Et Zc344-l5 Part1

BITS PilaniPilani Campus

Instrumentation and controlET ZC 341

1

Swapna Kulkarni

Lecturer,


Analog Signal Processing

BITS Pilani, Pilani Campus

• Signal conditioning refers to operations performed on

signals to convert them to a form suitable for interfacing

with other elements in the process control loop.

• Even in applications involving digital processing, some

type of analog conditioning is usually required before

analog –to- digital conversion is made.

Introduction


• Analog signal conditioning provides the operations

necessary to transform a sensor output into a form

necessary to interface with other elements of the

process control loop.

• It is possible to categorize signal conditioning into

several general types.

Principles of Analog Signal Conditioning


• This element could be a sensor or some other part of the

signal conditioning circuit, such as bridge circuit or

amplifier.

• Open circuit means that nothing is connected to the

output. Loading occurs when we do connect something,

a load, across the output, and the output voltage of the

element drops to some value,Vy<Vx. Different loads

result in different drops.

• Qualitatively: Thevenin’s voltage source in series within

output impedance.

Concept of loading

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Fig.2.2 the Thevenin equivalent circuit of a sensor allows easy

visualization of how loading occurs.

• This could be the input resistance of an amplifier.

• A current will flow, and voltage will be dropped across Rx. It is easy to calculate the loaded output voltage will thus be given by

• Objective :make RL >>Rx

Example


The unloaded o/p of the sensor is simply VT=(20mV/⁰C)50 ⁰C=1.0V.

Vout=10Vin=10*1.0V=10V

But this is wrong, because of loading!

Fig.2.3b shows the correct analysis. Here we see that there will be a voltage dropped across the output resistance of the sensor. The actual amplifier input voltage will be given by

Where VT=1.0V. So, Vin=0.67V. Thus,output of the amplifier is Vout=10(0.67)=6.7V

Question: An amplifier outputs a voltage that is 10 times the voltage on its input terminals. It has an input resistance of 10kΩ. A sensor outputs a voltage proportional to temperature with a transfer function of 20mV/⁰C. The sensor

has an output resistance of 5.0 kΩ,find the amplifier output.

Fig.2.3 If loading is ignored, serious

errors can occur in expected outputs

of circuits and gains of amplifiers.


• Bridge and divider circuits are two passive techniques

that have been extensively used for signal conditioning

• Bridge circuits are used primarily as an accurate means

of measuring changes in impedance.

• Such circuits are particularly useful when the fractional

changes in impedance are very small.

• Another common type of passive circuit involved in

signal conditioning is for filtering unwanted frequencies

from the measurement signal.

Passive Circuits

8


• It is quite common in the industrial environment to find signals that possess high-and/or low-frequency noise as well as the desired measurement data.

• Ex.: a transducer may convert temp. information into a dc voltage, proportional to temp.

• Because of the ever-present ac power lines, however, there may be a 60 Hz noise voltage impressed on the output that makes determination of the temp. difficult.

• A passive circuit consisting of a resistor and a capacitor often can be used to eliminate both high- and low-frequency noise without changing the desired signal information

Passive Circuits

9


Fig. 2.4 The simple voltage divider can often be used to convert resistance variation into voltage variation

• The elementary voltage divider often can be used to provide conversion of resistance variation into a voltage variation.

• The voltage of such divider is given by

VD=(R2*Vs)/(R1+R2) (2.2)

Where Vs=supply voltage;R1,R2=divider resistors

Either R1 or R2 can be the sensor whose resistance varies with some measured variable.

Divider Circuits

10


• It is important to consider the following issues when using a

divider for conversion of resistance to voltage variation:

1. The variation of VD with either R1 or R2 is nonlinear; that is,

even if the resistance varies linearly with the measured

variable, the divider voltage will not vary linearly.

2. The effective output impedance of the divider is the parallel

combination of R1 and R2. This may not necessarily be high,

so loading effects must be considered.

3. In a divider circuit, current flows through both resistors; that

is, power will be dissipated by both, including the sensor.

The power rating of both the resistor and sensor must be

considered.

Divider Circuits

11


• Solution:

a) The solution is given by eq. 2.2. For R2=4kΩ,we have

VD=(5V)(4kΩ)/(10+4)kΩ=1.43V;

For R2=12kΩ,VD=(5V)(12kΩ)/(10+12)kΩ=2.73V

b) Thus, the voltage varies from 1.43 to 2.73 V.

c) The range of output impedance is found from the parallel

combination of R1 and R2 for the min. and max. of R2. Simple

parallel resistance computation shows that this will be from

2.86 to 5.45 k Ω.

d) The power dissipated by the sensor can be determined most

easily from V²/R2 as the voltage across R2 has been calculated.

The power dissipated varies from 0.51 to 0.62mW.

Q. The divider circuit has R1 = 10.0 kΩ and Vs =5.00 V. Suppose R2 is a sensor

whose resistance varies from 4.00 to 12.0 kΩ as some dynamic variable varies

over a range. Then find (a) the minimum and maximum of VD, (b) the range of

output impedance, and (c) the range of power dissipated by R2.

12


Bridge Circuits

13


• It is used to convert impedance variation into voltage

variations.

• Adv.: it can be designed so the voltage produces varies

around zero.

• This means that amplification can be used to increase

the voltage level for increased sensitivity to variation of

impedance.

• Another application of bridge circuits is in the precise

static measurement of an impedance.

Bridge Circuits

14


Fig. 2.5 The basic dc Wheatstone bridge

• The simplest and most common bridge circuit is the dc Wheatstone bridge.

• This network is used in signal conditioning applications where a sensor changes resistance with process variable changes.

• The object labeled D is a voltage detector used to compare the potentials of points a and b of the network.

• For initial analysis, assume the detector impedance is infinite.

Wheatstone Bridge

15

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Upper Saddle River, New Jersey 07458

All rights reserved.

Va=VR3/(R1+R3) and Vb=VR2/(R2+R4)

∆V=Va-Vb

∆V= V(R3R2-R1R4)/(R1+R3)(R2+R4)


• In most modern applications, the detector is a very high

input impedance differential amplifier.

• In some cases , a highly sensitive galvanometer with a

relatively low impedance may be used, especially for

calibration purposes and spot measurement instruments.

Wheatstone Bridge


A particular combination of resistors can be found that

will result in zero difference and zero voltage across the

detector, i.e. , a null.

R3R2=R1R4 (2.8)

The application of Wheatstone bridges to process-control

applications using high-input impedance detectors.

Wheatstone Bridge

17


• The use of galvanometer as a null detector in the bridge

circuit introduces some differences in our calculations

because the detector resistance may be low and because we

must determine the bridge offset as current offset.

• When the bridge is nulled, eq 2.8 still defines the relationship

between the resistors in the bridge arms.

• Eq 2.7 must be modified to allow the determination of current

drawn by the galvanometer when a null condition is not

present.

• Easy way to determine the offset current is to first to find the

Thevenin equivalent circuit between points a and b of the

bridge.

Galvanometer Detector

18


• The Thevenin voltage is simply the open circuit voltage difference between points a and b of the circuit.

Consider the internal resistance of supply is negligible compared to the ridge arm resistances.

The offset current is

Fig 2.6 When a galvanometer is used for a null detector , it is convenient to use the Thevenin equivalent circuit of the bridge

Galvanometer

19

4231

4123

RRRR

RRRRVVTh

42

42

31

31

RR

RR

RR

RRRTh

GTh

ThG

RR

VI





• The resolution of the bride circuit is a function of

the resolution of the detector used to determine

the bridge offset. Thus, referring to the case

where a voltage offset occurs, we define the

resolution in resistance as that resistance

change in one arm of the bridge that causes an

offset voltage that is equal to the resolution of

the detector.

20

Bridge Resolution


• There are many effects that can change the

resistance of the long lead wires on a transient

basis, such as frequency, temperature, stress,

and chemical vapors. Such changes will show

up as a bridge offset and be interpreted as

changes in lead resistance are introduced

equally into two arms of the bridge circuit, thus

causing no effective change in bridge offset.

Lead Compensation

21


• Assume R4 is sensor and has been removed to a remote location (1),(2) & (3).

• Wire(3) is power lead and has no influence on the bridge balance condition.

• If wire(2) changes in resistance because of spurious influences, it introduces this change into the R4 leg of the bridge.

• Wire (1) is exposed to the same environment and changes by the same amount, but is in the R3 leg of the bridge.

Lead Compensation

22

Figure For remote sensor applications, this compensation system is used to avoid errors from lead resistance.





• Both R3 & R4 are identically changed, and thus eq. 2.8

shows that no change in the bridge null occurs.

• This type of compensation is often employed where

bridge circuits must be used with long leads to the active

element of the bridge.

Lead Compensation


• One disadvantage of the simple Wheatstonebridge is the need to obtain a null by variation ofresistors in bridge arms.

• A technique that provides for an electronicnulling of the bridge and that uses only fixedresistors (except as may be required forcalibration) can be used with the bridge. Themethod uses a current to null the bridge.

• A closed-loop system can even be constructedthat provides the bridge with a self-nulling ability.

Current Balance Bridge

24


Figure : The current balance bridge

• The standard Wheatstonebridge is modified by splittingone arm resistor into two R4and R5. A current, I is fed intothe bridge through thejunction of R4 and R5 asshown. We now stipulate thatthe size of the bridge resistorsis such that the current flowspredominantly through R5.This can be provided for byany of several requirements.The least restrictive is torequire.

R4>>R5

25






• Often, if a high – impedance null detector is used, therestriction of equation becomes

(R2 + R4)>>R5

• Assuming that either conditions of equation are satisfied,the voltage at point b is the sum of the divided supplyvoltage plus the voltage dropped across R5 from thecurrent, I.

• The voltage of point a is still given by equation. Thus, thebridge offset voltage given by


26

5

542

54 IRRRR

RRVVb

ba VVV

5

542

54

31

3 IRRRR

RRV

RR

VRV



27

• This equation shows that a null is reached byadjusting the magnitude and polarity of the current Iuntil IR5 equal the voltage difference of the first twoterms. If one of the bridge resistors changes, thebridge can be renulled by changing current, I. In thismanner, the bridge is electronically nulled from anyconvenient current source. In most applications, thebridge is nulled as some nominal set of resistancewith zero current. Changes of a bridge resistor aredetected as a bridge offset signal that is used toprovide the renulling current.


Figure Using the basic Wheatstonebridge for potential measurement

• A bridge circuit is also useful to measure small potentials at a very high impedance, using either a conventional Wheatstone bridge or a current balance bridge. This type of measurement is performed by placing the potential to be measured in series with the detector, as shown in figure.

Potential Measurement Using Bridges

28





axC VVV

Where Va is given by Equation and Vx is the potential to be measured. The voltage appearing across the null detector is

V = Vc-Vb= Vx+Va-Vb

• Vx can be measured by varying the bridge resistors to provide a null with Vx in the circuit and solving for Vxusing the null condition

29

5

542

54 IRRRR

RRVVb

And



042

4

31

3

RR

VR

RR

VRVx

• The potential is placed in series with the detector, andV is defined exactly as before. Now, however, Vb isgiven by equation, so the null condition becomes

• If the fixed resistors are chosen to null the bridge withI=0 and Vx = 0,then a very simple relationship betweenVx and the nulling circuit is:

30

Vx- I R5=0


05

542

54

31

3

IR

RRR

RRV

RR

VRVx


Employs an ac excitation, usuallya sine wave voltage signal. Theanalysis of bridge behavior isbasically the same as in theprevious treatment, butimpedances replace resistance.The bridge offset voltage then isrepresented as

voltageoffsetacE

ZZZZ

ZZZZEE

4231

4123

Figure : A general ac bridge circuit

AC Bridge

31

Where E = sine wave excitation voltage

Z1,Z2,Z3,Z4 = bridge impedances. Z3Z2=Z1Z4





AC Bridge

32

Figure : The ac bridge circuit and components for example

•An ac bridge employs impedance as shown in figure. Find the value of Rx and Cx when the bridge is nulled.





A capacitor has a purely reactive impedance which is inversely

proportional to the signal frequency. A capacitor consists of two

conductors separated by an insulator, also known as a dielectric.

At low frequencies a capacitor is open circuit, as no charge flows in the

dielectric.

A DC voltage applied across a capacitor causes charge to accumulate

on one side; the electric field due to the accumulated charge is the

source of the opposition to the current. When the potential

associated with the charge exactly balances the applied voltage, the

current goes to zero.

Driven by an AC supply, a capacitor will only accumulate a limited

amount of charge before the potential difference changes sign and

the charge dissipates. The higher the frequency, the less charge will

accumulate and the smaller the opposition to the current.

Capacitive reactance

http://en.wikipedia.org/wiki/Inversely_proportional

http://en.wikipedia.org/wiki/Inversely_proportional

http://en.wikipedia.org/wiki/Frequency

http://en.wikipedia.org/wiki/Electrical_conduction

http://en.wikipedia.org/wiki/Electrical_insulation

http://en.wikipedia.org/wiki/Dielectric

http://en.wikipedia.org/wiki/Open_circuit

http://en.wikipedia.org/wiki/Electrical_charge

http://en.wikipedia.org/wiki/Electric_field

http://en.wikipedia.org/wiki/Potential


Solution

Because the bridge is at null, we have

Z2Z3=Z1Zx

or

Example

34

x

xC

jRR

C

jRR

132

x

xC

jRRR

C

RjRR

1

12

32


35

x

xC

jRRR

C

RjRR

1

12

32

The real and imaginary parts must be independently equal, so that

01

32 R

RRRx

kR

k

kkR

x

x

2

1

12

And1

2

x

CRC

R

1 *(1 / 2 )

0.5

xC F k k

F


• To convert variations of resistance into variations of

voltage.

• This voltage variation is then further conditioned for

interface to an ADC or other system.

• The variation of bridge offset is nonlinear with respect to

any of the resistors.

• If the range of resistance variation is small and centered

about the null value , then the nonlinearity of voltage

versus resistance is small.

• Amplifiers can be used to amplify this voltage variation,

since it is centered about zero, to a useful range.

Bridge Applications

36


To eliminate unwanted noise signals from

measurements, it is often necessary to use

circuits that block certain frequencies or bands

of frequencies. These circuits are called filters. A

simple filter can be constructed from a single

resistor and a single capacitor.

RC Filters

37

Low pass RC filter

Figure : Circuit for the low-pass RC filter

• It is called low pass because it blocks high frequencies and passes low frequencies.

• A low pass filter had a characteristic such that all signals with frequency above some critical value are simply rejected.

• Practical filter circuits approach that ideal with varying degrees of success.

• When the ratio of o/p voltage to i/p voltage is one, the signal is passed without effect; When it is very small or zero, the signal is effectively blocked.

38


• The critical frequency is that frequency for which

the ratio of o/p to i/p voltage is approx. 0.707.

• In terms of the resistor and capacitor, the critical

frequency is given by

fc = (1/ 2ΠRC)

• The output to input voltage ratio for any signal

frequency an be determined graphically or can

be computed by

• ΙVout/VinΙ = 1/√[1+(f/fc)²]

Low pass filter

39


• A typical filter design is accomplished by finding

the critical frequency, fc that will satisfy the

design criteria.

• The following practical guidelines are offered

on this process.

1. Select a standard capacitor value in the F topF range.

2. Calculate the required resistance. If it is below1k or above 1M , try a different value ofcapacitor so that the required resistance fallswithin this range, which will avoid noise andloading problems.

Design Methods

40


3.If design flexibility allows, use the nearest standard valueof resistance to that calculated.

4. Always remember that components such as resistorsand capacitors have a tolerance in their indicatedvalues. This must be considered in your design. Quiteoften, capacitors have a tolerance as high as 20%.

5. If exact values are necessary, it is usually easiest toselect a capacitor, measure its value, and thencalculate the value of the required resistance. Then atrimmer resistor can be used to obtain the requiredvalue.

Design Methods

41


Figure : Circuit for the high-pass RC filter

• A high –pass filter passes high frequencies (no rejection) and blocks (rejects) low frequencies. A filter of this type can be constructed using a resistor and a capacitor, as shown in schematic of figure.

High-Pass RC Filter

42





• It is similar to low pass filter, the rejection is not

sharp in frequency but distributed over a range

around a critical frequency.

• The magnitude of Vout/Vin= 0.707 when the

frequency is equal to the critical frequency.

• An equation for the ratio of output to input

voltage as a function of the frequency for the

high filter is found to be

ΙVout/VinΙ=(f/fc)/√[1+(f/fc)²]

High pass RC filter

43


1. After the critical frequency is determined, the values of R and C

are selected. A number of practical issues limit the selection:

a) Very small values of resistance are to be avoided because they can

lead to large currents, and thus large loading effects. Similarly very

large capacitors should be avoided. In general , we try to keep the

resistance in the kΩ and above range and capacitors in the F or

less range.

b) Often, the exact critical frequency is not important, so that fixed

resistors and capacitors of approx. the computed values can be

employed . If exact values are necessary, it is usually easier to

select and measure a capacitor, then compute and obtain the

appropriate resistance using a trimmer resistance.

Practical consideration

44


2. The effective input impedance and output impedance of the RC filter may have an effect on the circuit in which it is used because of loading effects . If the input impedance of the circuit being fed by the filter is low, you may want to place a voltage follower between the filter output and the next stage. Similarly, if the input impedance of the feeding stage to the filter is high , you may want to isolate the input of the filter with a voltage follower.

3. It is possible to cascade RC filter in series to obtain improved sharpness of the filter cutoff frequency. However, it is important to consider the loading of one RC stage by another. The output impedance of the first stage filter must be much less than the input impedance of the next stage to avoid leading.

Practical Consideration

45


Active Circuits:

OP-Amp

• Inverting Amplifier

• Summing Amplifier

• Differential Amplifier

• Instrumentation Amplifier

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