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9. Signal processing
The output signal from the sensor of a measurement system has generally to be processed in some way to make it suitable for display or use in some control system. For example, the signal may be too small and have to be amplified, be analogue and have to be made digital, be digital and have to be made analogue, be a resistance change and have to be made into a current change, be a voltage change and have to be made into a suitable size current change, be a pressure change and have to be made into a current change, etc. All these changes can be referred to as signal processing. For example, the output from a thermocouple is a very small voltage, a few millivolts. A signal processing module might then be used to convert this into a larger voltage and provide cold junction compensation (i.e. allow for the cold junction not being at 0°C). Note that the term signal conditioning is sometimes used for the conversion of the output from a sensor into a suitable form for signal processing. The following gives some of the elements that are used in signal processing.
FIG. 57 Resistance to voltage conversion for a thermistor
9.1 Resistance to voltage converter
Consider how the resistance change produced by a thermistor when subject to a temperature change can be converted into a voltage change.
FIG. 57 shows how a potential divider circuit can be used. A constant voltage, of perhaps 6V, is applied across the thermistor and another resistor in series. With a thermistor with a resistance of 4.7 k-Ohm, the series resistor might be 10 k-Ohm. The output signal is the voltage across the 10 k-ohm resistor. When the resistance of the thermistor changes, the fraction of the 6 V across the 10 k-Ohm resistor changes.
The output voltage is proportional to the fraction of the total resistance which is between the output terminals. Thus:
output =[R /R+Rt ] V
where V is the total voltage applied, in FIG. 57 this is shown as 6 V, R the value of the resistance between the output terminals (10 kOhm) and Rx e the resistance of the thermistor at the temperature concerned. The potential divider circuit is thus an example of a simple resistance to voltage converter.
FIG. 58 Wheatstone bridge
FIG. 59 Potential drops across AB and AD
Another example of such a converter is the Wheatstone bridge. FIG. 58 shows the basic form of the bridge. The resistance element being monitored forms one of the arms of the bridge. When the output voltage Vo is zero, then there is no potential difference between B and D and so the potential at B must equal that at D. The potential difference across R1, i.e. V_AB, must then equal that across
R3, i.e. V_AD. Thus:
I1R1 = I2R2
We also must have the potential difference across R2, i.e. V_BC, equal to that across R4, i.e. V_DC. Since there is no current through BD then the current through R2 must be the same as that through R1 and the current through R4 the same as that through R3. Thus:
I1R2 – I2R4
Dividing these two equations gives:
The bridge is said to be balanced.
Now consider what happens when one of the elements has a resistance which changes from this balanced condition. The supply voltage Vs is connected between points A and C and thus the potential drop across the resistor R1 is the fraction R1/(R1 + R2) of the supply voltage (FIG. 59(a)). Hence:
Similarly, the potential difference across R3 (FIG. 59(b)) is:
Thus the difference in potential between B and D, i.e. the output potential difference Vo, is:
This equation gives the balanced condition when Vo = 0. Consider resistance R1 to be a sensor which has a resistance change, e.g. a strain gauge which has a resistance change when strained. A change in resistance from R1 to R1 + dR1 gives a change in output from Vo to Vo + dVo, where:
Hence:
If dR1 is much smaller than R1 then the denominator R1 + dR1 + R2 approximates to R1 + R2 and so the above equation approximates to:
With this approximation, the change in output voltage is thus proportional to the change in the resistance of the sensor. We thus have a resistance to voltage converter. Note that the above equation only gives the output voltage when there is no load resistance across the output. If there is such a resistance then the loading effect has to be considered.
As the output voltage is proportional to the bridge excitation voltage, voltage drops along the cables from the voltage supply and the cable resistance between resistors in the bridge can affect the output, this being a particular problem if temperature changes cause resistance changes in these cables. Three-wire compensation (FIG. 60(a)) can be used to help overcome the problem of cable resistance between, say, a temperature sensor and the bridge. FIG. 60(b) shows a four-wire form of compensation, there being two parallel, dummy, leads.
FIG. 60 Temperature compensation
Example:
A platinum resistance coil is to be used as a temperature sensor and has a resistance at 0°C of 100 ohm. It forms one arm of a Wheatstone bridge with the bridge being balanced at this temperature and each of the other arms also being 100 ohm, If the temperature coefficient of resistance of platinum is 0.0039 K^-1 what will be the output voltage from the bridge per degree change in temperature if the supply voltage is 6.0 V? The variation of the resistance of the platinum with temperature can be represented by:
where R, is the resistance at t °C, Ro the resistance at 0 C and a the temperature coefficient of resistance. Hence:
change in resistance = R1-R0= R0at
Thus, for a one degree change in temperature: change in resistance = 100 x 0.0039 x 1 = 0.39 ohm Since this resistance change is small compared to the 100 ohm, the approximate equation for the output voltage can be used. Hence, the change in output per degree change in temperature is:
9.2 Temperature compensation
The electrical resistance strain gauge is a resistance element which changes resistance when subject to strain. However, it will also change resistance when subject to a temperature change. Thus, in order to use it to determine strain, compensation has to be made for temperature effects.
One way of eliminating the temperature effect is to use a dummy strain gauge. This is a strain gauge identical to the one under strain, the active gauge, which is mounted on the same material as the active gauge but not subject to the strain. It is positioned close to the active gauge so that it suffers the same temperature changes. As a result, a temperature change will cause both gauges to change resistance by the same amount.
The active gauge is mounted in one arm of a Wheatstone bridge (FIG. 61(a)) and the dummy gauge in an opposite arm so that the effects of temperature-induced resistance changes cancel out.
Strain gauges are often used with other sensors such as diaphragm pressure gauges or load cells. Temperature compensation is still required. While dummy gauges could be used, a better solution is to»use four strain gauges with two of them attached so that the applied forces put them in tension and the other two in compression. The gauges, e.g. gauges 1 and 3, that are in tension will increase in resistance while those in compression, gauges 2 and 4, will decrease in resistance. The gauges are connected as the four arms of a Wheatstone bridge (FIG. 61(b)). As all the gauges and so all the arms of the bridge will be equally affected by any temperature changes the arrangement is temperature compensated. The arrangement also gives a much greater output voltage than would occur with just a single active gauge.
FIG. 61 Temperature compensation, (a) dummy gauge, (b) with four active
strain gauges
9.3 Thermocouple compensation
With a thermocouple, one junction should be kept at 0°G; the temperature can then be obtained by looking up in tables the e.m.f produced by the thermocouple. FIG. 62 illustrates what is required.
This keeping of a junction at 0 C, i.e. in a mixture of ice and water, is not always feasible or very convenient and the cold junction is often allowed to be at the ambient temperature. To take account of this a compensation voltage has to be added to the thermocouple. This voltage is the same as the e.m.f that would be generated by the thermocouple with one junction at 0 C and the other at the ambient temperature. We thus need a voltage which will depend on the ambient temperature. Such a voltage can be produced by using a resistance temperature sensor in a Wheatstone bridge. FIG. 63 illustrates this. The bridge is balanced at 0 C and the output voltage from the bridge provides the correction potential difference at other temperatures. By a suitable choice of resistance temperature sensor, the appropriate voltage can be obtained.
FIG. 62 Cold junction at 0V
FIG. 63 A Wheatstone bridge compensation circuit
The resistance of a metal resistance temperature sensor is given by:
where R, is the resistance at t C, Ro the resistance at 0 C and a the temperature coefficient of resistance. When there is a change in temperature:
change in resistance = Rt -Ro= R0at
The output voltage for the bridge, taking R1 to be the resistance temperature sensor, is given by:
This voltage must be the same as that given by the thermocouple with one junction at 0°C and the other at the ambient temperature. The thermocouple e.m.f e is likely to vary with temperature / in a reasonably linear manner over the small temperature range being considered, i.e. from 0 C to the ambient temperature. Thus we can write e = at, where a is a constant, i.e. the e.m.f produced per degree change in temperature.
Hence for compensation we must have:
and so the condition:
Example:
Determine the value of the resistance R2 in FIG. 63 if compensation is to be provided for an iron-constantan thermocouple giving 51 ^V/°C. The compensation is to be provided by a nickel resistance element with a resistance of 10 ohm at 0C and a temperature coefficient of resistance of 0.0067 K^-1 . Take the supply voltage for the bridge to be 2.0 V.
Using the equation developed above, aR2 = Ro(Vsa - a), then:
Hence R2 is 2617 ohm.
FIG. 64 Protection against high currents
FIG. 65 Zener diode: (a) current-voltage relationship, (b) protection circuit
9.4 Protection
An important element that is often present with signal processing is protection against high currents or high voltages. For example, sensors when connected to a microprocessor can damage it if high currents or high voltages are transmitted to the microprocessor. A high current can be protected against by the incorporation in the input line of a series resistor to limit the current to an acceptable level and a fuse to break if the current does exceed a safe level (FIG. 64). Protection against high voltages and wrong polarity voltages may be obtained by the use of a Zener diode circuit (FIG. 65). The Zener diode with a reverse voltage connected across it has a high resistance up to some particular voltage at which it suddenly breaks down and becomes conducting (FIG. 65(a)). Zener diodes are given voltage ratings, the rating indicating at which voltage they become conducting. For example, to allow a maximum voltage of 5 V but stop voltages above 5.1 V being applied to the following circuit, a Zener diode with a voltage rating of 5.1 V might be chosen. For voltages below 5.1 V the Zener diode, in reverse voltage connection, has a high resistance. When the voltage rises to 5.1 V the Zener diode breaks down and its resistance drops to a very low value. Thus, with the circuit shown in FIG. 65(b), with the applied voltage below 5.1 V, the Zener diode, in reverse voltage connection, has a much higher resistance than the other resistor and so virtually all the applied voltage is across the Zener diode. When the applied voltage rises to 5.1 V, the Zener diode breaks down and has a low resistance. As a consequence, most of the voltage is then dropped across the resistor, the voltage across the diode drops and so the output voltage drops. Because the Zener diode is a diode with a low resistance for current in one direction through it and a high resistance for the opposite direction, it also provides protection against wrong polarity.
To ensure protection, it is often necessary to completely isolate circuits so that there are no electrical connections between them. This can be done using an optoisolator. Such a device converts an electrical signal into an optical signal, transmits it to a detector which then converts it back into an electrical signal (FIG. 66). The input signal passes through an infrared light-emitting diode (LED) and so produces a beam of infrared radiation. This infrared signal is then detected by a phototransistor. To prevent the LED having the wrong polarity or too high an applied voltage, it is likely to be protected by the Zener diode circuit (of the type shown above in FIG. 65). Also, if there is likely to be an alternating signal in the input a diode would be put in the input line to rectify it.
FIG. 66 An optoisolator
FIG. 67 Signals: (a) analogue, (b) digital
9.5 Analog-to-digital conversions
The electrical output from sensors such as thermocouples, resistance elements used for temperature measurement, strain gauges, diaphragm pressure gauges, LVDTs, etc. is in analogue form. Microprocessors require digital inputs. Thus, where a microprocessor is used it has to be converted into a digital form before it can be used as an input to the microprocessor. The output from a microprocessor is digital. Most control elements require an analogue input and so the digital output from a microprocessor has to be converted into an analogue form before it can be used by them. Thus there is a need for analogue-to-digital converters (ADC) and digital-to-analogue converters (DAC). Microcontrollers are microprocessors with input and output signal processing incorporated on the same chip and often incorporate analogue-to-digital and digital-to-analogue converters.
An analogue signal (FIG. 67(a)) is one that is continuously variable, changing smoothly over a range of values. The signal is an analogue, i.e. a scaled version, of the quantity it represents. A digital signal increases in jumps, being a sequence of pulses, often just on-off signals (FIG. 67(b)). The value of the quantity instead of being represented by the height of the signal, as with analogue, is represented by the sequence of on-off signals.
Analogue-to-digital conversion involves a number of stages. The first stage is to take samples of the analogue signal (FIG. 68(a)). A clock supplies regular time signal pulses (FIG. 68(b)) to the analogue-to-digital converter and every time it receives a pulse it samples the analogue signal. The result is a series of narrow pulses with heights which vary in accord with the variation of the analogue signal (FIG. 68(c)). This sequence of pulses is changed into the signal form shown in FIG. 68(d) by each sampled value being held until the next pulse occurs. It is necessary to hold a sample of the analogue signal so that conversion can take place to a digital signal at an analogue-to-digital converter. This converts each sample into a sequence of pulses representing the value. For example, the first sampled value might be represented by 101, the next sample by Oil, etc. The 1 represents an 'on' or 'high' signal, the 0 an 'off or 'low' signal. Analogue-to-digital conversion thus involves a sample and hold unit followed by an analogue-to-digital converter (FIG. 69).
FIG. 69 Analogue-to-digital conversion
FIG. 68 (a) Analogue signal, (b) time signal, (c) sampled signal (d) sampled
and held signal.
To illustrate the action of the analogue-to-digital converter, consider one that gives an output restricted to three bits. The binary digits of 0 and 1, i.e. the 'low' and 'high' signals, are referred to as bits. A group of bits is called a word. Thus the three bits give the word length for this particular analog-to-digital converter. The word is what represents the digital version of the analogue voltage. The position of bits in a word has the significance that the least significant bit is on the right end of the word and the most significant bit on the left. This is just like counting in tens, 435 has the 5 as the least significant number and the 4 as the most significant number, the least significant number contributing least to the overall value of the 435 number. The position of the digit in a decimal number is significant; the least significant digit having its value multiplied by 10^0, the next by 10^1, the next by 10^2, and so on. Likewise, the position of bits in a binary word is significant; the least significant bit having its value multiplied by 2^0 the next by 2^1, the next by 2^2 and so on. For a binary word of n bits :
2^n-1,... , 2^3, 2^2, 2^1, 2^0
Most significant bit (MSB) | Least significant bit(LSB)
With binary numbers we have the basic rules: 0 + 0 = 0, 0 + 1 = 1, 1 + 1 = 10.
Thus if we start with 000 and add 1 we obtain 001. If we add a further 1 we have 010. Adding another 1 gives 011. With three bits in a word we thus have the possible words of:
000 001 010 011 100 101 110 111
There are eight possible words which can be used to represent the analogue input; the number of possible words with a word length of n bits is 2^n. Thus we divide the maximum analogue voltage into eight parts and one of the digital words corresponds to each. Each rise in the analogue voltage of (1/8) of the maximum analogue input then results in a further bit being generated. Thus for word 000 we have 0 V input. To generate the next digital word of 001 the input has to rise to 1/8 of the maximum voltage. To generate the next word of 010 the input has to rise to 2/8 of the maximum voltage. FIG. 70 illustrates this conversion of the sampled and held input voltage to a digital output.
Thus if we had a sampled analogue input of 8 V, the digital output would be 000 for a 0 V input and would remain at that output until the analogue voltage had risen to 1 V, i.e. 1/8 of the maximum analogue input. It would then remain at 001 until the analogue input had risen to 2 V. This value of 001 would continue until the analogue input had risen to 3 V. The smallest change in the analogue voltage that would result in a change in the digital output is thus 1 V. This is termed the resolution of the converter.
The word length possible with an analogue-to-digital converter deter mines its resolution. With a word length of n bits the maximum, or full scale, analogue input Vfs is divided into 2^n pieces. The minimum change in input that can be detected, i.e. the resolution, is thus FFS / 2^2. With an analogue-to-digital converter having a word length of 10 bits and the maximum analogue signal input range 10 V, then the maximum analogue voltage is divided into 2^10 = 1024 pieces and the resolution is 10/1024 = 9.8 mV. There are a number of forms of analogue-to-digital converter; the most commonly used being successive approximations, ramp and flash.
The flash form is much faster than either the successive approximations form or the ramp form. The term conversion time is used to specify the time it takes a converter to generate a complete digital word when supplied with the analogue input.
FIG. 70 Digital output from an ADC
Application---Analogue-to-digital converters are generally purchased as integrated circuits. FIG. 71 shows an example of the pin connections for the ZN439, a successive approximation form of ADC. Its specification includes: Resolution 8 bits
Conversion time 5 ms
Linearity error ± 14 LSB Power dissipation 150 mW
FIG. 71 The GEC Plessey ZN439E S-bit analogue-to-digital converter
Example
A thermocouple gives an output of 0.4 mV for each degree change in temperature. What will be the word length required when its output passes through an analogue-to-digital converter if temperatures from 0 to 100 C are to be measured with a resolution of 0.5°C? The full scale output from the sensor is 200 x 0.4 = 80 mV. With a word length n there are 2^n digital numbers. Thus this voltage will be divided into 2^n levels and so the minimum voltage change that can be detected is 80/2^n mV. For a resolution of 0.5 C we must be able to detect a signal from the sensor of+0.5 x 0.4 = 0.20 mV.
Hence:
0.20 = 80/ 2^n
and so 2^n = 400 and n = 8.6. Thus a 9-bit word length is required.
FIG. 72 Digital-to-analogue conversion
FIG. 73 The principle of a S-bit digital-to-analogue converter
9.6 Digital-to-analogue conversions
The input to a digital-to-analogue converter is a binary word and the output its equivalent analogue value. For example, if we have a full scale output of 7 V then a digital input of 000 will give 0 V, 001 give 1 V, ... and 111 the full scale value of 7 V. FIG. 72 illustrates this.
The basic form of a digital-to-analogue converter involves the digital input being used to activate electronic switches such that a 1 activates a switch and a 0 does not, the position of the 1 in the word determining which switch is activated. Thus when, with say a 3-bit converter, 001 is received we have a voltage of, say, 1 V switched to the output, when 010 is received we have 2 V, switched to the output, and when 100 is received we have 4 V switched to the output. Hence if we have the digital word 011 we have the least significant bit 001 switching 1 V to the output and the 010 bit 2 V to the output to give a summed output of 3 V (FIG. 73).
FIG. 74 The GEC Plessey ZN558D 8-bit latched digital-to-analogue converter
Application---FIG. 74 shows details of the GEO Plessey ZN558D 8-bit latched input digital-to-analogue converter. After the conversion is complete, the 8-bit result is placed in an internal latch until the next conversion is complete.
A latch is just a device to retain the output until a new one replaces it. The settling time is the time taken for the analogue output voltage to settle within a specified band, usually ± LSB/2, about its final value when the digital word suddenly changes.
Resolution 8 bits
Settling time 800 ns
Non-linearity 0.5% of full scale
Power dissipation 100 mW
Example
A microprocessor gives an output of an 8-bit word. This is fed through an 8-bit digital-to-analogue converter to a control valve which requires 6.0 V to be fully open. If the fully open state is to be indicated by the output of the digital word 11111111 what will be the change in output to the valve when there is a change of 1 bit? The output voltage will be divided into 2^8 intervals. Since there is to be an output of 6.0 V when the output is 2^8 of these intervals, a change of 1 bit is a change in the output voltage of 6.0/2^8 = 0.023 V.
9.7 Op-amps
FIG. 75 Pin connections for a 741 op-amp
FIG. 76 Inverting amplifier
The operational amplifier (op-amp) is a very high gain d.c. amplifier, the gain typically being of the order of 100 000 or more, which is supplied as an integrated circuit on a silicon chip. It has two inputs, known as the inverting input (-) and the non-inverting input (+). In addition there are inputs for a negative voltage supply, a positive voltage supply and two inputs termed offset null, these being to enable corrections to be made for the non-ideal behavior of the amplifier.
FIG. 75 shows the pin connections for a 741 type operational amplifier with the symbol for the operational amplifier shown superimposed. On the symbol the "+" sign indicates the non-inverting input and the "-" sign the inverting input.
The operational amplifier is a very widely used element in signal conditioning and processing circuits and the following indicates common examples of such circuits.
FIG. 76 shows the connections made to the amplifier when it is used as an inverting amplifier, such a form of amplifier giving an output which is an inverted form of the input, i.e. it is out of phase by 180°. The input is taken to the inverting input through a resistor R\ with the non-inverting input being connected to ground. A feedback path is provided from the output, via the resistor R1 to the inverting input. The operational amplifier has a very high voltage gain of about 100 000 and the change in output voltage is limited to about ±10 V. Thus the input voltage at point X must be between +0.0001 V and -0.0001 V. This is virtually zero and so point X is at virtually earth potential and hence is termed a virtual earth. With an ideal operational amplifier with an infinite gain, point X is at zero potential. The potential difference across R1 is (Vin - Vx). Hence, for an ideal operational amplifier with an infinite gain, and hence Vx = 0, the input potential Vin can be considered to be across R1. Thus:
The operational amplifier has a very high impedance between its input terminals, for a 741 this is about 2 M-ohm. Thus virtually no current flows through X into it. For an ideal operational amplifier the input impedance is taken to be infinite and so there is no current flow through X into the amplifier input. Hence, since the current entering the junction at X must equal the current leaving it, the current I1 through R1 must be the current through R2. The potential difference across R2 is (Vx - Vout ) and thus, since Vx is zero for the ideal amplifier, the potential difference across R2 is -Vout. Thus:
-Vout = I1R2
Dividing these two equations gives the ratio of the output voltage to the input voltage, i.e. the voltage gain of the circuit. Thus:
voltage gain of circuit =
The voltage gain of the circuit is determined solely by the relative values of R2 and R1. The negative sign indicates that the output is inverted, i.e. 180° out of phase, with respect to the input.
To illustrate the above, consider an inverting operational amplifier circuit which has a resistance of 10 k-ohm in the inverting input line and a feedback resistance of 100 k-ohm. The voltage gain of the circuit is: voltage gain of circuit =
FIG. 77 Non-inverting amplifier
FIG. 77 shows the operational amplifier connected as a non-inverting amplifier. Since the operational amplifier has a very high input impedance, there is virtually no current flowing into the inverting input. The inverting voltage input voltage is Vin . Since there is virtually no current through the operational amplifier between the two inputs there can be virtually no potential difference between them. Thus, with the ideal operational amplifier, we must have Vx = Vin. The output voltage is generated by the current I which flows from earth through R1 and R2. Thus:
But X is at the potential Vin, thus the potential difference across the feedback resistor R2 is (Vin - Vout ) and we must have:
Thus:
Hence:
voltage gain of circuit =
A particular form of this amplifier which is often used has the feedback loop as a short circuit, i.e. R2 = 0. The voltage gain is then just 1. However, the input voltage to the circuit is across a large resistance, the input resistance of an operational amplifier such as the 741 typically being 2 M-ohm. The resistance between the output terminal and the ground line is, however, much smaller, typically 75 ohm. Thus the resistance in the circuit that follows is a relatively small one compared with the resistance in the input circuit and so affects that circuit less. Such a form of amplifier circuit is referred to as a voltage follower and is typically used for sensors which require high impedance inputs such as piezoelectric sensors.
FIG. 78 Current-to-voltage converter
FIG. 79 Voltage-to-current converter
FIG. 78 shows how the standard inverting amplifier can be used as a current to voltage converter. Point X is the virtual earth. Thus any input current has to flow through the feedback resistor R2. The voltage drop across ^2 must therefore produce the output voltage and so Vout = -IR2. So the output voltage is just the input current multiplied by the scaling factor R2. The advantage of this method of converting a current to a voltage, compared with just passing a current through a resistor and taking the potential difference across it, is that there is a high impedance across the input and so there is less likelihood of loading problems.
Situations often arise where the output needs to be a current in order to drive perhaps an electromechanical device such as a relay or possibly give a display on a moving coil meter. A voltage-to-current converter is provided by the basic inverting amplifier circuit with the device through which the current is required being the feedback resistor (FIG. 79). Since X is a virtual earth, the potential difference across R1 is Vin and the current through it I1. Hence I1 = Vin/R1. The current through R2 is I1. Thus the input voltage has been converted to the current I1 through the feedback resistor, with the current being Vin/R1.
FIG. 80 shows how an operational amplifier can be used as a differential amplifier, amplifying the difference between two input signals. Since the operational amplifier has high impedance between its two inputs, there will be virtually no current through the operational amplifier between the two input terminals. There is thus no potential difference between the two inputs and therefore both will be at the same potential, that at X. The voltage V2 is across resistors R1 and R2 in series.
We thus have a potential divider circuit with the potential at the non-inverting input, which must be the same as that at X of Vx, as:
The current through the feedback resistance must be equal to that from V1 through R1. Hence:
This can be rearranged to give:
Hence, substituting for Vx using the earlier equation:
FIG. 80 Differential amplifier
The output is thus proportional to the difference between the two input voltages. Such a circuit might be used with a thermocouple to amplify the difference in e.m.f s between the hot and cold junctions. Suppose we require there to be an output of 1 mV/°C. With an iron-constantan thermocouple with the cold junction at 0 C, the e.m.f produced between the hot and cold junctions is about 53 uV/ C. Thus, for a 1 degree C temperature difference between the junctions, the above equation gives:
Hence we must have R2/R1 = 18.9. Thus if we take for R1 a resistance of 10 k-Ohm then R2 must be 189 k-Ohm.
The differential amplifier is the simplest form of what is often termed an instrumentation amplifier. A more usual form involves three operational amplifiers (FIG. 81). Such a circuit is available as a single integrated circuit. The first stage involves the amplifiers A1 and A2. These amplify the two input signals without any increase in the common mode voltage before amplifier A3 is used to amplify the differential signal. The differential amplification produced by A1 and A2 is (R1 + R2 + R3)/R1 and that produced by A3 is R5/R4 and so the overall amplification is the product of these two amplifications. The overall gain is usually set by varying the value of R1. Normally the circuit has R2 =R3, R4 = R6 and R5 = R7.
FIG. 81 Three op-amp form of an instrumentation amplifier
A charge amplifier provides an output voltage which is proportional to the charge stored on a device connected to its input terminals and is widely used with sensors employing piezoelectric crystals. Basically, a charge amplifier can be considered to be an op-amp with a capacitor in the feedback path (FIG. 82). The potential difference across the capacitor is (vx - Vout ) and since vx is effectively zero, being the virtual earth, it is -Vout . The charge on this capacitor is Cv_out and thus the current through it is the rate of movement of charge and so –Cdvout/dt. But this is the same current as supplied by the sensor and this is dq/dt, where q is the charge on sensor. Thus, -Cdvout/dt = dq/dt and the output voltage is -(1/C) times the charge on the sensor.
FIG. 82 Charge amplifier
9.8 Pressure-to-current converter
Control systems generally are electrical and so there is often a need to convert a pressure into an electrical current. FIG. 83 shows the basic principle of such a converter. The input pressure causes the bellows to extend and so apply a force to displace the end of the pivoted beam. The movement of the beam results in the core being moved in a linear variable differential transformer (LVDT). This gives an electrical output which is amplified. The resulting current is then passed through a solenoid. The current through the solenoid produces a magnetic field which is used to attract the end of the pivoted beam to bring the beam back to its initial horizontal position. When the beam is back in the position, the solenoid current maintaining it in this position is taken as a measure of the pressure input.
FIG. 83 Pressure-to-current converter
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