

• Define Loop Currents That Do NOT Share Current Sources – Even If It Means

ABANDONING Meshes

For Convenience, Begin by Using Mesh Currents Until Reaching Shared CURRENT Source as V-across an I-source is NOT Known• At That Point

Define a NEW Loop

To Guarantee That The New Loop Gives An Independent Equation, Must Ensure That It Includes Components That Are NOT Part Of Previously Defined Loops

1I 2I

3I



General Loop Approach cont. A Possible Approach

• Create a Loop by Avoiding The Current Source

The Eqns for Current Source Loops

mAI

mAI

4

2

2

1

The Eqns for 3rd Loop (3 Eqns & 3 Unknowns)

0)(1)(2)(21][6 13123233 IIkIIIkIIkkIV

The Loop Currents Obtained With This Method Are Different From Those Obtained With A SuperMesh• A SuperMesh used previously Defined Mesh Currents

3I

1I 2I



Example Find V Across R’s For Loop Analysis Note

• Three Independent Current Sources

• Four Meshes• One Current Source

Shared By Two Meshes

Mesh Equations For Loops With I-Sources

-+

1R2R

3R

4R

1SI 2SI

3SISV

1I 2I

3I4I

Careful Choice Of Loop Currents Should Make Only One Loop Equation Necessary• Three Loop Currents Can

Be Chosen Using Meshes And Not Sharing Any Source

33

22

11

S

S

S

II

II

II



Example Find V Across R’s cont.

KVL for I4 Loop

-+

1R2R

3R

4R

1SI 2SI

3SISV

1I 2I

3I4I

4V

1V

2V

3V

Solve For The Current I4 Using The ISj

• Now Use Ohm’s Law To Calc Required Voltages

0)(

)()(

443

1431324

RII

RIIIRIIVS

)( 43111 IIIRV

)( 1222 IIRV

)( 4233 IIRV )( 4344 IIRV

Note that Loop-4 does NOT pass thru ANY CURRENT-Sources• This AVOIDS the

UNknown potentials across the I-sources



Dependent Sources General Approach

• Treat The Dep. Source As Though It Were Independent

• Add One EquationFor The Controlling Variable

Example at Rt.: Mesh Currents Defined by Sources

Mesh-3 by KVL

k

VI

mAI

X

2

4

2

1

0)(1)(21 4313 IIkIIkkI x

Mesh-4 by KVL012)(1)(1 2434 VIIkIIk



Dependent Sources cont. The Controlling

Variable Eqns

In Matrix Form

Solve by Elimination or Linear-Algebra

)(2 13

24

IIkV

III

x

x

Combine Eqns,Then Divide by 1kΩ

mAIII

mAIII

III

mAI

122

823

0

4

432

432

321

1

12

8

0

4

2110

2310

0111

0001

4

3

2

1

I

I

I

I



Loop & Node Compared

Consider the Ckt

Find Vo by NODE Analysis• ID Nodes• Make a SuperNode

Vsrc to GND ][VV 44

SuperNode Constraint xVVV 221



Loop & Node Compared (2)

KCL at SuperNode

The SuperNode Eqn

Mult. By 1kΩ LCD

01

4

1112 131322

k

VV

k

VV

k

VV

k

VmA

The Node 3 KCL

VVVV

VVVVVVVV

6222or

24

321

131322

011

2 1323

k

VV

k

VVmA

Mult. By 1kΩ LCD

VVVV

VVVVV

22or

2

321

1323




The Controlling Var.

In SuperNode Eqn

Thus 3 Eqns in Unknowns V1, V2, V3

VVVV 3321

VVVV 22 321

2VVx

032 2121 VVVVV x

03 21 VV

Recall the GOAL

31 VVVo




Now by Loops

Start with 3 Meshes

The Mesh/Loop Eqns• Loop-1: mAI 21

0112 322 IIkkIVx

• Loop-3:1

I2

I

3I

Add a General Loop to avoid the Isrc

4I mAI 23

• Loop-2:

– Note I3 = –2 mA

• Loop-4:

04121 4143 VkIVIIIk x

– Note I1 = 2 mA




The Controling Var

As Before Vx = V2

And V2 is related to the net current From Node-2 to GND

By Net Current & Ohm’s Law

4311 IIIkVx

0112 322 IIkkIVx

SubOut Vx in Loop-2 & Loop-4 Eqns:

04121 4143 VkIVIIIk x

1I

2I

3I

4I

After Subbing Find:V84 4 kI

VkIkI 642 42




Summarize Loops

Using The Loop/Mesh Eqns Find

Recall the GOAL

General Comments• Nodes (KCL) are

generally easier if we have VOLTAGE Sources

• Loops (kVL) are generally easier if we have CURRENT Sources

mAI 21 mAI 12 mAI 23 mAI 24

2I

21kIVo



Thevenin’s & Norton’sTheorems These Are Some Of The Most Powerful

Circuit analysis Methods They Permit “Hiding” Information That Is

Not Relevant And Allow Concentration On What Is Important To The Analysis



Low Distortion Power Amp

From PreAmp(voltage )

To speakers

to Match Speakers And Amplifier One Should Analyze The Amp Ckt



Low Dist Pwr Amp cont

To Even STAND A CHANCE to Match the Speakers & Amp We Need to Simplify the Ckt• Consider a Reduced CIRCUIT

EQUIVALENT

Replace the OpAmp+BJT Amplifier Ckt with a MUCH Simpler (Linear) Equivalent +

-

R T H

V T H

• The Equivalent Ckt in RED “Looks” The Same to the Speakers As Does the Complicated Circuit



Thevenin’s Equivalence Theorem

LINEAR C IRC U ITM ay contain

independent anddependent sources

with their contro ll ingvariablesPART A



with their contro ll ingvariablesPART B

a

b_Ov

i

LINEAR C IRC U IT

PART B

a

b_Ov

i

THR

THv

PART A

Thevenin Equivalent Circuit for PART A

vTH = Thevenin Equivalent VOLTAGE Source

RTH = Thevenin Equivalent Series (Source) RESISTANCE



Norton’s Equivalence Theorem

Norton Equivalent Circuit for PART A

iN = Norton Equivalent CURRENT Source

RN = Norton Equivalent Parallel (Source) RESISTANCE







a

b_Ov

i

LINEAR C IRC U IT

PART B

a

b_Ov

i

NRNi

PART A



Examine Thevenin Approach

For ANY Part-B Circuit

The Thevenin Equiv Ckt for PART-A →




A NYPA RT B

a

b_Ov

i

iRvv THOCO R T H

i +

_OvOCv

+_• V-Src is Called the THEVENIN

EQUIVALENT SOURCE• R is called the THEVENIN

EQUIVALENT RESISTANCEPART A MUST BEHAVE LIKETHIS CIRCUIT



Examine Norton Approach

In The Norton Case

The Norton Equiv Ckt for PART-A →




A NYPA RT B

a

b_Ov

i

TH

O

TH

OCTHOCO R

v

R

viiRvv

SCTH

OC iR

v

SCiTHR

Ov

a

b

i

Norton• The I-Src is Called The NORTON EQUIVALENT SOURCE



Interpret Thevenin & Norton

This equivalence can be viewed as a source transformation problem. It shows how to convert a voltage source in series with a resistor into an equivalent current source in parallel with the resistor• SOURCE TRANSFORMATION CAN BE A GOOD TOOL

TO REDUCE THE COMPLEXITY OF A CIRCUIT

R T H

i +

_OvOCv

+_

Thevenin

SCiTHR

Ov

a

b

i

Norton

SC

OCNTH i

vRR In BOTH Cases



Source Transformations Source transformation is a good tool to reduce

complexity in a circuit ...WHEN IT CAN APPLIED• “IDEAL sources” are NOT good models for the

REAL behavior of sources– .e.g., A Battery does NOT Supply huge current When Its Terminals

are connected across a tiny Resistance as Would an “Ideal” Source

These Models are Equivalent When+

-

Im proved m odelfo r vo ltage source

Im proved m odelfo r current source

SVVR

SI

IRa

b

a

b

SS

IV

RIV

RRR

Source X-forms can be used to determine the Thevenin or Norton Equivalent• But There May be More Efficient Methods



Example Solve by Src Xform In between the terminals we

connect a current source and a resistance in parallel

The equivalent current source will have the value 12V/3kΩ

The 3k and the 6k resistors now are in parallel and can be combined

In between the terminals we connect a voltage source in series with the resistor

The equivalent V-source has value 4mA*2kΩ

The new 2k and the 2k resistor become connected in series and can be combined



Solve by Src Xform cont. After the transformation the

sources can be combined The equivalent current source

has value 8V/4kΩ = 2mA

1. Do another source transformation and get a single loop circuit

2. Use current divider to compute IO and then Calc VO using Ohm’s law

The Options at This Point



Or one more source transformation

eqeqeq IRV +-V e q

R e q R 3

R 4

0V

PROBLEM Find VO using source transformation

Norton

Norton

3 current sources in parallel and three resistors in parallel

0I

eqeqeq IRV eqeq

VRRR

RV

34

40

EQUIVALENT CIRCUITS

TH

TH

V

R



Source Xform Summary These Models are Equivalent

+-

Im proved m odelfo r vo ltage source

Im proved m odelfo r current source

SVVR

SI

IRa

b

a

bSS

SS

IV

VRI

RIV

RRR

Source X-forms can be used to determine the Thevenin or Norton Equivalent

Next Review Several Additional Approaches To Determine Thevenin Or Norton Equivalent Circuits



Determine the Thevenin Equiv.

vTH = OPEN CIRCUIT Voltage at A-B if Part-B is Removed and Left UNconnected

iSC = SHORT CIRCUIT Current at A-B if Voltage at A-B is Removed and Replaced with a Wire (a short)

Then by R = V/ISC

OCTH i

vR



Graphically...

NSC

OCTH

SCabNOCabTH

Ri

vR

iIivVv

1. Determine the Thevenin equivalent source

Remove part B andcompute the OPENCIRCUIT voltage abV

2. Determine the SHORT CIRCUITcurrent

Remove part B and compute the SHORTCIRCUIT current abI




a

b_

0v

SCi

abI

Second circuit problem

One circuit problem

_abV

LINEAR C IRC U IT

M ay containindependent and

dependent sourceswith their contro ll ing

variablesPART A

a

b_

OCv

0i

Then



Thevenin w/ Indep. Sources The Thevenin Equivalent V-Source is computed

as the open loop voltage The Thevenin Equivalent Resistance CAN BE

COMPUTED by setting to zero all the INDEPENDENT sources and then determining the resistance seen from the terminals where the equivalent will be placed

+-

a

b

T o P a rt BV S

R 1

R 2IS

a

b

R THR 2R 1



Thevenin w/ Indep. Sources cont

kRTH 3

“Part B” Since the evaluation of the Thevenin equivalent Resistance for INdependent-Source-Only circuits can be very simple, we can add it to our toolkit for the solution of circuits

“Part B”

kkkkRTH 4632



Thevenin Example Find Vo Using Thevenin’s

Theorem Identify Part-B (the Load)

Break The Circuit At the Part-B Terminals “PART B”

V6

k5

DEactivate 12V Source to Find Thevenin Resistance • Produces a SHORT



Thevenin Example cont. Note That RTH Could be

Found using ISC

By Series-Parallel R’s

Then by I-DividerSCI

kReq 5.7266

Then Itot

mAkVI tot 6.15.712

totI

mAmAISC 2.126

66.1

Finally RTH

kmAVIVR SCOCTH 52.16

• Same As Before



Thevenin Example cont.2 Finally the Thevenin

Equivalent Circuit

And Vo By V-Divider

][1)6(51

1VV

kk

kVO

][1V



Let’s do MQ-02e

Find: Vt & Rt

A

B



Thevenin Example Use Thevenin To

Find Vo Have a CHOICE on

How to Partition the Ckt• Make “Part-B” As Simple

as Possible

Deactivate the 6V and 2mA Source for RTH

kRTH 3

10422

“Part B”



Thevenin Example cont For the open circuit

voltage we analyze the circuit at Right (“Part A”)

Use Loop/Mesh Analysis

0)(246

2

211

2

IIkkIV

mAI

mAmAI

I3

5

6

26 21

Then VOC

][3/3243/20

*2*4 21

VVVV

IkIkV

OC

OC

Finally The Equivalent Circuit



CALCULATE Vo USING NORTON

PART B kRR THN 3

SCI

mAmAk

VII NSC 22

3

12

NINR

k4

k2

NN

NO I

kR

RkkIV

622

I

][3

4)2(

9

32 VmAkVO

COMPUTE Vo USING THEVENIN PART B

THV

023

12 mA

k

VTH kkRTH 43

+-

THR

THVk2

OV

][3

4)6(

72

2VVVO

612 THV



DEpendent & INdependent Srcs Find The Open Circuit Voltage And

Short Circuit Current Solve Two Circuits (Voc & Isc) For

Each Thevenin Equivalent Any and all the techniques may be

used; e.g., KCL, KVL, combination series/parallel, node & loop analysis, source superposition, source transformation, homogeneity

Setting To Zero All Sources And Then Combining Resistances To Determine The Thevenin Resistance is, in General, NOT Applicable!!




a

b_Ov

i

+-

THR

THV

a

b

OCTH VV

SC

OCTH I

VR



Example Recognize Mixed sources

• Must Compute Open Circuit Voltage, VOC, and Short Circuit Current, ISC

The Open Ckt Voltage

Use V-Divider to Find VX

XV bV

SSX VVRR

RV

3

2)2(

2

bXTH VVV

SX

SSXb

VV

VaRVaVRV

32 :as

)3/41()(2

THV

For Vb Use KVL

SSSXX

bXTH

VVRaVRaVV

VVV

)3/2)(21()2(

SbXTH VaR

VVV3

41

Solve for VTH

The Short Ckt Current• Note that Shorting a-to-b

Results in a Single Large Node

Now VTH = Vx − Vb



Example cont Need to Find Vx

KCL at Single Node

Then RTH

SCISingle node

XV

022

2

R

VVaV

R

V

R

VV sXX

XSx

aR

VV SX 24

3

Solving For Vx

KCL at Node-b for ISC

R

VVaVI SX

XSC 2

SSC VaRR

aRI

)2(4

41

3

)2(4 aRR

I

V

I

VR

SC

TH

SC

OCTH

The Equivalent Circuit

a

b

R TH

V TH

SVaR

3

41

3

)2(4 aRR

Va = Vb = VX



Illustration Use Thevenin to

Determine Vo

Partition Guidelines• “Part-B” Should be as

Simple As Possible• After “Part A” is replaced by

the Thevenin equivalent should result in a very simple circuit

• The DEpendent srcs and their controlling variables MUST remain together

Use SuperNode to Find Open Ckt Voltage

“Part B”

1V

Constraint at SuperNode

OCOC VVVV 1212 11

KCL at SuperNode

k 2 :Where

022

12

1

)()12( '

ak

V

k

V

k

aIV OCOCXOC



Illustration cont The Controlling

Variable

Solving 3 Eqns for 3 Unknowns Yields

k

VI OCX 2'

)1/2(4

36

)1/(4

36

kkkaVOC

Now Tackle Short Circuit Current

02

" k

VI AX

AV

At Node-A find

VA=0 → The Dependent Source is a SHORT• Yields Reduced Ckt

k 322||1



Illustration cont Using the Reduced Ckt

Now Find RTH

mAkk

VISC 18

2||1

12

k3

1

mA 18

V 6

SC

OCTH I

VR

Setting All Sources To Zero And Combining Resistances Will Yield An INCORRECT Value

Finally the Solution

VVV

VRkk

kV TH

TH

7

186

3333.2

1

11

1

0

0

)2( kaRTH

OCV

Note: Some ckts can produce NEGATIVE RTH



Numerical Example Find Vo using Thevenin

Define Part-A Find VOC using SuperNode

XI

Super node1V

KVL

THV

XI

06

)3(1

2 :KCL 11

k

VVmA

k

V

])[4/3(1 VV

Apply KVL

01000 1 VIV XTH

With The Controlling Variable Find

k

VI X 2

1

])[8/3( VVTH



Numerical Example cont Next The Short

Circuit Current

The ONLY Value That Satisfies the Above eqns

SCI1XI

11V

111 1000 XIV

1111

111 12

2 XXX kIkIVk

VI

00 11

1 VI X KCL at Top Node

• Recall Dep Src is a SHORT

mAkVmAISC 5.0)6/()3(1

MINUS3V

Using VOC & ISC k

4

3

ma 21

V 83

SC

OCTH I

VR

The Equiv. Ckt

+-V TH

R TH 1 k

2 k

+

V O

_

])[8/3()4/3(12

20 VV

mV375

750



Note on Example The Equivalent Resistance CanNOT Be

Obtained By Deactivating The Sources And Determining The Resistance Of The Resulting Interconnection Of Resistors• Suggest Trying it → Rth,wrong = 2.5 kΩ

– Rth,actual = 0.75 kΩ

Req



EXAMPLE: Find Vo By Thevenin Select Partition

Use Meshes to Find VOC

“Part B”

KVL for V_oc

In The Mesh Eqns

mAIV

I X 2;2000 2

'

1

2

'

21'

244 I

k

VkIIkV X

X

mAIIIkkIkIVX 4)(422 12111'

The Controlling Variable

By Dep. Src Constraint

Solve for VOC

VVmAkVOC 11342

Now KVL on Entrance Loop

][3*20 1 VIkVOC



Find Vo By Thevenin cont Now Find ISC

The Mesh Equations

The Controlling Variable

mAIV

I x 2;2000 2

"

1

0)(23 1 IIkV SC

)(*4 21" IIkVX

Solving for I1 Find Again

mAI 41 Find ISC by Mesh KVL

mAk

IkVISC 2

11

2

*23 1

kmA

V

I

VR

SC

OCTH 2

)2/11(

][11

Then Thevenin Resistance

Use Thevenin To Find Vo

VVkk

kV 433][11

62

60

THV

THR

2I

scI

1I



Illustration The Merthod for Mixed Sources

For the Short Ckt Current

+-

vS

R 1

R 2

V x

R 3

g m V x

a

b

L ine a r M o de l fo r T ra ns is to r

+-

V TH

R TH

a

b

THV

SCI

SC

OCTHOCTH I

VRVV ,

The Open Ckt Voltage

333RVgRIV xmRTH

SmTHSx vRR

RRgVv

RR

RV

21

23

21

2

SmxmSC vRR

RgVgI

21

2

3

21

2

321

2

R

RRR

vg

RRR

Rvg

I

VR

Sm

Sm

SC

OCTH



WhiteBoard Work

Let’s Work This Problem

2K

2K

2 K

2K 4m A6V

+

-

Vo

• Find Vo by Source Transformation



All Done for Today

More onSourceXforms



Thevenin Example Find Vo Using Thevenin’s Theorem

in the region shown, could use source transformation twice and reduce that part to a single source with a resistor.

Alternative: apply Thevenin Equivalence to that part (viewed as “Part A”)

Deactivating (Shorting) The 12V Source Yields

kRTH 4362

Opening the Loop at the Points Shown Yields

][8][1263

6VVVV THOC



Thevenin Example cont. Then the Original Circuit Becomes After “Theveninizing”

Apply Thevenin Again

For Open Circuit Voltage Use KVL

Result is V-Divider for Vo

Deactivating The 8V & 2mA Sources Gives

kR TH 41

1THV

VVmAkVTH 1682*41

VVV 8][1688

80



Thevenin Example Alternative Can Apply Thevenin only once to get a voltage divider

• For the Thevenin Resistance Deactivate Sources

For the Thevenin voltage Need to analyze this circuit

Find VOC by SuperPosition

“Part B” kRTH 8



Thevenin Alternative cont. Open 2mA Source To find

Vsrc Contribution to VOC

Short 12V Source To find Isrc Contribution to VOC

VVVOC 81263

61

VmAkkkVOC 8)2(*)362(2 Thevenin Equivalent of “Part A”

A Simple Voltage-Divider as Before



Example

+-

2SI

3SI 1SV

1SI1R

2R

3R

4R

1V2V

3V

4V

OV

R1 = 1k; R2 = 2k, R3 = 1k, R4 = 2kIs1 =2mA, Is2 = 4mA, Is3 = 4mA,

Vs1 = 12 V210 VVV

Need Only V1 and V2 to Find Vo

Known Node Potential

][12:@ 133 VVVV S

Now KCL at Node 1

021

][2

0:@

121

4

1

1

2111

k

V

k

VVmA

R

V

R

VVIV S

Find Vo

To Start• Identify & Label All Nodes• Write Node Equations• Examine Ckt to Determine

Best Solution Strategy

Notice



Example cont.

+-

2SI

3SI 1SV

1SI1R

2R

3R

4R

1V2V

3V

4V

OV

R1 = 1k; R2 = 2k, R3 = 1k, R4 = 2kIs1 =2mA, Is2 = 4mA, Is3 = 4mA,

Vs = 12 V

021

12

1][4

0:@

42212

2

42

3

32

1

1232

k

VV

k

V

k

VVmA

R

VV

R

VV

R

VVIV S

At Node 4

02

][4][2

0:@

24

2

24214

k

VVmAmA

R

VVIIV SS

To Solve the System of Equations Use LCD-multiplication and Gaussian Elimination

At Node 2



Example cont. The LCDs

021

][2 121

k

V

k

VVmA

021

12

1][4 42212

k

VV

k

V

k

VVmA

02

][4][2 24

k

VVmAmA

][423 21 VVV *2kΩ

*2kΩ]3252 421 VVVV

*2kΩ][442 VVV

(1)

(2)

(3)

Now Add Eqns (2) & (3) To Eliminate V4

][182][3642 2121 VVVVVV (4)

Now Add Eqns (4) & (1) To Eliminate V2

][11][222 11 VVVV

][5.14][182][11 22 VVVVV

][5.3][5.14][11210 VVVVVV

BackSub into (4) To Find V2

Find Vo by Difference Eqn



Complex SuperNode Write the Node Eqns Set UP

• Identify all nodes • Select a reference• Label All nodes

Nodes Connected To Reference Through A Voltage Source

Eqn Bookkeeping:• KCL@ V3

• KCL@ SuperNode • 2 Constraint Equations• One Known Node

+-

+ -

+-1R

2R

3R

4R

5R

6R

7R

supernode

1V

2V 3

V

4V

5V

Voltage Sources In Between Nodes And Possible Supernodes

• Choose to Connect V2 & V4



Complex SuperNode cont. Now KCL at Node-

3

+-

+ -

+-1R

2R

3R

4R

5R

6R

7R

supernode

1V

2V 3

V

4V

5V

Constraints Due to Voltage Sources

07

3

5

43

4

23

R

V

R

VV

R

VV

Now KCL at Supernode• Take Care Not to Omit

Any Currents

04

32

5

34

6

4

3

5

2

15

1

12

R

VV

R

VV

R

V

R

V

R

VV

R

VV

11 SVV 252 SVVV 345 SVVV

Vs1

Vs2 Vs3

5 Equations 5 Unknowns → Have to Sweat Details



Dep V-Source Example Find Io by Nodal Analysis

Notice V-Source Connected to the Reference Node

SuperNode Constraint

KCL at SuperNode

Mult By 12 kΩ LCD

VV 63

xVVV 221

212 3VVVVx 062)6(2 2211 VVVV

Controlling Variable in Terms of Node Voltage



Dep V-Source Example cont Simplify the LCD Eqn

By Ohm’s Law

mAk

V

k

VIo 8

3

24

9

121

VV

VV

VV

VVV

5.4

184

3 and

1833

1

1

12

21



Numerical Example Select Soln

Method• Loop Analysis

– 3 meshes– One current source

• Nodal Analysis– 3 non-reference

nodes– One super node

• Both Approaches Seem Comparable → Select LOOP Analysis– Specifically Choose

MESHES

Select Mesh Currents

Write Loop Eqns for Meshes 1, 2, 3 by KVL

+-

+

VO

_

6k

4k

2k2k

IS

VS

IS = 2mA, V S = 6V

Determine VO

P ro b lem 3.46 (6th Ed )

1I

2I

3I

SII 1

0)(2)(4 1232 IIkIIkVS06)(4)(2 32313 kIIIkIIk



Numerical Example cont

We Seek Vo, Thus Using Ohm’s Law Need only Find I3

Simplify: Divide Loop Eqns by 1kΩ• I Coeffs Become

NUMBERS• Voltages

Converted to mA

Note That I1 = IS and Sub into Loop Eqns

Substitute into The Remaining Loop Eqns• See Next Slide

+-

+

VO

_

6k

4k

2k2k

IS

VS

IS = 2mA, V S = 6V

Determine VO


3I

1I

2I



Numerical Example cont.2 The Loop-2 and

Loop-3 Eqns

Then by Ohm’s Law

+-

+

VO

_

6k

4k

2k2k

IS

VS

IS = 2mA, V S = 6V

Determine VO


3I

mAII

mAIII

mAIk

VII

S

S

28

323421028

-----------------------------

Add and342124

2)46(21

46

33

32

132

2I

VmAA

VkkIVO 7

48

7

866 3

1I



Find Vo- Compare Mesh vs. Loop Using MESH Currents Using LOOP Currents

Treat The Dependent Source As One More Voltage Source Mesh-1 & Mesh-2

0)(422 211 IIkkIVx 04)(22 121 kIIIkVx

0)(463 122 IIkkI 063)(22 221 kIIIkVx

Loop-1 & Loop-2



Compare Loop vs. Mesh cont. Using MESH Currents Using LOOP Currents

Now Express The Controlling Variable In Terms Of MESH or LOOP Currents

14kIVx )(4 21 IIkVx

Solving

3104

042

21

21

kIkI

kIkI

386

066

21

21

kIkI

kIkI



Compare Loop vs. Node cont. Using MESH Currents Using LOOP Currents

SolutionsmAImAI 5.1,3 21 mAImAI 5.1,5.1 21

][96 2 VkIVO Finally

Notice The Difference Between MESH Current I1 and LOOP Current I1 even Though They Are Associated With The Same Path

The Selection Of LOOP Currents Simplifies Expression for Vx and Computation of Vo



Dep Isrc Not Shared by Mesh Treat The Dependent

Source As A Conventional Source

Equations For Meshes With Current Sources

Express The Controlling Variable, Vx, In Terms Of Loop Currents

Then KVL on The Remaining Loop (I3)



Dep Isrc Not Shared by Mesh Asked to Find Only Vo

• Need Only Determine I3

The Dep Src Eqns

From KVL Eqn for I3

mAIIIIkV

kIVV

I

x

xx

42)(4

22000 21

21

11

mAIkIkI8

11238 313

Thus

VkIVO 4

336 3



Find Vo Using Mesh Analysis

Draw the Mesh Currents Controlling Variable In Terms Of Loop Currents

1I2I

Write KVL Mesh Eqns For Mesh-1 & Mesh-2

0)(4212 122 IIkkI

0)(422 211 IIkkIkI x2II x



Find Vo Using Mesh Analysis cont

Substitute & Collect Terms

Solve for I2

1I2I

1264

066

21

21

kIkI

kIkImAIkI 612624 22

][122 2 VkIVO

Finally Vo

122 1 kI



WhiteBoard Work


Find the OutPut Voltage, VO

1K1K 1K

12V

VO

+

-

1K

IO2IX IX



Outline of Theorem Proof Consider Linear Circuit → Replace vo with a SOURCE

If Circuit-A is Unchanged Then The Current Should Be The Same FOR ANY Vo (Source or Rat’sNest Generated)

Use Source SuperPosition• 1st: Inside Ckt-A OPEN all I-Src’s, SHORT All V-Srcs

– Results in io Due to vo

• 2nd: Short the External V-Src, vo

– Results in iSC Due to Sources Inside Ckt-A



Theorem Proof Outline cont. Graphically the Superposition

Then The Total CurrentSCi

All independent sources set to zero in A

Oi

O

OTH i

vR

SCO iii

SCTH

O

SCO

iR

vi

iii

or

Now DEFINE using V/I for Ckt-A

Then By Ohm’s Law



Theorem Proof Outline cont.2 Consider Special Case Where Ckt-B is an OPEN (i =0)

The Open Ckt Eqn Suggests

SCTH

OCOCO i

R

vvvi 0:0For

SC

OCTH

TH

OCSC i

vR

R

vi and

OCTHOTH

OC

TH

OSC

TH

O viRvR

v

R

vi

R

vi

Also recall

How Do To Interpret These Results?• vOC is the EQUIVALENT of a single Voltage Source

• RTH is the EQUIVALENT of a Single Resistance which generates a Voltage DROP due to the Load Current, i

OCv

Think y = mx + b

0i



Theorem Proof –Version 2

1. Because of the LINEARITY of the models, for any Part B the relationship between vO and the current, i, has to be of the form

2. Result must hold for “every valid Part B”

3. If part B is an open circuit then i=0 and...







a

b_Ov

i

bimvO

OCO vvb 4. If Part B is a short circuit then vO is zero. In this case

OCTHOTHSC

OCSC viRvorR

i

vmbim 0

(Linear Response)



Example Find Thevenin Equiv.

Find VTH by Nodal Analysis: Iout = 0

+-

a

b

T o P a rt BV S

R 1

R 2IS

THV

SCI

Part B is irrelevant. The voltage Vab will be the value of the Thevenin equivalent source.

SS

TH

SSTH

SS

TH

SSTHTH

IR

V

RR

RRV

IRR

RRV

RR

RV

IR

VV

RR

IR

VV

R

V

121

21

21

21

21

2

121

12

)11

(

0

For Short Circuit Current Use Superposition

When IS is Open the Current Thru the Short

11 RVI SSC

When VS is Shorted the Current Thru the Short

SSC II 2



Example – Find Thevenin cont

Find Thevenin Resistance

+-

a

b

T o P a rt BV S

R 1

R 2IS

THV

SCI

Find the Total Short Ckt Current

To Find RTH Recall

Then RTH In this case the Thevenin resistance

can be computed as the resistance from a-b when all independent sources have been set to zero• Is this a GENERAL Result?

1R

VII SSSC

SC

THTH I

VR

S

STH I

R

V

RR

RRV

121

21

21

21

RR

RRRTH



Numerical Example Find Vo Using

Thevenin’s Theorem

First, Identify Part-B

Deactivate (i.e., Short Ckt) 6V & 12V Sources to Find RTH

“PART B”

kkkRTH 26||3

THR



Numerical Example cont. Use Loop Analysis to

Find the Open Circuit Voltage

The Resulting Equivalent Circuit

OCVI

mAIVkI 2][189

][61230 VVkIV OCOC

OV

kRTH 2 k2

k4VVTH 6

][3)6(44

4VVVO

Finally the Output



+-

R TH

V TH

RRRRTH 5.13||3

THR

1I

2I

SII 10)(5 221 RIIIRVS

THVKVL

)(2 212 IIRRIVTH

Example

Xform

Deactivate Srcs for RTH

Use Loops for VTH



Example cont. OR, Use Superposition to Find

Thevenin Voltage First Open The Current Source

Next Short-Circuit the Voltage Source• Using I-Divider

Find Isrc Contribution by KVL

2321

211 SSTH

V

RRR

RRVV

R

2 R3 R

IS

+

V 2TH

_

1ISII

6

51

2I SII6

12

KVL

STH RIRIRIV2

12 21

2

22

21

SSTH

THTHTH

RIVV

VVV

Add to Find Total VTH



WhiteBoard Work


2K

2K

2 K

2K 4m A6V

+

-

Vo

• Find Vo by Source Transformation

bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

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