et438b-4.pptx

1Lecture Notes Part 4

ET 483b Sequential Control and Data Acquisition

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Measuring Instrument Characteristics

Static Characteristics

Error = (measured value) – (ideal value)

Ways of expressing instrument error

1.) In terms of measured variableExample ( + 1 C, -2 C)

2.) Percent of spanExample (0.5% of span)

3.) Percent of actual output Example (+- 1% of 100 C)

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Measuring Instrument Characteristics

The difference between the upper and lower measurement limits of an instrument define the device’s span

Span = (upper range limit) – (lower range limit)

Resolution is the smallest discernible increment of output. Average resolution is given by:

N

100 (%) Resolution Average

Where: N = total number of steps in span100 = normalized span (%)

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Instrument CharacteristicsExample: A tachogenerator (device used to measure speed) gives an output that is proportional to speed. Its ideal rating is 5 V/ 1000 rpm over a range of 0-5000 rpm with an accuracy of 0.5% of full scale (span) Find the ideal value of speed when the output is 21 V. Also find the speed range that the measurement can be expected to be in due to the measurement error.

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Example Solution

5

Determine the maximum output voltage

GnV maxmax

Where:Vmax = maximum output voltagenmax = maximum speedG = tachogenerator sensitivity (V/rpm)

V 25rpm) V/1000 5(rpm) 5000(V

rpm V/1000 5 G

rpm 5000n

max

max

Find ideal value of speed

Idealspeed

Find Vmax

rpm 4200V/rpm 0.005

V 21

rpm) V/1000 5(

V 21

G

Vn

V 21V

out ideal

out

Accuracy +-0.5% of full scale +-0.005(5000) = +-25 rpm

Speed range 4200+25 = 4225 rpm 4200-25 = 4175 rpm

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Span, Resolution, and SensitivityA 1200 turn wire-wound potentiometer measures shaft position over a range from -120 to +120 degrees. The output range is 0-20 volts. Find the span, the sensitivity in volts/degree, the average resolution in volts and percent of span.

V 01667.01200

V 20(V)resolution

span of % 0833.01200

100%(%)resolution

V/degree 0833.0240

020

span

VVysensitivit

240)120(120(span

minmax

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Repeatability and Accuracy

7

Repeatability - measurement of dispersion of a number of measurements (standard deviation)Accuracy is not the same as repeatability

Example+++++

+ +

++ +

+

+ ++

++

+++++

Not repeatableNot accurate

RepeatableNot accurate

RepeatableAccurate

Ideal Value

Reproducibility - maximum difference between a number of measurements taken with the same input over a time interval

Includes hysteresis, dead band, drift and repeatability

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Calibration CurvesDetermining the accuracy of a measuring instrument is called calibration. Measure output for full range of input variable. Input could be increased then decreased to find hysteresis. Repeat input to determine instrument repeatability.

Increasing Input Measurements

Input1i

0

10

20

30

40

50

60

70

80

90

100

Output1i

0.06

9.80

19.69

29.65

39.70

49.85

60.2

70.16

80.21

90.19

100.08

1Input2i

100

90

80

70

60

50

40

30

20

10

0

Output2i

100.08

87.24

77.26

66.22

57.12

46.80

34.70

25.73

16.75

8.83

0.01

Decreasing InputMeasurements

2

0 20 40 60 80 10020

0

20

40

60

80

100

Output1i

Output2i

,Input1iInput2

i

Plot the data

1

2

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Calibration Curve Characteristics

Hysteresis and Dead BandDifference between upscale and downscale tests called hysteresis and dead band

0 20 40 60 80 10020

0

20

40

60

80

100

Output1i

Output2i

,Input1iInput2

i

Hysteresis &Dead Band

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Calibration Curve Characteristics

LinearityIdeal instruments produce perfectly straight calibration curves. Linearity is closeness of the actual calibration curve to the ideal line.

Types of Linearity Measure

% Input

% O

utp

ut

Average up and down

scale values

Zero-basedline

Terminal-basedline

Least-squares

line

Least-squaresminimizes thedistance betweenall data points

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Dynamic CharacteristicsFirst order instrument responseFirst order model

transfer function s1

G

)s(C

)s(Cm

For step inputs

K)s(C

Step response)s1(s

KG)s(Cm

Where : Cm(s) = instrument output C(s) = instrument input G = steady-state gain of instrument

= instrument time constant

with K = step input size (1 for unit step) Exponentially

increasing function time

constant

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Dynamic CharacteristicsTime required toreach 63.2% offinal value is time constant, tt=2

Time required togo from 10% to90% of final valueis the rise time, tr

t90 – t10 = tr

0 2 4 6 8 10 12 14 160

20

40

60

80

100Sensor Step Response

Time (Seconds)

% O

utpu

t

63.2

63.2%

90%

10%

t90 = 4.57 S

t10 = 0.22 S

tr=4.57 S - 0.22 S=4.35 S

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Dynamic CharacteristicsTypical Instrument time constants

Bare thermocouple in air (35 Sec) Bare thermocouple in liquid (10 Sec)

Thermal time constant determined by thermal resistance RT

and thermal capacitance CT. t = RT∙CT

Example: A Resistance Temperature Detector (RTD) is made of pure Platinum. It is 30.5 cm long and has a diameter of 0.25 cm. The RTD will operate without a protective well. Its outside film coefficient is estimated to be 25 W/m2-K. Compute: a.) the total thermal resistance of the RTD, b.) the total thermal capacitance of the RTD, c.) The RTD thermal time constant.

Example Solution

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RTD

To signalConditioner

a.) Find the surface area of the probe to find RT

L=30 cm

D=0.25 cm

m 305.0cm 100

m 1cm) 5.30(L

m 0025.0cm 100

m 1cm) 25.0(d

23232621T

232

2622

1

m 104.2m 102.395m 1091.4AAA

m 102.395m) (0.305m) 0025.0(dLA

m 1091.44

)m 0025.0(

4

dA

ho = 25 W/m2-K

K/W 67.16R

)m 104.2()K W/m25(

1

Ah

1R

T

232To

T

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Example Solutionb.) Find the volume of the probe to find CT

mT SVC

Where: r = density of Platinum = 21,450 Kg/m3

V = volume of probeSm = specific heat of Platinum = 0.13 kJ/Kg-K

Find volume of cylinder

3622

m 10497.1m 305.04

)m 0025.0(L

4

dV

Now find the thermal capacitance

J/K 174.4C

kJ 1

J 1000)K-kJ/Kg 13.0()m 10497.1()Kg/m 450,21(C

SVC

T

363T

mT

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Example Solutionc.) Find the RTD time constant

S 6.69

J/K) (4.174K/W) 67.16(

CR

J/K 174.4C

K/W 67.16R

TT

T

T

0 100 200 300 4000

20

40

60

80

100RTD Response

Time (Seconds)

Per

cent

Out

put

63.2

5

RTD Response curve

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COMMON MODE VOLTAGES IN INSTRUMENTATION AND CONTROL

Common mode voltages are voltages that have the same magnitude and phase shift and appear at the inputs of a data acquisition system. Common mode voltages mask low level signals from low gain transducers.

Data recordingsystem

Vs

Vcmn

Vcmn

Induced voltage and noise

Sensor and signal conditioning source

Common mode voltages also appear on shielded systemsdue to differences between input potentials

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Common Mode Voltage Due to Inputs

Common mode voltage due to ground

2

VVV

VVV

cmg

d

-

+

V+

V-

VoVd

Differential Amp

Total common mode voltage Vcm = Vcmn+Vcmg

OP AMP differential inputs designed to reject common mode voltages. Amplify only Vd = V+ - V-.

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Common Mode Voltages and OP AMPs

19

Define: Ac = gain of OP AMP to common mode signals (designed to be low)

Ad = differential gain of OP AMP. Typically high (Ad = 200,000 for 741)

Ideal OP AMPs have infinite Ad and zero Ac

Common mode rejection ratio (CMRR) is a measure of quality for non-ideal OP AMPs. Higher values are better.

c

d

A

ACMRR

Where Ad = differential gainAc = common mode gain

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Common Mode Rejection

20

Common Mode Rejection (CMR) calculation

CMRRlog20A

Alog20CMR

c

d

CMR units are db. Higher values of CMR are better.

Example: A typical LM741 OP AMP has a differential gain of 200,000. The typical value of common mode rejection is 90 db. What is the typical value of common mode gain for this device

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Common Mode Rejection Example Solution

From problem statement Vd = 200,000 CMR = 90 db

c

d

A

Alog20CMR

Solve for Ac by using the anitlog

CMR A

Ad

c20

log Raise both sides to

powerof 10

10 20

CMRd

c

A

A

Solve for Ac

AAd

CMR c

10 20

Plug in values and get numericalsolution

c5.420

90 A32.6623,31

000,200

10

000,200

10

000,200 Common mode gain is

6.32 for typical LM741

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Difference Amplifier and Instrumentation Amplifiers

22

Characteristics of Instrumentation Amplifiers

- Amplify dc and low frequency ac (f<1000 Hz)- Input signal may contain high noise level- Sensors may low signal levels. Amp must have high

gain.- High input Z to minimize loading effects- Signal may have high common mode voltage with

respect to ground

Differential amplifier circuit constructed from OP AMPs are the building block of instrumentation amplifiers

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Basic Difference Amplifiers

23

Amplifies the difference between +/ - terminals

Input/output Formula

To simplify let R1 = R3 and R2 = R4

V R

RV

R

R

RR

RRV 1

1

22

1

4

34

12o

)VV(R

RV 12

1

20

Polarity of OP AMP input indicatesorder of subtraction

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Basic Difference AmplifiersPractical considerations of basic differential amplifiers

- Resistor tolerances affect the CMRR of OP AMP circuit. Cause external unbalance that decreases overall CMRR.

- Input resistances reduce the input impedance ofOP AMP

- Input offset voltages cause errors in high gainapplications

- OP AMPs require bias currents to operate

Difference Amplifier Loading Effects

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To minimize the loading effects of the OP AMP input resistors, their values should be at least 10x greater than the source impedance

Example: Determine the loading effects of differential amp Input on the voltage divider circuit. Compare the output predicted by differential amplifier formula to detailed analysis of circuit.

R1= 5kW

R3= 5kW

R2= 5kW

R410kW

R510kW

R6 =10kW

R7 =10kW

1 Vdc Assume no loading effects anduse the OP AMP gain formula

)VV(4R

6RV 120

VR22R12 V)VV()VV(

V -0.3333V 1k15

k5V

V 1k5k5k5

k5V

3R2R1R

2RV

2R

2R

+

I

-

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Difference Amplifier Loading EffectsFind the output ignoring the loading effects that the OP AMP has onthe voltage divider.

V 333.0333.0k 10

k 10V

V4R

6RV

0

2R0

Now solve the circuit and include the loading effects of the OP AMP input resistors. Use nodal analysis and check with simulation.

Remember the rules of ideal OP AMPs:

Iin = 0 and V+=V-

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Difference Amplifier Loading Effects

Solution using nodal analysis

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Difference Amplifier Loading Effects

Solve simultaneous equations and determine percent error due to loading

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Difference Amplifier Loading Effects

Results of operating point analysis in LTSpice

V1 =0.514 V

V2 =0.229 V

V0 =0.286 V

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Example: Measuring dc current with a current shunt

Dc motor draws a current of 3A dc when developing full mechanical power. Find the gain of the last stage of the circuit so that the output voltage is 5 volts when the motor draws full power. Also compute the power dissipation of the shunt resistor

12VdcI=3 A

0.1W

100 kW

100 kW

820 kW

820 kW

10 kW Rf = ?

0 - 5 Vdc

+

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Example: Measuring dc current with a current shunt

Example Solution

12VdcI=3 A

0.1W

100 kW

100 kW

820 kW

820 kW

+ 2.46 V

0.3

00

V

+ V

d

-

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Example Solution

10 kW Rf = ?

0 - 5 Vdc

2.46 V

Caution: Note the maximum differential Voltage specification of OP AMP. (30 V for LM741)

Rf is a non-standard value. Use 8.2 kΩ resistor and 5 kΩ potentiometer. Calibrate with 300 mV sourceUntil 5.00 V output is achieved

Compute powerdissipation at full loadI= 3 A so….

Use 1 Watt or greaterStandard value

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Example Solution Simulation

DC V 300.0mV

DC V 2.453 V

DC V 4.992 V

10V

-10V

Vo

R85k 43%

+

U1BLM324

12V

10V

+

U1ALM324

-10V

Ra3.9

R18.2k

R710k

R60.1 R5

100k

R4100k

R3820k

R2820k

Simulated with Circuit Maker (Student Version)