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SEQUENTIAL DIGITAL CONTROL AND DATA ACQUISITION
ET 438bDepartment of Technology Southern Illinois University Carbondale
et438b-1 1
DIGITAL CONTROL OF ANALOG SYSTEMS
et438b-1 2
Set point
Error
Controller
SensorsFeedback
PlantControlOutput
-
+
Summing (error generation)Integration (integral control)Differentiation (derivative control)Amplification (proportional control)
Analog Controllers
Controller implemented with analog electronics (OP AMPS)
DIGITAL CONTROL OF ANALOG SYSTEMS
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3
Digital Controllers
Analog-to-Digital
Converter
Controller
Digital–to- Analog
ConverterPlant
Control Output
SensorsFeed Back
Setpoint
+
-
Error
CharacteristicsAnalog Inputs and outputsContinuous signals converted to digital valuesController - Implemented with microprocessor System control variable modified by mathematical functions + - x /Result converted to analog signal by digital-to-analog conversion
ADVANTAGES AND CHALLENGES OF DIGITAL PROCESS CONTROL
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Advantages
Can implement complex control algorithms along with P-I-DSoftware-based controller Direct input of digital sensors
Challenges
Need Analog-to-Digital (A/D) Conversion- World is analogHigh speed sampling required for rapidly changing signalsPrecision of converted value. Infinite number of values mapped to a finite number of bitsMust reconstruct most signals to analog for output to analog world. Need DAC (digital-to-analog converters)
OVERVIEW OF THE DATA ACQUISITION PROBLEM
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5
Use computers to gather data, process data, and control system: Higher level control
Monitored Systems
Computer System with data acquisition
Analog Output
sDACs
Analog InputsADCs
Digital In/Out
Application Program
Transducers
(Sensors)Signal
ConditioningAnalog Input Signals
Analog Output Signals
Digital Signal
Conditioning
OVERVIEW OF THE SEQUENTIAL CONTROL PROBLEM
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6
Control of a staged processes with discrete steps.
Examples around the home
Washing Machines
Dishwashers
Time-driven sequentialprocesses
Cloths Dryers
All processes driven by timingof the events
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OVERVIEW OF THE SEQUENTIAL CONTROL PROBLEM
Event-Driven Sequential Processes
Next step of process can not take place until an external event occurs
Robotic Arms
Motion sequence depends of position of mechanical part
Sensors are switches thatIndicate position
Examples
CONTROL PROGRAMMING AND DATA ACQUISITION USING LABVIEW
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8
LabVIEW is a graphical programming language that allows rapid development of programs that:
Read analog input signal data
Process and store data
Display data and system status
Read switch input (digital) signal data
Write analog output signals
Write digital output signals for on/off control
LABVIEW EXAMPLE
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9
Compact Florescent Light Testing Controller Analog
inputs read lamp
currentsDigital outputs
control lamp operation
Plot monitors lamp voltage
over time
Analog samples processed to
give RMS V and I values
LABVIEW PROGRAMMING ENVIRONMENT
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10
LabVIEW is a graphical programming environment
Create user interface
here
Front Panel
Block Diagram
(back panel
Create program
here
Controls and Indicators
Palette
Palette changes to programming functions when you click on block diagram
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11
LABVIEW PROGRAMMING ENVIRONMENT
Tool Palette
Program Control
Run/Stop Pause
Operate Value
PointerText
WiringDebug Tools
LABVIEW PROGRAMMING ENVIRONMENT
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12
In LabVIEW Input variables = controls Output variables = indicators
Numeric Controls
Numeric Indicators
BooleanControls
BooleanIndicato
rs
LABVIEW PROGRAMMING ENVIRONMENT
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Other data types in LabVIEW
Integers(Signed and Unsigned) I32 I16 I8 U32 U16 U8
ArraysCollect data of the same type. 1-D and multi-D Indexing begins at 0
ClustersCollect data of dissimilar data types. Only include indicators or controls
Data Structures in LabVIEW Programming
StringsArrays of characters
LABVIEW PROGRAMMING ENVIRONMENT
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Data Types and Structures in LabVIEW Programming
Array of Float Point
Array Icon
Integer I32 and U32
Integer Icons
String Control
String Icon
Array of Integers
Integer Array Icon
Cluster
Cluster
LABVIEW PROGRAMMING STRUCTURES
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15
FOR Loop – Repeats enclosed instructions for a predetermined amount of iterations (N)
WHILE Loop – Repeats enclosed instructions until stop condition is met
Index, i, in both structures holds current iteration number
Graphical Programming Structures
StopConditio
n
LABVIEW PROGRAMMING STRUCTURES
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16
IF-THEN/CASE Executes enclosed based on logical test (TRUE/FALSE) or Index (CASE)
Graphical Programming Structures
Logical test connects
here
TRUE condition executes these items
FALSE condition executes these items
LABVIEW PROGRAMMING STRUCTURES
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17
Computed nodes, when written code is simpler
MathScript Node Write code in syntax similar toMatlab. Define I/O variable. Allows error checking from other blocks.
Error inError Out
Formula Node Write code in syntax similar toC . Define I/O variable likeMathScript
PROGRAMMING IN LABVIEW
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18
Define Input/Output Variables and Design User
Interface
Define and/or Select Data
Collection and Control Channels
and Tasks
Program Functionality
Control and Data Acquisition Programming
PROGRAMMING STEPS
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Define the I/O and design the interface
Place the programming blocks on back panel
Wire the programming blocks on back panel to make a functional program
See more programming examples on the course website and in D2L
DEVELOPING DATA ACQUISITION AND CONTROL PROGRAMS IN LABVIEW
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Overview of Data Acquisition and Control Program Structure
HardwareNI-6024NI 6221
Measurement and Automation
Explorer
LabVIEW Program
Define I/O Tasksand channels
DATA ACQUISITION IN LABVIEW
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21
Define Types of Signals to Measure
Connecting to the outside world with Measurement and Automation Explorer (MAX)
Analog input signalsTransducer inputs (V, I)Digital inputs signalsBinary inputs Switches
Create Measurement
Channels and Tasks Using (MAX)
Analog input signalsDifferential, Ground Referenced Digital inputs signalsPorts (8-bits)Digital lines (1-bit)
Access Channels and Tasks Using
DAQmax in LabVIEW
Analog/Digital Read/Write Single/Multi-sample
DATA ACQUISITION IN LABVIEW
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22
Reading Analog Inputs Reading Digital Inputs
Access the DAQ functions from the Measurement I/O choice on the programming palette
Polymorphic Virtual Instrument (VI). Click to change nature.
IDEAL OP AMPSet438b-1
23
Ideal OP AMP Model
Av = 200,000Ri = 1-2 MWR0 = 75 WI1 = I2 = 80 nAfc=1.5 MHz
Parameters
Idea
Voltage Gain: Av = infiniteInput Resistance: Rin= infiniteOutput Resistance: Ro = 0Input I: I1=I2=0Cutoff Frequency, fc: infinite
Typical (LM741)
Rin Rout
+
-+
-V1
V2
+Vcc
-Vcc
Vo
Vo= Av(V2-V1)
Av
I1
I2
NON-IDEAL OP AMP MODEL
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24
Non-idea OP AMP parameters and characteristics
Slew rate - maximum rate of change of output voltage for large changes in the input voltage. Typical value - 0.5 V/mS = 500,000 V/s
Output offset voltage - voltage on the output when both of the inputs are grounded. Typical value - 2 mV (LM741)
Gain-Bandwidth Product-rate of frequency roll-off for OP AMP without feedback. Frequency at which the open loop gain of the OP AMP is 1 (0 dB). Typical 1 MHz (LM741) Applies to small signal level changes.
NON-IDEA OP AMP PARAMETERS et438b-1
25
Determine the frequency limit due to slew rate limiting.
Assume sine input and determine the rate of voltage change, dV/dt
maxp
maxp
p
p
fV2
dtdv
f2V=dt
dv
)ft2cos()f2V(dt
dv)ft2sin(V = v(t)
Maximum rate of change in sine occurs at t=0, so set t=0 in derivative to find fmax
Example dv/dt = 500,000 V/s , Vp = 10 V
dvdtV
fp2
500 000
2 107958
max
,
( ) Hz
Found in OP AMP
Data Sheets
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26
NON-IDEA OP AMP PARAMETERS
Gain-Bandwidth Product=GBP
Open loop frequency response
GBP = (Gain)(Input Frequency) = 1 MHz Find max frequency for 20 dB gain: 1 MHz/10(20/20) = 1,000,000/10 = 100 kHzFind max frequency for 40 dB gain: 1 MHz/10(40/20) = 1,000,000/100 = 10 kHzFind max frequency for 60 dB gain: 1 MHz/10(60/20) = 1,000,000/1000 = 1 kHz
1 MHz100 kHz10 kHz1 kHz100 Hz10 Hz
100 dB
Input Frequency
80 dB
60 dB
40 dB
20 dB
Op
en L
oop
Ga
in
IDEAL VOLTAGE COMPARATORSet438b-1
27
V1
+Vcc
V2-Vcc
Vo
Open loop OP AMP operation
Ideal Voltage Comparator Operation
Operation Logic
When V1 ≥ V2, Vo = -Vsat
When V2 ≥ V1, Vo = + Vsat
Transition take placeexactly when voltagesare equal
PRACTICAL OP AMP OUTPUT LIMITSet438b-1
28
OP AMP outputs typically saturate at 80% of supply voltages ±Vcc
Output Voltage Vo=Av(V2-V1)
0 8 0 8. ( ) . ( ) V V Vcc o ccOutput Range
For a practical OP AMP with Av = 100,000 find the difference voltage that will cause output saturation. (±Vcc = 15 Vdc)
Vd = (V2-V1) so Vo/Vd = Av
0 8
0 8 15
100 0000 12
2 1
. ( )( )
. ( )
,.
V
AV V Vcc
Vd
mV = Vd
NON-IDEAL OP AMP COMPARATORS
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29
Assume ±15 Vdc = ±Vcc Av = 100,000. so Vsat = 0.8(± Vcc) = 0.8(±15) = ± 12 Vdc
From previous calculation Vd = 0.12 mV
so Vd = V2 - V1 which gives 0.12 mV = V2 - V1
Take V1 as the input voltage
sat021
12
VV then VV + mV 12.0
)VV( = mV 12.
Take V2 as the input voltage012
0122 1
1
. ( )
.
mV =
mV - V2
V V
V
NON-INVERTING COMPARATORS
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30
Circuit realization with OP AMPs
Logic: When Vin > Vref V0 = +Vsat
If Vin > 5 then V0 = +12 Vdc
+15 Vdc
-15 Vdc
V0
Vin
Vref
V=+10 Vdc 5 kW
5 kW
R1
R2
Find Vref using voltage divider
V VR
R R
V
ref
ref
2
1 2
105000
5000 500010 0 5 5 0( . ) . Vdc
Input/output plot
Vin
V0
-Vsat
+Vsat
-12 Vdc
+12 Vdc
Vref = +5 Vdc
Input to +
NON-IDEAL VOLTAGE COMPARATORS
et438b-1
31
Determine the voltage where output transition takes place with V2 as input
V mV
mV V then V = + V2
1 0 sat
012
0121
2
.
.
V
V
V1
+Vcc
Vin-Vcc
Vo
Input/Output Diagram
-Vsat
+Vsat
V2=Vin
V0
0.12 mV
V2=V1
Av < infinity produces small voltage error in comparator circuits
INVERTING COMPARATORS
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32
Use standard OP AMPs when slow transitions are expected in the input signal. (e.g. thermostat application). When higher speeds are needed use dedicated comparator IC. (LM311)
If Vin > 8 then V0 = -12 Vdc
+15 Vdc
-15 Vdc
Vin
Vref
V=+10 Vdc
2 kW
8 kWR2
R1
Logic: When Vin > Vref V0 = -Vsat
Input to -
V0
Find Vref using voltage divider
V VR
R R
V
ref
ref
2
1 2
108000
8000 200010 0 8 8 0( . ) . Vdc
Input/output plot
Vin
V0
-Vsat
+Vsat
-12 Vdc
+12 Vdc
Vref = +8 Vdc
Controller
SIMPLE DIGITAL CONTROL ON/OFF PROCESS CONTROL
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In on/off control error signal is binary in nature.
Process
Disturbances
Sensor
Final ControlElement
Controlledvariable
Comparator
Signal Conditionin
g
ControllerLogic
Setpoint
Errorsignal
Final control element is run at either 100% or 0%
Comparator is hardware or software that comparesthe sensor value to the desired value (setpoint) and then outputs a binary value
Manipulatedvariable
ON/OFF CONTROL EXAMPLE
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34
Home heating
Controller
Room
Heat Loss
BimetallicStrip
Furnace
Thermostat
Mechanical Scale
Setting
Fuel/FanElectric Control
Desired Temperature
Errorsignal
Furnacerun time
RoomTemperature
When room temperature falls below a preset temperature, the thermostat contacts activatethe furnace fan and fuel supply.
ON/OFF CONTROL APPLICATION CRITERIA
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35
1Precise control must not be required
2
Process must have sufficient internal storage capacity to allow final control element to supply the load while measurement is taken.
3Energy entering the load must be small compared to the stored energy in the process
ON/OFF CONTROLLER TIME PLOTS
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36
Controller output goesto 100% when the temperature falls belowset point value .
Example for furnaceshows furnace on whentemperature falls belowset point of 72 degrees
DIFFERENTIAL GAP CONTROLLER
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37
To improve the stability of an on/off controller a hysteresis is added to the comparator element This is called differential gap control
Logic - when measured variable goes above upper boundary final control element turns on. Remains on until variable falls below lower level Gap also known as dead zone. Typically 0.5-2.0% of full range.
Gap introduces a known control error but reduces cycling
DIFFERENTIAL GAP CONTROLLER TIME PLOT
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Furnace controller with3 degree differential gap
Temperature below Set=72degrees furnace on. Shut off temperature isSet1=75 degrees.
Controller Logic: IF Room Temp ≤ Set AND Furnace Output= 0 THEN Furnace Output =1 (T ↑) IF Room Temp ≤ Set1 AND Furnace Output= 1 THEN Furnace Output =0 (T↓)
Output depends on temperature and previous output state
COMPARATORS WITH HYSTERESIS
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39
Implementing Differential Gap Control with Comparators requireshysteresis
Hysteresis - the output depends on the input and the previous state of the output.
Vo
Vin
+
R1=5kR2= 5k
+10V
Inverting Comparator with Hysteresis
Vref
Analysis: assume that Vin < Vref V0 = 0.8Vcc
Vo =+ Vsat = 8 Vdc
Determine Vref from voltage divider formula
Vdc 4k5k5
k58V
RR
RVV
ref
21
2oref
Define this as the upper trip point (UTP)
COMPARATORS WITH HYSTERESIS
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40
Analysis continued Now assume that Vin > Vref
If Vin > Vref, then V0 = -Vsat = 0 Vdc
When Vin > 0, Vo = 0 Vdc and Vref = 0 Vdc Define as lower trip point (LTP)
Vo
Vin
+
R1=5kR2= 5k
+10V
Vref
Vo= 0 V
Vref = 0 V
Input/output Plot Comparator with Hysteresis
-VsatVin
V0
+10 Vdc
VUTP = +4 Vdc
VLTP= 0 Vdc
+Vs
at
Increasing Vin
Decreasing Vin
COMPARATORS WITH HYSTERESIS
et438b-1
41
Vref5V
Vo Vin +
+10V
R11k
R29k
Non-zero voltage referenceUse superposition to find the contributions to V1 from output Vo and Vref.
Circuit Analysis
Assume Vin < V1 V0 = +Vsat = 8 Vdc
Ground Vref and find contribution to V1 due to Vo
V1
Vo= 8 V
V 8.0k9k1
k18
RR
RV'V
21
1o1
Ground V0 and find contribution due to Vref
V VR
R R
k
k kref' ' .12
1 2
59
1 94 5
V
Final value when Vin > V1
V''
Vdc1
V V'
. . .1 1
4 5 0 8 5 3 Upper trip point value (UTP)
VUTP = 5.3 V
et438b-1
42
COMPARATOR WITH HYSTERESIS - ANALYSIS
Vref5V
Vo Vin +
+10V
R11k
R29k
Non-zero voltage reference
V1
Vo= 0 V
Assume that Vin > V1 so V0 = -Vsat = 0 V
V VR
R R'1 0
1
1 2
0
Vdc
Since Vo = 0 and Vref is grounded
Vdc 5.4k9k1
k95
RR
RV''V
21
2ref1
Now find contribution due to Vref
Lower trip point value (LTP)
V''
Vdc1
V V'
. .1 1
4 5 0 4 5By Superposition
VLTP = 4.5 V
Hysteresis voltage is the difference between the VUTP
and VLTP. In this case: 5.3 - 4.5 = 0.8 Vdc (hysteresis)
COMPARATOR WITH HYSTERESIS - ANALYSISet438b-1
43
Input/output Plot
-Vsat=0
+Vsat=+8 Vdc
Vin
VUTP = 5.3 Vdc
VLTP = 4.5 Vdc
V0
Increasing Vin
Decreasing Vin
Hysteresis voltage is the difference between the Vutp and Vltp. In this case: 5.3 - 4.5 = 0.8 Vdc (hysteresis)
LTPUTPh VVV
Vh
HYSTERESIS COMPARATOR TIME PLOT
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Xa: 0.000 Xb: 0.000 Yc: 5.333 Yd: 4.533
a-b: 0.000 c-d: 800.0m
freq: 0.000
X: 0.000 Offsets Y:-100.00mOffsets
Ref=Ground X=8.33m/Div
Y=voltage
d
c
baV(3)
V(4)
VUTP VLTP
Vin
Vout
Vh
COMPARATORS WITH HYSTERESIS
et438b-1
45
Design Equations and Procedure
2R1R
1RV2V
2R1R
2RV
2R1R
1RVV
2R1R
2RV
2R1R
1RVV
sath
refsatLTP
refsatUTP
Design Procedure Given: Vcc, Vh, VUTP, and R1,
1.) Find Vsat 2.) Use Equation 3 to find R23.) Use Equation 1 to find Vref
1
2
3
Assumesbipolar output voltage
DESIGN EXAMPLE
et438b-1
46
A temperature sensor has a gain of 20 mV/F. It will be used in an electronic thermostat system. Design a comparator with hysteresis circuit that will give a 4 degree F deadband for the thermostat control around a setpoint temperature of 72 degrees F. The comparator will use bipolar power supplies at +- 5Vdc. Interface the thermostat logic to a transistor driver (2N3904 hfe = 300) that will actuate a furnace control relay. The relay has a dc resistance of 250 ohms.
DESIGN CALCULATIONS
et438b-1
47
Vref
Vo Vin +
+ 5 V
R1R2
-5 V
DESIGN CALCULATIONS
et438b-1
48
Continued -1-
DESIGN CALCULATIONS
et438b-1
49
Continued -2-
Check Centering
DESIGN CALCULATIONS
et438b-1
50
Continued -3-
Vref=1.44 V
Vo Vin +
+ 5 V
R1=4.7kR2=582.8k
-5 V
Now include the transistor output stage
DESIGN CALCULATIONS
et438b-1
51
D11N4004
Q12N3904
24V
-5V
Vref1.44V
Vin +
+5V Rc250
Rb
R14.7k
R2583k
Find value of Rb to activate relay
Vce(SAT)+-
Vbe(SAT)
+
Find Ic assuming saturation
Ic
DESIGN CALCULATIONS
et438b-1
52
D11N4004
Q12N3904
24V
-5V
Vref1.44V
Vin +
+5V Rc250
Rb
R14.7k
R2583k
Vce(SAT)+-
Vbe(SAT)
+
Ic Relate Ic to Ib through hFE (also known as b, dc current gain)
Reduce hFE by a factor of 10 due to effects ofsaturation on dc gain
Apply KVL around the base-to-emitter circuit
Ib
D1 Suppresses
voltage spikes