CHAPTER 28 Direct Sensing_hUpdted_1
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CHAPTER 28 DIRECT SENSING (A2)
a) show an understanding that an electric sensor consists of a sensing device and a circuit that provides an output voltage b) show an understanding of the change in resistance with light intensity of a lightdependent resistor (LDR) c) sketch the temperature characteristic of a negative temperature coefficient thermistor d) show an understanding of the action of a piezo-electric transducer and its application in a simple microphone e) describe the structure of a metal-wire strain gauge f) relate extension of a strain gauge to change in resistance of the gauge g) show an understanding that the output from sensing devices can be registered as a voltage h) recall the main properties of the ideal operational amplifier (op-amp) i) deduce, from the properties of an ideal operational amplifier, the use of an operational amplifier j) show an understanding of the effects of negative feedback on the gain of an operational amplifier k) recall the circuit diagrams for both the inverting and the non-inverting amplifier for single signal input l) show an understanding of the virtual earth approximation and derive an expression for the gain of inverting amplifiers m) recall and use expressions for the voltage gain of inverting and of non-inverting amplifiers n) show an understanding of the use of relays in e lectronic circuits o) show an understanding of the use of light-emitting diodes (LEDs) as devices to indicate the state of the output of electric circuits p) show an understanding of the need for calibration where digital or analogue meters are used as output devices
28-1
When you approach an automated door, an detects your presence and causes the door to open. Sensors are used in any electronic system that responds to an external change. Examples of such electronic systems: smoke alarm, sensors inside incubator, triple sensor watch, robotic machines in vehicle assembly plants, and etc.
A sensor supplies a signal to a , which can operate an such as an indicator lamp or a relay. The processing unit may receive signals from more than one sensor and it is programmed to respond according to the signals it receives.
Sensing device
Processing unit
Output device
28-2
An electronic sensor consists of a external change by producing: o
o
, which responds to an
a change of p.d. because it is in a resistance changes (as in a or a ); a p.d. directly as in
and its or a .
Using a potential divider o consists of two resistors in series of resistances R 1 and R2, connected to a source of fixed p.d. V s as shown in Fig. 1.
Fig. 1
o
Fig. 2
Because the current through the resistors is equal to the source p.d. divided by the total resistance of the two resistors,
o
The p.d. across each resistor is equal to its resistancethe current. Therefore, the p.d. across R1,
the p.d. across R2,
28-3
o o o
o
One of the resistors is a sensing device such as thermistor. The other resistor is a fixed resistance. When increases, the current decreases so the p.d. across decreases. Therefore, p.d. across the sensing device increases.
A in Fig. 1. o
o
o
uses a
as the sensing device, as shown
Thermistors that have a resistance that falls with increase of temperature are said to have a negative temperature coefficient and are referred to as NTC thermistors. Fig. 2 show how the resistance of an NTC thermistor decreases as the temperature increases. Notice that: the variation of resistance with temperature is non-linear the rate of change of resistance with temperature decreases as the temperature increases increases In Fig. 1, when temperature of the thermistor increases, its resistance decreases. As a result, the current in the circuit increases so the p.d. across the fixed resistance (=current resistance of the fixed resistance) must therefore increase. The p.d. across the thermistor therefore decreases because it is equal to the source p.d minus the p.d. across the fixed resistor. Since the output p.d. in Fig. 1 is across the thermistor, the output p.d. therefore decreases. Notice that if the output p.d. had been across the fixed resistor, resistor, the output p.d. would have increased i ncreased..
28-4
A uses a light-dependent resistor (LDR) as the sensing device. The resistance of an LDR decreases non-linearly when the intensity of light incident on it increases. Typically, the resistance of an
LDR decreases from 1 MΩ in darkness to about 100 Ω in sunlight. o
Fig. 3 shows an LDR in a potential divider.
Fig. 3
o
o
Fig. 4
When the light intensity is increased, the resistance of the LDR decreases (Fig. 4) so the current in the circuit increases. As a result, the p.d. across the fixed resistor increases. This means that the p.d. across the LDR decreases and so the output p.d. decreases. When the light intensity is decreased, the current in the circuit decreases so the p.d. across the fixed resistor decreases. Hence the p.d. across the LDR increases so the output p.d. increases.
A is any device that is designed to convert energy from one form to another. o o
A piezo-electric transducer generates generates a p.d. when it is squeezed. The piezo-electric material contains positive and negative ions, which are held together by the electrostatic forces they exert on each other.
28-5
Fig. 5 Tetrahedral silicate unit.
o
o
o
o
Fig. 6 Quartz – complex structure made up of a large number of repeated tetrahedal silicate units
In the normal unstressed state of the crystal, the centre of charge of the positive ions coincides with the centre of charge of the negative ions. When pressure is applied to the crystal, the crystal changes shape by a small amount and the centres of the positive and the negative charge no longer coincide. A voltage is generated across the crystal. This is known as the . The magnitude of the voltage that is generated depends on the magnitude of the pressure on the crystal, and its polarity depends on whether the crystal is compressed or expanded. The greater the pressure is, the larger the p.d. that is generated. A piezo-electric transducer can therefore be used directly as a pressure sensor or a force sensor. Piezo-electric ceramics ceramics are used in pressure sensors. Force sensors, microphones and gas lighters.
A piezo-electric rod, about 1 mm in diameter, of length 10 mm squeezed by a force of 10 N between its ends, can generate a p.d. in excess of 100 V.
28-6
Fig. 7 Piezo-electric effect in quartz o
o
o
A sound wave consists of a series of compressions and rarefactions. If a sound wave is incident on a piezo-electric crystal, then a voltage will be produced across the crystal that varies in a similar way to the variation in pressure of the sound wave. To detect the voltages, opposite faces of the crystal are coated with a metal and electrical connections are made to these metal films. Since the voltages generated are small, they are amplified. The crystal and its amplifier may be used as a simple microphone for converting sound signals into electrical signals.
Fig. 8 Microphone and its symbol
A takes advantage of the change in resistance of a metal wire as its length and cross-sectional area change. o When stretched, a metal wire becomes narrower narrower and longer; both these changes increase the electrical resistance. o When compressed, a metal wire becomes shorter and wider; as long as it does not buckle, these changes decrease its electrical resistance.
28-7
Fig. 9 Metal wire strain gauge (made by sealing a length of very fine wire in a small rectangle of thin plastic)
o
The resistance R of the wire in the strain gauge is given by the equation:
where L is the length of the wire, A is its area of cross-section is its resistivity.
Suppose the wire is stretched so its length increases from to , causing its resistance to increase i ncrease from to . Therefore,
assuming the area of cross-section does not change significantly.. significantly
Hence the increase of its resistance
, which means
that its increase of resistance is proportional to its change of length . As the strain on wires in the strain gauge is defined as , it follows that the increase of resistance of
the strain gauge is directly proportional to the strain. The strain gauge could be connected as the sensing device in a circuit such as Fig. 10. The output p.d. could be recorded using a data logger. If the output p.d. is found to change over time, the strain gauge must have been gradually stretched more if its resistance increased or stretched less if its resistance decreased. 28-8
Fig. 10 o
Strain gauge is used by an engineer to study crack line of building. By sticking a strain gauge over the crack the engineer can measure the resistance many days or even years later and see if there has been any movement.
MJ2010 P42 Q10(a)
(a)
State the name name of an electrical electrical sensing device that will will respond respond to changes in (i) length, strain gauge (ii)
pressure. transducer/piezo-electric/q transducer/p iezo-electric/quartz uartz crystal
ON09P41Q9(a)&(b)
A metal wire strain gauge is firmly fixed across a crack in a wall, as shown in Fig. 9.1, so that the growth of the crack may be monitored.
28-9
(a)
Explain why, as the crack becomes wider, the resistance of the strain gauge increases.
The resistance of wire,
as crack widens, increases and decreases so resistance increase i ncreases. s. (b)
The strain gauge has an initial resistance resistance of 143.0 Ω and, after being fixed in position across the crack for several weeks, the resistance is
found to be 146.2 Ω.
The change in the area of cross-section of the strain gauge wire is negligible. Calculate the percentage increase in the width of the crack. Explain your working.
..
100
.. ..
2. 2.24 24
100 00
Once the output of the sensing device is in the form of a voltage, it may require further change (processing) before it is used to control an output device. The basis of many circuits used for processing the sensor voltage is the operational amplifier or op-amp. An operational amplifier is an integrated circuit that contains about 20 transistors together with resistors and capacitors. It is referred to as an
‘integrated circuit’ because all the components are formed on a small slice of a semiconductor into a circuit. The whole of the integrated 28-10
circuit is encapsulated. The symbol for an op-amp and some of its
connections are shown in Fig. 11. It is referred to as an ‘operational amplifier’ because the circuit can easily be made to carry out different
operations. These operations include: o acting as a switch when a voltage reaches a certain level o amplifying direct voltages o amplifying alternating voltages o comparing two voltages and giving an output that depends on the result of the comparison. The op-amp has two inputs, o inverting input (-) o non-inverting input (+) For many applications, the positive and the th e negative power supplies are 6 V or 9V.
Fig. 11 An operational amplifier and its symbols.
Properties of an ideal op-amp o
(no current enters or leaves either of the inputs). Connecting the op-amp to the potential divider would change the p.d across the components in the potential divider. If the op-amp connection (the input) has infinite resistance (or, more strictly, for alternating voltages, infinite impedance) then connecting it to the potential divider will not affect the potential differences. The input impedance of an op-amp may be as high as 10 12 Ω, but 10 16 Ω is typical.
28-11
o
o
(the whole of the output voltage is seen across the load connected to the output). If the output connection had some resistance, then the voltage produced in the op-amp (the output voltage) would be divided between the output and the resistor connected to the output load. Zero output resistance (or zero output impedance) means that all of the output voltage will be seen across the resistor connected to the output load. i.e. if the op-amp has zero output resistance it is acting just like an electrical battery with zero internal resistance. (this means that when there is only a very small input voltage, the amplifier will and the output will have the same value as the supply voltage). The voltage gain, or simply gain of an amplifier is a measure of how many times the output voltage is greater than the input voltage.
When there are no components connected between the output and the input of the op-amp (no feedback loop), the gain that is measured is said to be the open-loop gain. The output of the op-amp cannot be greater than the supply voltage (from energy conservation) and so, if a very small voltage is applied to the input and the gain is infinite, then the output will be at the supply voltage. The output cannot be any greater, even for a larger input signal. The amplifier is said to be saturated.
28-12
o
(all frequencies are amplified equally). If an alternating voltage is applied to the input, then the output will have the same frequency but larger amplitude. The range of frequencies that are amplified by the same amount (the input signals of different frequencies that all have the same gain) is known as the bandwidth. Note: the bandwidth of an op-amp is the range of frequencies that will produce an output voltage without a big attenuation.
o
When the input signal is changed, then the output signal will also change. The slew rate is a measure of the time delay between the changes to the input and output. (or how fast the op-amp can respond to changes in the input ) A high slew rate implies a short time delay. With an infinite slew rate there is no delay.
The operational amplifier as a o When an op-amp is incorporated in an electrical circuit, it is usually connected to a dual power supply so that the output voltage can be either positive or negative. Such a power supply can be represented as two sets of batteries, as shown in Fig. 12,
Fig. 12 The op-amp used as a comparator
28-13
o
The link between the two batteries is referred to as the ‘zero volt line’ or ‘earth’ and it provides the reference from which the voltages on the inputs and the output are measured. The output Vout of of this circuit is given by
where
o
is the voltage at the non-inverting input, is the voltage at the inverting input, is the open-loop gain of the amplifier.
Consider case 1, where, non-inverting input, is 0.95 V, inverting input, is 0.94 V,
10 supply voltages are 6 V By substituting into the equation, the output voltage , 10 0.9 0.95 0.9 0.94 1000 000 V. From energy considerations, these answers are impossible since the output voltage can never exceed the power supply voltage. The amplifier is saturated and the output voltage will be +6 V. o
Consider case 2, where, 3.65 3.6522 V 3.65 3.6544 V supply voltages are 6 V
10 3.65 3.6522 3.65 3.6544 200 V o
From energy considerations, these answers are impossible since the output voltage can never exceed the power supply voltage. The amplifier is saturated and the output voltage will be -6 V. So,
If
, the output is 28-14
o
If
, the output is –
the polarity of the output depends on which input is the larger.
The circuit of Fig. 12 is called a comparator because it compares the voltages applied to the non-inverting and the inverting inputs and then gives an output that depends on whether or .
ON07P4Q8
(a)
Fig. 8.1 shows a circuit incorporating an ideal operational amplifier (opamp).
The voltage applied to the inverting and the non-inverting inputs are and respectively. State the value of the output voltage when (i)
, -9 V
(ii)
. +9 V
(b)
The circuit of Fig. 8.2 is used to monitor monitor the input voltage .
28-15
At point A, a potential of 5.0 V is maintained. At point B, a potential of 3.0 V is maintained. Complete Fig. 8.3 by indicating with a tick √ the light -emitting diodes (LEDs) that are conducting for the input voltages shown. Also, mark with a cross (X) those LEDs that are not conducting.
+2.0 +4.0 +6.0
Red LED X
√ √
Green LED X X
√
Fig. 8.3
(c)
The input voltage in (b) is provided by a sensor circuit. (i) Complete Fig. 8.4 to show a sensor circuit that will provide a voltage output that increases as the temperature of the sensor decreases. Show clearly the output connections from the circuit.
Output V
28-16
(ii)
Explain the operation of the sensor circuit. as temperature decreases, thermistor resistance p.d. across thermistor =
increases.
as increases, output increases
When a circuit incorporating an op-amp is used as a comparator, it is usual to connect each of the two inputs to a potential divider, as shown in Fig. 13. o One potential divider provides a fixed voltage at one input while the other potential divider provides a voltage dependent on light intensity. o In Fig. 13, the resistors of resistance R will give rise to a constant voltage of o
at the inverting input.
The LDR, of resistance is connected in series with a fixed resistor of resistance F.
Fig. 13 The op-amp used as a comparator to monitor illumination
o
o
o
If (that is, the LDR is in darkness), then output is positive.
and the
If (that is, the LDR is in daylight), then output is negative.
and the
The output can be made to operate an output device such as LED.
28-17
o
o
It can be seen that by suitable choice of the resistance F, the comparator gives an output, either positive or negative, that is dependent on light intensity. The light intensity at which the circuit switches polarity can be varied if the resistor of resistance F is replaced with a variable resistor. The LDR can be replaced by other sensors to provide alternative sensing devices. For example, use of a thermistor can provide a frost-warning device.
For the circuit shown below, the resistance of the thermistor T is 8 kΩ at a temperature of 15 °C. What are and , the potentials at the inverting and non-inverting inputs? And what happens when the temperature falls so that
the resistance of T rises above 10 kΩ?
p.d. at A
9 5 . 0 V
p.d. at B
9 4 . 5 V
the op-amp acts as a comparator and, since is larger than , the output will be the highest voltage that the op-amp can produce, in this case +9 V. The thermistor T is a negative temperature coefficient thermistor and so its
resistance rises sharply and eventually becomes larger than 10 kΩ. Suppose it becomes 12 kΩ. Then, 9 4 . 1 V p.d. at A Now smaller that and the op-amp output voltage is the lowest it can
provide, near the negative supply voltage, in i n this case -9 V. The switch from +9 V to –9 V is quite sudden because of the large open-loop voltage gain. The value of the temperature when the output voltage switches from +9V to 9V can be altered by adjusting the resistance of the resistor in series with the thermistor. 28-18
Operational amplifiers and feedback f eedback o is a process whereby a fraction of the output of any device is fed back to the input, so as to assist in the control of the device. Much of the movement of humans is controlled by feedback. If a person wishes to pick up an object, for example, then the person stretched out a hand while, at the same time, looking at the hand and the object. The visual signal from the eye is fed back to the brain to provide control for the hand. This feedback is a continuous process refining the position of the hand relative to the object. o
For an amplifier circuit, the basic arrangement is as shown in Fig. 14.
Fig. 14 An amplifier circuit with feedback
o
A fraction of the output signal is fed back and added to the input signal . The amplifier has open-loop gain and it amplifies whatever voltage there is at its input. So,
the amplifi amplifier er input input and this gives
1 the overall voltage gain (or simply gain) ⁄ of the amplifier is given by
1 28-19
o
o
Note: if the fraction is negative, then 1 is greater than unity and the overall gain of the amplifier circuit is less than the open-loop gain of the operational amplifier itself. This is known as .
Negative feedback seems, at first sight, to defeat the object of an amplifier. However, the reduction in amplification is a small price to pay for the benefits. These benefits: 1) increased bandwidth (the range of frequencies for which the gain is constant) 2) less distortion of the output (output signal is exactly For higher frequency, the voltage gain is the same as input signal) small. This means that the amplification signal of 10 Hz to 10 Hz is unstable. The 3) greater stability (the gain is more stable, not affected lower frequency amplification is very by changes in temperature, etc) much higher than the high frequency. 4) the output resistance (impedance) can be low and the input resistance (impedance) high. o Negative feedback can be achieved achieved by feeding part of the output of the op-amp back to the inverting input, see Fig. 15. 5
Fig. 15 Negative feedback with an op-amp
28-20
o
A circuit for an inverting amplifier incorporating an op-amp is shown in Fig. 16.
Fig. 16 An inverting amplifier
o o o
o
o
o
An input signal is applied to the input resistor . Negative feedback is applied by means of the resistor . The resistors and act as a potential divider between the input and the output of the op-amp. In order that the amplifier is not saturated, the two input voltages must be almost the same. To understand how the inverting amplifier works, we need to understand the concept of the . In this approximation, the potential at the inverting input (-) is very close to 0 V. 1) The op-amp multiplies the difference in potential between the inverting and non-inverting inputs, and , to produce the output voltage . Because the open-loop voltage gain is very high, the difference between and must be almost zero. 2) The non-inverting input (+) is connected to the zero volt line so 0. Thus must be close to zero and the inverting input (-) is almost at earth potential. Point is known as a is very close to 0 V.
. It cannot actually be 0 V but it
28-21
o
The input impedance of the op-amp itself is very large and so there is no current in either the non-inverting or the inverting inputs. So all the current from, or to, the input signal to the circuit must pass through the feedback resistor to the output, as shown in Fig. 17.
Fig. 17 Feedback current o
o
Because the inverting input is at zero volts, a positive input gives rise to a negative output and vice versa. This is why the arrangement arrangement is given the name inverting amplifier. Referring to Fig. 17, since the input resistance of the op-amp is infinite, current in current in and ..
.. ..
The potential at P is zero (virtual earth) and so
0 0 The overall voltage gain of the amplifier circuit is given by, voltage gain o
o
The negative sign shows that when the input voltage is positive then the output voltage is negative and when the input is positive the output is negative. It the input voltage is alternating then there will be a phase difference of 180° or rad between the input and the output voltages.
28-22
o
The circuit for a non-inverting amplifier incorporating an op-amp is shown in Fig. 18.
Fig. 18 A non-inverting amplifier o o
o
o
o
The input signal is applied to the non-inverting input. Negative feedback is provided by means of the potential divider consisting of resistors and . As long as the op-amp is not saturated, the potential difference between the inverting (-) and non-inverting (+) is almost zero. So . Since the non-inverting input (+) is connected to the input voltage, . Thus, . The two resistors and form a potential divider. The total voltage across and is and the voltage across alone is
. o
The current in the two resistors can be written as:
voltage gain
o
1
The non-inverting amplifier produces an output voltage that is in phase with the input voltage. When the input voltage is positive, so is the output voltage. 28-23
MJ10P42Q9
(a)
Negative feedback may be used in amplifier circuits. State (i)
what is meant by negative feedback. Fraction of the output (signal) is added to the input (signal); Out of phase by 180° between input and output voltages.
(ii)
two effects of negative feedback on an amplifier incorporating an operational amplifier (op-amp) ( op-amp) Less distortion/increases bandwidth/gain is more stable/reduces gain
(b)
Fig. 9.1 is a circuit for an amplifier that is used with a microphone. The output potential difference is 4.4 V when the potential at point P is 62 mV. Determine, (i) the gain of the amplifier Gain (ii)
.
. 71
the resistance of the resistor R.
1 71
1 (c)
71 ⇒ 1.7 10 Ω
The maximum potential produces by the microphone microphon e at point P on Fig. 9.1 is 95 mV. The power supply for the operational amplifier may be either 5 V or 9 V.
28-24
State which power supply should be used. Justify your answer quantitatively. For the amplifier not to saturate, Maximum output = 7 1 9 5 1 0 Power supply should be 9 V.
6.7 V
MJ09P4Q10
(a)
By reference reference to an amplifier, explain what is meant by negative negative feedback. (part of) the output is added to/returned to /mixed with the input and is out of phase with the input/fed to inverting input.
(b)
An amplifier circuit incorporating an ideal operational amplifier (opamp) is shown in Fig. 10.1. The supply for the op-amp is 9.0 V. The amplifier circuit is to have a gain of 25. Calculate the resistance of resistor R.
251 5 kΩ
State the value of the output of the amplifier in (b) for input voltages of (i) -0.08 V -2 V (ii) +4.0 V +9 V
(c)
Circuits incorporating op-amps produce an output voltage. This output voltage can be used to operate warning lamps, digital meters, motors etc. However, the output of an op-amp cannot exceed a current of more than . Otherwise, the op-amp would be destroyed. In fact, op-amps generally contain an output resistor so that, should the output be ‘shorted’, the op -amp will not be damaged. 28-25
The sensing circuit may be required to switch on or off an appliance that requires a large current; e.g. an electric motor. The switching on or off of a large current by means of a small current can be achieved using a relay. The o
o
It is an electromagnetic switch that uses a small current to switch on or off a larger current. The symbol for a relay is shown in Fig. 19.
Fig. 19 A relay o
o
o
o
When a current passes through the coil of the electromagnet, the iron armature is attracted on to the iron core of the electromagnet. electromagnet. The armature turns about the pivot and closes the gap between the switch contacts. In this way, a small current can be used to switch on a much greater current. For example, when the ignition switch of a car is turned on, a small current passes through the electromagnet coil so the relay switch closes. This allows a much greater current to pass through the starter motor. The connection of a relay to the output of an op-amp circuit is shown in Fig. 20.
28-26
Fig. 20 The connection of a relay to an op-amp
The diode D 1 conducts only when the output is positive with respect to earth and thus the relay coil is energized only when the output is positive. When the current in the relay coil is switched off, an e.m.f. is induced (back e.m.f.) across the terminal in the coil. This e.m.f. could be large enough (because of a very rapid fall in magnetic flux within the coil) to damage the op-amp. A diode D2 is connected across the coil to protect the opamp from this back e.m.f. When the output of the op-amp is positive, the diode D2 is reverse-biased and will not conduct any current. When the op-amp is switched off, the diode D 2 is forward-biased forward-biased and will conduct current. The induced voltage is the coil causes the bottom of the coil to be more positive than the top of the coil. (D2 creates an
easy path for the current to circulate until the relay’s inductance has lost all its energy, through wire and diode. MJ10P42Q10
(a)
State the name name of an electrical electrical sensing device that will will respond respond to changes in (i) (ii)
length, strain gauge pressure. transducer/quartz transducer/qua rtz crystal/piezo-ele crystal/piezo-electric ctric 28-27
(b)
A relay is sometimes sometimes used as the output of a sensing circuit. The output of a particular sensing circuit is either +2 V or -2 V. On Fig. 10.1, draw symbols for a relay and any other necessary component so that the external circuit is switched on only when the output from the sensing circuit is +2 V.
o
The LED is a semiconductor device that is robust, reliable and dissipates much less power than a filament lamp; commonly used
as ‘indicators’ because they have a low power consumption. consumption. o
They are available in different colours including red, green, yellow and amber.
Fig. 21 Symbol for a light-emitting diode (LED) o o
o
The LED emits light only when it is forward-biased. A resistor is frequently connected in series with an LED so that, when the LED is forward biased (the diode is conducting), the current is not so large as to damage the LED. A typical maximum forward current for an LED is 20 mA. Furthermore, the LED will be damaged if the reverse bias voltage exceeds about 5 V.
28-28
Fig. 22 Circuit using two diodes to indicate whether the output from an op-amp is positive or negative with respect to earth.
o
o
o
When the output is positive with respect to earth, diode D 1 will conduct and emit light. Diode D 2 will not conduct because it is reverse biased. If the polarity of the output changes, then D2 will conduct and emit light and D1 will not emit light. The state of the output can be seen by which diode is emitting light. The diodes can be chosen so that they emit light of different colours.
ON09P41Q10
The circuit of Fig. 10.1 may be used to indicate temperature change.
28-29
The resistance of the thermistor T at 16°C is 2100Ω and at 18°C, the resistance
is 1900Ω. Each resistor P has a resistance of 2000Ω. Determine the change in the states of the light-emitting diodes R and G as the temperature of the thermistor changes from 16°C to 18°C.
At 16°C, 1.00V, 0.98V ( So, at 16°C, output is positive.
)
Diode R is ‘on’ and diode G is ‘off’. As temperature rises, diode R goes ‘off’ and diode G goes ‘on’. ) MJ08P4Q9
A block diagram for an electronic sensor is shown in Fig. 9.1.
(a)
Complete Fig. 9.1 by labeling the remaining boxes.
(b)
A device is to be built that will emit a red light when its input is at +2V. When the input is at -2V, the light emitted is to be green. (i)
On Fig. 9.2, draw a circuit diagram of the device.
(ii)
Explain briefly the action of this device. correct polarity for diode to conduct identified hence red LED conducts when input (+) or vice versa. 28-30
o
o
o
o
o
An LED may be used to indicate whether an output is positive or negative. If the output is from a comparator, then LEDs can give information as to, for example, whether a temperature is above or below a set value. However, the LED does not give a value of the temperature reading. Many sensors, for example, a thermistor or an LDR, are non-linear. It was seen that the sensor could be connected into a potential divider circuit so that the output of the potential divider varied with some property, for example temperature or light intensity. This variable voltage could be measured using an analogue or a digital voltmeter. The reading on the voltmeter would vary with the property being monitored. However, However, the reading on the voltmeter would not vary linearly with change in the property. In order that the property can be measured, a calibration curve is required. The reading on the voltmeter is recorded for known values of the property X (i.e. the physical property that changes its resistance). For example: temperature a thermistor can be used to measure temperature u sed to measure light intensity an LDR can be used a strain gauge can be used to monitor strain A graph is then plotted showing the variation with the property X of the voltmeter reading. The value of the property X can then be read from the graph for any particular reading on the voltmeter.
Fig. 23 A calibration curve
28-31
MJ12P41Q10
A student designs an electronic sensor that is to be used to switch on a lamp when the light intensity is low. Part of the circuit is shown in Fig. 10.1.
(a)
State the name of the component labeled X on Fig. 10.1. [1] light-dependentt resistor (allow LDR) light-dependen (b) On Fig. 10.1, draw the symbols for (i) two resistors to complete the circuit for the sensing device, [2] two resistors in series between +5V line and earth midpoint connected to inverting input of op-amp (ii) a relay to complete the circuit for the processing unit. [2] relay coil between diode and earth switch between lamp and earth (c)(i) (c)(i) State the purpose of the relay. [1] switch on/off mains supply using a low voltage/current output (ii) Suggest why the diode is connected to the output of the operational amplifier (op-amp) in the direction shown. [2] relay will switch on for one polarity of output (voltage) switches on when output (voltage) is negative
28-32
MJ12P42Q9
An operational amplifier (op-amp) may be used as part of the processing unit in an electronic sensor. (a) State four properties of an ideal operational amplifier. [4] infinite input impedance/ impedance/resistance resistance zero output impedanc impedance/resistance e/resistance infinite (open loop) gain infinite bandwidth infinite slew rate (b) A comparator circuit incorporating an ideal op-amp is shown in Fig. 9.1.
The variation with time t of the input potential V IN is shown in Fig. 9.2.
On the axes of Fig. 9.2, draw a graph to show the variation with time of the output potential VOUT. [3] square wave 180° phase change amplitude 5.0V 28-33
(c)The output potential V OUT is to be displayed using two light-emitting diodes (LEDs). A diode emitting red light is to indicate when V OUT is positive and a diode emitting green light is to be used to indicate when VOUT is negative. Complete Fig. 9.3 to show the connections of the two LEDs to the th e output of the op-amp. Label each LED with the colour of light that it emits. [3]
correct symbol for LED diodes connected correctly between and earth diodes identified correctly
ON13P43Q9
(a)
State three properties of an ideal operational amplifier (op-amp). [3]
(b)
An amplifier circuit is shown in Fig. 9.1.
(i) Calculate the gain of the amplifier circuit. . gain 1
10
[2]
.
(ii) The variation with time of the input potential is shown in Fig. 9.2. 28-34
On the axes of Fig. 9.2, show the variation with time of the output potential . [3] straight line from (0, 0) toward 1.0V, 10V horizontal line at 9.0V to 2.0V correct +9.0V -9.0V (and correct shape to 0)
ON10P43Q10
(a) (b)
State three properties of an ideal operational amplifier (op-amp). [3] A circuit incorporating an ideal op-amp is to be be used to indicate whether a door is open or closed. Resistors, each of resistance R, are connected to the inputs of the opamp, as shown in Fig. 10.1.
The switch S is attached to the door so that, when the door is open, the switch is open. The switch closed when the door is closed. (i) Explain why the polarity of the output of the op-amp changes when the switch closes. [3] with switch open, is less (positive) than output is positive with switch closed, is more (positive) than so output is negative 28-35
(ii)
A red light-emitting diode (LED) is to be used to indicate indicate when the door is open. A green LED is to indicate when the door is closed. On Fig. 10.1, 1. draw symbols for LEDs to show how they are connected to the output of the op-amp. [1] diodes connected correctly between output and earth 2. identify the green LED with the letter G. [1] green identified correctly
ON11P42Q9
(a)
(b)
State two effects of negative feedback on the gain of an amplifier incorporating an operational amplifier (op-amp). [2] reduced gain/increase stability/greater bandwidth bandwidth or less distortion An incomplete circuit diagram of a non-inverting amplifier using an ideal op-amp is shown in Fig. 9.1.
(i)
(ii)
Complete the circuit diagram of Fig. 9.1. Label the input and the output. [2] connected to midpoint between resistors. resistors. clear and input to clear Calculate the resistance of resistor R so that the non-inverting amplifier has a voltage gain of 15. [2] gain 1 151
860Ω (c)
On Fig. 9.2, draw a graph to show variation with input potential of the output potential VOUT. You should consider input potentials in the range 0 to +1.0 V. 28-36
(d)
The output of the amplifier circuit of Fig 9.1 may be connected to a relay. State and explain one purpose of a relay. [2] relay can be used to switch a large current/voltage output current of op-amp is a few mA/very small
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