Lecture 4A SignalConditioning DCBRIDGE SV

8/10/2019 Lecture 4A SignalConditioning DCBRIDGE SV

1/48

Signal Conditioning for

Electronic Instrumentation

1

MCT 3332 : Instrumentation and Measurements

Dr. Hazlina Md Yusof

Department of Mechatronics Engineering

International Islamic University Malaysia


2/48

2

Content:

Introduction

Bridge Circuits

Op-Amp circuits for Inverting

Non-inverting amplifier High impedance buffer

Differential amplifier

Instrumentation amplifier Active filters


3/48

Analog Signal Conditioning

1. Introduction

Signal conditioning is used to process an output

signal from a sensor to make it suitable for the nextstage of operation.

3


4/48


Signal conditioning is applied when signal:-

is too small and needs to be amplified

contains interference which need to be attenuated

is non linear and needs to be linearized

Need to be differentiated/integrated

in analogue form and digital form is needed

in the form of resistance change needs to be converted into

current change

in the form of voltage change needs to be converted intocurrent change

4


5/48


Advantages of signal conditioning

to prepare the signal to be in a suitable form for interfacing with

other elements in the process-control loop.

In analog conversions, the conditioned output signal is still in

an analog representation of the variable.

Even in applications involving digital processing, some type of

analog conditioning is usually required before analog-to-

digital conversion is made.

5


6/48


Principles of analog signal conditioning

sensor measures a variable/measurand by convertingthe information about that variable into a dependentsignal of either electrical or mechanical in nature.

analog signal conditioning provides the necessaryoperations to transform the sensor output into a formnecessary to interface with other elements.

the effect of the signal conditioning is defined by atransfer function

6


7/48

7Measurement Sub-system

A/D board



8/48


Signal Conditioning is categorized into several types:

(a) Linearization

(b) Conversion(c) Filtering and Impedance Matching

(d) Concept of Loading

8


9/48


(a) Linearization

a sensor output varies nonlinearlywith a process variable.

a linearization circuit will condition

the sensor output to be linear withthe process variable.

generally, these types of circuits aredifficult to design and operate onlywithin narrow limits.

9


10/48


Modern approach :

i. provide the nonlinear signal as input to a computer and

perform the linearization using software.

ii. virtually any nonlinearity can be handled in this manner

and, with the speed of modern computers, can be done inreal time.

10


11/48


(b) Conversion

Circuits can be used to convert the resistance change to a :

(i) voltage signal

(ii) current signal (i.e. bridge)

11


12/48


(c) Filtering and Impedance Matching

(i) Filtering

high-pass, low-pass, or notch filtersare used to attenuate

unwanted signals from the process control loop.

filtering can be accomplished by, passivefilters; resistors, capacitors, and inductors activefilters, using amplifier and feedback.

(ii) Impedance matching

active and passive circuit networks are employed to provide

impedance matching to maximum power transfer from source to

load

12


13/48


(d) Concept of Loading

the loading of one circuit byanother is important.

loading occurs when something is

connected to an open circuit

(across the output), and the output

voltage of the element drops to

some value, Vy < Vx.

Different loads result in different

drops.

13

RLis connected across the output of

the element. When a current flow,and voltage will be dropped across

The loaded output voltage will be

The voltage that appears across theload, instead of Vx is reduced by the

voltage dropped across the internal

resistance.

xL

xxy

RR

R1VV


14/48


Problem 1

An amplifier outputs a voltage that is ten times the

voltage on its input terminals. It has an input

resistance of 10k. A sensor outputs a voltage

proportional to temperature with a transfer function

of 20mV/C. the sensor has an output resistance of

5.0k . If the temperature is 50 C, find the

amplifier output.

14


15/48


3. Passive circuits

bridge and divider circuits are two passive

techniques that is used for signal conditioning bridge circuits are used for measuring changes in

impedance.

a passive circuit consists of a resistor and capacitor

can be used to eliminate high- and low-frequencynoise without changing the desired signal

information.

15


16/48


(a) Divider Circuits

The voltage of such a divider isgiven by the well-known

relationship

(2.2)

where

VS, = supply voltage

R1, R2=divider resistors16

21

2

RR

VRV

S

D


17/48


Problem 2

The divider of figure has R1=10.0kandVs=5V. Suppose R2is a sensor whoseresistance varies from 4.00 to 12.0k as some dynamic variable varies over arange. Then find

(a) The maximum and minimum of VD(b) The range of output impedance

(c) The range of power dissipated by R2

17


18/48


Issues to be considered for converting resistance tovoltage :

1. the variation of VDwith either R1or R2is nonlinear,

even if the resistance varies linearly with themeasured variable.

2. the effective output impedance of the divider is theparallel combination ofR2and R1.

3. in a divider circuit, current flows through bothresistors; power will be dissipated by both, includingthe sensor.

18


19/48


(b) Bridge Circuits

Bridge Circuit is a null method, operates on the principleof comparison. That is a known (standard) value isadjusted until it is equal to the unknown value.

used to convert impedance variations into voltagevariations.

advantages : it can be designed so the voltage produced varies around

zero. amplification can be used to increase the voltage level for

increased sensitivity to variation of impedance.

19


20/48


20

Bridge Circuit

DC Bridge(Resistance)

AC Bridge

Inductance Capacitance Frequency

Schering Bridge Wien BridgeMaxwell Bridge

Hay BridgeOwen BridgeEtc.

Wheatstone Bridge

Kelvin BridgeMegaohm Bridge


21/48

Analog Signal ConditioningWheatstone bridge

A Wheatstone bridge is an electrical circuit used to

measure an unknown electrical resistance by balancing

two legs of a bridge circuit, one leg of which includes the

unknown component.

Its operation is similar to the original potentiometer. It was invented by Samuel Hunter Christie in 1833 and

improved and popularized by Sir Charles Wheatstone in

1843.

One of the Wheatstone bridge's initial uses was for thepurpose of soils analysis and comparison

21


22/48


Wheatstone Bridge simplest and common bridge circuit .

used in signal-conditioning applicationswhere a sensor changes resistance withprocess variable changes.

Dis a high-input impedance differential

amplifiervoltage detector used tocompare the potentials of the points aand bof the network.

22


23/48


Instrument for making

comparison in measurements

Operates on null-indication

principle

Indication is independent of the

calibration of the indicating device

Able to achieve higher degree of

accuracy

Used in control circuits

One arm contains a resistive

element that is sensitive to thephysical parameterbeing

controlled

23


24/48


Accurate and reliable

Heavily used in industry

Accuracy of 0.1% compared to ordinary ohmmeter for resistance

measurement

Bridge used to determine an unknownresistor. Vary one of the remaining resistors until the current through the null

detector (usually a galvanometer other sensitive current meters ) is

zero.

This indicates that the bridge is in balanced condition.24

GC D

A

B

R1

R3

R2

R4

I4

I1

I3

I2Ratio arms

Standard arm


25/48


25


26/48

Analog Signal ConditioningWheatstone Bridge

26



27/48

Analog Signal ConditioningWheatstone bridgeMeasurement Errors

27

1. Limiting error of the known resistorsUsing 1st order approximation:

2. Insufficient sensitivity of Detector3. Changes in resistance of the bridgearms due to the heating effect (I2R) or

temperatures4. Thermal emf or contact potential in

the bridge circuit5. Error due to the lead connection3, 4 and 5 play the important role in the

measurement of low value resistance



28/48

Example

In the Wheatstone bridge circuit, R3is a decade

resistance with a specified in accuracy 0.2% and R1 and

R2= 500 0.1%. If the value of R3at the null position is

520.4 , determine the possible minimum and maximumvalue of RX

28



29/48

Example

A Wheatstone bridge has a ratio arm of 1/100 (R2/R1). At

first balance, R3 is adjusted to 1000.3 . The value of Rx

is then changed by the temperature change, the new value

of R3 to achieve the balance condition again is 1002.1 .Find the change of Rx due to the temperature change.

29



30/48


Sensitivity of the Wheatstone Bridge

At an unbalanced condition, current that passes through the

galvanometer will cause a deflection of the pointer.

The amount of deflection is a function of the sensitivity of thegalvanometer.

Deflection, D per unit current.

S = millimeters or degrees or radians

A A A

The total deflection D is then becomes,

D = S x I (I in A) 30


31/48


Sensitivity of the Wheatstone Bridge A galvanometer is use to detect an unbalance condition in Wheatstone bridge. Its

sensitivity is governed by: Current sensitivity (currents per unit defection) and internal

resistance.

consider a bridge circuit under a small unbalance condition, and apply circuit analysis

to solve the current through galvanometer

To calculate the current that passes the galvanometer Use TheveninsTheorem to find the Thevenin equivalent voltage seen by the

galvanometer.

31

A l Si l C diti i


32/48


Galvanometer Detector Usage of a galvanometer as a

null detector in the bridgecircuit introduces difference incalculation because :

the detector resistance may below but it is still there

the bridge offset as current offsetmust be determined first.

32


33/48

33



34/48

Example

Figure below show the schematic diagram of a Wheatstone bridge with

values of the bridge elements. The battery voltage is 5 V and its internal

resistance negligible. The galvanometer has a current sensitivity of 10

mm/Aand an internal resistance of 100 . Calculate the deflection of

the galvanometer caused by the 5- unbalance in arm BC

34


35/48

Example

The galvanometer in the previous example is replaced by one

with an internal resistance of 500 and a current sensitivity of

1mm/A. Assuming that a deflection of 1 mm can be observedon the galvanometer scale, determine if this new galvanometer is

capable of detecting the 5- unbalance in arm BC

35


36/48


Example 3

Calculate the current, IG

through the galvanometer in the

circuit. Given

E=6V, R1=1k,R2=1.6k,R3=3.5k,

R4=7.5k , RG=200,

36

Ca

A

B

D

E

R1

R3

R2

R4

b


37/48


Bridge Resolution

is used to determine the bridge offset.

viewed as an overall accuracy of the instrument.

R represents the uncertainty in the resistor used in the givenbridge and error signal detector (i.e. across terminal a and b).

37


38/48


Slightly Unbalanced Wheatstone Bridge

38


39/48

Example

Circuit in Figure (a) below consists of a resistor Rv which is sensitive to

the temperature change. The plot of R vs Temp. is also shown in Figure

(b). Find (a) the temperature at which the bridge is balance and (b) The

output signal at Temperature of 60C.

39



40/48


Kelvin Bridge

40

RY

=1

2

3

Low resistance Bridge: R

x

< 1



41/48


Kelvin Double Bridge: 1 to 0.00001

41



42/48

42


Kelvin Double Bridge: 1 to 0.00001


43/48

Kelvin Bridge

Example 4

If in the figure given, the ratio of Rato Rbis 1000, R1=5

and R1=0.5R2,what is the value of Rx?

43

A l Si l C diti i


44/48

Analog Signal ConditioningLead Compensation

used where changes in lead resistance

are introduced equally to two (both)

arms of the bridge circuit without

causing effective changes in bridge

offset.

assume R4is the sensor

wire 3 is the power leadhas no

influence on bridge balance condition changes in wire 2 will give some

changes to R4.

wire 1 is exposed to the same

environment (i.e. lead) and changed by

the same amount

R3and R4are changed identically, no

change in the bridge; null occurs.

this type of compensation is used with

long leads to the active element of the

bridge.

44


45/48

Analog Signal ConditioningCurrent Balanced Bridge

a technique that provides an electronicnulling of the bridge (i.e. zeroing network)

and uses only fixed resistors by using

current

this method is used to overcome the

disadvantage of a Wheatstone Bridge

where a null by variation of resistors is

needed.

the Wheatstone bridge is modified by

splitting one arm resistor into two, R4and

R5.

current, I is fed into the bridge throughjunction R4 &R5.

R4R5 (1)

if a high impedance null detector (high

Rm) is used, Eq. 1 will be

(R2+R4)R5 (2)

45


46/48


5

542

54

b IRRRR

)RV(RV

46

Vbis the sum of the divided supply voltage + the voltage dropped across R5from the current, I

(3)

the voltage at point a is

the bridge offset voltage is V = Va-Vb

(4)

31

3

aRR

VRV

5

542

54

31

3IR

RRR

)RV(R

RR

VRV



47/48


Problem 4

A current balance bridge as

shown has resistors R1=R2=10k

R4= 950 R3=1.0kR5=50

and a high impedance nulldetector. Find the current

required to null the bridge if R3

changes by 1 . The supply

voltage is 10V.

47

A l Si l C diti i


48/48


Bridge Controlled Circuits

If the ratio of R2to Rvequals the ratio ofR1and R3then the error signal is zero

If the physical parameter changes, Rvwillchange. the bridge becomes unbalanced

in most control applications, the measuredand control parameter is corrected, restoringRvto the value that creates a null conditionat the output of the bridge.

the Rvvalue can be calculated using thebalanced bridge equation.

due to small value of Rv, the error signal isquite lowneeds to be amplified beforebeing used for control purposes

48

Documents

Lecture 4A SignalConditioning DCBRIDGE SV