Analog Signal Conditioning

Chapter 2

Analog Signal Conditioning

Signal Conditioning

• Signal conditioning is the operation

performed on the signal to convert them to

a form suitable for interfacing with other

elements in the process control.elements in the process control.

Signal Conditioning

• Signal conditioning can be categorized

into 6 types

– Signal-level and bias changes

– Linearization– Linearization

– Conversions

– Filtering and impedance matching

– Concept of loading

Signal-level and bias changes

• The method to adjust the level

(magnitude) and bias (zero value) of

voltage signal

• For example• For example

Signal conditioning circuit0.2 V – 0.6 V 0 V – 5 V

Zero shift0.2 V – 0.6 V 0 V – 5 V

Amplification0 V – 0.4 V

Amplification

Linearization

• Often, the

characteristic of a

sensor is nonlinear

• Special circuit were • Special circuit were

devised to linearize

signals

• Modern approach is

to use computer

software to linearize

Conversion

• The circuit to covert one form of signal or

physical values into the other form

– Resistance to voltage

• Typical conversion is to convert resistance • Typical conversion is to convert resistance

or voltage to 4 to 20 mA and convert back

to voltage at the receiving end

• Thus, voltage-to-current and current-to-

voltage circuits are essential

Digital Interface

• The use of computer is process control

requires the conversion of analog to digital

signal

– ADC– ADC

– DAC

Filtering

• Some signals input are spurious (contain

more than 1 frequency)

• It is necessary to filter the frequency

matched with the devicesmatched with the devices

– The electric line frequency is 50 Hz

– The transient of motor is kHz

• Example

– Highpass, lowpass, bandpass filter

Impedance Matching

• Connecting the sensors or process control

element with different impedance causes

signal reflection

• The network or circuit to match impedance • The network or circuit to match impedance

thus to reduce signal reflection

Concept of Loading

• When the sensor or circuit is connected to

load, this will introduce the uncertainty in

the measurement (amplitude of voltage)

• The output voltage is calculated using

voltage division as

1

Ly x

L x

xx

RV V

R R

RV

R R

=

+

= −

+

• Output voltage is reduced by the voltage

drop

• To reduce the uncertainty,

x

L xR R

+

L xR R≥

Example

An amplifier outputs a voltage that is 10 times the

voltage on its input terminals. It has an input resistance

of 10 kΩ. A sensor outputs a voltage preoperational to

temperature with a transfer function of 20 mV/°C. The

sensor has an output resistance of 5.0 kΩ. If the

temperate is 50 °C, find the amplifier output.

Sensor50 °C ? V

Amplification

Signal Conditioning: Passive

Element• Signal conditioning circuit with element R,

L, and C are

– Divider circuits

– Bridge circuits– Bridge circuits

– RC filter

Divider circuits

• Useful to convert resistance into voltage

• The voltage of the divider is given as

2

1 2

D s

RV V

R R=

+

sV

DV

1R

2R

1 2

supply voltage

, divider resistors

sV

R R

=

=

• It is important to consider the following

issues

1. The variation of VD with either R1 or R2 is

nonlinear

2. The effective output impedance of the

divider is the parallel combination of R1 and

R2.R2.

3. The current flows to both R1 and R2. The

power rating of both resistors should be

considered

Example

The divider shown 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 variables 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

sV

DV

1R

2R

Bridge Circuit

• Bridge circuits are used to convert

impedance variations into voltage

variations.

• Application of bridge circuits is in precise • Application of bridge circuits is in precise

static measurement of an impedance

Wheatstone Bridge

• Wheatstone bridge is represented below

• The potential difference ∆V between

points A and B is simply

Where

A BV V V∆ = −

3RV V= 3

1 3

AV VR R

=+

4

2 4

B

RV V

R R=

+

Bridge supply voltageV =

• The voltage difference between A and B is

• The equation above can be reduced to

3 4

1 3 2 4

VR VRV

R R R R∆ = −

+ +

• At particular combination of resistance

values, the voltage difference is zero

( )( )3 2 1 4

1 3 2 4

R R R RV V

R R R R

−∆ =

+ +

3 2 1 4R R R R=

Advantage of

Wheatstone Bridge• It can be used as a resistance sensor that

eliminates the supply voltage offset or

changes.

• The null still maintains• The null still maintains

Galvanometer detector

• Galvanometer is used as a null detector in

the Wheatstone bridge to detect the

condition of the Wheatstone bridge

• And it is required to determine the current • And it is required to determine the current

offset instead of voltage offset


• Galvanometer is represented as an

resistance RG

• The equivalent circuit of the Wheatstone

bridge with Galvanometer is bridge with Galvanometer is


• The current offset is determined by

Where

thG

th G

VI

R R=

+

( )( )3 2 1 4

1 3 2 4

1 3 2 4

1 3 2 4

Thevanin eqivalent voltage of the Wheatsone bridge

Thevanin eqivalent resistance of the Wheatsone bridge

th

th

V

R R R RVR R R R

R

R R R R

R R R R

=

−=

+ +

=

= ++ +

Bridge resolution

• The resolution of the bridge is determined

by the resolution of the detector

• We can convert the resolution of the

detector to find the smallest resistance detector to find the smallest resistance

change in the bridge

Example

A bridge circuit has resistance of R1 = R2 =

R3 = 2.00 kΩ and R4 = 2.05 kΩ and a 5.00

V supply. If the galvanometer with a 50.0

Ω internal resistance is used for a

detector, find the offset current.detector, find the offset current.

Example

A bridge circuit has R1 = R2 = R3 = R4

=120.0 Ω resistance and a 10.0 V supply.

Clearly, the bridge is null. Suppose a 3½

digit DVM on a 200 mV scale will be used

for the null detector. Find the resistance for the null detector. Find the resistance

resolution for measurements of R4.

Lead Compensation

• In many process control, the bridge circuit

may be located at far distance

• The resistance are chosen to compensate

the resistance of the leadthe resistance of the lead

Current Balance Bridge

• Current balance is

the way to obtain a

null in a quick

time.

• Electronic nulling

with fixed resistor

• One arm of the

Wheatstone

Bridge is modified

as following

• The resistance is splited into R4 and R5

• The current is fed into R5

• We want the current to flow to R5

predominantly by select

54 RR >>

• Now, the voltage at point b is

4 55

2 4 5

( )b

V R RV IR

R R R

+= +

+ +

54 RR >>

• Thus, the offset voltage is

( )4 535

1 3 2 4 5

a bV V V

V R RVRIR

R R R R R

∆ = −

+= − −

+ + +

• Which shows that the null can be achieved

by adjusting magnitude and current I

Example

A current balance bridge has resistors R1

= R2 = 10 kΩ , R4 = 950 Ω, R3 = 1 kΩ, R5

= 50 Ω and a high impedance null

detector. Find the current required to null

the bridge if R3 changes by 1 Ω. The the bridge if R3 changes by 1 Ω. The

supply voltage is 10 V

Potential Measurement using

Bridge• The bridge can be

used to measure

potential

• The potential to be • The potential to be

measured is

placed in series

with the detector

as shown in the

figure

• The voltage at point C is

• The voltage appearing across the null

detector is

c x aV V V= +

• The potential Vx can be measured by

varying the bridge till null and solve for Vx

c b x a bV V V V V V∆ = − = + −

3 4

1 3 2 4

0x

RV VRV

R R R R+ − =

+ +

• The current balance bridge can also be

used for potential measurement

( )4 535

1 3 2 4 5

0x

V R RVRV IR

R R R R R

++ − − =

+ + +

Example

A bridge circuit for potential measurement

nulls when R1 = R2 = 1 kΩ, R3 = 650 Ωand R4 =500 Ω with a 10 V supply. Find

the unknown potential

AC bridge

• The bridge concept can be applied to the

impedance match

• The bridge offset is

where

( )( )3 2 1 4

1 3 2 4

Z Z Z ZE E

Z Z Z Z

−∆ =

+ +

1 2 3, 4

sine wave excitation voltage

, , bridge impedance

E

Z Z Z Z

=

=

Example

An ac bridge employs impedance as

shown. Find the value of Rx and Cx when

the bridge is nulled

Bridge Application

• Convert variations of

resistance into voltage

• The relationship of the

bridge is non linear for bridge is non linear for

large scale range of R

• Linear near the null

condition

RC Filter

• To eliminate unwanted noise signals from

measurement, it is needed to use filter

circuit

A filter is a circuit that is

designed to pass signals

with desired frequencies

and reject or attenuate

others.

Filters

Background:

. Filters may be classified as either digital or analog.

. Digital filtersDigital filters are implemented using a digital . Digital filtersDigital filters are implemented using a digital

computer

or special purpose digital hardware.

. Analog filtersAnalog filters may be classified as either passive or

active and are usually implemented with R, L, and C

components and operational amplifiers.

Filters

Background:

. An active filteractive filter is one that, along with R, L, and

C

components, also contains an energy source, components, also contains an energy source,

such

as that derived from an operational amplifier.

. A passive filterpassive filter is one that contains only R, L, and

C components. It is not necessary that all three be

present. L is often omitted (on purpose) from

passive filter design because of the size and cost

of inductors – and they also carry along an R that

must be included in the design.

Passive Analog Filters

Background: Four types of filters Four types of filters -- “Ideal”“Ideal”

lowpasslowpass highpasshighpass

bandpassbandpass bandstopbandstop

Background: Realistic Filters:Realistic Filters:

lowpasslowpass highpasshighpass


bandpassbandpass bandstopbandstop


Low Pass Filter Consider the circuit below.

R

CVI VO

+

_

+

__

1( ) 1

1( ) 1OV jw jwCV jw jwRC

RijwC

= =++

Low pass filter circuit


Low Pass Filter

0 dB

ω

Bode.-3 dB

1

ω

ω0

1/RC

1/RC

Linear Plotx

0.707

Passes low frequencies

Attenuates high frequencies


High Pass Filter Consider the circuit below.

C

RVi V

+

+

_RVi VO

_

_

( )1( ) 1

OV jw jwRCRV jw jwRC

RijwC

= =++

High Pass Filter


High Pass Filter

0 dB

. -3 dB

BodePasses high frequencies

.

0 ω

ω

1/RC

1/RC

1/RC

10.707

Linear

Attenuates low frequencies

x


Bandpass Pass Filter Consider the circuit shown below:

C L

RViVO

++

_RVi

__


Bandpass Pass FilterWe can make a bandpass from the previous equation and select

the poles where we like. In a typical case we have the following shapes.

0 dB

-3 dB. . Bode

ω

ω0

-3 dB

ωlo

ωhi

.

. .

.10.707

Bode

Linear

ωlo

ωhi

RLC Band stop Filter

Consider the circuit below:

R

L

C

+

VO

+

_Vi

=)(sGv C

_

=)(sGv

VinVO

C

Rfb

+

_

+

_

Rin

Basic Active FiltersLow pass filter

Basic Active Filters

RinC

Vin

Rfb

VO

+

_

+

_

High pass


Vin

R1

R1

C1

C2

R2R2

Rfb

Ri

V

+

+_

Band pass filter

VO_


R1

R1

C1

Rfb

Band stop filter

Vin

C2

R2Ri

VO

+

_

+

_

Low-pass RC filter

• The simple circuit for low-pass filter is

shown below

• It passes low frequency and rejects high

frequency

• The frequency response of the low-pass

filter is shown below

• The critical frequency is the frequency for

which the ratio of the output to the input

voltage is 7.07

• The output to input ratio is determined by

1

2cf

RCπ=

• The output to input ratio is determined by

1/22

1

1

out

in

c

V

Vf

f

= +

Design Method

• To design a filter is to find fc satisfied the

criteria

– Select a stand capacitor value in the µF to pF

rangerange

– Calculate the required resistance value, if R <

1 kΩ or R > 1 MΩ, pick another capacitor

– Consider device tolerance

– If exact value is required, use trimmer resistor

Example

A measurement signal has a frequency <

1 kHz, but there is unwanted noise at

about 1 MHz. Design a low-pass filter that

attenuates the noise to 1%. What is the

effect on the measurement signal at its effect on the measurement signal at its

maximum of 1 kHz?

High-pass RC Filter

• High-pass filter passes high frequencies

and rejects low frequencies.

• The circuit for RC high-pass is shown

belowbelow

• The ratio of output voltage to input voltage

of the high pass filter is

1/22

cout

f

fV

V

=

1/2

2

1in

c

Vf

f

= +

Example 2.12

Pulses for a stepping motor are being

transmitted at 2000 Hz. Design a filter to

reduce 60 Hz noise but reduce the pulses

by no more than 3 dB.

RC Filter Consideration

• Very small resistance should be avoided because

it can lead to large current and loading effect

• If input impedance of the circuit fed by the filter is

low, a voltage follower circuit is neededlow, a voltage follower circuit is needed

• The output impedance of the filter must be much

less than the input impedance of the next stage

circuit

Example

A 2 kHz data signal is contaminated by 60

Hz of noise. Compare a single-stage and

a two stage high-pass RC filter for

reducing the noise by 60 dB. What effect

does each have on the data signal?does each have on the data signal?

Example

Suppose we require the first stage of the

last example to use a capacitor of C =

0.001 µF. Find the appropriate value of

resistance, R. Suppose these same

values are used for the second stage. values are used for the second stage.

How much further attenuation occurs at 2

kHz because of loading? What output

impedance does the series filter present?

Assume Vin source resistance is very

small.

Band-pass Filter

• Band-pass filter passes frequencies in a

certain band and rejects frequencies

below and above the band

• The ratio of output to input voltage is

( )2

22 21

out H

inH

H L L H

L

V f f

VR

f f f f f fR

=

− + + +

where

HH

L

LL

H

CRf

CRf

π

π

2

1

2

1

=

=

Example

A signal-conditioning system uses a

frequency variation from 6 kHz to 60 kHz

to carry measurement information. There

is considerable noise at 120 Hz and at 1

MHz. Design a band-pass filter to reduce

the noise by 90%. What is the effect on

the desired passband frequency.

Band-reject Filter

• Band-reject filter blocs specific range of

frequencies

• Normally, it is difficult to realize the band-

reject filter with passive RC elements

• The design of active circuit is easier

• One special RC band-reject filter is notch

filter

• The notch frequency is determined by

• The frequencies for which the output is down 3 dB from the pass band are given by

( )RCfff ccn π2/1 where785.0 ==

cH

cL

ff

ff

57.4

187.0

=

=

Example

A frequency of 400 Hz prevails aboard an

aircraft. Design a twin-T notch filter to

reduce the 400 Hz signal. What effect

would this have on voice signal at 10 to

300 Hz? At what higher frequency is the 300 Hz? At what higher frequency is the

output down by 3 dB

Operational Amplifier

• A active device integrated R, L, C,

transistor, diode into single IC chip

• An op amp is an active circuit element

designed to perform mathematical designed to perform mathematical

operations of addition, subtraction,

multiplication, division, differentiation, and integration.

• Op amps are commercially available in

integrated circuit packages

• A typical one is the eight-pin dual in-line

package (or DIP),

• The circuit symbol for the op amp is the

triangle as shown,

• The op amp has two inputs and one

output. The inputs are marked with minus

(−) and plus (+) to specify inverting and

noninverting inputs, respectively. noninverting inputs, respectively.

• The equivalent model for op-am is shown

2 1dv v v= −

o dv Av=

Ideal Inverting Amplifier

• Consider the circuit in figure

• With feedback

– The summing point voltage is equal to the (+)

op amp input level.

– No current flow through the op amp input

terminal because of infinite impedance

• The current at the summing point is

• By Ohm’s law

1 2 0I I+ =

0in outV V

R R+ =

• Thus, the response of the op amp is

1 2

0R R+ =

2

1

out in

RV V

R= −

Design Rules

• Rule 1 Assume that no current flows

through the op amp input terminals – that

is, the inverting and noninverting terminals

• Assume that there is no voltage difference • Assume that there is no voltage difference

between the op amp input terminals

Non Ideal Effect

• Final open-loop gain. The gain is defined

as the slope of the voltage-transfer function

2V V∆= ≈

– For typical op amp, Vsat ~ 10 V, ∆V ~ 100 µV,

so A ~200,000

( )2 1

2out satV VA

V V V

∆= ≈∆ − ∆

Non Ideal Effect

• Finite input impedance

• Nonzero output impedance

• The summing current at the summing

point givespoint gives

1 2 3 0I I I+ + =

• Nodal voltage law on the summing point gives

• The output voltage related to op amp gain is

1 2

0in s o s s

in

V V V V V

R R z

− −+ − =

is

• Combine the two equation2

o so s o

V VV AV z

R

−= −

2

1

1

1o in

RV V

R µ

= − −

Where

2 2

2 1

2

1 1o

in

o

z R R

R R z

zAR

µ

+ + +

=

+

Typically, µ is very small compared to unity.

Op Amp in Instrumentation

• Voltage follower

• Inverting amplifier

• Noninverting amplifier

• Differential instrumentation amplifier• Differential instrumentation amplifier

• Voltage-to-current converter

• Current-to-voltage converter

• Integrator

• Differentiator

• Linearization

Voltage Follower

• The figure shows the voltage follower

circuit.

• The input impedance of the voltage

follower is high

Inverting Amplifier

• Inverting amplifier gives reverse polarity at

the output

• Variation of inverting amplifier is summing

amplifier amplifier

2 21 2

1 3

out

R RV V V

R R

= − +

Noninverting Amplifier

• Noninverting Amplifier gives the output in

the same polarity with the input

1 2

0in in outV V V

R R

−+ =

2

1

1out in

RV V

R

= +

Example

Design a high-impedance amplifier with a

voltage gain of 42.

Differential Amplifier

• Can be used to measure the difference

between two voltages

( )out d a bV A V V= −

Where A is the

differential gain and both

Va and Vb are voltage

with respect to the

ground

Common Mode Rejection

• Common mode signal is the signal that

common to both inputs

• A good differential amplifier should amplify

only the differential input.only the differential input.

( )2

a bout a b c

V VV A V V A

+ = − +

• The common mode rejection ratio is the

ratio of the differential gain to the

common-mode gain

d

c

ACMRR

A=

• The larger CMRR, the better the

differential amplifier

• The output voltage of the differential

amplifier is

( )22 1

1

out

RV V V

R= −

Example

A sensor output a range of 20.0 to 250 mV

as a variable varies over its range.

Develop signal conditioning so that this

become 0 to 5 V. The circuit must have

very high input impedancevery high input impedance

Instrumentation Amplifier

• Differential amplifier with high input

impedance and low output impedance

• One disadvantage of this differential circuit

is that changing gain requires changing 2

pairs of resistors

• A more common differential amplifier that

the gain can be changed is shown below

• The gain can be changed by adjusting RG

• The output voltage is given by

( )312 1

2

21out

G

RRV V V

R R

= + −

Example

A bridge circuit for which R4 varies from

100 Ω to 102 Ω. Show how an

instrumentation amplifier could be used to

provide an output of 0 to 2.5 V. Assume

that R2 = R3 = 1 kΩ and that R1 = 100 kΩthat R2 = R3 = 1 kΩ and that R1 = 100 kΩ

Voltage-to-Current Converter

• Signals are normally transmitted as a

current, specifically 4-20 mA

• The circuit should sink the current into

different load without losing voltage different load without losing voltage

information

• The circuit of voltage-to-current is shown

below

• The output of the current related to input

voltage then is

• Provided that

2

1 3

in

RI V

R R= −

( )1 3 5 2 4R R R R R+ =

• The maximum load resistance is

where

( )4 5 3

3 4 5

sat

m

ml

VR R R

IR

R R R

+ −

=+ +

maximum load resistance

op amp saturation on voltage

maximum current

ml

sat

m

R

V

I

=

=

=

Example

A sensor outputs 0 to 1 V. Develop a

voltage-to-current converters so that this

becomes 0 to 10 mA. Specify the

maximum load resistance if the op amp

saturates at ±10 Vsaturates at ±10 V

Current-to-Voltage Converter

• At the receiving end of the process-

control, the current is converted back to a

voltage

outV IR= −

Integrator

• The configuration is shown below

0

1

in out

out in

V dVC

R dt

V V dtRC

+ =

= − ∫

• If the input voltage is constant (Vin = K),

the output will be linear ramp voltage

out

KV t

RC= −

Example

Use an integrator to produce a linear ramp

voltage rising at 10 V per ms

Differentiator

• The differentiator is shown below

0in out

inout

dV VCdt R

dVV RC

dt

+ =

= −

Linearization

• Op amp can be used to linearize the

relationship between input and output by

placing the non linear element in the

feedbackfeedback

• The summation of the current is

( ) 0inout

VI V

R+ =

( )

input voltage

input resistance

nonlinear variation of current with voltage

in

out

V

R

I V

=

=

=

• Solve for Vout

inout

VV G

R

=

( )Inverse funtion which is a linear functioninout

VG I V

R

=

• If a diode is placed in the feedback

then

( ) 0outV

outI V I eα=

V0

0

0

0

1ln

out

out

Vin

V in

inout

VI e

R

Ve

RI

VV

RI

α

α

α

+ =

= −

= −

Analog Signal Conditioning

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