Transcript Data Acquisition System

Data Acquisition System
Data Acquisition System


The purpose of any data acquisition system is to
acquire analog signals and present them to the
MCU in a form that can be manipulated.
The main components of any general data
acquisition system consists of the following:
1. Transducers (sensors)
2. Analog Multiplexer
3. Signal Conditioning (Amplification, Filtering, ..)
4. Sample and Hold Circuit
5. Analog to Digital Converter
6. Microcomputer System
7. Digital to Analog Converter
8. Actuator
2
Data Acquisition Components


Transducers (sensors) convert variable processes such
as pressure, temperature into electrical signals such as
voltage or current.
Signal conditioning:
1.
2.
3.



Isolation & buffering: protection from dangerous voltages (i.e.
interfacing an MCU to a 110V or 220V)
Amplification: Need full scale signal for conversion
(transducers usually provide very small signals millivolts)
Bandwidth limiting: Low pass filter to limit range (noise)
Sample and Hold: used to keep signals constant while
converting an analog signal to digital
A/D and D/A are interfaces to the MCU to the outside
world
Actuators: interfaces to activate motors, switches, e.t.c
3
Signal Paths of a DAS
Humidity
Temp
Pressure
……
Real
Analog
Signal
Mux
Conditioning
Transducer
World
(sensors)
Measurand
Actuator
D/A
Conv
MCU
A/D
Sample and
Conv
Hold Circuit
4
Signal Conditioning: Op-Amps
Inverting
Non-inverting




Operational Amplifiers are electronic devices used to
amplify signals, filters, signal conditioning, e.t.c.
It consists of three terminals (two inputs, one output)
OP-AMPs require dc power to operate (+VCC, -Vcc)
OPERATION: OP-AMPS are designed to sense the
difference between the voltage signals applied to its
two input terminals (V2 – V1) , multiply this by a
number A , and cause the resulting voltage A (V2 –
V1) to appear at output terminal 3.
5
Operational Amplifiers:
Rules for Analysis
+12V
+12V
Inverting
Non-inverting
-12V
-12V
1.
2.
3.
4.
Voltage ranges are bounded by supply voltage i.e +/- 12V
Input currents are zero because input impedance is large
(infinite input impedance for ideal op-amp)
Positive feedback or no feedback drives output to
saturation (i.e. +12V and -12V)
A Virtual short circuit exists between the two input
terminals (i.e. whatever voltage is at terminal 2 will
automatically appear at terminal 1) due to infinite gain A.
6
Inverting OPAMP
7
Analysis of Inverting OPAMP
8
Inverting Adder
1.
2.
3.
4.
According to virtual ground rule, voltage at inverting and
non-inverting inputs are 0.
According to Ohm’s law: i1 = V1/R1, i2 = V2/R2, …
All these currents sum together to produce the current i;
that is i = i1 + i2 + i3 (will be forced to flow through Rf)
The output voltage V0 = 0 – iRf = -iRf
9
Binary Weighted DAC
1.
2.
3.
The binary weighted DAC is the simplest converter.
This DAC creates currents that are proportional to the
weight of each code bit.
The inverting OP AMP can produce an output voltage
which is linear combination of several input voltages.
10
Binary Weighted DAC

By using input resistors which scale by a factor of 2, a
summing Op Amp can produce an output which follows a
binary pattern.
11
Binary Weighted DAC

By using switches on the input resistors, a summing
Op Amp can produce an output which is a binary number
(representing which switches are closed) times a
reference voltage.
Conversion
Factor 12
Binary Weighted DAC





CF = - (RF/R0)VB is called the conversion factor or
quantum interval.
The quantum Interval is defined as the change in output
signal for a unit change of input code, 0000, 0001, 0010
The largest voltage that can be produced by an n-bit
binary DAC is Vmax = (2n-1)CF
NOTE: CF will be set to the maximum allowable value
which is limited by the maximum output of the OP-AMP
(VOM)
The largest conversion factor is calculated as
CFMAX = (VOM)/(2n-1)
13
Binary Weighted DAC: Example

A 4-bit binary weighted DAC has values of RF = 10K,
R0=360K
1.
2.

1.
2.
3.
What is the value of R1, R2, R3
Calculate
(a) the conversion factor when VB = -8V
(b) and the maximum output of the system.
Solution:
R0 = 2R1 = 4R2 = 8R3,
So R1 = 180K, R2 = 90K, R3 = 45K
CF = (-RF x VB)/R0 = -(10)(-8)/360 = 0.222
Maximum output of this system
emax = (2N-1) x CF = 15 x 0.222 = 3.33V
14
A-to-D Conversion Samples in the Time Domain



When an analog signal is to be represented by a digital code  the
range of analog signal is divided into levels (quantization levels) 
Pulse Code Modulation (PCM)
Using level numbers to represent analog signals leads to
Inaccuracy  Quantization Error (shaded region)
Sampling theory states that if samples are taken of an analog
signal, the original analog signal can be recovered if  fs > 2B
15
A Simple A/D Converter
Components: (a) Counter (b) D/A (c) Comparator (d) Latch

I.
II.
III.
IV.
V.

The counter counts from 0 to 2n-1
The counter drives the input of the D/A converter
The output of the D/A is compared to the input voltage vin
When the output of the comparator switches logic this means that the
generated voltage of D/A passed the input voltage.
The comparator then latches the counter output to the latch.
Problem with Slope A/D?
 it takes 2n clock cycles for conversion
16
Recall: Staircase A/D Converter


Problem with Slope A/D  it takes 2nclock
cycles for conversion
Can we do better!
17
Tracking Converter

Is an improved version of the staircase ADC



If VA increases over previously converted value the
counter will count up
If VA decreases over previously converted value the
counter will count down
Conversion time is Not Constant
18
Successive Approximation A/D




Is based on intelligent trial-and-error method
When start signal is asserted the controller instructs the sequencer to

Place a `1’ in MSB position of Register.

All other bits remain at `0’ level
A comparator compares two voltage values on its two inputs.

If guess is low then the next MSB is set to `1’

If guess is high then that bit is set to `0’ and the next MSB is set to `1’
Requires N clock periods for N-bit converter.
19
Successive-Approximation Example


6-bit A/D with range
0V – 5V
Step Size =
5/26 =
0.078
20
Successive-Approximation Example


4.16 – 4.141 = 0.019
Error is smaller
than ½ LSB
Digital output is 110101
21
Successive Approximation A/D


A Successive Approximation A/D can give the wrong output if the
voltage changes during a conversion
A Track/Hold or Sample/Hold circuit is needed to hold the input
voltage constant during conversion.
22
Sample and Hold Circuit


Circuit  Capacitor & Switch
It works by charging the capacitor to the input voltage, then
disconnecting the capacitor from the input voltage during
conversion.
23
Concepts and Terminology


Analog signals have a ratio-metric range
These range from a low value to a high value
1.
2.
3.

Binary Code representation for Analog (0V to 5V)
1.
2.
3.

0V to 5V
-2.5V to 12.5V
4mA to 20mA
0V  $00
5V  $FF
2.5V  $80 (middle of analog range)
If the analog range starts from a nonzero minimum
value it is usually represented by an “offset binary
code”
•
e.g. -2.5 V - +12.5V, then -2.5 == $00
24
Concepts and Terminology

Definitions :Assume an analog signal ranges from 0V-5V, and
number of bits used in A/D converter is 3
1.
Offset  Minimum Analog Value  0V
2.
Span  Max to Min Analog Value  5V-0V = 5V
3.
Step Size, Quantum Interval  Span/2n
4.
Resolution  Refers to the number of bits in ADC and also
to smallest analog change corresponding to a change in a bit
in the digital number.
5.
Conversion:

Analog Number = (Digital Number x Step Size) + Offset

Digital Number = (Analog Number – Offset)/Step Size
6.
Dynamic Range (DR) = (Range or Span)/Resolution = 2n
The dynamic range determines the required word size of an
ADC.
25
Example

A 6-bit DAC has an analog output range of -2.5 to 5.0V.
Calculate the analog output when input is 010101 (i.e.
decimal 21)

Solution:
1.
2.
3.
4.
Offset = -2.5
Span = 5 – (-2.5) = 7.5V
Step Size = 7.5V/26 = 10/64 = 0.1172V
Analog Output = (digital number x step size) + offset
= (21 x 0.1172) – 2.5V = -0.039V
26
Example


A temperature sensor has a measurement range of -10 to 140C.
The output range is -2.5 to +5mv. It has a resolution of 0.5C.
1.
Determine the # of bits in ADC?
2.
What is the reading of the ADC when the output of the sensor is
-1.0 mv?
Solution:
1.
Span = 140 – (-10) = 150 C (or) 5 – (-2.5mv) = 7.5mv
2.
Dynamic Range = 150 C / 0.5C = 300
DR = 300 = 2n  n = log2 (300) = 8.22  9 bits
1.
2.
Step Size = Span /2n = 7.5mv/(29) = 0.01464
Digital Number = (Analog Number – Offset)/Step Size
= (-1.0mv – (-2.5mv))/0.01464
= (102)10 = (001100110)2
27
CPU12 ADC Subsystem: Features


The CPU12 has no DAC!
Features of the ADC module include:
1. Eight multiplexed input channels
2. ADC based on successive approximation
3. 8-bit Resolution
4. Single or continuous conversion
5. Conversion complete flag with CPU interrupt request.
6. Selectable ATD clock
28
CPU12 A/D




The HC12 has a sample/hold amplifier built in.
The Successive Approximation A/D converter has speeds up to several
million samples per second.
VRH is the highest voltage the A/D converter can handle
VRL is the lowest voltage the A/D converter can handle
29
CPU12 A/D
Channel 0
Analog
Multiplexer
The HC12 uses
an analog
multiplexer to
allow eight input
pins to connect
to the A/D
converter
30
CPU12 A/D: Control and Status Registers
31
CPU12 A/D: Control Registers



ADPU  1: Powers up the A/D ( if ADPU  0, then AD Port
General Purpose). Stabilization Period!
SCAN: Perform one conversion or continuous
MULT: Conversion of sequential channels or single channel
ADPU
SCAN
MULT
32
CPU12 A/D: Channels
33
CPU12 A/D: ATDCTL2



Could either use Polling or Interrupt Driven I/O
If ASCIE  1, then interrupt requests are enabled.
The ASCIF interrupt FLAG is set = 1, when conversion
sequence is complete.
ASCIE ASCIF
34
CPU12 A/D: ATDCTL4
Conversion Time = Initial Sample Time + Transfer
Time + Final Sample time + Resolution Time
SMP1 SMP0 PRS4 PRS3 PRS2 PRS1 PRS0
35
CPU12 A/D: ATDCTL5


The S8CM bit in ATDCTL5 is used to select conversion
sequences of either 8 or 4 conversions.
If set to  0, then 4 conversion sequences are selected and
remember that the result is stored in ADRH2 and not ADRH6 if
CC=1, CB=1, CA =0 (0110)
AN6
ADRxH2
S8CM
36
CPU12 A/D: Status Registers

SCF (Sequence Complete Flag)
1.
2.

In single conversion sequence mode, SCF set at end of conversion
In continuous conversion mode, SCF is set at end of first conversion
CCF0-CCF7 (Conversion Complete Flags)
1.
2.
Each ATD channel has a CCF flag which is set at the end of the
conversion on that channel.
CCF flag can be cleared by reading STATUS REGISTER #1 and then
reading the result register of that channel.
SCF
CCF7
CCF0
37
CPU12 A/D: LAB
38
Using the ATD to Measure Signal
; -------------------------------------------------------------------;
Main Program
;--------------------------------------------------------------------ORG $7000
; 16K on board RAM
MAIN:
BSR INIT
; branch to init ATC
BSR CONVERT ; Branch to conversion
DONE:
BRA DONE
; Branch to Self
;----------------------------------------------------------------------;
Subroutine INIT
;----------------------------------------------------------------------INIT:
LDAA #$80
; Power up ATC
STAA ATDCTL2
; ATD Flags clear normally & Disable Interrupt
BSR DELAY
; Delay (100 us)
LDAA #$00
; Select continue conversion in BGND Mode
STAA ATDCTL3
; Ignore FREEZE in ATDCTL3
LDAA #$01
; Select Final sample time = 2 A/D clocks
STAA ATDCTL4
; Prescaler = Div by 4
RTS
39
Using the ATD to Measure Signal
;----------------------------------------------------------------------;
Subroutine DELAY 100 us
;----------------------------------------------------------------------; Delay required for ATD converter to stabilize (100 usec)
DELAY:
LDAA #$C8
; Load Acc with “100 usec delay value”
DECA
; Decrement ACC
BNE
DELAY ; Branch if not equal to Zero
RTS
; Return from subroutine
40
Using the ATD to Measure Signal
; -------------------------------------------------------------------;
Subroutine Convert
; set-up ATC, make single conversion and store result to a memory location
; Configure and start A/D conversion
; Analog Input Signal: on Port AD6
; Convert: using single channel, non-continuous
; The result will be located in ADR2H
;--------------------------------------------------------------------CONVERT
LDAA
#$06
; Initialize ATD  SCAN = 0, MULT=0,
;
; S8CM = 0, PAD6, Write clears Flag
STAA
ATDCTL5
; 4 Conversions on a Single Conversion
; sequence
WTCONV: BRCLR ATDSTATH,#$80,WTCONV; Wait for Sequence Complete Flag
LDD
RTS
ADR2H
; Loads Conversion result (ADR2H)
; into Accumulator
; Return from Subroutine
41
Summary





Data acquisition systems consists of several
components (i) sensors (ii) signal conditioning, (iii)
Analog to Digital Converters, (iv) MCU, (v) Digital to
Analog converters.
Engineers must justify the usage of each component in
the path and consider (i) Cost, (ii) error
Operational Amplifiers are at the heart of Signal
conditioning circuits, ADCs, DACs and therefore
analysis of these circuits is very important.
Determining the # of bits required by ADC or DAC is
important and knowing the sampling frequency is also
as important to recover the original signal.
There are several types of ADCs available (next topic)
42
43
Example


Given an analog signal with range -5V to +5V and an
8-bit ADC
1. Determine: Offset, Span, Step Size, %Resolution
Solution:
1.
2.
3.
4.
Offset = -5V
Span = 5 – (-5) = 10V
Step Size = 10V/28 = 10/256 = 39.1mv
% Resolution = 0.391%
44