Signal Conditioning Circuits :Power Supplies

คณะวศวกรรมศาสตรมหาวทยลยธรรมศาสตร

Measurement and Instrumentation

Signal Conditioning Circuits

:Power Supplies

Electronic measurement system

Transducer

Conditioning

circuits

Amplifier

Recorder

Data

processor

Controller

Power

supply

Command

generator

For engineering analysis

- Graphs or Tables

Process control

Power Supplies Some transducers need the power supply in order to convert the

sensed information from the sensor into the electrical signal.

Power supplies for instruments1. Battery supplies2. Line voltage supplies

TransducerPower

supply

Signal Conditioning

Battery supplies

widely used and inexpensive voltage declines after used voltage regulating circuit required

Voltage regulating circuits

Output voltage = Zener voltageEbatt. Eo

Simplest and lease expensive voltage regulator

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 Regulator

Power line

220 Vac at 50 HzLow voltage

Vac

Unregulated

Vdc Regulated

Vdc

220V

ac

Transformer 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 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/p

Vrms = Vp/2

Voltage regulator is an electrical regulator designed to automatically maintain a constant voltage level

Three pin 12 V DC voltage regulator IC.

คณะวศวกรรมศาสตรมหาวทยลยธรรมศาสตร

Measurement and Instrumentation


:Bridge Circuits

Bridge circuits

have 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 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 conditionthat the current flow through the galvanometer, Ig = 0, the bridge is in abalanced condition.

For a balanced bridge:

If the resistor R1 is a transducer, a change inresistance, associated with a change in somephysical variable, causes an unbalance of thebridge.

R2 R4R1 R3

=

1. Null Method

Bridge’s analysis

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, one of the other arms of the bridge can be adjusted to null the circuit and determine resistance.

Ig = 0

1. Null Method

TransducerWheatstone bridge

Null MethodDM DR1 DR2

(unknown) (known)

Advantages: 1. no need to know the input voltage.2. current detector required (Galvanometer) to detect Ig

Disadvantages: 1. sensitivity depending on the galvanometer2. uncertainty in the measured resistance depending on the input

voltage

Uncertainty, uR 1/Ei

The resistance R2 must be calibrated such that the adjustments

directly indicate the value of R1

2. Deflection Method

Bridge’s analysis

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 resistance.

𝐸𝑜 = 𝐸𝑖 𝑅1

𝑅1 + 𝑅2−

𝑅3

𝑅3 + 𝑅4

2. Deflection Method

R1R1 = R2 = R3 = R4 = R

TransducerWheatstone bridge

Deflection MethodDM d R d Eo

𝛿𝐸𝑜

𝐸𝑖 =

𝛿𝑅/𝑅

4 + 2(𝛿𝑅/𝑅)

EEo

For identical resistors,

Under balanced conditions (Ig = 0),

In an unbalanced condition,

Eo = 0

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: 1. high capability of the galvanometer required.2. known and regulated input voltage required.

In case of the low impedance galvanometer, Rg

𝐼𝑔 = 𝐸𝑖

𝛿𝑅/𝑅

4(𝑅 + 𝑅𝑔)

𝐸𝑜 = 𝐸𝑖

𝛿𝑅/𝑅

4(1 + 𝑅/𝑅𝑔)

A certain temperature sensor experiences a change in electrical resistance with temperature according to the equation

Example 1 Null method

𝑅 = 𝑅𝑜 1 + 𝛼(𝑇 − 𝑇𝑜)

where R = sensor resistance [W]

R0 = sensor resistance at the ref. temperature T0 [W]

T = measuring temperature [°C]

a = constant = 0.00395 °C-1

Given:1) R3 = R4 = 500 W

2) R0 = 100 W at T0 = 0°C

Determine the value of R2 that would balance the bridge at 0 °C

R1

R3 = R4 = 200 W

SolutionUnder balanced condition,

R2 = R1 (R4 /R3 ) = R1

60

80

100

120

140

160

0 20 40 60 80 100

R2

Temperature [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

𝛿𝐸𝑜

𝐸𝑖 =

𝛿𝑅/𝑅

4 + 2(𝛿𝑅/𝑅)

At deflection state,

A change in the resistance of the sensor can be

obtained from

dEo = 0.569 V

We obtain

Therefore, the temperature of the sensor is

By Kirchoff’s law, the current flowing through R1 is

given by

The power dissipated from the sensor is

d R = 25.67 ohms

100 + 25.67 = 100 1 + 0.00395 𝑇 − 0

𝑇 = 65℃

𝐼1 = 𝐸𝑖

1

𝑅1 + 𝛿𝑅 + 𝑅2= 44.3 𝑚𝐴

𝑃1 = 𝐼12 𝑅1 + 𝛿𝑅 = 0.25 𝑊

𝑅1 = 𝑅0 + 𝛿𝑅 = 𝑅0[1 + 0.00395(𝑇 − 𝑇0)]

0.0

0.2

0.4

0.6

0.8

1.0

0 20 40 60 80 100

dE 0

[V]

Temperature [C]

R1

Ei = 10 V

R

RR

EEo

SolutionUnder balanced condition,

𝛿𝐸𝑜

𝐸𝑖 =

𝛿𝑅/𝑅

4 + 2(𝛿𝑅/𝑅)

𝑅 = 𝑅0 + 𝛿𝑅 = 𝑅0[1 + 0.00395(𝑇 − 𝑇0)]

Types of Wheatstone’s bridges

Quarter bridge

3

4

2

1 2

R

R

R

RR K =

2-wire circuit analysis

Initially, where

If then

RRRRR ==== 4321

01 DR

02

4

1

=

R

Rvv K

so

D=

R

RRvv K

so

2

4

1 1

Influence of lead resistances

3

4

2

1

R

R

RR

RR

K

K =


Initially, where

If then

RRRRR ==== 43210=ov

01 DR

D=

K

soRR

Rvv 1

4

1

3

4

2

1

R

R

R

R=


Initially, where and

If then and

04321 RRRRR ====

TRRRR T D=D= a001

0=ov

D=

1

,4

1

R

Rvv T

sTo

Temperature effects

04321 TTTTT ====

TTT D= 01

DD=

1

1

4

1

R

RRvv T

so

3

4

2

1

R

R

R

R=

Half brigde circuit analysis

Initially, where and

If then and

04321 RRRRR ====

TRRRRR T D=D== a0021

0=ov

04

1

21

, =

D

D=

R

R

R

Rvv TT

sTo

Compensation of Temperature effects

04321 TTTTT ====

TTTT D== 021

D

D=

2

2

1

1

4

1

R

R

R

Rvv so


Half bridge

Benefits- Compensation of interference effects, especially temperature effects

- Increase of the static sensitivity


Full bridge

Benefits- Compensation of interference effects, especially temperature effects

- Increase of the static sensitivity

คณะวศวกรรมศาสตรมหาวทยลยธรรมศาสตร Measurement and Instrumentation


:Amplifiers

Amplifiers Is 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

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10

t

vo

0

0.2

0.4

0.6

0.8

1

0 2 4 6 8 10

t

vi

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)

2. Differential amplifiersare amplifiers which amplify two input signals. All voltages are referenced to the circuit's ground point. (used for bridge circuits)

Vi Vo

Vs

G

Vi,2 Vo

Vs

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 > 107W

3. Low output impedance Zo < 100 W4. two input ports, a noninverting and an inverting input, and one output port

Inverting terminal

Non-inverting terminal

Vo = G (Vnon-inv - Vinv)

(+5V)

(-5V)

http://upload.wikimedia.org/wikipedia/commons/9/97/Op-amp_symbol.svg

http://upload.wikimedia.org/wikipedia/commons/9/97/Op-amp_symbol.svg

Inverting amp. Non-inverting amp.

Summing amp.

IntegratorDifferentiator

Voltage Follower

𝐺𝑐 =𝑉𝑜𝑉𝑖

≅ −𝑅𝑓

𝑅1

Inverting Amplifier:

Rf

𝑖1 + 𝑖𝑓 = 𝑖𝑎

Circuit analysis

𝑖𝑓 =𝑉𝑜 − 𝑉𝑎

𝑅𝑓 𝑖𝑎 =

𝑉𝑎𝑅𝑎

𝑖1 =𝑉𝑖 − 𝑉𝑎

𝑅1

Sum of currents at Node A

Rf

R1

Ra

Vi

Vo

Vai1

if

iaA

Output voltage of the inverting amplifier

𝑣𝑎 = −𝑉𝑜𝐺

Non-inverting Amplifier:

𝐺𝑐 =𝑉𝑜𝑉𝑖

≅ 1 +𝑅𝑓

𝑅1

Non-inverting

amplifier

iV ico VGV =

Summing Amplifier:

Summing

amplifier

321 ,, VVV

=

3

3,

2

2,

1

1,

R

V

R

V

R

VRV

iii

fo

Voltage follower Amplifier:

Voltage

follower

iV ico VGV =𝐺𝑐 =

𝑉𝑜𝑉𝑖

≅𝐺

1 + 𝐺

Differentiator

DifferentiatoriV

dt

dVCRV i

fo 1=

Inverting

amplifier

11

1

CRR

RG

f

f

c == dt

dVV i

diff =

Integrator

IntegratoriV

=

t

i

f

o dtVCR

V01

1

Inverting

amplifier

f

f

c CRR

RG 1

1

===

t

idtVV0

int



:Filters

Filters In 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 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

RC filter circuits

R

C

C

R

Filters

Unfiltered signal with low frequency

variation and high frequency noise

High-pass filter

High-pass filter employed to

remove low frequency variation

Low-pass filterFiltered signal

Low-pass filter employed to remove

high frequency noise

High-pass RC filters

High-pass filter permits only frequencies above the cutoff frequency to pass.

𝑉𝑜 =𝜔𝑅𝐶

1 + (𝜔𝑅𝐶)2 𝑉𝑖 𝑒

𝑗𝜔𝑡+𝜙

𝜙 =𝜋

2− 𝑡𝑎𝑛−1(𝜔𝑅𝐶)

where

𝑉𝑜 ~ 0 if w 0

Consider a summation of voltage drops around the loop

𝑉𝑖 −𝑞

𝐶− 𝑅𝑖 = 0

If and𝑉𝑖 = 𝑉𝑖 𝑒𝑗𝜔𝑡 𝑞 = 𝑖

𝑡

0

𝑑𝑡

The output voltage Vo is

High pass

Passband

Stopband

Low-pass filters Low-pass filter permits only frequencies below the cutoff frequency to pass while

blocking the passage of frequency information above the cut off frequency.

where

𝑉𝑜 ~ 0 ถา w

The output voltage is

𝑉𝑜 =1

1 + (𝜔𝑅𝐶)2 𝑉𝑖 𝑒

𝑗𝜔𝑡−𝜙

𝜙 = 𝑡𝑎𝑛−1(𝜔𝑅𝐶)

Passband

Stopband

Cut-off frequency is 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| = wCRC/[1+(wCRC)2]1/2 = 0.02 wC = 5/RC fC = 5/(2pRC)

|Vo|/|Vi| = 1/[1+(wCRC)2]1/2 = 0.02 wC = 0.203/RC fC = 0.203/(2pRC)

Active filters

Operational amplifiers are employed to construct active filters where select frequencies can be attenuated and the signal amplified during the filtering process.

741+

-R1

R2

C2

Vi Vo

741+

-R1

R2

Vi Vo

C1

Low-pass active filter High-pass active filter

𝑓𝑐 =1

2𝜋𝑅2𝐶2 𝑓𝑐 =

1

2𝜋𝑅1𝐶1



:Modulators and Demodulators

Amplitude Modulators Amplitude 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 signal(sinusoidal, transient, or random)

Carrier signal(sinusoidal, transient, or random)x

𝑉𝑖(𝑡) = 𝑣𝑖sin(𝜔𝑖𝑡) 𝑉𝑐(𝑡) = 𝑣𝑐sin(𝜔𝑐𝑡)

𝜔𝑐 ~ 10 − 100 𝜔𝑖

𝑉𝑜 = 𝑉𝑖 sin 𝜔𝑖𝑡 (𝑉𝑐 sin 𝜔𝑐𝑡)

𝑉𝑜 =𝑉𝑖𝑉𝑐

2 cos 𝜔𝑐−𝜔𝑖 𝑡 − cos 𝜔𝑐+𝜔𝑖 𝑡

Modulated signal

𝜔𝑐 = 4000 𝐻𝑧

In general, noise signals occur at 50-Hz

𝜔𝑚 = 3950 𝐻𝑧 𝑎𝑛𝑑 4050 𝐻𝑧

𝜔𝑖 = 50 𝐻𝑧

Amplitude Modulators

Advantages(1) Stability(2) Low power(3) Noise suppression

Demodulation is the process that the transducer signal is separated from the carrier signal.

Rectifier Filter

Assignment #3

1. Shown that the output voltage of the summing amplifier be expressed by the equation

2. Consider the Wheatstone bridge. Suppose R3 = R4 = 200 W, 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?

b) If the bridge is operated in a balanced condition in order to measure x, determine the relationship between R2 and x.

𝑉𝑜 = −𝑅𝑓 𝑉𝑖,1

𝑅1+

𝑉𝑖,2

𝑅2+

𝑉𝑖,3

𝑅3


Fundamental of Sampling, Data

Acquisition and Digital Devices

Sampling concepts

Conversion of analog signals to digital code is extremely importantin any instrument system that involves digital processing of the analogoutput signals from the signal conditioners.

Continuous

Signal

Discrete

Signal

“discrete time series”

Discrete time series

Suppose the transducer signal is measured repeatedly at successive sample time increments dt

𝑦(𝑟𝛿𝑡) = 𝑦 𝑡 𝛿(𝑡 − 𝑟𝛿𝑡) 𝑟 = 1, 2, 3, … , 𝑁

sample rate

total sample period

fs = 1/dt [Hz]

Sample rate

The sample rate has a significant effect on perception and reconstruction of the continuous analog signal in the time domain

fs > 2 fm

*** s = sampling

m = measuring

fm = 10 Hz (sine wave analog signal) fs = 100 Hz

fs = 27 Hz fs = 12 Hz

d t < 1/(2 fm )

Alias Frequencies

When 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”.

Original signal10 Hz 2 Hz

sample rate 12 Hz

Interpreted signal

is the diagram that is used to

find the alias frequency.

Example

f = 10 Hz

fN = 6 Hz

Folding Diagram

[f/fN]y-axis= 10/6 = 1.67

0.33

[f/fN]x-axis = 0.33 Hz

1.67

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 sample rate for the signal

fN = 0.5 fs

Signal = Signal [f < fN ] + Signal [f > fN ]

Anti-aliasing filter

Example Consider a complex periodic signal in the form

-15

-10

-5

0

5

10

15

0 20 40 60 80 100 120 140

y(t)

t

ym

10 sin ()

0.5 sin()

1.2 sin()

𝑦𝑚 𝑡 = 10 sin 50𝜋𝑡 + 0.5 sin 150𝜋𝑡 + 1.2 sin(250𝜋𝑡)

Example (cont.) If the signal is sampled at 100 Hz

Observe all frequency content in the sampled signal

fN = 0.5 fs = 0.5 x 100 Hz = 50 Hz

f1 = 25 Hz f2 = 75 Hz f3 = 125 Hz

fN < fm Alias frequency

𝑦𝑚 𝑡 = 10 sin 50𝜋𝑡 + 0.5 sin 150𝜋𝑡 + 1.2 sin(250𝜋𝑡)

-15

-10

-5

0

5

10

15

0 20 40 60 80 100 120 140

y(t)

t

ys

ym

10 sin ()

Example (Cont.) Sampled signal

Alias frequency = 25 Hz

𝑦𝑠 𝑟𝛿𝑡 = 10 sin 50𝜋𝑟𝛿𝑡 + 0.5 sin 50𝜋𝑟𝛿𝑡 + 1.2 sin(50𝜋𝑟𝛿𝑡)

Example (cont.) Use a low-pass filter with the cut-off frequency > 50 Hz

𝑦𝐴 𝑟𝛿𝑡 = 10 sin 50𝜋𝑟𝛿𝑡

-15

-10

-5

0

5

10

15

0 20 40 60 80 100 120 140

y(t)

t

ym

yA = 10sin()

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.

fs > 2 fm

mT1 = Nd t

Note: N = 2M: M = # of sample points

Amplitude spectra

Example

Amplitude = 10Fundamental period T1 =1/ 100 s

(a) m = N dt / T1 = 2.56

(b) m = N dt / T1 = 10.24

(c) m = N dt / T1 = 8

No leakage

amplitude leakage

amplitude leakage

𝑦 𝑡 = 10cos(2𝜋100𝑡)

(8-bit)

d f = (Nd t)-1 = fs/N = 39 Hz

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 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.

Signal flow scheme for an automated data-acquisition system

Data-Acquisition System components

A/D

converter

Signal Conditioning: Filters and

Amplification

Filters1. Analog filters are used to control the frequency content of the signal being sampled.

Example: Anti-alias filter2. Digital filters, which are software-based algorithms, are effective for signal analysis after sampling

Example: moving-averaging scheme (for removing noise)

𝑦𝑖∗ = (𝑦𝑖−𝑛 + ⋯ + 𝑦𝑖−1+𝑦𝑖 + 𝑦𝑖+1 + ⋯ + 𝑦𝑖+𝑛)/(2𝑛 + 1)


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.

0-50 V signal 0-10 V A/D converter

Voltage divider

for signal attenuation

R1 = 40 kW

R2 = 10 kW


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 W, the current signal would be

converted into the voltage signal

1 – 5 V


Amplification

MultiplexerWhen 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, is given by

𝑄 =𝐸𝐹𝑆𝑅

2𝑀

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 =

0.5 to 1000

The minimum detectable voltage

when set at maximum gain is

Q = 2.44 mV / 1000 = 2.44 mV3-bits resolution

Single-ended connection Differential-ended connection

Suitable 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

Analog signal input connections

Example

A signal from the strain gage with the sensitivity = 2.5mV/me 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 me, specify appropriate G for amplifier, filter type and cut-off frequency.

VO

Vi

+ -

Strain gage (Rg) , S = 2.5mV/me

signal

Resolution = 10/212 = 2.44 mV

Amplifier

Signal Range = (1 me - 500 me) x 2.5 mV/me = 2.5 mV – 1.25 mV

G = 1000: Amplified Signal = (2.5mV – 1.25 mV) x 1000 = 2.5 mV – 1.25 V

G = 3000: Amplified Signal = (2.5mV – 1.25 mV) x 3000 = 7.5 mV – 3.75 V

Sample rate, fs = 1000 Hz

Filter

Nyquist frequency , fN = 1000/2 = 500 Hz

*** Low-pass filter with a cut-off frequency = 500 Hz required

*** Differential-ended connections required

Final Examination on OCT 8, 2011 (1300-1600)

Closed books and closed notes

Non-programmable calculators are permitted

Electronic dictionaries are not permitted

7 questions (10 marks each)

Data Logger

คออปกรณอเลคทรอนคสซงท ำหนำทบนทกและจดเกบขอมลทตรวจวดจำกเครองมอวดหรอเซนเซอรตำงๆ ปจจบน Data logger ใชพนฐำนของระบบประมวลผลแบบดจตอล สวนประกอบทส ำคญคอ ไมโครโปรเซสเซอร หนวยควำมจ ำ และเซนเซอร สำมำรถแบงออกไดเปน 2ลกษณะคอ

- General purpose types

- Specific devices

ขอด 1. ท ำงำนดวยตวเอง (Stand-alone)2. เคลอนยำยงำย (Portable)3. บนทกขอมล 24 ชวโมง4. สำมำรถใชกบ DC power

http://www.dataq.com/images/products/data-logger/gl200_500w.jpg

ตวอยาง Data logger

ตวอยาง ขอมลจ าเพาะ Data logger

ตวอยาง ขอมลจ าเพาะ Data logger (ตอ)

Grounds, Shielding, and connecting

wires การเชอมตอสายสญญาณในอปกรณอเลกทรอนคสอาจเปนสาเหตของการเกดสญญาณรบกวน (noise) ในระบบ

สญญาณรบกวนจะมผลมากในสญญาณทมก าลงต า (< 100 mv)

วธปองกนสญญาณรบกวน

1. ใชสายสญญาณใหสนทสด

2. ไมใหสายสญญาณอยใกลแหลงสญญาณรบกวน

3. ใชชล (shield) ปองกนและตอสายลงดน (ground)

4. บดสายเปนเกลยว

Ground and Ground loop

Ground คอจดอางองในวงจรไฟฟาใชในการเปรยบเทยบแรงดนไฟฟา โดยทวไป Ground จะเปนเสนทางเดนของกระแสลงสพนดน ( earth ground) ซงมคาแรงดนไฟฟาเทากบศนย

Ground potential อาจมคาตางกนขนกบจดเชอมตอ (เชน อปกรณ อาคาร และพนดน)

สายดนทความยาวมากๆ อาจเปนท าหนาทเหมอนกบ เสาอากาศ (Antennae)

Signal

ground

Chassis

ground

Earth

ground

Ground Symbols

Ground loop

Ground loop เปนสาเหตมาจากการเชอมตอ ground ของหลายวงจรเขารวมกน โดยGround potential ของวงจรทตอรวมกนมคาไมเทากน

ผลตางของศกยไฟฟาทเกดใน Ground loop สงผลใหเกดการไหลของกระแสในสายสญญาณ น าไปสสญญาณรบกวนในระบบ

Signal current

Ground current

วธปองกน ground loop มการตอสายดนเพยง 1 จด (single-point ground) ตดตง Isolation transformer ( 1:1 power transformer AC signal ผานไดเทานน) ใช Shield ปองกนสญญาณรบกวนหากสญญาณสงมก าลงต า

การตอ Shield ปองกนทไมถกตอง การตอ Shield ปองกนทถกตอง(ตอสาย ground ของShield ปองกนทปลายดานสายสงสญญาณดานเดยว)

Signal Conditioning Circuits :Power Supplies

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