ADDITIONAL ANALYSIS TECHNIQUES DEVELOP THEVENIN’S AND NORTON’S THEOREMS These are two very powerful analysis tools that allow us to focus on parts of a

ADDITIONAL ANALYSIS TECHNIQUES

DEVELOP THEVENIN’S AND NORTON’S THEOREMSThese are two very powerful analysis tools that allow us tofocus on parts of a circuit and hide away unnecessary complexities

MAXIMUM POWER TRANSFERThis is a very useful application of Thevenin’s and Norton’s theorems

These are some of the most powerful analysis results to be

discussed.They permit to hide information

that is not relevant and concentrate in what is important to

the analysis

THEVENIN’S AND NORTON’S THEOREMS

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Low distortion audio power amplifier

From PreAmp(voltage ) To speakers

TO MATCH SPEAKERS AND AMPLIFIERIT IS MUCH EASIER TO CONSIDER THISEQUIVALENT CIRCUIT!

TO MATCH SPEAKERS ANDAMPLIFIER ONE SHOULD ANALYZETHIS CIRCUIT

+-

R T H

V T H

REPLACE AMPLIFIERBY SIMPLER “EQUIVALENT”

Courtesy of M.J. Renardson

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

THEVENIN’S EQUIVALENCE THEOREM

Resistance Equivalent Thevenin

Source Equivalent Thevenin

TH

TH

R

v

LINEAR C IRC U IT

PART B

a

b_Ov

i

THR

THv

PART A

Thevenin Equivalent Circuit

for PART A







a

b_Ov

i

NORTON’S EQUIVALENCE THEOREM

Resistance EquivalentNorton

Source EquivalentNorton

N

N

R

i

LINEAR C IRC U IT

PART B

a

b_Ov

i

NRNi

PART A

Norton Equivalent Circuit

for PART A

OUTLINE OF PROOF

2. Result must hold for “every valid Part B” that we can imagine

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 nimvO *

3. If part B is an open circuit then i=0 and... OCvn4. If Part B is a short circuit then Vo is zero. In this case

OCTHO viRv

OCSC vim *0 THSC

OC Ri

vm







a

b_Ov

i

This is the Thevenin equivalentcircuit for the circuit in Part A

OCTHO viRv For ANY circuit in Part B

The voltage source is called the THEVENIN EQUIVALENT SOURCE

The resistance is called the THEVENIN EQUIVALENT RESISTANCE

R T H

i +

_OvOCv

+_

PART A MUST BEHAVE LIKETHIS CIRCUIT




A NYPA RT B

a

b_Ov

i

THEVENIN APPROACH

Norton Approach

SCiTHR

Ov

a

b

i

Norton

TH

O

TH

OCTHOCO R

v

R

viiRvv




A NYPA RT B

a

b_Ov

i

SCTH

OC iR

v

Source Equivalent Norton SCi

Part Afor tionRepresenta

Equivalent Norton

R T H

i +

_OvOCv

+_

Thevenin

TH

OCSC R

vi

This equivalence can be viewed as a source transformation problemIt shows how to convert a voltage source in series with a resistorinto an equivalent current source in parallel with the resistor

SCiTHR

Ov

a

b

i

Norton

ANOTHER VIEW OF THEVENIN’S AND NORTON’S THEOREMS

SOURCE TRANSFORMATION CAN BE A GOOD TOOL TO REDUCE THECOMPLEXITY OF A CIRCUIT

EXAMPLE: SOLVE BY SOURCE TRANSFORMATION

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

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

In between the terminals we connect a currentsource and a resistance in parallel

In between the terminals we connect a voltagesource in series with the resistor

The equivalent source has value 4mA*2k

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

After the transformation the sources can be combined

The equivalent current source has value 8V/4kand the combined current source has value 4mA

Options at this point

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

2. Use current divider to compute I_0 and thencompute V_0 using Ohm’s law

A General Procedure to Determine the Thevenin Equivalent

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

SC

OCTHOCTH i

vRvv ,




a

b_

0v

SCi

abI

Second circuit problem

Resistance Equivalent Thevenin

circuitshort aby

replaced is BPart if b - a throughcurrent

CurrentCircuit Short

removed is BPart if b-aat oltage v

ltageCircuit vo Open

SC

THTH

SC

TH

i

vR

i

v

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

AN EXAMPLE OF DETERMINING THE THEVENIN EQUIVALENT

+-

a

b

T o P a rt BV S

R 1

R 2IS

Part B is irrelevant.The voltage V_ab will be the value of theThevenin equivalent source.

What is an efficient technique to compute theopen circuit voltage?

THV

Now for the short circuit currentLets try source superposition

SCI

1R

VII SSSC

When the current source is open the current through the short circuit is

1

1

R

VI SSC

When the voltage source is set to zero,the current through the short circuit is SSC II 2

To compute the Thevenin resistance weuse

SC

THTH I

VR

S

STH I

R

V

RR

RRV

121

21

For this case the Thevenin resistance can be computed asthe resistance from a - b when all independent sources have beenset to zero

SSTH

SS

TH

SSTHTH

IRR

RRV

RR

RV

IR

VV

RR

IR

VV

R

V

21

21

21

2

121

12

)11

(

0

NODEANALYSIS

21

21

RR

RRRTH

Determining the Thevenin Equivalent in Circuits with Only INDEPENDENT SOURCES

The Thevenin Equivalent Source is computed as the open loop voltage

The Thevenin Equivalent Resistance CAN BE COMPUTED by setting to zero all the sourcesand 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

“Part B”

kRTH 3

“Part B”

kRTH 4

Since the evaluation of the Theveninequivalent can be very simple, we can add it to our toolkit for thesolution of circuits!!

V6

k5

“PART B”

][1)6(51

1VV

kk

kVO

EXAMPLE

In the region shown, one could use source transformation twice and reduce that part toa single source with a resistor.

... Or we can apply Thevenin Equivalenceto that part (viewed as “Part A”)

kRTH 4 For the open loop voltagethe part outside the regionis eliminated][8][12

63

6VVVTH

The original circuit becomes...

And one can apply Thevenin one more time!

kR TH 41

1THV

For open loop voltage use KVL

VVmAkVTH 1682*41

...and we have a simple voltage divider!!

VVV 8][1688

80

COMPUTE Vo USING THEVENIN

Or we can use Thevenin only once to get a voltage divider

“Part B”

For the Thevenin resistance

For the Thevenin voltage we have to analyze thefollowing circuit METHOD??

Source superposition, for example

Contribution of the voltage source

VVVOC 81263

61

Contribution of the current source

VmAkkVOC 8)2(*)22(2

Simple Voltage Divider

Thevenin Equivalent of “Part A”

kRTH 8

COMPUTE 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 THEVENINPART B

THV

023

12

mA

k

VTH

kkRTH 43

+-

THR

THVk2

OV

][3

4)6(

72

2VVVO

MAXIMUM POWER TRANSFER

From PreAmp(voltage ) To speakers

+-

R T H

V T H

The simplest model for aspeaker is a resistance...

+-

R T H

VT H SPEAKERM O DEL

BASIC MODEL FOR THE ANALYSIS OF POWER TRANSFER

http://angelfire.com/ab3/mjramp/index.html

Courtesy of M.J. Renardson

MAXIMUM POWER TRANSFER

+-

SO U R CE

(L O A D )

R TH

V TH

R L

LV

THLTH

LL

L

LL V

RR

RV

R

VP

;

2

2

2 THLTH

LL V

RR

RP

For every choice of R_L we have a different power.How do we find the maximum value?

Consider P_L as a function of R_L and find the maximum of such function

4

22 2

LTH

LTHLLTHTH

L

L

RR

RRRRRV

dR

dP3

Set the derivative to zero to find extreme points.For this case we need to set to zero the numerator

02 LLTH RRRTHL RR *

The maximumpower transfer theorem

The load that maximizes the power transfer for a circuit is equal to the Thevenin equivalent resistance of the circuit.

The value of the maximumpower that can betransferred is

TH

THL R

VP

4(max)

2

ONLY IN THIS CASE WE NEED TO COMPUTE THE THEVENIN VOLTAGE

Documents

ADDITIONAL ANALYSIS TECHNIQUES DEVELOP THEVENIN’S AND NORTON’S THEOREMS These are two very powerful analysis tools that allow us to focus on parts of a