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EQUIVALENT FORELEMENT+SOURCE
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
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
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
4
3
1.96 2R
R DESIGN EQUATIONS!
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