Slide 1

ET 438B Sequential Control and Data Acquisition

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Slide 2

Angular and Linear Position Sensors
Simple linear displacement
transducers: Slide
Potentiometer

Simple angular transducers: Rotary
Potentiometers, multi-turn or single,
Wire-wound or carbon composite

Slider
Es

L
x

Eout
E out 

E out 

x
L


L

Eout

Es

Es
Es
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Slide 3

Wire-wound potentiometers do not have infinite resolution
due to wire diameter

Resistance varies with
position

Cross
section
Number of turns per unit length determines resolution of
the potentiometer
Resolution (%) = 100/N
Where
N = number of turns in the pot.
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Slide 4

Es

Req

Rp = total potentiometer R
aRp = fraction of potentiometer R
from wiper to ground

a


E0  
 Es
 1  a  r  a2  r 
Rp
r 
RL

Rp
+
aRp

RL

E0
-

RL is load on potentiometer. Must
include in calculation
R L || a  R p

Solve using voltage division

Loading error LE defined as = aEs -Eout
 a 2  r  (1  a ) 
%L E  
 100
 1  a  r  (1  a ) 
Rp
r 
RL

R eq  (1  a )  R p  R L || a  R p
et438b-5a.pptx

Percent Loading Error
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Slide 5

Problem Statement
A linear wire-wound potentiometer has a total
resistance of 50,000 ohms The voltage output of the
potentiometer will be converted to a digital value by a
12 bit ADC. Determine minimum number
of turns required so that the resolution of the
potentiometer does not exceed the quantization error of
the ADC.
What percent loading error can be expected if the input
resistance of the ADC is 100 kW and the range of
measurement is from 20% to 80% of the potentiometer
value?
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Slide 6

Find the resolution of the 12-bit ADC then convert it to a percentage.
Assume 1 volt range
Q . E. 

VL S B
2

VL S B 

VF S
2

n



1V
2

12



1
4096

 2 .4 4 1  1 0

et438b-5a.pptx

4

V

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Slide 7

Loading error depends on position of potentiometer
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Slide 8

Note: dots indicates
instantaneously positive
voltages in transformers

Secondary

Primary
Ls1

Vp

Vp must be AC typical
frequency 50k - 15 kHz

Vs1

Lp

Core Motion causes Vs1
and Vs2 to change
Ls2

Vs2

Lp and Ls1 coupled Vs1 Increasing

Movable
Core

Lp and Ls2 coupled Vs2 Increasing

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Slide 9

Coils connected so voltages subtract
V out +

Output V

0.06-

0

+

0.06
Core Position

-

+ core
position

V out opposite phase
core coupling
Lp and Ls1

et438b-5a.pptx

core neutral

core coupling
Lp and Ls2

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Slide 10

Three parts to system 1.) Control Transmitter 2.) Control Transformer
3.) Control Differential
ac
source

Control
transmitter

Control
transformer

Vout
+
-

Vout

time

Sinusoidal output with maximum value from
this position
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Slide 11

ac
source

Control
transmitter

Control
transformer

a
c

-

+

Vout

Output voltage is zero when rotors in this position

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Slide 12

ac
source

Control
transmitter

Control
transformer

a
c

Vout

+

Output voltage 180 degrees out of phase from initial position

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Slide 13

Output voltage magnitude is a function of the angular
displacement between the control transmitter and the
control transformer
V o u t  (E m c o s (  )) s in (  t )

Where

Em = the maximum amplitude of excitation
 = the angular position difference between
the transmitter and the transformer rotor
 = frequency of excitation voltage

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Slide 14

Control differential transformer can be added to add
constant phase shift.
V o u t  (E m c o s (    d )) s in (  t )

Sending and receiving devices will have a constant
angular difference given by the value of d.
System can be used to maintain a constant position
difference between two shafts. It can also be used to
keep two shafts synchronized.

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Slide 15

A syncho system operates at a frequency of 400 Hz. The maximum
amplitude of the transformer rotor voltage is 22.5 V. Determine the ac
error signal produced by each of the following pairs of angular
displacements
a.) = 90° d=0°
c.) = 135° d=-15°
At 400 Hz

b.) = 60° d=0
d.) = 100° d=-45°

  2   ( 400 Hz)  2570 rad/s

a.)

b.) V  ( 22 . 5  cos( 60  0 )) sin( 2570  t )
0

V 0  ( E m  cos(    d )) sin(   t )

V 0  11 . 25 sin( 2570  t )

E m  22 . 5
V 0  ( 22 . 5  cos( 90  0 )) sin( 2570  t )  0
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Slide 16

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Slide 17

Velocity Measurements
Angular and Linear Velocity
Angular Velocity Measurement Methods
- dc tachometer
- ac tachometer
- optical tachometer
DC tachometer
N

Dc generator using
permanent magnets for
field flux

S

Magnetic Field

Field Poles

Brushes and
Commutators
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Slide 18

Tachometer produces a dc voltage that is proportional to
angular velocity
Proportionality constant = emf constant KE (V/rpm)
KE depends on the construction of tachometer

KE 

2  R  B  N  L
60

E  K E s 

30  K E  


emf
constant
Induced
emf

Where
E = tachometer output (V)
KE = emf constant
s = angular velocity (rpm)
 = angular velocity (rad/s)
R = average radius (m)
B = flux density (Wb/m2)
N = number of conductors
L = length of conductor in field (m)

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Slide 19

Example: A dc tachometer has the following parameters
R = 0.03 m
N = 220
L = 0.15 m
B = 0.2 Wb/m2
Find KE and the output voltage at the following speeds s = 1000,
2500, and 3250 rpm
KE 

For s = 1000 rpm

2 R  B  N  L
60



2  ( 0 . 03 m)(0.2 Wb/m

K E  0 . 0207 V/rpm

2

)( 220 )( 0 . 15 m)

60

E  K E s
E  0 . 0207 V/rpm  1000 rpm   20 . 7 V

For s= 2500 rpm
And 3250 rpm

E 2  0 .0 2 0 7 V / rp m (2 5 0 0 rp m ) = 5 1 .8 V
E 3  0 .0 2 0 7 V / rp m (3 2 5 0 rp m ) = 6 7 .3 V
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Slide 20

Ac Tachometers Construction : a 3-phase alternator with a 3-phase
rectifier to convert the output to dc. Must have constant field excitation Permanent magnet field.

Non-linear at low speeds due to the forward drop of diodes Limited to lower
speed ranges due to this. Range 100-1
Ac Tachometers can also produce a variable frequency output that has a
constant voltage.
Use frequency-to-voltage conversionet438b-5a.pptx
to get proportional dc

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Slide 21

Digital encoder
attached to shaft
produces a sequence
of pulses.

3 light
sources

3 light
sensors

Encoder tracks
Inner – locates home
Middle - gives direction info
Phase shift between outer
and middle tracks
Middle leads or lags outer
based on direction
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Slide 22

Optical Tachometers
Pulses counted over time interval. Counter is then reset.
The number of pulses counted in interval is proportional to
the angular velocity

Formulas

s
C 

60  C
N  Tc
s  N  Tc
60

Where
s = shaft speed (rpm)
N = number of pulses per shaft
revolution
C = total count during time period
Tc = counter time interval (sec)

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Slide 23

Incremental encoder produces 2000 pulses/rev.
a.) determine count produced by shaft speed of 1200
rpm with a count interval of 5 mS
b.) determine the speed measured at a count of 224
for a timer interval of 5 mS
a.) C 

s  N  Ts



1200

rpm   2000   0 . 005 s 

60

 200

60

Number of counts for 1200 rpm speed

b.)

s

60  C
N  Ts



60  224



2000  ( 0 . 005 s)

 1344 rpm

Speed for 224 counts
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Slide 24

Definition - acceleration is rate of change of velocity
a 

Sensing Methods

dv
dt

 lim it

v
t

Newton’s Law f = Ma
Where: f = force acting on body
M = mass of body
a = acceleration

Can measure acceleration by measuring force required to accelerate
known mass.
For angular acceleration, differentiate the velocity
signal from velocity sensors
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Slide 25

Sensing Methods
For angular acceleration, differentiate the velocity signal
from velocity sensors
For sampled data systems, derivative is given by difference of
readings

at 

v
t

Where
at = acceleration at time t
vt-1 = velocity sample at t-1
vt = velocity sample at t
Ts = sampling time



v t 1  v t
Ts

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Slide 26

Springs

General Structure
K/2

Mass

M

Accelerometer attached to device under
test and experiences same acceleration
as measured object. Motion of mass
damped by viscous fluid around mass
Theory of operation same for
Integrated devices.

K/2

Springs
Displacement
Sensor (LVDT)
et438b-5a.pptx

Applications
Gaming
Automotive
Appliance control
Disk Drive protection
Bearing Condition Monitoring
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Slide 27

Accelerometers

Accelerometer forms a second order mechanical system.
Accelerometer in action

For small deflections,
spring force offsets
force due to acceleration so....

Displacement
caused by
acceleration

Kx = Ma
Which gives
x = (M/K)a
so acceleration is
proportional to displacement
f = Ma

Direction of
acceleration
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Slide 28

Acceleration is dynamic measurement, interested in
a(t), acceleration as function of time.
Consider second order response

fo 
 

1

K

2

M
b

2

Where fo = resonant frequency of accelerometer (Hz)
4K M
 = damping ratio
K = spring constant (N/m)
fa = max. frequency of a(t)
M = mass (Kg)
b = damping constant (N-s/m)

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Slide 29

Accuracy of response related to resonant frequency of
device
Must have sufficient range to measure changes
if fa << fo then measurements will be accurate
if fa >>fo then the mass does not have time to react
incorrect measurement
fa = fo then displacement is greatly exaggerated
incorrect measurement
Make fo at least 2.5 times greater than fa max for < 0.5% error

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Slide 30

The accelerometer shown below has the following specifications:
M = 0.0156 kg
K = 260 N/m
b = 2.4 N-s/m
xmax = +- 0.3 cm

K/2

b

K/2

M

Find the following
a.) maximum acceleration that can be
measured
b.) resonant frequency
c.) damping ratio
d.) maximum frequency that can be used
with 0.5% error or less

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Slide 31

Accelerometer Example
M = 0.0156 kg K = 260 N/m
b = 2.4 N-s/m xmax = +- 0.3 cm

K/2

a.) k  x  M  a 
b

M

a max 

k  x max
M

 a max

( 260 N/m)  0 . 003 m 

 50 m/s

2

0 . 0156 kg

b.)
f0 

K/2

c.)


1
2

k



b



M
2

4k M



1
2



260 N/m

 20 . 6 Hz

0 . 0156 kg
( 2 . 4 N - s/m )

2

4 ( 260 N/m)  (0.0156 kg)

 0 . 595

d.) Maximum frequency of a(t) with 0.5% error is…..
f 0 / 2 .5

f max  20 . 6 Hz/2.5  8.25 Hz
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Slide 32

All force measurement based on force balance
f = M∙a
Where : f = force (N) M=Mass (kg) a=acceleration (m/s2)
Null Balance – unknown force off set by know weight (beam balance)
When weights equal
scale balanced

Weight= f

M∙a

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Slide 33

Displacement – displacement of elastic material with unknown
balancing force determines measurement (Spring Scale) (Strain gage
load cell)
F

Elastic properties of beam determines
Displacement due to external force

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Slide 34

Strain gages turn changes in displacement into changes in electric
resistance. Fine wire bonded to plastic base.
Bonded strain gages measure strain at specific location on deformable
body. Cemented to material and change length when body deforms.
L

Change in length = L+L
Define strain:



L
L

Where L = change in length due to force
(m)
L = unforced length (m)
L

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Slide 35

Gage factor determines the sensitivity of the sensor
R
G  R
L
L

Typical values of strain gage parameters:
G: 2 to 4
R: 50 to 5000 ohms
L: 0.5 to 4 cm (unstrained)

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Slide 36

Define Stress

S

f
A

Where
S  stress (N/m )
2

f  force (N)
A  area (m )
2

Stress and strain related through the modulus of elasticity for a given
material

E 

S


Where E = modulus of elasticity (N/m2 )
S = stress (N/m2)
 = strain (m/m)
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Slide 37

Stress depends on the geometry of the object. Consider
the cantilever beam (diving board) below.
L

f

Strain Gage

S

A

Combining previous equations
R

 6G L 

f
2

R
bh E 

bh

2

Where:

A
Section
AA

6f L

b

h

S = stress (N/m2)
L= distance from fixed end (m)
f = applied force (N)
b = width of beam (m)
h = height of beam (m)

Normalized gage resistance change proportional
to force
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Slide 38

L

f

Strain Gage

A

Length =L = 2 m
Width =b = 20 cm
Height=h = 6 cm

A
Section
AA

Find the applied force

The beam structure
shown has the following
parameters

b

h

et438b-5a.pptx

It is made from
Aluminum with a
modulus of elasticity of
E = 6.9x1010 N/m2
The strain gage has an
unstrained R=100 Ω
G=3
R =0.073 Ω
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Slide 39

Define the variables
10

L  2 m

E  6.9  10

 Nm

b  20  cm

R  100  W

h  6  cm

 R  0.073  W

Restate the formula
 2

G  3

R

 6G L 

f
2

R
bh E 

Solve for force

f 



2



R b  h  E
( R 6 G L)

Substitute values and simplify
0.073W
  0.20m
  ( 0.06m
 )  6.9 10  N m
2

f 

10

 2



100 W  6 3 2 m
f  1007N
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Slide 40

Survey of Types
Resistance Temperature Detector (RTD)-change in resistance of pure metals
relates to temperature. Key features: wide temperature range, high accuracy,
excellent repeatability, good linearity. Need constant current source and other
electronics to produce output signal
Thermistor – temperature-sensitive semiconductor. Resistance inversely
proportional to temperature. Increasing temperature causes decreasing
resistance. Key features: high sensitivity, small size, fast response., narrow
temperature range Not recommended in applications requiring high accuracy.
Thermocouple – junctions of two dissimilar metals produces small (mV) voltages
when placed at different temperatures. Magnitude of voltage depends on
temperature difference. Key features: small size , low cost, rugged, wide
measurement range. Limitations: noise pickup, low signal levels, high minimum
span (40° C.)
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Slide 41

Survey of Types
Integrated Circuit Temperature Sensors –precision solid-state devices with
linear output to temperature. Directly calibrated to various temperature scales (
Celsius, Fahrenheit, Kelvin, Rankine). Key features: calibrated output voltages,
linear scaling, low voltage and current draws, good linearity. Typical devices
LM34, LM 35. Limitations: low temperature range typically (-55 C to 155 C)

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Slide 42

Construction: Coil of Nickel or Platinum wire in protective tube. Need
to detect small changes in resistance.
Platinum –high accuracy, linearity, and cost
Nickel –moderate cost, higher output than Platinum
Typical Relationship
R  R o (1  a 1  T  a 2  T )
2

Where: R = resistance at given Temp, T C
Ro = resistance at 0 degrees C
T = temperature
a1, a2 = constants

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Slide 43

Finding constants from data
Typical values for Platinum

Use table values to find constants
Ro, a1 and a2. Substitute data into previous
equation.





Resistance (W)

100  R o 1  a 1 0  a 2 0

0

100.00

109 . 9  100 1  a 1 25  a 2 25

25

109.90

50

119.80

75

129.60

100

139.30

T(°C)

2


139 . 9  100 1  a 100  a
1

 R o  100 W

  100 1  a 25  a 635 
100   100 1  a 100  a 10000 
2

1

2

2

2

1

2

Simplify equations and solve simultaneously
0 . 099  a 1 25  a 2 625
0 . 399  a 1100  a 2 10000

a 1  0 . 00395

Mainly linear

a 2  4  10

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Slide 44

Must convert RTD resistance changes into usable voltage or current
signals
Direct Methods (2 and 4 wire) - Constant current supplied to RTD
and voltage measured across it. The 4 wire method removes lead-wire
error.
Bridge Methods (2 and 3 wire) –Use dc bridge to convert resistance
changes into voltage changes. Three wire method removes most leadwire resistance error.

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Slide 45

Iin≈0

is
Rs

is

is

+
vs
-

Voltage
To
Current
Converter

4-20 mA
output

Design circuit using Platinum RTD

Zin very high
Iin approx. zero

1.) constant current source is = 1 mA
2.) Instrumentation amp. G=50
3.)Summing amp
4.) V-to-I converter 4 mA @ 0 C
20 mA @ 100 C
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Slide 46

5.1 V regulated source available. Remember the V-to-I converter . Use LM324\
quad OP AMPs.
R1

12V

Vs-12V
U1A
LM324

R2
1k

RTD
Rs
100
Vs+

+

D1
1N751

Is

12V

I s  1 mA
V in  5 . 1 V
Io 

V in
R1

 R1 

V in
Is



5 .1 V
1 mA

 5 .1 k W

Now design the Instrumentation
amplifier using the two stage
amplification circuit

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Slide 47

Av1

12V
+

Vs+

G=50 . Divide gain between
the two stages.

U1B
LM324

Av2
-12V
R3

R8

R5
R4

R9

Av1 = 5
Av2 = 10

R6

-12V

+

-12V

U1D
LM324
12V

Vo

R7
Vs-

+

 2R 3
 R 6 
Vo  
 1 
 ( Vs  )  ( Vs  ) 
 R5
 R 8 

U1C
LM324

12V

 2R 3

A v1  
 1  Let R3  10 k W
 R5


Gain formula
 2R 3
 R 6 
Vo  
 1 
 ( Vs  )  ( Vs  ) 
 R5
 R 8 

Where R3=R4 R8=R9 and R6=R7

 2 (10 , 000 )

5
 1
R5


et438b-5a.pptx

 R5 

20,000

 5 kW

4

Use 5.1 k W standard value

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Slide 48

Stage 2 gain is 10

 R6 
A v1  
 Let R6  56 k W
 R8 
56,000
 56 k W 
10  
 5 .6 k W
  R8 
R
8
10



Now determine the span of the RTD output voltage
@ 0 C (0.001 A)(100 W )  0 . 1V


@ 100 C (0.001 A)(139.3 W )  0 . 1393 V


V o (min)  50 ( 0 . 1 V)   5.0 V

After
Amp

V o(max)  50 ( 0 . 1393 V )   6 . 965 V

Desired output span 16 mA – 4 mA
Input span -6.965 - -5.0
Size the R of the V-to-I converter based on span ratio
1
R



I

o (max)

 I o (min)



( V in (max)  V in (min) )



 20

mA  4 mA



( 6 . 965 V  5 . 0 V )



16 mA

R 

1.965 V
et438b-5a.pptx

1.965 V

 123 W

16 mA

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Slide 49

Now determine the offset voltage for the 0 degree input to the converter.
Set Io to minimum and compute Vin using the computed R.
12V

Zero Set

Vb
R14
10k 37.25%

Vo

R10
470k

R11
470k

Span Set

R12
470k

R15
500 24.3%

-12V

123 W

-12V

RL
100

Io
+

U2A
LM324
12V

+

Vin

U2B
LM324
12V

V b  Vin (max)  Vin  (  1)(  5 . 0  0 . 492 V)

Io 

V in
R

 V in  I o  R

V in  ( 0 . 004 A)(123 W )  0 . 492 V

@ 4 mA

U2A inverting V
b
summer
Find value of Vb
et438b-5a.pptx

 4 . 508 V

49


Slide 50

Thermister Characteristic Curve

Therm is ter R eis t ance (O hm s )

6000

Photoresistor
has similar
characteristic

4000

2000

0
0

20

40

60

80

100

120

140

160

Temperature (Degrees C)

Very Non-linear- High sensitivity in (0-40 C range)
et438b-5a.pptx

50


Slide 51

Therm is ter R e is tance (O hm s )

Semi-log plot (y axis logarithmic) of thermister data

1  10

4

1  10

3

Approximately
linear over small
range 40-100 C

Thermiste r Characteristic Curve

Log(R)=mT+b

100

10
0

20

40

60

80

100

120

140

160

Temperature (Degrees C)

et438b-5a.pptx

51


Slide 52

Linearizing Thermister Characterisitcs
Vcc
Volta ge Divider Output Curve
10

Thermister

R=2000 ohms

RT
Vo


R
V o  
R  RT


 V cc



O utput V ol tage (V o)

R

8

6

4

Changing value of R
changes output curve
shape

2
0

20

40

60

80

100

120

140

160

Temp eratu re (Degrees C)

et438b-5a.pptx

52


Slide 53

Junctions of dissimilar metals produces voltage
proportional to temperature
Iron
V

Constantan

0˚C
Heat

Cold Junction
et438b-5a.pptx

ANSI Standard Type J
-190-800 ˚C
53