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LEARNING BY APPLICATION
FIND THE THEVENIN EQUIVALENT FOR THE UNKNOWN
ELEMENT USING A RESISTOR AND A VOLTMETER

2.4V
EQUIVALENT FOR
ELEMENT+SOURCE 
VTH  2.4V
(measured
in open
circuit)
 1.6V 
2.4V
RTH  2k
0.8mA
MEASURED ACROSS
TEST RESISTOR
http://www.wiley.com/college/irwin/0470128690/animations/swf/5-23.swf
DESIGN EXAMPLE
Implement the fine/coarse adjustment
VTUNE
Possible Circuit
1
1
   VCOARSE    VFINE
2
 20 
• Sum of terms suggests superposition
• gains less than one suggest voltage divider
Circuits for
superposition
R || R1
1
VTUNE _ C
R || R2
1 VTUNE _ F 



( R || R1 )  R2 20
VCOARSE ( R || R2 )  R1 2 VFINE
INFINITE POSSIBLE SOLUTIONS.
USE OTHER CRITERIA PLUS
ENGINEERING JUDGMENT
e. g ., R  1k  (reasonable choice)
 R1  900, R2  9k 
DESIGN EQUATIONS
2 EQS AND THREE UNKNOWNS!
DESIGN EXAMPLE
DESIGN AN ATTENUATOR PAD
DESIGN EQUATIONS
RTH _ IN
RTH _ IN  RTH _ OUT  50
VOUT
1

VS
10
 R2   R1 || ( R2  RL )  50
RTH _ OUT  R2   R1 || ( R2  RS )  50
1
RL  RS  Dependent Eqs!
VTH 
VOUT 
R1
V
R1  R2  RS S
RL
1
VTH  VTH
RL  RTH _ OUT
2
 VOUT
R1
1

VS
2 R1  R2  RS

R1
1
1

2 R1  R2  RL 10
2
SOLVING THE EQUATIONS YIELDS
R1  20.83 , R2  33.33
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 EQUATION
VO  3VS  2000 I S
[VO in Volts , I S in mA]
ANALYSIS OF THE REQUIREMENTS
• sum of voltage and current
• gains larger than one
• inverting
ANALYSIS OF PROPOSED CIRCUIT
VA  0 (infinite gain)
Proposed solution
VO
VS
@ A: 
 IS 
0
R2
R1
OTHER METHODS
• superposition
• Norton (see book)
R2
VO   R2 I S  VS
R1
 R2  2000
ANALYSIS OF SOLUTION
• 2k is standard resistor
• 667 is 1k||2k
• uses standard components!
 R2  3 R1  R1  667
DESIGN EXAMPLE
USE A SERIES RESISTOR WITH EACH FAN TO
SENSE CURRENT
PROVIDE AN INDICATION OF TOTAL AIRFLOW
VF  100 I F FCFM  200 I F
CONSTRAINTS
• VOLTAGE DROP ON SENSING RESISTOR CANNOT
EXCEED 2% OF NOMINAL 24V FAN VOLTAGE
• 1V = 50CFM FOR THE INDICATOR
Design of Indicator
Adder
inverter
Design of sensor
24  0.48
R
IF 
 235.2mA V
100
R
0.48  RSENSE I F
VO  4 VSENSE 
R3
RSENSE  2.04
R
 1.96  2
2
R
P  I R  0.11W
SENSE
4
SENSE
F
SENSE
3
IF 
SENSE
RSENSE
F
 0.0102 FCFM
200 CFM
Inverter
R5  R6
R4
R
0.0102FCFM || FCFM  50  VO  1  1  4 0.0102  50
R3
R3
DESIGN EQUATIONS!
LEARNING BY DESIGN
CURRENT OVERLOAD SENSOR
VS
Vbatt  12V
50
v  VS
51
KCL@v  :
VS  v VA  v

0
1k
50k
VS
v  v
v
Vbatt
2
THIS POINT MUST GO
HIGH WHEN CURRENT
EXCEEDS 9A
VA  50(VS  VS )  50Vsense 50  Rsense  9( A)  6V  Rsense  0.0133
DESIGN REQUIREMENT
I batt  9 A  VA  6V
LEARNING BY DESIGN
VS
DESIGN EQUATIONS
 R 
R1
R
9
0  1  2  (0.5)  2 Vref
R
R1
1
 R1 

5
 R2 
Vref  V
R2
5  1   (1.0)  Vref
9
R
R

1
1
CHOOSE R1  10k
DESIGN REQUIREMENT
INPUT
OUTOUT
0.5  VS  1.0V
0  VO  5V
DETERMINE Vref , R1, R2
KCL @ v-
GENERATES Vref AND ISOLATES
VOLTAGE DIVIDER
VO  VS Vref  VS

0
R2
R1
 R 
R
VO  1  2 VS  2 Vref
R1
 R1 
Analyzing circuit using superposition