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.

Signal Conditioning • Signal conditioning can be categorized into 6 types – Signal-level and bias changes – 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 0.2 V – 0.6 V

Signal conditioning circuit

0V–5V

0.2 V – 0.6 V

Zero shift

0 V – 0.4 V

Amplification

0V–5V

Linearization • Often, the characteristic of a sensor is nonlinear • 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 or voltage to 4 to 20 mA and convert back to voltage at the receiving end • Thus, voltage-to-current and current-tovoltage circuits are essential

Digital Interface • The use of computer is process control requires the conversion of analog to digital signal – ADC – DAC

Filtering • Some signals input are spurious (contain more than 1 frequency) • It is necessary to filter the frequency matched 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 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  RL  Vy = Vx   R R + x   L  Rx  = Vx 1 −   RL + Rx 

• Output voltage is reduced by the voltage drop • To reduce the uncertainty, RL ≥ Rx

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.

50 °C

Sensor

Amplification

?V

Signal Conditioning: Passive Element • Signal conditioning circuit with element R, L, and C are – Divider circuits – Bridge circuits – RC filter

Divider circuits • Useful to convert resistance into voltage

• The voltage of the divider is given as VD = Vs

R2 R1 + R2

Vs

R1

Vs = supply voltage VD

R2

R1 , R2 = divider resistors

•

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. 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 Vs

R1 VD R2

Bridge Circuit • Bridge circuits are used to convert impedance variations into voltage variations. • 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 ∆V = VA − VB

Where

VA = V

R3 R1 + R3

R4 VB = V R2 + R4

V = Bridge supply voltage

• The voltage difference between A and B is VR3 VR4 ∆V = − R1 + R3 R2 + R4

• The equation above can be reduced to R3 R2 − R1 R4 ∆V = V ( R1 + R3 )( R2 + R4 )

• At particular combination of resistance values, the voltage difference is zero R3 R2 = R1 R4

Advantage of Wheatstone Bridge • It can be used as a resistance sensor that eliminates the supply voltage offset or changes. • 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 offset instead of voltage offset

Galvanometer detector • Galvanometer is represented as an resistance RG • The equivalent circuit of the Wheatstone bridge with Galvanometer is

Galvanometer detector • The current offset is determined by IG =

Where

Vth Rth + RG

Vth = Thevanin eqivalent voltage of the Wheatsone bridge R3 R2 − R1 R4 =V ( R1 + R3 )( R2 + R4 ) Rth = Thevanin eqivalent resistance of the Wheatsone bridge R1 R3 R2 R4 = + R1 + R3 R2 + R4

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 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.

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 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 R4 >> R5

• Now, the voltage at point b is V ( R4 + R5 ) Vb = + IR5 R2 + R4 + R5

• Thus, the offset voltage is ∆V = Va − Vb V ( R4 + R5 ) VR3 = − − IR5 R1 + R3 R2 + R4 + R5

• 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 supply voltage is 10 V

Potential Measurement using Bridge • The bridge can be used to measure potential • The potential to be measured is placed in series with the detector as shown in the figure

• The voltage at point C is Vc = Vx + Va

• The voltage appearing across the null detector is ∆V = Vc − Vb = Vx + Va − Vb

• The potential Vx can be measured by varying the bridge till null and solve for Vx R3V VR4 Vx + − =0 R1 + R3 R2 + R4

• The current balance bridge can also be used for potential measurement V ( R4 + R5 ) VR3 Vx + − − IR5 = 0 R1 + R3 R2 + R4 + R5

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 Z 3 Z 2 − Z1Z 4 ∆E = E ( Z1 + Z3 )( Z 2 + Z 4 )

where E = sine wave excitation voltage Z1 , Z 2 , Z 3, Z 4 = bridge impedance

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 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 filters are implemented using a digital computer or special purpose digital hardware.

.

Analog 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 filter is one that, along with R, L, and

C components, also contains an energy source, such as that derived from an operational amplifier. A passive 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 - “Ideal”

lowpass

highpass

bandpass

bandstop

Passive Analog Filters Background:

Realistic Filters:

lowpass

highpass

bandpass

bandstop

Passive Analog Filters Low Pass Filter Consider the circuit below.

+

+

R

VI

C

VO

_

_ Low pass filter circuit

1 VO ( jw) jwC = 1 Vi ( jw) R+ jwC

=

1 1+ jwRC

Passive Analog Filters Low Pass Filter 0 dB -3 dB

.

Bode

ω

1/RC

Passes low frequencies Attenuates high frequencies

1 0.707

0

x

1/RC

Linear Plot

ω

Passive Analog Filters High Pass Filter Consider the circuit below.

+ Vi

_

+

C R

VO _

High Pass Filter

VO ( jw) R = 1 Vi ( jw) R+ jwC

=

jwRC 1+ jwRC

Passive Analog Filters High Pass Filter 0 dB

.

-3 dB

Passes high frequencies

Bode 1/RC

Attenuates low frequencies 1/RC

ω

1 x.

0.707

Linear

0

1/RC

ω

Passive Analog Filters Bandpass Pass Filter

+ Vi

_

C

Consider the circuit shown below:

+

L R

VO _

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

ωlo 1

.

0.707

. ωhi

Bode

ω

.

Linear

0

ωlo

ωhi

ω

RLC Band stop Filter Consider the circuit below:

+ Vi

_

+

R L

VO

Gv (s) = C

_

Basic Active Filters Low pass filter

C Rfb

+

Vin _

Rin

+

VO _

Basic Active Filters High pass Rfb C R in +

Vin _

+

VO _

Basic Active Filters Band pass filter C2

R1 C1 +

Vin _

R1

R2 R2

Rfb Ri +

VO _

Basic Active Filters Band stop filter C1 R1 R1 Rfb R2 +

Vin

_

C2

Ri +

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 1 fc = 2π RC

• The output to input ratio is determined by Vout 1 = 2 1/2 Vin   f   1 +      f c  

Design Method • To design a filter is to find fc satisfied the criteria – Select a stand capacitor value in the µF to pF range – 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 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  f     fc 

Vout = 2 1/2 Vin   f   1 +      f c  

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 • 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. 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 Vout = Vin

fH f

(f

2

− fH fL

)

2

2

  RH   2 +  f L + 1 +  fH  f RL    

where 1 fH = 2πRL C L 1 fL = 2πRH C H

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 bandreject 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 f n = 0.785 f c

where f c = 1 / (2πRC )

• The frequencies for which the output is down 3 dB from the pass band are given by f L = 0.187 f c f H = 4.57 f c

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 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 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

vd = v2 − v1 vo = Avd

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 I1 + I 2 = 0

• By Ohm’s law Vin Vout + =0 R1 R2

• Thus, the response of the op amp is Vout

R2 = − Vin R1

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 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 ∆Vout 2Vsat A= ≈ ∆ (V2 − V1 ) ∆V

– For typical op amp, Vsat ~ 10 V, ∆V ~ 100 µV, so A ~200,000

Non Ideal Effect • Finite input impedance • Nonzero output impedance • The summing current at the summing point gives I1 + I 2 + I 3 = 0

• Nodal voltage law on the summing point gives Vin − Vs Vo − Vs Vs + − =0 R1 R2 zin

• The output voltage related to op amp gain is  Vo − Vs Vo = AVs −   R2

• Combine the two equation R2  1  Vo = −   Vin R1  1 − µ 

  zo 

Where  zo   R2 R2  +  1 +  1 + R2   R1 zin   µ=  zo   A+  R2  

Typically, µ is very small compared to unity.

Op Amp in Instrumentation • • • • • • • • •

Voltage follower Inverting amplifier Noninverting 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

Vout

 R2 R2  = −  V1 + V2  R3   R1

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

Vin Vin − Vout + =0 R1 R2

Vout

 R2  = 1 +  Vin  R1 

Example Design a high-impedance amplifier with a voltage gain of 42.

Differential Amplifier • Can be used to measure the difference between two voltages Vout = Ad (Va − Vb )

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.

Vout

 Va + Vb  = A(Va − Vb ) + Ac   2  

• The common mode rejection ratio is the ratio of the differential gain to the common-mode gain Ad CMRR = Ac

• The larger CMRR, the better the differential amplifier

• The output voltage of the differential amplifier is Vout

R2 = (V2 − V1 ) R1

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 RG • The output voltage is given by

Vout

 2 R1  R3  = 1 +   (V2 − V1 ) RG   R2  

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

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 information

• The circuit of voltage-to-current is shown below

• The output of the current related to input voltage then is I =−

R2 Vin R1 R3

• Provided that R1 ( R3 + R5 ) = R2 R4

• The maximum load resistance is  Vsat  ( R4 + R5 )  − R3  Im   Rml = R3 + R4 + R5

where Rml = maximum load resistance Vsat = op amp saturation on voltage I m = maximum current

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 processcontrol, the current is converted back to a voltage

Vout = − IR

Integrator • The configuration is shown below

Vin dVout +C =0 R dt Vout

1 Vin dt =− ∫ RC

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

K =− t RC

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

Differentiator • The differentiator is shown below

dVin Vout C + =0 dt R Vout

dVin = − 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

• The summation of the current is Vin + I (Vout ) = 0 R Vin = input voltage R = input resistance I (Vout ) = nonlinear variation of current with voltage

• Solve for Vout Vout

V G  in  R

 Vin  = G   R

  = Inverse funtion I (Vout ) which is a linear function 

• If a diode is placed in the feedback I (Vout ) = I 0 eαVout

then Vin + I 0 eαVout = 0 R αVout

e

Vout

Vin =− RI 0  Vin  = ln  −  α  RI 0  1