Chapter 5 Signal Conditioning

Chapter 5

Signal Conditioning

Measurement and Instrumentation / EEE 2223By :

Elya Mohd Nor and Azrena Abu Bakar 1

A major concern in transducer design is simply identifying some material that changes in a physical parameter being sensed.

The amplitude and linearity of the transducer output signal are usually not first-order concerns in transducer design, but they are of considerable concern in data acquisition system.

Therefore, signal conditioning to increase amplitude and improve linearity is usually necessary in data acquisition systems.

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Signal Conditioning

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Temperature Sensor Curve

Thermistor Thermocouple

RTD

1. Signal-Level and Bias ChangeExample 1-

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Principles of Analog Signal Conditioning

An IR pyrometer provides an output voltage varies from 0.2 to 0.6 V as a temperature in the annealing of glasses changes from 0oC to 600oC.

However, the transmitter to which this transducer output must be connected require a voltage that varies from 0 to 5V, for the same variation of the process variable.

What to do? How to do it??

We perform the required signal conditioning by first changing the 0V to occur when the transducer output is 0.2V.This is done by bias adjustment, or zero shift.

Now we have a voltage that varies from 0V to 0.4V, so we need to make the voltage larger.

If we multiply the voltage by 12.5, the new output will vary from 0 to 5V (12.5 x 0.4V).This is called amplification, and 12.5 is called the gain.

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Solution to Example 1

2. Improve Linearity

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Thermistor resistance versus temperature is highly nonlinear and usually has a negative slope.

Vapor temperature measurement

2. Improve LinearityExample 2-

Suppose a thermistor varies nonlinearly with a process variable.

A linearization circuit is required to conditioned the transducer/sensor output so that a voltage was produced which was linear with the process variable.

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3. ConversionsExample 3 –Suppose a metal resistance varies nonlinearly with the temperature inside the oven. In order to measure the oven temperature, we need a circuit to convert the resistance change to a voltage or current signal.

This is done by bridges when the fractional resistance change is small and/or amplifiers whose gain varies with resistance.

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4. Signal Transmission An important type of conversion is associated with the standard of transmitting signals as 4-to-20mA current levels in wire.

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Why current transmission than voltage?


5. Digital Interface The use of computers in measurement requires

conversion of analog data into a digital format by integrated circuit devices called ADC.

Analog signal conversion is needed to adjust the analog measurement signal to match the input requirements of the ADC.

Example : ADC need a voltage between 0 to 5V, but sensor provides signal that varies from 30 to 80 mV. Signal conversion circuits must be developed to interface the output requirement to the required ADC.

6. FilteringIndustrial environments in which data acquisition

systems are often placed tend to introduce spurious interference signals into the acquisition system.

These unwanted signals are noise, and often due to 60-Hz or 400-Hz power line interference caused by inductive load changes such as starting motors.

A filter is designed to reduce such interference.

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In summary, the requirements for signal conditioning in instrumentation are for:-

1. Signal-level and bias change2. Linearization3. Signal conversion4. Filtering5. Impedance matching

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One of the most important concerns in analog signal conditioning is the loading of one circuit by another.

This introduces uncertainty in the amplitude of a voltage as it is passed through the measurement process.

If this voltage represents some process variable, then we have many uncertainty in the value of the variable.

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Concept of Loading

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The Thévenin equivalent circuit of a sensor allows easy visualization of how loading occurs.

The voltage that appear across the load is reduced by the voltage dropped across the internal resistance.

How to reduce the effect of loading?? – by making RL much larger than Rx (that is RL >> Rx).

Example 4

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 proportional to temperature with a transfer function of 20 mV/oC.The sensor has an output resistance of 5.0 kΩ.If the temperature is 50oC, find the amplifier output.

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Solution

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If loading is ignored, serious errors can occur in expected outputs of circuits and gains of amplifiers.

Common passive circuits that can provide some of the required signal conditioning operations are:

1. Divider circuit2. Bridge circuit3. RC filters

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Divider Circuit

To provide conversion of resistance into a voltage variation.

VD = supply voltageR1,R2 = divider resistors

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

The most frequently performed type of signal conditioning is a level or amplitude change.

A change in signal level may require either attenuation or amplification of the input signal.

Attenuation is accomplished with a resistive voltage divider network.

Amplification requires an active device such as a transistor or an op-amp.

It is common practice to use both attenuation and amplification for signal-conditioning purposes in many electronic test instruments, such as electronic multimeter and oscilloscopes in order to provide multiple ranges.

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Divider Circuit in Signal-level change

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Bridge Circuit (Wheatstone Bridge)

To convert impedance variations into voltage variations.

• This circuit 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 potentials of points a and b of the network.

• Potential of point a with respect to c : Va

• Potential of point b with respect to c : Vb

• When there is zero difference and zero voltage across the detector-it is called nullR3R2 = R1R4

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Bridge Circuit

If a Wheatstone bridge nulls with R1=1000 Ω, R2=842 Ω, and R3=500 Ω, find the value of R4. Assume the detector impedance is infinite.

Answer : 421 Ω

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Example 5

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Bridge Circuit

Va

Vb

∆V = Va - Vb

The resistor in a bridge are given by R1=R2=R3=120 Ω and R4=121 Ω. If the supply is 10 V, find the voltage offset. Assume the detector impedance is infinite.

Answer : -21 mV

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Example 6

• Whenever Wheatstone bridge is assembled and resistors are adjusted for a detector null, the resistor values must satisfy the indicated equality [R3R2 = R1R4].

• The null is maintained, even when the supply voltage drifts or changes.

• In modern applications, the detector is a very high-input impedance differential amplifier.

• In the calibration purposes and spot measurement instruments, a highly sensitive galvanometer with low impedance may be used.

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Bridge Circuit

• In a impedance with low impedance, when the bridge is in unbalanced condition, current flows through the galvanometer detector causing a deflection of its pointer. This current is known as offset current.

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Galvanometer detector

Example 7A bridge circuit has resistance of R1=R2=R3=2.00 kΩ and R4=2.05 kΩ and a 5.00 V supply. If a galvanometer with a 50.0-Ω internal resistance is used for a detector, find the offset current.

How to do it?

The easiest way to determine this offset current is to find the Thevenin’s equivalent circuit between point a and b of the bridge (with the detector removed).

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Bridge resolution is where a minimum resistance change that must occur before the detector indicates a change in offset voltage.

It is an overall accuracy of the instruments.

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Bridge Resolution

Example 8

A bridge circuit has R1=R2=R3=R4=120.0-Ω resistances and a 10.0-V supply. Clearly, the bridge is nulled. Suppose a 3½-digit DVM on a 200-mV scale will be used for the null detector. Find the resistance resolution for measurement of R4.

Answer : On a 200-mV scale, the DVM measures from 000.0 to 199.9 mV, so the smallest change is 0.1 mV, or 100 µV.The smallest change in resistance that can be measured is 0.0048 Ω

A bridge offset of +100µV is caused by a reduction of R4. it follow that a bridge offset of -100µV would be casued by an increase in R4.

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Lead Compensation – in many process-control applications, a bridge circuit may be located at some distance from the sensor whose resistance changes are to be measured.

sensor

Wire 3 is the power lead, 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 exposed to the same environment and changes by the same amount, but is in the R3 leg of the bridge.

Effectively, both R3 and R4 are identically changed, and thus R3R2 = R1R4 shows that no change in the bridge null occurs.

In the past, many process-control applications used a feedback system in which the bridge offset voltage was amplified and used to drive a motor whose shaft altered a variable resistor to renull the bridge.

Such a system does not suit the modern technology of electronic processing because it is not very fast, is subject to wear, and generates electric noise.

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Current Balance Bridge

A technique that provides for an electronic nulling of the bridge and that uses only fixed resistors can be used with the bridge.

This method uses a current to null the bridge.

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

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Current Balance Bridge

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Current Balanced Bridge

Example 9

A current balance bridge, as shown below has resistors R1=R2=10kΩ, R3=1kΩ, R4=950Ω, and R5=50Ω and a high-impedance null detector. Find the current required to null the bridge if R3 changes by 1Ω. The supply voltage is 10V.

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ac Bridges

Z2Z3 = Z1Z4

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ac Bridges

1. Primary application of bridge circuits in modern process-control signal conditioning is to convert variations of resistance into variations of voltage.

2. This voltage variation is then further conditioned for interface to an ADC or other system.

3. It is thus important to note that the variation of bridge offset is nonlinear with respect to any of the resistors.

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Bridge Applications

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Bridge Applications

4. If a sensor has an impedance that is linear with respect to the variable being measured, such linearity is lost when a bridge is used to convert this to a voltage variation.

5. Figure below shows how ∆V varies with R4 for a bridge with R1=R2=R3=100Ω and V = 10V. Note the nonlinearity of ∆V with R4 as it varies from 0 to 500Ω.

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Bridge Applications

6. If the range of resistance variation is small and centered about the null value, then the nonlinearity of voltage versus resistance is small.

7. Figure shows that when R4 is between 90 to 110Ω, then the variation of ∆V with R4, on an expanded scale, is relatively linear.

Amplifiers can be used to amplify this voltagevariation.

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a) Bridge off-null voltage is clearly nonlinear for large-scale changes in resistance. (b) However, for small ranges of resistance change, the

off-null voltage is nearly linear.

RC FilterWhat is a filter?A filter is a circuit that passes a certain band of frequencies while attenuating the signals of other frequencies.

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Passive FilterRC Filter

1. Low-pass filter2. High-pass filter3. Band-pass filter4. Band-reject filter

Active FilterOp-Amps with Gain and Feedback

Filter

Circuit for Low-Pass Filter

Response of the low-pass RC filter.

Circuit for High-Pass Filter

Response of the high-pass RC filter.

Circuit for Band - Pass Filter

Response of the band-pass RC filter.

A band-pass RC filter can be made from cascaded high-pass and low-pass RC filters.

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Circuit for Band – Reject Filter

Filter performance is described in terms of output voltage to input voltage (Vo/Vin) at different frequencies and is expressed as

dB = 20 log (Vo/Vi)

The signal source for data acquisition system is a transducer.

The maximum bandwidth of most transducers is 10 Hz. Thus, filtering noise from these transducer signals is done using low-pass filter.

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Comments of Filter

A measurement signal with amplitude 540-mV has a frequency of <1kHz. There is unwanted noise at about 1 MHz. A filter is needed to eliminate the noise from the signal. Identify the filter type. Why?

Answer :- Low-pass filter- Because the low-pass filter blocks high frequencies and

passes only the low frequencies to the next system.- Thus, the filter the measurement signal of <1 kHz to

pass and block the unwanted signal at 1 MHz.

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Example 10

1. 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.

2. A 2-kHz data signal is contaminated by 60Hz of noise.

3. A signal-conditioning system uses a frequency variation from 6kHz to 60 kHz to carry measurement information. There is considerable noise at 120 Hz and at 1 MHz. Design a bandpass filter to reduce the noise by 90%.

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Other examples when filters are needed.

4. A frequency of 400 Hz prevails aboard an aircraft. Design a twin-T notch filter to reduce the 400-Hz signal.

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Other examples when filters are needed.

An op-amp is a circuit consist of resistors, transistors, diodes, and capacitors. It requires bipolar power supplies +Vs and -Vs, with respect to ground. We concerned only with its input and output signals.

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Operational Amplifiers

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Op-Amp Characteristics

The output voltage, Vout is plotted versus the differential input voltage (V2-V1).

When V2 is much larger than V1, the output is saturated at –Vsat.

When V1 is much larger than V2, the output is saturated at +Vsat.

There is a narrow range of differential input voltage, ∆V where output change from +Vsat to –Vsat.

For most op-amps, this input voltage range is less than a milivolt, whereas the saturation voltages are typically on the order of 10V.

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The schematic symbol and response of an op amp.

The input impedances are very high, typically exceeding 1MΩ, whereas the output impedance is very low, typically less than 100 Ω.

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Op-Amp Characteristics

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The op amp inverting amplifier.

Summing point

I1 + I2 = 0

• Voltage follower• Inverting amplifier

- Summing amplifier• Non-inverting amplifier

- Differential Instrumentation Amplifier- Differential amplifier- Instrumentation amplifier

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Op-Amp in Instrumentation

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The op amp voltage follower. This circuit has unity gain but very high input impedance.

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The op amp summing amplifier.

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Example 11

Develop an op-amp circuit that can provide an output voltage related to the input voltage by Vout = 3.4Vin + 5

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A noninverting amplifier.

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Differential Instrumentation Amplifier

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Most common configuration of Instrumentation Amplifier

Example 12 Figure below shows a bridge circuit for which R4 varies from 100Ω to 102Ω. Show how an instrumentation amplifier could be used to provide an output of 0 to 2.5V.

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The basic differential amplifier configuration.

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An instrumentation amplifier includes voltage followers for input isolation.

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V-to-I Convertor

1. Because signals in process control are most often transmitted as a current, specifically 4 to 20 mA, it is often necessary to employ a linear voltage-to-current convertor.

2. Such a circuit must be capable of sinking a current into a number of different loads without changing the VtoI transfer characteristics.

Common measurement transmission in the industrial environment

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A sensor outputs 0 to 1 V. Develop a voltage-to-current convertor so that this becomes 0 to 10 mA. Specify the maximum load resistance if the op amp saturates at ±10V.

Determine the value of R3 and R4 if we want to convert the 0 to 1V to 4 to 20 mA?

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Example 13

At the receiving end of the process-control transmission system, we often need to convert the current back into voltage. This can easily be done by this circuit.

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I –to-V Convertor

Books

1. Foster, A.C. (1995). Electronic Instruments and Measurement. Prentice-Hall.

2. Curtis, D.J. (2003). Process Control Instrumentation Technology. Prentice-Hall.

3. Liptak, B.G. (2003). Instrument Engineer’s Handbook. 4th Edition. CRC Press.

References

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Chapter 5 Signal Conditioning

Documents

Transcript of Chapter 5 Signal Conditioning