Section VII 28 Direct Sensing

7/28/2019 Section VII 28 Direct Sensing

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Direct Sensing 28. Direct sensing

Content

28.1 Sensing devices 28.2 The ideal operational amplifier

28.3 Operational amplifier circuits

28.4 Output devices

Learning outcomes

Candidates should be able to:

(a) show an understanding that an electronic 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 light-dependent 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


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Direct sensing

This topic deals with how circuits incorporating sensing device/s thatdetect changes in the physical properties produce output voltages thatwill operate an output

An electronic sensor or circuit may be thought of as made up of 3parts namely; a sensing device, into a processing unit and to an outputdevice


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Sensing circuit - example

thermistor

LDR

Strain gauge

piezoelectric

Potential

divider

circuit

LED

Relay

switch


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e.g. Thermal pot

SENSING DEVICE THERMISTOR

PROCESSING UNIT POTENTIAL DIVIDER CIRCUIT

OUTPUT VOLTAGE LED 1 (RED) Boil

- LED 2 (YELLOW) warm


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Sensing devices

There are simple sensors and complex sensors

Simple sensors e.g. those detecting temperature in a thermostat or asensor detecting light levels in an automatic switch for a lamp

Complex sensors e.g. those detecting whether a car wheel is skiddingin a braking system and the circuit controlling not only the brakes butalso to stop the skidding and to stop further skidding

The sensing device is the 1st

stage of any electronic sensor and is themeans by which whatever is to be detected or monitored is convertedinto an electrical signal i.e. a voltage

Ideally, the generated voltage should be proportional to the externalchange causing it

Sensing devices detect many different types of changes e.g. changes

in length, mass, weight, force, velocity, acceleration, current, voltage,light intensity, temperature, sound level, humidity, pressure, strain,magnetic field

We shall consider 4 sensing devices light dependent resistor(LDR),thermistor, piezo-electric transducer and the metal wire strain gauge


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LDR properties

Resistance

Lux

Symbol is as above

It consists of 2 metal grids or electrodes that intersect each other whose space between isfilled with a semiconductor material e.g. cadmium sulphide doped with copper

When light is incident on the semiconductor material, the number of electrons in the

semiconductor that are free to conduct increases

Light intensity is measured in lux which is a unit used to measure the light powerincident per unit area of a surface

The higher the intensity of light on the LDR, the greater the number of electrons that can

move freely hence as intensity of light increases, the resistance of the LDR decreasesi.e. more current, same p.d. R decreases

Typical LDR data: (a) normal room lighting 450 lux, LDR resistance 900 ohms

(b) sunlight 28000 lux, LDR resistance 100 ohms


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Light-dependent resistor - LDR


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Thermistor

Symbol is as above

Resistance of most metals increases to a certain extent with rise in

temperature i.e. positive temperature coefficient(ptc)

Resistance ofthermistor also changes with temperature but with negativetemperature coefficient(ntc)

Thermistors manufactured in various shapes and sizes including rods, discs

and beads, are of 2 types

(i) positive temp coefficient, ptc - R increases with temp rise

(ii) negative temp coefficient, ntc - R decreases with temp rise Ptc made of ceramic or semiconductor

Ntc made of semiconductor and metal oxides

Typical thermistor data: (a) 20Cthermistor resistance - 1800 ohms

(b) 40Cthermistor resistance - 900 ohms


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Thermistor


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Thermistor

- Linear or non-linear?

- Variation with temperature of a typical thermistor is non-linear and in fact

is approximately exponential over a limited range of temperature

- Which part of the graph shows resistance changes more with temperature?


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Piezo-electric effect

Piezo-electric devices contain a crystal which can expand and compresswhen external pressure is varied e.g. quartz

The crystals structure is such that the centre of positive charges coincideswith the centre of negative charges when not stressed

When expanded, both centres will not coincide.

When compressed, the centres will be in the opposite direction as comparedto under expansion.

The separation results in a voltage across the crystal surface and this effectis known as the piezo-electric effect


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Piezo-electric effect


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Piezo-electric transducer

A piezo-electric device/transducer is a sensor that detects differences inpressure (sound wave)

Variation in pressure will result in an a.c voltage

The magnitude of the voltage generated depends on the magnitude of thepressure on the crystal and the polarity depends on whether the crystal iscompressed or expanded i.e. whether the pressure is greater than or lessthan the ambient pressure

A transducer is any device that converts energy from one form to another The piezoelectric transducer converts mechanical energy (vibration) intoelectrical energy in the form of a.c voltages

It also can convert electrical voltages back to vibration.

Hence it acts as a receiver as well as an emitter.

To detect the voltages, opposite faces of the crystal are coated with a metal

(silver) and electrical connections are made to these metal films and sincethe voltages are very small they are amplified

The crystal and its amplifier may be used as a simple microphone forconverting sound signals into electrical signals


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Strain Gauge

In engineering, it is frequently required to test for the strains experienced

in different parts of structures This is achieved by using a length of very fine wire embedded and sealed

into a small rectangle of thin plastic known as a strain gauge, as shownbelow

Its resistance changes when the gauge is stretched. The wire since it is

strained increases in length together with the plastic. As the lengthincreases the cross sectional area decreases causing an increase in theelectrical resistance of the wire


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Strain Gauge


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Strain Gauge

R = L/A

If the wire increases in length by only a small amount L, then the changein the cross-sectional area can be assumed to be negligible and the new

resistance is given by

(R + R) = (L +L) /A,

By subtracting the above 2 expressions, the change in resistance R is

given by R = L/A hence R L since it is assumed that A is a constant

and is also a constant

Since strain is proportional to the extension, then the strain is alsoproportional to the change in resistance


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Processing units

The change in the physical property of a sensing device must beprocessed in some way before the change can be displayed or

measured or is used to control

This is carried out by the processing unit which is some form of anelectrical circuit that is connected to the sensing device which provides

a voltage at its output If the voltage is small, it is usually necessary to amplify it or it may

require further processing before the voltage can be used to control an

output device

2 such circuits are the potential divider circuit and the operational

amplifier circuit

P i l di id i i f


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Potential divider as a processing unit of

sensing device

A change in resistance of a sensing device results in a change of voltage

across the sensor if the sensor is connected to the circuit and forms onepart of a potential divider

V = R/(R+F) E assuming internal resistance is negligible

As the resistance of the sensing device increases, the output voltageincreases


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Example

A potential divider consists of a battery of e.m.f. 6.00 V and negligible internal

resistance connected in series with a resistor of resistance 120 ohms and avariable resistor of resistance 0 to 200 ohms. Determine the range of potential

difference that can be obtained across the fixed resistor.

Solution

When the variable resistor is 0 ohm,

p.d. across fixed resistor is = 6.00 V

When the variable resistor is 200 ohms,

p.d. across fixed resistor is = 120/(120 + 200) x 6.00 = 2.25 V

Hence the range of potential differenceacross the fixed resistor is = 2.25 V to 6.00 V


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Operational amplifier(op-amp) as a processing unit

Any small change in voltage can be amplified using an electrical circuitincorporating a circuit called an operational amplifier (op-amp).

It is referred to as an operationalamplifierbecause the circuit can easily be madeto carry out different operations e.g. acting as a switch, amplifying direct voltages,amplifying alternating voltages, comparing 2 voltages

A key component in electronics is the transistor, which is a semiconductor. It canamplify or act as a switch

An integrated circuit(IC) is a package that may have thousands of transistors and

other components formed on a single chip of silicon An op-amp contains about 24 transistors together with resistors and capacitors.

It is referred to as an integrated circuit because all the components are formed on asmall slice of a semiconductor(e.g. silicon or germanium) with external connectionsto enable the op-amp to be connected into a circuit and the whole assembly isencapsulated

An op-amp and some of its connections or terminals and its symbol is as shownbelow:


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Op-amp - actually how it looks


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Op-amp how it looks on the inside


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Terminals of the op-amp

The output voltage from a sensing circuit can be joined to any of the 2inputs of an op-amp

+ve input non-inverting input

-ve input inverting input

+ve and ve input terminals do not mean that the voltage connected to these

terminals must be +ve and ve accordingly, but whether they invert or notthe output voltage

If V+ > V- the output is positive (+)

If V+ < V- the output is negative (-)

If V+ = V- the output is zero

If only one input is in use and the other is at zero voltage,

A positive voltageon the inverting input causes a negative output voltage

A positive voltageon the non-inverting inputcauses a positive output voltage


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cont

The power supply for an op-amp is a 3-terminal DC supply made up of 0 Vearth and typically +/- 9 V. This a called a dual, or split, power supply.

The op-amp is powered by the dual power supply. Current that flows withinthe op-amp comes from the power supply, not from inputs to the op-ampinput terminals

Power supply lines connect to the power supply

The common link between the two sets of batteries or power supplies is termedthe zero-volt, or earth line. This forms the reference line from which all inputand output voltages are measured


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cont

Output terminal connects to any output device, e.g. LED, relay, bulb


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5 main properties of an idealop-amp

There are 5 main properties of an ideal op-amp:1) Infinite input impedance

2) Zero output impedance

3) Infinite open-loop gain

4) Infinite bandwidth

5) Infinite slew rate

In reality op-amps are not ideal.


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Input impedance

Impedance is similar to resistance but it is more than that.

While a capacitor does not have resistance in a d.c. circuit, in an a.c. circuit ithas reactance(similar to resistance, also in ).

The combined resistance of a resistor and reactance of a capacitor is known as

impedance(in ).

Input impedancerefers to the impedance between the two input terminals


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Infinite input impedance

Infinite input impedance means no current enters or leaves either of theinputs

This means that connecting for example the output of a potential divider to theinput of an op-amp will not affect the potential differences

An ideal op-amp should have infinite input impedance.

When a sensing circuit is joined to an op-amp, ideally no current should flowinto the op-amp through the input terminals otherwise, the voltage in thesensing circuit will be altered

This is analogous to a voltmeter where a voltmeter should have infiniteresistance so that when joined across a component, it should draw very littlecurrent and thus no change occurs to the voltage it tries to measure

In reality, the input impedance is in the range 1061012 ohms


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Output impedance

Output impedance refers to the impedance inside the op-amp near theoutput terminal

An ideal op-amp should have zero output impedance so that all the outputvoltage is seen across the load connected to the output

This is analogous to a battery of zero internal resistance where all the

voltage supplied by the battery should fall across the external circuit In reality, the output impedance is about 102 ohms


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Gain

Gain is defined as the ratio of output voltage to input voltage and the symbol is

A i.e. how many times the output is greater than the inpute.g. if the output is 9V and the input is 0.1 V, gain A = 9/.1 = 90

The way an op-amp is connected in a circuit will produce 2 different types of gain

(a) open loop gain, symbol Aoand

(b) closed loop gain, A

Open-loop Closed-loop


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Infinite open-loop gain

An op-amp actually amplifies the difference between the two inputs to the

op-amp, (V+ - V- ). That is why it is sometimes known as a differentialamplifier.

So Ao= Vout/ (V+- V- )

hence Vout= Ao (V+- V- )

When there are no components connected between the output and theinput, the gain that is measured is said to be open-loop gain

(Closed-loopgain will be examined later)

An ideal op-amp should have infinite open-loop gain which means thatwhen there is only a very small input voltage, the output will have the samevalue as the supply voltagesince the output of the op-amp cannot be greater

than the supply voltage(from energy conservation) and the amplifier is saidto be saturated

In reality, the open-loop gain is not constant but is in the order of 10 5 fordirect voltages(dc)


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Bandwidth

When a.c. voltage is connected to the two inputs of an op-amp, the outputvoltage is different for different frequencies of the ac voltage.

The gain varies with the frequency, f

Below is a graph of open-loop gain Ao versus frequency f

The range of frequencies over which the gain is constant is known as thebandwidth


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Infinite bandwidth

Infinite bandwidth means that all frequencies of the input areamplified equally

This means that if an ac signal is applied to the input, then the output

will have the same frequency but a larger amplitude

In reality, the bandwidth is finiteand not infinite


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Slew rate

When the input signal is changed, the output signal will also change

The output voltage may not be in sync with the input voltage. When the input

voltage has changed with time, the output voltage may not have changed yet

An ideal op-amp should have infinite slew rate meaning no delay betweenchanges in the input and consequent changes in the output.

A high slew-rate implies a short time delay

In reality, the slew-rate is about 10 V s-1 i.e. 107 V s-1


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Op-amp characteristics


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The op-amp as a comparator

An op-amp actually amplifies the difference between the two inputs to theop-amp, (V+ - V- ). That is why it is sometimes known as a differential

amplifier. So Ao= Vout/ (V+- V- ) hence Vout= Ao (V+- V- ) Consider the case where the non-inverting input V+ is 0.95 V, the

inverting input V- is 0.94 V, open-loop gain A0 = 105, and the supply

voltages at 6 V

Using Vout= Ao (V+- V- ) = 1000 V which from energy considerations is

impossible as the output voltage can never exceed the power supplyvoltage.

This means that the amplifier is saturated and the output voltage will be +6 V, so if V+ > V- the output is + Vsupply and if V- > V+ the output is Vsupply

Because this circuit compares the voltages applied to the non-inverting

and the inverting inputs and then gives an output it is called a comparator


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comparator When a circuit incorporating an op-amp is used as a comparator, it is usual to connect

each of the inputs to a potential divider

+ 6 V

10 k R R

- -

+ +

R R output

- 6 V

The 2 resistors of equal resistance provide an input voltage of 3 V at the inverting input

When the LDR is in darkness its resistance is greater than 10 k, hence the voltage at

the non-inverting input is greater than 3 V and the output will be at 6 V.

In daylight the resistance of the LDR will be less than 10 k and the voltage at the non-

inverting input will be less than 3 V and the output will switch to - 6 V

The output therefore depends on the level of light illumination and the values of the

resistors in the potential divider. In practice one of the 2 resistors would be a variable

resistor

+

-


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Feedback

Feedback 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

As humans, much of our learning is through feedback;

e.g to pick up an object we stretch out our hand and at the same

time look at the hand and object, and the visual signal is fed back

to the brain to provide control for the hand

This is a continuous process of refining the position of the handrelative to the object

Vin Voutadd

feedbackfraction

gain Ao


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Op-amp and feedback

For an amplifier circuit a fraction of the output signal Vout is fed backand added to the input signal Vin. Since the amplifier has open-loop

gain Ao, it amplifies whatever voltage there is at the input, soVout = Ao x (amplifier input) = Ao x (Vin + Vout) which gives

Vout(1-Ao) = Ao x VinRearranging, Vout/Vin= Ao/(1-Ao) and

Vout/Vinis called the voltage gain

If the fraction is negative, then (1-Ao) is greater than unity and theoverall gain of the amplifier circuit is less than the open-loop gain ofthe op-amp itself. This process is known as negative feedback

Negative feedback can be achieved by feeding part of the output of theop-amp back to the inverting input

Although negative feedback reduces the gain of an amplifier, it has anumber of benefits: Increased bandwidth(the range of frequencies for which the gain is

constant)

Less distortion of the output(output signal directly prop to input signal)

Greater stability(less affected by environment in the circuit)


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Concept of virtual earth

Since the open-loop gain of the op-amp is very large, then the input

voltages at the non-inverting and inverting inputs must be almost the

same.

The non-inverting input is connected to the zero-volt line(earth line) and

therefore the input is zero

This means that the inverting input must be very nearly zero volts(or

earth) and so the point between Rin and Rf is known as a virtual earth

The input impedance of the op-amp is infinite which means that there is

no current in either the inverting or the non-inverting inputs, meaning that

all the current from or to the input signal to the circuit must passthrough the feedback resistor to the input


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Inverting op-amp configuration

Vin either positive or negative, is fed to V- of op-amp while V+ is earthed. Applicable to ac and dc voltage as well

A part or whole of the output is fed back to the inverting input of the op-amp by means of the resistor Rf. This is called negative feedback becausethe signal being fed back partly cancels the input signal

The resistors Rin and Rf act as a potential divider between the input and theoutput of the op-amp

This feedback voltage superimposes with the original input voltage,

resulting in a smaller value. The output is therefore also smaller

Inverting op amp with negative feedback


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Inverting op-amp with negative feedback

Current in Rin= current in Rf , hence p.d. across Rin = p.d. across RfRin RfwhereRin andRfare the resistances

The potential at P is zero (since virtual earth) and so

Vin0 = 0 Vout

Rin Rf

Hence the overall voltage gain of the amplifier circuit is given byVoltage gain A = Vout = - Rf or Vout = - VinRf

Vin Ri n Rin

The gain does not depend on the characteristics of the op-ampbut only on the 2 resistors,and since the output is the inverse of the input, the amplifier is referred to as an invertingop-ampand the output voltage is 180o out of phase() with the input voltage


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Non-inverting op-amp with negative feedback

For this configuration, the input voltage is applied directly to the non-inverting input

Negative feedback is achieved by means of the potential divider resistorsR1 and Rf

Based on Kirchhoffs 1st law, If = I1

Vf/Vout = R1/(R1 + Rf)

therefore Vout/Vf = (R1 + Rf)/R1but Vf is approximately equal to Vintherefore Vout/Vin = (R1 + Rf)/R1 = 1 + Rf/R1and the output is in phasewith the input

i.e. Voltage gain, A = 1 + RfR1 Vin Vout

Rf

R1

-

+

Vf

Al i d i i f i i


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Alternative derivation for non-inverting op-

amp

Vf = R1/(Rf+R1) x Vout.(1) Voltage input between the 2 inputs = VT

therefore VT = VinVf............(2)

Now Vout = A0 x VT

therefore VT = Vout/A0(3)

Equating (2) & (3), we get V0/A0 = VinVfand rearranging, Vf= VinV0/A0........(4)

Now substituting (1) into (4), we get R1/(Rf+ R1)Vout = VinV0/A0

Rearranging, Vin = V0(R1/(R1 + Rf)) + V0/A0

= V0[(R1/(R1 + Rf) + 1/A0]since A0 > 10

5, 1/A0 is negligible

Therefore A = V0/Vin = (R1 + Rf)/R1

= 1 + Rf/R1


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Example

The magnitude of the gain of an inverting amplifier is 25. The supplyvoltage to the operational amplifier is 9 V and the non-inverting input is

at earth potential.

Calculate the output voltage of the amplifier circuit for an input voltage at

the inverting input of:

(a) + 40 mV(b) - 80 mV

(c) - 1.2 V

Solution

(a) - 1.0 V

(b) + 2.0 V

(c) + 9.0 V (amplifier is saturated)


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Op-amp as a summing/adder circuit

Inverting op-amp with negative feedback can be used as a summing or addercircuitwhere a number of input voltages can be added up.


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cont..

In a mixer for example, several input from microphones, guitars and keyboards

can be added in a summing op-amp and amplified to produce a sound track. Three input voltages V1, V2 and V3 send currents I1, I2 and I3 through three

resistances R1, R2 and R3 respectively.

I1 = V1/R1, I2 = V2/R2 and I3 = V3/R3 If= 0 Vout = - Vout

Rf Rf Since current does not flow into the op-amp

If= I1 + I2 + I3

Vout/ Rf = V1/R1+ V2/R2 + V3/R3 hence the output is the sum of the 3 inputvoltages i.e. a summing circuit or an adder circuit

R1, R2 and R3 can be varied to increase the amplification of an input signal.

If Rf > R1 the signal through R1 (e.g. a keyboard) will have a greater effect thanother inputs.


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Output devices

Circuits incorporating op-amps produce an output voltage

If this output voltage is connected across some form of resistor, there will bea current from the output of the op-amp to the resistor, which can then be

used to operate output devices such as relays, warning lamps e.g LEDs,digital/analogue meters, motorsetc.

Output current usually cannot exceed about 25 mA otherwise the op-amp

would be destroyed. Op-amps ICs are designed and built to contain an outputresistor so that, should the output be shorted, the op-amp will not be

damaged.

R l i h d i


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Relay switch as an output device In order that electronic circuits e.g. an op-amp may be used

to switch on and off appliances that require large currents tooperate them, a relaymay be used.

A relay is an electromagnetic switch that uses a smallcurrent to switch on or off a larger current. The smallcurrent energises an electromagnet that operates contacts,switching on or off the larger current.(symbol for relay onright)

The diode D1 conducts only when the output is positive withrespect to earth and thus the relay coil is energised only whenthe output is positive. When the current in the relay coil isswitched off, a back e.m.f. is generated in the coil that coulddamage the op-amp. A diode D2 is connected across the coilto protect the op-amp from this back e.m.f.

The light emitting diode(LED) as an output


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The light emitting diode(LED) as an output

device There are many instances when all that is required of a sensor is a visible indicator

i.e. a lamp when a high or low is detected i.e. op-amp is saturated

Filament lamps not only dissipate a large amount of power(>100 mA) but also are

not very robust on multiple operations

The light emitting diode(LED) which is available in many colours, is asemiconductor device that is very robust, reliable and dissipates little power

A LED emits light only when it is forward biased and the typical maximum

current is 20 mA

A LED may be damaged if the reverse-bias voltage exceeds approximately 5 V

(called breakdown voltage) but is dependent on the device

A LED with an in-built resistor to protect the LED is often used to indicate the state

of output of an op-amp


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Example

A LED can take a current of 0.01 A, which produces a 2 V drop across

it. If an op-amp circuits output voltage is 9 V, what is the value of theprotection resistor that should be used?

Solution

R = 7/0.01 = 700 ohms


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Digital and analogue meters as output devices

When the output of an op-amp does not saturate, the output voltage

can be used to indicate the magnitude of whatever is being sensed e.g.level of fuel in a tank

A digital or analogue voltmeter for example connected between the op-

amp output and earth will indicate the output voltage

The output voltage will be proportional to the input to the processing

unit which is dependent on the magnitude of whatever is being sensed

As the digital or analogue voltmeter reads a potential difference andnot the quantity that is being monitored and measured, the voltmeterreading is unlikely to vary linearly with the change in this quantity and

hence the voltmeter needs to be calibrated

A calibration curve of for example the temperature against thevoltage needs to be plotted so that we can read off the temperaturecorresponding to a voltage reading directly


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Question

i


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Exercise

The LED's L1 and L2 emit light when the output from the appropriateoperational amplifier is positive and high. When the thermistor T, which has anegative temperature coefficient, is at 70C, the potential difference acrossthe resistor R is 3.5 V.

(i) Explain why, when the thermistor is at 70 C, L1 emits light and L2 does

not.(ii) The temperature of the thermistor is raised and there is a change of state

of one or more of the LED's. State and explain what change isobserved.

(iii) Suggest one use for the above circuit

Section VII 28 Direct Sensing

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