[email protected] ENGR-44_Lec-04-3a_Thevenin-Norton.ppt 1 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Licensed

[email protected] • ENGR-44_Lec-04-3a_Thevenin-Norton.ppt1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

Chp 5.3aChp 5.3aTheveninTheveninNortonNorton



Thevenin’s & Norton’sTheoremsThevenin’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 AmpLow Distortion Power Amp

From PreAmp(voltage )

To speakers

to Match Speakers And Amplifier One Should Analyze The Amp Ckt



Low Dist Pwr Amp contLow 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 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 TheoremThevenin’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 RESISTANCE



Norton’s Equivalence TheoremNorton’s Equivalence Theorem

Norton Equivalent Circuit for PART A

iN = Norton Equivalent CURRENT Source

RN = Norton Equivalent Parallel RESISTANCE







a

b_Ov

i

LINEAR C IRC U IT

PART B

a

b_Ov

i

NRNi

PART A



Outline of Theorem ProofOutline 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

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.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 Then By Ohm’s Law



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

The Open Ckt Case Suggests

SCTH

OCOCO i

R

vvvi 0:0For

SC

OCTH

TH

OCSC i

vR

R

vi and

iRvvR

v

R

vi

R

vi THOCO

TH

OC

TH

OSC

TH

O

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



Theorem Proof –Version 2Theorem 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

nimvO

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

OCTHOTHSC

OCSC viRvorR

i

vmnim 0

(Linear Response)



Examine Thevenin ApproachExamine 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 THEVENINEQUIVALENT RESISTANCE

PART A MUST BEHAVE LIKETHIS CIRCUIT



Examine Norton ApproachExamine 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 & NortonInterpret 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 TransformationsSource 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 Example Solve by Src Xform 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.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 SummarySource 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.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...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



Example Example Find Thevenin Equiv. 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 contExample – 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



Thevenin w/ Indep. SourcesThevenin 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 contThevenin w/ Indep. Sources cont

kRTH 3

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

“Part B”

kkkkRTH 4632



Thevenin Example 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



Thevenin Example cont.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.2Thevenin Example cont.2 Finally the Thevenin

Equivalent Circuit

And Vo By V-Divider

][1)6(51

1VV

kk

kVO

][1V



Thevenin Example 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.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 Thevenin Example Alternative 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.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



Thevenin Example 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 contThevenin 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



Numerical Example 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.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



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 VVO

COMPUTE Vo USING THEVENIN PART B

THV

023

12

mA

k

VTH kkRTH 43

+-

THR

THVk2

OV

][3

4)6(

72

2VVVO

612 THV



+-

R TH

V TH

RRRRTH 5.13||3

THR

1I

2I

SII 10)(5 221 RIIIRVS

THVKVL

)(2 212 IIRRIVTH

ExampleExample

Xform

Deactivate Srcs for RTH

Use Loops for VTH



Example cont.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

_

1I SII6

51

2I SII6

12

KVL

STH RIRIRIV2

12 21

2

22

21

SSTH

THTHTH

RIVV

VVV

Add to Find Total VTH



WhiteBoard WorkWhiteBoard Work

Let’s Work This Problem

2K

2K

2 K

2K 4m A6V

+

-

Vo

• Find Vo by Source Transformation

Documents

[email protected] ENGR-44_Lec-04-3a_Thevenin-Norton.ppt 1 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Licensed