Signal Conditioning Circuits :Power Suppliescharnnarong.me.engr.tu.ac.th/charnnarong/My...
Transcript of Signal Conditioning Circuits :Power Suppliescharnnarong.me.engr.tu.ac.th/charnnarong/My...
Chapter 3 Signal conditioning systems
1
Signal Conditioning Circuits:Power Supplies
คณะวศวกรรมศาสตรมหาวทยลยธรรมศาสตร
Measurement and Instrumentation
Electronic measurement systemElectronic measurement system
TransducerPowersupply
Conditioningcircuits
Amplifier
Recorder
Dataprocessor
For engineering analysisFor engineering analysis-- Graphs or TablesGraphs or Tables
ControllerCommandgenerator
Process controlProcess control
Chapter 3 Signal conditioning systems
2
Power SuppliesSome transducers need the power supply in order to convert the sensed information from the sensor into the electrical signal.
Power supplies for instruments
1. Battery supplies
TransducerPowersupply
Signal ConditioningSignal Conditioning
2. Line voltage supplies
Battery supplieswidely used and inexpensive
voltage declines after used
voltage regulating circuit required
Voltage regulating circuits
Output voltage = Zener voltageEbatt. Eo
Simple voltage regulator
Chapter 3 Signal conditioning systems
3
Zener diodeis a type of diode that permits current not only in the forward
direction like a normal diode, but also in the reverse direction if the voltage is larger than the breakdown voltage known as "Zener voltage”
(Breakdown voltage)
P
N
Line voltage suppliestake the (high) voltage from a wall outlet and lower its into the desired
(low) voltage. The (low) alternating voltage is converted to direct voltage and is then regulated.
RectifierTransformer RegulatorPower line
220 Vac at 50 HzLow voltage
Vac Unregulated
Vdc Regulated
Vdc
Vac
220V
Chapter 3 Signal conditioning systems
4
TransformerTransformer is a device that transfers electrical energy from one circuit to another through inductively coupled conductors – The transformer’s coils.
Vs
Vp
Ns
Np=
Step-up transformer
Step-down transformer
Ns
Np1>
Ns
Np1<
Rectifier Rectifier is an electrical device that converts alternating current (AC), which periodically reverses direction, to direct current (DC), which is in only one direction
Vdc = Vav = 2Vp/π
Vrms = Vp/√2
Chapter 3 Signal conditioning systems
5
Voltage regulatorVoltage regulator is an electrical regulator designed to automatically maintain a constant voltage level
Three pin 12 V DC voltage regulator IC.
Signal Conditioning Circuits:Bridge Circuits
คณะวศวกรรมศาสตรมหาวทยลยธรรมศาสตร
Measurement and Instrumentation
Chapter 3 Signal conditioning systems
6
Bridge circuitshave been devised for measuring capacitance, inductance, and most often for measuring resistance. A purely resistance bridge, called a “Wheatstone bridge”, provides a means for accurately measuring resistance, and for detecting small changes in resistance.
Galvanometer
Bridge’s analysisBridge’s analysisThe basic arrangement for a bridge circuit is shown in the figure below. A
dc voltage is applied as an input across nodes A to D, and the bridge forms aparallel circuit arrangement between these two nodes. The currents flowingthrough the resistors R1 to R4 are I1 to I4 , respectively. Under the condition
that the current flow through the galvanometer, Ig = 0, the bridge is in abalanced condition.
For a balanced bridge:
R2 R4R1 R3
=
If the resistor R1 is a transducer, a change inresistance, associated with a change in somephysical variable, causes an unbalance of thebridge.
1 3
Chapter 3 Signal conditioning systems
7
1. Null Method
Bridge’s analysis
R1 is the resistance of the transducer R1 is the resistance of the transducer
R2 is the resistance of the adjustable resistor
R1Principle: If the resistance R1 varies with changes in the measured variable,
Ig = 0
one of the other arms of the bridge can be adjusted to null the circuit and determine resistance.
1. Null Method
TransducerWheatstone bridge
ll h dΔM ΔR1 ΔR2
The resistance R2 must be calibrated such that the adjustments directly indicate the value of R1
Transducer Null Method(unknown) (known)
Advantages: 1. no need to know the input voltage.
2. current detector required (Galvanometer) to detect Ig
Disadvantages: Disadvantages: 1. sensitivity depending on the galvanometer
2. uncertainty in the measured resistance depending on the input voltage
Uncertainty, uR ∝ 1/Ei
Chapter 3 Signal conditioning systems
8
2. Deflection Method
Bridge’s analysis
R1 is the resistance of a transducer R1 is the resistance of a transducer
R1
Principle: a voltage measuring device is used to measure the voltage unbalance in the bridge (the voltage drop from B to C) as a n indication of the change in resistanceof the change in resistance.
1
1 2
3
3 4
2. Deflection Method
TransducerWheatstone bridge
Deflection MethodΔM δ R δ Eo
R1R1 = R2 = R3 = R4 = R
EEo
For identical resistors,
Under balanced conditions (Ig = 0),
E = 0
/4 2 /
o
In an unbalanced condition,
Eo = 0
Chapter 3 Signal conditioning systems
9
Advantage: can be used to measure time-varying signals, but the voltage measuring device must have the frequency response not less than that of the sensor.
Disadvantage: Disadvantage: 1. high capability of the galvanometer required.
2. known and regulated input voltage required.
In case of the low impedance galvanometer, Rg
//4
/4 1 /
A certain temperature sensor experiences a change in electrical resistance with temperature according to the equation
Example 1 Null method
1
where R = sensor resistance [Ω]
R0 = sensor resistance at the ref. temperature T0 [Ω]
T = measuring temperature [°C]
α = constant = 0.00395 °C-1
Given:Given: 1) R3 = R4 = 500 Ω
2) R0 = 100 Ω at T0 = 0°C
Determine the value of R2 that would balance the bridge at 0 °C
Chapter 3 Signal conditioning systems
10
R1
R = R = 200 Ω
SolutionUnder balanced condition,
R2 = R1 (R4 /R3 ) = R1R3 R4 200 Ω
120
140
160
R2
60
80
100
0 20 40 60 80 100Temperature [C]
Example 2 Deflection methodConsider a deflection bridge, which initially has all arms of the bridge equal to 100 ohms, with the temperature sensor described in Example 1. The input voltage to the bridge is 10V. If the temperature of R1 is changed such that the bridge output is 0.569 V, what is the temperature of the sensor? How much current flows through the sensor and how much power must it dissipate?
Solution
At deflection state,
δE = 0 569 V
/4 2 /
A change in the resistance of the sensor can be obtained from
δEo = 0.569 V
Chapter 3 Signal conditioning systems
11
We obtain
Therefore, the temperature of the sensor is
δ R = 25.67 ohms
1 0 0 1 0.00395 0
By Kirchoff’s law, the current flowing through R1 isgiven by
100 25.67 100 1 0.00395 0
65
The power dissipated from the sensor is
11
1 244.3
1 12
1 0.25
R1 R
EEo
SolutionUnder balanced condition,
/4 2 /
0 0 1 0.00395 0
0.6
0.8
1.0
0 [V
]
Ei = 10 V
RR 0 0 1 0.00395 0
0.0
0.2
0.4
0 20 40 60 80 100
δE 0
Temperature [C]
Chapter 3 Signal conditioning systems
12
Signal Conditioning Circuits:Amplifiers
คณะวศวกรรมศาสตรมหาวทยลยธรรมศาสตร Measurement and Instrumentation
AmplifiersIs a device that scales the magnitude of an analog input signal according to the relation
Input voltage, Vi Output voltage, Vo
Supply voltage, Vs
Vo = G x Vi where G = Gain
11
0
0.2
0.4
0.6
0.8
0 2 4 6 8 10
t
vo
0
0.2
0.4
0.6
0.8
0 2 4 6 8 10
t
vi
Chapter 3 Signal conditioning systems
13
Types of amplifier1. Single-ended amplifiers
are amplifiers which amplify a single input signal. Both input and output signal voltages are referenced a common connection in the circuit called ground (used for potentiometer circuits)ground. (used for potentiometer circuits)
2. Differential amplifiers
are amplifiers which amplify two input signals. All voltages are referenced to the circuit's ground point. (used for bridge circuits)
Vs Vs
Vi Vo
G
Vi,2 Vo
G
Vi,1
Vo = G x Vi Vo = G (Vi,1- Vi,2)
Single-ended amplifiers Differential amplifiers
Operational Amplifiers (op-amp)An “op-amp” is a high-gain electronic voltage amplifier with a differential input and, usually, a single-ended output.
Characteristics of op-amps
1. High internal gain G = 105 to 106
2. High input impedance Zi > 107 Ω
3. Low output impedance Zo < 100 Ω
4. two input ports, a noninverting and an inverting input, and one output port
Non-inverting terminal(+5V)
Inverting terminal
Non-inverting terminal
Vo = G (Vnon-inv - Vinv)
(-5V)
Chapter 3 Signal conditioning systems
14
Summing amp
Inverting amp. Non-inverting amp.
Summing amp.
IntegratorDifferentiator
Voltage Follower
Inverting Amplifier:
Rf
Circuit analysisRf
R1Vi
Vo
Va
if
A
1
Sum of currents at Node A
Ra
i1 iaA
1
11
Output voltage of the inverting amplifier
Chapter 3 Signal conditioning systems
15
Non-inverting Amplifier: 11
,1
1
,2
2
,3
3 Summing Amplifier:
1
1
1
Voltage follower:
Differentiator:
Integrator:
***
1 0Integrator:
Signal Conditioning Circuits:Filters
คณะวศวกรรมศาสตรมหาวทยลยธรรมศาสตร Measurement and Instrumentation
Chapter 3 Signal conditioning systems
16
FiltersIn many instrumentation applications, the (dynamic) signal from the transducer is combined with noise or some other parasitic signal. These undesired signals can be eliminated with a filter that is designed to attenuate th i i l b t t it th t d i l ith t di t ti the noise signals but transmit the transducer signal without distortion.
Two filters that utilize passive components and are commonly employed in signal conditioning include (1) High-pass RC filter and (2) Low-pass RC filter
R C
RC filter circuits
C R
Filters
HighHigh--pass filterpass filter
Unfiltered signal with low frequency variation and high frequency noise
High-pass filter employed to remove low frequency variation
LowLow--pass filterpass filterFiltered signal Filtered signal
Low-pass filter employed to remove high frequency noise
Chapter 3 Signal conditioning systems
17
High-pass RC filtersHigh-pass filter permits only frequencies above the cutoff frequency to pass.
Consider a summation of voltage drops around the loop
| |
0
If and | | 0
The output voltage Vo is High pass
Passband
1 2| |
21
where
~ 0 if ω 0 Stopband
Low-pass filtersLow-pass filter permits only frequencies below the cutoff frequency to pass while blocking the passage of frequency information above the cut off frequency.
where
The output voltage is
11 2
| |
Passband
~ 0 ถา ω ∞
1
Stopband
Chapter 3 Signal conditioning systems
18
Cut-off frequencyis defined as the frequency at which the ratio of the amplitudes of the output to the input has a magnitude of 0.02 (2% error).
For examples,
(1) Hi-pass filter
(2) Lo-pass filter
|Vo|/|Vi| = ωCRC/[1+(ωCRC)2]1/2 = 0.02 ωC = 5/RC fC = 5/(2πRC)
|Vo|/|Vi| = 1/[1+(ωCRC)2]1/2 = 0.02 ωC = 0.203/RC fC = 0.203/(2πRC)
Active filtersOperational amplifiers are employed to construct active filters where select frequencies can be attenuated and the signal amplified during the filtering processprocess.
741+
-R1
R2
C2
Vi V
741+
-R1
R2
Vi V
C1
i Voi Vo
Low-pass active filter High-pass active filter 1
2 2 2
12 1 1
Chapter 3 Signal conditioning systems
19
Signal Conditioning Circuits:Modulators and Demodulators
คณะวศวกรรมศาสตรมหาวทยลยธรรมศาสตร Measurement and Instrumentation
Amplitude ModulatorsAmplitude modulation is a signal conditioning process in which the signal from a transducer is multiplied by a carrier signal of constant frequency and amplitude.
Transducer signalTransducer signal(sinusoidal, transient, or random)(sinusoidal, transient, or random)
Carrier signalCarrier signal(sinusoidal, transient, or random)(sinusoidal, transient, or random)x
sin sin
~ 10 100
Chapter 3 Signal conditioning systems
20
sin sin
2cos cos
Modulated signal
2
4000
In general, noise signals occur at 50-Hz
3950 4050
50
Amplitude ModulatorsAdvantages
(1) Stability
(2) Low power(2) Low power
(3) Noise suppression
Demodulation is the process that the transducer signal is separated from the carrier signal.
RectifierRectifier FilterFilter
Chapter 3 Signal conditioning systems
21
Assignment #31. Shown that the output voltage of the summing amplifier be expressed by the
equation
2. Consider the Wheatstone bridge. Suppose R3 = R4 = 200 Ω, R2 = variable
calibrated resistor, and R1 = transducer resistance = 40x + 100
a) When x = 0, what is the value of R2 required to balance the bridge?
,1
1
,2
2
,3
3
b) If the bridge is operated in a balanced condition in order to measure x, determine the relationship between R2 and x.
Fundamental of Sampling, Data Acquisition and Digital Devices
คณะวศวกรรมศาสตรมหาวทยลยธรรมศาสตร Measurement and Instrumentation
Chapter 3 Signal conditioning systems
22
Sampling conceptsConversion of analog signals to digital code is extremely important
in any instrument system that involves digital processing of the analogoutput signals from the signal conditioners.p g g
ContinuousSignal
DiscreteSignal
“discrete time series”
Discrete time seriesSuppose the transducer signal is measured repeatedly at successive sample time increments δt
1 2 31, 2, 3, … ,
total sample period
sample rate
Chapter 3 Signal conditioning systems
23
fs = 1/δt [Hz]
Sample rate
The sample rate has a significant effect on perception and reconstruction of the continuous analog signal in the time domain
f 2 f
fm = 10 Hz (sine wave analog signal) fs = 100 Hz
fs > 2 fm
*** s = samplingm = measuring fs = 27 Hz fs = 12 Hz
δ t < 1/(2 fm )
Alias FrequenciesWhen the sample rate is less than 2fm , the higher frequency content of the analog signal will take on the false identity of a lower frequency in the
resulting discrete series A false frequency is called an “alias frequency”resulting discrete series. A false frequency is called an alias frequency .
sample rate 12 Hz
Original signal10 Hz 2 Hz
Interpreted signal
Chapter 3 Signal conditioning systems
24
is the diagram that is used to find the alias frequency.
Folding Diagram
Example
f = 10 Hz
fN = 6 Hz[f/fN]y-axis= 10/6 = 1.67
[f/fN] 0 33 Hz
1.67
0.33
[f/fN]x-axis = 0.33 Hz
falias = 0.33 x fN = 2 Hz
Nyquist frequency and Anti-Aliasing
Nyquist frequency is the highest frequency that can be coded at a given sampling rate in order to be able to fully reconstruct the signal,
Anti-aliasing is a process that remove signal content at and above Nyquist frequency by use of a low-pass filter prior to sampling and use an appropriate
fN = 0.5 fs
frequency by use of a low pass filter prior to sampling and use an appropriate sample rate for the signal
Signal = Signal [f < fN ] + Signal [f > fN ]
Anti-aliasing filter
Chapter 3 Signal conditioning systems
25
ExampleConsider a complex periodic signal in the form
10 sin 50 0.5 sin 150 1.2 sin 250
0
5
10
15
y(t)
ym
10 sin ()
0.5 sin()
1.2 sin()
‐15
‐10
‐5
0 20 40 60 80 100 120 140
t
Example (cont.)If the signal is sampled at 100 Hz
f = 0 5 f = 0 5 x 100 Hz = 50 Hz
Observe all frequency content in the sampled signal
fN 0.5 fs 0.5 x 100 Hz 50 Hz
10 sin 50 0.5 sin 150 1.2 sin 250
f1 = 25 Hz f2 = 75 Hz f3 = 125 Hz
fN < fm Alias frequency
Chapter 3 Signal conditioning systems
26
Example (Cont.)Sampled signal
10 sin 50 0.5 sin 50 1.2 sin 50
0
5
10
15
t)
ys
ym
10 sin ()
Alias frequency = 25 Hz
‐15
‐10
‐5
0
0 20 40 60 80 100 120 140
y(t
t
Example (cont.)Use a low-pass filter with the cut-off frequency > 50 Hz
10 sin 50
0
5
10
15
0 20 40 60 80 100 120 140
y(t)
ym
yA = 10sin()
‐15
‐10
‐5
0 20 40 60 80 100 120 140
t
Chapter 3 Signal conditioning systems
27
Amplitude Ambiguity
The sampling signal can be completely reconstructed from a discrete time series, if
(1) Total sample period remains an integer multiple of the fundamental period, T1
(2) Sample rate meets the sampling criterion
mT1 = Nδ t(2) Sample rate meets the sampling criterion.
fs > 2 fm
Note: N = 2M: M = # of sample points
Example
Amplitude = 10Fundamental period T =1/ 100 s
10cos 2 100
(8-bit)
δ f = (Nδ t)-1 = fs/N = 39 Hz
Fundamental period T1 =1/ 100 s
(a) m = N δt / T1 = 2.56
(b) m = N δt / T1 = 10.24
amplitude leakage
amplitude leakage
Amplitude spectra Amplitude spectra
(c) m = N δt / T1 = 8
No leakage
amplitude leakage
Chapter 3 Signal conditioning systems
28
Selecting sample rate and data number
For an exact discrete representation in both frequency and amplitude of analog signal, both the number of data points and the sample rate should be chosen based on the and the sample rate should be chosen based on the following criteria: (1) fs > 5 fm (2) long total sample periods (large N) and (3) use of anti-alias filter.
Data-Acquisition System (DAS)
is the portion of a measurement system that quantifies and stores data. A typical signal flow scheme is shown in the following diagram.
Chapter 3 Signal conditioning systems
29
Data-Acquisition System components
A/D converter
Signal flow scheme for an automated data-acquisition system
Signal Conditioning: Filters and Amplification
Filters1. Analog filters are used to control the frequency content of the signal being sampled.
Example: Anti-alias filter
2. Digital filters, which are software-based algorithms, are effective for signal analysis after sampling effective for signal analysis after sampling
Example: moving-averaging scheme (for removing noise)
1 1 / 2 1
Chapter 3 Signal conditioning systems
30
Signal Conditioning: Filters and Amplification
AmplifiersAll DASs are input range limited; that is, there is a minimum value and a maximum value of signal. Some transducer signals will need amplification or attenuation prior to conversion. Most DASs contain on-board instrumentation amplifiers.
Voltage divider
0-50 V signal 0-10 V A/D converter
for signal attenuation
R1 = 40 kΩR2 = 10 kΩ
Signal Conditioning: Filters and Amplification
Shunt circuitsAn A/D converter requires a voltage signal at its input. It is straightforward to convert current signals into voltage signals using a shunt resistor.
In general, the standard current signal has a range of
4 – 20 mA
Using a shunt resistor = 250 Ω, the current signal would be converted into the voltage signal 1 – 5 V
Chapter 3 Signal conditioning systems
31
Signal Conditioning: Filters and AmplificationMultiplexer
When multiple input signal lines are connected by a common throughout line to a single A/D converter, a multiplexer is used to switch between connections, one at a time.
A/D Convertersconvert the analog signals into the digital signals with conversion rates typically up to the 1 kHz – 10 MHz range . The input resolution, Q which depends on the number of bits of the converter, , p ,is given by
2
where EFSR = Full scale voltage range
M = resolution in bits
Example A12 bit A/D converter with input signal range of 0-10V. The resolution
Q = 10/212 = 2.44 mV
The amplifier permits signal conditioning with gains from G = M = resolution in bits conditioning with gains from G = 0.5 to 1000
The minimum detectable voltage when set at maximum gain is
Q = 2.44 mV / 1000 = 2.44 μV3-bits resolution3-bits resolution
Chapter 3 Signal conditioning systems
32
Analog signal input connections
Single-ended connection Differential-ended connection
Suitable only when all of the analog Allow the voltage difference betweenSuitable only when all of the analog signals can be made relative to the common ground point.
Allow the voltage difference between two distinct input signals to be measured
ExampleVO+ -signal
Vi
Strain gage (Rg) , S = 2.5μV/με
A signal from the strain gage with the sensitivity = 2.5μV/με is connected to DAS which has the 12-bit A/D converter, an input range of ± 5V, and sample rate of 1000 Hz. For an expected measurement range of 1-500 με, specify appropriate G for amplifier, filter type and cut-off frequency.
Chapter 3 Signal conditioning systems
33
Resolution = 10/212 = 2.44 mV
Amplifier
Signal Range = (1 με - 500 με) x 2.5 μV/με = 2.5 μV – 1.25 mV
G = 1000: Amplified Signal = (2.5μV – 1.25 mV) x 1000 = 2.5 mV – 1.25 V
G = 3000: Amplified Signal = (2.5μV – 1.25 mV) x 3000 = 7.5 mV – 3.75 V
Sample rate, fs = 1000 Hz
Filter
s
Nyquist frequency , fN = 1000/2 = 500 Hz
*** Low-pass filter with a cut-off frequency = 500 Hz required
*** Differential-ended connections required
Chapter 3 Signal conditioning systems
34
Final Examination on OCT 8, 2011 (1300-1600)
Closed books and closed notes
Non-programmable calculators are permittedp g p
Electronic dictionaries are not permitted
7 questions (10 marks each)
Data Loggerคออปกรณอเลคทรอนคสซงทาหนาทบนทกและจดเกบขอมลทตรวจวดจากเครองมอวดหรอเซนเซอรตางๆ ปจจบน Data logger ใชพนฐาน
ป ป ของระบบประมวลผลแบบดจตอล สวนประกอบทสา คญคอ ไมโครโปรเซสเซอร หนวยความจา และเซนเซอร สามารถแบงออกไดเปน 2 ลกษณะคอ
- General purpose types
- Specific devices
ขอด 1. ทางานดวยตวเอง (Stand-alone)
2. เคลอนยายงาย (Portable)
3. บนทกขอมล 24 ชวโมง
4. สามารถใชกบ DC powerhttp://www.dataq.com/images/products/data-logger/gl200_500w.jpg
Chapter 3 Signal conditioning systems
35
ตวอยาง Data logger
ตวอยาง ขอมลจาเพาะ ตวอยาง ขอมลจาเพาะ Data loggerData logger
Chapter 3 Signal conditioning systems
36
ตวอยาง ขอมลจาเพาะ ตวอยาง ขอมลจาเพาะ Data loggerData logger ((ตอตอ))
Grounds, Shielding, and connecting wiresการเชอมตอสายสญญาณในอปกรณอเลกทรอนคสอาจเปนสาเหตของการเกดสญญาณรบกวน (noise) ในระบบ
สญญาณรบกวนจะมผลมากในสญญาณทมกาลงตา (< 100 mv)
วธปองกนสญญาณรบกวน
1. ใชสายสญญาณใหสนทสด
2 ไมใหสายสญญาณอยใกลแหลงสญญาณรบกวน2. ไมใหสายสญญาณอยใกลแหลงสญญาณรบกวน
3. ใชชล (shield) ปองกนและตอสายลงดน (ground)
4. บดสายเปนเกลยว
Chapter 3 Signal conditioning systems
37
Ground and Ground loopGround คอจดอางองในวงจรไฟฟาใชในการเปรยบเทยบแรงดนไฟฟา โดยทวไป Ground จะเปนเสนทางเดนของกระแสลงสพนดน ( earth ground) ซงมคาแรงดนไฟฟา เทากบศนย
Ground potential อาจมคาตางกนขนกบจดเชอมตอ (เชน อปกรณ อาคาร และพนดน)
สายดนทความยาวมากๆ อาจเปนทาหนาทเหมอนกบ เสาอากาศ (Antennae)
Ground Symbols
Signalground
Chassisground
Earthground
Ground Symbols
Ground loop
Ground loop เปนสาเหตมาจากการเชอมตอ ground ของหลายวงจรเขารวมกน โดยGround potential ของวงจรทตอรวมกนมคาไมเทากน
ผลตางของศกยไฟฟาทเกดใน Ground loop สงผลใหเกดการไหลของกระแสในสายสญญาณ นาไปสสญญาณรบกวนในระบบ
Signal current
Ground current
Chapter 3 Signal conditioning systems
38
วธปองกน ground loopมการตอสายดนเพยง 1 จด (single-point ground)
ตดตง Isolation transformer ( 1:1 power transformer AC signal ผานไดเทานน)
ใช Shield ปองกนสญญาณรบกวนหากสญญาณสงมกาลงตา
การตอ Shield ปองกนทไมถกตอง การตอ Shield ปองกนทถกตอง(ตอสาย ground ของShield ปองกนทปลายดานสายสงสญญาณดานเดยว)