LEARNING BY APPLICATION FIND THE THEVENIN EQUIVALENT FOR THE UNKNOWN ELEMENT USING A RESISTOR AND A...

LEARNING BY APPLICATION FIND THE THEVENIN EQUIVALENT FOR THE UNKNOWN ELEMENT USING A RESISTOR AND A VOLTMETER

MEASURED ACROSSTEST RESISTOR

circuit)

open in

(measured

V4.2

VVTH 4.2

mA8.0 V6.1

V4.2 kRTH 2

EQUIVALENT FORELEMENT+SOURCE

DESIGN EXAMPLE Implement the fine/coarse adjustment

1 1

2 20TUNE COARSE FINEV V V

• Sum of terms suggests superposition• gains less than one suggest voltage divider

Possible Circuit

_ 2

2 1

|| 1

( || ) 2TUNE C

COARSE

V R R

V R R R

_ 1

1 2

|| 1

( || ) 20TUNE F

FINE

V R R

V R R R

DESIGN EQUATIONS2 EQS AND THREE UNKNOWNS!

INFINITE POSSIBLE SOLUTIONS.USE OTHER CRITERIA PLUS ENGINEERING JUDGMENT

1 2

. ., 1 (reasonable choice)

R 900 , 9

e g R k

R k

Circuits forsuperposition

http://www.wiley.com/college/irwin/0470128690/animations/swf/5-23.swf

DESIGN EXAMPLE DESIGN AN ATTENUATOR PAD

DESIGN EQUATIONS_ _

50TH IN TH OUT

R R 1

10OUT

S

V

V

_ 2 1 2|| ( ) 50

TH IN LR R R R R

_ 2 1 2|| ( ) 50

TH OUT SR R R R R

Dependent Eqs!L S

R R

1

1 2

TH S

S

RV V

R R R

_

1

2L

OUT TH TH

L TH OUT

RV V V

R R

1

1 2

1

2OUT S

S

RV V

R R R

1

1 2

1 1

2 10L

R

R R R

1

2

SOLVING THE EQUATIONS YIELDS

1 220.83 , 33.33R R

Analysis of Solution• requires special, high accuracy resistors• small resistance may imply large power dissipated• may require large power rating to avoid heating

DESIGN EXAMPLE DESIGN A CIRCUIT TO REALIZE THE EQUATION3 2000 [ , ]

O S S O SV V I V inVolts I in mA

ANALYSIS OF THE REQUIREMENTS• sum of voltage and current• gains larger than one• inverting

Proposed solution

ANALYSIS OF PROPOSED CIRCUIT

2 1

0 (infinite gain)

@ : 0

A

O S

S

V

V VA I

R R

2

2

1

O S S

RV R I V

R

OTHER METHODS• superposition• Norton (see book)

ANALYSIS OF SOLUTION• 2k is standard resistor• 667 is 1k||2k• uses standard components!

22000R 2 1 1

3 667R R R

DESIGN EXAMPLE USE A SERIES RESISTOR WITH EACH FAN TOSENSE CURRENTPROVIDE AN INDICATION OF TOTAL AIRFLOW

100F F

V I 200CFM F

F I

• VOLTAGE DROP ON SENSING RESISTOR CANNOT EXCEED 2% OF NOMINAL 24V FAN VOLTAGE• 1V = 50CFM FOR THE INDICATOR

CONSTRAINTS

Design of sensor

24 0.48235.2

1000.48

F

SENSE F

I mA

R I

2.04SENSE

R 2 0.11

SENSE F SENSEP I R W

Design of Indicator

5 6R R0.0102

200SENSE

SENSE SENSE F CFM CFM

RV R I F F

Inverter

Adderinverter

4 4 4

3 3 3

0.0102 || 50 1 1 0.0102 50O SENSE CFM CFM O

R R RV V F F V

R R R

DESIGN EQUATIONS!296.13

4 R

R

LEARNING BY DESIGNCURRENT OVERLOAD SENSORSV

SV

v

SVv51

50

vv

0501

:

k

vV

k

vV AS

KCL@v

senseSSA VVVV 50)(50

2battV

VVAI Abatt 69 TREQUIREMEN DESIGN

0133.06)(950 sensesense RVAR

THIS POINT MUST GO HIGH WHEN CURRENT EXCEEDS 9A

VVbatt 12

VVVV OS 500.15.0 OUTOUT INPUT

TREQUIREMEN DESIGN

21,, RRVref DETERMINE

SV

012

R

VV

R

VV SrefSO

- v@KCL

refSO VR

RV

R

RV

1

2

1

21

ref

ref

VR

R

R

R

VR

R

R

R

1

2

1

2

1

2

1

2

)0.1(15

)5.0(10

EQUATIONS DESIGN

VV

R

R

ref 9

5

91

1

LEARNING BY DESIGN

kR 101CHOOSE

GENERATES Vref AND ISOLATESVOLTAGE DIVIDER

Analyzing circuit using superposition