Signal Conditioning Circuits :Power Suppliescharnnarong.me.engr.tu.ac.th/charnnarong/My classes/AU351... · 2016-02-25 · Signal Conditioning Circuits:Power Supplies ... are amplifiers

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


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


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


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


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


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


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


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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


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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


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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


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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]


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


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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)


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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


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



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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


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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


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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


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Signal Conditioning Circuits:Modulators and Demodulators


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


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


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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



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


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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


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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


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


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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


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


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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.


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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


30


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


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


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


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.


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


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


35

ตวอยาง Data logger

ตวอยาง ขอมลจาเพาะ ตวอยาง ขอมลจาเพาะ Data loggerData logger


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


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


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)

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Signal Conditioning Circuits :Power Suppliescharnnarong.me.engr.tu.ac.th/charnnarong/My classes/AU351... · 2016-02-25 · Signal Conditioning Circuits:Power Supplies ... are amplifiers