Analog Signal Conditioning

Chapter 2

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.

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 circuit

Zero shift

0.2 V – 0.6 V 0 V – 5 V

Amplification 0 V – 0.4 V

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

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 devices

matched with the devices

– The electric line frequency is 50 Hz – The transient of motor is kHz

• Example

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

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 L y x L x x x R V V R R R V R R   = _{ } +     = _ − _ +  

• Output voltage is reduced by the voltage drop

• To reduce the uncertainty,

x L x R R  ₊    L x R ≥ 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.

Sensor

50 °C ? V

Signal Conditioning: Passive

Element

• Signal conditioning circuit with element R, L, and C are

– Divider circuits – Bridge circuits – Bridge circuits – RC filter

Divider circuits

• The voltage of the divider is given as 2 1 2 D s R V V R R = + s V D V 1 R 2 R 1 2 supply voltage , divider resistors s V R R = =

• It is important to consider the following issues

1. The variation of V_D 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 V_D (b) the range of output impedance, and (c) the range of power dissipated by R₂ s V D V 1 R 2 R

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

Wheatstone Bridge

• The potential difference ∆V between points A and B is simply

Where A B V V V ∆ = − 3 R V =V 3 1 3 A V V R R = + 4 2 4 B R V V R R = + Bridge supply voltage

• The voltage difference between A and B is

• The equation above can be reduced to

3 4 1 3 2 4 VR VR V R R R R ∆ = − + +

• At particular combination of resistance values, the voltage difference is zero

( 1 3 32)( 21 4 4) R R R R V V R R R R − ∆ = + + 3 2 1 4 R 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

Galvanometer detector

• Galvanometer is represented as an resistance R_G

• The equivalent circuit of the Wheatstone bridge with Galvanometer is

Galvanometer detector

• The current offset is determined by Where th G th G V I R R = + ( 1 3 32)( 21 4 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 R V R 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 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

5

4 R

R >>

• Now, the voltage at point b is

4 5 5 2 4 5 ( ) b V R R V IR R R R + = + + + 5 4 R R >>

• Thus, the offset voltage is ( 4 5) 3 5 1 3 2 4 5 a b V V V V R R VR IR 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 a

V =V +V

• The potential Vx can be measured by

varying the bridge till null and solve for Vx

c b x a b V V V V V V ∆ = − = + − 3 4 1 3 2 4 0 x R V VR V R R R R + − = + +

• The current balance bridge can also be used for potential measurement

( 4 5) 3 5 1 3 2 4 5 0 x V R R VR V IR R R R R R + + − − = + + +

Example

A bridge circuit for potential measurement nulls when R1 = R2 = 1 kΩ, R3 = 650 Ω

and 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 ( 1 3 32)( 21 4 4) Z Z Z Z E 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 C_x 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

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

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”

lowpass

lowpass highpasshighpass

bandpass

Background: Realistic Filters:Realistic Filters:

lowpass

lowpass highpasshighpass

Passive Analog Filters

bandpass

Passive Analog Filters

Low Pass Filter _{Consider the circuit below.}

R C VI V_O + _ + _ _ 1 ( ) 1 1 ( ) 1 O V jw jwC V jw_{i R} jwRC jwC = = + +

Passive Analog Filters

Low Pass Filter

0 dB ω Bode

.

-3 dB 1 ω ω 0 1/RC 1/RC Linear Plot x 0.707

Passes low frequencies

Passive Analog Filters

High Pass Filter _{Consider the circuit below.}

C R V_i V + + _ R V_i V_O _ _ ( ) 1 ( ) 1 O V jw R jwRC V jw_{i R} jwRC jwC = = + +

Passive Analog Filters

High Pass Filter

0 dB

.

-3 dB

Bode Passes high frequencies

. 0 ω ω 1/RC 1/RC 1/RC 1 0.707 Linear

Attenuates low frequencies

Passive Analog Filters

Bandpass Pass Filter Consider the circuit shown below:

C L R V_i VO + + _ R V_i _ _

Passive Analog Filters

Bandpass Pass Filter

We 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 ω 0 -3 dB ω_lo ω_hi .

.

.

. 1 0.707 Bode Linear ω_lo ω_hi

RLC Band stop Filter

Consider the circuit below:

R L C + V_O + _ V_i = ) (s G_{v C} _ = ) (s G_v

V_in V_O C R_fb + _ + _ R_in

Basic Active Filters

Basic Active Filters R_in C V_in R_fb VO + _ + _ High pass

Basic Active Filters V_in R₁ R₁ C₁ C₂ R₂ R₂ R_fb R_i V + + _

Band pass filter

V_O

Basic Active Filters

R₁

R₁ C₁

R_fb

Band stop filter

V_in C₂ R₂ Ri V_O + _ + _

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 2 c f RC π =

• The output to input ratio is determined by

1/2 2 1 1 out in c V V f f =  _{ }   _{+  }   _{ }   

Design Method

• To design a filter is to find fc satisfied the criteria

– Select a stand capacitor value in the µF to pF range

range

– Calculate the required resistance value, if R < 1 kΩ or R > 1 MΩ, pick another capacitor

– Consider device tolerance

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 below

• The ratio of output voltage to input voltage of the high pass filter is

1/2 2 c out f f V V       =  21/2 1 in c V f 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 needed

low, 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?

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 V_in source resistance is very

Band-pass Filter

• Band-pass filter passes frequencies in a certain band and rejects frequencies

• The ratio of output to input voltage is

(

₂

)

2 2 ₂ 1 out H in H H L L H L V f f V R f f f f f f R =  _{ }  − + _ +_ + _{ }     where H H L L L H C R f C R f 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

( RC) f f f_n = 0.785 _c where _c =1/ 2π c H c L f f f f 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.

• The equivalent model for op-am is shown 2 1 d v = v −v o d v = 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 0 I + I = 0 in out V V R + R =

• Thus, the response of the op amp is

1 2 0 R + R = 2 1 out in R V 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

Non Ideal Effect

• Final open-loop gain. The gain is defined as the slope of the voltage-transfer

function

2

V V

∆

= ≈

– For typical op amp, Vsat ~ 10 V, ∆V ~ 100 µV, so A ~200,000 ( 2 1) 2 out sat V V A V V V ∆ = ≈ ∆ − ∆

Non Ideal Effect

• Finite input impedance

• Nonzero output impedance

• The summing current at the summing point gives

point gives

1 2 3 0

• Nodal voltage law on the summing point gives

• The output voltage related to op amp gain is 1 2 0 in s o s s in V V V V V R R z − − + − = is

• Combine the two equation

2 o s o s o V V V AV z R  −  = _{−  }   2 1 1 1 o in R V V R µ   = − _{ } −  

Where 2 2 2 1 2 1 o 1 in o z R R R R z z A R µ     + _ + + _      =   +    

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 2 1 2 1 3 out R R V V V R R   = − _ + _  

Noninverting Amplifier

• Noninverting Amplifier gives the output in the same polarity with the input

1 2 0 in in out V V V R R − + = 2 1 1 out in R V 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 b

V = A V −V

Where A is the

differential gain and both Va and Vb are voltage with respect to the

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 b out a b c V V V = A V −V + A _ + _  

• The common mode rejection ratio is the ratio of the differential gain to the

common-mode gain d c A CMRR A =

• The larger CMRR, the better the differential amplifier

• The output voltage of the differential amplifier is ( ) 2 2 1 1 out R V 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 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 R_G • The output voltage is given by

( ) 3 1 2 1 2 2 1 out G R R V 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 kΩ that 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 R I V R R = − ( ) 1 3 5 2 4 R R + R = R R

• The maximum load resistance is where ( 4 5 ) 3 3 4 5 sat m ml V R R R I R 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 V

Current-to-Voltage Converter

• At the receiving end of the

process-control, the current is converted back to a voltage

out

Integrator

• The configuration is shown below

0 1 in out out in V dV C R dt V V dt RC + = = −

∫

• If the input voltage is constant (Vin = K), the output will be linear ramp voltage

out

K

V t

RC

Example

Use an integrator to produce a linear ramp voltage rising at 10 V per ms

Differentiator

• The differentiator is shown below

0 in out in out dV V C dt R dV V RC dt + = = −

Linearization

• Op amp can be used to linearize the

relationship between input and output by placing the non linear element in the

feedback feedback

• The summation of the current is ( ) 0 in out V I V R + = ( ) input voltage input resistance

nonlinear variation of current with voltage

in out V R I V = = =

• Solve for Vout

in out V V G R   =     ( )

Inverse funtion which is a linear function

in out V G I V R  _{ =}    

• If a diode is placed in the feedback then ( ) 0 out V out I V = I eα V 0 0 0 0 1 ln out out V in V in in out V I e R V e RI V V RI α α α + = = −   = _− _  