Manual for D1750.pdf

An Introduction to Transducers and Instrumentation

Curriculum Manual IT02

©2007 LJ Create. This publication is copyright and no part of it may be adapted or reproduced in any material form except with the prior written permission of LJ Create.

Lesson Module: 17.50 Version 0 Issue: ME706/F

IT02 An Introduction to Transducers and Instrumentation Curriculum Manual Addendum Sheet

Addendum Sheet

Please note that the following warning label has now been added to the D1750 trainer. This is to indicate the area of moving parts, and that fingers should be kept clear.

Keep fingers clear of all moving parts

Technical Publications Department LJ Create

IT02 An Introduction to Transducers and Instrumentation Curriculum Manual Contents

Chapter

Contents

Pages

Introduction

............................................................................................. i - iv

Basic Control Systems

Chapter 1

Basic Control Systems Equipment and Terms Used ..........1 - 16

Input Transducers

Chapter 2

Positional Resistance Transducers ...................................17 - 32

Chapter 3

Wheatstone Bridge Measurements ...................................33 - 52

Chapter 4

Temperature Sensors ........................................................53 - 76

Chapter 5

Light Measurement...........................................................77 - 98

Chapter 6

Linear Position or Force Applications............................99 - 114

Chapter 7

Environmental Measurement .......................................115 - 126

Chapter 8

Rotational Speed or Position Measurement .................127 - 152

Chapter 9

Sound Measurements ...................................................153 - 162

Output Transducers

Chapter 10

Sound Output................................................................163 - 172

Chapter 11

Linear or Rotational Motion.........................................173 - 190

Display Devices

Chapter 12

Display Devices............................................................191 - 208

Signal Conditioning Circuits

Chapter 13

Signal Conditioning Amplifiers ...................................209 - 234

Chapter 14

Signal Conversions.......................................................235 - 252

Chapter 15

Comparators, Oscillators and Filters ............................253 - 270

Chapter 16

Mathematical Operations .............................................271 - 290

Closed Loop Control Systems

Chapter 17

Control System Characteristics ....................................291 - 300

Chapter 18

Practical Control Systems ............................................301 - 334

An Introduction to Transducers and Instrumentation IT02 Contents Curriculum Manual

Appendices

Appendix A

Using a Multimeter ...................................................... 335 - 340

Appendix B

The Oscilloscope ......................................................... 341 - 362

IT02 An Introduction to Transducers and Instrumentation Curriculum Manual Introduction

Introduction

Introduction This comprehensive course of study is based on a single panel Transducer and Instrumentation Trainer, the DIGIAC 1750. The D1750 unit provides examples of a full range of input and output transducers, signal conditioning circuits and display devices. The unit is self-contained and enables the characteristics of many individual devices to be investigated, building to form complete closed loop systems. As each item is introduced there is a description of the principles of the device, together with practical exercises to illustrate its characteristics and applications. The treatment is non-mathematical and little previous knowledge is assumed, although it is expected that students will have a basic knowledge of electrical circuits and units, and electronic components and devices. It is the intention that at the end of this course the student will, with the knowledge gained, be able to select suitable components and interconnect them to form required closed-loop systems. Although the course has been laid out progressively it is sometimes necessary to make use of a device before a full investigation has been carried out. For instance, in order to investigate any input transducer, an input signal may be needed. This signal may be provided by one of the output transducers not yet covered. Also signal conditioning and display devices will be needed from an early stage. In the event of any difficulty, it is recommended that the student should skip forward to the relevant section to obtain further information.

i

An Introduction to Transducers and Instrumentation IT02 Introduction Curriculum Manual

Test Instruments A digital multimeter will be required when working through this module. The meter must have ranges to cover at least: DC voltage: DC current: Resistance:

200mV to 20V 1mA to 100mA 10Ω to 10MΩ

To complete the exercises you will need to be familiar with connecting, setting the range and obtaining readings from multimeters. If you are not familiar with the use of these instruments please refer first to Appendix A before carrying out any exercises. Note that some of the exercises in Chapter 4 (Temperature Measurement) will require the use of two multimeters. Some examinations of voltage waveforms will be required using an oscilloscope. You will be expected to be able to make the necessary adjustments and settings to obtain time related sketches of the waveforms examined. Recommendations for the settings of the various controls will be given where appropriate. Again, if you are not familiar with this instrument or the applications of it, please refer to Appendix B before attempting the relevant exercise. A function generator will be required to provide sinewave and square wave inputs to some circuits. This should have a range of frequencies covering at least 10Hz 1MHz, and output of 20Vp-p (with an internal attenuator to allow amplitude settings), and an output impedance of 50 Ω. The output lead should be terminated in standard 4mm banana plugs for ease of connection directly to the D1750 Trainer panel.

ii

IT02 An Introduction to Transducers and Instrumentation Curriculum Manual Introduction

The Module Power Supplies The D1750 Transducer and Instrumentation Trainer contains all of the power supplies needed to make it operate. You can switch these power supplies ON and OFF with the Power Supplies switch located on the rear panel.

Making Circuit Connections During each Practical Exercise in this manual, you will be asked to make circuit connections using the 4mm Patching Cords. Whenever you make (or change) circuit connections, it is good practice to always do so with the Power Supplies switch in the OFF position. You should switch the Power Supplies ON only after you have made, and checked, your connections. Remember that the Power Supplies switch must be ON in order for you to be able to make the observations and measurements required in the Exercise. At the end of each Exercise, you should return the 'Power Supplies' switch to the 'OFF' position before you dismantle your circuit connections.

Your Workstation Depending on the laboratory environment in which you are working, your workstation may, or may not, be computer managed. This will affect the way that you use this laboratory manual. If you are in any doubt about whether your workstation is computer managed, you should consult your instructor.

iii

An Introduction to Transducers and Instrumentation IT02 Introduction Curriculum Manual

Using this Manual at a Computer Managed Workstation In order to use this curriculum manual at a computer managed workstation you will require one of the following items: ♦ ♦

D3000 Hand-held Data Terminal. A personal computer (PC) that has been installed with computer managed student workstation software.

If you are working in a computer managed environment for the first time, you should first read the operating information that has been provided with your computer managed workstation. This tell you how to: ♦ ♦ ♦ ♦

Log onto the management system and request work. Make responses to questions in a computer managed environment. Hand in your work when completed. Log off at the end of your work session.

Whenever you see the symbol in the left-hand margin of this Curriculum Manual, you are required to respond to questions using your computer managed workstation. You should also record your responses so that you can review them at any time in the future. The following D3000 Lesson Module is available for use with this Curriculum Manual: D3000 Lesson Module 17.50

Using this Manual at a Workstation that is

not Computer

Managed

Whenever you see the symbol in the left-hand margin of this Curriculum Manual, you are required to answer a question. If your workstation is not computer managed, you should record your answer so that it can be subsequently marked by your instructor.

Good luck with your Studies.

iv

IT02 Basic Control Systems Equipment and Terms Used Curriculum Manual Chapter 1

Chapter 1 Basic Control Systems Equipment and Terms Used

Objectives of this Chapter

Having studied this Chapter you will be able to: 

State the difference between open loop and closed loop systems.



Write the expression for the overall gain of a negative feedback closed loop system.



Calculate the overall gain of a negative feedback closed loop system from given information.



List the basic components of a closed loop system and explain their functions.



Explain the meaning of terms associated with control system equipment.

1

Basic Control Systems Equipment and Terms Used IT02 Chapter 1 Curriculum Manual

1.1

Open Loop System Figure 1.1 represents a block diagram of an open loop system. A reference input, or command signal, is fed to an actuator which operates on the controlled variable to produce an output.

Reference I/P (Command Signal)

Actuator

Controlled Variable

O/P

Fig 1.1

The output magnitude depends on the magnitude of the reference input signal but the actual output magnitude for a particular input may not remain constant but may vary due to changes within or exterior to the system. For example, in a simple room heating application, a heater set for a certain output will result in a certain room temperature. The actual temperature will depend on the ambient temperature outside the room and also whether the doors and windows are open or closed.

2


1.2

Closed Loop System Figure 1.2 shows a basic block diagram of a closed loop control system. With this system, the output magnitude is sensed, fed back and compared with the desired value as represented by the reference input. Any error signal is fed to the actuator to vary the controlled variable to reduce this error. Reference I/P Error Detector

Feedback signal

Actuator

Controlled Variable

O/P

Sensor

Fig 1.2

The system thus tends to maintain a constant output magnitude for a fixed magnitude input reference signal. The feedback signal is effectively subtracted from the reference signal input to obtain the error signal and hence the system is referred to as a negative feedback system. The magnitude of the reference signal required for a particular output magnitude for a closed loop system will be greater than that required for open loop operation because the negative feedback reduces the overall gain of the system.

3


1.3

Gain in an Open Loop System

Gain G

Input Vi

Output Vo

Fig 1.3

Output Vo = G Vi

1.4

Gain = G

Gain in a Closed Loop System Error (Vi - HVo)

Input Vi

Feedback (HVo)

Gain G

Output Vo

Attenuator H

Fig 1.4

H = the fraction of the output fed back to the input The error signal = Vi - HVo The output Vo = G(Vi - HVo) = GVi - GHVo Vo + GHVo = GVi Vo(1 + GH) = GVi Vo G = Vi 1 + GH i.e.

4

Gain =

G 1 + GH


The gain is therefore reduced, and, if the gain G is very large, the formula simplifies to:G 1 Gain = = GH H If the gain of the amplifier (G) is high then the overall system gain is dependent only on the feedback fraction H.

1.5

Examples

1.5a

(i)

An amplifier has a gain (G) of 15 and a feedback loop with an attenuation 1 fraction (H) of . 30 Vo The loop gain of the system will be: Vi 15 15 G = = = 10 1 1 1 + GH 1 + 15 1+ 30 2

(ii)

An amplifier with a gain of 100 has 10% negative feedback (H = 0.1). Vo The loop gain of the system will be: Vi G 100 = = 100 = 9.1 1 + GH 1 + 100 × 0.1 1 + 10 1 Note that = 10, which is very nearly the same as the loop gain. H

An amplifier with a gain (G) of 20 has a feedback fraction (H) of 0.15. The loop gain of the system will be:

a 3

b

4

c

5

d

6.67

5


1.6

Practical Closed Loop Control System Figure 1.5 shows a block diagram of a practical closed loop control system. This shows signal conditioning blocks in the signal paths between the error detector and the actuator and between the sensor and the error detector.

Reference I/P

Error Detector

Signal Conditioning Signal Conditioning

Actuator

Controlled Variable

O/P

Sensor

Signal Conditioning

Display

Fig 1.5

It also shows a display which indicates the magnitude of the output variable and includes a signal conditioning block in the display path. Signal conditioning may consist of signal amplification, attenuation or linearising, waveform filtering or modification, conversion from analog to digital form, or may be a matching circuit. These may be necessary to convert the output from one circuit into a form suitable for the input to the following circuit, or to improve the system accuracy.

6


1.7

Controlled Variables For a particular industrial process there may be more than one controlled variable and each of the controlled variables will have its own closed loop control system. The controlled variable may be:-

Position (angular or linear) Temperature Pressure Flow rate Humidity Speed (angular or linear) Acceleration Light level Sound level The control system may operate using pneumatic, hydraulic or electrical principles and the sensors used for the measurement of the controlled variable must provide an output signal in a form suitable for the system in use. This will normally involve a conversion from one energy system to another and devices used to accomplish this energy conversion are referred to a TRANSDUCERS. Sensors and actuators are both forms of transducer, sensors representing input transducers and actuators representing output transducers. The DIGIAC 1750 unit is an electrical system and includes a full range of sensors, actuators, signal conditioning circuits and display devices. Used with this manual, the unit will introduce the student to the basic principles and characteristics of a comprehensive range of transducers and their application to practical closed loop control systems.

7


1.8

Glossary of Terms - Transducers Transducer:

A device which converts information from one energy system to another.

Sensor:

A device which senses, or measures, the magnitude of system variables. Normally also convert thetransducers. measured quantity into another energy system andthey hence they are also

Actuator:

A device which accepts an input in one system and converts it into another energy system, which is normally mechanical. These devices are also transducers.

Specification:

Data specifying the performance capabilities and requirements of equipment.

Accuracy:

The error present in a measurement as compared to the true value of the quantity.

Sensitivity:

The ratio of the output of a device compared to the magnitude of the input quantity.

Resolution:

The largest change in the input that produces no detectable change in the output; for example, the degree to which a system can distinguish

Range: Bandwidth:

between adjacent values or settings. A statement of the values over which the device can be used and within which the accuracy is within the stated specification. The range of input signal frequencies over which a device or circuit is capable of being operated while providing an output within its stated specification.

Transfer function: The mathematical relationship between two variables that are related. Normally the relationship between the input and output of a system. Linear:

A relationship between two quantities that have a constant ratio; for example, a graphical straight line relationship.

Non linear:

A relationship between two quantities that cannot be described by a linear relationship.

Linearity: Response time:

A measure of the deviation of a measurement from an ideal straight line response of the same measurement over the same range. The time taken for the output to reach, or be within a rated percentage of, a new final value, after the input has been changed.

9


1.9

Glossary of Terms - Signal Conditioning Circuits Amplifier:

A circuit having an input and output that are related linearly and with the output greater than the input. The circuit may operate on both DC and AC circuits

Offset:

For amplifier, input zero,these the output may not be zero. ThisaisDC referred to aswith the the offset. With amplifiers, a control is provided and labeled: "Offset" or "Set Zero" to set the output to zero with the input zero, before the amplifier is used.

Gain:

The ratio of output to input for a circuit.

Attenuator:

A circuit having an input and an output that are related linearly and having an output less than the input.

AC Amplifier:

An amplifier that will amplify alternating signals only.

Differential Amplifier: A voltage amplifier having two inputs and where the output voltage magnitude is proportional to the difference in voltages between the two inputs.

10

Summing Amplifier:

A voltage amplifier having multiple inputs, the output being proportional to the sum of the various applied inputs.

Inverter:

A voltage amplifier having the polarity of the output the reverse of the input. The output magnitude may be the same as the input (gain of -1), or there may be voltage gain associated with the polarity reversal.

Power Amplifier:

An amplifier with a large current output capability.

Buffer Amplifier:

An amplifier having unity gain (output = input), and having a high input impedance and a low output impedance.

Comparator:

A circuit having two inputs A & B and an output that can be in one of two possible states depending on the magnitudes of the inputs. With input A greater than B, the output will be in one state (possibly high voltage). With input A less than B, the output will be in the alternative state (low voltage).

Oscillator:

A circuit producing an alternating output at a particular frequency.

Alarm Oscillator:

A circuit having an input and an output. With the input magnitude below a certain level, the output is zero. When the input exceeds the threshold the output is an alternating voltage.


The transfer characteristic of a non-linear device for increasing input voltages may be different from the characteristic for decreasing input voltages. The result is a 'hysteresis loop,' as shown in figure 1.7(a) below. For a switching circuit, the term 'hysteresis' normally refers to the input switching voltages. The input voltage to cause switching for rising input voltages is arranged to be higher

Hysteresis:

than that to produce switching for falling input voltages (see figure 1.7(b) below). The difference between the input voltages is referred to as the hysteresis.

Output

Output for fallinginput

Rising Voltage I/P Output for risinginput

Hysteresis Voltage

Switchin Levels

Falling Voltage

Input (a)

(b)

Fig 1.7

Latch:

A circuit having two possible output states depending on the magnitude of the input voltage. When operated with the input level sufficient to change the output to its alternative state, the output is held (or latched) in this state irrespective of the subsequent magnitude of the input voltage.

Filter:

Circuit designed to allow signals of a selected frequency range to pass through and stop all others.

Low Pass Filter:

A circuit allowing low frequency signals to pass while blocking the passage of higher frequencies.

High Pass Filter:

A circuit allowing high frequency signals to pass while blocking the passage of lower frequencies.

11


12

Band Pass Filter:

A circuit allowing signals over a selected range of frequencies to pass while blocking the passage of signals at both lower and higher frequencies.

Full-Wave Rectifier:

A circuit converting an alternating waveform into a unidirectional or DC waveform.

V/F Converter:

A circuit converting a DC input voltage to an alternating voltage, the frequency being dependent on the magnitude of the DC input voltage.

F/V Converter:

A circuit converting an alternating input voltage to a direct voltage output, the output voltage magnitude being proportional to the frequency of the input voltage.

V/I Converter:

A circuit converting an direct input voltage into an output current, the current magnitude depending on the input voltage.

I/V Converter:

A circuit converting an input current into an output voltage, the voltage magnitude being dependent on the magnitude of the input current.

Integrator:

A circuit having an output voltage that is proportional to the product (input voltage x time).

Differentiator:

A circuit having an output voltage that is proportional to the rateof-change of the input voltage.

Sample and Hold:

A circuit with input and output. In the sample state, the output voltage is equal to and follows any changes in the input voltage. In the hold state, the output voltage is held at the value of the input signal at the instant the "hold" signal was initiated.

Ultrasonic:

A signal at a frequency above the normal audio range and hence inaudible to the human ear (normally >16kHz).


Student Assessment 1

Reference I/P 1

2

Actuator

4

3

O/P

Sensor

5 Practical Closed Loop Control System

6 Fig 1

1.

2.

3.

4.

In Fig 1 block numbered 1 is a:

a

signal conditioner

b

display

c

controlled variable

d

error detector


a

signal conditioner

b

display

c

controlled variable

d

error detector


a

signal conditioner

b

display

c

controlled variable

d

error detector

A closed loop control system has an open loop gain G and a negative feedback factor H. The expression for the overall gain of the system with feedback applied is: H G 1+ GH a b 1 + GH c d 1+ GH 1 + GH G

Continued ...

13


Student Assessment 1 Continued ...

5.

An electrical control system produces an output of 50V for a reference input of 1V under open loop conditions. If a fraction 0.1 of the output is fed back as negative feedback to the input, the input reference voltage now required to produce the same output of 50V is:

a 6.

7.

8.

b

8.33V

c

6V

a

waveform modification

b

error detection

c

amplification

d

attenuation

The term accuracy refers to the:

a

conversion of energy from one form to another

b

largest change of input that produces no change in the output

c

error present in a measurement compared to the true value

d

mathematical relationship between the input and output of a system

A device has an output that is variable in 250 equal increments and has a maximum output of 10V. The resolution of the device is:

0.25V

b

0.1V

c

0.04V

d 0.4V

A filter which passes all signals at frequencies above 10kHz is called a:

a

high pass filter

b

band pass filter

c

low pass filter

10. An example of a band pass filter response is one which will:

14

d 5V

Which of the following is NOT a function of a signal conditioning circuit?

a 9.

10V

a

pass all signals at frequencies below 10kHz

b

pass all signals at frequencies above 12kHz

c d

pass all signals at frequencies below 10kHz and above 12kHz pass all signals at frequencies below 12kHz and above 10kHz

d

band stop filter



11. An electrical comparator is a circuit with:

a

one input and one output

b

two inputs and one output

c

one input and two outputs

d

two inputs and two outputs

12. A comparator with hysteresis is one which will:

a

hold the output state after the input voltages are removed

b

change output state at precisely the same differential between the input voltages

c

change output state at different rising input voltage to falling input voltage

d

change output state if the input signals are within the correct range of frequencies

13. A comparator with latch is one which will:

a

hold the output state after the input voltages are removed

b

change output state at precisely the same differential between the input voltages

c

change output state at different rising input voltage to falling input voltage

d

change output state if the input signals are within the correct range of frequencies

14. An open-loop control system is one in which:

a

there is not feedback from output back to input

b

negative feedback is applied from output to input

c

positive feedback is applied from output to input

d

there is no connection between input and output

15


16

IT02 Curriculum Manual

Positional Resistance Transducers Chapter 2

Chapter 2 Positional Resistance Transducers


Having studied this Chapter you will be able to: 

Describe the basic construction of rotary and slider variable resistors.



State that the resistance section may be either a carbon track or wirewound.



Describe the difference between a logarithmic and a linear track.



Draw the basic characteristics of output voltage against variable control setting.



Describe the effect on the output voltage of loading the output circuit.



Compare the application of a carbon track variable resistor to the wirewound type.

Equipment Required for

• •

DIGIAC 1750 Transducer and Instrumentation Trainer. 4mm Connecting Leads.

this Chapter

•

Digital Multimeter.

17


2.1


Variable Resistor Construction A variable resistor consists of a "track" having a fixed overall resistance with a "wiper" which can be moved to make contact with any point along the track. In the carbon type, the total track resistance is varied by adjusting the proportion of non-conducting material to carbon in the compound during manufacture. This will produce a track of constant resistance along its length, so that any section of the track will have the same resistance as any other similar section. The track will be linear. Variable resistors intended for use in audio applications, where subjective appreciation of sound amplitude (loudness) is proportional to logarithmic scales, are made with similar logarithmic (non-linear) scales. The resistance along the track is not a linear relationship, increasing with the square of the rotation of the spindle, or movement of the slide wiper (R ∝ S2, where S is the setting of the wiper). A close approximation is made to the ideal logarithmic characteristic by using three or four sections of track with different resistance slopes. Non-Linear variable resistors are not suitable as positional transducers and are therefore not included on the DIGIAC 1750 Trainer facilities. The track can be laid out on a rotary or a straight base, as in Fig 2.1. CarbonTracks

Wiper

Carbon Track

Wirewound Carbon Track

Wiper

Wiper

Wire Coil

Metal strip Connections

Slider Type

Connections

Connections

Rotary Types

Fig 2.1

For higher power applications the track may be wire wound, with the wiper making contact with the top edge of a coil of resistance wire.

18


2.2


Linear Variable Resistor Characteristics A variable resistor can be used to provide a variable voltage. A steady voltage is applied across the ends of the fixed track. The wiper then picks off a variable voltage at the contact point with the track (with respect to the end of the track). Used in this way the variable resistor is called a potentiometer.

+5V

10 9 8 7 6 5 4 3

0V

+5V

Variable Contact Setting

O/P

O/P

2 1

0V

1 2 3 4 5 6 7 891 0 Variable Contact Setting

Fig 2.2

With a dual polarity voltage source, the polarity and magnitude of the output voltage will depend on the direction of movement of the wiper from its central position, as shown in Fig 2.3.

+5V

0V

10 9 8 7 6 5 4 3

-5V

2 1

+5V

Variable Contact Setting

O/P 0V 1 2 3 4 5 6 7 891 0 Variable Contact Setting

O/P -5V

Fig 2.3

Note that the position of the variable resistor spindle (or slider) setting is indicated by the output voltage from the potentiometer.

19


2.3


Practical Exercise Variation of Output Voltage with Setting of Rotary Potentiometer C +12V

CARBON TRACK 5 4

B

Digital Meter

C

6 7

3

8

2

9

B A

10

1

V

100k

V 0V 0V

Digital Meter +12V

A

SchematicDiagram

PhysicalLayoutDiagram

Fig 2.4 

Locate the 100k Ω variable resistor on the DIGIAC 1750 Trainer (bottom left-hand corner). Connect the circuit as shown in Fig 2.4 using the power supply facilities at the bottom of the panel and the 20V DC range of a digital multimeter.



Set the 100k Ω rotary resistor control fully counter-clockwise to setting 1 as shown in Fig 2.4. Note that the dial is not marked with numbers on the printed panel. These numbers have been shown in Fig 2.4 to make it easier to follow these instructions and collate results.



After ensuring that the voltage adjustment is correctly set switch ON the power supply (switch on the rear of the unit just above the main power socket).



Note the output voltage as indicated on the digital multimeter and record in Table 2.1.

Control Setting

1

Output Voltage Table 2.1

20

2 V



V

3 V

4 V

5 V

6 V

7 V

V

8 V

9

10

V

Set the rotary control to "2" and repeat the reading, recording the result in again Table 2.1.





Repeat the reading and recording for all other settings of the rotary control.



From the results recorded in Table 2.1 plot the characteristic of the 100k Ω variable resistor on graticule of Graph 2.1 below. 12 11 10 Output 9 Voltage (volts) 8

7 6 5 4 3 2 1 0

1

2

3

4

5

6

7

8 91 0 resistor setting

Graph 2.1 Characteristic of a Linear Rotary Carbon Potentiometer

Note that it is not easy to be precise with your setting of the variable resistor and this may result in the plotted points not following a smooth relationship. You should draw the best compromise to show the characteristic as you believe that it should be. At the ends of the track the wiper comes into contact with the terminal connections to the track, causing non-linearity at both ends. From setting 2 through setting 9 the variation of voltage should be fairly linear. Voltage across this section (V9 - V2) = Voltage per division ( 2.3a

V9

− V2

9-2

)=

V V

Enter your voltage per division. 

Switch OFF the power supply.

21


2.4


Practical Exercise Variation of Output Voltage with Setting of Slide Potentiometer C

SLIDE

+5V

C B

1

B

V

-5V

2

4

5

6

7

8

9

10

A

10k

V

Digital Meter

Digital Meter

0V

A

3

-5V

0V

SchematicDiagram

+5V


Fig 2.5 

The 10k Ω slide potentiometer on the DIGIAC 1750 Trainer is just above the rotary potentiometers. Connect the circuit as shown in Fig 2.5 using the power supply facilities at the bottom of the panel and the 20V DC range of your digital multimeter.



Set the 10k Ω slide resistor control to the left to setting 1 as shown in Fig 2.5. Note that the marked numbers are again not on the printed panel.



Switch ON the power supply.



Note the output voltage as indicated on the digital multimeter and record in Table 2.2.

Control Setting

1

Output Voltage

2 V

V

3 V

4 V

5 V

6 V

7 V

V

8 V

9

10

V

Table 2.2  

22

Set the control to "2" and repeat the reading. Repeat the readings for all other settings of the slide control, recording the result in Table 2.2.





From the results recorded in Table 2.2 plot the characteristic of the 10k Ω slide resistor with dual polarity supply on graticule of Graph 2.2 below. +5

Output +4 Voltage (volts) +3 +2 +1 0

-1 -2 -3 -4 -5

1

2

3

4

5

6

7

8

9

10

resistor setting

Graph 2.2 Characteristic of a Linear Slide Carbon Potentiometer 



Switch OFF theand power supplysupply and remove potentiometer the power panels.the connections between the slide Use the digital multimeter on a suitable range (20k Ω) to measure the resistance between terminal A and wiper B with the wiper set to position 9: Resistance R9 =



k

Move the wiper to position 2 and repeat the resistance measurement: Resistance R2 =

k

Resistance between settings 9 & 2 = R 9 - R2 = Voltage between settings 9 & 2 = V9 - V2 = Voltage per k 2.4a

=

V9 -V=2 (R 9 -R 2 )k Ω

k V

V/k

Enter your voltage per k

23


2.5


Effect of Loading Consider a 10kΩ variable resistor connected to a 10V supply with the wiper in its central position. There will be a resistance of 5k Ω from the wiper to each end of the track (Fig 2.6(a)). If a 5kΩ fixed resistor is connected across the output then it will be in parallel with the lower half of the potentiometer (Fig 2.6(b)) and will draw current through the upper half of the potentiometer. This causes a higher voltage drop across the upper half of the track than the lower half (Fig 2.6(c)).

+10V

+10V

5k Ω

+10V

5k Ω

6.67V

5k Ω

3.33V O/P

5V O/P

5k Ω 0V

5k Ω

5k Ω

0V

(a)

2.5k Ω

3.33V

0V

(b)

(c)

Fig 2.6

Another way of looking at this is that the shunting effect of the 5k Ω load resistor is to reduce the total resistance of the lower half to 2.5k Ω (Fig 2.6(c)). Only one third of the applied voltage will be dropped across the lower half and two thirds across the upper. The variations of resistance as the wiper is moved will be quite complex and the voltage at the output will be non-linear.

24


2.6


Practical Exercise Effect of Loading on the Potentiometer Output Voltage With the power supply switched OFF and no connections made to any components, measure the resistance of the 100k Ω rotary variable resistor between contact A and the wiper as it is set to the marked points on its scale. Use a suitable scale (200k Ω) on your digital multimeter and record the results in Table 2.3 overleaf in the row marked "Load Resistance".



The 100kΩ resistor is to be used as a load resistance across the output of a 10k Ω position sensing variable resistor. Connect the circuit as shown in Fig 2.7 but initially leave out the lead from contact C of the 100k Ω resistor to contact B of the 10k Ω so that the load is not connected across the output.



C +12V 5

CARBON TRACK 6 C

WIREWOUND TRACK 6 5 C

7

4

Omit lead at first

7

4

3

8

B

3

8

B

2

9

A

2

9

A

B 10k Ω

1

10

1

100k

10

V

Digital Meter

100k Ω 0V

0V

V Digital Meter

10k +12V

A

SchematicDiagram

Physical LayoutDiagram

Fig 2.7 

Switch the power supply ON and adjust the 10k Ω rotary resistor to give an output of 6V. Do not re-adjust this setting during the rest of this exercise.



Set the 100k Ω resistor fully clockwise (10) and connect the missing lead from contact C of the 100kΩ resistor to contact B of the 10k Ω so that the load is connected across the output of the positional sensor (10kΩ resistor).



Note the output voltage and record in Table 2.3.

25


Control Setting

10

Output Voltage

9 V

Load Resistance

8

V

kΩ

7

V

kΩ


V

kΩ

6

5

V kΩ

V

4 V

kΩ

3

V

kΩ

V

kΩ

2

1

V kΩ

kΩ

kΩ

Table 2.3 

Change the setting of the 100k load resistor and record the effect as the load resistor is set to each marked position in Table 2.3.



From the information in Table 2.3, plot the characteristic of Output Voltage against Load Resistance on the graticule of Graph 2.3 below: 7 Output 6 Voltage (volts) 5

4 3 2 1 0

10

20

30

40

50

60

70

80

90

100

Load Resistance (k

Graph 2.3

C

MOVING COIL METER

+12V CARBON TRACK 6 5 C 7

4

8

3

B 10k Ω

9

2 1

V 100k Ω 0V

26

A

10k

Digital Meter

+

V 0V

0V

+12V

A

SchematicDiagram

Fig 2.8

Digital Meter

10

B

5 -10


L J

0

5 +10



Do not alter the setting of the 10k

resistor.

With the Load Resistance (100k Ω resistor) removed from circuit connect the panel mounted Moving Coil Meter as in Fig 2.8 and switch ON the power supply.





Note the effect on the output voltage reading of having the analog type meter connected in circuit as well as the digital multimeter. Multimeter voltage reading with the Moving Coil Meter connected =

Compare this reading with the results on the characteristic curve of Graph 2.3 and read off the graph the loading resistance presented by the Moving Coil Meter to the output:



Loading resistance of the Moving Coil Meter = 2.6a

V

k

Enter your value of the loading resistance of the Moving Coil Meter in k .

What you have observed here is a problem which can be very misleading if you are not aware of the difficulties of using a low impedance meter to take measurements in a high impedance circuit. The problem can be overcome by using a Buffer Amplifier. MOVING COIL METER 5

CARBON TRACK 6 C

5

BUFFER #1

7

4

-10

3

8

2

9 1

I/P A

+10

+ +VIN

-

10k

0V 0V

5

O/P

B

10

0

+12V

V

L J

Fig 2.9 

Modify the circuit to include Buffer #1 as in Fig 2.9 and note the effect on the output voltage as indicated by both meters.

Output voltage = 2.6b 

V (digital)

V (analog)

Enter your value of analog output voltage with Buffer #1 in circuit in volts.

Switch OFF the power supply. 27


2.7


Resolution Resolution has been defined as the largest change in the input which does not cause a change in the output. Alternatively it can be defined as the smallest change in input which does cause a change in output. For the carbon track resistor this value is very small the individual particles of carbon are tiny and variations of resistance can besince considered to be infinitely small. The resolution for a wirewound resistor is not so good, since, as the wiper is moved, it has to jump from one turn of the wire coil to the next. The output voltage therefore increases in steps equal to the applied voltage divided by the number of turns if the wiper only makes contact with one turn at a time. Wiper

1

2

3

4

5

(a)

1

2

3

4

5

(b)

1

2

3

4

5

(c)

Fig 2.10

This may not be quite the case, since the wiper may make contact with two or more turns at once as in Fig 2.10(b). The mathematical treatment of this will depend on the thickness of the wire (power rating) and the size of the wiper contact (current rating). Multi-turn wirewound tracks will largely overcome this problem.

2.8

Comparison of Carbon with Wirewound Track Carbon

Cheap Good Resolution Can be made miniature

28

Wirewound

High Current Ratings Durability (Reliability)


2.9


Practical Exercise Servo Potentiometer A special positional potentiometer is mounted on the experiment board which has a very large arc of turning, approaching 360 °. It is called a Servo Potentiometer.

SERVO POTENTIOMETER

+5V

2.2kΩ

O/P

20kΩ V

-5V

0V

O/P

Schematic Diagram

V

Layout Diagram

0V

Fig 2.11

To bring the potentiometer scale into contact with the drive wheel on the shaft, press and release the mounting at the point arrowed in Fig 2.11. The potentiometer can then be turned manually with the shaft, using one of the large wheels, such as the Hall Effect Sensor Disk. The potentiometer can be turned directly from the dial, manually, if preferred. The ±5V input voltages to the Servo Potentiometer are connected internally. 

Connect a digital multimeter on the 20V DC range to the output of the potentiometer as shown in Fig 2.11.



Turn the potentiometer to find the maximum positive output voltage position. Note the value of this voltage and the angle, as given on the potentiometer dial, in the first column of Table 2.4 overleaf.

29


150

Control Dial Setting Output Voltage

V

V

120

V

90

V


60

V

360 0

30

V

V

330 -30

V

300 -60

270 -90

V

V

240 -120

V

210 -150

V

V

Table 2.4 

Rotate the dial in steps of 30 ° clockwise from the maximum voltage position (beginning with 150°), noting the output voltage at each step and recording the values in Table 2.4.



At the final step note the angle from the dial setting and the value of the maximum negative voltage setting.



From the information recorded in Table 2.4, draw the characteristic of the output voltage/dial setting of the Servo Potentiometer on the graticule provided below:

Output Voltage

+5 +4 +3 +2 +1

-180 -150 -120 -90 -60 -30 30 60 90 120 150 180 -1 -2

Dial Setting

-3 -4 -5

Graph 2.4 Characteristic of the Servo Potentiometer

2.9a

30

Enter the dial setting in degrees for the maximum positive output voltage.



Student Assessment 2 1.

2.

A log. scale (logarithmic) potentiometer is most suitable for use as a:

a

linear positional transducer

b

linear voltage controller

c

sound volume controller

d

rotary positional transducer

A variable resistor is stated to be wirewound. It is most likely to be:

a

linear because the resistance will be constant along the wire's length

b

linear as long as it is of the slide (not rotary) type

c

logarithmic because of the inductance of the coiled wire

d

non-linear because of the difficulty of making wire of a constant resistance 5

6 7

4

8

3

9

2 10

1

3.

A linear potentiometer (as in Fig 1 above) has control settings of 1-10 and is connected to a +9V, 0, -9V supply, with the -9V connection made to the end marked setting "1". The voltage at the setting marked "6" will be:

a 4.

-1V

b

0V

c

+1V

d +1.8V

A 10k linear potentiometer is connected to a 12V supply. With the output circuit loaded by a 5k load, the output voltage in the mid position (A) and maximum setting (B) will be:

a 5.

Fig 1

A = 6V, B = 12V b

A = 3V, B = 12V c

A = 4V, B = 8V

d A = 4V, B = 12V

A wirewound variable resistor consists of 1200 turns of wire. Assuming that the wiper only makes contact with one turn at any time, and the applied voltage across the track is 10V, the resolution in terms of the output voltage will be:

a

8.3mV

b

10mV

c

12mV

d 16.7mV

Continued...

31



Student Assessment 2 Continued...

100kΩ A=1

10V

A

B

100kΩ

EMFSource

BufferAmplifier

Fig 2

6.

7.

The circuit of Fig 2 consists of a source of EMF of 10V with an output resistance of 100k . The Buffer Amplifier has a gain of 1 and an input impedance of 100k and is not initially connected to the source. What voltage would you expect to get at: A - with a digital multimeter with an input resistance of 10M (A1), A - with a moving coil meter with a resistance of 10k (A2), and - using the same moving coil meter as in A2 above but connected to the supply via the buffer amplifier (B):

a

A1 = 10V, A2 = 1V, B = 10V

b

A1 = 9.9V, A2 = 0.9V, B = 10V

c

A1 = 9.9V, A2 = 0.91V, B = 5V

d

A1 = 10V, A2 = 1V, B = 0.83V

Which of the following is NOT an advantage of the carbon track variable resistor when compared with the wirewound type?

a

32

cost

b

reliability

c

small size

d resolution


Wheatstone Bridge Measurements Chapter 3

Chapter 3 Wheatstone Bridge Measurements


Equipment Required for this Chapter


State the principles of the basic Wheatstone Bridge circuit for resistance measurement.



Describe the term "null balance".



State and apply the expression for calculating an unknown resistance from the Bridge values at balance.



Discuss the factors affecting the resolution and accuracy of measurements.



Discuss the reason for the three-wire resistance circuit.



Apply null methods to voltage measurements.



Make resistance and voltage measurements using the DIGIAC 1750 facilities.

• • •

DIGIAC 1750 Transducer and Instrumentation Trainer. 4mm Connecting Leads. Digital Multimeter.

33


3.1


Wheatstone Bridge Circuit Fig 3.1 shows the basic Wheatstone Bridge circuit, consisting of four resistors and a sensitive center zero meter connected to a DC source.

I I2

I1 R1 DC Supply

Im

R3

R2 G R4

Fig 3.1

R1, R2 & R3 are accurate, close tolerance, resistors. R3 is variable and calibrated over its full range. R4 is the unknown resistor to be measured.

3.2

Null Balance During measurement, R3 is adjusted until there is no current (Im) flowing in the galvanometer circuit. The galvanometer current is zero or "null". Under these conditions, the bridge is said to be "balanced". Hence the term "null balance". The purpose of the galvanometer is to "detect"the presence of the null condition. From the known values of R1, R2 & R3 at balance, the value of R4 can be calculated from:R2 R4 = x R3 R1 The ratio of the values of resistors R2:R1 sets the range, so that values of the unknown resistor R4 which are larger or smaller than the variable resistor R3 can be measured. There is no limit to the range of values which can be measured. Any inaccuracy in the values of the ratio arm resistors R1 & R2, and also in the standard variable resistor R3, will result in errors in the measured value of R4. Since no current flows in the "null introduced by this part of the circuit.

34

detector"branch

at balance no error can be


3.3


Deriving the Formula With no current in the galvanometer circuit, the voltages at either end of it must be the same. This means that the voltages across R1 & R2 must be the same and similarly those across R3 & R4. With no current in the galvanometer, the current in R1 must be the same as that in R3 and the current in R2 must equal that in R4. If current I1 flows in R1 & R3 and current I 2 flows in R2 & R4:I1R1 = I2R2..................................................................... (i) I1R3 = I2R4.................................................................... (ii) Dividing:

(i) ÷ (ii) I 1 R1

=

I1 R3 ∴

R1

I 2 R4 =

R3 ∴

R4

I 2 R2

=

R2 R4 R2 R1

× R3

The unknown resistance R4 depends on the ratio R2:R1 and the value of R3 at balance. The resistors R1 and R2 are normally referred to as the "ratio arms"of the bridge. Note 1. The value of the supply voltage or the magnitude of the currents flowing in the resistors does not affect the result. This means that the supply voltage need not be stabilized, and that the circuit currents can be kept to low values for a component where the self heating effect of the current flowing could affect the result. 2. The galvanometer current accuracy is unimportant, since, under balanced

conditions, the current in it is zero. The main characteristics required for the galvanometer are a low resistance and a high sensitivity so that a small deviation of voltage from zero produces a large scale reading.

35


3.4


The Three Wire Resistance Measuring Circuit With some resistance transducer circuits, the transducer may be situated a relatively large distance from the bridge circuit, and the resistance of the connecting leads may be significant and could affect the results. For these situations the three wire connection arrangement is used.

R1

R2

DC Supply

G

R1

Long leads 2-wire

R3

DC Supply

R2 G

Long leads 3-wire

R3

R4

(a)

R4

(b)

Fig 3.2

Fig 3.2 (a) shows the circuit with a resistance transducer R4 situated remotely from the bridge and connected via two wires. The resistance of these wires will be included in the measurement of R4. Fig 3.2 (b) shows the three wire arrangement. One of the wires to the transducer is now included in the R2 circuit and the other is in the R4 circuit. The resistance of both circuits will therefore be increased equally and the effect on the balance condition will be minimized, provided that the resistances of R2 and R4 are of similar magnitudes. The extra wire in the galvanometer circuit will have no effect on the reading, since there is no current flowing in it at the balance condition.

36


3.5


The DIGIAC 1750 Facilities Fig 3.3 shows the Wheatstone Bridge layout provided with the DIGIAC 1750 unit.

WHEATSTONE BRIDGE

Fine Setting

D

unlock

12k

C

lock 3

A OUT

B

Coarse Setting

Switch

IN 1V

0V

Rx

Fig 3.3

A high quality 10-turn potentiometer fulfills the functions of the resistors R1 & R3 for resistance, or a potentiometer for voltage measurements. The track resistance of 10kΩ has a maximum non-linearity of 0.25%. The "Fine" dial is calibrated 0 - 100 in steps of 2, and the "Coarse" reading is calibrated 0 - 10, thus enabling readings to be estimated from the dial with a discrimination of 1:1000, representing a resolution of 10Ω. Reading the dial: If the number in the window (coarse setting) is 3 and the fine setting is on 74, then the dial reading is 374. The resistance between the 0V terminal and A (the wiper) is 10Ω x 374 = 3.74kΩ.

A close-tolerance 12kΩ resistor (R2) and an unknown resistor Rx (R4) are provided for resistance measurement. A switch open circuits the unknown resistor Rx to allow the measurement of other unknown resistors which can be connected between socket C and the 0V terminal. An accurate standard voltage of 1V is available at socket B. The moving coil meter can be used as a center zero indicating instrument. Since it is arranged as a 10V voltmeter its sensitivity is insufficient for a direct application as a galvanometer. This problem can be overcome by using a differential amplifier followed by a high gain DC amplifier from the signal conditioning circuits.

37


3.6


Practical Exercise Measurement of Resistance Fig 3.4 shows the layout diagram required for setting up the null detector.

DIFFERENTIAL AMPLIFIER O/P B

-

A

+

A-B

MOVING COIL METER

5

AMPLIFIER #2 I/ P

-10

O/ P

0

5 +10

+ -

.5

+

.4

OFFSET

.7

.8

.3

1 100 10

.6

.9

.2

GAIN COARSE

.1

1.0

0V

L J

GAIN FINE

Fig 3.4

Initially the amplifier and meter configuration which forms the sensitive galvanometer must be set up so that zero input produces zero output when the gain is set to maximum. 

Connect the meter and amplifiers as shown in Fig 3.4 with the + & - inputs to the Differential Amplifier short circuited so that the input is zero. Set the Amplifier #2 GAIN COARSE control to 10 and the GAIN FINE to 1.0.



Switch the power supply ON and adjust the OFFSET control so that the moving coil meter indicates approximately zero. Then set the GAIN COARSE control to 100 and re-adjust the OFFSET control for zero output precisely.

You will find that this adjustment is very sensitive. That is why you were instructed to obtain an approximate setting with the gain set to 10 first. Note The setting of the offset control may require adjustment as the temperature of the unit varies during use and it is advisable to use the above procedure to check and re-adjust as necessary at regular intervals.

38



WHEATSTONE BRIDGE D

+5V DIFFERENTIAL AMPLIFIER

12k

C A B

3

OUT

B

IN 1V

A-B

A

Rx

0V

O/P

+

MOVING COIL METER

5

AMPLIFIER #2 I/ P

-10

O/ P

0

5 +10

+ -

.5

+

.4 1 100 10

OFFSET

GAIN COARSE

.6 .7

.8

.3

.9

.2 .1

0V

1.0

L J

GAIN FINE

Fig 3.5 

With the switch on the Wheatstone bridge circuit set to IN (connecting the unknown resistor in circuit) set the Amplifier #2 GAIN COARSE control to 10 and connect the circuit as shown in Fig 3.5.



Adjust the control of the 10-turn variable resistor so that the moving coil meter reading is approximately zero, then set the GAIN COARSE control to 100. Finally adjust the 10-turn resistor control accurately for zero meter (null) reading to balance the bridge.

Reading the dial: If the number in the window (coarse setting) is 3 and the fine setting is on 74, then the dial reading is 374. 

Note the resistor dial reading (overleaf).

This represents the resistance R3 in the theoretical circuit considered earlier.

39



=

Dial reading Resistance R3 = 10 x dial reading

=

Resistance R1 = 10,000 - R3

=

Resistance R2 = 12,000Ω Unknown resistance Rx =

3.6a

R2 R1

x

R3

=

Enter your value for the unknown resistor Rx in k

Carry out further resistance measurements on the 10kΩ slide variable resistor to obtain familiarity with the equipment and its adjustments as follows: 



Set the Wheatstone Bridge switch to OUT to remove the unknown resistor Rx from the circuit. Connect the 10k Ω Slide variable resistor terminals A & B to the Wheatstone Bridge circuit connections C & 0V. With the 10k

Ω

resistor control set to maximum, measure its resistance as

follows:1. Check that the amplifier offset is set correctly and adjust if necessary.





2.

With Amplifier #2 GAIN COARSE control set to 10, obtain an approximate balance by adjusting the 10-turn resistor.

3.

Set Amplifier #2 GAIN COARSE control to 100 and obtain final balance. Note the dial reading and enter the value in Table 3.1.

Repeat the procedure to measure the resistance of the 10k Ω resistor for all settings from 9 through 1, recording the dial readings at balance in Table 3.1. Calculate the resistance corresponding with each reading, recording the results in Table 3.1. R2 is still 12kΩ.

Note: Since the quoted accuracy of the 10-turn variable resistor is 0.25%, this represents 1 part in 400. There is no reason for giving results to any more than four significant figures. 

40




10kΩ Resistor Setting

Dial Reading at Balance

R3 (10 x Dial)

10

R1 (10kΩ - R3)

R4 =

R2 R1

x

R3

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

9 8 7 6 5 4 3 2 1 Table 3.1

3.6b

Enter your value for the 10k

variable resistor at the setting 5 in k .

C

1kΩ B A Fig 3.6

Note that a 1kΩ resistor is connected in series with the wiper of all potentiometers on the D1750 Trainer. This prevents damage to the potentiometer in the event of back-driving the output with a voltage, which could otherwise cause a heavy current to flow as the wiper is moved towards terminal A.

41


3.7


Measurement of Voltage Method 1 A calibrated variable resistor, standard voltage source and galvanometer are required, these being connected as shown in Fig 3.7.

+ Unknown Voltage

G

Rt

-

R

+ -

Standard Voltage

Fig 3.7

The position of the slider of the variable resistor is adjusted until the circuit is balanced with no current flowing in the galvanometer. Under these conditions, the voltage across the R section of the variable resistance is equal to the value of the standard voltage supply. The unknown voltage is proportional to the total resistance of the variable resistor Rt and the section resistance R, and can be calculated from:Unknown voltage =

Rt R

x

Standard voltage

The method has disadvantages:1.

The unknown voltage source is loaded by the variable resistor and hence the voltage may be affected.

2.

The method only allows measurement of voltages greater than the standard voltage.

This method of measuring potential is the srcin of the term "potentiometer" for a variable resistor. Early models of this measuring instrument were made of a highly accurate, close tolerance, resistance wire which was stretched between terminals on a scaled background. It was known as a Slide-Wire Potentiometer.

42


3.8


Practical Exercise Measurement of Voltage Using Method 1 WHEATSTONE BRIDGE D


DIFFERENTIAL

7

4 3

8

2

9 1

C

B

B

0V

V

OUT

B

-

IN

A

+

A-B

10k 1V

0V

AMPLIFIER O/P

A 3

A

10

12k

Rx

+5V

MOVING COIL METER AMPLIFIER #2

5

I /P

-

O/P .5

+

.4 1 100 10

OFFSET

GAIN COARSE

.6

0

5 +10

+

.7

.3

.8

.2

.9 .1

-10

1.0

GAIN FINE

0V

L J

Fig 3.8 

First set the OFFSET control of Amplifier #2 using the same procedure used in Practical Exercise 3.6: Switch ON the power supply and with the Differential Amplifier inputs shorted together and Amplifier #2 GAIN FINE set to 1.0, adjust the OFFSET for approximately zero output with the GAIN COARSE set to 10. Adjust finally for zero with the GAIN COARSE set to 100.



Connect the circuit as shown in Fig 3.8 and set the switch on the Wheatstone Bridge circuit to OUT to disconnect the 12k Ω ratio arm resistor and the unknown resistor Rx from the circuit.



Set the Amplifier #2 GAIN COARSE to 10 and set the output from the 10k Ω wirewound resistor to 4V as indicated by the digital meter. This represents the "unknown" voltage.



Adjust the 10-turn resistor for approximate balance and then obtain final balance with Amplifier #2 GAIN COARSE set to 100.

43





Note the dial reading at balance, enter the value in Table 3.2 and calculate the value of the unknown voltage from:Unknown voltage =

1000 Dialreadin 1000

= Dialreadin 

x

x

Standard voltage

1V

Repeat the procedure with the "unknown" voltage input set to each of the values indicated in Table 3.2, recording the readings and calculating the voltages for each value. "Unknown" Voltage 4.0 3.5 3.0 2.5 2.0 1.5 1.0


Calculated Voltage V V V V V V V

Table 3.2

3.8a

Enter your dial reading with the "unknown" voltage set to 2.5V.

The method has the disadvantage of loading the unknown voltage source and this can be demonstrated as follows:



44

Set the "unknown " voltage to 2.0V and obtain balance conditions. Now remove the connection from the output of the wirewound resistor (socket B) to the Wheatstone bridge (socket D) and note the revised value of the unknown voltage as indicated by the digital voltmeter.



"Unknown" Voltage:

3.8b

When connected to the bridge =

V

Disconnected from the bridge =

V

Enter your value of the "Unknown" Voltage when disconnected from the bridge in V.

3.9

Measurement of Voltage Method 2 This method requires an additional DC source of voltage with a magnitude exceeding the maximum value of the unknown voltages to be measured and another variable resistor Rs. The schematic diagram is shown in Fig 3.9.

Rs

+ G

Rt R

Standard + Voltage

+

Unknown Voltage

Fig 3.9

For measurement of voltages less than the standard voltage , the slider of the variable resistor is set to its maximum position and, with the galvanometer connected to the standard voltage source, the value of Rs is adjusted until there is no current flowing in the galvanometer and the circuit is balanced.

45



The full resistance Rt is then calibrated to represent the value of the standard voltage. To measure an unknown voltage, the galvanometer is connected to the unknown voltage and the slider position is again adjusted for circuit balance. The section R at balance represents the magnitude of the unknown voltage. Unknown voltage =

R Rt

x

Standard voltage

For the measurement of voltages higher than the standard voltage , the variable resistor can be calibrated against the standard voltage with the slider set to a position lower than the maximum setting. This setting will now represent a magnitude equal to the standard voltage. Balance with an unknown voltage is obtained as before and the unknown voltage calculated from :Unknown voltage =

R(unknown connected) R(standardconnected)

x

Standard voltage

With this method, no current is taken from the unknown voltage source at balance and hence the circuit is not loaded. The voltage obtained should therefore be accurate, within the limits of accuracy of the variable resistor.

46



3.10 Practical Exercise Measurement of Voltage Using Method 2 You should be familiar with the procedures for initially setting the amplifier offset and balancing the bridge circuit by now. Instructions for the procedures will not therefore be repeated in this exercise. Measurement of Voltages Less Than the Standard Voltage. 

Carry out the OFFSET initializing procedure and then connect the circuit as indicated in Fig 3.10, using the 100k Ω variable resistor as Rs (Fig 3.9) in the supply circuit of the additional DC source. Note that the output of the 10kΩ wirewound variable resistor is not connected initially. This will be used as the source of the "unknown" voltage.

WHEATSTONE BRIDGE D

AMPLIFIER #2

12k

C

I/P

DIFFERENTIAL AMPLIFIER O/P

A

-

OUT

3

B

B

-

A

+

IN 1V

0V

O/ P .5

+

Rx

OFFSET

.8

.3

.9

.2

GAIN COARSE

.6 .7

.4 1 100 10

A-B

1.0

.1

GAIN FINE

MOVING COIL METER 5

CARBON TRACK 6 C


7

4

8

2

9 1

B A

10

100k

5

7

4

3

3

8

2

9 1

10

-10

B A

10k

5 +10

+ 0V

0V

0

+5V

L J

V

Fig 3.10 

Set the 10-turn resistor to its maximum setting (1000) and adjust the setting of the 100kΩ resistor for balanced conditions, i.e. null indication on the moving coil (M.C.) meter. Set Amplifier #2 GAIN COARSE control to 10 initially and then finally to 100 during the balancing.

47



When completed, the 10-turn resistor has been calibrated so that full scale reading of 1000 represents a voltage of 1.000V. 

 

Replace the 1.0V reference voltage source (from the Wheatstone Bridge circuit) with the “unknown” voltage output of the 10kΩ wirewound variable resistor, by moving the lead that is connected to socket A of the Differential Amplifier FROM socket B of the Wheatstone Bridge circuit TO socket B of the 10kΩ wirewound variable resistor. Set the "unknown" voltage to 0.25V as indicated on the digital multimeter. Adjust the control of the 10-turn resistor for balance and note the dial reading for this balance condition. This reading will represent the unknown voltage directly in mV. Record the value in Table 3.3 and compare with the reading indicated by the digital multimeter.

"Unknown" Voltage Input

0.25V


0.40V

mV

0.60V

mV

0.70V

mV

mV

0.80V mV

0.95V mV

Table 3.3 

Repeat the procedure for other "unknown" voltage inputs given in Table 3.3. 1000 Dial 900 Setting

800 700 600 500 400 300 200 100 0

Graph 3.1

48

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7 0.8 0.9 1.0 "Unknown" Voltage


Plot the characteristic of Dial Reading against "unknown" input voltage on the graticule provided.



3.10a


Read from your Graph 3.1 and enter the Dial Setting corresponding to an input voltage of 0.53V. Measurement of Voltages Greater Than the Standard Voltage. 

Remove the lead from socket C of the 10k Ω wirewound resistor to socket B of the 100kΩ resistor to remove the 1V supply.



Replace the 100k Ω resistor used for calibration with the 10k Ω slider unit and apply the +12V supply to this and the 10kΩ wirewound instead of the +5V.



Set the control dial of the 10-turn resistor to setting 0100 and connect the A socket of the Differential Amplifier back to socket B of the Wheatstone Bridge as shown in Fig 3.10.



Adjust the 10k Ω slider resistor control setting for bridge balance. When completed, the 10-turn resistance has been calibrated so that a dial reading of 0100 represents a voltage of 1.00V and a maximum dial reading of 1000 will represent a voltage of 10V.



Remove the 1.0V reference voltage source from socket A of the Differential Amplifier and connect the "unknown" voltage from socket B of the 10kΩ wirewound resistor to socket A of the Differential Amplifier.



Apply various "unknown" voltages in the range 0 - 10V to the circuit . Note the dial reading for balance for each input voltage setting and enter the values in Table 3.4.

"Unknown" Voltage Input Dial Reading at Balance Measured Voltage Table 3.4

(Volts)

1

2

V

3

V

4

V

6

V

8

V

9

V

V

49



Loading Effect 

Set the "unknown" input voltage to 5V and note the voltage change on the digital meter when the lead to the Differential Amplifier is removed. "Unknown" Voltage:

3.10b

When Connected to the bridge

=

V

Disconnected from the bridge

=

V

Enter the apparent change of the "Unknown" Voltage in mV when the amplifier is disconnected from the bridge.

The slight loading effect is due to the input resistance of the Differential Amplifier.

Notes:

........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ 50



Student Assessment 3

1000Ω +1V

100Ω G

0V Rx Unknown

853Ω

Fig 1

1.

2.

For the circuit of Fig 1, the name of the circuit is:

a

wirewound potentiometer

b

Wheatstone Bridge

c

slide wire potentiometer

d

carbon track variable resistor

When the circuit of Fig 1 is balanced, the value of the unknown resistor Rx is:

a 3.

4.

b

853

Ω

c

1172

Ω

d 8.53k

Ω

For the circuit of Fig 1, if the supply voltage was increased to +2V the effect on the balance condition would be to:

a

double the resistance value

b

half the resistance value

c

half the error in the measurement

d

make no change at all

For the circuit of Fig 1, which of the following components would NOT affect the accuracy of the measurement?

a 5.

85.3 Ω

100 Ω resistor

b

853 Ω resistor

c

1000 Ω resistor

d

galvanometer G

A resistance transducer is situated at a distance from the measuring bridge and is connected to it via just two wires, each of which has a resistance of 10 . If the resistance of the transducer is 120 , the bridge reading will be:

a

110 Ω

b

120

Ω

c

130

Ω

d 140

Ω

Continued ...

51



Student Assessment 3 Continued ... 6.

If the circuit connection described in question 5 used a three-wire system, the bridge reading would be:

a

110 Ω

b

Ω

120

c

Ω

130

d 140

Ω

Rs

7438Ω

+

+ G

V1

V1

2562 Ω

1.5V

+ -

R

Fig 2

7.

V2

V3

Fig 3

Fig 2 voltage, Fig 3 resistance both voltage

b d

Fig 3 voltage, Fig 2 resistance both resistance

When the circuit in Fig 2 is balanced, the value of V1 will be:

a 9.

+ +

The circuits of Fig 2 and Fig 3 are for measuring:

a c 8.

G

Rt

5.85V

b

4.35V

c

3.9V

d 0.52V

The circuit of Fig 3 is calibrated against a standard voltage (V2) of 1V to a dial setting of 0200. The dial has a discrimination of 1:1000. To be able to measure the maximum unknown voltage the value of the additional supply (V1) must be:

a

<1V

b

>1V

c

=2V

d >5V

10. Which of the following is NOT an advantage of the circuit of Fig 3 compared to that of Fig 2?

a c

52

measures higher and lower voltages does not load the voltage source tested

b d

measures resistance draws no current from the source tested


Temperature Measurement Chapter 4

Chapter 4 Temperature Measurement




Describe the characteristics of an IC temperature sensor.



Describe the construction and characteristics of a Platinum RTD resistance transducer.



Describe the construction and characteristics of an NTC thermistor.



Discuss the characteristics of NTC thermistor bridge circuits.



Describe the construction and characteristics of a thermocouple.



Deduce temperatures from a voltage reading across a transducer.

• • •

DIGIAC 1750 Transducer and Instrumentation Trainer. 4mm Connecting Leads. 2 x Digital Multimeters.

• •

Stopwatch (not supplied). Scientific Calculator (not supplied).

53


4.1


The DIGIAC 1750 Temperature Transducer Facilities Fig 4.1 shows the layout of the temperature transducer facilities of the DIGIAC 1750 unit. The active transducers are mounted within a clear plastic enclosure which contains a heater.

REFERENCE THERMOCOUPLE (+ LM335)

TYPE 'K' THERMOCOUPLE

IC TEMP SENSORS EXT.

O/P

O/P

+ CLEAR PLASTIC ENCLOSURE

INT.

REF +

+

B

HEATER

O/P

O/P A NTCTHERMISTORS HEATER ELEMENT

PLATINUMR.T.D. I/P

Fig 4.1

The heated enclosure is provided to raise the temperature of the sensor transducers to allow measurements to be taken during experiments. In the case of the NTC thermistors and the thermocouples, an additional, separate unit is mounted outside the heated enclosure. The externally mounted sensors are made available for comparison between ambient (room) temperature and the temperature within the enclosure. The externally mounted "K" type thermocouple is contained within a package in contact with an IC temperature sensor (LM335) to act as a thermometer with voltage output. This will be used in many of the experiments as the reference (REF) thermometer. Note: It is important to give the unit time to cool down between the experiments. This will allow the enclosure to return to ambient temperature.

54


4.2


The IC Temperature Sensor This is an integrated circuit containing 16 transistors, 9 resistors and 2 capacitors contained in a transistor type package. The device reference number is LM335 and it provides an output of 10mV/°K. Measurements of the output voltage therefore indicate the temperature directly in degrees Kelvin (°K). For example, at a temperature of 20°C (293°K) the output voltage will be 2.93V. The circuit arrangement provided with the IC Temperature Sensor on the DIGIAC 1750 unit is shown in Fig 4.2. +5V

1kΩ

1kΩ Ext. O/P Int.

LM 355 0V

Socket for connection of external LM335

Fig 4.2

A 2-pin socket is provided for the connection of an external LM335 unit if desired. Note An LM335 unit is mounted on the Type "K" Thermocouple panel, external to the heated enclosure and fitted in a heat sink together with another type "K" thermocouple, its output being available from the REF socket on that panel. The output from this can be used as an indication of the ambient temperature outside the heated enclosure, and that from the INT. socket in Fig 4.2 indicates the temperature within the heated enclosure.

The output from the REF socket does not give an accurate value of the room (ambient) temperature when the heater is in use, due mainly to heat passing along the PCB by conduction from the heater. An LM335 remotely mounted or some other method is necessary if accurate measurement of ambient temperature is required.

55


4.3


Practical Exercise Characteristics of an LM335 IC Temperature Sensor TYPE 'K' THERMOCOUPLE

IC TEMP SENSORS EXT.

O/P

O/P

+

INT.

REF +

+

B

O/P

O/P

V

A N T C T H E R M IS T O R S

P L A T IN U M R . T . D .

HEATER ELEMENT

I/P

+1 2V

0V

Fig 4.3 

Connect a voltmeter to the circuit (as shown in Fig 4.3), switch the power supply ON and note the output voltage, this (x100) representing the ambient temperature in °K. Record the value in Table 4.1.



Connect the +12V supply to the heater input socket and note the voltage reading every minute until the value stabilizes. Record the values in Table 4.1. (Note °C = °K - 273.)

Time (minutes)

0

Voltage

1 V V V

2 V

3 V

4 V

5 V

6 V

°K

Temperature

°C Table 4.1

4.3a

Enter your temperature reading in C after 5 minutes. 

56


7

8

9 V

10 V

V



Exercise 4.3 illustrates the characteristics of the LM335 transducer, indicates the maximum temperature rise possible using the heater supplied at 12V, and also gives you an idea of the time scale required for the unit to reach stable conditions.

4.4

The Platinum RTD (Resistance Temperature Detector) Transducer Laser Trimmed Platinum Film

Ceramic Substrate

Gold Contact Plates Connections

Fig 4.4

The construction of the Platinum RTD Transducer is shown in Fig 4.4, consisting of a thin film of platinum deposited on a ceramic substrate and having gold contact plates at each end that make contact with the film. The platinum film is trimmed with a laser beam to cut a spiral for a resistance of 100Ω at 0°C. The resistance of the film increases as the temperature increases. It has a positive temperature coefficient (PTC). The increase in resistance is linear, the relationship between resistance change and temperature rise being 0.385Ω/°C. Rt = Ro + 0.385t

where

Rt = resistance at temperature t°C Ro = resistance at 0°C (= 100Ω)

Normally, the unit would be connected to a DC supply via a series resistor and the voltage developed across the transducer is measured. The current flow through the transducer will then cause some self heating, the temperature rise due to this being of the order of 0.005°C/mW dissipated in the transducer.

57



The very simple electrical circuit arrangement of the DIGIAC 1750 unit is as shown in Fig 4.5. O/P

RTD

0V

Fig 4.5

The white dot signifies that this is a PTC, not NTC (negative temperature coefficient) type of resistor which would have a black dot. In the practical exercise you will connect the platinum RTD in series with a high resistance to a DC supply and measure the voltage drop across it. Due to the small variation of resistance, the current change will be negligible and the voltage drop across the transducer will be directly proportional to its resistance.

Notes: ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

58


4.5


Practical Exercise Characteristics of a Platinum RTD Transducer

IC TEMP SENSORS

SLIDE

C

EXT. B

O/P INT. +

1

2

3

4

5

6

7

8

9

10

10k

A

+

O/P

V PLATINUM R.T.D. HEATER ELEMENT

I/P

0V

+5 V

Fig 4.6 

Set the slider of the 10k Ω carbon resistor to mid-way and connect the circuit as shown in Fig 4.6, with the digital multimeter set to its 200mV or 2V DC range.



Switch ON the power supply and adjust the slider control of the 10k Ω resistor so that the voltage drop across the platinum RTD is 108mV (0.108V) as indicated by the digital multimeter.

This calibrates the platinum RTD for an assumed ambient temperature of 20°C, since the resistance of the RTD at 20°C will be 108Ω. Note that the voltage reading across the RTD in mV is the same as the RTD resistance in Ω, since the 0.108 current flowing must be = 1mA. 108 Note: If the ambient temperature differs from 20°C, the voltage can be set to the correct value for this ambient temperature if desired:

1.

Set the voltmeter to its 20V range and measure the INT output from the IC Temperature Sensor to obtain the ambient temperature: Voltage x 100 = temperature in °K Temperature in °C = °K - 273

2.

RTD resistance = 100 + 0.385 x °C. Set the voltage drop across the RTD for this value.

59





Connect the voltmeter, set to its 20V DC range, to the INT output of the IC Temperature Sensor. This represents the RTD temperature (voltage x 100 = temperature in °K). Record the temperature in the first column of Table 4.2.



Connect a second voltmeter, set to its 200mV range, to measure the voltage output from the RTD transducer. This voltage (in mV) is equal to the RTD resistance (in Ω). Record the resistance in the first column of the table.



Connect the +12V supply to the Heater Element input and record the RTD temperature (in °K) and RTD resistance (in Ω) after each of the times given in the table.

Time (minutes)

0

1

2

3

4

5

6

7

8

9

10

°K

RTD Temperature

°C

RTD Resistance

Ω

Ω

ΩΩΩΩΩΩΩ

Ω

Ω

Table 4.2 

Convert the RTD Temperature into °C (°K - 273) and add to Table 4.2.



Plot the graph of RTD resistance ( Ω) against temperature (°C) on the axes provided. Extend your graph down to cover 0°C. 130 128 126 RTD 124 Resistance 122 120 118 116 114 112 110 108 106 104 102 100 98 0

Graph 4.1

60

10

20

30

40 50 60 RTD Temperature o C

70


4.5a


Enter the total change in the resistance of the RTD Transducer over the temperature range 20-50 C in

4.5b

Is the resistance/temperature characteristic linear?

Yes

4.5c

.

or

No

Enter your estimated (extrapolated) resistance of the RTD Transducer from the graph at 0 C.

During the exercise, the current flowing was of the order of 1mA. Since the applied voltage was +5V, the total circuit resistance was therefore of the order of 5kΩ (which you can see from the setting of the 10k Ω slider resistor). The variation of resistance of the RTD Transducer therefore had little effect on the circuit current and hence the voltage drop across it represented the resistance value reasonably accurately. The current of 1mA in the RTD represents a very low power dissipation in the RTD. The self-heating effect would produce a temperature rise of 0.02°C. 

Calculate the power dissipation in the RTD Transducer at a temperature of 50°C when the standard circuit current of 1mA flows in it.

Power dissipation in the RTD Transducer =

4.5d

W

Enter your calculated value of the power dissipated in the RTD Transducer in W. 


61


4.6


The NTC (Negative Temperature Coefficient) Thermistor The thermistor (thermally sensitive resistor) is manufactured with the intention that its value will change with temperature. Unlike a normal resistor, a large coefficient of resistance (change of resistance with temperature) is desirable. Some are made with resistance which increases with temperature (positive temperature coefficient, PTC) or decreases (negative temperature coefficient (NTC). They are made in rod, disc or bead form. The construction of a typical NTC thermistor is shown in Fig 4.7(a), consisting of an element made from sintered oxides of metals such as nickel, manganese and cobalt, with contacts made to each side of the element.

+5V Th1

active disc element

B O/P A

Th2 Contacts

0V

Construction

Electrical Circuit

(a)

(b)

Fig 4.7

As the temperature of the element increases, its resistance falls, the resistance/ temperature characteristic being non-linear. The resistance of the thermistors provided with the DIGIAC 1750 unit is of the order of 5kΩ at an ambient temperature of 20°C (293°K). Two similar units are provided, one being mounted inside the heated enclosure. This is connected to the +5V supply and designated A. The other is mounted outside the heated enclosure. It is connected to the 0V (ground) line and is designated B. The circuit arrangement is shown in Fig 4.7(b).

62


4.7


Practical Exercise Characteristics of an NTC Thermistor The resistance of the NTC thermistor varies over a wide range for the temperature range available within the heated enclosure. For this reason the method used to measure the resistance in Exercise 4.5 cannot be used this time. If resistance readings are to be taken at regular intervals of 1 minute, the readings must be obtained very quickly. The method selected connects the thermistor in series with a calibrated resistor to the +5V supply. For each reading, the variable resistor is adjusted until the voltage at the junction of the thermistor and resistor is half of the supply voltage. For this setting there will be the same voltage drop across the thermistor and the resistor and, since the same current flows in each, their resistances must be equal. Hence the value of the resistance read from the calibrated resistor scale is the same as the resistance of the thermistor.

+5V n.t.c. 1k Ω

Calibrated Variable

2.5V

V +

0V

+

Schematic Diagram B

WHEATSTONE BRIDGE D

O/P A 12k

C A

OUT

B

3

NTC THERMISTORS HEATER ELEMENT

I/P

IN 1V

V

Rx

0V

Fig 4.8

63





Connect the circuit as shown in Fig 4.8. Set the switch on the Wheatstone bridge circuit to OUT to disconnect the 12kΩ and Rx resistors from the circuit, and set the calibrated variable resistor dial reading to approximately 500.



Switch the power supply ON and adjust the resistor control until the voltage indicated by the voltmeter is 2.5V.



Connect a second voltmeter, set to the 20V DC range, to measure the INT output of the IC Temperature Sensor. This represents the temperature (voltage x 100 = temperature in °K).



Record the values of temperature (in °K) and dial reading in the first column of Table 4.3.

Time (minutes)

0

1

2

3

4

5

6

7

8

9

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

kΩ

10

°K

Temperature (from IC Transducer)

°C

Dial Reading for 2.5V Thermistor Resistance (10 x Dial reading + 1k Ω) Table 4.3

kΩ



Connect the +12V supply to the Heater Element input socket and, at oneminute intervals, note the value of the temperature (in °K) and the dial reading to produce 2.5V across the resistance. Record in Table 4.3.



Convert the temperature measurements into °C (°K - 273) and add to the table.



For each of the dial readings recorded in Table 4.3, calculate the thermistor resistance and record in the table. Note: There is a 1kΩ resistor in the output lead of the variable resistance, so the thermistor resistance will be 10 x Dial reading + 1k .



Plot the graph of thermistor resistance against temperature on the axes provided.

Due to the shape of the response characteristic, the device is not suitable for applications where an accurate indication of temperature is required.

64



7k Ω

6k Ω Thermistor Resistance 5k Ω

4k Ω

3k Ω

2k Ω

1k Ω 0

10

20

30

40 50 60 Temperature o C

70

Graph 4.2

4.7a

From your graph, enter the resistance of your thermistor at a temperature of 35 in k .

Thermistors are used in very many electronic circuit applications for the control of currents and voltages as equipment temperatures vary. As transducer sensors they are more suitable for applications in protection and alarm circuits where an indication of temperature threshold is required. Some thermistors are available which have a rapid change of resistance when the temperature exceeds a certain value. 


65


4.8


Two Thermistor Bridge Circuits When used for alarm or protection circuits, two thermistors would normally be used, these being connected in a bridge circuit as shown in Fig 4.9.

R

Th1 A

DC Supply

O/P B

R

Th2

Fig 4.9

The two resistors R have the same resistance as the "cold" resistance of the thermistors. When cold, there will be no output at the connections AB because the bridge will be balanced under this condition. As the temperature rises, the resistance of both thermistors will decrease. The potential of connection A will rise and that of connection B will fall, giving a larger output than would be obtained with a circuit using only one thermistor.

4.9

Practical Exercise Characteristics of NTC Bridge Circuits Two bridge circuits will be investigated, one containing only one thermistor (Th1) and the other, two.

RV2

DC Supply

66

Th1 (A)

Digital Multimeter

V

RV1

Th2 (B)

Fig 4.10

10kΩ

10kΩ

RV3

10kΩ



Since the three branches to be used are all in parallel (Fig 4.10) they can be connected at the beginning and brought into operation simply by moving the null detector (digital multimeter). Note that the second thermistor (Th2) is not contained within the heated enclosure and will therefore not be subjected to the same heating effect as Th1. The circuit will not be as efficient as can be expected from one in which both thermistors are mounted in the same temperature environment. Variable resistors, RV2 & RV3 are adjusted to balance the branch "cold" resistances (approximately 5kΩ) to give 2.5V at the center-tap, and RV1 is also adjusted for 2.5V at the wiper. The circuit will then be ready for heating measurements.

WHEATSTONE BRIDGE D 12k

C A

TYPE 'K' THERMOCOUPLE OUT

B

3

-

IN 1V

0V

O/P +

Rx

REF + SLIDE

B 1

2

3

4

5

6

+

C

7

8

9

10

NTC THERMISTORS

7

4

8

2

9 10

10k

HEATER ELEMENT

I/P

V

3

1

A

A

10k


+5V

B O/P

B A 0V

Fig 4.11

Th1, the 10kΩ 10-turn resistor (RV3) and the 10k Ω wirewound resistor (RV1) form the bridge circuit with one active thermistor. Th1, the 10kΩ 10-turn resistor (RV3), Th2 and the 10kΩ carbon slide resistor (RV2) form the bridge with two active thermistors.

67




Connect the circuit as shown in Fig 4.11 and set the switch on the Wheatstone Bridge circuit to OUT.



Switch the power supply ON and adjust the 10k Ω wirewound resistor (RV1) so that the voltmeter reading is 2.5V. The fixed branch of the bridge is now set for center balance.



Connect the voltmeter between socket B of the 10k Ω wirewound resistor (RV1) and NTC thermistor A. Adjust the 10kΩ 10-turn resistor (RV3) on the Wheatstone Bridge circuit for a voltage reading of zero.



Now connect the voltmeter between socket B of the 10k Ω wirewound resistor (RV1) and NTC thermistor B. Adjust the 10kΩ carbon slider resistor (RV2) for an output voltage of zero.



Both bridges are now set for zero output with the thermistors at ambient temperature.



Connect the voltmeter, set to the 20V DC range, to measure the INT output of the IC Temperature Sensor. This represents the temperature (voltage x 100 = temperature in °K). Record the value in the first column of Table 4.4.

Time (minutes) Temperature (IC Temperature Transducer) Bridge Output


0

1

2

3

4

5

6

7

8

9

10

°K °C

1 active NTC 2 active NTC

V V V

V

V

V

V

V

V V V

V V V

V

V

V

V

V

V V V

Table 4.4

68



Connect a second voltmeter, set to the 2V DC range, between socket A of the NTC and socket B of the 10k Ω wirewound resistor (RV1). This is the bridge output for 1 active NTC thermistor. Record your measured voltage in the table.



Move one of the connections for the second voltmeter from the 10k Ω wirewound resistor (RV1), to socket B of the 10kΩ slide resistor (RV2). This is the bridge output for 2 active NTC thermistors. Again record your measured voltage in the table.


Now connect the 12V supply to the heater input and, at one-minute intervals, record the temperature (in °K) from the first voltmeter, and the bridge outputs for the circuit with 1 active NTC and 2 active NTC thermistors (by changing the connections for the second voltmeter as you did before).






Draw graphs of output voltage against temperature for the two bridge circuits on the same axes provided (Graph 4.3): 2.4 2.2 Output Voltage 2.0 (volts) 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0

0

10

20

30

40 50 60 Temperature o C

70

Graph 4.3

4.9a

From your graph, enter the output voltage for the one transducer bridge at a temperature of 50 in V.

4.9b

From your graph, enter the output voltage for the two transducer bridge at a temperature of 50 in V.

Note that the output with two active thermistors is greater than that with only one thermistor. However, if both active thermistors were at the same temperature, the output voltage would be twice that for one active thermistor. 4.9c

Is the relationship linear ?

Yes 

or

No

Switch OFF the power supply. 69



4.10 Type "K" Thermocouple Chromel Wire Spot-welded "Hot" junction Alumel Wire

"Cold" junction

Output Voltage

Fig 4.12

Fig 4.12 shows the construction of a thermocouple, consisting of two wires of different materials joined by welding together at one end. For the type "K" thermocouple the two materials are alumel and chromel. With this arrangement, when the ends that are joined together are heated, an output voltage is obtained between the other two ends. The ends that are joined together are referred to as the "hot" junction and the other ends are the "cold" junction. The magnitude of the output voltage depends on the temperature difference between the "hot" and "cold" junctions and on the materials used. For the type "K" thermocouple the output voltage is fairly linear over the temperature range 0-100°C and of magnitude 40.28 µV/°C difference between the "hot" and "cold" junctions. Two thermocouples are provided with the DIGIAC 1750 unit, one being mounted within the heated enclosure, this being the active unit which will have its "hot" and "cold" junctions at different temperatures in operation. The unit an is mounted outside the heated enclosure incorporatedofinthe a heat other sink with LM335 IC Temperature Sensor so thatand theistemperature "cold" junction of the active thermocouple can be measured.

70



The second thermocouple is connected in series with the first with the wires of the same material connected together. This ensures that the connections to the output circuit are made from the same material which eliminates the possibility of an EMF being introduced into the circuit by connections between different materials. The second thermocouple does not contribute to the output voltage because its "hot" and "cold" junctions are maintained at the same temperature. The circuit arrangement is as shown in Fig 4.13.

+5V

Active Thermocouple

Ref

+

LM 335

O/P

Inactive Thermocouple

-

0V

Fig 4.13

Due to the low output voltage of the thermocouple, amplification is required. An amplifier gain of 200 will give readings within one range of the digital multimeter. During operation, the temperature of the "cold" junction varies, due mainly to heat conduction from the heater along the PCB and the junction is in effect "floating". This is a common occurrence with thermocouple installations where the thermocouple leads are short. To overcome the problem, extra leads of the same material or different materials having the same thermoelectric properties are used to extend the "cold" junction to a point where a steady temperature can be maintained. These cables are referred to as "compensating cables".

71



4.11 Practical Exercise Characteristics of a "K" Type Thermocouple INSTRUMENTATION AMPLIFIER O/P


O/P

+

B

-

A

+

x100 AMPLIFIER O/P I/P

A-B

+100V IN

REF

+

AMPLIFIER #1 O/P

I/P

-

HEATER ELEMENT

.5

+

.4 1 100 10

I/P

GAIN COARS E

.7 .8

.3

.9

.2 .1

OFFSET

.6

1.0

GAIN FINE V

0V

Fig 4.14 



Connect the circuit as shown in Fig 4.14, set the voltmeter to the 2V DC range and set Amplifier #1 GAIN COARSE to 10 and GAIN FINE to 0.2. Switch the power supply ON and set the Amplifier #1 OFFSET control as follows: 1) Temporarily disconnect the inputs to the Instrumentation Amplifier. 2) Short circuit together the Instrumentation Amplifier input connections. 3) Adjust the OFFSET control for zero indication on the voltmeter.



Re-connect the Thermocouple outputs to the Instrumentation Amplifier as shown in Fig 4.14. The measured voltage should still be zero with the "hot" and "cold" junctions at the same temperature.



Connect a second voltmeter, set to the 20V DC range, to the INT. socket of the IC Temperature Sensor. This represents the hot junction temperature inside the heated enclosure (voltage x 100 = temperature in °K). Record this in Table 4.5.



72

Reconnect the second voltmeter to the REF output socket in the Type ‘K’ Thermocouple block. This represents the cold junction temperature outside the enclosure (voltage x 100 = temperature in °K). Record this in Table 4.5.


Temp. °K


0

1

2

3

4

mV

mV

mV

mV

mV

5

6

7

8

9

10

mV

mV

mV

mV

Hot Junction (INT.) Cold Junction (REF) Difference

Thermocouple O/P

mV

mV

Table 4.5 



Connect the +12V supply to the heater and at 1 minute intervals, record the thermocouple output voltage (displayed on the first voltmeter), and the voltages representing the hot and cold thermocouple junction temperatures (by changing the connections for the second voltmeter as you did before). Construct the graph of thermocouple output voltage against temperature difference between the "hot" and "cold" junctions on the axes provided.

320 300 280 260 240 220

Output 200 Voltage (mV) 180 160 140 120 100 80 60 40 20 0

0

5

10

15

20

25

30

35

Temperature Differenceo C

Graph 4.4

73


4.11a

Is the characteristic linear?

Yes

4.11b


or

No

Deduce from your graph and enter the relationship in mV/ C. 


The actual value of the transfer characteristic will depend on the gain provided by the amplifier system at the settings used, which can be adjusted to calibrate the system as desired.

Notes: ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

74




The output voltages you would expect to obtain from an LM335 Temperature Transducer at temperatures of (i) 0 C, (ii) 50 C & (iii) -20 C, are:

a c 2.

6.

119 Ω

b

138

Ω

c

150

5.78°C

b

15.2 °C

c

The resistance of an NTC thermistor is 5k resistance will be:

a 5.

(i) 0V, (ii) +0.50V, (iii) -0.20V (i) 2.73V, (ii) 3.23V, (iii) 2.53V

less than 5k Ω

b

equal to 5k Ω

at 0 C and 138

Ω

A platinum RTD Transducer has resistance of 100 temperature when its resistance is 115.2 will be:

a 4.

b d

A platinum RTD Transducer has resistance of 100 resistance at 50 C will be:

a 3.

(i) 0V, (ii) -0.50V, (iii) +0.20V (i) 2.93V, (ii) 3.43V, (iii) 2.73V

d 188

at 0 C and 138

30.4 °C

at 100 C. Its

Ω at 100 C. The

d 40 °C

at 20 C. If its temperature is i ncreased, its

c

greater than 5k Ω d infinite

A thermocouple gives an output of 40µV/ C difference in temperature between the "hot" and "cold" junctions. The output voltages expected for the junction temperatures of (i) "cold" 0 C "hot" 50 C (ii) "cold" 20 C "hot" 70 C (i) "cold" 50 C "hot" 50 C will be:

a

(i) 2V, (ii) 2V, (iii) 0V

b

(i) 2mV, (ii) 2.8mV, (iii) 2mV

c

(i) 2mV, (ii) 2mV, (iii) 0V

d

(i) 200 µV, (ii) 280µV, (iii) 0V

A thermocouple gives an output of 40µV/ C difference in temperature between the "hot" and "cold" junctions. The amplifier gain required to enable the thermocouple circuit to produce an output of 1V for a temperature difference of 100 C between the "hot" and "cold" junctions is:

a

25

b

40

c

250

d 500

Continued ...

75




4kΩ

+5V

Th1 O/P

0V Th2

4kΩ

Fig 1

7.

8.

The characteristics of the devices Th1 & Th2 shown in Fig 1 are such that as the temperature is increased:

a

the voltage across them will rise

b

their resistance will increase

c

their resistance will fall

d

the current through them will decrease

Two thermistors Th1 & Th2 having resistance 4k

at 20 C and 1k

at 60 C are

connected in the bridge circuit shown in Fig 1. The output voltage from the circuit at a temperature of 20 C will be:

a 9.

0V

b

2.5V

c

4V

d 5V

Two thermistors Th1 & Th2 having resistance 4k at 20 C and 1k at 60 C are connected in the bridge circuit shown in Fi g 1. The output voltage from the circuit at a temperature of 60 C will be:

a

1V

b

2V

c

3V

d 4V

10. The characteristics of an IC Temperature Sensor give an output voltage of:

a

76

1mV/°C

b

10mV/ °K

c

100mV/°C

d 1V/ °K


Light Sensors Chapter 5

Chapter 5 Light Sensors




Discuss the characteristics of a filament lamp.



Describe the construction and characteristics of a photovoltaic cell.



Describe the construction and characteristics of a phototransistor.



Describe the construction and characteristics of a photoconductive cell.



Describe the construction and characteristics of a PIN photodiode.

• • • •

DIGIAC 1750 Transducer and Instrumentation Trainer. 4mm Connecting Leads. Digital Multimeter. Opaque box to cover the clear plastic enclosure.

77


5.1


The DIGIAC 1750 Opto-Transducer Facilities Fig 5.1 shows the arrangement of the opto-electronic (light) transducers provided on the DIGIAC 1750 Trainer. The opto-sensors are contained within a clear plastic enclosure and can be illuminated by a lamp which is placed centrally. CLEAR PLASTIC ENCLOSURE

P.I.N. PHOTODIODE

PHOTOVOLTAIC CELL

O/P

O/P

+

+

O /P

LAMP

PHOTOCONDUCTIVE CELL LAMP FILAMENT

O /P

PHOTOTRANSISTOR I/P

Fig 5.1

All semiconductor devices are sensitive to light falling upon them. That is why the devices (diodes, transistors, IC's) are contained within opaque encapsulations, to prevent light getting at the active materials. With some devices, the main effect of light irradiation will be to increase their conductivity (reduce their resistance). In others either an EMF is generated or currents are released to flow in an external circuit.

78


5.2


The Incandescent Lamp The light source to be used in the experiments is a tungsten filament lamp. The filament glows more brightly as the power feeding the lamp is increased. Two factors will be affected as the lamp voltage is increased: 1.

The temperature of the filament is proportional to the input power. Power varies with the square of the voltage, and is also affected by the resistance of the lamp, which increases as the filament temperature increases (it has a positive temperature coefficient).

2.

The spectral response of the lamp varies with the filament temperature. At low temperatures the light is in the infra-red region of the visible spectrum and the light output gradually increases in frequency (red → orange → yellow . . . ) as the temperature is raised.

These factors make it difficult to be too precise about the response of the sensors which will be investigated. In order to determine the response of the filament lamp an acceptable reference must be established. The photovoltaic cell is a linear device, the output short circuit current being directly proportional to the luminous flux (lux) being received.

79


5.3


Practical Exercise The Filament Lamp

MOVING COIL METER

0 5

-10

5

+10

+ P . I. N . P H O T O D IO D E

-

L J

0V

P H O T O V O L T A IC C E L L

O/P

O/P +

+

O/ P WIREWOUND TRACK 6 5 C

POWER AMPLIFIER

7

4 3

8

2

9 1

O/P

B

O/ P

PHOTOCONDU CTIV E CELL

I/P

I/P

Digital Multimeter

A

10

P HOTOTRA NSISTOR

A

LAMP FILAMENT

10k

+12V

0V

Fig 5.2

Lamp filament voltage (volts) Lamp filament current (mA) Lamp powerfilament (mW) Lamp resistance ( Ω) Table 5.1

80



Connect the circuit as shown in Fig 5.2 with the digital multimeter connected as an ammeter on the 200mA range in between the power amplifier and the lamp filament socket. Switch ON the power supply.



Set the 10k Ω wirewound resistor to minimum for zero output voltage (on the moving coil meter) from the power amplifier.



Take readings of lamp filament current as indicated on the digital multimeter as the lamp voltage is increased in 1V steps. Record the results in Table 5.1. 0

1

2

3

4

5

6

7

8

9

10





Calculate the corresponding values of lamp filament power (V xI) and resistance (V÷I), recording the results in Table 5.1.



Plot the graphs of lamp power and resistance against applied voltage on the graticule provided. 800

200

750

190

700

180

Lamp 650 Power (mW) 600

170

550

150

500

140

450

130

400

120

350

110

300

100

250

90

200

80

150

70

100

60

50

50

Lamp Resistance

160

0 0

1

2

3

4

40 5 6 7 8 9 10 Lamp Voltage (volts)

Graph 5.1

5.3a

From your graph estimate and enter the lamp voltage necessary to give a power dissipation of 250mW.

5.3b

From your graph estimate and enter the resistance of the lamp filament when the applied voltage is 4.5V.




81


5.4


Photovoltaic Cell A photovoltaic cell is one which generates an EMF when light falls onto it.

P-type Depletion Layer

N-type

Fig 5.3

One of the regions is made very thin (about one millionth of a meter, 1µm). Light can easily pass through this without much loss of energy. When the light reaches the junction, at the depletion layer, it is absorbed and the released energy creates hole-electron pairs which diffuse across the junction. The thin layer, which is only lightly doped, rapidly becomes saturated and charge carriers can be released into an external circuit to form a current, pushed around the circuit by the force (electro-motive force, EMF, electron-moving-force) of the surplus of charge carriers released by the energy absorbed. -200

-150 Anode Current -100 (microamps) -50

10,000 lux

2,000 lux

Anode Voltage (volts)

Symbol

+0.5

Fig 5.4

Note that the anode current is shown as negative because the internal current inside any source of EMF must flow with opposite polarity to the external current, the electrons arriving at the anode returning to the cathode inside the photo-cell.

82



The lux referred to in Fig 5.4 is the unit of incident light (light arriving at the cell).

O/P

0V Circuit Arrangement

Fig 5.5

Characteristics of Photovoltaic Cell Type MS5B

Open circuit voltage (in sunlight)

500mV

Short circuit current (in sunlight)

10mA

Peak spectral response wavelength 840nm (IR) Note: IR = infra red Response time

10 µ s

Table 5.2

If the output of the cell is short circuited there will be no output voltage at all, since this will all be dropped internally across the resistance of the cell. The short circuit output current obtained will vary from zero to maximum according to the incident light. The device can be used either as a voltage source or as a current source and is inherently a linear device. To increase the output voltage, cells may be connected in series. Parallel connection allows a greater current to be drawn. When used as an energy source they are known as Solar Cells. Note:

For the characteristic to be linear it is necessary for the light output of the lamp to be of constant light frequency (spectral color) and for the light output (in lux) to be directly proportional to the power input.

83


5.5


Practical Exercise The Photovoltaic Cell

MOVING COIL METER

0

5

5 +10

-10 +

P .I .N . P H O T O D I O D E

-

L J

0V

P HO T O V O LT A I C C E LL

O/P

O/P

+

+

O /P WIREWOUND TRACK 6 5 C

POWER AMPLIFIER

7

4

8

3

1

PH OT OC O ND UC TI V E C E LL

A

PH O TO TRA NS IS T OR

LAMP FILAMENT

I/P

I/P

9

2

O/P

B

O/P

A

10

10k

+12V

0V

Fig 5.6

Lamp filament voltage (volts) Short Circuit Output Current Open Circuit Output Voltage Table 5.3

84



Connect the circuit as shown in Fig 5.6 with the digital multimeter (ammeter) on the 2mA range to measure the short circuit current between the Photovoltaic Cell output and Ground. Fit an opaque box over the Clear Plastic Enclosure to exclude all ambient light.



Switch ON the power supply and set the 10k Ω wirewound resistor to minimum for zero output voltage from the power amplifier.



Take readings of Photovoltaic Cell Short Circuit Output Current as indicated on the digital multimeter as the lamp voltage is increased in 1V steps. Record the results in Table 5.3. 0

1

2

3

4

5

6

7

8

9

10

µA

µA

µA

µA

µA

µA

µA

µA

µA

µA

µA

V

V

V

V

V

V

V

V

V

V

V





Switch OFF the power supply, set the multimeter as a voltmeter to read the Open Circuit Output Voltage. Switch ON the power supply and repeat the readings, adding the results to Table 5.3.



Plot the graphs of Photovoltaic Cell Short Circuit Output Current and Open Circuit Output Voltage against Lamp filament voltage on the graticule provided. 1100

1.10

1050

1.05

1000

1.00

950

0.95

900 Photovoltaic Cell 850 Short Circuit Output Current 800 (µA)

0.90

750

0.75

700

0.70

650

0.65

600

0.60

550

0.55

500

0.50

450

0.45

400

0.40

350

0.35

300

0.30

250

0.25

200

0.20

150

0.15

100

0.10

50

0.05

Photovoltaic Cell Open Circuit Output Voltage 0.80 (volts)

0.85

0 0

1

2

3

0 4 5 6 7 8 9 10 Lamp Filam ent Voltage (volts)

Graph 5.2

85


5.5a


From your graph estimate and enter the short circuit output current in A when the Lamp filament voltage is 7.5V.

5.5b

Are the graphs linear?

Yes 

5.6

or

No


The Phototransistor The construction and circuit used are shown in Fig 5.7. The device is an NPN three layer semiconductor device similar to a normal transistor, the regions being called emitter (e), base (b) and collector (c).

+V

c Incident Light

N P N Lens

Case

Incident Light

eb

R Output c

b

e 0V

Fig 5.7

The device differs from the normal transistor in allowing light to fall onto the base region, focused there by a lens. The circuit connection is shown in Fig 5.7, the collector being connected to the positive of a DC supply via a load resistor R. The base connection is not used in this circuit but is available for biasing to change the threshold level. With no light falling on the device there will be a small leakage current flowing due to thermally generated hole-electron pairs and the output voltage from the circuit will be slightly less than the supply voltage due to the voltage drop across the load resistor R.

86



When light falls on the base region the leakage current increases. With the base connection open circuit, this current flows out via the base-emitter junction and is amplified by normal transistor action to give a large change in the collector leakage current. With increased current flowing in the load resistor R, the output voltage reduces and is dependent on the light falling on the device. Vout = V - Iceo R where: V = Supply voltage, Iceo = Collector leakage current, R = Collector load resistance. Fig 5.8 shows the circuit arrangement for the DIGIAC 1750 unit.

O/P

c b

e 0V

Fig 5.8

The main characteristics of the device are: Type

MEL12

Collector Current

Dark

100nA

(Vce = 5V)

Typical ambient

3.5mA

Table 5.4

87


5.7


Practical Exercise Characteristics of a Phototransistor

MOVING COIL METER

5

0

P . I . N. P HO T O D I O D E

P HO T O V O L T A I C C E LL

O/P

O/P

5 +10

-10

+

+

+

O/ P

O /P

-

L J

0V

PH O TO CO NDUCT IV E C ELL

PH O T OT RA NSI ST O R

LAMP FILAMENT

I/P SLIDE


C

V

B

POWER AMPLIFIER

7

4 3

8

2

9 1

O/P

B

1

2

3

4

5

6

7

8

9

10

A

10k

I/P A

10

10k

+12V

0V

+5V

Fig 5.9

88



Connect the circuit as shown in Fig 5.9 and set the 10k Ω carbon slider control to minimum setting (1) so that the Phototransistor load resistance is approximately 1kΩ (protection resistor only).



Connect the digital multimeter on the 20V DC range to measure the Phototransistor output voltage. Fit the opaque box over the Clear Plastic Enclosure to exclude all ambient light.






Take readings of Phototransistor output voltage as indicated on the digital multimeter as the lamp voltage is increased in 1V steps. Record the results in Table 5.5.


Lamp filament voltage (volts) Phototransistor Output Voltage

0


1 V

2 V

3 V

4 V

5 V

6 V

7 V

8 V

9 V

10 V

V

Table 5.5 

Plot the graph of Phototransistor Output Voltage against Lamp filament voltage on the graticule provided.

5.5 5.0 Phototransistor 4.5 Output Voltage 4.0 (V) 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0

0

1

2

3 4 5 6 7 8 9 10 Lamp Filament Voltage (volts)

Graph 5.3

5.7a

From your graph estimate and enter the filament input voltage when the Phototransistor output voltage is 2.5V.

5.7b

As the filament input voltage increases the phototransistor output voltage 'levels out' at approximately:

a 4.5-5.5V 

b

3-4V

c

1.5-2.5V

d

0-1.0V


89


5.8


The Photoconductive Cell, LDR Fig 5.10 shows the basic construction of a photoconductive cell, consisting of a semiconductor disc base with a gold overlay pattern making contact with the semiconductor material. The circuit arrangement for the DIGIAC 1750 unit is also shown.

O/P

Cadmium Sulfide disk

Contact

Contact 0V Gold Contact Fingers

Circuit Arrangement

Fig 5.10

The resistance of the semiconductor material between the gold contacts reduces when light falls on it. With no light on the material, the resistance is high. Light falling on the material produces hole-electron pairs of charge carriers and reduces the resistance. Out of the various semiconductor materials available, a cadmium sulfide photoconductive cell is used on the DIGIAC 1750 unit because it responds to light with a range of wavelengths similar to those of the human eye (400-700nm). An alternative name for this device is the Light Dependent Resistor, LDR.

Cell Resistance Peak Spectral Response

Dark 1MΩ

Ambient (typ.) 400Ω 530nm

Table 5.6

When is removed from This the device, the hole-electron slow totime. reform and thelight response is sluggish. is indicated by the large pairs fallingare response

90


5.9


Practical Exercise Characteristics of a Photoconductive Cell

MOVING COIL METER

5

0

P . I . N. P HO T O D I O D E

P H O T O V O LT A I C C E L L

O/P

O/P

5 +10

-10

+

+

+

O/P

O /P

-

L J

0V

PH O TO C ONDUCT IV E C E LL

PHOT OT RANSI ST O R

LAMP FILAMENT

I/P

V SLIDE WIREWOUND TRACK 6 5 C

C B

POWER AMPLIFIER

7

4 3

8

2

9 1

O/P

B

1

2

3

4

5

6

7

8

9

10

A

10k

I/P A

10

10k

+12V

0V

+5V

Fig 5.11 

Connect the circuit as shown in Fig 5.11 and set the 10k Ω carbon slider control to setting 3 so that the Photoconductive Cell load resistance is approximately 3kΩ.

+5V

3kΩ

0V

Output LDR

Fig 5.12 

Connect the digital multimeter on the 20V DC range to measure the Photoconductive Cell output voltage. Fit the opaque box over the Clear Plastic Enclosure to exclude all ambient light.




91





Take readings of Photoconductive Cell output voltage as indicated on the digital multimeter as the lamp voltage is increased in 1V steps. Record the results in Table 5.7.

Lamp filament voltage (volts)

0

Photoconductive Cell Output

1 V

2 V

3 V

4

5

V

V

6 V

7 V

8 V

9 V

10 V

V

Table 5.7 

Plot the graph of Photoconductive Cell Output Voltage against Lamp filament voltage on the graticule provided.

5.5 5.0 Photoconductive 4.5 Cell Output 4.0 Voltage (V) 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0

0

1

2


Graph 5.4

5.9a

From your graph estimate and enter the lamp filament voltage when the circuit output voltage is 3V. 

92




5.10 The PIN Photodiode Fig 5.13 shows the construction of the PIN photodiode.

Light

+

Light

N

Lens Depletion Layers Hole

Intrinsic (I) Region

I P

Electron

Electron/Hole pairs generated in the I region Contacts

Fig 5.13

This differs from a standard PN photodiode by having a layer of intrinsic (pure) silicon, the I region, between the normal P and N regions. The main improvement of the introduction of the I region is a reduction in the capacitance of the junction, resulting in a faster response time which can be as high as 0.5ns. The device can be operated in one of two ways: (a) as a photovoltaic cell, measuring the voltage output, and (b) by amplifying the output current and converting it into a voltage.

O/P

0V

Sensitivity

0.55A/W

Current Characteristic

2856KnA/lx

Response Time

3.5ns

Peak Spectral Response

850nm (IR)

Characteristics of a BPX65 PIN Diode

Fig 5.14

Fig 5.14 shows the circuit arrangement and characteristics for the PIN Diode mounted on the DIGIAC 1750 unit.

93



5.11 Practical Exercise Characteristics of a PIN Photodiode

MOVING COIL METER

5

0

-10

P .I .N . P HO T O D I O D E

P H O TO V O LT A I C C E L L

O/P

O/P

5 +10 +

+

+

O/ P

O/ P

-

L J

0V

P HOTO CO N DU CTIVECE LL

CURRENT AMPLIFIER WIREWOUND TRACK 6 5 C 7

4 3

8

2

9 1

10

P HO TO TRAN SISTO R

LAMP FILAMENT

I/P AMPLIFIER #1 I/ P

POWER AMPLIFIER O/P

B

O /P

O/P I/P 104 I IN

-

+

I/P

.4 1 100 10

A

10k

.5

+12V

OFFSET

GAIN COARSE

.6 .7 .8

.3

.9

.2 .1

1.0

GAIN FINE

V 0V

Fig 5.15

94



Connect the circuit as shown in Fig 5.15, using the Current Amplifier to measure the current output of the PIN Photodiode.



Use the digital multimeter on the 20V DC range to measure the output voltage of Amplifier #1. Fit the opaque box over the Clear Plastic Enclosure to exclude all ambient light.






Set the GAIN COARSE of Amplifier #1 to 10 and set the GAIN FINE to 1.0. Check that the OFFSET is giving zero output for zero input and adjust if necessary.



Take readings of Amplifier #1 output voltage as indicated on the digital multimeter as the lamp voltage is increased in 1V steps. Record the results in Table 5.8 in the row labeled PIN Photodiode Current Amp. O/P.


Lamp filament voltage (volts) PIN Photodiode

0

Current Amp. O/P


1

2

3

4

5

6

7

8

9

10

V

V

V

V

V

V

V

V

V

V

V

V

V

V

V

V

V

V

V

V

V

V

PIN Photodiode Output Voltage Table 5.8 

Transfer the Current Amplifier input and output connections to the Buffer Amplifier, so that the Buffer Amplifier replaces the Current Amplifier in the circuit. This will allow you to measure the output voltage of the PIN Photodiode.



Take readings of PIN Photodiode amplified Output Voltage as the lamp voltage is again increased in 1V steps. Record the results in Table 5.8 in the row labeled PIN Photodiode Output Voltage.



Plot the graphs of PIN Photodiode Current Amplifier Output Voltage and Buffered Output Voltage against Lamp filament voltage on the graticule provided.

4.5 4.0

PIN Photodiode Output Voltage 3.5 (V) 3.0

2.5 2.0 1.5 1.0 0.5 0

0

1

2


Graph 5.5

95


5.11a


From your graph, estimate and enter the lamp filament voltage when the circuit output voltage is 2V with the Current Amplifier connected.

5.11b

From your graph, estimate and enter the lamp filament voltage when the circuit output voltage is 2V with the Buffer Amplifier connected.

5.11c

Are the two graphs similar shapes?

Yes 

or

No


Notes: ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

96




2.

3.

4.

As the input voltage is varied, the light output from a filament lamp is:

a

directly proportional to the input voltage

b

of constant color temperature for tungsten

c

a non-linear relationship

d

reduced as the voltage is increased

The Solar Cell is a form of:

a

photovoltaic cell

b

light dependent resistor

c

photoconductive cell

d

phototransistor

A photovoltaic cell gives an output of 0.5V for a certain illumination level and is capable of a current output of 5mA. With two identical units connected (i) in series or (ii) in parallel the output capability will be:

a

(i) V = 0.5V, (ii) I = 5mA

b

(i) V = 1.0V, (ii) I = 5mA

c

(i) V = 0.5V, (ii) I = 10mA

d

(i) V = 1.0V, (ii) I = 10mA

A phototransist is connected tocollector a 10V DC supplyisvia a 2kTheload resistor. For one level of ambientor illumination the current 2mA. collector voltage will be:

a 5.

b

4V

c

6V

d 8V

For the phototransistor in question 4 above, if the level of ambient illumination is doubled and assuming that the device has a linear relationship, you would expect the collector voltage to be:

a 6.

2V

2V

b

4V

c

6V

d 8V

The Light Dependent Resistor (LDR) is a form of:

a

photovoltaic cell

b

PIN photodiode

c


d

phototransistor

Continued ...

97




8.

9.

As the level of illumination on a photoconductive cell is increased:

a

its resistance is increased

b

a voltage is generated

c

the junction temperature is increased

d

its resistance is reduced

The greatest advantage of a PIN photodiode over an ordinary photodiode is:

a

simpler construction

b

less semiconductor regions

c

greater capacitance

d

faster response times

Examining the characteristics of all of the opto-sensors seems to indicate that:

a

there is no similarity between any of them

b

the lamp has very little light output at low driving voltages

c

all of the circuits give an increasing output voltage as the illumination is increased

d

all of the circuits gave a decreasing output voltage as the illumination was increased

10. The characteristics of the PIN photodiode are most similar to those of the:

98

a

photovoltaic cell

b

tungsten filament lamp

c


d

phototransistor


Linear Position or Force Applications Chapter 6

Chapter 6 Linear Position or Force Applications



Describe the construction, principle and characteristics of a Linear Variable Differential Transformer (LVDT).



Describe the construction and characteristics of a linear variable capacitor.



Describe the construction and characteristics of a strain gauge.


•

• • •

DIGIAC 1750 Transducer and Instrumentation Trainer. 4mm Connecting Leads. Digital Multimeter. Oscilloscope.

99


6.1


The Linear Variable Differential Transformer (LVDT) The construction and circuit arrangement of an LVDT are as shown in Fig 6.1. It consists of three coils mounted on a common former and having a magnetic core that is movable within the coils.

Secondaries

Primary Coil Former

A

Primary A

B

B

Motion

Core

Magnetic Core

Secondaries Connections

Fig 6.1

The center coil is the primary and is supplied from an AC supply. The coils on either side are secondary coils and are labeled A & B in Fig 6.1. Coils A & B have equal number of turns and are connected in series opposing so that the output voltage is the difference between the voltages induced in the coils. Fig 6.2 shows the output obtained for different positions of the magnetic core.

AC Input A

AC Input B

A

C or e

A

Cor e

Output

+

Output

+

I/P

+

I/P

I/P

0

0

0

-

-

-

+

+

O/P

+

O/P

0

O/P

0

-

0

-

(a) Fig 6.2

B

Core

Output

100

AC Input B

-

(b)

(c)



With the core in its central position as shown in Fig 6.2(b) there should be equal voltages induced in coils A & B by normal transformer action and the output voltage would be zero. In practice this ideal condition is unlikely to be found, but the output voltage will reduce to a minimum. With the core moved to the left as shown in Fig 6.2(a), the voltage induced in coil A (Va) will be greater than that induced in coil B (Vb). There will therefore be an output voltage Vout = (Va - Vb) and this voltage will be in phase with the input voltage as shown. With the core moved to the right as shown in Fig 6.2(c) the voltage induced in coil A (Va) will be less than that induced in coil B (V b) and again there will be an output voltage Vout = (Va - Vb) but in this case the output voltage will be antiphase with the input voltage. Movement of the core from its central (or neutral) position produces an output voltage. This voltage increases with the movement from the neutral position to a maximum value and then may reduce for further movement from this maximum setting. Note that the phase will remain constant on either side of the neutral position. There is no gradual change of phase, only an abrupt reversal when passing through the neutral position. An amplitude only measurement of the output voltage, such as that provided by a meter, gives an indication of movement from the neutral position but will not indicate the direction of that movement. Used in conjunction with a phase detector, an output can be obtained that is dependent on both magnitude and direction of movement from neutral position. The oscilloscope gives both phase and magnitude indications. Fig 6.3 shows the circuit arrangement and device characteristics of the DIGIAC 1750 unit.

Turns per coil

I/P O/P

75

Inductance per coil 68 µ H Output Voltage

10mV/mm from neutral

Mechanical Travel 15mm 0V

LVDT Characteristics

Fig 6.3

101


6.2


Practical Exercise Characteristics of a Linear Variable Differential Transformer

LVDT

A.C. AMPLIFIER O/P

VARIABLE CAPACITOR

40kHz FILTER

FULL WAVE RECTIFIER

O/P

I/P

I/P

O/P I/P VIN

I/P

O /P

I /P

V

10 1000 100

O/ P GAIN

0V

40kHz OSCILLATOR

MOVING COIL METER

O/P 5

AMPLIFIER #1 I /P

-10

O/ P

0

5 +10

+ -

.5

+

.4 1 100 10

OFFSET

GAIN COARSE

.6 .7

.8

.3

.9

.2 .1

1.0

0V

L J

GAIN FINE

Fig 6.4

In this exercise you will measure the rectified output using the digital multimeter on the 20V DC range and also amplify and measure it using the M.C. analog meter, as this gives a better impression of the variation of output voltage with core position.

102



Connect the circuit as shown in Fig 6.4 with the digital multimeter on the 2V DC range to monitor the output of the Full-Wave Rectifier. Switch ON the power supply.



Set the A.C. Amplifier gain to 1000.



Set the GAIN COARSE control of Amplifier #1 to 100 and GAIN FINE control to 0.2. Check that the OFFSET control is set for zero output with zero input and adjust if necessary.



Adjust the core position by rotating the operating screw to the neutral position. This will give minimum output voltage. Note the value of this voltage from the digital multimeter and record in Table 6.1.



Rotate the core control screw in steps of 1 turn for 4 turns in the clockwise direction (when viewing the control from the left-hand side of the D1750 unit) and record your results in Table 6.1. Then turn the control screw in the counter clockwise direction, again recording the results in Table 6.1.



Core position (turns from neutral)

Output Voltage

-4

Digital meter

-3

-2

-1

0

+1

+2

+3

+4

V

V

V

V

V

V

V

V

V

V

V

V

V

V

V

V

V

V

Analog meter

Table 6.1 

Plot the graph of output voltage from the analog meter readings against core position on the axes provided.

10 Output 9 Voltage (volts) 8

7 6 5 4 3 2 1 0 -4

-3 -2 -1 0 +1 +2 +3 Core Position (turns from neutral)

+4

Graph 6.1

6.2a

6.2b

Enter your minimum voltage reading from the digital multimeter in mV.

Enter your voltage reading from the M.C. analog meter when the core is turned 2 turns out (-2) from the neutral position in V.




103


LVDT


VARIABLE CAPACITOR

A.C. AMPLIFIER O/P

40kHz FILTER

I/P

O/P I/P

I/P

O/ P

I/ P

10 1000

O /P

GAIN 100 40kHz OSCILLATOR

TP1

0V

TP2

0V

O/P

CH .1

C H .2

OSCILLOSCOPE

Fig 6.5 

Change the circuit to that shown in Fig 6.5 to observe the effect of the polarity ofchange in the output. Note panel that test are provided at the bottom the DIGIAC 1750 Trainer for points connection of oscilloscope probes.

104



Note: for the LVDT considered here, unless the two secondary coils are identical, there will be non-perfect coupling between each secondary coil and the primary coil, resulting in a frequency-dependent phase shift in the output voltage (relative to the input voltage).



Set up the oscilloscope as follows: lock the timebase to CH.1, trigger selector to AC CH.1 Amplifier on AC input, 50mV/div CH.2 Amplifier on AC input, 0.5V/div timebase to 5µs/div position both traces on the center horizontal line of the display



Switch ON the power supply and vary the core position through its full range and observe the effect on the output voltage as seen on CH.2 of the oscilloscope display.



Adjust the timebase fine control to give 1½ cycles of displayed waveform.





Sketch the oscilloscope waveforms when the core is turned 2 turns in (+2) from the neutral position on the graticule provided.

Waveform Sketch 6.1

6.2c

a CH.1

The waveform sketch, for perfectly coupled coils, would look most like: CH.2

b

CH.1

c CH.2

CH.1

d

CH.1

CH.2

CH.2



Switch OFF the power supply and reset the timebase fine control to the calibrated position.

Notes: ................................................................................................................................................................. ................................................................................................................................................................. .................................................................................................................................................................

105


6.3


The Linear Variable Capacitor Any capacitor consists of two conducting plates separated by an insulator which is referred to as the dielectric. The capacitance of the device is directly proportional to the cross sectional area that the plates overlap and is inversely proportional to the separation distance between the plates. A variable capacitor can therefore be constructed by varying either the area of plates overlapping or the separation distance.

Plated Brass Sleeve (fixed plate)

I/P

O/P

length l

Metal Slug (moving plate)

10kΩ

Spring

0V Contacts

(a)

(b)

Fig 6.6

Fig 6.6(a) shows the construction of the capacitor fitted in the DIGIAC 1750 unit, being fitted at the end of the coil former of the LVDT. This uses the magnetic slug core as the moving plate of the capacitor. The fixed plate consists of a brass sleeve fitted around the coil former. The capacitance magnitude depends on the length ( l) of the slug enclosed within the brass sleeve, the capacitance increasing with increase of length l. Fig 6.6(b) shows the circuit arrangement in the DIGIAC 1750 unit. The main characteristics of the unit are: Capacitance (minimum)

25pF

Capacitance (maximum)

50pF

Mechanical travel

15mm

Table 6.2

106


6.4


Practical Exercise Characteristics of a Variable Capacitor Transducer

LVDT

A.C. AMPLIFIER O/P

VARIABLE CAPACITOR

I/P

40kHz FILTER

FULL WAVE RECTIFIER

O/P I/P

O/P I/P VIN

I/P

O /P

I /P

10 1000 100

O /P GAIN WHEATSTONE BRIDGE D

40kHz OSCILLATOR

B

DIFFERENTIAL AMPLIFIER O/P A-B

O/P

0V

+

A

12k

C A 3

OUT

B

IN 1V

+5V

0V

V

AMPLIFIER #1 I/ P

O/ P

Rx -

.5

+

.4 1 100 10

OFFSET

GAIN COARSE

.6 .7 .8

.3

.9

.2 .1

1.0

GAIN FINE

Fig 6.7

The purpose of the Differential Amplifier is to provide a reference to give zero output voltage at any desired value of input voltage. The reference voltage is adjusted by the setting of the 10-turn potentiometer. 

Connect the circuit as shown in Fig 6.7 with the digital multimeter on the 20V DC range connected to the output of Amplifier #1.



Set the capacitor moving plate fully out to the minimum capacitance position, and then turn it back in until the marker on the operating control is first at the top. Now the device is near to its minimum capacitance position.



Set the AC Amp gain to 1000.



Switch ON the power supply and set the GAIN COARSE control of Amplifier #1 to 100 and GAIN FINE control to 0.4. Check that the OFFSET control is set for zero output with zero input and adjust if necessary.



Adjust the 10-turn potentiometer on the Wheatstone Bridge panel to give zero (as near as possible) output from Amplifier #1 (as close to 0V as possible) as indicated on the digital multimeter.



Turn the operating screw inwards in steps of 1 turn clockwise to increase the capacitance and at each step note the output voltage and enter the value in Table 6.3.

107


Approximate Capacitance Turns of screw


25pF ←Screw full out, minimum 0

1

2

3

4

50pF

Screw full in, maximum→ 5

6

7

8

9

10

0

Output Voltage

V

V

V

V

V

V

V

V

V

V

V

Table 6.3 

Plot the graph of output voltage against core positions above on the axes provided: 3.6 3.4 3.2 3.0 2.8 2.6 2.4 Output 2.2 Voltage (volts) 2.0

1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0

2

3

45678

9 10 Core Position (turns in )

Graph 6.2

6.4a

Enter the output voltage when the core is in position 4 above in V.

6.4b


Yes

108

or

No


6.5


The Strain Gauge Transducer Fine Resistance Wire

Fine Resistance Wire

(b) Sensitive Axis

(c) (a) Fig 6.8

Fig 6.8 shows the construction of a strain gauge, consisting of a grid of fine wire or semiconductor material bonded to a backing material. When in use, the unit is glued to the beam under test and is arranged so that the variation in length under loaded conditions is along the gauge sensitive axis (Fig 6.8(a)). Loading the beam increases the length of the gauge wire and also reduces its cross-sectional area (Fig 6.8(c)). Both of these effects will increase the resistance of the wire. +5V

Load Platform

Strain Gauges

O/P

+ Beam 0V

(a)

(b)

Fig 6.9

The layout and circuit arrangement for the DIGIAC 1750 unit is shown in Fig 6.9. Resistors are electro-deposited on a substrate on a contact block at the right-hand end of the assembly.

109



The gauge is normally connected in a Wheatstone Bridge arrangement with the bridge balanced under no load conditions. Any change of resistance due to loading unbalances the bridge and this is indicated by the detector (Galvanometer).

Standard

Strain

Dummy

Active

Resistance

Gauge

Gauge

Gauge

DC Supply

D Standard Resistance

Dummy

D Standard Resistance

Standard

(a)

Active D

Standard

Active

(b)

Dummy

(c)

Fig 6.10

Fig 6.10(a) shows the basic Wheatstone Bridge arrangement with one strain gauge transducer. This circuit is liable to give inaccurate results due to thermal changes. A variation of temperature will also produce a change of resistance of the gauge and this will be interpreted as a change of loading. To correct for this an identical gauge is used and connected in circuit as shown in Fig 6.10(b). This gauge is placed near to the other gauge but is arranged so that it is not subjected to any loading. Any variation of temperature now affects both gauges equally and there will be no thermal effect on the bridge conditions. The gauge subjected to loading is referred to as the active gauge and the other is called the dummy gauge. The output from the circuit is small and to increase this, four gauges are normally used with two active gauges and two dummies as shown in Fig 6.10(c). The DIGIAC 1750 uses two active gauges formed along the axis of the beam and two dummies formed at right angles to these. The main characteristics of the device are: Load capacity

Non-linearity

0.10%

Maximum deflection

0.5mm

Hysteresis

0.03%

Sensitivity

25µV/g

Creep

0.05%

Table 6.4

110

100g


6.6


Practical Exercise Characteristics of a Strain Gauge Transducer

STRAIN GAUGE

INSTRUMENTATION AMPLIFIER O/P

O/P

B

-

+

A

+

LOAD


A-B

+100VIN

MOVING COIL METER 5

AMPLIFIER #1 I/ P

-10

O/ P

0

5 +10

+ -

.5

+

.4 1 100 10

OFFSET

GAIN COARSE

.6 .7

.8

.3

.9

.2 .1

1.0

0V

L J

GAIN FINE

Fig 6.11

You will need ten similar weights, such as ten equal value coins, to increase the loading in regular steps. 

Connect the circuit as shown in Fig 6.11 and set Amplifier #1 GAIN COARSE control to 100.



Switch ON the power supply and with no load on the strain gauge platform, adjust the offset control of Amplifier #1 so that the output voltage is zero.



Place all ten of your weights on the load platform and adjust the GAIN FINE control to give an output voltage of 7.0V as indicated on the moving coil meter. Note that this value of output voltage should cover all ranges of coins within the setting of the GAIN FINE control.



Place one weight (coin) on the load platform and note the output voltage. Record the value in Table 6.5 overleaf.



Repeat the process, adding further weights one at a time, noting the output voltage at each step and recording the values in Table 6.5.

111


Number of coins

0

1

2

3

4


5

6

7

8

9

10

0

Output Voltage

V

V

V

V

V

V

V

V

V

V

V

Table 6.5 

Plot the graph of output voltage against number of coins on the axes provided:

Output 7 Voltage 6.5 (volts) 6 5.5 5 4.5 4 3.5 3 2.5 2

1.5 1 0.5 0 0

1

2

3

45678

9 10 Number of coins

Graph 6.3

a

112

6.6a

Enter the output voltage obtained with four coins on the platform.

6.6b

Your characteristic sketch is most similar to:

b

c

d




A Linear Variable Differential Transformer (LVDT) has two secondary coils A and B. When the core is moved to be nearest to coil A the voltages induced will be:

a c 2.

250mV aiding

b

500mV opposing

c

500mV aiding

d

1.0V opposing

A Linear Variable Differential Transformer (LVDT) has its core centralized under the primary and the voltage induced in one of the secondary coils is 500mV. The secondary output voltage will be:

a 4.

b greater in coil A than in coil B d reduced to zero

A Linear Variable Differential Transformer (LVDT) has its core centralized under the primary and the voltage induced in one of the secondary coils is 500mV. The voltage induced in the other coil will be:

a 3.

the same in both coils greater in coil B than in coil A

0V

b

250mV

c

500mV

d 1.0V

The output voltage of a Linear Variable Differential Transformer (LVDT) is taken to a full-wave rectifier. As the core is moved through the LVDT from one end to the other the output voltage of the full-wave rectifier will:

a

reverse polarity as the core passes through the central (neutral) position

b

remain constant at all positions of the core

c

gradually reduce from a maximum positive to zero

d

reduce to minimum and then return to maximum positive again

Continued ...

113




The Linear Variable Capacitor mounted on the DIGIAC 1750 Trainer varies in capacitance by changing:

6.

a b

the nature of the dielectric varying the distance between the plates

c

varying the effective cross-sectional-area of the plates

d

the reactance of the capacitor due to a change in frequency

From the information given in the notes about the main characteristics of the Linear Variable Capacitor, as the slug is moved the capacitance varies by about:

a 7.

25pF/mm

b

15pF/mm

c

1.67pF/mm

d 0.6pF/mm

When using a circuit similar to that investigated in Practical Exercise 6.4, as the slug of a Variable Capacitor is moved in towards the middle of the sleeve the output voltage:

a 8.

remains the same

b

increases

c

changes polarity

d reduces

The main disadvantage of using a single Strain Gauge in a Wheatstone Bridge sensing circuit is that it:

9.

a

has poor sensitivity

b

gives a non-linear response

c

is affected by temperature

d

cannot be null-balanced

The strain gauge bridge used in the DIGIAC 1750 Trainer has:

a c

one active and one dummy strain gauges four active strain gauges

b d

two active and two dummy strain gauges two active and four dummy strain gauges

10. The difference between an active and a dummy strain gauge is that:

114

a

they are made of different materials

b

an active gauge is longer than a dummy

c

an active gauge is shorter than a dummy

d

they are mounted in different directions


Environmental Measurements Chapter 7

Chapter 7 Environmental Measurements




Describe the construction and characteristics of an air flow transducer.



Describe the construction and characteristics of an air pressure transducer.



Describe the construction and characteristics of a humidity transducer.

• • •


115


7.1


The Air Flow Transducer

RTD and Heater

RTD Unheated

Fig 7.1

Fig 7.1 shows the construction of an Air Flow Transducer, consisting of two RTD's (Resistance Temperature Dependent) mounted in a plastic case. One of the devices has an integral heating element incorporated with it and the other is unheated. The operation of the device uses the principle that when air flows over the RTD's, the temperature of the heated unit will fall more than that of the unheated unit. The temperature difference will be related to the air flow rate which will in turn affect the resistance of the RTD's. With the DIGIAC 1750 Trainer, the transducers are enclosed in a clear plastic container and provision is made for air to be pumped over the device. Fig 7.2 shows the electrical circuit arrangement and main characteristics of the device in the DIGIAC 1750 Trainer.

+5V

Heater power

-

+ RTD

RTD

Heater

Fig 7.2

116

1.7k Ω

Output voltage (-). Pump OFF

2.1V

Output voltage (+). Pump OFF

1.7V

Voltage change (airflow)

0.06V

Table 7.1

0V

1W

Output impedance


7.2


Practical Exercise Characteristics of an Air Flow Transducer

AIR FLOW SENSOR

PRESSURE

FLOW

AIR PRESSURE SENSOR

-

-

O/P

O/P

PUMP OFF

+

ON

V

MOVING COIL METER

+

5

AMPLIFIER #1 0V

INSTRUMENTATION AMPLIFIER O/P B

I /P

5 +10

+ -

.5

+

.4 1 100 10

A-B

A

0

-10

O/P

+ OFFSET

GAIN COARSE

.6 .7

.8

.3

.9

.2 .1

1.0

0V

L J

GAIN FINE

Fig 7.3 

Connect the circuit as shown in Fig 7.3 and set the GAIN COARSE control of Amplifier #1 to 10 and GAIN FINE control to 1.0. Check that the pump control is set to OFF.



Set the digital multimeter to the 20V range.



Switch ON the power supply and allow the temperature to stabilize.









Adjust the OFFSET control of Amplifier #1 for zero output continuously during this time, setting the GAIN COARSE control to 100 when stabilized conditions are approached. Set the Flow/Pressure control to FLOW. Check that the OFFSET control is set for zero output voltage. Use the digital multimeter to note the voltages at the - and + outputs from the transducer, then note the Amplifier #1 output voltage displayed on the Moving Coil Meter. Record the values in Table 7.2 overleaf.

117





Switch the pump ON and note the voltages again when conditions have stabilized, recording the values in Table 7.2 Pump OFF

Transducer - Output Voltage Transducer + Output Voltage Amplifier #1 Output Voltage

Pump ON

V

V

V

V

0

V

Table 7.2

The RTD's have a positive temperature coefficient.

7.2a

Which output is connected to the heated RTD, - a or + b ?



Switch OFF the power supply and the pump.

Notes: ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ 118


7.3


The Air Pressure Transducer Fig 7.4 shows the construction of an air pressure transducer and also shows the electric circuit arrangement of the DIGIAC 1750 unit. The device consists of an outer plastic case which is open to the atmosphere via two ports. Within this case is an inner container from which the air has been evacuated and a strain gauge Wheatstone bridge circuit is fitted on the surface. +5V Backing Plate -

Strain Gauge

O/P

Ports

Contacts

+

Vacuum Cavity 0V

Construction

Electrical Circuit

Fig 7.4

The air pressure in the outer container will produce an output from the bridge and variation of the pressure will produce a variation of this output. The transducer output can be calibrated and may be called an absolute pressure transducer. Provision is made for air to be fed to the unit from the pump. The main characteristics of the device are: Type

SPX200AN

Sensitivity (typical)

300µV/kPa

Temperature coefficient

1350ppm/°C

Output Voltage Pump OFF (-) Output Voltage (+) Pump ON

2.48V

Voltage difference Pump OFF Voltage difference Pump ON Output impedance

35mV 39mV 1.6kΩ

2.51V

Table 7.3

119


7.4


Practical Exercise Characteristics of an Air Pressure Transducer

AIR FLOW SENSOR

PRESSURE

FLOW

AIR PRESSURE SENSOR

-


-

O/P

O/P

PUMP OFF

+

ON

+

B

-


A-B

A

+100VIN

+


5

I/P

-

O/P .5

+

.4 1 100 10

OFFSET

.6

GAIN COARSE

0

5 +10

+

.7

.3

.8

.2

.9 .1

-10

1.0

GAIN FINE

0V

L J

Fig 7.5 

Connect the circuit as shown in Fig 7.5 and set the Amplifier #1 GAIN COARSE control to 10 and GAIN FINE control to 0.3. Ensure that the pump

switch is set OFF. 

Switch ON the power supply and adjust the OFFSET control of Amplifier #1 for zero output voltage. The unit is now calibrated zero for the current value of the atmospheric pressure.



Set the Flow/Pressure control to PRESSURE and then switch the pump ON. The output voltage from the Amplifier #1 will increase. Note the value of this voltage.

Output voltage (Pump ON) = 7.4a

V

Enter your value of output voltage with the Pump ON in V.

Note that a large amplification is required due to the low magnitude of the device output. 

120

Switch OFF the power supply and the pump.


7.5


The Humidity Transducer Fig 7.6(a) shows the construction of a humidity transducer, consisting of a thin disc of a material whose properties vary with humidity. Each side of the disc is metalized to form a capacitor.

Capacitor Plates

I/P

O/P

Dielectric Disc

47kΩ 0V Contacts

(a)

(b)

Fig 7.6

Variation of humidity of the surrounding air alters the permittivity and/or thickness of the dielectric material, changing the value of the capacitor. The unit is housed in a perforated plastic case. Fig 7.6(b) shows the electrical circuit arrangement for the DIGIAC 1750 unit. The unit is connected in series with a resistor with the output taken from the resistor. With an alternating voltage applied to the input, the output voltage will vary with humidity due to the variation of capacitance of the transducer.

121



The main characteristics of the device are: Type

90001

Capacitance (25°C, 45%R/H)

122pF ± 15%

Sensitivity

0.4pF/%RH

Humidity Range

10%-90% RH

Table 7.4

Note: R/H is Relative Humidity,

Amb ie nt H umid it y Sat urated Air

x

100%.

The device is slow to respond fully to humidity changes, taking in the order of minutes, but this will normally be of no consequence in practice since natural changes in humidity are very slow. The variation of output voltage from the circuit is only a small percentage of the output and this is difficult to detect. In the practical exercise you will use signal processing circuits which are available on the DIGIAC 1750 to convert the output to asmall DC signal, out the standing DC level andTrainer thus enable amplification of the voltagebalance changes.

Notes: ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

122


7.6


Practical Exercise Characteristics of a Humidity Transducer

40kHz OSCILLATOR

A.C. AMPLIFIER O/P

HUMIDITY SENSOR I/P

O/P

40kHz FILTER

FULL WAVE RECTIFIER

O/P

O/P

I/P I/P

O/P

I/P

V

10 1000 100


GAIN SLIDE

C

2

3

4

5

6

7

8

9

10

0V

B

-

A

+

A-B

B 1

A

10k

MOVING COIL METER

5

AMPLIFIER #1 +5V

VIN

I/ P

-10

O/ P

0

5 +10

+ -

.5

+

.4 1 100 10

OFFSET

GAIN COARSE

.6 .7

.8

.3

.9

.2 .1

1.0

0V

L J

GAIN FINE

Fig 7.7 

Connect the circuit as shown in Fig 7.7, setting the AC Amplifier gain control to 10 and the Amplifier #1 GAIN COARSE control to 10 and GAIN FINE to 1.0.



Switch ON the power supply, remove the leads from the Differential Amplifier inputs and connect a short circuit between them. Adjust the OFFSET control of Amplifier #1 for zero output. Switch GAIN COARSE to 100 and make a final adjustment.



Replace the connections to the inputs of the Differential Amplifier and adjust the control of the 10k Ω carbon resistor for zero output from Amplifier #1. It may be advisable to set the coarse gain to 10 initially and then back to 100 finally during this process.

The bridge circuit is now balanced for the ambient conditions, the Differential Amplifier input from the 10kΩ variable resistor balancing that from the rectifier.

123





Note the output voltage from the rectifier circuit as indicated by the digital voltmeter. Output Voltage

Digital Meter

Ambient Conditions After Breathing

Moving Coil Meter 0 V

V

V

V

Table 7.5 

Now place your mouth near the humidity transducer and breath on it for a short time. The reading indicated by the Moving Coil Meter will change slowly.



Note the maximum value of the voltage and also the reading of the digital voltmeter.

Considering the readings obtained. Which meter do you consider gives a better indication of the voltage changes: a the Digital Multimeter, or 7.6a

b the Moving Coil Meter?

Enter your answer, a or b .

The time taken for the output voltage to return to zero after reaching the maximum voltage illustrates the slow response of the device to humidity changes. Time taken for output to return to zero =

Was the time taken for recovery less than 10 minutes a , or more b ? 7.6b

Enter your answer, a or b . 


Note: It is advisable to check the OFFSET of Amplifier #1 at regular intervals in case there has been any drift. This can be checked by just removing both of the input connections from the Differential Amplifier. The OFFSET control can then be adjusted if necessary. The ambient humidity conditions should not change during the test, but should a change occur, the bridge output will not return to zero.

124




The operating principle of the Air Flow Transducer relies on the use of:

a 2.

3.

strain gauges

b

RTD's

c

a capacitor

d a pressure pump

The Instrumentation Amplifier used in these experiments is a form of:

a

differential amplifier

b

bandpass filter

c

AC amplifier

d

summing amplifier

In Practical Exercise 7.2 (Air Flow Transducer Characteristics) the moving coil meter was balanced to zero at the start of the experiment using the:

a

10-turn variable resistor on the Wheatstone Bridge panel

b

10kΩ carbon slider potentiometer

c

balancing inputs on the Instrumentation Amplifier

d

offset control on Amplifier #1

4.

The operating principle of the Air Pressure Transducer relies on the use of: a strain gauges b RTD's c a capacitor d a pressure pump

5.

The output from an Air Pressure Transducer device is derived from a:

a 6.

series resistor

b

series capacitor

c

bridge circuit

d

x100

amplifier

At the start of the Air Pressure Transducer Characteristic experiment the output of the device is calibrated zero against:

a

relative humidity

b

ambient temperature

c

atmospheric pressure

d

ambient illumination

Continued ...

125




The operating principle of the Humidity Transducer relies on the use of:

a 8.

b

RTD's

c

a capacitor

d a pressure pump

The output from a humidity detector circuit varies between DC values of 3.50V and 3.52V over its full humidity range. Which of the following signal processing circuits would be necessary to provide an output range from 0 - 10V DC?

a 9.

strain gauges

AC amplifier

b

oscillator

c

DC amplifier

d 40 kHz filter

In the Humidity Transducer investigation, the DC component of the full-wave rectifier circuit was balanced out using:

a

10-turn variable resistor on the Wheatstone Bridge panel

b

10kΩ carbon slider potentiometer

c

balancing inputs on the Instrumentation Amplifier

d

offset control on Amplifier #1

10. The device investigated in this chapter with the slowest response time was the:

126

a

Air Flow Sensor

b

Air Pressure Sensor

c

Strain Gauge

d

Humidity Sensor


Rotational Speed or Position Measurements Chapter 8

Chapter 8 Rotational Speed or Position Measurements




Describe the construction, principles and application of Slotted Opto Transducers for counting and speed measurement.



Describe the construction, principles and application of Reflective Opto Transducers and Gray Coded Disc for position measurement.



Describe the construction, principles and application of Inductive Transducers for speed measurement.



Describe the construction, principles and application of Hall Effect Transducers to speed and positional measurement.



Describe the construction, principles and application of a Tacho-Generator to speed measurement.

• • •


127


8.1


The Slotted Opto-Transducer +5V

+12V

Slotted Aluminum Disc

Spindle

680Ω Infra-red LED

Photo-transistor

Case

O/P

47kΩ Contacts

470Ω

0V

(a)

(b)

Fig 8.1

Fig 8.1(a) shows the construction of a slotted opto transducer, consisting of a gallium arsenide infra-red LED and silicon phototransistor mounted on opposite sides of a gap in the case, each being enclosed in a plastic case which is transparent to infra-red radiations. The gap between them allows the infra-red beam to be broken when a solid object is inserted. The collector current of the phototransistor is low when the infra-red beam is broken and increases when the beam is admitted. Positive voltage pulses are obtained from the emitter circuit of the phototransistor each time the beam is admitted and hence the device generates pulses which are suitable for counting rotations. A slotted aluminum disc connected to the motor shaft assembly rotates in the transducer gap in the DIGIAC 1750 unit and an LED is provided to indicate when the slot position allows the beam to be admitted. Fig 8.1(b) shows the electrical circuit arrangement for the DIGIAC 1750 unit The main characteristics of the device are: Type

Output Voltage (beam broken)

0.1V

Output Voltage (beam admitted)

4.9V

Table 8.1

128

K8102


8.2


Practical Exercise Characteristics of a Slotted Opto Transducer

DC MOTOR

I/P

SLOTTED OPTO SENSOR

COUNTER/ TIMER I/P

O/P

TIM E

R E SET

O/P

FR E E R U N

C O U NT

1s

MOVING COIL METER WIREWOUND TRACK 6 5 C

5 POWER AMPLIFIER

7

4 3

8

2

9 1

10

B

O/P

0

5 +10

-10

V +

I/P

A

-

10k

0V +12V

L J

0V

Fig 8.2



Connect the circuit as shown in Fig 8.2 and set the 10k Ω wirewound resistor control fully counter-clockwise for zero output voltage.






Rotate the shaft by hand using the large aluminum disc provided with the Hall effect device. Note and record in Table 8.2 the output voltage from the Slotted Opto Transducer output socket and also the state of the indicating LED: (a) with the beam broken by the aluminum disc, and (b) with the beam admitted through the slot in the aluminum disc. Beam Broken Output Voltage

Beam Admitted V

V

LED - ON/OFF Table 8.2

129


Set the Timer/Counter to COUNT and FREE RUN. The display should show zero. If not, press RESET.






Rotate the shaft backwards and forwards by hand so that the slot in the aluminum disc passes between the opto transducer.



Note the counter display, this should increment by 1 each time the slot is in line with the transducer beam. This illustrates the use of the opto transducer for counting applications.



Now adjust the 10kΩ wirewound resistor control to give a drive voltage to the motor of 2V as indicated by the Moving Coil Meter. The motor should operate and rotate the shaft. The counter value will increment once for each revolution of the shaft and can be used to measure the shaft speed:



Motor Drive Voltage (volts) Shaft Speed (rev/sec) Shaft Speed (rev/min)

Press the RESET button and hold down. With a watch, stop watch if available, release the reset button at a suitable time and note the count value after one minute. This value represents the shaft speed in revolutions per minute (rev/min). Record the value in the last row of Table 8.3. 2

3

4

5

6

7

8

9

10

Table 8.3 



Repeat with a motor drive voltage of 3V and add the result to Table 8.3. Set the COUNTER/TIMER FREE RUN/1s switch to 1s (1 second). Set the 10k Ω resistor to give a motor drive voltage of 4V. Press the RESET button of the counter. The counter now counts for one second and the count value is "frozen" at the end of this time. The count displayed represents the number of revolutions per second of the shaft. Press RESET again. The displayed value should correspond with the previous value. Record the value in Table 8.3 in the shaft speed (rev/sec) row.

130





Repeat the procedure with the other motor drive voltages shown in Table 8.3 and for each setting note the shaft speed in rev/sec as displayed by the counter and add to the table. Switch OFF the power supply.



Multiply each recorded rev/sec value by 60 to give the shaft speed in revolutions per minute (rev/min or rpm) and add to the last row of Table 8.3.



Plot the graph of motor speed in rev/min against drive voltage on the axes provided:

3000 2800 Motor Speed 2600 (rev/min) 2400 2200 2000 1800 1600 1400 1200 1000 800 600 400 200 0

2

3

4

5

6

7 8 9 10 Motor Drive Voltage (volts)

Graph 8.1

8.2a

From your graph deduce and enter the motor drive voltage needed to give a speed of 1800 rev/min in V. Keep the motor drive circuits connected for later experiments.

131


8.3


The Reflective Opto Transducer

Gray Code Disc Drive Shaft

Infra-red LED

Phototransistor

Motherboard Reflective Opto-Sensors

PlanView

SideView(elevation)

Fig 8.3

Fig 8.3 shows the construction of a reflective opto transducer, consisting of an infra-red LED and phototransistor. This is similar to the slotted opto transducer, but in this device the components are arranged so that the beam is reflected back if a reflective surface is placed at the correct distance. A non reflective surface breaks the beam. Three separate units are provided with the DIGIAC 1750 unit, being mounted in line vertically. The reflective surface is a Gray-coded disc, which is fixed approximately 4mm from the transducers. With the beam not reflected the output from the phototransistor emitter is low. When the beam is reflected the output is high. Three LED's are provided to indicate when the beam is reflected from the respective transducer unit. The output A is the least significant bit (LSB) and C is the most significant bit (MSB). The Gray code is used for the encoded disc rather than normal binary because only one digit changes state at any boundary with this code and this minimizes any possibility of error in identifying the actual position when at a segment boundary.

132



The arrangement of the Gray-coded disc and the respective LED outputs is shown in Fig 8.4. Gray Code Disc 4

5 A

P osi t i on

C

B

A

B

0

0

0

0

1

0

0

1

2

0

1

1

3

0

1

0

4

1

1

0

5

1

1

1

6

1

0

1

7

1

0

0

3

6 C

2

7

0

1

Fig 8.4

The dark areas break the beam and produce a low output from the associated transducer and the bright areas reflect the beam and produce a high output. The DIGIAC 1750 unit operates as a rotational angular position transducer but similar principles can be used for linear position applications. Slotted opto devices could be used with a transparent disc (transparent where the above disc is reflective).

A B C 01234567

Linear Position Fig 8.5

Fig 8.5 shows a linear Gray-coded track, the A track is the LSB and C the MSB. The resolution provided with a 3-bit code (3 opto devices) is poor but this can be improved by increasing the number of devices and tracks.

133



Note the Gray code pattern: START

REPEATS

LSB

A

1 unit length '0'

2 unit lengths '1'

2 unit lengths '0'

MSB

B C

2 unit length '0' 4 unit length '0'

4 unit lengths '1' 8 unit lengths '1'

4 unit lengths '0' 8 unit lengths '0'

Table 8.4

The electrical circuit arrangement for the DIGIAC 1750 unit is shown in Fig 8.6:

+12V

+5V

C

B

A

C

B

0V

Fig 8.6

The main characteristics of the device are:

Type Output Voltage (beam broken)

Output Voltage (beam admitted) Table 8.5

134

K8711 0.5V

5V

A


8.4


Practical Exercise Characteristics of Reflective Opto Transducers and Gray Code Disc

REFLECTIVE OPTO SENSORS

C

0V

B

A

V

Fig 8.7 

Connect the circuit as shown in Fig 8.7 with the digital multimeter on the 20V DC range.



Switch ON the power supply and rotate the drive shaft by hand to alter the LED states.



Rotate the shaft until it is in the position with all LED's OFF. Use the digital multimeter to measure the voltage at each of the outputs and record in Table 8.6.

Output A B

Output Voltage LED OFF LED ON V

V

V

V

V

V

C Table 8.6 

Turn the shaft until all LED's are ON and repeat the readings, recording the results again in Table 8.6.

135





With the shaft initially in the position with all LED's OFF, rotate the shaft counterclockwise, when looking at the coded side of the disc, and note the state of the LED's at each change of state. Denote an LED OFF as logic state 0 and LED ON as logic state 1.



Record the values in Table 8.7. Position

C

B

A

0 1 2 3 4 5 6 7

Table 8.7

Check the sequence against that shown in the table in Fig 8.4.

8.4a

Enter the voltage at the 'B' output when the LED is ON.

8.4b

Enter the voltage at the 'B' output when the LED is OFF.

8.4c

The code which you have recorded for step 6 in the form C B A is:

a 110 

136

b

011


c

101

d

none of these


8.5


The Inductive Transducer

Thick slotted disc

10kΩ O/P I/P

1mH Ferrite bobbin

Coil

Fig 8.8

Fig 8.8 shows the construction and electrical circuit arrangement for the Inductive Transducer provided with the DIGIAC 1750 unit. This consists of a 1mH inductor and a slotted aluminum disc fitted to the drive shaft which rotates above the inductor. The inductance of the unit varies with the position of the slot. With an aluminum disc the inductance increases with the slot positioned directly above the inductor. If a magnetic disc was used, the inductance would decrease for the condition when the slot was above the inductor. Note that, if unscreened, an inductor will be liable to pick up any stray interference, such as that which may be generated by the motor commutator switching. This can generate spurious short duration output pulses which may need to be suppressed by using a low pass filter. The main characteristics of the device (in circuit under the disc) are: Inductance

(under slot)

1mH

Inductance change

(under disc)

7µH

Output voltage

(under slot)

6.9mV

Output voltage change

(under disc)

2mV

Table 8.8

137


8.6


Practical Exercise Characteristics of an Inductive Transducer

40kHz OSCILLATOR

INDUCTIVE SENSOR

O/P

A.C. AMPLIFIER O/P

40kHz FILTER

I/P

FULL WAVE RECTIFIER

O/P

O/P

I/P

I/P VIN

I/P

10 1000 100

O/P GAIN

+5V WHEATSTONE BRIDGE D

DIFFERENTIAL AMPLIFIER O/P 12k

C

B

-

A

+

A


A-B OUT

B

3

+100VIN

IN 1V

Rx

0V

MOVING COIL METER

5

AMPLIFIER #1 I/ P

-10

O /P

0

5 +10

+ -

.5

+

.4 1 100 10

OFFSET

GAIN COARSE

.6 .7

-

.3

.8

.2

.9 .1

1.0

0V

L J

GAIN FINE

Fig 8.9 

Connect the circuit as shown in Fig 8.9. Set the AC Amplifier gain to 1000 and Amplifier #1 GAIN COARSE to 10 and GAIN FINE to 1.0. Set the drive shaft with the disc slot in the top vertical position.



Remove the leads from the input to the Differential Amplifier, short the inputs together and switch ON the power supply.





138

Adjust the

OFFSET control of Amplifier #1 for zero output.

Replace the leads to the input of the Differential Amplifier and adjust the control of the 10kΩ 10-turn resistor so that the meter reading is again zero. The control setting will be critical with such high overall amplifier gains.



Check the zero reading and then rotate the motor shaft to obtain the maximum output voltage when the slot is immediately above the Inductive Sensor. Note the value of this voltage:



Output voltage with slot over the inductor =

8.6a

V

Enter your maximum value of output voltage in V.

This indicates an application of inductive transducers to proximity detection of metallic objects. The device can also be used for counting or speed measurement applications. Switch OFF the power supply. Retain your circuit, but remove the Moving Coil Meter from the output of Amplifier #1 and then add the circuits of Fig 8.10.



LOW PASS FILTER O/P

DIFFERENTIATOR O/P

I/P

I/P

T I/P

DC MOTOR

100ms 10ms 1s

100ms 1s 10ms

TIME CONSTANT

To the output of Amplifier#1 O/P

8

2

9 1

10

COUNTER/ TIMER

B

TIME

R ESET

POWER AMPLIFIER

7

3

TIME CONSTANT

I/P

WIREWOUND TRACK 6 5 C 4

O/P

dVIN dt

C O UN T

FREE RUN

1s

MOVING COIL METER

I/P

5

A

-10

10k

0

5 +10

+ +12V

0V

0V

J L Fig 8.10

139





Set the motor speed to zero.



Set the TIME CONSTANT switches of the Low Pass Filter and the Differentiator to 1s and set the counter to COUNT and 1s.



Switch on the power supply.



Apply 2V input to the DC motor so that the shaft rotates slowly. Press the counter reset button several times and note the displayed value, this represents the speed in rev/sec.

Speed of the shaft recorded with the Inductive Sensor = 

Remove the lead from the o/p of the Low Pass Filter to the Differentiator and take the lead from the input of the Low Pass Filter and connect it to the Differentiator input. Press the Counter RESET button several times and observe the result. If the result is zero, then refer to the re-calibration procedure described in the next point and repeat the counts with and without the Low Pass Filter. When a reading has been observed restore the Low Pass Filter back into the circuit by moving the lead back and adding the connection between the Low Pass Filter and Differentiator.

Speed of the shaft recorded without Low Pass Filter =

140



Re-calibrate the Inductive Sensor circuit by removing the lead from the MC meter to the Power Amplifier and connecting it between the MC meter and the output of Amplifier #1. Adjust the control of the 10KΩ 10-turn resistor so that the meter reading is zero. Then reconnect the MC meter to the Power Amplifier.



Remove the Counter input lead from the Differentiator output and connect it to the output from the Slotted Opto Transducer. Press the counter reset button and note the displayed reading which also represents the shaft speed. Compare these value with the value obtained from using the Inductive Sensor.



Repeat the two measurements for the motor input voltages and complete Table 8.9 on the next page.



Motor Voltage Shaft Speed (rev/sec)

2V

4V

7V

10V

Inductive Transducer Slotted Opto Transducer

Table 8.9

8.6b

When the Low Pass Filter was removed from circuit the effect on the Counter readings was to:

a produce a constant count each time b

make no change

c reduce the count

increase the count

d

You will note that a considerable amount of signal conditioning has been required for the inductive transducer unit due to the small output voltage available and also the problem of the susceptibility of the counter to voltage spikes.

Notes:

........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................

141


8.7


The Hall Effect Transducer Magnetic Field Thick Aluminum Disc +5V Drive Shaft

Hall Effect Voltage VH -

Current 634SS2

O/P

+

N S

Hall Effect Sensor 0V

Embedded Magnet Motherboard

Fig 8.11

Fig 8.11 shows the layout and electrical circuit arrangement of the Hall Effect Transducer assembly fitted to the DIGIAC 1750 Trainer and illustrates the Hall Effect principle. Hall Effect Principle When current flows through the flat slice of semiconductor at right-angles to a magnetic field there is a force on each individual electron which tends to move it in one particular direction (the motor principle).

The current is pushed to one side of the slice. The surplus of electrons on one side of the slice means that this side is negatively charged, resulting in an EMF across the slice (the Hall voltage VH) which is at right-angles to both the current and the magnetic field. The value of this voltage is directly proportional to the strength of the magnetic field. The transducer provided on the DIGIAC 1750 Trainer also contains an active silicon semiconductor device to increase the output voltage and provide differential outputs, one going more positive and the other more negative (less positive). The main characteristics of the device are: Output voltage (+) (no field) Output voltage (-) (no field) Output voltage change Output voltage change (under magnet) Table 8.10

142

1.75-2.25V 1.60V 7.5-10.6mV/mT 380mV


8.8


Practical Exercise The Characteristics of a Hall Effect Transducer

DC MOTOR

I/P

HALL EFFECT SENSOR

COUNTER/ TIMER

B

I/P

TI ME

FR EE RU N

A-B

-

O/P

DIFFERENTIAL AMPLIFIER O/P -

+

A O/P

R ESET

C OU N T

1s

+

POWER AMPLIFIER

7

3

8

2

9 1

10

MOVING COIL METER

V


O/P

B

I/P

5

AMPLIFIER #1 I/P

A

0

5 +10

-10

O/P

+

10k

-

+12V

.5

+

.4 1 100 10

0V OFFSET

.6 .7

-

.3

.8

.2

.9

GAIN COARSE

.1

1.0

0V

L J

GAIN FINE

Fig 8.12 

Connect the circuit as shown in Fig 8.12. Set the Amplifier #1 GAIN COARSE control to 10, GAIN FINE to 0.8 and the motor drive voltage to zero. Switch ON the power supply.



Set the drive shaft position so that the magnet in the Hall effect disc is horizontal (to one side) so that there is no magnetic field cutting the Hall effect device.



Adjust the OFFSET control of Amplifier #1 for zero output indication on the Moving Coil Meter.



Note the output voltage from the - and + output sockets of the Hall Effect device with the digital voltmeter directly on the Hall Effect Sensor panel and also from the Moving Coil Meter. Record the results in Table 8.11. Magnetic Field None Maximum

Digital Multimeter Output Voltage (-) Output Voltage (+) V

Moving Coil Meter 0 V V

V

V

V

Table 8.11

143





Rotate the disc so that the magnet is directly above the Hall effect device. This position will be indicated by the maximum output voltage.



Note the voltages again and record in Table 8.11.

These readings illustrate the basic characteristics of the Hall Effect device and indicate its application measurement applications.to proximity detection. It is also suitable for speed 

With the output of Amplifier #1 connected to the Counter/Timer input set the controls for COUNT and 1s.



Transfer the digital multimeter to the output of the Power Amplifier and apply an input voltage of 2V to the motor so that the shaft rotates slowly. Press the Counter RESET button and note the displayed value, this representing the shaft speed in rev/sec. Record the result in Table 8.12



Remove the input to the counter from Amplifier #1 and connect it to the output of the Slotted Opto Transducer unit. Press the counter "reset" button and note the displayed value, this being the shaft speed for comparison with the previous reading. Add the value to Table 8.12. Motor Voltage Shaft Speed (rev/sec)

2V

4V

7V

10V

Hall Effect Transducer Slotted Opto Transducer Table 8.12 

8.8a

Repeat the procedure for the other values of motor drive voltage given in Table 8.12 for comparison. Switch OFF the power supply.

Enter your value of output voltage from the (+) O/P of the Hall Effect sensor with no magnetic field.

8.8b

Enter your value of output voltage from the (+) O/P of the Hall Effect sensor with maximum magnetic field.

Hall Effect devices are available for proximity detection, linear or angular displacement, multiplier and current or magnetic flux density measurement applications.

144


8.9


The DC Permanent Magnet Tacho-Generator Fig 8.13 shows the construction and electrical circuit arrangement of the DC Permanent Magnet Tacho-Generator fitted to the DIGIAC 1750 Trainer. This consists of a set of coils connected to a commutator which rotate inside a permanent magnet stator.

N

+12V

Coils

O/P

Commutator S

Brushes

Permanent Magnet Stator

M 0V

Armature

-12V

Connections

Fig 8.13

The rotating assembly is called the armature. With the coils rotating, an alternating EMF is generated in them. The commutator converts this to DC. The magnitude of the generated EMF is proportional to the rate of cutting flux and therefore to the rotational speed. The polarity depends on the direction of cutting flux and therefore on the direction of rotation. The diodes are fitted to limit any voltage spikes that may be generated by the commutation process (i.e. conversion from AC to DC) to a maximum of ±12V. The main characteristics of the device are: Open circuit voltage (12V to motor)

10.5V

Short circuit current (12V to motor)

750mA

Output impedance

39Ω

Output noise

200mV p-p

Table 8.13

145



8.10 Practical Exercise Characteristics of a Permanent Magnet DC Tacho-Generator


0

I/P

5

O/ P

-

.5

+

.4 1 100 10

OFFSET

GAIN COARSE

+10

+

.7 .8

.2

.9

0V

1.0

GAIN FINE

DC MOTOR

I/P

.6

.3

.1

5

-10

L J

SLOTTED OPTO SENSOR

TACHOGENERATOR

O/P

O/P

O/P


POWER AMPLIFIER

7

4 3

8

2

9 1

B

O/P

COUNTER/ TIMER I/P

T IM E

F R E E R UN

I/P

A

10

10k

V +12V

R ES ET

C O UNT

0V

Fig 8.14 



Connect the circuit as shown in Fig 8.14. Set the

COUNTER/TIMER controls to COUNT and 1s.

Set Amplifier #1 GAIN COARSE control to 10 and GAIN FINE to 0.1. 

146


1s





Apply an input to the motor and set the shaft speed to 5 rev/sec (Note: Table 8.3 and Graph 8.1 may help) as indicated by the counter after pressing the RESET button. Note the output voltages indicated on the Moving Coil Meter and record the values in Table 8.14. Shaft Speed (rev/sec)

5

Output Voltage (Moving Coil Meter)

10

V

20

V

30

V

40

V

V

Table 8.14 

Repeat the procedure for the other shaft speed settings indicated in Table 8.14.



Draw the graph of output voltage against shaft speed on the axes provided.

10 Output 9 Voltage 8 (volts) 7

6 5 4 3 2 1 0

0

5

10

15

20

25 30 35 40 Shaft Speed (rev/sec)

Graph 8.2

8.10a


Yes

8.10b

or

No

From your graph estimate and enter your recorded output voltage from the digital multimeter when the shaft speed is 25 rev/sec.

147



Calibration of the Moving Coil Meter to Indicate Speed Directly.

The scale to be used is 20V represents 2000 rev/min (100 rev/min/V).

MOVING COIL METER

1000 0

-10

500 5 -7

0

1500 5 2000 +8 +10

rev/min

+ 0V

L J

Fig 8.15 

Transfer the connection of the Moving Coil Meter from the input of Amplifier #1 to the output of Amplifier #1. Set the GAIN FINE control to just a little above 0.3.

148



Apply a low input to the motor and set the shaft speed to 5 rev/sec (300 rev/min) as shown on the Counter after pressing RESET. Adjust the OFFSET control of Amplifier #1 to set the Moving Coil Meter reading to -7V (Fig 8.15).



Change the motor drive voltage to set the shaft speed to 30 rev/sec (1800 rev/min) as shown on the Counter after pressing RESET. Adjust the GAIN FINE control of Amplifier #1 so that the Moving Coil Meter indicates +8V (Fig 8.15).



Repeat both of the above settings and adjustments as often as necessary to make both of them correct (changing one of them will have altered the other. Some anticipation may be helpful). The meter will then be calibrated as shown in Fig 8.15.





Use the calibrated Moving Coil Meter to set the motor speed as shown in Table 8.15.



Calculate the corresponding speed in rev/sec and then check at each setting against those obtained from the Opto Transducer and Counter. Shaft Speed (rev/min) Calculated Shaft Speed (rev/sec) Shaft Speed from Counter (rev/sec)

600

1000

1200

1600

Table 8.15

8.10c

Which was easier for setting the motor speed, the calibrated Moving Coil Meter a , or the Counter b ?

8.10d

Enter your motor speed as indicated on the Counter in rev/sec when the motor speed was set to 1000 rev/min.

8.10e

Enter your motor speed as indicated on the Counter in rev/sec when the motor speed was set to 1200 rev/min. 

Switch OFF the power supply

149




For a slotted opto transducer to count revolutions of a shaft it requires a:

a 2.

7.

150

non-magnetic disc d magnetic disc

500 µV

b

50mV

c

5V

d 12V

240

b

600

c

720

d 1440

0 1 1, 0 1 0

b

0 0 1, 0 1 0

c

1 1 0, 1 1 1

d 1 1 1, 1 0 1

The number of outputs generated by a 3-bit Gray code system is :

a 6.

c

Which of the following could NOT be in sequence (one after the other) for a Gray code output?

a 5.

reflective disc

A shaft speed of 24 rev/sec corresponds to a motor speed in rev/min (rpm) of:

a 4.

b

The output voltage generated by the slotted opto transducer used in your experiments was in the order of:

a 3.

slotted disc

3

b

6

c

8

d 10

Comparing a slotted magnetic disc to a slotted aluminum disc, the effect on the inductance of the slot passing over an inductive transducer is to:

a

increase the inductance in both cases

b

increase the inductance with a magnetic disc but decrease it with aluminum

c

decrease the inductance in both cases

d

decrease the inductance with a magnetic disc but increase it with aluminum

The purpose of the low pass filter used with the inductive transducer was to:

a

remove interference pulses

b

respond to low revolutions only

c

respond to high revolutions only

d

display only low revolution counts




9.

The purpose of the differential amplifier in the inductive transducer experiment was to:

a

respond to both increases or decreases in transducer output voltage

b

balance out the steady DC component of the rectifier output

c

invert the polarity of the signal from the transducer

d

sharpen up the output pulses from the transducer

An output voltage is generated by a Hall effect device when:

a

a slot in an aluminum disc passes over it

b

a magnetic field passes through it

c

it is exposed to light radiations

d

an external EMF is applied

10. The expected change in output voltage of an activated Hall effect transducer of the type fitted on the DIGIAC 1750 Trainer is of the order of:

a

±500µV

b

±500mV

c

±5V

d

±12V

11. The magnitude of the output voltage generated by a permanent magnet tacho-generator is dependent on:

a

rate of cutting flux

b

direction of rotation

c

frequency of interference pulses

d

number of commutator segments

12. The purpose of the commutator in a permanent magnet tacho-generator is to:

a

increase the magnitude of the output voltage

b

increase the frequency of the output voltage

c

stabilize the input voltage

d

convert the output from AC into DC

151



Notes:

........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................

152


Sound Measurements Chapter 9

Chapter 9 Sound Measurements



Describe the construction and characteristics of a dynamic microphone.



Describe the construction and characteristics of an ultrasonic receiver and transmitter.




• • • •

Compares the various methods of measuring sound signals.

DIGIAC 1750 Transducer and Instrumentation Trainer. 4mm Connecting Leads. Oscilloscope. 12 inch (30cm) ruler (not supplied).

153


9.1


The Dynamic Microphone

Fig 9.1

The construction of the dynamic microphone is shown in Fig 9.1(a), consisting of a coil attached to a thin diaphragm, the coil being suspended in the field of a permanent magnet. The diaphragm moves in response to any vibration in the air caused by sound and moves the coil in the magnetic field. An alternating EMF is induced in the coil, the magnitude and frequency of which is proportional to the sound vibrations. The electrical circuit for the device provided with the DIGIAC 1750 unit is shown in Fig 9.1(b). A resistor is fitted to provide a load matched to the output impedance of the microphone. The main characteristics of the device are: Output impedance

Typically 200 - 500Ω

Frequency response (-3dB)

100Hz - 10kHz

Output voltage

5mV (normal maximum)

Table 9.1

154


9.2


Practical Exercise Characteristics of a Dynamic Microphone It is most unlikely that your laboratory will include a broadband constant output audio generator system/loudspeaker amongst its facilities. Even if it did, a full acoustic booth would be required, and the noises generated would be unacceptable for other laboratory users. We are therefore not able to test the full dynamic range of a microphone, either for frequency or amplitude. It is therefore necessary for us to limit the investigation to a review of the measurement techniques that can be adopted, and it is these which will be examined, rather than the microphone itself.

A.C. AMPLIFIER O/P

MICROPHONE

I/P

FULL WAVE RECTIFIER O/P I/P

O/P

VIN 10 1000 100

L.E.D. BARGRAPH DISPLAY

GAIN I/P AMPLIFIER #1 I/P

-

O/P .5

+

OFFSET

.6

GAIN COARSE

MOVING COIL METER

.7

.4 1 100 10

DRIVER I.C.

.8

.3

5

.9

.2

-10

1.0

.1

0

5 +10

GAIN FINE

+ AMPLIFIER #2

-

I/P

O/P

0V -

.5

+

.4 1 100 10

OFFSET

GAIN COARSE

.6

L J

.7 .8

.3

.9

.2 .1

TP1

0V

1.0

GAIN FINE

Oscilloscope CH.1

Fig 9.2

155



In this exercise three different forms of monitoring device will be investigated. The response time of digital multimeters is too slow to make any record of the signals at all, due to the transient nature of sound. 

Connect the circuit as shown in Fig 9.2. Set the AC Amplifier gain control to 1000 and the Amplifier #1 GAIN COARSE to 1 and GAIN FINE to 0.4.

The LED bargraph display has an excellent response time and requires 0.5V for each bar, 5V to light the whole display. This type of device is often used on HI-FI systems. 

Switch ON the power supply. Check the OFFSET control of Amplifier#1 for zero with the Moving Coil Meter temporarily connected to its output. Note the display on the Bargraph when the bench is tapped with the finger.



Tap the case of the 1750 unit and observe the effect on the Bargraph display.



Change the GAIN COARSE of Amplifier #1 to 10 and the FINE GAIN to 1.0 then talk, cough, sing or whistle near the unit. You will find that the bargraph will respond to any sound made, but needs more gain for speech or whistling.

A Moving Coil Meter is frequently used by sound (audio) engineers to indicate peak power (PPM, peak power meter), but requires a rectifier and amplifier since the moving coil meter only responds to DC, and its movement is slow to respond due to inertia and damping. 

The Moving Coil Meter is connected to the AC Amplifier output via the Full Wave Rectifier and Amplifier #2. Set the GAIN COARSE to 10 and GAIN FINE to 1.0, and zero the indication of the meter using the OFFSET control. Tap the baseboard so that all LED's of the bargraph are lit and note the maximum reading of the Moving Coil Meter. Maximum voltage indication given by the Bargraph is 5V. Maximum voltage output (Moving Coil Meter) =

9.2a

156

V

Enter your maximum voltage indicated by the Moving Coil Meter in V.



Without any doubt, the oscilloscope is the most versatile device for monitoring sound, since it is able to give an indication of frequency, waveform and magnitude of signals and is very sensitive, even to small signals. 

Set the timebase of the oscilloscope to 2ms/div and the CH.1 Y1 amplifier to 1V/div.



Generate various types of sound and observe the display on the oscilloscope. Note that sound engineers, to save their embarrassment, will often count say from one through ten and back again - to test a microphone circuit. It may be necessary to vary the Y amplifier setting to obtain the most satisfactory displayed waveform.



9.2b

Change the timebase setting to 0.5ms/div. and try whistling two different notes, one low pitch and the other high, and observe the effect on the number of cycles (frequency) of the displayed waveform.

The most commonly observed waveform for all types of sound is a:

a square wave

9.2c

sinewave

c

triangular wave d

irregular wave

The steadiest, purest waveform was produced by:

a tapping

9.2d

b

b

talking

c

coughing

d

whistling

The effect of a high-pitched note compared to a low note was to produce:

a the same number of cycles

b

less cycles, lower frequency

c more cycles, higher frequency

d

no difference at all

157


9.3


The Ultrasonic Transmitter/Receiver O/P

Fine Wire Mesh

I/P

Case Cavity Diaphragm

Piezo Ceramic Element

Ultrasonic Receiver

Ultrasonic Transmitter 0V

0V

Contacts

(a)

(b)

Fig 9.3

The construction of both ultrasonic devices and their electrical circuit arrangements for the DIGIAC 1750 unit are shown in Fig 9.3. The receiver and transmitter are almost identical and consist of a slice of ceramic material with a small diaphragm fixed to it, inside the case of the unit. The operation of the receiver relies on the principle that certain ceramic materials produce a voltage when they are stressed. This is referred to as the piezo-electric principle. Vibration of the diaphragm stresses the ceramic material and produces an output voltage. The reciprocal applies to the transmitter. An applied alternating voltage produces stress which causes the ceramic slice to vibrate. The dimensions of the components are arranged so that there is resonance (best response) at around 40kHz. This is above the audible range (maximum 20kHz) and is therefore referred to as ultrasonic. The ceramic slice is arranged in four quarters which are connected in series for the receiver and in parallel for the transmitter. The main characteristics of the devices are: Receiver Peak resonance (typical)

40kHz

Directional angle Impedance Output amplitude Table 9.2

158

Transmitter

30° 30kΩ 5-60mV

500Ω


9.4


Practical Exercise Characteristics of an Ultrasonic Transmitter/Receiver

ULTRASONIC TRANSMITTER

40kHz OSCILLATOR O/P

I/P

A.C. AMPLIFIER O/P

ULTRASONIC RECEIVER

O/P

40kHz FILTER

I/P

LOW PASS FILTER O/P

FULL WAVE RECTIFIER

O/P I/P

I/P

O/P I/P VIN

10 1000 100

100ms 10ms 1s

GAIN

TIME CONSTANT L.E.D. BARGRAPH DISPLAY I/P

AMPLIFIER #1 I/P

-

DRIVER I.C.

O/P .5

+

.4 1 100 10

OFFSET

GAIN COARSE

MOVING COIL METER

.6 .7

.3

.8

.2

.9 .1

1.0

GAIN FINE

5 -10

0

5 +10

+ 0V

L J

Fig 9.4 

Connect the circuit as shown in Fig 9.4. Set the AC Amplifier gain control to 1000 and Amplifier #1 GAIN COARSE control to 10 and GAIN FINE to 0.5. Switch the Low Pass Filter time constant to 100ms.



Switch ON the power supply and adjust Amplifier #1 OFFSET to give zero



Note the bargraph display as you move your hand or any other object over the ultrasonic devices. The display should respond, indicating the receipt of a signal of frequency 40kHz by the ultrasonic receiver.

output on the Moving Coil Meter.

159




9.4a


Place a small book (approximately 6 inches (15cm) × 4 inches (10cm) or other flat object 3 feet (90cm) above the Ultrasonic Transducers. Slowly move the object closer to the transducers, watching the output reading on the bargraph display, until the object is covering the transducers.

At which of the following positions is the maximum output obtained? a Object in contact with the Ultrasonic Tranducers.

b

Object 4 inches (10cm) above the unit.

c

Object 12 inches (30cm) above the unit.

d

Object 3 feet (90cm) above the unit.

Remove any other equipment from the vicinity so that you have free access to the ultrasonic transmitter/receiver area. 

9.4b

Increase the Amplifier #1 GAIN FINE control to 1.0. Hold a thin object such as a pencil approximately 6 inches (15cm) above the Ultrasonic Transducers, move it horizontally and vertically and note the effect on the output response. This indicates how critical the direction angle is for the device.

Is the position of the reflector critical?

Yes 

9.4c

or

No

Put a sheet of paper over the Ultrasonic Transducers to intercept the path and move your hand up and down above the transducers.

Does the beam pass through a piece of paper?

Yes

or

No

In this exercise the received signal has been amplified, rectified, filtered (to remove all unwanted frequencies) and then amplified again to operate the display. Pulsed ultrasonic devices can be used for distance measurement to reflecting surfaces by measurement of the time between the transmission and return of the pulsed signal.

160




2.

The dynamic microphone operates due to:

a

a change of resistance

b

the piezo electric effect

c

the Doppler effect

d

a conductor cutting magnetic flux

The output impedance of the microphone fitted on the DIGIAC 1750 Trainer is:

a 3.

6.

8.

Ω

c

200 - 500

Ω

d 1500 - 1200

Ω

digital multimeter b

oscilloscope

c

LED bargraph

d moving coil meter

rectifier

b

low pass filter

c

band pass filter

d

40kHz oscillator

Ultrasonic Receivers and Transmitters rely for their operation on:

a

a change of resistance

b

the piezo electric effect

c

the Doppler effect

d

a conductor cutting magnetic flux

The peak resonance of the ultrasonic receiver and transmitter used on the D1750 unit is at a frequency of:

a 7.

100 - 150

The Moving Coil Meter cannot respond to sound signals without a:

a 5.

b

The most versatile test instrument for monitoring sound signals is the:

a 4.

25 - 75 Ω

20kHz

b

30kHz

c

40kHz

d 60kHz

Which audio transducer on the D1750 unit has the highest impedance?

a

Ultrasonic receiver

b

Ultrasonic transmitter

c

Microphone

d

Low pass filter

The frequency response (-3dB) of the microphone used on the D1750 unit is:

a

0 - 40KHz

b

0 - 10KHz

c

100Hz - 40KHz

d

100Hz - 10KHz

161



Notes: ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...........................................................................................................................................

162


Sound Output Chapter 10

Chapter 10 Sound Output




Describe the construction and characteristics of a moving coil loudspeaker.



Describe the construction and characteristics of a buzzer.

• • • • • •

DIGIAC 1750 Transducer and Instrumentation Trainer. 4mm Connecting Leads. BNC to 4mm Connecting Lead. Digital Multimeter. Oscilloscope. Function Generator.

163



10.1 The Moving Coil Loudspeaker

Paper Cone I/P

Contacts

100Ω Loudspeaker

Frame S

Permanent Magnet

S N

Coil

0V

(a)

(b)

Fig 10.1

The construction of a moving coil loudspeaker is shown in Fig 10.1(a). It is similar to the moving coil microphone. The permanent magnet, coil and diaphragm are much the same but in this device the diaphragm is attached to a large paper cone supported by a frame. The cone is free to move with the coil. Alternating currents flowing in the coil cause it react with the magnetic field and move in and out. With applied currents at frequencies in the audible range, the cone movement will cause a variation of pressure in the surrounding air particles and produce sound waves that are audible to the human ear. If a speaker is placed in a vacuum, there are no air particules, so the movement of the cone does not produce any sound. The electrical circuit of the device fitted to the DIGIAC 1750 unit is shown in Fig 10.1(b). The 100Ω resistor is fitted to limit the maximum power dissipation to 100mW, half of the rated value for the loudspeaker. The main characteristics of the device fitted to the DIGIAC 1750 unit are: Impedance

8Ω

Power rating

200mW rms.

Frequency response (-3dB)

400-5000Hz

Table 10.1

Note that the speaker response is well below the maximum frequency detectable by the human ear (approximately 20kHz).

164



10.2 Practical Exercise Characteristics of a Moving Coil Loudspeaker Function Generator


POWER AMPLIFIER

LOUDSPEAKER

7

4 3

8

2

9 10

1

B

O/P I/P

I/P

A

10k MICROPHONE

x100 AMPLIFIER

0V O/P

O/P I/P +100VIN

TP1

0V

CH.1

TP2

Oscilloscope

0V

CH.2

Fig 10.2



Connect the circuit as shown in Fig 10.2 and switch ON the power supply.



Set the 10kΩ wirewound variable resistor to position 5 on its scale (see Fig10.2).



Set the oscilloscope timebase initially to 1ms/div, CH.1 Y amplifier to 5V/div and CH.2 Y amplifier to 0.2V/div.



Set the function generator to 200Hz sinewave output and adjust the amplitude control to maximum and then adjust the 10kΩ wirewound resistor to give a signal input of 10Vp-p (2 div.) as seen on CH.1 of the oscilloscope. The signal input level of 10Vp-p is to be carefully maintained for tests at all frequencies.

The microphone and its amplifier will pick up all of the background sounds and interference the laboratory. Try to ignore these in taking your readings at so lower signal levels.inYou will be contributing to other peoples background noises, try to keep yours to a minimum.

165




Frequency (Hz)

200


Take readings at each of the frequencies given in Table 10.2, ensuring that the input signal remains constant at 10Vp-p. 300

400

500

600

700

800

900

1k

2k

3k

Output Voltage Vpeak-to-peak Table 10.2 

One of your readings should have been much greater than any of the rest. Return to this frequency and use the fine frequency control on the function generator to peak the signal to maximum. Record the value in Table 10.3.



Ensure that the timebase controls are in the calibrated settings and measure the number of divisions taken for one complete cycle. Record in Table 10.3.



Measure the frequency as follows, adding the results to Table 10.3:

One cycle 6.7 div.

Peak Signal Amplitude Vp-p

Number of divisions

Time for one cycle (Τ) ms

Frequency f= 1 Τ

Hz

Table 10.3

The time for one cycle is calculated by multiplying by the timebase setting, for example 6.7 x 0.2ms = 1.34ms. 1 The reciprocal of this gives the frequency: 13 . 4 × 10 − 3 = 746Hz. Note that this example has been chosen to be different from the result which you should get.

166



The frequency which you calculate is the natural resonant frequency of the loudspeaker. The response curve of the loudspeaker has a very pronounced peak at this frequency. It is caused by the dimensions of the loudspeaker cone, largely the cone diameter. 

Plot the response of the loudspeaker on the axes provided. A logarithmic scale is used for frequency because this matches the response of the ear. 0.5 Output Voltage (Vp-p)

0.4

0.3

0.2

0.1

100

200

500 700 1k

2k

5k 7k 10k frequency (Hz)

Graph 10.1

If this type of loudspeaker was used for music output then the response of the electronic driving circuit would need to be shaped to compensate for the response. This would be done by boosting both the lower and higher frequencies. If used as an alarm generator then it would be best to choose the resonant frequency for greatest efficiency, to generate the loudest sound output from a given power input. 10.2a

Enter your calculated loudspeaker resonant frequency in kHz. 


167



10.3 The Buzzer The construction of the buzzer used in the DIGIAC 1750 unit is shown in Fig 10.3(a).

I/P

Diaphragm Magnet

Spring

Iron core

Circuit board

Coil 0V Contacts

(a)

(b)

Fig 10.3

A small transistorized oscillator circuit feeds an alternating EMF to an iron cored coil. The alternating magnetic field produced by the coil attracts and repels a small permanent magnet attached to a spring. This magnet vibrates against a diaphragm and creates a loud noise. In control system applications the device is used as an alarm indication. The electrical circuit of the device is shown in Fig 10.3(b). The diode is fitted to prevent damage to the transistorized circuit if the supply is connected with incorrect polarity. The polarity of the input supply should be positive. The rated voltage is 12V. The main characteristics of the device fitted to the DIGIAC 1750 unit are: Supply voltage

8V

12V

16V (max.)

Supply current

15mA

-

30mA

400Hz

-

Output frequency Output sound level Table 10.4

168

-

70dBA at 7.87" (20cm)



10.4 Practical Exercise Characteristics of a Buzzer


POWER AMPLIFIER

MICROPHONE

BUZZER

7

4 3

8

2

9

A

I/P

O/P

I/P

A

10

1

O/P

B

10k 0V

COUNTER/ TIMER

+12V A.C. AMPLIFIER O/P I/P

DIFFERENTIATOR O/P I/P

T

I/P

TIME

dVIN dt R E SE T

10 1000 100

GAIN

F RE E R U N

100ms 1s 10ms

COUNT

1s

MOVING COIL METER

TIME CONSTANT

5 -10

0

5 +10

+ 0V

L J

Fig 10.4



Connect the circuit as shown in Fig 10.4. Set the control of the 10k Ω resistor for zero output voltage (fully counter-clockwise). Connect the digital multimeter as an ammeter on the 20/200mA range between the output of the power amplifier and the buzzer to monitor the buzzer current. Set the A.C. Amplifier to 1000 and the Differentiator to 1s.

Note: When you first switch on, there may be readings on the counter immediately, due to background noise being picked up by the microphone and processed by the Counter. The readings should be ignored as they will not affect the experiment results.

169





Switch ON the power supply and adjust the 10kΩ resistor to increase the voltage applied to the buzzer. Note the voltage on the Moving Coil Meter at which the buzzer begins to operate. Press RESET on the Counter to read the buzzer frequency.

The buzzer begins to operate at 

V at a frequency of

Hz

Alter the setting of the 10k Ω resistor to increase the voltage applied to the buzzer to 4V, 6V, 8V and then 10V as given in Table 10.5. Record the current and frequency at each step. Voltage Current Frequency

4V

6V

8V

10V

12V

mA

mA

mA

mA

mA

Hz

Hz

Hz

Hz

Hz

Table 10.5 

Transfer the positive lead of the digital multimeter from the output of the Power Amplifier to the +12V socket to bypass the 10k Ω resistor and Power Amplifier and apply the full 12V directly to the buzzer. Record the current and frequency again in Table 10.5.



170


10.4a

Enter your minimum voltage for buzzer to operate.

10.4b

Enter your current reading in mA when the applied voltage is 10V.

10.4c

Enter your frequency reading in Hz when the applied voltage is 12V.




From this chapter or a previous one, identify a device that emits a sound wave of constant frequency in the audio range when a DC voltage is applied:

a c 2.

3.

4.

b d

ultrasonic transmitter buzzer

From this chapter or a previous one, identify a device that emits a pressure wave at higher than audio frequencies:

a

microphone

b

ultrasonic transmitter

c

loudspeaker

d

buzzer

From this chapter or a previous one, identify a device that emits sound waves over a wide range of audio frequencies:

a

microphone

b

ultrasonic transmitter

c

loudspeaker

d

buzzer

The maximum sound wave frequency which is detectable by a human ear is approximately:

a 5.

microphone loudspeaker

20Hz

b

5kHz

c

20kHz

d 40kHz

The reason for the diode fitted in series with the buzzer in the DIGIAC 1750 Trainer is to:

a

rectify the applied AC

b

protect the electronic circuit against wrong polarity

c

convert the output to DC

d

prevent interference spikes being generated by the buzzer

Continued ...

171




The resonant frequency of the loudspeaker fitted in the DIGIAC 1750 Trainer is nearest to:

a 7.

172

0-200Hz

b

300Hz-950Hz

c

1kHz-1.5kHz

d 2kHz-3kHz

A loudspeaker is fed with a 1kHz signal and placed in a vacuum. The effect would be:

a

no sound because a vacuum has no air particles

b

very reduced sound because of the box enclosing the vacuum

c

increased frequency (pitch) of the note produced by sound waves in a vacuum

d

an impure tone due to the signal waveform producing the sound


Linear or Rotational Motion Chapter 11

Chapter 11 Linear or Rotational Motion




Describe the construction and characteristics of a DC solenoid.



Describe the construction and characteristics of a DC relay.



Describe the construction and characteristics of a DC solenoid air valve.



Describe the construction and characteristics of a DC permanent magnet motor

• • •


173



11.1 The DC Solenoid Coil

+12V

Soft iron core/Actuator shaft I/P

Return spring

End stop

0V

-12V

Case

(a)

(b)

Fig 11.1

The construction of a DC solenoid is shown in Fig 11.1(a), consisting of a soft iron core and actuator shaft which is free to move inside a coil. When the coil is energized, the soft iron core is attracted inside the coil and is held in position. When the coil is de-energized, the core returns to its neutral position under the action of a return spring. The voltage required to attract the core into the coil will be less than the rated value and will depend on the load applied to the actuator shaft. The voltage at which the core is pulled in by the coil is referred to as the pull-in voltage. With the coil energized and the core attracted, if the coil voltage is reduced gradually, when the voltage has fallen sufficiently the core will return to its neutral position under the action of the spring. This voltage is referred to as the drop out or release voltage. The release voltage will be much less than the pull-in voltage. Fig 11.1(b) shows the electrical circuit arrangement of the device fitted to the DIGIAC 1750 Trainer. When the coil is de-energized a large EMF can be induced in the coil, the magnitude depending on the inductance and the rate of change of current. Diodes are provided to limit the induced voltage to a maximum of ±12V. The main characteristics of the coil fitted to the DIGIAC 1750 Trainer are: Resistance

50Ω

Pull-in voltage

6V

Coil rating

12V/3W

Release voltage

1V

Table 11.1

174



11.2 Practical Exercise Characteristics of a DC Solenoid

SOLENOID WIREWOUND TRACK 5

4

6

C

7

3

8

2

9 10

1

POWER AMPLIFIER

B

O/P

A

I/P

I/P

A

10k MOVING COIL METER 0V

+1 2V

5 -10

0

5 +10

+ 0V

L J

Fig 11.2



Connect the circuit as shown in Fig 11.2 and set the 10k Ω resistor for zero output voltage (control fully counter clockwise). Connect the digital multimeter as an ammeter on the 200mA range in between the Power Amplifier and the Solenoid.



Switch ON the power supply and rotate the 10kΩ resistor control to gradually increase the voltage applied to the solenoid coil. Note the voltage at which the iron core of the solenoid is attracted fully into the coil. This value is the pull-in voltage. Record this voltage and the current in Table 11.2 overleaf.

Note: The core will start to move at a lower value than the pull-in voltage, the actual pull-in voltage will be the value when you hear the click, as the core aligns itself inside the coil. In this position you will find a distinct resistance to pushing the actuator back towards its neutral position.

175



Unloaded Readings Pull In Drop-out (Release)

Voltage

Loaded

Current

Voltage

Current

V

mA

V

mA

V

mA

V

mA

Table 11.2 



11.2a

With the coil energized and the core in its pulled in position, slowly reduce the coil applied voltage and note the value at which the core returns to its neutral position, the drop-out or release voltage. Record voltage and current again in Table 11.2. Repeat the process with your finger against the actuator shaft to exert a little load and note the voltage and current required for pull in and release.

With the actuator shaft loaded the effect on the voltage and current required for pull-in was:

11.2b

a voltage increased, current reduced

b

both voltage and current reduced

c voltage reduced, current increased

d

both voltage and current increased

With the actuator shaft loaded the effect on the voltage and current required for drop-out (release) was:

a voltage increased, current reduced

b

both voltage and current reduced

c voltage reduced, current increased

d

both voltage and current increased



176




11.3 The DC Relay +12V N.O.

I/P (COM.) N.C.

0V Common Coil Contacts

N.O.

N.C.

-12V

(a)

(b)

Fig 11.3

The construction of a DC relay is shown in Fig 11.3(a). It consists of a coil with an iron core which has a soft iron armature attached to a spring which holds it just above the core. Changeover contacts are attached to the spring and with the armature in its rest position it makes contact with one of the terminals. This is referred to as the normally closed (N.C.) contact. With the coil energized, the core will be magnetized and attract the soft iron armature. The spring is moved, which breaks the connection to the N.C. terminal and makes the contact to the other terminal. This terminal is referred to as the normally open (N.O.) contact. With this construction, the contacts will bounce for a short period each time they close or open (make or break) and this can cause problems with some circuits. The problem can be overcome by using an electronic debounce circuit or a time delay prior to checking the contact state after operation. Fig 11.3(b) shows the electrical circuit arrangement of the device fitted to the DIGIAC 1750 Trainer. The diodes limit any induced voltages to a maximum of approximately ±12V, as for the solenoid device. The main characteristics of the device fitted to the DIGIAC 1750 Trainer are: Coil rated voltage

12V

Coil resistance Coil operating voltage Coil release voltage

1.8V

Operate/release time

5ms

320Ω

Contact rating

12V, 1A

7.5V

Lifetime cycles

5x106

Table 11.3

177



11.4 Practical Exercise Characteristics of a DC Relay

P . I . N . P H O TO D I O D E

P HO TO V O LT A I C CE LL

O/P

O/P

+

+

O/ P

O/ P MOVING COIL METER

P HO TO CON D UCTIV E CELL +12V

LAMP FILAMENT


8

2

9 1

10

O/P B A

10k

5

I/P

-10

RELAY

POWER AMPLIFIER

7

4

P HOTO TRA NSISTO R

A

N.O.

5 +10

+ -

I/P I/P

0

0V

L J

N.C.

0V

Fig 11.4



178

Connect the circuit as shown in Fig 11.4 and set the 10k Ω resistor control for zero output voltage.



Switch ON the power supply. The relay will be in its de-energized state. Note the state of the Lamp. Lamp ON means that the contacts are closed. Lamp OFF means that the circuit is broken because the contacts are open.



The relay coil will have pull-in and release voltage characteristics similar to those for a solenoid.



Determine the pull-in and release voltages and currents for this device by gradually increasing and decreasing the applied voltage. Record the results in Table 11.4 opposite.



Note when a change of state of the Lamp connected to the N.O. contact occurs.



Move the lamp connection to the N.C. terminal and observe the effect on the lamp switching. Add to Table 11.4.



Lamp state ON/OFF when connected to: Voltage Pull In Drop-out (Release)

Current

V

mA

V

mA

N.O. contact

N.C. contact

Table 11.4

11.4a

Enter the pull-in voltage for your relay in V.

11.4b

Enter the drop-out current for your relay in mA. 


Notes: ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. .................................................................................................................................................................

179



11.5 The Air Valve Inlet port

Coil Return spring +12V I/P

Inlet valve

0V Exhaust valve

Cylinder port

(a)

-12V

Exhaust port

(b)

Fig 11.5

Fig 11.5(a) shows the construction of the device fitted to the DIGIAC 1750 Trainer. It is similar to the solenoid considered previously, but the soft iron core now operates on two valves, the inlet and the exhaust valves. With the coil de-energized the core is held, by the return spring, in the position with the inlet valve closed and the exhaust valve open. In this position the cylinder port is connected to the exhaust port outlet. When the coil is energized, the core is attracted and held in the position with the exhaust valve closed and the inlet valve open. In this position the inlet port is connected to the cylinder port In the DIGIAC 1750 Trainer, the inlet port is connected to the pump and the cylinder port is connected to a pneumatic actuator. With the pump ON, the pneumatic actuator will be operated when the coil is energized and illustrates the principle of electrical control of pneumatic devices. The electrical circuit arrangement of the device fitted to the DIGIAC 1750 Trainer is shown in Fig 11.5(b). The main characteristics of the device are: Rated voltage

12V

Coil resistance Coil pull-in voltage

140Ω 8.3V

Coil release voltage

1.7V

Table 11.5

180



11.6 Practical Exercise Characteristics of an Air Valve

AIR FLOW SENSOR

PRESSURE

FLOW

AIR PRESSURE SENSOR

O/P

ON

+

A


MOVING COIL METER POWER AMPLIFIER

7

5

O/P

3

8

2

9

B A

10

1

O/P

PUMP OFF

+

4

AIR VALVE I/P

-

-10

I/P

0

5 +10

+

10k

0V

+ 12V

0V

L J

Fig 11.6



Connect the circuit as shown in Fig 11.6. Set the 10k Ω resistor control for zero output voltage (fully counter clockwise) and set the pump control (Air Pressure/Flow Sensor panel) to PRESSURE.



Switch ON the power supply and then switch the pump ON. The coil is deenergized in this state, the inlet valve is closed, and the pneumatic actuator will not operate.



Adjust the resistor control to apply 10V to the solenoid coil. The coil will be energized, the inlet valve will open and the exhaust valve will close. The pump pressure will be applied to the pneumatic actuator. Observe the effect on the actuator.



Reduce the voltage and observe the effect on the pneumatic actuator



Switch the pump OFF. Observe the effect on the operation of the pneumatic actuator with no air pressure when the solenoid voltage is raised and lowered.

181


11.6a


With the pump OFF the pneumatic actuator:

a still operates but more slowly

b

operates at a higher voltage

c does not operate at all

d

operates in reverse

The Air Valve solenoid will have pull-in and release voltages and currents as for any solenoid. To determine these values for the device: 

With the pump switched OFF, increase and decrease the applied voltage gradually and note the voltages at which switching occurs. You will hear a faint click when the device switches. Voltage Pull In Drop-out (Release)

Current

V

mA

V

mA

Table 11.6 11.6b

Enter the pull-in voltage for your Air Valve solenoid in V.

11.6c

Enter the drop-out current for your Air Valve solenoid in mA.

11.6d

Using your pull-in figures of voltage and current, calculate and enter the resistance of your Air Valve solenoid in



182


.



11.7 The DC Permanent Magnet Motor

N

Coils Commutator S

Brushes

Permanent Magnet Stator

Armature Connections

Fig 11.7

The construction of a permanent magnet DC motor is shown in Fig 11.7. The unit is identical with the tacho-generator unit but for a motor, a DC supply is fed to the armature coils. Current flowing in the armature coils sets up a magnetic field which reacts with the field of the permanent magnet to produce a force causing the armature to rotate. The force acting on the armature is proportional to the current flowing. When the armature rotates, an EMF is induced in the coils, in exactly the same way as in the tacho-generator. The self-induced EMF opposes the applied voltage and is referred to as the back EMF. The armature accelerates until the speed is such as to produce a back EMF ( e) equal to the applied voltage (V) less the voltage dropped across the armature resistance rai.

V = e + rai The speed with no load on the shaft is thus roughly proportional to the applied voltage. When a load is applied to the shaft, the speed will tend to fall, reducing the back EMF. More current flows from the supply and the current self-adjusts to the value that produces a torque (turning force) just sufficient to balance the load torque. The speed will fall slightly with load due to the increase in voltage lost across the armature coils due to the higher current.

183



The electrical circuit arrangement of the device fitted to the DIGIAC 1750 Trainer is shown in Fig 11.8.

+12V L2

I/P

M C2 L1 O/P

1Ω

C1 -12V

Fig 11.8

The 1Ω resistor is fitted in series with the armature to allow monitoring of the armature current by measurement of the voltage dropped across it. Since the resistor is 1Ω, voltages measured across it in mV will directly correspond to currents in mA. The diodes limit any voltage spikes to a maximum of approximately ±12V. Capacitor C1 provides some noise filtering at the output and the combination L1, L2 and C2 reduces radiation of radio frequency noise. The main characteristics of the device fitted to the DIGIAC 1750 Trainer are: DC resistance No load current (12V applied) Stall current (12V applied) Shaft speed (no load, 12V applied)

120mA 1.93A 2400 rev/min (max.)

Starting torque

7 Ncm/A

Torque constant

3.5 Ncm/A

Time constant Efficiency Table 11.7

184

6.2Ω

19.6ms 82% (max.)



11.8 Practical Exercise Characteristics of a DC Permanent Magnet Motor

DC MOTOR

I/P

SLOTTED OPTO SENSOR

COUNTER/ TIMER I/P

O/P

R E SET

O/P

TIM E

F R EE R U N

C O U NT

1s

MOVING COIL METER WIREWOUND TRACK 6 5 C

5 POWER AMPLIFIER

7

4

8

3 2 1

V 10

9

B

-10

+10

+

I/P

A

0V

0V

5

O/P

10k

-12V

0

L J

+12V

Fig 11.9 

Connect the circuit as shown in Fig 11.9. Set the 10k Ω resistor control for zero output voltage, (control fully counter clockwise), and set the counter controls to COUNT and 1s.



Switch ON the power supply and set the voltage applied to the motor, as indicated by the Moving Coil Meter, to 10V. The motor should run at a high speed. Allow it to run for a short time and then note the reading of the digital voltmeter. This reading in mV represents the current in mA taken by the motor, since it is the voltage dropped across a 1Ω resistor.



Press the Counter RESET button and note the displayed Counter value. This represents the motor speed in rev/sec. Record the values in Table 11.8 overleaf.



Repeat the procedure, noting the speed and current readings for motor applied voltages of 8V, 6V, 4V and 2V and record the values in Table 11.8.

185


Motor Applied Voltage


10V

Armature Current

8V mA

6V mA

4V mA

2V mA

mA

Speed (rev/sec.) Speed (rev/min.) Table 11.8 

Multiply the speed in rev/sec by 60 to convert to rev/min and add the results to Table 11.8.



Slowly reduce the applied voltage until the motor just stops turning and observe the effect on the voltage and the current. Stopped voltage =

V

Stopped current =

mA

Construct the graph of speed in rev/min. against applied voltage and armature current on the axes provided:



Applied Voltage (volts) 3000

0

1

2

3

4

5

6

7

8

9

101

11

2

10

20

30

40

50

60

70

80

90

100 110 120

2800

Motor Speed (rev/min)

2600 2400 2200 2000 1800 1600 1400 1200 1000 800 600 400 200 0

130 140 150 160 170 180 190 200

Armature Current (mA) Graph 11.1

186


11.8a


From your graph deduce and enter the armature current needed to give a speed of 1500 rev/min in mA.

11.8b

Examine the two graphs. Which of the following statements is most true?

a they are identical b shaft speed is directly proportional to the applied voltage c shaft speed is directly proportional to the armature current d both characteristics are non-linear

11.8c

As the applied voltage is very slowly reduced, the moment the motor stops the effect on the armature current is to:

a reduce it directly with the applied voltage b show a sharp increase due to loss of back EMF c fall immediately to zero as torque stops d increase slightly due to reduction in armature resistance 

Set the applied voltage to 7V and note the armature current taken and the shaft speed when the motor is unloaded. Record in Table 11.9. Applied Voltage = 7V Armature current

Unloaded

mA

Loaded

400mA

Shaft speed (rev/sec) Table 11.9 

Now place your left hand near the Hall effect disc with the finger nails down and touching the baseboard of the DIGIAC 1750 Trainer. Move your fingers

gently so that comes between the Hall effect disc and theforward baseboard and your exertsmiddle a smallfinger load on the motor. 

Vary the pressure of the load so that the current is approximately 400mA (0.4V reading on the digital voltmeter) and then note the shaft speed by pressing the Counter RESET button. Record in Table 11.9.

187


11.8d


Enter your speed in rev/sec with the motor loaded and the armature drawing a current of about 400mA.

Set the control of the 10k Ω resistor to the zero output voltage position. Disconnect socket C of the 10kΩ resistor from the +12V supply and re-



connect it to the -12V supply. 11.8e

The direction of rotation is:

a

11.8f

the same as before

b

reversed

Is the same speed range possible?

Yes

or

No

The characteristics are typical for this size of machine, larger machines would not have such a large drop in speed with load. 

188





2.

3.

For any solenoid in which a soft iron core is drawn inside a coil, the pull-in current:

a

and release current will be the same

b

will be greater than the release current

c

will be less than the release current

d

will depend on the applied voltage

A pneumatic actuator, which is used to indicate gas pressure, is controlled by an electrically operated air valve in a similar configuration to that used on the D1750 unit. The actuator will operate:

a

if there is both an electrical input and adequate gas pressure.

b

if there is an electrical input but there is no gas pressure.

c

if there is adequate gas pressure but there is no electrical input.

d

if neither the electrical input nor adequate gas pressure is present.

A relay has changeover contacts marked N.C. and N.O. When the relay is de-energized a lamp will operate if it is connected to a supply via:

a 4.

5.

neither of them

b

N.C.

c

N.O.

d either of them

The term N.C. applied to the contacts of a relay means:

a

Not Contacted

b

Not Changeover

c

Normally Closed

d

Neither Connection

A solenoid has a coil rated for operation at 10V. You might expect the pull-in and dropout voltages to be:

a

pull-in = 12V, drop-out = 5V

b


c


d


Continued ...

189




For the Air Valve investigated in this Chapter, if the coil is de-energized the air valve will:

a c 7.

8.

b d

rise immediately fall back slowly

The DC Permanent Magnet Motor investigated in this Chapter is:

a

completely unlike the tacho-generator

b

identical to the tacho-generator

c

similar to the tacho-generator but with the commutator reversed

d

similar to the tacho-generator but with more coils

A DC Permanent Magnet Motor with an armature resistance of 5 , runs at 1200 rev/min. when the applied voltage is 6V. The speed for 12V applied is likely to be:

a 9.

not be affected drop down immediately

600 rev/min

b

1200 rev/min

c

1800 rev/min

d

2400 rev/min

The voltage applied to the DC Motor in question 8 above is slowly reduced, the motor stops when the applied voltage is 2V. The armature current will then be:

a

20mA

b

50mA

c

100mA

d 400mA

10. If the DC motor in question 8 is loaded without any increase of applied voltage, the effect is likely to be:

190

a

reduced speed and increased armature current

b

reduced speed and armature current

c

increased speed and reduced armature current

d

increased speed and armature current


Display Devices Chapter 12

Chapter 12 Display Devices




Describe the characteristics and application of the Counter/Timer.



Describe the characteristics and application of the LED Bargraph Display.



Describe the characteristics and application of the Moving Coil Meter.



State and calculate the requirement to extend the voltage range of a Moving Coil Meter.



Select a suitable device for a particular voltage measurement.

• • • •

DIGIAC 1750 Transducer and Instrumentation Trainer. 4mm Connecting Leads. Digital Multimeter. Stopwatch (not supplied).

191



12.1 The Timer/Counter

Reset

1 second Delay

FREE RUN COUNT

100Hz Oscillator

TIME

I/P

Decade Counter

Decoder/ Driver

Decade Counter

Decoder/ Driver

Decade Counter

Decoder/ Driver

1s

+5V

&

&

Fig 12.1

A system logic diagram of the Counter/Timer facility provided with the DIGIAC 1750 unit is shown in Fig 12.1. The output display uses three 7-segment LED's. The unit can be used in three ways: 1.

Time measurement, with the controls set to TIME and FREE RUN.

2.

Counting (pulses), with the controls set to COUNT and FREE RUN.

3.

Frequency (pulses/sec), with the controls set to COUNT and 1s.

In addition, with some signal conditioning, it can be used for voltage measurement. The main characteristics of the unit are: Input impedance Input voltage levels (TTL) Timing intervals Timing accuracy Table 12.1

192

1MΩ +5V max. 10ms 5%


Time

TIME


and FREE RUN

With the input at TTL logic level "1", (+5V), the display increments at 10ms intervals, or every 1 second. With the input at logic level "0" (0V), the 100 displayed value is held. The unit will therefore display the time in hundredths of a second that the input is held at logic level "1". Note that with a 3-digit display, the maximum count is 999 and hence one complete cycle from 0-999 will represent 1000 x 10ms = 10s.

Counting

COUNT

and FREE RUN

The count increments by 1 each time the input voltage level changes from TTL logic level "0" to level "1", i.e. on receipt of a positive edge of a pulse of amplitude 5V. Set in this way the Counter counts input pulses and displays the total. With the 3-digit display the maximum count will be 999.

Frequency

COUNT

and 1s

The unit counts the number of positive pulses at TTL logic level "1" that are received at the input in a period of one second, following a RESET of the Counter, thus giving the count rate in pulses per second, or the frequency in Hz.

Note that you have already used the Counter/Timer to count the number of pulses received in one minute and to measure frequency in pulses/sec.

193



12.2 Practical Exercise Time Measurement and Counting

COUNTER/ TIMER I/P

T IM E

F R EE R UN

AMPLIFIER #2 I/ P

-

R ES ET

O/ P .5

+

.4 1 100 10

OFFSET

GAIN COARSE

.6 .7

.3

.8

.2

.9 .1

C O U NT

1s

MOVING COIL METER

1.0

GAIN FINE

5 -10

0

5 +10

+ 0V

L J

Fig 12.2

Time Measurement 

Connect the circuit as shown in Fig 12.2 and switch ON the power supply. With the Amplifier #2 GAIN COARSE control set to 100 and GAIN FINE to 1.0, adjust the OFFSET control for +5V output. Switch the GAIN COARSE control to 1. The output voltage will drop to nearly zero.



Set the Counter/Timer controls to TIME and FREE RUN and press the RESET button. The display should show zero.



Switch the Amplifier #2 coarse gain control to 100. The counter display should increment at 10ms ( 1 sec.) intervals. Return the GAIN COARSE 100 control to 1, the display will be held. This illustrates the application of the unit to time measurement, the display indicating the number of 10ms intervals (or the time in hundredths of a second) that the input is held at



With Amplifier #2 GAIN COARSE set to 1, RESET the Counter display to zero. Switch Amplifier #2 GAIN COARSE to 100 and note the time taken for the count to complete one cycle from 0 to 999 and back to 0.

+5V.

194


12.2a


The time taken for a complete cycle of counting is:

a 10ms

b

100ms

c

1s

d

10s

Use the timer facility to time some operations and obtain practice in its use, such as the time taken for you to verbally count from zero through to 250, or to write down a long word.



Counting Pulses

With the circuit still as shown in Fig 12.2, set the Counter/Timer controls to COUNT and FREE RUN and RESET the display to zero.





12.2b

Switch Amplifier #2 GAIN COARSE control from 1 to 100 and back to 1.

The count increments by:

a 1

b

10

c

100

d

1000



Repeat the process, you will find that the count increments for each change of the gain from 1 to 100, or on the application of a +5V pulse to the counter



Remove the Counter input lead from the output of Amplifier #2 and touch it on the +5V supply socket.



Return the Counter input lead back to the output of Amplifier #2 and, with the GAIN COARSE set to 100, alter the OFFSET control to give zero output. Slowly raise the setting again and watch the Counter display for a response.



Note the threshold level on the Counter input from the indication on the Moving Coil Meter.

input.

Threshold voltage level =

12.2c

V

Enter the threshold level for an increment of the count in V. 


195



12.3 Practical Exercise Frequency Measurement

WHEATSTONE BRIDGE D

DIFFERENTIATOR O/P V/F CONVERTER O/P

12k

C A

OUT

B

3

I/P

I/P

T

dVIN dt

IN

V

1V

0V

COMPARATOR

+5V

OFF ON HYSTERESIS

B

-

A

+

100ms 1s 10ms

Rx

O/P

TIME CONSTANT

COUNTER/ TIMER I/P

TI M E

RES ET

C O UNT

F R E E R UN

1s

Fig 12.3

The connection of the +5V supply places the 12k Ω fixed resistor in series with the 10kΩ 10-turn resistor to make low voltage settings easier. Switch the unknown resistor Rx OUT. A Voltage to Frequency (V/F) Converter is available in the signal conditioning circuits. This unit converts a DC voltage input to a pulsed output of frequency 1kHz/volt of input. For example, an input of 0.6V will produce an output frequency of 0.6kHz or 600Hz. The pulses from the V/F Converter are unsuitable to be fed directly to the input of the Counter/Timer. The Differentiator and Comparator are used to shape the pulses from the V/F Converter, so that they may be detected by the Counter/Timer.

196



Connect the circuit as shown in Fig 12.3 and switch ON the power supply. Set the Counter controls to COUNT and 1s. Set the Differentiator TIME CONSTANT to 1s and switch OFF the Comparator HYSTERESIS.



Set the 10k Ω 10-turn resistor output voltage to 0.1V, press the Counter RESET button and note the displayed reading. Enter the value in Table 12.2.


Input Voltage to V/F Converter Counter Display (Hz)

0.1


0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Table 12.2 

Repeat the procedure for the other voltage settings shown in Table 12.2 and record the displayed values that are obtained following the pressing of the reset button. The accuracy in the calibration of the V/F converter will affect the readings as will your accuracy in setting the voltages and also the accuracy of the 1s delay in the Counter/Timer.

12.3a

Enter your Counter Display reading for an input voltage of 0.5V.

In this exercise the V/F converter was used purely as a means of obtaining a variable frequency. However, the method used also illustrates the application of the unit to voltage measurement. The displayed Counter readings represent the voltage in mV, as can be seen from Table 12.2. The maximum voltage range is limited by the frequency capability of the counter and the number of digits in the display. The voltage range can be extended by attenuating the input to the V/F converter using the additional circuits shown in Fig 12.4. Also, carefully note the change of the voltage feed to the 10kΩ 10-turn resistor.

WHEATSTONE BRIDGE D V/F CONVERTER 12k

C

O/P

A OUT

B

3

I/P

IN

V

1V

Rx

0V

BUFFER #1

SLIDE

C

O/P

B

I/P +VIN 1

2

3

4

5

6

+12V

7

8

9

10

10k

A

0V

Fig 12.4

197



The Buffer Amplifier is used to reduce the loading on the 10k Ω 10-turn resistor. The circuit will be calibrated so that a counter display of 600 represents a voltage of 6V. 

Connect the additional circuitry shown in Fig 12.4, to the V/F converter input. The V/F Converter, Differentiator, Comparator and Counter/Timer remain connected as shown in Fig 12.3. Set the output control of the 10kΩ slide resistor for zero output (to the left).



Set the output voltage from the 10-turn resistor to 6V as indicated by the digital voltmeter and then slowly adjust the 10kΩ slide resistor until the Counter display indicates 600 after the RESET button is pressed. You will find that the setting of the resistor control is very sensitive, it is possible to set accurately but if it is too difficult, set the value as near as you can. The unit is now calibrated.



Input Voltage

Set the 10-turn resistor control in steps to each of the other voltage values indicated in Table 12.3. Note the Counter displayed value after pressing the reset button. Record the values in Table 12.3. 1

2

3

4

Counter Display

5

6

7

8

9

9.5

600

Table 12.3

12.3b

Enter your displayed reading for an input voltage of 3.5V.

Using this principle, the counter could be calibrated for any desired voltage range. AC measurements are possible if the Full-Wave Rectifier circuit is added and the unit re-calibrated for RMS values. 

198




12.4 The LED Bargraph Display

Light pipe

LED Chip

Connections

Fig 12.5

The construction of the Bargraph device is shown in Fig 12.5, consisting of 10 separate light emitting diodes (LED's) fitted in a 20-pin package. The light from each diode is collected by a light pipe and appears at the top surface as a red bar. A dedicated IC driver chip controls the device and provision is made for adjusting the voltage levels required for adjacent LED's to light. With the device as fitted to the DIGIAC 1750 unit the voltage level between adjacent LED's is 0.5V and hence the minimum voltage for all LED's to light is 5.0V. The device has a high input impedance, a low time constant, and is suitable for indication of an approximate and rapidly varying voltage level, but the resolution is low. The main characteristics of the device are: Input impedance

1MΩ

Input voltage range

±35V

Accuracy Segment overlap

2% 1mV

Table 12.4

The unit is adjusted so that an input of +5V just lights the last LED.

199



12.5 Practical Exercise Characteristics of an LED Bargraph Display


L.E.D. BARGRAPH DISPLAY

7

4 3

8

2

9 1

I/P

B A

10

DRIVER I.C.

V

10k

+12V

0V

Fig 12.6 

Connect the circuit as shown in Fig 12.6. Set the 10k Ω Wirewound resistor control for zero output voltage (fully counter clockwise).



Switch ON the power supply. Adjust the resistor control to increase the voltage applied to the bargraph unit gradually and note the voltage values at which each LED lights. Record the values in Table 12.5.

LED number

1

Input Voltage

2 V

3 V

4 V

5 V

6 V

7 V

8 V

9 V

10 V

V

Table 12.5

12.5a

Enter the input voltage required just to light LED number 7 in V. 

12.5b

Vary the voltage rapidly by rotating the control quickly in both directions and note how the display follows. Repeat the procedure, this time noting the display on the digital meter. Switch OFF the power supply.

When the voltage was varied rapidly:

a both displays responded immediately to the variations b only the LED Bargraph was able to follow the variations precisely c only the digital multimeter was able to follo w the variations precisely d neither display responded at all

200



12.6 The Moving Coil Meter 0

5 -10

Scale

5 +10

Pointer Radial magnetic field

Soft-iron core Permanent Magnet

Hairspring

N

S

Pivot

+

V

0V

Coil

Connections

Fig 12.7

The construction and electrical circuit arrangement of the moving coil meter fitted to the DIGIAC 1750 unit are shown in Fig 12.7. Using the connections + and -, the voltage difference between any two points in a circuit can be measured. By connecting the - socket to 0V, the voltage of any point with respect to 0V (ground) can be measured using the + connection. The moving coil meter consists of a coil suspended between the poles of a permanent magnet with a pointer attached to the coil which moves over the meter scale. The coil is held in its center position by two hairsprings. A set zero screw is attached to one of the hairsprings for adjustment of the pointer position to zero with no voltage applied to the meter. When current is fed to the coil via the hairsprings, a force is produced by interaction between the current in the coil and the permanent magnetic field, and the coil rotates. The direction of rotation depends on the direction of the current through the coil (Fleming's Rule), and the amount of rotation depends on the magnitude of the current flowing. The coil rotates until the force produced by the current is balanced by the force exerted by the hairsprings. The coil is wound on an aluminum former. When the coil rotates, an EMF is induced in this former, similar to the back EMF induced in the armature coils of a DC motor. This produces a current and a force opposing the motion of the coil (Lenz's Law).

201



The coil movement is thus damped and allows the pointer to take up its final position, after a step change of current, with the minimum of oscillation (or hunting) occurring. The meter movement is a damped control system and this effect together with the inertia of the coil system limits the response speed of the pointer. The hairsprings are fine to allow a large angular movement and high sensitivity. The amount of coil current needed for full-scale deflection (f.s.d.) will be determined by the tension of the hairsprings. The current flow in the meter circuit must be limited to this value of current. When used as a voltmeter, a series resistor (called a multiplier) is fitted to limit the current to the value required to produce full-scale deflection. For instance, if the f.s.d. current for a particular meter is 1mA, then the value of the multiplier 1 (series resistor) must be or 1kΩ for each volt (1kΩ/V) to be represented 1 × 10 − 3 by full-scale deflection. This figure (1kΩ/V) is known as the sensitivity of the meter. From this figure it is possible to calculate the loading resistance of a meter when it is operated on any voltage range. A 10V voltmeter using a 1mA f.s.d. meter would require a multiplier of 10 x 1k Ω = 10kΩ. Many analog multimeters are based on a 50 µA meter movement (50µA f.s.d.). 12.6a

Calculate and enter the sensitivity of a 50 A meter movement in k /V.

The main characteristics of the meter fitted to the DIGIAC 1750 unit are: Full-scale current

±1mA

Sensitivity

1kΩ/V

Total voltmeter resistance

20kΩ

Accuracy Table 12.6

202

± 1-2%



12.7 Practical Exercise Characteristics of a Moving Coil Meter MOVING COIL METER WIREWOUND TRACK 6 5 C 4 3

8

2

9 1

0

5

7

5

-10

B

+10

+ A

10

Set Zero -

10k

L J

V 0V +12V

-12V

Fig 12.8 

Connect the circuit as shown in Fig 12.8. Set the resistor control to its central position and check that the Moving Coil Meter pointer is at zero. Adjust the Set Zero screw (Fig 12.8) if necessary to set the pointer to zero. Use only the correct small screwdriver for this task.



Switch ON the power supply. Set the resistor output voltage to 0V as indicated by the digital multimeter and note the voltage indicated by the Moving Coil Meter. Enter the value in Table 12.7.

Digital Meter

V

Moving Coil Meter

-10

V

-8

V

-6

V

-4

V

0

+2

-2

V

+4 V

+6 V

+8 V

+10 V

V

Table 12.7 

Repeat the procedure for all positive values of voltage listed in Table 12.7.



Repeat the procedure for the negative values of voltage indicated in Table 12.7, but setting up with the Moving Coil Meter and reading the digital multimeter. Record the results in Table 12.7. Switch OFF the power supply.

12.7a

Which meter gives the greatest resolution (indicates the smallest change)?

12.7b

Which meter is the easiest to use to set up a selected voltage?

a Digital Multimeter

a Digital Multimeter

b

b

Moving Coil Meter

Moving Coil Meter

203



12.8 Practical Exercise Extending the Voltage Range of a Moving Coil Meter MOVING COIL METER CARBON TRACK 6 5 C 4


7

4

3

8

2

9 1

10

B

3

8

A

2

9 1

100k

5

7

B A

10

V

5 +10

+ -

10k

0V +12V

0

-10

-12V

L J

Fig 12.8

The voltage range of a moving coil meter can be increased by adding a resistor in series with it to extend the existing multiplier. 

Connect the 100k Ω variable resistor in series with the Moving Coil Meter as shown in Fig 12.8. Note that the ±12V supplies are being used together as a single-ended +24V supply.



Switch ON the power supply and use the 10k Ω variable resistor to set the voltage to 10V as indicated on the digital multimeter.



Adjust the 100k Ω variable resistor so that the Moving Coil Meter reads +5V.



Keep re-adjusting both settings until they are correct.

When completed, the Moving Coil Meter is calibrated for a voltage range of ± 20V. 

Check this by setting the voltage to 20V (digital multimeter) and note the Moving Coil Meter scale reading. Switch OFF the power supply.

Moving Coil Meter scale reading with 20V applied = 

12.8a

204

V

Isolate the 100k Ω Carbon Track Resistor from the circuit and use your digital multimeter on an Ohms (Resistance) range to measure the resistance of the part of the 100k Ω variable resistor which was connected into circuit.

Enter your measured value of the 100k

variable resistor in k .



12.9 Practical Exercise Comparison of Voltage Display Devices

L.E.D. BARGRAPH DISPLAY I/P

DRIVER I.C.

+12V

MOVING COIL METER


5

7

4

8

3

9

2 1

0

-10

10

10k

5 +10

B

+ A

V

-12V

0V

L J

Fig 12.9



Connect the circuit as shown in Fig 12.9. All three voltage display devices are connected in circuit for comparison of their characteristics.






12.9a

Vary the output voltage slowly over the range 0V through +5V and back to 0V and note the meter indications.

From your observations, which of the following is true?

a the Bargraph is too slow to follow the variations b the Digital Multimeter responds to the changes accurately and immediately c both the Moving Coil Meter and the Ba rgraph follow the variations well d the Moving Coil and Digital Meters give the best response

205




12.9b


Vary the output voltage over the same range rapidly and note the readings of the Moving Coil Meter and Bargraph.

The fastest response came from the:

a Moving Coil Meter 

b

Bargraph

Increase the input voltage from 0V to +3V, with the 3V indicating LED of the Bargraph just on, and note the readings of all the meters. Record the results in Table 12.8 Voltage indications

Bargraph

All three devices in circuit Moving Coil Meter removed

3V (sixth bar)

Digital Multimeter

Moving Coil Meter V

V

V

Table 12.8

12.9c

Enter your reading from the Moving Coil Meter in V. 

12.9d

Enter your reading from the Digital Multimeter in V when the Moving Coil Meter is removed from circuit.

12.9e

The loading caused by the Moving Coil Meter can be improved by:

a only using it on low voltage ranges b

only using it on AC

c using a full-wave rectifier for DC

using a buffer amplifier



206

Remove the lead to the + connection of the Moving Coil Meter thus disconnecting it from the circuit. Note and record in Table 12.8, the revised readings of the Digital Multimeter and Bargraph.


d



Student Assessment 12 The following questions all relate to devices fitted to, or recommended for use with, the DIGIAC 1750 Trainer. 1.

For the Counter/Timer unit to measure a period of time, the TIME/COUNT and switches are set to:

FREE RUN/1s

2.

3.

a

TIME & FREE RUN

b

COUNT & FREE RUN

c

TIME & 1s

d

COUNT & 1s

For the Counter/Timer unit to count input pulses, the TIME/COUNT and FREE RUN/1s switches are set to:

a

TIME & FREE RUN

b

COUNT & FREE RUN

c

TIME & 1s

d

COUNT & 1s

For the Counter/Timer unit to measure frequency, the TIME/COUNT and FREE RUN/1s switches are set to:

a c 4.

b d

TIME & 1s

COUNT & FREE RUN COUNT & 1s

The Counter/Timer display reads 254 when the unit is used to measure time. The amount of time elapsed is:

a 5.

TIME & FREE RUN

254 µs

b

254ms

c

2.54s

d 25.4s

The LED Bargraph is displaying six LED's ON. The input voltage is in the range:

a

2.50-3.45V

b

3.0-3.45V

c

2.75-3.25V

d

6.00-6.45V

Continued ...

207




7.

The device with the fastest response time for measurement of voltage is the:

a

Moving Coil Meter

b

Digital Multimeter

c

Counter/Timer

d

Bargraph

The f.s.d. current of the Moving Coil Meter is:

a 8.

b

±500µA

c

±1mA

d

±10mA

The Moving Coil Meter has a multiplier (series resistor) to range it for 10V. The value of additional multiplier for a range of 30V is:

a 9.

±50µA

10k Ω

b

20k

Ω

c

30k

Ω

d 50k

Ω

The best device for monitoring a rapidly, randomly varying voltage would be:

a

digital multimeter

b

moving coil meter

c

bargraph

d

oscilloscope

10. The best device for monitoring a slowly varying, precise voltage measurement (such as that found when adjusting a voltage setting to, say, 3.7V) would be:

a

digital multimeter

b

moving coil meter

c

bargraph

d

oscilloscope

11. The best device for monitoring instantaneous repetitive variations of voltage at very high frequencies would be:

a

digital multimeter

b

moving coil meter

c

bargraph

d

oscilloscope

12. The best device for voltage measurements in a high impedance circuit:

208

a

digital multimeter

b

moving coil meter

c

bargraph

d

oscilloscope


Signal Conditioning Amplifiers Chapter 13

Chapter 13 Signal Conditioning Amplifiers


















Describe the characteristics and application of DC amplifiers. Explain the term "Offset" and the need for offset control. Describe the characteristics and application of an AC amplifier. Describe the characteristics and application of a power amplifier. Describe the characteristics and application of a current amplifier. Describe the characteristics and application of a buffer amplifier. Describe the characteristics and application of an inverter amplifier. Describe the characteristics and application of a differential amplifier.

• • •


• • •

Oscilloscope. Function Generator. BNC to 4mm connecting lead.

209



13.1 DC Amplifiers

I/P

O/P

Fig 13.1

The symbol used for a DC amplifier is shown in Fig 13.1. The device consists of directly coupled amplifiers (without coupling capacitors) which are therefore capable of amplifying both DC and AC signals. There may be many active devices (transistors) in a DC amplifier such as the types of Integrated Circuit (IC) Operational Amplifier (Op Amp) chosen for the DIGIAC 1750 Trainer. The ratio of the output signal voltage to the input signal voltage is referred to as the voltage gain of the circuit (Av). With the input to these amplifiers at zero, the output should be zero, but there could be a small value of voltage. This is more of a problem with high gain circuits and an offset control may be provided to counteract the effect. This control is adjusted with zero input, to set the output voltage to zero. Given data for an amplifier normally specifies the input offset voltage for the device. This represents the difference in voltage at two input connections that may be required to produce zero output voltage. The second input connection is not accessible for the DC amplifiers provided with the DIGIAC 1750 Trainer although an offset control is provided for Amplifier #1/2 connected internally. Various DC amplifier circuits are provided with the DIGIAC 1750 Trainer, but only three are specifically designed for amplification applications, these being:

210

1.

Amplifier #1 having a variable preset gain over the range of 0.1 to 100 approximately. This amplifier is provided with an "offset" control.

2.

Amplifier #2 which is identical to Amplifier #1.

3.

X100 Amplifier

which has a fixed gain of 100 and has no offset control.



The requirements for an ideal amplifier are:

High input impedance to prevent loading the signal source. Low output impedance to ensure good transfer of signal to any succeeding stage and prevent loss of signal. High gain to reduce the number of amplifier stages required. Broad bandwidth to ensure that all required signals for a given band of frequencies are passed without attenuation. Low distortion so that only the amplitude of the signal is altered (high fidelity). Low noise factor to reduce the introduction of unwanted signals or interference. Stability. No tendency to self- (spurious) oscillation. These requirements apply to any type of amplifier, not just to DC amplifiers.

Amplifiers can be connected in cascade (one after another), to increase the overall gain, if required. Note The output voltage that can be provided by a DC amplifier cannot exceed the value of its supply voltage. In the case of the DIGIAC 1750 Trainer the output voltage is limited to a maximum of approximately ±10V.

The main characteristics of these devices are: Amplifier #1/2 Input signal voltage (max.)

X100

Amplifier

12V

12V

0.1 - 100

100

Voltage gain error (max.)

± 30%

± 4%

Output noise voltage (typ.)

10mV

10mV

fully adjustable

± 30mV

100kΩ

101kΩ

Voltage gain (nominal)

Output offset voltage (max.) Input impedance Table 13.1

211



13.2 Practical Exercise Characteristics of DC Amplifiers

SLIDE

BUFFER #1

C

O/P

B I/P 1

2

3

4

5

6

7

8

9

10

A

10k

+VIN

MOVING COIL METER

5

AMPLIFIER #1 +5V

I/ P

-10

O/ P

0

5 +10

+ 0V

V

-

.5

+

.4 1 100 10

-5V OFFSET

GAIN COARSE

.6 .7

.8

.3

.9

.2 .1

1.0

0V

L J

GAIN FINE

Fig 13.2 

Connect the circuit as shown in Fig 13.2 with the Amplifier #1 in circuit. Set the GAIN COARSE control to 100 and GAIN FINE to 1.0 for both amplifiers, Amplifier #1 and Amplifier #2. Note that Buffer #1 is needed so that the OFFSET adjustment does not affect the input voltage.

212



Switch ON the power supply. Set the 10k Ω variable resistor mid-way for exactly zero volts output as indicated by the digital multimeter. Adjust the OFFSET control of Amplifier #1 so that the output voltage is zero (or as near as it is possible to get to zero).



Increase the input voltage positively and note the output voltage. This increases to saturation quickly and then remains at this maximum value for further increase of input voltage. Record the value of this saturation voltage at the Moving Coil Meter in Table 13.2.



Repeat for the negative saturation voltage, recording again in Table 13.2.



Set the input voltage so that the output voltage is between +7 and +8V (Moving Coil Meter) and use the digital multimeter to note the values of the Amplifier #1 input and output voltages. Record the results in Table 13.2.





Outp ut volta ge

), this representing the maximum gain Inp ut vo ltage with positive polarity possible for the amplifier. Add this to Table 13.2.

Calculate the gain (

Gain (Av) set to x = 100 1001.0 Saturation voltage Input voltage Output voltage

Positive

Amplifier #1 Negative V mV V

Positive V

Vm

Amplifier #2 Negative V

Vm V

V mV

V

V

Voltage gain (Av) Table 13.2 

13.2a

Repeat with the 10k Ω variable resistor adjusted to give between -7 and -8V, to determine the gain of the amplifier for negative polarity input signals.

Enter your value of positive polarity gain for Amplifier #1 with maximum gain settings.

13.2b

Enter your value of negative polarity gain for Amplifier #1 with maximum gain settings.

This dual-polarity operation signifies that the amplifier is capable of amplifying AC signals as well as DC voltages. 

13.2c

Replace Amplifier #1 in the circuit with Amplifier #2 and repeat the procedures to adjust the OFFSET and to determine its maximum positive and negative gain values.

Enter your value of positive polarity gain for Amplifier #2 with maximum gain settings.

13.2d

Enter your value of negative polarity gain for Amplifier #2 with maximum gain settings.

213





Reset both Amplifier #1 and Amplifier #2 GAIN COARSE control to 1 and GAIN FINE to 0.1 for minimum amplifier gain.



With Amplifier #1 in circuit and an input voltage of +4V approximately, note and record the values of the input and output voltages in Table 13.3. Gain (Av) set to x = 0.1 10.1

Positive

Amplifier #1 Negative

Input voltage V V

V

V

Output voltage V V

V

V

Positive

Amplifier #2 Negative

Voltage gain (Av) Table 13.3

13.2e



Reset the input to -4V and repeat the readings, recording the results in Table 13.3.



Change to Amplifier #2 and repeat the readings for both polarities.



Using the values of input and output voltages given in Table 13.3 calculate the gain for each of the four conditions and add the results to the table.

Enter your value of positive polarity gain for Amplifier #1 with minimum gain settings.

13.2f

Enter your value of negative polarity gain for Amplifier #2 with minimum gain settings. 

Replace Amplifier #2 with the X100 Amplifier. Temporarily ground the input and note the output voltage with zero input voltage (the output offset voltage) using the digital multimeter.Use the same 0V patch panel as you use for the digital multimeter. Note that there is no offset control with this amplifier. The offset is adjusted to an acceptably low figure during production.

214



Nominal Gain (Av) = 100

X100

Positive

Saturation voltage Input voltage Output voltage

Amplifier Negative

V mV V

V Vm V

Voltage gain (Av) Table 13.4

13.2g



Repeat the procedures to measure the saturation voltages and the input and output voltages with the output set to a value between±(7-8)V. Record the values in Table 13.4.



Calculate the gain for both polarities and add these to Table 13.4.

Enter your calculated value for the gain of the X100 Amplifier with positive polarity.

13.2h

Enter your calculated value for the gain of the X100 Amplifier with negative polarity. 

Compare the results with the amplifier specifications given earlier.




215



13.3 The AC Amplifier The symbol for an AC amplifier is the same as for a DC amplifier.

I/P

O/P

Fig 13.3

The AC amplifier provided with the DIGIAC 1750 Trainer is a two-stage IC amplifier which has three fixed gain settings, 10, 100 and 1000. The mimic diagram on the DIGIAC 1750 Trainer shows the capacitors in the input and output circuits. These capacitors remove any DC level and hence there is no offset problem with an AC amplifier. Two of the main aspects of amplifiers are in conflict with each other, gain and bandwidth. As the gain of an amplifier is increased its bandwidth will be reduced. It is common to specify a gain bandwidth product for an amplifier. For instance, an amplifier with a gain bandwidth product of 106 could have a gain of 100 with a bandwidth of 104 or 10kHz, or a gain of 1000 with a bandwidth of 1kHz. This is why the amplifier on the DIGIAC 1750 Trainer is a 2-stage circuit; to get a bandwidth of 16kHz (covering the full audio band) and a gain of up to 1000. When the gain is switched to 100 (or 10) the bandwidth will be increased. The main characteristics of the device are: Input voltage (max.) Bandwidth (-6dB, gain = 1000) Maximum gain at 40kHz Output noise voltage (gain = 1000)

±12V

10Hz - 16kHz 225 100mV

Table 13.5

A high proportion of the output noise will be found to be stray pick-up of the output of the 40kHz oscillator which is adjacent to the AC Amplifier.

216



13.4 Practical Exercise Characteristics of the AC Amplifier From Function Generator A.C. AMPLIFIER O/P

SLIDE C

I/P

B 1

2

3

4

5

6

7

8

9

10

A

10k

10 1000 100

GAIN 0V

TP1

0V

TP2

0V

Oscilloscope

C H. 1

CH.2

Fig 13.4



Construct the circuit of Fig 13.4. Set the slider of the 10kΩ variable resistor to mid-way. This is to operate as a fine amplitude control on the input signal.



Switch the output of the Function Generator to a 1kHz sinewave. Switch the oscilloscope timebase to 0.5ms/div, Y1 amplifier (CH.1) to 10mV/div and the Y2 amplifier (CH.2) to 5V/div.



Switch ON the power supply and adjust the Function Generator output amplitude control to obtain 20Vp-p output from the AC Amplifier as Ω slider variable resistor indicated on CH.2 of the oscilloscope. Use the 10k for the final adjustment if necessary. Measure the input amplitude (CH.1) and record in Table 13.6.

Switch the AC Amplifier to maximum gain, 1000.

Gain setting Output voltage Input voltage

1000

100

10

20Vp-p

20Vp-p

20Vp-p

mVp-p

mVp-p

Vp-p

Amplifier gain Table 13.6

217


13.4a




Switch the AC Amplifier gain to 100 and repeat the setting of the output voltage to 20Vp-p and again measure the input signal amplitude, changing the Y1 amplifier setting as required. Record the result in Table 13.6.



Switch the AC Amplifier gain to 10 and repeat the setting and measurement.



Calculate the amplifier gain ( Outp ut volta ge ) for each setting of the gain Inp ut voltage switch and add the results to Table 13.6.

Enter your value of gain at 1kHz with the AC Amplifier set to x100. 

Change the Function Generator frequency to 40kHz and the oscilloscope timebase setting to 5µs/div, switch the AC Amplifier gain to 1000 and repeat the setting and measurement. Input voltage for 20Vp-p output =



mVp-p

Calculate the amplifier gain at 40kHz. Amplifier gain at 40kHz =

13.4b

Enter your value of gain at 40kHz with the AC Amplifier set to x1000. 

218




13.5 The Power Amplifier The symbol for a power amplifier is again the same as that for any DC amplifier.

I/P

O/P

Fig 13.5

The main characteristic of a power amplifier is the capability of a large power output. In order to do this the output impedance of the amplifier must be very low in order to provide a heavy current to a load without loss of output voltage across the output impedance. The components used must also be capable of dissipating the heat generated in high current circuits. The device provided with the DIGIAC 1750 Trainer has unity gain and a maximum output current of the order of 1.5A. The main characteristics of the device are as follows: Input voltage (max.)

±12V

Input impedance

100kΩ

Output current Output power (limited by power supply) Upper -3dB frequency

1.5A 9W 10.6kHz

Table 13.7

219



13.6 Practical Exercise Application of a Power Amplifier

POWER AMPLIFIER O/P I/P

From Function Generator SLIDE

A.C. AMPLIFIER O/P

C

LOUDSPEAKER

I/P

B I/P 1

2

3

4

5

6

7

8

9

10

10k

A

10 1000 100

GAIN 0V

TP1

0V

TP2

0V

Oscilloscope

CH.1

CH.2

Fig 13.6

220



Connect the circuit of Fig 13.6.



Switch ON the power supply and adjust the Function Generator to give a sinewave input at 1kHz to the AC Amplifier. Increase the amplitude to give maximum undistorted output from the amplifier.



Connect the Loudspeaker directly to the output of the AC Amplifier and observe the effect on the output waveform.


13.6a


With the Loudspeaker connected directly to the output of the AC Amplifier the waveform looks like:

a

b



13.6b

c

d

Transfer the output of the AC Amplifier to the input of the Power Amplifier. Transfer the oscilloscope CH.2 connection to the output of the Power Amplifier. Finally connect the output of the Power Amplifier to the Loudspeaker.

With the Loudspeaker connected via the Power Amplifier the waveform looks like:

a

b



c

d


Note that you have already used the Power Amplifier for DC applications when driving the lamp for opto-electronic experiments and for driving the motor for rotating motion investigations.

221



13.7 The Current Amplifier and Buffer Amplifier

I/P

O/P

Fig 13.7

The symbol for a current amplifier is once more the same as for any DC amplifier. The amplifier converts an input current to an output voltage. The device provided with the DIGIAC 1750 Trainer is intended for use with the P.I.N. photodiode, giving an output voltage 10,000 times the input current. An input current of 1mA (max.) will provide 10V (max.) at the output. The main characteristics of the Current Amplifier are shown in Table 13.8 below. The symbol for a buffer amplifier is again as shown in Fig 13.7. These amplifiers have a high input impedance and a low output impedance and are inserted in the circuit between a device having a high output impedance and one having a low input impedance to prevent loading, as shown in Fig 13.8. Device 1 (High output impedance)

Buffer

Device 2 (Low input impedance)

Fig 13.8

The characteristics are similar to those of the Power Amplifier but they have a much lower output current capability, (of the order of 20mA maximum for the devices provided with the DIGIAC 1750 Trainer). Two buffer amplifiers are provided with the DIGIAC 1750 Trainer, Buffer #1 and Buffer #2 and their main characteristics are shown in Table 13.9.

Input current (max) Transfer ratio Table 13.8

222

1mA 10,000V/A

Input voltage (max.)

±12V

Input impedance

100kΩ

Input offset voltage

300µV

Voltage gain Table 13.9

1.0



13.8 Practical Exercise Characteristics and Applications of Current and Buffer Amplifiers

SLIDE

CURRENT AMPLIFIER

C

O/P B 1

2

3

4

5

6

7

8

9

10

I/P 104 I IN

A

10k

MOVING COIL METER +5V

BUFFER #1


4 3

8

2

9

B

10k

0

5 +10

-10

I/P +VIN

+ -

A

10

1

5 O/P

V

0V

L J

0V

Fig 13.9 

Connect the circuit as shown in Fig 13.9 with the Buffer Amplifier out of circuit initially. Set the 10kΩ wirewound resistor for zero output (control fully counter clockwise) and the 10kΩ slider resistor for maximum resistance (slider to right).



Switch ON the power supply and set the output voltage from the 10k Ω wirewound resistor to 1V as indicated by the digital voltmeter.



Vary the slider resistor control from maximum resistance to minimum and note the reading of the digital voltmeter. You will note that it falls due to the increased current loading. Note the lowest value. 10k

slider resistance minimum voltage =

V

The current has varied from 0.1mA to 1.0mA approximately but this has been sufficient to produce the voltage drop above. The Buffer Amplifier can be used to reduce this loading effect.

223







Disconnect the lead between socket B of the Wirewound Track potentiometer and socket A of the Slide potentiometer. Connect socket B of the Wirewound Track to the input socket of Buffer #1. Connect the output socket of Buffer #1 to socket A of the Slide potentiometer. Buffer #1 is now connected between the Wirewound Track potentiometer and the Slide potentiometer.



13.8a

With the 10k Ω slider control at maximum (slider to right) set the voltage as indicated by the digital voltmeter to 1.0V. Vary the 10k Ω slider control over its full range and note the reading of the digital voltmeter.

The variation of voltage with the Buffer Amplifier in circuit was:

a >0.5V

b

<0.5V, >0.3V

c

<0.3V, >0.1V

d

virt ually nil



Check that the output from the 10k Ω wirewound resistor is still 1.0V and then remove the digital multimeter from the circuit, switch to a 2mA range and reconnect it as an ammeter into the circuit between the 10k Ω slider resistor and the Current Amplifier to monitor the input current.



Set the 10k Ω slider resistor control to each of the settings indicated in Table 13.10 and for each setting note the input current and the output voltage for the Current Amplifier.

Resistor setting

10

Input current Output voltageV

8 mA

V

V

6 mA

V

V

mA

4 mA

2 mA

1 mA V

Table 13.10 

224

Plot the graph of Output voltage against Input current for the Current Amplifier.



10 9 Output Voltage 8 (volts)

7 6 5 4 3 2 1 0

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7 0.8 0.9 1.0 Input current (mA)

Graph 13.1

13.8b

Is the graph linear?

Yes

13.8c

or

No

For each 0.1mA of input current the output voltage of the Current Amplifier changes by approximately:

a 0.1V

b

0.5V

c

1.0V

d

2V

This exercise has illustrated the characteristics of the current amplifier and the application of a buffer amplifier for circuits requiring a low output current. 


225



13.9 The Inverter Yet again the symbol is the same as for any amplifier.

I/P

I

O/P

Fig 13.10

The inverter amplifier, as the name implies, reverses the polarity of the voltage applied to the input, either DC or AC. The device provided with the DIGIAC 1750 Trainer has a voltage gain of unity. One aspect of all IC amplifiers which has not been mentioned before is the slew rate. This imposes a limitation on alternating signals on the rate at which the output voltage can change with respect to time. You can have either a small signal voltage at a high frequency or a larger signal voltage at a lower frequency. This is not quite the same thing as the gain/bandwidth product which was introduced earlier, as you will see from the experiment which follows. The main characteristics of the device are: Input voltage (max.) Voltage gain

-1.0

Input impedance

100kΩ


300µV

Slew rate (typ.) Table 13.11

226

±12V

0.15V/µs



13.10 Practical Exercise Characteristics of an Inverter

INVERTER O/P I/P -VIN

V -5V

0V

+5V

Fig 13.11 

Connect the circuit as shown in Fig 13.11.



Switch ON the power supply. With the Inverter input connected to the +5V supply note the value of the output voltage in Table 13.12. Inverter input

+5V

Inverter output

-5V V

V

Table 13.12 

13.10a

Transfer the Inverter input to the -5V supply and again note the value of the output voltages.

The polarity of the output voltage is:

a the same as the input 

b

opposite to the input


The output voltage magnitude may not be identical with the input due to the offset voltage. No facility for adjusting this has been provided.

227


40kHz OSCILLATOR


SLIDE

INVERTER

C

O/P B 1

2

3

4

5

6

0V

7

8

9

10

A

10k

TP1

0V

O/P I/P -VIN

TP2

0V

Oscilloscope CH.1

CH.2

Fig 13.12 

Connect the circuit as shown in Fig 13.12. Switch ON the power supply.



Set the oscilloscope timebase to 5 µs/div. and both Y amplifiers (CH.1 & CH.2) to 0.5V/div.



Adjust the control of the 10k Ω slider resistor to give an input voltage of 1Vp-p.



Sketch the input and output (Output 1) waveforms on the graticule provided:

Input

Output 1

Output 2

Waveform Sketch 13.1

228


13.10b



Change the Y2 (CH.2) amplifier to 1V/div and increase the setting of the 10kΩ slider resistor until the full effect of the slew rate is observed.



Add a sketch of the output (Output 2) waveform.

Your sketched waveforms are most similar to:

a

c

b



13.10c


Check the slew rate (

voltage ti me ( µs)

d

) against the specification given earlier.

Enter the value of peak-to-peak output voltage at which the slewing first occurred with an input signal of 40kHz. 



13.10d

Replace the input to inverter with a 5kHz sinewave output from the Function Generator. Increase the amplitude of the signal until slewing again begins to occur. Note the maximum peak-to-peak value of the undistorted output signal.

Enter the value of peak-to-peak output voltage at which the slewing first occurred with an input signal of 5kHz.

229



13.11 The Differential Amplifier

B

-

A

+

I/P's

O/P

Fig 13.13

The symbol for a differential amplifier is shown in Fig 13.13. The amplifier has two inputs which can be driven by separate signals. It is called differential because the output voltage depends on the difference in voltages applied to the two inputs. If the two inputs are driven by the same signal in phase then theoretically there should be no output. There will, however, be a small output the amount being determined by the common mode gain, which is designed to be as near to zero as possible. For the device provided on the DIGIAC 1750 Trainer, the output voltage is given by (VA - VB). Two differential Amplifier". amplifier circuits are provided, the basic secondfunctions being labeled "Instrumentation This carries out the same as the differential amplifier but has an improved (reduced) common mode gain. The main characteristics of the devices are: Differential Amplifier Input voltage (max.)

±12V

Differential gain

1.0

Common mode gain (max.)

0.02

0.006

Input impedance (input A)

200kΩ

100kΩ

Input impedance (input B) Table 13.13

230

Instrumentation Amplifier

100kΩ



13.12 Practical Exercise Characteristics of a Differential Amplifier

SLIDE

C B

1

2

3

4

5

6

7

8

9

10


+5V

7

4

-5V

3

8

2

9 1


A

10k

B

-

A

+

A-B

B

V

A

10

10k

0V

Fig 13.14



Connect the circuit as shown in Fig 13.14 and switch ON the power supply.



Moving the digital voltmeter lead as necessary, set the voltage at input A of the Differential Amplifier to -3V and input B also to -3V and note the resulting output voltage. Record the value in Table 13.14.

Step

1

2

3

4

5

6

7

8

Input B voltage

-3V

+1V

+4V

+2V

0V

+4.5V

+2V

-2.7V

Input A voltage

-3V

+1V

+4V

+4V

+3V

+2.2V

-3V

+3.6V

Output voltage V V

VV

V

V

V

V

Table 13.14 

Repeat the procedure for each of the other pairs of inputs in Table 13.14 and record the output voltage again.

231


13.12a

Enter your output voltage reading for Step 1 in V.

13.12b


13.12c


13.12d


13.12e


13.12f


13.12g

In Steps 1, 2 & 3 the amplifier was working in:

a differential mode

b

inverting mode

c common mode

d

opposing mode



232






2.

3.

4.

5.

The term "input offset voltage" applied to a DC amplifier means the voltage:

a

at the input with no signal applied

b

at the output with no signal applied

c

needed at the input for zero output voltage

d

needed at the input with signal applied

The main difference between an AC amplifier and a DC amplifier is that the:

a

AC amplifier does not suffer from offset

b

only the AC amplifier can amplify AC signals

c

DC amplifier will not pass low frequency signals

d

gain of an AC amplifier will be greater

The type of amplifier which will pass AC signals is:

a

the DC amplifier only

b

the AC amplifier only

c

either the DC or the AC amplifier

d

neither the DC nor the AC amplifier

The purpose of a buffer amplifier is to:

a

remove DC signals

from AC signals

b

increase the signal level at low frequencies

c

convert positive polarity signals into negative polarity signals

d

reduce loading on a high impedance source

Comparing a buffer amplifier to a power amplifier, the buffer amplifier will:

a

only operate on DC

b

deliver less current

c

only operate on AC

d

restrict the bandwidth

Continued ...

233




A signal source has an open-circuit EMF of 10V and an output impedance of 100k . If this was connected directly to a 10k load the voltage delivered to the load would be:

a 7.

5V

c

1.1V

d <1V

AC

b

buffer

c

inverter

d differential

A current amplifier has a transfer ratio of 5000V/A. If the input current is 2mA the output voltage will be:

a 9.

b

For the situation referred to in question 6 above the type of amplifier which would restore the voltage delivered to the load would be:

a 8.

10V

2.5V

b

5.0V

c

10V

d 10000V

A current amplifier has a transfer ratio of 5000V/A. If the output voltage is 5V the input current will be:

a

10mA

b

5mA

c

2.5mA

d 1mA

10. A differential amplifier has inputs VA connected to the non-inverting (-) input and VB connected to the inverting (+) input. The amplifier has unity (1.0) gain. If the input voltages are VA = +5.0V, VB = +2.0V then the output voltage will be:

a

+3V

b

+7V

c

-7V

d +10V

11. A differential amplifier has inputs VA connected to the non-inverting (-) input and VB connected to the inverting (+) input. The amplifier has unity (1.0) gain. If the input voltages are VA = -5.0V, VB = +2.0V then the output voltage will be:

a

+3V

b

+7V

c

-7V

d +10V

12. A differential amplifier has inputs VA connected to the non-inverting (-) input and VB connected to the inverting (+) input. The amplifier has unity (1.0) gain. If the input voltages are VA = +5.0V, VB = -5.0V then the output voltage will be:

a

234

0V

b

+5V

c

-5V

d +10V


Signal Conversions Chapter 14

Chapter 14 Signal Conversions




Describe the characteristics of a voltage to current converter (V/I).



Describe the characteristics of a current to voltage converter (I/V).



Describe the characteristics of a voltage to frequency converter (V/F).



Describe the characteristics of a frequency to voltage converter (F/V).



Describe the characteristics of a full wave rectifier.

• • • •

DIGIAC 1750 Transducer and Instrumentation Trainer. 4mm Connecting Leads. Digital Multimeter. Oscilloscope.

235



14.1 Voltage to Current Converter The voltage to current converter converts an input voltage to an output current. The device operates as a constant current source within the limits of the supply voltage. As an example of this, if 20mA is supplied to a load of 50 Ω, then the voltage dropped across the load is: 20x10-3 x 50 = 1.0V. With the V/I converter supplied from +12V DC this is no problem. If, however, the load resistance is increased to 1k Ω, then the voltage across the load at 20mA would be: 20x10-3 x 1000 = 20V, which the device would be unable to provide from a +12V supply. A simple block diagram is used to represent the V/I Converter on the DIGIAC 1750 Trainer. The standard symbol for a constant current source is given in Fig 14.1.

0-24mA

Fig 14.1

The main characteristics of the device fitted to the DIGIAC 1750 Trainer are: Input voltage range

0-1.5V

Output current range (max.)

0-24mA

Transfer ratio

16mA/V

Table 14.1

236



14.2 Practical Exercise Characteristics of a Voltage to Current Converter

B

O/P

O/P A

+5V

N T C T H E R M IS T O R S 0V

P L A T I NU M R . T . D .

HEATER ELEMENT SLIDE

I/P C B

1

2

3

4

5

6

7

V8

9

10

10k

V/I CONVERTER

A

O/P I/P

A

Fig 14.2

Note that a second meter is shown as an ammeter connected between the output of the V/I Converter and the load (the heater element on the thermal transducer panel). If a second instrument is available then the measurements will be simplified. The instructions will be given assuming that this is not the case. 

Connect the circuit as shown in Fig 14.2 and set the 10k Ω resistor for zero output voltage (slider to left).



Switch ON the power supply. Set the input voltage to the V/I converter to 0.5V.



Remove the digital multimeter from the circuit, range it as an ammeter (up to 25mA will be needed), and reconnect it in between the output of the V/I Converter and the load. Measure the load current and record the result in Table 14.2. Restore the digital multimeter as a voltmeter in the srcinal position as shown in Fig 14.2.

Input voltage Output current

0V

0.5V mA

1.0V mA

1.5V mA

mA

Table 14.2

237





Repeat the procedure for input voltage settings of 1.0V and 1.5V and record the results in Table 14.2. Keep the multimeter connected as an ammeter monitoring the load current after the final reading.



Connect the input of the V/I Converter to 0V (ground) and note the effect on the output current. Record the result in Table 14.2.



Plot the characteristic of output current against input voltage for the V/I Converter on the axes provided:

25 Output Current 20 (mA) 15 10 5 0

0.5

1.0 1.5 Input Voltage (V)

Graph 14.1

14.2a

Is the output current proportional to the input voltage?

Yes 

or

No

Calculate the Transfer Ratio from any pair of voltage and current readings. Transfer Ratio =

14.2b

mA/V

Enter your value of the Transfer Ratio in mA/V. 

Restore the input of the V/I Converter to terminal B of the 10k Ω slider resistor and the input voltage to 1.5V. Transfer the digital multimeter to the output of the V/I Converter. First unplug the load and note the effect on the output voltage of the V/I Converter. Then connect the Lamp Filament on the opto-transducer panel as the load and note the voltage again.

238


14.2c


Is the output voltage constant?

Yes 

or

No


14.3 Current to Voltage Converter The current to voltage converter converts an input current to an output voltage and is thus the converse of the voltage to current converter. The V/I and I/V Converters provided with the DIGIAC 1750 Trainer are arranged to have parameter values that are the reciprocal of each other. This means that the pair of devices could be used to send a voltage down a long wire without attenuation, since the current which is launched into the transmission line at one end must also appear at the termination (except in the unlikely case of leakage current, which can be restricted by good insulation).

Input Voltage

V/I

I

I/V

Converter

Converter

Output Voltage

Fig 14.3

The actual voltage on the transmission line is irrelevant unless it tries to be greater than the supply feeding the V/I Converter. The main characteristics of the I/V converter are: Input current range Output voltage range Transfer ratio

0-24mA (100mA max.) 0-1.5V (6V max.) 62.5mV/mA

Table 14.3

239



14.4 Practical Exercise Characteristics of a Current to Voltage Converter

SLIDE

V/I CONVERTER

C

I/V CONVERTER

O/P B

V 1

2

3

4

5

6

0V

7

8

9

10

I/P

O/P I/P

A

10k +5 V

Fig 14.4



Connect the circuit as shown in Fig 14.4. Set the 10k Ω slider resistor for zero output voltage.






Set the input voltage to the V/I converter to 0.5V. Transfer the digital multimeter to the output of the I/V Converter and note the output voltage.



Repeat the procedure for input voltage settings of 1.0 and 1.5V and enter the values in Table 14.4.

Record the value in Table 14.4.

Input voltage (V/I) Output voltage (I/V)

0

0.5 V

1.0 V

1.5 V

V

Table 14.4 

14.4a

Enter your value of output voltage when the input voltage was 1.0V.



240

Transfer the input of the V/I converter to 0V (ground) and note and record the output voltage from the I/V Converter in Table 14.4.




14.5 Voltage to Frequency Converter This device converts an input voltage to an output frequency, the frequency being proportional to the input voltage. The circuit is based on a dedicated (designed for the job) IC type LM331. The output waveform is in the form of short duration (approximately 60 µs) negativegoing pulses, the repetition rate of which can be controlled over a very wide range. The negative excursion duration remains constant as the frequency is increased. This limits the overall time period of the output waveform to about 85 µs, or a frequency of just under 12kHz. The pulse shape is degraded at frequencies above about 10.5kHz. The Counter/Timer facility has a limited range, having only a 3-digit display, but it is better for counting pulses at very low frequencies. The oscilloscope gives a very good display of the waveform and can also be used for measurement of higher frequencies. The main characteristics of the device provided with the DIGIAC 1750 are: Type

Input voltage (max.)

LM331

12V

Transfer ratio

1kHz/V

Maximum frequency

10kHz

Non-linearity (typ.)

0.024% full scale

Non-linearity (max.)

0.14%

Table 14.5

241



14.6 Practical Exercise Characteristics of a Voltage to Frequency Converter WHEATSTONE BRIDGE D

V/F CONVERTER O/P

C

12k

IN T dV dt

I/P

A OUT

B

3


100ms 1s 10ms

IN

V

1V

0V

TIME CONSTANT

Rx COUNTER/ TIMER

COMPARATOR +12V

OFF ON HYSTERESIS

TP1

0V

B

-

A

+

I/P

T IME

F RE E R U N

O/P

RE SE T

COUNT

1s

CH.1 Oscilloscope

Fig 14.5

The Counter/Timer is range. used as frequency meter measure the lowershaping output frequencies, within its Thea Differentiator and to Comparator are pulse circuits to enable the V/F Converter output to trigger the Counter/Timer. An oscilloscope is used to monitor the output waveform and to determine frequencies above the range of the Timer/Counter. 

Connect the circuit as shown in Fig 14.5. Set the Differentiator control to 1s, the Counter controls to COUNT and 1s, the Comparator HYSTERESIS OFF and the 10kΩ 10-turn resistor to zero.



Switch ON the power supply and set the input voltage to 0.2V. Press the RESET button of the Counter and note the displayed value, which represents the frequency output of the V/F converter. Record the value in Table 14.6.

Input Voltage (volts) Output frequency (Hz) Table 14.6

242

0.2

0.4

0.6

0.8

1.0





Repeat the procedure for input voltage settings of 0.4, 0.6, 0.8V and 1.0V recording the output frequency values in Table 14.6.



Continue with further increased values of input voltage if possible while the Counter/Timer unit is registering the frequency correctly. The unit may operate beyond 1kHz, this being signified by the count going through 999. When the frequency is too high for the counter, the display will only reach a low value and not pass through 999.



Reset the frequency to 1kHz (1.0V input) and turn your attention to the oscilloscope. Disconnect the feed to the Differentiator, since the loading effect will degrade the output waveform of the V/F Converter.



Set the oscilloscope timebase to 0.2ms/div and ensure that the variable control is in its calibrated position. Set the Y1 amplifier (CH.1) to 2V/div. You should have a stable trace of 2/3 negative-going pulses of about 5V (2.5 div) amplitude.



Measure the time taken for one cycle along the X axis (for instance, one cycle covering 2.8 div. would be 2.8 x 0.2ms = 0.56ms) and record this in Table 14.7. Take the reciprocal of this to convert to frequency (for instance, 1 0.56 × 10-3 = 1786Hz or 1.79kHz).

Input voltage (volts)

1

Time for one cycle Frequency = 1

2

ms

ms

3

4 µs

5 µs

6 µs

7 µs

8 µs

9 µs

10 µs

µs

(kHz)

T

Table 14.7 

14.6a

Take measurements and calculations at each of the other input voltages listed in Table 14.7, changing the oscilloscope timebase setting as necessary.

Is the output frequency proportional to the input voltage?

Yes

14.6b

or

No

Calculate and enter the Transfer Ratio in kHz/V. 


243



14.7 Frequency to Voltage Converter This device converts an input frequency to an output voltage. Each input pulse triggers a monostable multivibrator to generate a constant period pulse which pumps one packet of charge into a reservoir capacitor. The voltage across the capacitor is therefore dependent on how many pulses are received each second. For the unit provided with the DIGIAC 1750 Trainer, the parameters are arranged to be reciprocal to those of the V/F converter. A communication channel would again be possible with frequency as the transmission medium. The main characteristics are: Input frequency (max.)

10kHz

Transfer ratio

1V/kHz

Time constant

100ms

Settling time

0.7s

Accuracy

± 0.1%

Output ripple

10mV

Output impedance

100kΩ

Table 14.8

244



14.8 Practical Exercise Characteristics of a Frequency to Voltage Converter WHEATSTONE BRIDGE D

V/F CONVERTER

F/V CONVERTER

O/P C

A

12k

O/P I/P

OUT

B

3

I/P

IN

V

1V

Rx

0V

+5V

Fig 14.6 

Connect the circuit as shown in Fig 14.6. Switch ON the power supply and set the input voltage to the V/F converter to 1.0V. Note the value of the output voltage from the F/V converter and record the value in Table 14.9.

Input voltage (V/F) Output voltage (F/V)

1

2

3

V

V

4 V

5 V

V

Table 14.9 

14.8a

Repeat the procedure for input voltage settings of 2, 3, 4 and 5V.

Enter your output voltage when the input voltage is 3V. 

You will see from the specification that the output impedance of the F/V Converter is 100kΩ. If you measure the output voltage using the M.C. meter the reading will be affected by the low loading impedance. Try it with the output voltage set 5V, recording the results in Table 14.10. Instrument

Digital Multimeter only

Output voltage

V

M. C. Meter only

M. C. Meter via Buffer #1 V

V

Table 14.10

14.8b

Enter your output voltage reading when using the M. C. Meter via Buffer #1. 


245



14.9 The Full Wave Rectifier The full wave rectifier converts a sinewave AC input into a series of unidirectional positive half cycles as shown in Fig 14.7. The negative half cycles are inverted so that the output is always of one polarity.

+

+ Full Wave

0

Rectifier

0

Fig 14.7

With an input DC signal of either polarity the output is always positive, the magnitude of the output being the same as that of the input signal. In the case of an input consisting of an AC waveform riding on a DC component, the output waveform will be a mixture of the input components, the negative components being inverted to be positive. If the DC component of the input is greater than the AC component then the same waveform will appear at the output, but always with positive polarity, irrespective of the polarity of the input. The circuit is active, containing two operational amplifiers; not just a full-wave diode bridge, since this cannot be adjusted to compensate for losses. It is not intended for delivery of DC power. Measurements of AC quantities using DC instruments are possible with accuracy using Full Wave Rectifiers. The main characteristics of the device provided with the DIGIAC 1750 Trainer are: Input voltage Output voltage error Table 14.11

246

12V (max) 2% (typ.), (6% max.)



14.10 Practical Exercise Characteristics of a Full Wave Rectifier with DC Applied

FULL WAVE RECTIFIER

O/P I/P

VIN

0V

-5V

+5V

Fig 14.8

Connect the circuit as shown in Fig 14.8. Switch ON the power supply and note the values of the input and output voltages for the Full Wave Rectifier with +5V applied to the rectifier input. Record the output voltage in Table 14.12.



Input voltage Output voltage

+5V

-5V V

V

Table 14.12

Transfer the input of the Full Wave Rectifier to the -5V supply and repeat voltage readings, recording the output voltage in Table 14.12 again.



14.10a

Is the output polarity the same for both input polarities?

Yes

14.10b

or

No

Is the output magnitude approximately the same as the input?

Yes 

or

No


247



14.11 Practical Exercise Characteristics of a Full Wave Rectifier with AC Applied

SLIDE

40kHz OSCILLATOR

C

O/P B 1

2

3

4

5

6

7

8

9

10

A

10k

A.C. AMPLIFIER O/P I/P

0V FULL WAVE RECTIFIER

O/P

V

I/P 10 1000 100

VIN

GAIN TP1

0V

CH.1

TP2

Oscilloscope

0V

CH.2

Fig 14.9

14.11a



Connect the circuit as shown in Fig 14.9. Set the gain of the AC Amplifier to 10.



Set the oscilloscope timebase to 5 µs/div and both Y amplifiers to 1V/div.



Switch ON the power supply and adjust the slider of the 10k Ω resistor so that the amplitude of the output of the AC Amplifier (CH.1) is the same as that of the 40kHz Oscillator (CH.2).



Switch the selectors on your Y amplifiers between DC and AC. Any movement of the waveform on the screen means that there is a DC component. If there is no DC component the waveform will not move.

When switching between AC and DC inputs to the oscilloscope Y amplifiers, the waveform(s) having a DC component is/are: a the 40kHz Oscillator b the AC Amplifier

c both of them

248

d

neither of them





Transfer CH.2 of the oscilloscope from the output of the 40kHz Oscillator to the output of the Full Wave Rectifier.



Sketch the input and output waveforms of the Full Wave Rectifier on the graticule provided, marking in the amplitude of the waveforms:


14.11b

Your waveforms (CH.1 at the top) are most like:

a

b



c

d

Record the DC value of the Full Wave Rectifier output from the digital multimeter reading, then switch OFF the power supply. DC value of the Full Wave Rectifier output =

14.11c

V

The reason for the difference between the amplitude of the output waveform and the measured DC value on the multimeter is:

a they are both DC quantities b the multimeter responds to the average of the waveform c the input waveform is measured in RMS d the input waveform is measured in peak-to-peak

249




A Voltage to Current Converter, supplied from a 12V supply has a Transfer Ratio of 12mA/V. If the input voltage was 1.2V, the short-circuited output current would be:

a 2.

b

10mA

c

12mA

d 14.4mA

A Voltage to Current Converter, supplied from a 12V supply has a Transfer Ratio of 12mA/V. If the input voltage was 1.2V, the voltage across a 75 load would be:

a 3.

1mA

0.52V

b

1.08V

c

1.2V

d 5.2V

A Voltage to Current Converter, supplied from a 12V supply has a Transfer Ratio of 12mA/V. If the input voltage was 1.2V, the voltage across a 1k load would be:

a

1.2V

b

8V

c

12V

d 14.4V

4.

A Voltage to Current Converter with Transfer Ratio of 10mA/V is connected in cascade to a Current to Voltage Converter having a Transfer Ratio of 0.25V/mA. With an input voltage of 1.5V applied to the input to the V/I Converter, the output voltage from the I/V converter will be:

5.

A Voltage to Current Converter with Transfer Ratio of 10mA/V is connected in cascade to a Current to Voltage Converter. In order for the output voltage from the I/V Converter to be the same as the input to the V/I Converter the Transfer Ratio of the I/V Converter would need to be:

a

a 6.

250

1V/mA

b

b

2.5V

0.1V/mA

c

c

3.75V

10mV/mA

d 5.25V

d 1mV/mA

A Voltage to Frequency Converter has a Transfer Ratio of 0.5kHz/V. The input voltage required for an output frequency of 3.5kHz will be:

a 7.

1.5V

1.75V

b

3.5V

c

4.0V

d 7.0V

The Voltage to Frequency Converter referred to in question 6 is to be cascaded with a Frequency to Voltage Converter. In order for the output voltage of the system to be the same as the input voltage, the Transfer Ratio of the Frequency to Voltage Converter will need to be: a 0.5V/kHz b 0.5kHz/V c 2.0V/kHz d 3.5V/kHz



Student Assessment 14 Continued ... The following questions refer to a Full Wave Rectifier similar to the one provided on the DIGIAC 1750 Trainer. 8.

A Full Wave Rectifier has an input voltage of -5V DC. The output will be:

a

-5V

b

0V

0V

c

+5V

0V

(a)

d

±2.5V

0V

(b)

(c )

Fig 1

9.

The waveform of Fig 1(a) is applied to the input of a Full Wave Rectifier. The output waveform will be:

a0V

b 0V

c

d 0V

0V

10. The waveform of Fig 1(b) is applied to the input of a Full Wave Rectifier. The output waveform will be: a c d b 0V

0V 0V

0V

11. The waveform of Fig 1(c) is applied to the input of a Full Wave Rectifier. The output waveform will be: a c d b 0V

0V 0V

0V

251



Notes: ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

252


Comparators, Oscillators and Filters Chapter 15

Chapter 15 Comparators, Oscillators and Filters



Describe the characteristics of a comparator.



Explain the effect of hysteresis on the operation of a comparator.



Describe the characteristics of an alarm oscillator.



Explain the term "latch" applied to an alarm oscillator.






Describe the characteristics of an electronic switch. Describe the characteristics of a 40kHz oscillator.



Describe the characteristics of band pass filters.



Describe the characteristics of low pass filters.

• • • • • •

DIGIAC 1750 Transducer and Instrumentation Trainer. 4mm Connecting Leads. Digital Multimeter. Oscilloscope. Function Generator. BNC to 4mm connecting lead.

253



15.1 The Comparator

B

-

A

+

I/P's

O/P

Fig 15.1

The symbol for a comparator is shown in Fig 15.1. It is the same as for a differential amplifier but the characteristics of the comparator are different. The differential amplifier investigated in Chapter 13 had unity gain. The output voltage was the simple mathematical difference between inputs A and B. The gain of a comparator is very high, so that only a very small difference between the two inputs will cause the output to saturate at a voltage near to the supply voltage, with either polarity. The comparator therefore has two possible output voltage states: 1.

with input voltage A more positive than B, the output is a maximum positive.

2.

with input voltage A more negative than B, the output is a maximum negative.

Only the very slightest variation between the inputs causes the output voltage to change from one state to the other and the circuit is therefore susceptible to noise variations. To overcome this problem, the circuit is modified so that the voltage at A must rise to a threshold value above B for switching to occur. Similarly, with the voltage falling, the voltage at A must fall to a different threshold value below B before the circuit switches back. This is referred to as hysteresis and the difference in the voltages is referred to as the hysteresis voltage.

254



This is illustrated in Fig 15.2.

Threshold Voltages

A Voltage B Voltage

+V

Rising

+V

Input Voltage

Falling

0V

0V

+V

+V

0V

Output 0V Voltage

-V

-V

N oH y s t e r e s i s

W i t hH y s t e r e s i s

Fig 15.2

With no hysteresis and voltage A varying, the output changes state frequently. With hysteresis the output does not change state for small variations of voltage around theoflast switching theswitching circuit. voltage, a large change of voltage is required to cause The circuit with hysteresis does not respond to any noise with a voltage amplitude less than the hysteresis voltage. The main characteristics of the device provided with the DIGIAC 1750 Trainer are: Input voltage (max.)

±12V


9mV

Output Voltage (no load) Hysteresis voltage (switch ON)

(-11.8) to (+12)V 4.2V

Table 15.1

255



15.2 Practical Exercise Characteristics of a Comparator

+12V

5

CARBON TRACK 6 C


7

4

8

2

9 10

B A

3

8

2

9 1

100k 0V

COMPARATOR OFF ON HYSTERESIS

7

4

3

1

0V

B A

10

10k

B

-

A

+

MOVING COIL METER

+5V

5

-5V

V

O/P

-10

BUFFER #1 O/P

0

5 +10

+

I/P +VIN

0V

L J

Fig 15.3

256



Connect the circuit as shown in Fig 15.3. Ensure that the Comparator HYSTERESIS switch is set to OFF. Set the controls of both resistors fully counter clockwise.



Switch ON the power supply. The voltage at input B will be 0V, that at A will be -5V and the output will be approximately -12V.



Gradually rotate the control of the 10k Ω resistor clockwise, making the voltage at input A (V A) less negative. Note the voltage at which the output voltage switches polarity with V A rising (VR). Record the value of V R in Table 15.2. Record also in Table 15.2 the Comparator output saturation voltage above threshold with V A rising.



Continue to increase input V A and observe the effect on the output voltage above switching.


No Hysteresis


VB= 0V

Output Saturation Voltage

V B = +4V

VA

VA

VA rising (VR)

V

V

V

VA falling (VF)

V

V

V

Table 15.2

15.2a



Reduce V A and note the value at which the output voltage switches back to a negative value with VA falling (VF). Note the value of the comparator output saturation voltage below threshold with VA falling.



Repeat the procedure with input B set to +4V, noting the switching voltages at input A. The comparator output voltage values will not alter so there is no need to record them.

Enter your measured value of the threshold voltage VA when the voltage at the B input VB = +4V. 

Set the HYSTERESIS switch in the ON position and repeat the procedure for voltage settings at the B input of 0V and +4V.

With Hysteresis

VB= 0V

Output Saturation Voltage

V B = +4V

VA

VA

VA rising (VR)

V

V

V

VA falling (VF)

V

V

V

Table 15.3 

15.2b


Calculate and enter your measured hysteresis voltage in V.

The circuit will have similar characteristics for all settings of the input voltage at B. Alternatively, the voltage at A may be set and that at B varied. The value of the hysteresis voltage can be set in the design stage to any desired value by adjusting the circuit component values.

257



15.3 The Alarm Oscillator The alarm oscillator consists of two stages.

Voltage Reference Comparator I/P

+

Alarm Oscillator

O/P

Fig 15.4

The input circuit is a comparator which is followed by the oscillator. With the input voltage low, the comparator output prevents the oscillator from operating. Oscillations only occur when the input voltage exceeds a level that is decided by the circuit component values. With the "latch" switch in the OFF position, the oscillator will be ON or OFF depending on whether the input voltage is above or below the threshold level. With the "latch" switch in the ON position, the oscillator is latched ON by the input voltage exceeding the threshold. It remains ON continuously, even if the input voltage is reduced below threshold, until the power supply is turned off. The unit is used as an alarm indication when the value of a controlled parameter exceeds a pre-determined level. The main characteristics of the device provided with the DIGIAC 1750 Trainer are: Input voltage (max.)

12V

Trip voltage (threshold)

2.3V

Oscillator frequency Output impedance Table 15.4

258

540Hz 4kΩ



15.4 Practical Exercise Characteristics of an Alarm Oscillator


T

COUNTER/ TIMER

dVIN dt

I/P

TIME

100ms 1s 10ms

RE S E T

TIME CONSTANT WIREWOUND TRACK 6 5 C

OFF

8

2

9 10

V

B

1s

LOUDSPEAKER

ON LATCH

3

1

CO UN T

ALARM OSCILLATOR

7

4

F RE E RU N

O/P

I/P

I/P

A

10k 0V

+5 V

Fig 15.5 

Connect the circuit as shown in Fig 15.5. Set the Alarm Oscillator LATCH switch to OFF and turn the 10kΩ resistor control fully counter clockwise. Switch the Counter to COUNT and 1s, and the Differentiator to 1s.



Switch ON the power supply and rotate the resistor control slowly clockwise to gradually increase the input voltage to the Alarm Oscillator. Note the input voltage threshold at which oscillations start. Record the threshold level in Table 15.5. Start Threshold

Without latch

Stop Threshold

Oscillator Frequency

V

V

V

V

With latch Table 15.5

Hz

259





Increase the voltage to maximum and note the effect on the oscillator output.



Now gradually reduce the input voltage and record the voltage threshold at which the oscillations stop in Table 15.5.



Set the latch switch to ON and repeat the procedure, noting the input voltage at which the oscillations start and then noting the effect of reducing the input voltage to zero.





Press the RESET button on the Counter to determine the oscillation frequency and add this to Table 15.5. Switch the power supply OFF and then ON again to observe the effect. Repeat the start and stop actions.

Note: The output sound level will be low due to the high output impedance of the oscillator. This can be increased if necessary by feeding the loudspeaker via the power amplifier, but this is not advisable in the laboratory situation.

15.4a

Enter your voltage threshold level with the LATCH OFF.

15.4b

Enter your oscillation frequency from the Counter reading (in Hz). 

260




15.5 The Electronic Switch A simplified diagram of the Electronic Switch is given in Fig 15.6.

+12V

O/P

Voltage Reference

I/P

+

Comparator

Fig 15.6

The series PNP transistor operates as a switch. When the input voltage to the Comparator (inverting input) is low the Comparator output is high and the transistor is switched off. If the input voltage is taken above the threshold established bybase-emitter the referencejunction voltage,ofthe output switches low and forward biases the theComparator switching transistor to turn it on and supply voltage to the load. The maximum permissible output current is limited by the parameters of the series switching transistor. The main characteristics of the device provided with the DIGIAC 1750 Trainer are: Input voltage (max.) Trip voltage Output current (max.)

12V +2.1V 1A

Table 15.6

261



15.6 Practical Exercise Characteristics of an Electronic Switch

O/ P

O/ P

P H OT OCONDU CT I V E CE L L

P H OT OT R A N S I ST OR

LAMP FILAMENT

I/P


ELECTRONIC SWITCH +12V

7

4

8

3

B

9

2 1

A

10

V

10k

0V

O/P

I/P

+5V

Fig 15.7 



Connect the circuit as shown in Fig 15.7. Set the resistor control fully counter clockwise. Switch ON the power supply and note the output voltage from the electronic switch. Record in Table 15.7.

Output voltage with input below trip

Input trip voltage rising

V

Output voltage with input above trip

V

V

Input trip voltage falling

V

Table 15.7

15.6a

262



Transfer the meter to the Electronic Switch input and increase the input voltage gradually and note the value of input voltage at which switching occurs and also the value of the output voltage after switching. Add these to Table 15.7.



Now gradually reduce the input voltage and note and record the value when the circuit switches off. Switch OFF the power supply.

Enter your value of trip voltage (input rising).



15.7 40kHz Oscillator

Colpitts Oscillator

Emitter Follower Buffer

Output

Fig 15.8

This nominally 40kHz oscillator produces a sinusoidal output of suitable frequency for use with some of the AC driven transducers provided with the DIGIAC 1750 Trainer. The Colpitts oscillator uses an LC tuned circuit with two capacitors in the feedback loop, giving good stability of oscillation frequency and amplitude. The effective component values are L = 1mH, C = 15nF giving a design oscillation frequency of: 1 fosc = = 41.09 kHz 2π L C The buffer gives a low output impedance and prevents loading of the oscillator which might cause frequency shifting. The main characteristics of the device are: Output frequency range

37-46kHz

Output frequency (typ.)

41kHz

Output amplitude

6Vp-p

Output impedance

1.1kΩ

Table 15.8

263



15.8 Practical Exercise Characteristics of a 40kHz Oscillator

40kHz OSCILLATOR

TP1

0V

O/P

Oscilloscope CH.1

SLIDE

C B

1

2

3

4

5

6

7

8

9

10

A

10k

0V

Fig 15.9 

Connect the circuit of Fig 15.9 with the variable resistor slider to the right for maximum resistance. The slider resistor will not be used initially.



Set the oscilloscope amplifier to 1V/div. timebase to 5 µs/div (calibrated) and the Y1 (CH.1) 




Note the amplitude of the 40kHz Oscillator output and the time taken for one cycle. Record these in Table 15.9. Open circuit amplitude Vp-p

Time taken for one cycle µs

Frequency kHz

Output Impedance kΩ

Table 15.9 

264

Calculate the reciprocal of the time taken for one cycle (the time period) to obtain the frequency and add this to Table 15.9.



Measurement of the Output Impedance

Ro

E

R

Fig 15.10



Connect socket B of the 10k Ω slider resistor to the output of the 40kHz Oscillator and reduce its value until the output amplitude of the oscillator falls to half of the open circuit value. You may find it convenient to change the setting of the Y1 Amplifier to 0.5V/div to do this measurement. The display amplitude will then be the same as before. When this is done the voltage dropped across the 10kΩ slider resistor (R in Fig 15.10) is the same as the output impedance of the 40kHz Oscillator (Ro). Since the two resistances are in series, the current through them must be the same, so their resistances must be the same. This is a standard technique for measurement of output impedance.



Switch OFF the power supply, disconnect the 10k Ω slider resistor from circuit (without changing the setting) and measure the resistance of the section used with your digital multimeter as ohmmeter. Add the result to Table 15.9.

15.8a

Enter your measured value of the frequency in kHz.

15.8b

Enter your measured value of the output impedance in k .

265



15.9 Filters

LPF

BPF

BSF

HPF

Fig 15.11

There are four main classifications of filter, specified by the range of frequencies passed: 1. Low pass filter, LPF, passing all frequencies below the design (cut-off) value. 2. Band pass filter, BPF, passing those frequencies within the design range. 3. Band stop filter, BSF, passing those frequencies outside the design range. 4. High pass filter, HPF, passing all frequencies above the design (cut-off) value. The symbols used to represent the four types are shown in Fig 15.11. The cut-off frequency is sometimes called the break or corner frequency and is the frequency at which the output first falls to -3dB (0.707V max) from the mid-band. Only a bandpass and a low pass filter are provided with the DIGIAC 1750 Trainer. The main characteristics of these are: Bandpass Filter Lower cut-off frequency (typ.)

39.5kHz

-

Upper cut-off frequency (typ.)

42.5kHz

16, 1.44 or 0.14Hz

Time constants

-

10ms, 100ms or 1s

Input impedance

10kΩ

1MΩ

Output impedance Input voltage (max)

10kΩ -

12V

Table 15.10

266

Low Pass Filter



15.10 Practical Exercise Characteristics of a Bandpass Filter The very low cut-off frequencies of the Low Pass Filter make it difficult to investigate the response because of the demands which would be made on the function generator ranges. This investigation is therefore limited to the 40kHz Bandpass Filter.

Function Generator SLIDE

C

40kHz FILTER O/P

B 1

2

3

4

5

6

7

8

0V

9

10

I/P

A

10k TP1

0V

TP2

0V

Oscilloscope CH.1

CH.2

Fig 15.12



Connect the circuit of Fig 15.12. The 10k Ω slider resistor is being used to provide a convenient monitoring point for the input signal rather than for signal amplitude adjustment. Set it to about scale point 7.



Set the oscilloscope timebase to 5 µs/div (calibrated), the Y1 (CH.1) amplifier to 1V/div and the Y2 (CH.2) amplifier to 0.5V/div. Inject a sinewave signal of large amplitude at about 40kHz.






Adjust the fine frequency control of the function generator to peak the output of the 40kHz Filter to maximum as seen on CH.2 of the oscilloscope, then adjust the amplitude to 2.5V peak-to-peak (5 div.) using either the function generator amplitude control and/or the 10k Ω slider resistor. If you are unable to obtain 2.5Vp-p from your function generator then the investigation can be carried out with any convenient lower value but this may result in some interference with the output signals.

267





Calculate the time for one cycle from the oscilloscope display and record this in Table 15.11. Peak response

Upper cut-off

Lower cut-off

Time period Frequency

µs

µs

µs

kHz

kHz

kHz

Table 15.11 

Without any further adjustment to amplitude, increase the frequency from the function generator until the amplitude of the CH.2 waveform is reduced to 3.5 div. This is a reduction of -3dB (0.707V) from the maximum value and corresponds to the upper cut-off frequency.



Calculate the time for one cycle again from the oscilloscope display and record this in Table 15.11.



Reduce the frequency back through the peak and carry on until the amplitude again falls to 3.5 div. at the lower cut-off frequency. Again record the time for one cycle in Table 15.11.



Take the reciprocal of the three time periods to find the center frequency and the upper and lower cut-off frequencies.

15.10a

Enter your upper cut-off frequency in kHz.

15.10b

Enter your lower cut-off frequency in kHz. 

268




Student Assessment 15 Input A Input B

Inputs

(a)

(b)

Inverting Input

-

Non-inverting Input

+

(c)

(d) Fig 1

The following questions 1 - 5 relate to the waveforms given in Fig 1. 1.

Input A is applied to the non-inverting (+) input and Input B to the inverting (-) input of a differential amplifier which has unity gain. The output waveform is given by:

a 2.

b

waveform (b)

c

waveform (c)

d none of these

Input A is applied to the non-inverting (+) input and Input B to the inverting (-) input of a comparator which has latch but no hysteresis. The output waveform is given by:

a 3.

waveform (a)

waveform (a)

b

waveform (b)

c

waveform (c)

d waveform (d)

Input A is applied to the non-inverting (+) input and Input B to the inverting (-) input of a comparator which has hysteresis. The hysteresis levels are shown by the two horizontal dotted lines on the input waveform diagrams. The output waveform is given by: a waveform (a)

b

waveform (b)

c

waveform (c)

d waveform (d)

Continued ...

269




Input A is applied to the non-inverting (+) input and Input B to the inverting (-) input of a comparator which has both latch and hysteresis. The hysteresis levels are shown by the twoishorizontal waveform given by: dotted lines on the input waveform diagrams. The output

a 5.

7.

b

waveform (b)

c

waveform (c)

d waveform (d)

Input A is applied to the inverting (-) input and Input B to the non-inverting (+) input of a comparator which has no hysteresis. The output waveform is given by:

a 6.

waveform (a)

waveform (a)

b

waveform (b)

c

waveform (c)

d

none of these.

An oscillator is a circuit which gives:

a

a square wave output

b

a sinewave output only

c

an alternating output from a DC input

d

a larger AC output than input

The function of the electronic switch investigated in this chapter is to control:

a b

small signal circuits from a large input voltage power circuits from a small input switching voltage

c

main power switching from a low DC source

d

the power output from a DC motor

(a)

(b)

(c)

(d)

Fig 2

8.

Fig 2(a) represents a:

a

270

high pass filter

b

band stop filter

c

low pass filter

d

band pass filter


Mathematical Operations Chapter 16

Chapter 16 Mathematical Operations




Describe the characteristics of a summing amplifier.



Describe the characteristics of an integrator.



Describe the characteristics of a differentiator.



Describe the characteristics and application of a "sample and hold" circuit.

• • • • • •

DIGIAC 1750 Transducer and Instrumentation Trainer. 4mm Connecting Leads. Digital Multimeter. Oscilloscope. Function Generator. BNC to 4mm connecting lead.

271



16.1 The Summing Amplifier I F RF V1 V2 V3

R1

I1

R2

I2

R3

I3

VO

+ VG

Summing Amplifier

+ Inverter

Fig 16.1

The gain of an operational amplifier is typically one million. To keep within saturation limits the input voltage must therefore be less than one millionth of the output voltage, or a few microvolts. The input voltage is so low that the input is known as the a virtual ground (VG) (Fig 16.1). The input impedance of the operational amplifier is very high, typically measured in MΩ. With an input voltage in µV and an input impedance in M Ω, the input current to the Op Amp is non-existent, or at least negligibly small. From Kirchhoff's Laws, the current(s) into a junction must be the same as the current(s) out of the junction, so, since there is no current flowing into the Op Amp, the feedback current (IF) must be equal to the sum of the three input currents (I1, I2 & I3). VO V V V = 1 + 2 + 3 +.... RF R1 R2 R3 If all of the resistors are made the same size, then they cancel out in the equation leaving: VO = V1 + V2 + V3 + . . . . The output voltage is the sum of the three input voltages. However, since the inverting input has been used it will be of opposite sign or polarity, so an inverter has been added to restore the srcinal polarity. Other input branches may be added. The main characteristics for the device provided are: Input voltage (max.) Voltage gain Output voltage (max.) Table 16.1

272

±12V 1.0 (VA + VB + VC) ±10V



16.2 Practical Exercise Characteristics of a Summing Amplifier

SLIDE

A

C

O/P B

1

5

2

3

4

5

6

7

8

9

CARBON TRACK 6 C

3

8

2

9 1

V

4

B A

10

A

10k

A B

A+B+C

C


B

7

4

10

SUMMING AMPLIFIER

3

8

2

9 1

100k

MOVING COIL METER

C

7

10

10k

0

5 B A

5 +10

-10 + -

-5V

0V

+ 5V

0V

L J

Fig 16.2 

Connect the circuit as shown in Fig 16.2. Set the variable resistors to their central positions.



Switch ON the power supply and adjust the controls of the three resistors to vary the output voltage. Note that variation of any of the input voltages affects the output voltage. You will find that increase of input voltage will increase the output voltage up to a certain maximum (saturation) after which any further increase of input does not increase the output any more.



Determine this maximum (saturation) output voltage. Maximum possible output voltage =



V

Set the Summing Amplifier input voltages to the values indicated in the first row of Table 16.2. Note the expected output voltage and also note and record the actual output voltage obtained in Table 16.2.

273



A

Inputs (volts) B

C

1

+1

+1

+1

2

+2

+1

+3

3

+2

+4

+3

4

-3

+4

+2

5

-3

-2

-2

6

+3

+5

+4

7

+3

-5

+4

8

-3.5

+2.7

-1.4

Output -(A + B + C) Voltage +3V

V V

V

V

V

V

V

V

V

V

V

V

V

V

V

Table 16.2



274

Repeat the procedure for the other settings listed in Table 16.2 to verify that the output voltage is the sum of the input voltages as long as you keep within the saturation limits.

16.2a

Enter your measured output voltage with the inputs of row 2.

16.2b


16.2c


16.2d




16.3 The Integrator i i

R

Vin

CF

VO +

Fig 16.3

An integrator is a circuit having an output voltage that is proportional to the average of the input voltage multiplied by units of time. In mathematical terms this is referred to as the integral of the voltage. Note that, in the feedback path, the resistor has been replaced by a capacitor, since the voltage across a capacitor at any time depends on the amount of current that has been flowing and the time for which it has flowed. Expressed in mathematical terms: Vc =

1 C

i. dt , where the symbol

∫

means the integral of . . .

∫

The feedback current ( i in the above equation) is fixed by the input voltage Vin V and the input resistor R (Fig 16.3), i = in . Substituting this into the equation: R Vo =

1 C

∫

Vin R

. dt

=

1 CR

∫ V . dt in

The output voltage is the integral of the input voltage, multiplied by a factor, 1 . CR With the input voltage constant, the output voltage will increase linearly with time. The time taken for the output voltage to reach the input voltage is referred to as the time constant of the circuit and is equal to CR seconds (from the equation). The maximum possible value of the output voltage is limited by the supply to the saturation voltage of approximately ±11V for the device provided.

275



The main characteristics of the device provided with the DIGIAC 1750 Trainer are: Time constants (switchable) Input impedance Gain error

100ms, 1s & 10s 10kΩ 1%

Table 16.3

Notes:

........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ 276



16.4 Practical Exercise Characteristics of an Integrator

WHEATSTONE BRIDGE D

A

MOVING COIL METER

C

12k OUT

B

3

I/P

IN 1V

V 0V

5

INTEGRATOR O/P

1 VIN dt T

Rx RESET

0

5

-10

+10

+ -

1s 10s 100ms

TIME CONSTANT

L J

0V

+5V

Fig 16.4 

Connect the circuit as shown in Fig 16.4. Set the Integrator time constant switch to 1s.



Switch ON the power supply. Set the input voltage to 1V. Press and hold the RESET button. This sets the output voltage to 0V. Release the RESET button and you will note that the output voltage increases and will reach a maximum value after approximately 12 seconds. Note this maximum value using the 20V digital meter. Maximum output voltage =



16.4a

V

Press the RESET button and release it to allow the output voltage to increase from 0V again. Remove the Integrator input lead when the voltage reaches approximately 5V.

When the input lead is removed from the Integrator the output voltage:

a immediately falls to zero

b

gradually reduces towards zero

c remains where it was

d

increases to supply voltage



Replace the input lead and observe the effect on the output voltage.

277


16.4b


When the input lead is replaced the Integrator output voltage:

a continues to increase from where it was b gradually reduces towards zero c starts again from zero d immediately increases directly to supply voltage The Timer facility of the DIGIAC 1750 Trainer will now be introduced. This allows you to accurately determine the time taken to reach any given voltage. The system will be made entirely automatic by using another facility of signal conditioning circuits, the Comparator.

MOVING COIL METER WHEATSTONE BRIDGE D

5

A B

3

I/P

OUT IN 1V

V 0V

+10

-

1s 10s 100ms

RESET

5

+

1 VIN dt T

Rx

0

-10

INTEGRATOR O/P

12k

C

0V

L J

TIME CONSTANT SLIDE

B

1

2

3

4

5

6

COMPARATOR

C

7

+12V

8

9

10

10k

A

OFF ON HYSTERESIS

B

-

A

+

O/P

COUNTER/ TIMER I/P

TIM E

RESET

C O UNT

F R EE R UN

1s

Fig 16.5

Note that the non-inverting input of the Comparator is taken to a positive reference voltage, the value of which is determined by the setting of the 10k Ω slider resistor. If this is set to 10V then the Comparator will give a high output until the output of the Integrator (which is connected to the inverting input of the Comparator) exceeds 10V, when the Comparator output will go low. While the Comparator output is high the Timer is enabled and will count in hundredths of a second. The moment the output of the Integrator goes above the Comparator reference voltage (in this case 10V) the Comparator output goes low and stops the Timer.

278





Construct the additional circuit of Fig 16.5, noting that the supply voltage to the variable resistors has been changed to +12V.



Reset the input voltage to 1V.



Ignore the Timer function for the moment. Press the Integrator RESET button and, the secondoutput hand voltage of a clock or watch, the timeon after it thatusing the Integrator reaches 10V note as indicated thereleasing Moving Coil Meter. This enables the circuit time constant to be determined. The input voltage is 1V. The output voltage should reach 1V after one time constant and should reach 10V after 10 time constants. The time constant can therefore be determined by dividing the time taken by 10. Record the results in row 1 of Table 16.4.

Switched time constant

Input Voltage (i)

Reference voltage (ii)

1

1s

1V

10V

2

100ms

1V

10V

3

100ms

0.2V

5V

4

10s

5V

2V

Number of time constants (iii) 10

Time taken to reach ref. (iv)

Calculated time constant (v) s

s

s

ms

s

ms

s

s

Table 16.4 

Switch the Timer to TIME and FREE RUN. If necessary press RESET to zero the display.



Move the digital multimeter to terminal B of the 10k Ω slider resistor and adjust the reference voltage to 10V.



Press the Timer RESET to zero the display. Re-adjust the Integrator input voltage to 1V, set the time constant to 100ms and VERY BRIEFLY press its RESET button. You must not hold the RESET button down or the Timer will be counting too soon. Observe the effect on the Timer. This will count up from zero until the output voltage of the Integrator exceeds the reference voltage applied to the Comparator. The display will be in hundredths of a second. For example, a display of 487 represents 4.87 seconds.

279





Repeat the test a few times to become familiar with the action. Zero the Timer each time. Record the result in row 2 of Table 16.4.



Calculate the time constant as follows: The number of time constants is the reference voltage divided by the applied voltage: (iii) = (ii) (i) The measured time constant is the time taken to reach the reference voltage divided by the number of time constants: (v) = (iv)

(iii)

Add the calculated time constant to Table 16.4.

16.4c



With the Integrator time constant still at 100ms, change the input voltage (10kΩ 10-turn resistor) to 0.2V and the reference voltage (10k Ω slider resistor) to 5V and repeat the test and calculation. Remember to zero the Timer each time. Record the results in row 3 of Table 16.4.



Change the Integrator time constant to 10s, the reference voltage (10k Ω slider resistor) to 2V and the input voltage (10kΩ 10-turn resistor) to 5V and repeat the test. Record the result in row 4 of Table 16.4.



Calculate the time constant and add to Table 16.4.




Enter your measured value of the time constant (in seconds) from row 1 of Table 16.4 when switched to 1s.

16.4d

Enter your measured value of the time constant (in ms) from row 3 of Table 16.4 when switched to 100ms.

16.4e

280

Enter your measured value of the time constant (in seconds) from row 4 of Table 16.4 when switched to 10s.



16.5 The Differentiator A simple differentiator is shown in Fig 16.6.

Input

Input

Output

Voltage across capacitor Output across resistor

Fig 16.6

The output voltage is proportional to the rate of change of the input voltage. Examine the waveforms of Fig 16.4. Initially the capacitor is uncharged and there is similarly no voltage across the resistor. When the input voltage suddenly rises to a positive value the capacitor voltage cannot change instantaneously so the full applied voltage appears across the resistor. Current flows and the capacitor charges. As the voltage rises across the capacitor it must fall across the resistor, until the capacitor is fully charged. The time taken for this will depend on the size of the resistor (controlling the charging current) and the size of the capacitor (how much charge is needed to raise the capacitor voltage). One time constant is the time it would take for the capacitor to fully charge to the applied voltage if the initial current could be maintained. Obviously the current must reduce as the voltage across the resistor reduces, so the rate of charge falls away. In theory it never reaches full charge. However, for all practical purposes full charge is reached after 5 time constants. The time constant is calculated from the value of the capacitor in farads multiplied by the value of the resistor in ohms:

time constant t = CR seconds

281



Note that for long time constants such as 1s, using a 1µF capacitor (typically the largest value non-electrolytic capacitor) the value of the resistor would need to be 1MΩ. Non-electrolytic capacitors are needed so that the capacitor can be charged with negative polarity. The high value of resistor raises the problem of a very high output impedance for the circuit. If any load was applied to the differentiator the operation would be seriously affected. To overcome this problem an active differentiator circuit is used on the DIGIAC 1750 Trainer, consisting of an active differentiator Op Amp followed by a unity gain buffer stage. Note that a sudden change of input voltage produces a similar change at the output, the amplitude of this being limited by the saturation voltage of the differentiator active circuits. With the input voltage then held constant, the output voltage falls exponentially, the rate of fall depending on the circuit time constant, the initial rate of fall aiming at a time span equal to the time constant. A steadily on changing input voltage results involtage. a constant output voltage, the value depending the rate-of-change of the input The main characteristics of the device provided with the DIGIAC 1750 Trainer are: Input voltage (max.) Input voltage rate of change (max.)

10-3V/µs

Output saturation voltage (typ.)

±12V

Output noise (time constant 1s)

50mV

Table 16.5

282

±12V



16.6 Practical Exercise Characteristics of a Differentiator INTEGRATOR O/P

WHEATSTONE BRIDGE D I/P

1 VIN dt T

12k

C A

T

dVIN dt

OUT

B

3


IN 1V

V 0V

RESET

1s 10s 100ms

100ms 1s 10ms

TIME CONSTANT

Rx

TIME CONSTANT

MOVING COIL METER

5

+12V

-10

0

5 +10

+ 0V

L J

Fig 16.7 

Connect the circuit as shown in Fig 16.7. Set the time constant controls of the Integrator and Differentiator to 1s. The Moving Coil Meter is used to monitor the change of voltage at the Integrator output.



Switch ON the power supply. Set the input voltage to the integrator to 1V, then transfer the digital multimeter to the output of the Differentiator. Press and then release the RESET button on the Integrator and note the output voltage from the Differentiator. The Integrator output voltage will be changing at 1V/s for approximately 11s and the output from the Differentiator should remain constant during this time. Note the output voltage. Output voltage =

V

16.6a

Enter your value of output voltage during the time the Integrator is charging.

16.6b

When the integrator voltage reaches its maximum value the Differentiator output voltage: a falls sharply to zero

c increases to maximum

b

remains constant at the former value

d

falls slowly to zero

283



Function Generator SLIDE

C B

1

2

3

4

5

6

7

8

9

10


T

dVIN dt

A

10k

100ms 1s 10ms

TIME CONSTANT 0V

TP1

0V

TP2

0V

Oscilloscope CH.1

CH.2

Fig 16.8 

Change to the circuit of Fig 16.8. Set the function generator to a 30Hz square wave output. Set the 10kΩ slider resistor to mid-way. Switch the oscilloscope timebase to 5ms/div chop mode, the Y1 amplifier (CH.1) to 0.5V/div and Y2 amplifier (CH.2) to 2V/div.



Set the Differentiator time constant to 10ms and adjust the signal input (function generator amplitude control and/or 10kΩ slider resistor) to give an input signal (CH.1) of 1Vp-p.



Sketch the two waveforms on the graticule provided with the input at the top:


284





Compare these waveforms with the theoretical waveforms given in the previous section (16.5).



The Differentiator will almost certainly be loading the function generator output to some extent and changing the waveform. Remove the lead to the Differentiator input and observe the effect on the function generator output waveform. This distortion is very common and, as you can see from the output waveform, does not seriously affect the operation of a differentiator.

16.6c

The output waveform from the function generator with the differentiator disconnected is most like: a b

16.6d

c

d

Your Waveform Sketch 16.1 is most like:

a

b

c

d

Notes: ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. .................................................................................................................................................................

285



16.7 A Sample and Hold Circuit

Input Droop

SAMPLE

Output

O/P

+

I/P

Hold Sample

Hold Sample

(a)

Sample

(b)

Fig 16.9

This circuit allows the value of an input signal at any instant of time to be stored on command and held for processing. In the sample mode (SAMPLE button pressed), the instantaneous value of the input signal is tracked at the output. When the SAMPLE button is released the circuit enters the hold mode and the value of the input at that instant is held as a charge on a capacitor, Fig 16.9(a). The capacitor voltage will fall gradually with time as the capacitor discharges through leakage paths and this fall in voltage is referred to as droop. Fig 16.9(b) illustrates the characteristics during sample and hold periods of operation. The circuit is normally used in connection with analog to digital conversion of a varying signal. The signal would be sampled frequently and, during the hold time, the value is digitally encoded. The main characteristics of the device provided with the DIGIAC 1750 Trainer are: Input voltage range (max.) Input time constant Droop rate Table 16.6

286

±12V 1ms 10mV/minute



16.8 Practical Exercise Characteristics of a Sample and Hold Circuit Function Generator WIREWOUND TRACK 5

C

7

I/P

MOVING COIL METER

O/P

8

2

9

5

-10

A

10

0

5

B

3

1

SAMPLE AND HOLD

6

4

10k

+10

+

SAMPLE

0V

L J

0V TP1

0V

TP2

Oscilloscope CH.1

0V

CH.2

Fig 16.10 

Connect the circuit as shown in Fig 16.10. Set the function generator output to 40Hz sinewave with high amplitude. Switch the oscilloscope timebase to 5ms/div, Y1 amplifier (CH.1) to 10V/div, chop mode (near the top of the screen) and Y2 amplifier (CH.2) to 2V/div, DC input (near the middle).



Switch ON the power supply and adjust the amplitude of the signal (function generator amplitude control and/or 10kΩ wirewound resistor) to give an input of 20Vp-p. If your function generator does not give 20Vp-p then use the AC Amplifier (GAIN = 10) to boost the signal input. Move CH.1 of the oscilloscope to the output of the AC Amplifier.



Press and release the SAMPLE button to catch a sample of the input voltage to the circuit. Note that while the SAMPLE button is pressed the input signal appears at the output (CH.2 of the oscilloscope). When released a random sample is captured and appears as a DC voltage at the output as indicated by the meter. Try several times and record the results in Table 16.7. 1

Output voltage

2 V

3 V

4 V

5 V

6 V

7 V

8 V

9 V

10 V

V

Table 16.7

16.8a

Is the output truly random?

Yes

or

No

287




A summing amplifier supplied from a 12V power supply has positive unity gain and three inputs. The value of the output voltage for input voltages +2v, +4V, +1V will be:

a 2.

+15V

b

+2V

b

+3.4V

b

9V

b

12V

b

+9.8V

c

+12V

d -9.8V

-2V

c

+12V

d -12V

-5.6V

c

-12V

d -7V

6V

c

3V

d 1V

6V

c

3V

d 1V

An integrator supplied from a 10V power supply has a time constant of 1s. A constant voltage of +2.5V is applied to the input. With the output initially zero, the output voltage after 5s will be: a 12.5V b

288

d -7V

An integrator supplied from a 10V power supply has a time constant of 1s. A constant voltage of +3V is applied to the input. With the output initially zero, the output voltage after 2s will be:

a 7.

+8V

An integrator supplied from a 10V power supply has a time constant of 1s. A constant voltage of +3V is applied to the input. With the output initially zero, the output voltage after 1s will be:

a 6.

c

A summing amplifier supplied from a 12V power supply has positive unity gain and three inputs. The value of the output voltage for input voltages -6.3v, +4.5V, -5.2V will be:

a 5.

+7V

A summing amplifier supplied from a 12V power supply has positive unity gain and three inputs. The value of the output voltage for input voltages -5v, +4V, +3V will be:

a 4.

b

A summing amplifier supplied from a 12V power supply has positive unity gain and three inputs. The value of the output voltage for input voltages +6v, +4V, +5V will be:

a 3.

+3.5V

10V

c

7.5V

d 5V




An integrator supplied from a 10V power supply has a time constant of 1s. A constant voltage of +3V is applied to the input. With the output initially +3V, the output voltage after a further 2s will be:

a 9.

12V

b

10V

c

9V

d 6V

A differentiator has a time constant of 2s. If a voltage is applied to the input which is increasing steadily at a rate of 4V/s, the output voltage will be:

a

steadily decreasing at a rate of 4V/s

b

constant at 4V

c

constant at 8V

d

steadily increasing at a rate of 4V/s

10. A square wave is applied to the input of a differentiator which has a time constant which is short compared to the time period of the applied square wave. The waveform at the output will be:

a

b

c

d

Continued ...

289



Student Assessment 16 Continued ... 6 Input Voltage (volts)

4

4Vp-p AC

2 +4V DC

30

60

90 time (µ s)

120

Fig 1

11. The waveform of Fig 1 is applied to the input of a "sample-and-hold" circuit similar to that on the DIGIAC 1750 Trainer. If the SAMPLE button is pressed and held down the output will be:

a

the DC component of the input only

b

the same as the input

c

the AC component of the input only

d

zero

12. The waveform of Fig 1 is applied to the input of a unity gain "sample-and-hold" circuit. If the hold function is engaged 70 s after the start of the cycle the output voltage will be:

a

290

2V

b

3V

c

4V

d 5V


Control Systems Characteristics Chapter 17

Chapter 17 Control Systems Characteristics


Having studied this Chapter you will be able to: 

Describe the characteristics of an ON/OFF system.



Describe the characteristics of a Proportional system.



Describe the characteristics of an Integral system.



Describe the characteristics of a Derivative system.



Explain that a practical system may incorporate Proportional, and (or Derivative components and be referred to Integral as a 3-term PID) controller.

291



17.1 A Basic ON/OFF Closed Loop System A controlled variable is any physical system which we may wish to control, such as a heated environment (hot water tank), lighting level (PIR controlled lighting), mechanical systems (speed, position or direction, linear or rotational), and many more. For instance, the modern airplane is full of electrical control systems. An error is any difference between a desired result and an actual result. In an electrical control system the output is converted into an electrical quantity by a transducer. Fig 17.1 shows a simple closed loop control system, the error detector detecting the difference between the actual and the desired value of the controlled variable.

Error Detector Reference Input

Error

+

Controlled Variable

Controller

-

Output

Feedback

Fig 17.1

The output of the controlled variable (the transducer) is compared with a reference input (command input) and an error signal is fed to the controller which initiates an actuating signal to alter the state of the controlled variable and reduce the error, ideally to zero. In an ON/OFF system the controller will have only two states: 1.

With the value of the controlled variable less than that desired, the controller output is maximum.

2.

292

With the value of the controlled variable greater than that desired, the controller output is zero.



This method of control is suitable for systems having inertia (a long time constant) such as the temperature control of a room, using a heater. The method might give characteristics as illustrated in Fig 17.2. Initially, the heater is ON and the temperature rises exponentially from its ambient state. When the desired temperature is reached, the heater is switched OFF.

Maximum

Actual

Temperature Reference

ON

OFF

ON

OFF

ON Time

Fig 17.2

The temperature will continue to rise or overshoot for a time due to the residual heat in the heater, but will eventually fall, the rate of the fall increasing with time. When the temperature has fallen below the desired value, the heater will again be switched ON but the temperature will continue to fall for a time before the heater has any effect. The resulting characteristic will be as shown in Fig 17.2, with the temperature varying continuously between two limits, provided that there is no change in the operating conditions, such as heat loss variations or a change in the thermostat setting (command input).

293



17.2 Proportional Control With this system of control, the output from the controller is proportional to the magnitude of the error signal (not just ON or OFF). Controller output = Kp x Error

where Kp is the proportional gain of the controller The characteristics of the system depend on the value of Kp. For large values of gain in the feedback loop the characteristics are similar to those for ON/OFF control. For small values of gain the system will be sluggish and very slow to respond.

Output

High gain - underdamped Critical damping

Reference

Low gain -overdamped

Time Fig 17.3

Fig 17.3 shows the characteristics of proportional control in response to a step input (or sudden change) and illustrates that a high gain results in a rapid response but produces an overshoot of the desired reference setting, together with oscillations about the reference setting. Medium gain results in a slower response with minimum overshoot and oscillations.

Low gaintheresults in asetting. slow response with no oscillations but possibly never reaching reference

294



The term damping is used to cover the inertia or friction of a feedback system. Characteristics such as those for high gain in Fig 17.3 are referred to as underdamped and for low gain, overdamped. A response which rises most rapidly to the reference with no overshoot is referred to as critically damped. The degree of damping is normally referred to in terms of the damping ratio, which is given the Greek symbol (Zeta). Critical damping has damping ratio of 1.0. For underdamping the damping ratio is less than 1.0 and for overdamping, greater than 1.0.

Position

Input

Underdamped Output

Velocity lag

time

Fig 17.4

Fig 17.4 shows the response of a proportional control system to an input varying with time (ramp input). The output tends to follow the input but, due to inertia within the system, the error between the input and output quantities has to increase to a threshold before there is sufficient actuating signal to produce a variation of the output. The output will thereafter follow the input but will lag behind the input, this being referred to as velocity lag. The magnitude of the lag will depend on the gain of the system, the friction and the output loading. There may be oscillations in the output characteristic as shown dotted, depending on the system gain.

295



These characteristics mean that pure proportional control is unsuitable for applications where the input may vary with time. In addition the system has some disadvantages with constant input (command) conditions. Consider the system operating with a set input and with the output at the reference setting so that there is no error. Under these conditions there will be no controller output. A load imposed on the output will produce a change of output state. An error signal will be produced to counteract this and reduce the error, but the output will not now be at the desired reference state. The error introduced will vary with the loading imposed on the output. Proportional control on its own is therefore unsuitable for control applications. In practice, due to saturation effects within the system, the controller output will be proportional to the error only over a part of the full range.

Output +

Proportional band -

+

Error

Fig 17.5

This is illustrated in Fig 17.5. The range over which the output is proportional to the error is referred to as the proportional band.

296



17.3 Integral Control Integral control can be used to eliminate any error present between the reference and actual output setting. An integrator produces an output that is proportional to input time and hence, if the error signal is fed via an integrator circuit, its output will increase with time. With this output fed to the system controller, an actuating signal will be produced to reduce the error, the time taken depending on the integrator time constant.

Error Detector Reference Input

Error

+

Integrator

Controller

-

Controlled Variable

Output

Feedback Reference Input Output time Error

Integrator output

Fig 17.6

Fig 17.6 illustrates the operation of integral control for ramp input conditions. While there is an error, the integrator output increases. This output, fed to the controller, produces an actuating signal to correct the error. When the error has been reduced to zero, the integrator output remains constant, thus compensating for the velocity error that would have been present without the integral control. Any further error, however caused, will be automatically compensated, provided the output required is within the capacity of the integrator circuit. Normally, the integral control would be combined with proportional control, the proportional control being the main control and leaving the integral control for final adjustments of the output setting.

297



17.4 Derivative (or Differential) Control Friction losses in a system produce damping and thus allow operation under proportional control with a higher system gain, but the introduction of friction represents a power wastage and increases the time taken to reach stable conditions following any disturbance. The same effect can be produced using an adder fed with derivative control, by feeding back a signal that is proportional to the rate-of-change of the output or the rate-of-change of the error signal. This is illustrated in Fig 17.7. Error Detector Reference Input

Error

+ -

Adder +

Controlled Variable

Controller -

Output

Differentiator

Feedback

Input (i)

Output(ii)

time Error (iii) Rate-of-change of output (differential) (iv) Actuating signal (v)

Fig 17.7

Error (iii) = Input (i) - Output (ii)

Rate-of-change of output (iv) = slope of Output (ii) Actuating signal (v) = Error (iii) - Rate-of-change of Output (iv)

17.5 PID Controller A practical system incorporating some elements of Proportional, Integral and Derivative components may be referred to as a 3-term, or PID, controller. 298




2.

3.

4.

5.

An ON/OFF control system is one in which the:

a

error detector is switched ON or OFF

b

controlled variable is switched ON or OFF

c

controlled variable is continuously varied

d

controlled variable is switched ON when the control system is OFF

A suitable application for a simple ON/OFF control system would be for:

a

temperature environment control

b

motor speed control

c

rotational position control

d

light level control

The term proportional control means that the controller output is proportional to the:

a

error signal

b

supply voltage

c

reference frequency

d

rate of change of the output

The term integral control means relating the output of the controller to the:

a

input amplitude only

b

input

c

rate-of-change of input

d

frequency of the input

time

The term derivative control means feeding back to the error detector a signal proportional to the output:

a 6.

x

amplitude

b

polarity

c

rate-of-change

d frequency

When a load is applied to a system with proportional control, the output may have:

a

greater amplitude

b

a continual error

c

less range of response

d

a slower response

Continued ...

299




0V

7.

Fig 1

0V

0V

Fig 4

0V

0V

0V

The waveform of Fig 2 is applied to the input of an integral controller. The output waveform will be:

a

Sat. 0V

9.

Fig 3

The waveform of Fig 1 is applied to the input of an integral controller. The output waveform will be: a Sat. c Sat. d Sat. b Sat. 0V

8.

0V

Fig 2

b

Sat. 0V

c

Sat. 0V

d

Sat. 0V

The waveform of Fig 3 is applied to the input of a derivative controller. The output waveform will be: Sat. Sat. Sat. Sat. a c d b 0V

0V

0V

0V

10. The waveform of Fig 4 is applied to the input of a derivative controller. The output waveform will be: Sat. Sat. Sat. Sat. a c d b 0V

0V

0V

11. A 3-term PID control system is one which uses:

300

a

post-integral differentiation in the feedback loop

b

partially integral design in three blocks

c

combined proportional, integral and derivative systems

d

a proportion of input derived feedback

0V


Practical Control Systems Chapter 18

Chapter 18 Practical Control Systems



Describe the characteristics of an ON-OFF temperature control system.



Describe the characteristics of a light controlled ON-OFF system.



Describe the characteristics of a positional control system having: proportional, proportional + integral, proportional + derivative and proportional + integral + derivative control.




• • • •

Describe the characteristics of a speed control system.

DIGIAC 1750 Transducer and Instrumentation Trainer. 4mm Connecting Leads. Digital Multimeter. Calculator (not supplied).

301



18.1 Practical Exercise Characteristics of an ON/OFF Temperature Control Syst em


IC TEMP SENSORS

P . I. N . P H O T O D I O D E

EXT.

-

P HO T O V O LT A I C C EL L

O/P

O/P

O/P +

O/P

+

INT.

REF +

+

O/ P

O/ P

+ PHOT OC ONDU CT IV E C ELL

B

O/P

PHOT OT RANS ISTOR

LAMP FILAMENT

I/P

O/P A N T C T HE R M I ST O R S


P LA T I N U M R . T . D .

HEATER ELEMENT

WHEATSTONE BRIDGE D

2

9

IN

V

1V

0V

10

10k

-

B

Rx

B

-

A

+

ELECTRONIC SWITCH

+12V

O/P

O/P

I/P

COUNTER/ TIMER

INVERTER

O/P

O/P I/P

+100VIN

O/P I/P

+

x100 AMPLIFIER

I/P

V/F CONVERTER

A-B

A

dVIN

I/P

TIM E

FREE RUN

COUNT

1s

-VIN

100ms 1s 10ms

TIME CONSTANT

A


OUT

B

T dt

OFF ON HYSTERESIS

B

12k

A

I/P

8

1

C

DIFFERENTIATOR O/P

3

0V

COMPARATOR

7

4

I/P

3

+5V

TEMPERATURE INDICATOR

RESET

Fig 18.1

The shaded area within the broken line is a digital thermometer indicating temperature in increments of 0.1°C. The internal Temperature Sensor is an integrated circuit which gives an output of 10mV/°K, so the output at an average room temperature of 20 °C will be 2.93V. The 10-turn potentiometer on the Wheatstone Bridge panel is adjusted to give 2.73V to the inverting input of the Differential Amplifier. The output from the Differential Amplifier will therefore be 0.01V/ °C, or 0.2V at 20°C. The V/F Converter gives an output of 1kHz/V, so an input of 0.2V will give an output of 200Hz (200 pulses in one second).

302



The Differentiator, x100 Amplifier and Inverter shape the pulses to be compatible with the Counter/Timer input, which will therefore display 200 for a temperature of 20°C, or the temperature in tenths of a degree. A display of 213 = 21.3 °C. 

Connect the circuit as shown in Fig 18.1. Switch the comparator HYSTERESIS OFF and set the 10kΩ resistor control fully counter clockwise. Set the Differentiator to 1s and the Counter/Timer controls to COUNT and 1s.



Remove the output lead from the Electronic Switch while you carry out the initial setting up.



Switch ON the power supply and adjust the 10k Ω 10-turn potentiometer for a voltage of 2.73V on the inverting input of the Differential Amplifier. This will set up the digital thermometer to display the ambient temperature in °C. Press the RESET button on the Counter/Timer each time you need to obtain a temperature reading.



Transfer the voltmeter to the output of the IC Temperature Sensor and note the output voltage (You may need to remove one of the leads while you do this). I.C. Temperature Sensor output voltage =



Transfer the voltmeter again to the output of the 10k Ω resistor and set the output voltage to a value 0.2V above the output value obtained from the IC Temperature Sensor. This sets the reference temperature of the system to 20°C above the ambient temperature. Reference voltage setting =



V

V

Restore the output lead to the Electronic Switch to start the heating process. Note the temperature-time characteristic of the system by noting the displayed temperature and the heater state (whether ON or OFF) at time intervals of 30s (0.5 minute).

Note: The heater state will be indicated by the lamp, lamp ON = heater ON and lamp OFF = heater OFF. Enter the details in Table 18.1.

303


Time (minutes) Heater State ON/OFF Temperature

0.5

1.0

1.5

2.0

2.5


3.0

3.5

4.0

4.5

5.0

5.5

6.0

6.5

7.0

7.5

8.0

8.5

°C

Table 18.1 

Plot the temperature-time characteristic on the axes provided:

Temperature o C 60

55 50 45 40 35 30 25 20 15 10 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0 8.5 9.0 time (minutes)

Graph 18.1

18.1a



Shade in blocks at the bottom of your graph to represent when the Heater was switched ON.



Mark in a line on your graph to represent the reference temperature setting.

Enter your value of the temperature variation (maximum - minimum) around the reference temperature (in C).

18.1b

From your recorded Reference voltage setting on the previous page enter your reference temperature setting (aiming temperature) in C.

304





If time permits add an alarm circuit to the system. The alarm is to operate if the temperature exceeds 30 °C above the ambient temperature.



Select suitable components from the devices available with the DIGIAC 1750 unit, connect, and check the operation of the system by simulating a fault. Do this by disconnecting the feedback from the Temperature Sensor to input B of the Comparator. Finally, switch OFF the power supply.

Notes: ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. ................................................................................................................................................................. .................................................................................................................................................................

305



18.2 Practical Exercise Characteristics of a Light Controlled ON/OFF System A system is to operate a solenoid. The solenoid is to be ON with the light level low and is to be automatically turned OFF when the light level exceeds a preset level.

P . I . N . P H O T O D IO D E

P HOT OV O L T A I C C E L L

O/P

ELECTRONIC SWITCH

O/P +

+

V

O /P

O/ P

+12V

COMPARATOR OFF ON HYSTERESIS

B

-

A

+

O/P

I/P

O/P

SOLENOID P HOTOCONDU CTIV E CELL

P HOTOTR AN SISTOR

LAMP FILAMENT

POWER AMPLIFIER

I/P O/P SLIDE

C

I/P I/P

B 1

2

3

4

5

6

7

8

9

10

A

10k

MOVING COIL METER

0 CARBON TRACK 6 5 C


7

4

8

2

9

B A

10

1

0V

+5V

3

8

2

9 1

100k

5

5

+10

7

4

3

-10

10

10k

B A

+ 0V

L J

+12V

Fig 18.2 





306

Connect the circuit shown in Fig 18.2. Switch the Comparator HYSTERESIS OFF and set the resistor controls as follows:fully counter clockwise for the carbon track, fully to the left for the slide, fully clockwise for the wirewound track. Switch ON the power supply (the Solenoid should energize).





Move the slide resistor to the right so that the Solenoid is just de-energized. This represents the preset conditions for operating the system with the lighting at the ambient level.



Move your hand over the Photoconductive Cell. You will note that the Solenoid will change its state as the lighting level falls due to your shadow (the Solenoid energizes, indicating that the electronic switch is closed).

With no hysteresis in the Comparator circuit, only a small drop in lighting level is required to produce the change. Introduction of some hysteresis would increase the lighting change required, but the hysteresis provided with the Comparator is too great for this application and would operate as a latch. 

Cover the opto-sensor clear plastic enclosure with an opaque box to exclude all ambient light. The Solenoid should immediately energize as the light level is reduced.



With the voltage applied to the lamp filament at 0V (control of the lamp voltage is via the 100k Ω carbon track resistor) as indicated on the Moving Coil Meter, move the slide resistor further to the right until the Solenoid changes state.

Lamp Filament Voltage Slide Resistor Setting

0

1

2

3

4

5

6

7

8

9

10

Table 18.2 

Adjust the lamp filament voltage to each of the settings given in Table 18.2, and repeat the procedure noting the slide resistor setting required for a change of state of the Solenoid. Record the results in Table 18.2.

307





Plot the graph of slide resistor setting against lamp voltage on the axes provided. 10 Slide Resistor Setting

9 8 7 6 5 4 3 2 1 0

1

2

3

4

5

6

7

8

9

10

Lamp Filament Voltage (volts)

Graph 18.2

This exercise has illustrated the use of an ON-OFF lighting control system. The slide resistor can be set to any value, within the range noted, to produce circuit switching at any desired value of lighting level.

18.2a

From your graph deduce and enter the slide resistor setting corresponding to a lamp filament voltage of 5V.



308




18.3 Practical Exercise Characteristics of a Positional Control System - 1 Proportional Control SERVO POTENTIOMETER

DC MOTOR

I/P POWER AMPLIFIER O/P

DIFFERENTIATOR O/P

I/P

I/P

T O/P



7 8

3

B

9

2 1

A

10

10k

B


A-B

O/P

A B

INVERTER

+

A

SUMMING AMPLIFIER

1s 10s 100ms

TIME CONSTANT

A-B

+

A

TIME CONSTANT 0V

1 V dt T IN

RESET

100ms 1s 10ms

V

INTEGRATOR O/P I/P

4

dVIN dt

0/P

A+B+C

C O/P

I/P -5V

-VIN

+5V

AMPLIFIER #1

AMPLIFIER #2

I/P

-

O /P .5

+

.4 1 100 10

OFFSET

-

.7

O /P

.1

.4 1 100 10

.9

.2

.5

+

.8

.3

GAIN COARSE

.6

MOVING COIL METER

I/P

1.0

GAIN FINE

OFFSET

GAIN COARSE

.9

.2 .1

0

5 +10

.7 .8

.3

5 -10

.6

+ -

1.0

GAIN FINE

0V

L J

Fig 18.3

Study the diagram. The proportional control section runs across the middle of the diagram. The 10kΩ wirewound resistor is the command input. The function of the Differential Amplifier is to inject a step input voltage later in the investigation. The step voltage is generated by Amplifier #2 offset voltage, which is the only purpose for including this amplifier. You will see that it does not need an input for this purpose. Integral control will be added later by connecting the Integrator in between the Error Detector (the Instrumentation Amplifier) and the Summing Amplifier. Derivative control will also be added later via the Summing Amplifier. The Inverter in between the Differentiator and the Summing Amplifier is to provide negative feedback.

309



The Summing Amplifier combines all of the control systems as required. 



Connect the circuit as shown in Fig 18.3. This circuit is arranged for proportional control only. Press the left hand side of the mounting plate of the Servo Potentiometer and then release it to engage with the drive shaft.





Set Amplifier #1 GAIN COARSE control to 10 and GAIN FINE to 0.1 to give an overall gain of 1.0. Remove the power connection to the Motor. Switch ON the power supply.



Set Amplifier #2 GAIN COARSE control to 100 and GAIN FINE to 1.0 and adjust the OFFSET control for an output of +3V. Return the GAIN COARSE control to 1. The output voltage should fall to near zero volts. Note that since this +3V step is fed into the system via the inverting input of the Differential Amplifier the actual step injected will be -3V.



Transfer the Moving Coil Meter to terminal B of the 10k Ω wirewound resistor. Adjust the setting of the 10k Ω resistor control to its central position to give 0V output.



Zero the setting of the Servo Potentiometer dial against the pointer.



Transfer the Moving Coil Meter to the output of the Power Amplifier and adjust Amplifier #1 OFFSET to give 0V. Restore the Motor power connection.



Rotate the 10k Ω wirewound resistor control slowly over its full travel. The Motor drive shaft and the Servo Potentiometer dial should rotate and follow the movement of the command input, although the system may be sluggish and there will be a lag before the Servo Potentiometer starts to follow the input setting. This is because the system gain is low, Amplifier #1 gain being set to 1.0. Amplifier #1 Gain = 1.0 Maximum Dial Reading (degrees)

Table 18.3

310

Positive

Negative


18.3a


Enter your maximum Servo Potentiometer dial reading with positive input voltage in degrees when Amplifier #1 gain is 1.0. 

Return the 10k Ω resistor to its central position. Set Amplifier #1 GAIN FINE to 0.5 (overall gain 5) and repeat the procedure. With this higher setting of the gain control the Servo Potentiometer should follow the input closely for no load on the drive shaft and it should be possible to obtain the full travel of the wirewound track resistor in both directions. Rotate the input control slowly when nearing the end of the travel or the Servo Potentiometer contact may overshoot and pass the end of the track, causing the drive shaft to rotate continuously. If this occurs, return the 10k resistor quickly to its central position.



Control Setting

Note the full range of travel of the Servo Potentiometer against the setting of the 10kΩ wirewound resistor command input. Record the results in Table 18.4. 1

2

3

4

5

0V

Servo-Potentiometer Dial Reading (deg.)

6

7

8

9

10

0/ 360

Table 18.4 180 Servo Potentiometer Dial Reading 150 (degrees) 120 90 60 30 0/360 330 300 270 240 210 180 1

2

3

4

5

6

7

8 91 0 Control Setting

Graph 18.3

311


18.3b




Plot the graph of Dial Reading against Control Setting on the axes provided on the previous page (Graph 18.3).



Repeat the readings in the reverse direction and compare the dial readings obtained with the previous readings recorded in Table 18.4.

Read from your graph and enter the approximate Control Setting for a Servo Potentiometer dial reading of 90 . 

Set Amplifier #1 GAIN FINE to 1.0 and use the input command control to return the Servo Potentiometer dial reading to 0°.



Move the Servo Potentiometer dial by rotating the Hall effect disc by hand and note the total range (for example +20° to -10° = 30°, it may not be symmetrical) over which the dial can be moved without the system responding and moving the dial back. This value represents a deadband over which the system does not respond. Record the result in Table 18.5. Amplifier #1 Gain

10 x 1.0 = 10

10 x 0.5 = 5

10 x 0.1 = 1

Deadband (deg.) Table 18.5 

18.3c

Repeat the procedure for Amplifier #1 GAIN FINE settings of 0.5 and 0.1, adding the results to Table 18.5.

The effect of system gain on the deadband is:

a high gain gives a small deadband b high gain gives a large deadband c gain makes no difference to deadband d there is no change in deadband below a gain of 5

18.3d

312

Enter your value of deadband in degrees when the gain of Amplifier #1 is 5.







Set Amplifier #1 GAIN FINE to 0.1. Switch the GAIN COARSE control of Amplifier #2 from 1 to 100 and note the effect on the output shaft position. Return the control to 1 and again note the effect. Repeat the procedure several times. Take care not to touch the OFFSET control when you are doing this as the setting is very critical.



18.3e

Repeat the procedure with the Amplifier #1 GAIN FINE set 0.5 and then 1.0.

The effect of gain on the response to a step input was that the speed of response was:

18.3f

18.3g

a greatest with high gain

b

slowest with high gain

c the same for any value of gain

d

no response with gain less than 2.5

The effect of gain on the response to a step input was that overshoot was:

a greatest with high gain

b

least with high gain

c the same for any value of gain

d

no response with gain less than 2.5

A proportional control system which has low gain in the feedback loop will have:

a large deadband and underdamped response to a step input b small deadband and overdamped response to a step input c large deadband and overdamped response to a step input d small deadband and underdamped response to a step inp ut



Switch OFF the power supply, but:

Keep the circuit connected if possible for the following Exercises.

313



18.4 Practical Exercise Characteristics of a Positional Control System - 2 Proportional + Integral Control SERVO POTENTIOMETER

DC MOTOR


DIFFERENTIATOR O/P

I/P

I/P

T O/P



7 8

3

B

9

2 1

A

10

10k

B


A-B

O/P

A B

INVERTER

+

A

SUMMING AMPLIFIER

1s 10s 100ms

TIME CONSTANT

A-B

+

A

TIME CONSTANT 0V

1 VIN dt T

RESET

100ms 1s 10ms

V

INTEGRATOR O/P I/P

4

dVIN dt

0/P

A+B+C

C O/P

I/P -5V

-VIN

+5V

AMPLIFIER #1

AMPLIFIER #2

I/P

O /P .5

-

+

.4 1 100 10

OFFSET

.6

.5 .7

-

+

.1

.4 1 100 10

.9

.2

GAIN COARSE

O/P

.8

.3

MOVING COIL METER

I /P

1.0

GAIN FINE

OFFSET

GAIN COARSE

.9

.2 .1

0

5 +10

.7 .8

.3

5 -10

.6

+ -

1.0

GAIN FINE

0V

L J

Fig 18.4 

314

If necessary, re-connect the circuit as shown in Fig 18.4 (without the Integrator output connected initially). Re-check the settings as follows:



Remove the power connection to the Motor. Zero the setting of the Servo Potentiometer dial against the pointer. Ensure that the potentiometer is engaged with the drive shaft.



Set Amplifier #1 GAIN COARSE control to 10 and GAIN FINE to 0.1 to give an overall gain of 1.0.






Connect the Moving Coil Meter temporarily to terminal B of the 10k resistor and check the setting to its central position to give 0V output.

Ω




Transfer the Moving Coil Meter back to the output of the Power Amplifier and check the adjustment of Amplifier #1 OFFSET to give 0V.



Transfer the Moving Coil Meter to the output of Amplifier #2, set the GAIN COARSE control to 100 and GAIN FINE to 1.0 and check the adjustment of the OFFSET control for an output of +3V. Return the GAIN COARSE control to 1.



18.4a


This control will again be used to introduce a step input. Restore the power connection to the Motor. With the Integrator time constant set to 1s, press and hold the RESET button, connect the Integrator output lead to the Summing Amplifier input as shown by the arrow in Fig 18.4 and then release the RESET button.



In the event of continuous rotation of the Motor shaft in the following tests, immediately return the Amplifier #2 GAIN COARSE switch to 1 and then hold the Integrator RESET button until the shaft becomes stationary.



Note the effect on the output Servo Potentiometer dial reading when a step input is applied by switching Amplifier #2 GAIN COARSE to 100 and then back to 1. Watch the long term effect on the Integrator output voltage (on the digital voltmeter) and on the dial setting.

The Integrator output voltage:

a goes negative and then quickly returns to zero volts b goes positive and then quickly returns to zero volts c goes quickly to a positive voltage and then remains constant d is still varying after two minutes 18.4b

The Servo Potentiometer Dial Setting:

a moves slowly and smoothly towards a target b moves rapidly towards a target, overshoots and stays there c moves rapidly towards a target, overshoots and then returns d moves slowly towards a target in jerking steps 

Repeat the procedure with the Amplifier #1 GAIN FINE set 0.5 (overall gain of 5) and then 1.0 (overall gain of 10) and respond to the following questions:

315


18.4c


With Amplifier #1 gain set to 5 compared to the gain at 1, the effect on speed of response was:

a faster 18.4d

18.4e

b

slower

c

the same

d

no rotation

With Amplifier #1 gain set to 5 compared to the gain at 1, the effect was:

a more overshoot

b

less overshoot

c no overshoot

d

continuous oscillation

With Amplifier #1 gain set to 10 compared to the gain at 5, the effect on speed of response was:

a faster 18.4f

b

slower

c

the same

d

no rotation

With Amplifier #1 gain set to 10 compared to the gain at 5, the effect was:

a more overshoot

b

less overshoot

c no overshoot

d

continuous oscillation



With Amplifier #1 GAIN COARSE set to 10 and GAIN FINE to 1.0, repeat the procedure with the time constant set to 10s and then 100ms and note the effect.



18.4g

Count the number of overshoots for each of the time constant settings (if possible).

The time constant which gave the smallest number of overshoots was:

a 1s 

18.4h

10s

c

100ms

d

all the same

With the time constant switched to 100ms and Amplifier #1 GAIN FINE set to 0.5, note the effect of displacing the output from its stable position manually by moving the Hall effect disc about 10° on the dial and then releasing it.

A very short time constant may result in:

a no overshoot on return

b

some overshoot on return

c build up to continuous running

d

slow response



316

b

Switch OFF the power supply but keep the circuit connected if you can.



18.5 Practical Exercise Characteristics of a Positional Control System - 3 Proportional + Derivative Control SERVO POTENTIOMETER

DC MOTOR


DIFFERENTIATOR O/P

I/P

I/P

T O/P



7 8

3

B

9

2 1

B

-

A

+

10k

SUMMING AMPLIFIER

1s 10s 100ms

RESET

-

A

+

O/P A

TIME CONSTANT INVERTER

A-B

TIME CONSTANT 0V

1 VIN dt T

B

B A-B

A

10


100ms 1s 10ms

V

INTEGRATOR O/P I/P

4

dVIN dt

0/P

A+B+C

C O/P

I/P -VIN

-5V

+5V

AMPLIFIER #1

AMPLIFIER #2

I/ P

-

O/ P .5

+

.4

OFFSET

.7

.1

.5

+

.4 1 100 10

.9

.2

GAIN COARSE

-

O/ P

.8

.3

1 100 10

.6

MOVING COIL METER

I/P

1.0

GAIN FINE

OFFSET

GAIN COARSE

.9

.2 .1

0

5 +10

.7 .8

.3

5 -10

.6

+ -

1.0

GAIN FINE

0V

L J

Fig 18.5 

If you still have the circuit connected then remove the lead from the Integrator output to the Summing Amplifier and connect the output from the Inverter to the Summing Amplifier as shown in Fig 18.5. Otherwise connect the circuit as shown. Re-check the settings as follows:



Remove the power the Motor. Zerothat the the setting of the Servo Potentiometer dial connection against thetopointer. Ensure potentiometer is engaged with the drive shaft.

317




18.5a








Transfer the Moving Coil Meter temporarily to terminal B of the 10k resistor and check the setting to its central position to give 0V output.



Transfer the Moving Coil Meter temporarily to the output of the Power Amplifier and check the adjustment of Amplifier #1 OFFSET to give 0V.



Transfer the Moving Coil Meter back to the output of Amplifier #2, set the GAIN COARSE control to 100 and GAIN FINE to 1.0 and check the adjustment of the OFFSET control for an output of +3V. Return the GAIN COARSE control to 1.



Restore the power connection to the Motor.

Ω



Set the Differentiator time constant to 1s and note the output Servo Potentiometer response to a step input of +3V applied by changing Amplifier #2 gain control from 1 to 100 and then back to 1.



Repeat the procedure and note the response for Differentiator time constant settings of 100ms and 10ms.

After the step input has been applied in both directions the Servo Potentiometer dial reading:

a returns accurately to zero setting b may not return to zero setting c returns nearest to zero setting with a time constant of 1s d overshoots when the time constant is long 

18.5b

With the Differentiator time constant set to 10ms, note the effect of manually moving the output from its stable position by about quarter of a turn with the Hall effect disc.

Is there any tendency to re-adjust and correct the error?

Yes

318

or

No


18.5c


Derivative control affects the response:

a when the output is in the steady state b during change of output only c when there is no error signal d only when the command input is steady

18.5d

Increasing the time constant of the differentiator:

a causes the output to change more quickly b increases the damping effect c corrects any steady state error d results in faster response 

Switch OFF the power supply, but Keep the circuit connected if possible for the next exercise.

319



18.6 Practical Exercise Characteristics of a Positional Control System - 4 Proportional + Integral + Derivative Control

SERVO POTENTIOMETER

DC MOTOR


DIFFERENTIATOR O/P

I/P

I/P

T O/P

dVIN dt

0/P 100ms 1s 10ms

V

INTEGRATOR O/P I/P



7

4

8

3

B

9

2 1

B

-

A

+

10k

0V

1 VIN dt T

SUMMING AMPLIFIER

1s 10s 100ms

RESET

O/P A

TIME CONSTANT

B

B

-

A

+

A-B

A

10


INVERTER

A-B

TIME CONSTANT

A+B+C

C O/P

I/P -VIN

-5V

+5V

AMPLIFIER #1

AMPLIFIER #2

I/ P -

O/P .5

+

.4

OFFSET

.1

.5

+

.4 1 100 10

.9

.2

GAIN COARSE

-

.7

O/ P

.8

.3

1 100 10

.6

MOVING COIL METER

I/ P

1.0

GAIN FINE

OFFSET

GAIN COARSE

.9

.2 .1

0

5 +10

.7 .8

.3

5 -10

.6

+ -

1.0

GAIN FINE

0V

L J

Fig 18.6 

Re-construct the circuit of Fig 18.6 if necessary, making sure that the output of the Inverter is connected to the input of the Summing Amplifier but do not connect the Integrator to the Summing Amplifier at this stage. Re-check the settings as follows:



Remove the power the Motor. Zerothat the the setting of the Servo Potentiometer dial connection against thetopointer. Ensure potentiometer is engaged with the drive shaft.

320











Connect the Moving Coil Meter temporarily to terminal B of the 10k resistor and check the setting to its central position to give 0V output.



Transfer the Moving Coil Meter back to the output of the Power Amplifier and check the adjustment of Amplifier #1 OFFSET to give 0V.



Transfer the Moving Coil Meter to the output of Amplifier #2, set the GAIN COARSE control to 100 and GAIN FINE to 1.0 and check the adjustment of the OFFSET control for an output of +3V. Return the GAIN COARSE control to 1. This control will again be used to introduce a step input.




Ω



Press the Integrator RESET button and then connect the Integrator output to the Summing Amplifier input. Set Amplifier #1 GAIN COARSE to 10 and GAIN FINE to 1.0.



Note and record in Table 18.6 the effect of applying a 3V step input to the system with all the possible combinations of Integrator and Differentiator time constants to note their effect and determine the combination giving optimum response, possibly with one small overshoot.

321


Test

Integrator time constant

1

Differentiator time constant


Continuous Response time running YES/NO SLOW/MEDIUM/FAST

Number of Oscillations

1s

2

10s

100ms

3

10ms

4

1s

5

1s

100ms

6

10ms

7

1s

8

100ms

9

100ms 10ms

Table 18.6 

18.6a

Check your best results against each other, referring to the question below.

Which of the following is the most desirable characteristic for a positional control system?

a

Continuous running

b Slow response time, no oscillations c

Fast response time, no oscillations

d Fast response time, two or more oscillations 

Switch OFF the power supply, but

Keep the circuit connected if possible for the next exercise.

322



18.7 Practical Exercise Characteristics of a Positional Control System - 5 Use of Velocity Feedback from a Tachogenerator SERVO POTENTIOMETER

DC MOTOR

I/P

TACHOGENERATOR

POWER AMPLIFIER O/P I/P

O/P

O/P

0/P



7

4

8

3

B

1

-

B

9

2

A

10

1 VIN dt T

A-B

+

A

I/P

-

B A-B

10k

INTEGRATOR O/P


+

A

+5V

SUMMING AMPLIFIER

1s 10s 100ms

RESET

- 5V

O/P A

TIME CONSTANT

0V

B

SLIDE

INVERTER

C

A+B+C

C O/P

B I/P 1

2

3

4

5

6

7

8

AMPLIFIER #1

9

10

A

10k

-VIN

MOVING COIL METER

AMPLIFIER #2

0 I/ P -

O/ P .5

+

.4

OFFSET

-

.7

O/ P

.1

.5

+

.4 1 100 10

.9

.2

GAIN COARSE

I/P

.8

.3

1 100 10

.6

1.0

GAIN FINE

OFFSET

GAIN COARSE

-10

5

5

+10

.6 .7

+ .8

.3

.9

.2 .1

1.0

GAIN FINE

0V

L J

Fig 18.7  

Construct the circuit shown in Fig 18.7. If you have retained the former circuit, remove both of the connections to the Differentiator and add the connection to the Tachogenerator shown in Fig 18.7. Also add connections from socket B of the slider resistor to the Inverter input, and from socket A of the slider resistor to 0V. The slider resistor is used to vary the magnitude of the velocity feedback from the Tachogenerator. Set its control initially fully to the left, that is, with no feedback. The system is equivalent to the previous 3-term PID system.

323



Re-check the settings as follows: 

Remove the power connection to the Motor. Zero the setting of the Servo Potentiometer dial against the pointer. Ensure that the potentiometer is engaged with the drive shaft.






Set the Integrator time constant to 10s.






With the Moving Coil Meter connected to the output of Amplifier #2, set the GAIN COARSE control to 100 and GAIN FINE to 1.0 and check the adjustment of the OFFSET control for an output of +3V. Return the GAIN COARSE control to 1.



Transfer the Moving Coil Meter temporarily to terminal B of the 10k resistor and check the setting to its central position to give 0V output.



Reset the Integrator.



Ω

Transfer the Moving Coil Meter to the output of the Power Amplifier and check the adjustment of Amplifier #1 OFFSET to give 0V.






Note the output response to a +3V step input for various settings of the 10kΩ slider resistor control to verify that similar responses to those previously can be obtained. Note: allow the servo potentiometer dial to return to zero after each step input is applied then removed (manually turning the Hall Effect disc using the supplied Load Simulator if necessary). Also, reset the Integrator before each new +3V step input is applied.



Note and record in Table 18.7 opposite the effect of applying a +3V step input to the system with all the possible combinations of Integrator time constants and settings of the 10kΩ slider resistor (remembering to zero the servo potentiometer dial and resetting the integrator between applications of +3V step inputs).

Note: in cases where overshoot occurs, count the number of oscillations before steady state is achieved, and in cases where undershoot occurs, estimate the initial movement as a percentage of the steady state value, by dividing the initial angle swept (A) by the final angle swept (B) and multiplying by 100 to give the percentage (C).

324


Test

Integrator time constant

10kΩ Slider Resistor Setting

1

2

2

4

3

10s

8

5

10

6

2

7

4 1s

Response time

YES/NO

SLOW MEDIUM FAST

Overshoot Number of Oscillations (if any)

Undershoot Angles Swept (A) (B) (C)

6

9

8

10

10

11

2

12

4

13

Continuous running

6

4

8


100ms

6

14

8

15

10

Table 18.7

18.7a

All control systems that have a fast response will overshoot and oscillate before reaching their final value, regardless of damping.

Yes 

or

No


325



18.8 Practical Exercise Characteristics of a Speed Control System DC MOTOR

I/P

SLOTTED OPTO SENSOR

POWER AMPLIFIER

TACHOGENERATOR DIFFERENTIATOR O/P I/P

O/P

T

I/P O/P

dVIN dt

O/P

100ms 1s 10ms

O/P

TIME CONSTANT COUNTER/ TIMER

AMPLIFIER #1 I/P

V

O/P I/P

-

OFFSET


.8

.3

8

2

9

B A

10

1

B

10k

I/P

MOVING COIL METER

1 VIN dt T

5 -10

A-B

+

A 0V

+1 2 V

1s 10s 100ms

RESET

TIME CONSTANT C

INVERTER

O/P

A

C O/P

B

0

5 +10

SUMMING AMPLIFIER

B

SLIDE

1s

GAIN FINE INTEGRATOR O/P


COUN T

1.0

.1

7

4 3

R ES ET

.9

.2

GAIN COARSE

FREE RUN

.7

.4 1 100 10

TIME

.6

.5

+

A+B+C

+ 0V

L J

I/P 1

2

3

4

5

6

7

8

9

10

A

10k

-VIN

Fig 18.8 

Connect the circuit as shown in Fig 18.8 (with the integral and derivative control components NOT initially connected to the Summing Amplifier).



Set Amplifier #1 GAIN COARSE control to 10 and GAIN FINE to 0.1, the Integrator time constant to 1s, the Differentiator time constant to 10ms, the Counter/Timer controls to COUNT and 1s and both resistor controls to minimum, fully counter clockwise or to the left.



Press the mounting plate of the Servo Potentiometer to disengage it from the drive shaft and thus minimize wear on the unit.

The 20V digital voltmeter is used to monitor the Motor current, indicating the volt Ω

drop across a 1 resistor. The indicated voltage represents current in amperes. The Moving Coil Meter is used to monitor the drive voltage to the Motor. The Counter/Timer is used to monitor the Motor shaft speed.

326





Remove the feedback connection from the Tachogenerator to the Differential Amplifier so that the circuit is operating in open loop. Switch ON the power supply and set the 10k Ω wirewound resistor control so that the Motor speed is 15 rev/s as indicated by the Counter/Timer (after pressing the RESET button). The Motor voltage required is of the order of 4V

Hall effect disc

Motor Load Simulator

Fig 18.9 

Load the Motor by placing the Load Simulator vertically on the baseboard and forwardcan to apply on the disc. You will then find moving that theit Motor easily pressure be stopped, andHall theeffect Motor current increases.

18.8a

Enter your Motor current when the Motor is stalled (in A). 

Repeat the procedure with Amplifier #1 GAIN FINE settings of 0.5 and 1.0. You will find that the amplifier gain only affects the setting of the 10k Ω wirewound resistor control but has no effect on the Motor characteristic.



Re-connect the Tachogenerator feedback connection to the Differential Amplifier so that the system is operating in closed loop. Set Amplifier #1 GAIN COARSE control to 10 and GAIN FINE to 0.1 and the Motor speed to 15 rev/s. This will require the same voltage as previously.



Load the Motor as before. You will find that the torque is greater and the current and voltage applied to the Motor will increase. Note the values of Motor voltage and current with the Motor stationary and record in Table 18.8.

327


Amplifier #1 gain Motor voltage


10 x 0.1 = 1

10 x 0.3 = 3

10 x 0.4 = 4

8-10

8-10

V

V

V

A

A

A

rev/s

rev/s

Motor current Motor speed Table 18.8

18.8b



Increase the GAIN FINE setting to 0.3 and re-adjust the speed to 15 rev/s. Load the Motor until its applied voltage is 8-10V. The Motor will probably still rotate. Record the Motor current and speed.



Repeat the procedure with the GAIN FINE set to 0.4 and initial speed to 15 rev/s, recording the results again in Table 18.8.

Enter your value of current when the Amplifier gain is 3 and the Motor is loaded so that the applied voltage is 8-10V.

18.8c

Enter your recorded Motor speed in rev/s when the Amplifier gain is 4 and the Motor is loaded so that the applied voltage is 8-10V.

With closed loop control, the amplifier gain obviously affects the characteristic, increase of gain increasing the torque available. On no-load the Motor may be very noisy at this low speed setting if the gain is increased much above 0.4, due to small errors producing large power fluctuations. 

With Amplifier #1 GAIN FINE set to 0.1 and the Integrator time constant set to 1s, press and hold the Integrator RESET button, connect the Integrator output to the Summing Amplifier and then release the RESET button. Transfer the digital multimeter to the output of the Integrator.



Set the Motor speed to 15 rev/s on no-load and then load the Motor until the Motor voltage is 8-10V and maintain this loading as constant as possible.

You will note that the Motor speed initially drops, then the Integrator output voltage increases. The Motor speed then increases again. The Integrator output voltage then remains constant if the loading is kept constant. 

328

Note and record the speed after loaded conditions have settled down with the Integrator output voltage risen to about 8.5-9.0V.



Motor Speed recovers to 

rev/s

Release the load and immediately press RESET on the Counter to read the Motor speed. Record the Motor speed immediately after releasing the load. Initial recovery speed =

18.8d

rev/s

Enter the initial recovery speed in rev/s when the load is first released.

After releasing the load the speed initially rises and then the Integrator output falls gradually and the speed is reduced to the preset value of 15 rev/s again. 

Restore the loading and then take note of the time for the Integrator output voltage to recover to the unloaded voltage after the load is released. Recovery time on removing the load =

18.8e

Enter the recovery time with an Integrator time constant of 1s in seconds: 

18.8f

s

Set the Integrator time constant to 100ms and repeat the process.

The recovery time with an Integrator time constant of 100ms is:

a shorter

b

longer

c

Fast Integral Speed

the same

d

zero

Integral Control

Medium Integral

Slow Integral

High gain

Low gain

Proportional Control

Open loop

LOAD time

Fig 18.9 

Set the integrator time constant back to 1s increase the Amplifier #1 GAIN FINE control to 0.3 and repeat the process. You will note that the 329



characteristics are similar but the response times are shorter due to the higher gain of the system. The characteristics of the system are shown in Fig 18.9. Introduction of derivative control affects the rate of response to transient conditions in the same way as for the positional control system. 

Connect the derivative output from the Inverter to the Summing Amplifier. Set the Differentiator time constant to 100ms, the Integrator time constant to 100ms, the 10kΩ slider resistor control to the left, so that the derivative feedback is zero and Amplifier #1 GAIN COARSE control to 10 and GAIN FINE to 0.3.



Set the Motor speed to 15 rev/sec on no-load and then very briefly increase the slider to 10, then back to 1 on the slider scale.

You will note that with derivative feedback the Motor operation becomes noisy. This is due to the voltage spikes generated by the Tachogenerator during the commutation process, the Differentiator differentiates these and produces large outputs, making the direct feedback of the derivative signal unsatisfactory. This is a common problem with derivative feedback systems where there may be noise on the signal, being differentiated. 

To overcome this problem, feed the output from the Differentiator to the 10kΩ slider resistor via the Low Pass Filter to remove the high frequency spikes. Set the Low Pass Filter time constant to 10ms. You will find that the 10kΩ slider resistor control can now be adjusted over its full range giving full control over the magnitude of the derivative feedback with a much smaller increase in noise.

Derivative feedback makes a very small change to the characteristics of the speed control system. 

330

Move the 10k Ω slider fully to the left. Apply the Load to the Hall Effect disc briefly and heavily (so that it only just turns) for less than a second, then release it.



When the load is released the motor should be heard to greatly increase in speed before settling back to the steady state value. 

Set the Differentiator to 1s and move the 10k around 3-4 and repeat the procedure.

Ω

slider resistor control to

When the load is released, the motor should return to its steady state speed with much greater control, without greatly increasing in speed. When the load is removed, the output voltage of the Summing Amplifier should reduce then oscillate around its steady state value (approximately 4.5V) before becoming stable. This oscillation is due to the overshoot of the differentiator, then the integrator and differentiator, trying to increase the speed of the shaft back to its steady state value. 

Repeat the loading of the motor with derivative feedback and watch the analog M.C. meter for oscillation as the system returns to its steady state speed.

The effect of derivative feedback on the system is small due to the system's slow response. For derivative feedback to be effective the time constant of the differentiator must be matched to the time constant of the system. 


331




A room heating system consists of an electric heater having a constant output when ON, this being controlled by a thermostat, the contacts operating at 70 C. The range of temperature you expect measureand if adoors temperaturetime characteristic measured for the room, thetowindows remaining closed would be:were

a 2.

20 - 70 °C

4.

332

c

50 - 90 °F

d 65 - 75 °C

c

b

d

In setting up a system you wish to combine the signals from three different sources together to generate an output. The signal conditioning circuit from those provided on the DIGIAC 1750 Trainer which you would select for this purpose is the:

a

Differential Amplifier

b

Instrumentation Amplifier

c

Summing Amplifier

d

Integrator

One of the three signals referred to in question 3 above has the wrong polarity. The signal conditioning circuit from those provided on the DIGIAC 1750 Trainer which you would select for to overcome this problem is the:

a 5.

50 - 70 °C

For the system described in question 1 above, a plot of the temperature characteristic when the system is tested when operating normally would look like:

a

3.

b

Differentiator

b

Inverter

c

Buffer

d Integrator

A step input is often used to test the response of a control circuit. This term refers to:

a

an abrupt change of voltage level of either polarity

b

a smoothly changing input voltage

c

repetitive sudden variations of voltage, ON and OFF at regular intervals

d

a steady increase of voltage from 0V to +3V




Derivative control feeds back a signal which is proportional to the output signal:

a

rate of change

Gray Code Inputs

b

amplitude

c

frequency

d polarity

Control Photovoltaic Cell +5V

A Zero Reference

Differential Amplifier

VR

M

Alarm +

0V

Fig 1

7.

8.

A control circuit is required which will always return a rotational position system to the zero setting of a 3-output Gray-code disc (such as the one on the DIGIAC 1750 Trainer), that is with the three outputs all zero. A suitable circuit for carrying this out is given in Fig 1. Selecting signal conditioning circuits from the DIGIAC 1750 Trainer, the Control block might contain:

a

proportional, derivative and integral controls

b

a summing amplifier

c

three DC amplifiers

d

differentiator, integrator and inverter

The component marked A in the circuit of Fig 1, selected from the signal conditioning circuits of the DIGIAC 1750 Trainer is a:

a 9.

Fig 2

power amplifier

b

AC amplifier

c

Buffer

d current amplifier

The effect of changing the setting of variable resistor VR in Fig 2 above might be to make the Alarm sound:

a

above a higher light threshold

b

louder

c

below a higher light threshold

d

intermittently

333



Notes: ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ...........................................................................................................................................

334


Using a Multimeter Appendix A

Appendix A Using a Multimeter

Units and Quantities There are three basic quantities to be considered in an electrical circuit: 1.

An EMF is applied to the circuit to provide the force or pressure which causes the current to flow around the circuit. This EMF is measured in volts.

2.

The current consists of a quantity of electrons which travel around the circuit in a given time. This current is measured in amps (amperes).

3.

As the current flows around the circuit it meets up with opposition due to the resistance of the circuit or its component parts. This resistance is measured in (ohms).

Multimeters The term multimeter derives from the ability to use one instrument for a multitude of different measurements. One instrument is capable of taking measurements of all three of the above quantities, and switches are provided for a wide range of values of each quantity, from the very small (µ - micro or m - milli) to the large (k - kilo or M - Mega). Also both direct current and voltage (DC) and alternating current and voltage (AC) measurements can be taken with the same instrument. ●

Examine the instrument(s) which you have available and familiarize yourself with the range switch(es), display and connection sockets/terminals.

= DC

= AC

335



Types of Meters There are two basic types of instrument, those which give a digital display of the reading, and those in which a pointer is moved across a scale by an angle which is analogous to the quantity being measured.

POWER OFF

ON

V 200m

2

V 20

200 1000 750 200

20 2

200m

20M

200µ 2m

2M

20m 10A 200m

200K 20K

Ω

A

2

2K 2 200m

200 200 200µ 20m 10A

2m

A

V-Ω

COM

A

(a) Digital Multimeter

10A

(b) Analog Multimeter

Fig A.1

The digital instrument will be found to be more convenient for taking static readings of a quantity, their accuracy tends to be very good, and it is less likely that you will make a mistake in reading the quantity. The analog instrument, on the other hand, has advantages when reading quantities which are subject to change during adjustments or otherwise. The load (in terms of current drawn) presented by the meter to the circuit under test also varies.

336



Reading the Analog Scale

Fig A.2

The instrument scale represented above might refer to a meter with ranges 50µA, 250µA, 2.5mA, 10mA, 25mA, 100mA, 250mA, 1A, & 5A and a selection of voltage ranges. Assuming that the 2.5mA scale has been selected then the scale can be read directly in milliamps. The pointer is between 1.5 & 2.0, so the reading lies between these limits. There are five divisions between 1.5 and 2 on the scale so each one represents a value of 0.1. The pointer is between the second and third divisions so the reading is between 1.7 & 1.8, or 1.7+. It is possible to make an estimate (guess) as to how far it lies between the two divisions, but it is advisable not to go any further than to say 0.05 (half way), although I am sure that you will try. So a reasonable reading of the scale would be 1.75mA. If the selected range is 100mA then the 0-10 scale is used and the pointer is half way between 6 & 8. The scale reading gives us 7. The scale factor is determined by dividing the full-scale marked value into the range value, 100mA ÷ 10 = 10mA. Multiply the reading by this factor: 7 x 10mA = 70mA. If the selected range had been 50µA then the 0 - 50 scale should be used and the pointer is half way between 30 and 40. The scale reading gives us 35. The scale factor is 35 x 1µA = 35µA. This is a major disadvantage of the analog multimeter. It is relatively easy to make an error in interpreting the scale and range settings. This factor alone is responsible for many people preferring the digital instrument. Try interpreting for yourself on the assumption that you have selected the 250V range. You should have arrived at a reading of 175V.

337



Testmeter Connections 1.

Voltage Readings

POWER OFF

ON

V 200m

2

V 20

200 1000 750 200

20 2

200m

20M

200 µ 2m

2M

(a)

20m 10A 200m

200K 20K

Ω

A

2

2K 2 200m

200 200

2m

200 µ 20m 10A

A

V-Ω

COM

A

10A

POWER OFF

ON

V

V 20

200m

200 1000 750 200

20

2

2

200m

20M

(b)

200 µ 2m

2M

20m A 10A 200m

200K 20K

Ω

2 2K 2 200m

200 200 200 µ 20m 10A

2m

A

V-Ω

COM

A

10A

Fig A.3 - (a) DC and (b) AC Voltmeter Connections

The voltage appears across the component. Therefore the meter must be connected in parallel with (or across) the component to measure the volt drop across it with the circuit still connected to the supply. Note that this is therefore the easiest of readings to be taken, since it involves no disconnections and is taken with the supply still connected. Ensure that the correct type AC or DC is selected, and always start with the highest range and work down unless you have every reason to expect a reasonably lower voltage. You will never damage a meter by connecting it to a lower voltage than it is adjusted to display.

338




Current Readings

POWER OFF

ON

V 200m

2

V 20

200 1000 750 200

20 2

200m

20M

200µ 2m

2M

(a)

20m A 10A 200m

200K 20K

Ω

2

2K 2 200m

200 200 200µ 20m 10A

2m

A

V-Ω

COM

A

10A

POWER OFF

ON

V 200m

2

V 20

200 1000 750 200

20 2

200m

20M

(b)

200µ 2m

2M

20m A 10A 200m

200K 20K

Ω

2

2K 2 200m

200 200 200µ 20m 10A

2m

A

V-Ω

COM

A

10A

Fig A.4 - (a) DC and (b) AC Ammeter Connections

The current flows around the circuit so it must be broken to allow the meter to be connected in series with the component under test. The circuit current then also flows through the meter and it can give an indication of how much this current is. This is often very inconvenient in practice, since it is not always easy to break into a circuit in the way required.

339




Resistance Readings

POWER OFF

ON

V 200m

2

V 20

200 1000 750 200

20 2

200m

20M

200µ 2m

2M

20m A 10A 200m

200K 20K

Ω

2

2K 2 200m

200 200 200µ 20m 10A

2m

A

V-Ω

COM

A

10A

Fig A.5 Ohmmeter Connection

It is essential that the resistor to be checked should be isolated from the power supplies and also desirable, when possible, from the remainder of the circuit. Analog Multimeter - A battery in the instrument applies a voltage to the resistor under test and then the instrument measures the current which flows. Since the battery voltage is known the current flowing can be calibrated into resistance. The scale is not linear since resistance is inversely proportional to current, zero resistance resulting in maximum current. A zeroing control is provided to allow for variation of the battery EMF with ageing. Digital Multimeter - The instrument contains a constant current generator, this current being fed to the resistor under test. The instrument measures the voltage dropped across the resistor and converts this to resistance. Since resistance is directly proportional to voltage this is a linear function and conversion to a digital display of resistance is simple.

340


The Oscilloscope Appendix B

Appendix B The Oscilloscope

How it Works Your understanding of the operation of this most valuable item of test equipment will be greatly enhanced if you have at least a superficial knowledge of its fundamentals.

glass envelope screen focus

cathode grid

Y plates

X plates

electron beam

heater

Fig B.1

The heater, made of tungsten wire, raises the temperature of the cathode, which is a nickel alloy cylinder coated with a mixture of oxides. The heated cathode emits electrons which are attracted by the high potentials on succeeding electrodes to form a divergent electron stream or beam. The electric field of the focus assembly accelerates the electrons in the beam and converges them so that they all meet at one spot at the screen. The internal face of the screen is coated with phosphorescent materials which glow when bombarded by the electron beam.

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The grid, which surrounds the cathode, allows control of the number of electrons leaving the cathode, and therefore the strength of the electron beam, and the intensity or brightness of the spot. The group of electrodes which generate the beam are known collectively as the electron gun. The screen is the faceplate of a glass envelope, which encloses all of the electrodes. This envelope is evacuated so that there are no gas atoms to impede the free movement of the electrons in the beam. Any voltage (potential gradient) across the Y plates will cause the beam to be deflected up or down as it passes through. The X plates will have a similar effect in the horizontal direction. The oscilloscope is therefore capable of drawing graphs with conventional X and Y axes. The inputs to X and Y channels must be in the form of voltages which can be applied to the plates. The primary purpose of the oscilloscope is to allow us to examine electrical waveforms in a circuit which are readily obtainable in the form of voltage (Y) against time (X). The Y drive is therefore already in the correct form - a voltage. The time scale for the X axis is provided as a function of the oscilloscope's circuitry known as the timebase. This generates a voltage which is steadily changing with time. The time is adjustable by front panel controls. The waveform necessary for this purpose has a sawtooth shape.

voltage stroke or scan

flyback

time

Fig B.2

The faceplate is scanned from left to right, relatively slowly, during which time the waveform to be examined is applied to the Y plates. The flyback is rapid and the Y signal is suppressed so that it cannot interfere with the forward display.

342



Practical Oscilloscope It is now time to examine the layout of the front panel of a typical oscilloscope and its controls. These may seem a little awe-inspiring at first, but you will find that you can easily master them. All oscilloscopes have the same basic functions. If the instrument which you have available is substantially different from that shown pictorially here, then you will find controls which perform the same functions, although they may sometimes have slightly different labels on them. Start by setting all controls to known initial conditions as follows:

POWER

X-Y

on/off

TIME/DIV. SLOPE +/-

X-POS.

ms 2

AT/NORM. µs

.5 .2 .1

50

5

20

10

TR

5

50

Y-POS. I

.5

X-MAG.

X10

CAL. 0.2V

COMPONENT

30pF 400Vp-p max.

.5 .1

50

5

20

VAR. 2.5:1

CAL.

5

INV.I

.1

2

50

CH.II

5

20

DC AC GD

10

10 20

mV CHI/II

V DUAL

10

ADD

1MΩ 400Vp-p max.

CHOP

HOR. INP.

20

2V TESTER

TRIG.I/II

Y-POS.I I

.2

1

2

V

100Vp-pmax.

CAL.

VOLTS/DIV.

.2

10

1MΩ

TRIG. INP.

.5

µs

ms EXT.

VOLTS/DIV. 1

CH.I

1

200

AC DC HF LF LINE

TRIGGER SELECTOR

DC AC GD

2

100

FOCUS

LEVEL

10

20

INTENS.

5

mV

30pF

Fig B.3

The arrowed rectangles and squares are push-on push-off buttons. ●

Ensure that they are all in their out positions. There are several round buttons in various colors with an indicator line on them. Turn all of these so that the line is pointing vertically upwards. This does not apply to the focus control. The pointed triangle on some colored knobs is a calibration indicator. The coarse setting on the outer switch is only correct when this arrowhead points to the left. Set them this way now.

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3

1 POWER

X-Y

on/off

TIME/DIV. SLOPE +/-

X-POS.

ms 2

.5

AT/NORM. µs

.2 .1

50

5

20

10 TR

10

20

5

50

INTENS.

Y-POS. I

.5

DC AC GD

.5

20 10

VAR. 2.5:1

5

CAL.

INV.I

CHI/II

TRIG.I/II

DUAL

DC AC GD

20

5

V

CH.II

50

10

mV

2

.1

2

10 20

Y-POS.I I

.2

1 50

V

CAL.

VOLTS/DIV.

.1

2

5

30pF 400Vp-p max.

100Vp-pmax.

µs

ms EXT.

.2

1

TRIG. INP.

.5

VOLTS/DIV.

2

1MΩ

1

200

AC DC HF LF LINE

TRIGGER SELECTOR

CH.I

2

100

FOCUS

LEVEL

10

ADD

1MΩ 400Vp-p max.

CHOP

HOR. INP.

20

4

5

mV

30pF

4

Fig B.4

Adjust the controls shown in Fig B.4 as follows: ● ●

1 2

TRIGGER SELECTOR to the upper (AC) position. Y amplifier inputs both to the lower (GD) position.

●

3

TIMEBASE set upwards to the 1ms/div position.

●

4

Y AMPLIFIER sensitivity both counterclockwise to the 20V/div position.

Note that the lower panel in Fig B.4 above contains the controls for two Y amplifiers. There is provision to operate the oscilloscope with either one or two traces (graphs) so that two waveforms of the same frequency (or harmonically related) can be observed at the same time. This is achieved by switching the electron beam from one trace position to the other and, at the same time, switching the inputs to the Y plates. The upper panel contains the controls for the screen and for the timebase settings. You will also see some controls marked TRIG or TRIGGER. These are to maintain a stable trace. More will be said about this function later.

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Operation You are now ready to power up. ●

Locate the power switch ( 1 in Fig B.5 below) and switch ON.

1

4

POWER

X-Y

on/off

X-POS.

TIME/DIV. SLOPE +/-

2

.5

.2 .1

AT/NORM. µs

50

5

20 10

10

TR

2

ms

INTENS.

20

5

50

2

100

FOCUS TRIGGER SELECTOR

AC DC HF LF LINE

LEVEL

1

200

.5

EXT.

TRIG. INP. 100Vp-p max.

µs

ms CAL.

3 Fig B.5

After a brief warm-up period you will find that you have a line across the screen caused by the spot moving from left to right across the screen under the influence of the internal timebase. ●

Adjust the brightness or intensity 2 to give a line of minimum intensity for comfortable viewing.

●

Adjust the focus 3 to give the sharpest line.

●

Adjust the X POSITION 4 to centralize the line across the screen.

●

Switch the TIMEBASE selector (see Fig B.4) fully counterclockwise to the 200ms/div position.

If you have a watch or clock available with a second hand, time how long it takes for - say - five passes across the screen. You should find that it takes about ten seconds for five scans.

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Timebase Examine the timebase control switch. This is pointing at 200ms/div. There are ten divisions across the screen. Count them. So it takes 10 x 200ms for one scan. 2000ms is 2 seconds, so 5 x 2 = 10 seconds for five scans. Turn the inner variable control clockwise. See that the spot speeds up. It is possible to set the speed to anything that you want (within limits) but you only know what speed it is when the pointer is to the left (the calibrated position). ●

Return it counter-clockwise.

Look to the left of the tip of the pointer and you will see a C (for calibrated) under a dot. There is one of these symbols to the left of each of the variable controls, including the two on the lower panel, to indicate the calibration position. ●

Switch the timebase selector to 100ms/div. Note that the spot now travels across the screen in about one second. Gradually increase the speed.

When you get to 20ms/div the spot has become a short line. This is due to two factors, one being the afterglow of the phosphor (which takes a small time to die away) and the other is the persistence of vision (where our eyes retains an image for a small period of time). This latter is what makes it possible for us to see apparently moving pictures on a television screen from a rapid sequence of still pictures. At 10ms/div the spot becomes a continuous line with a small amount of flicker as our eyes still try to follow the individual movements of the spot. Beyond this all we see is a steady line. When the timebase setting is increased to the maximum of 0.5µs/div the screen is being scanned in five millionths of a second (5µs). It is still accurate and linear.

346



Frequency Measurement Please note that if it takes 5µs (millionths of a second) for one trace and the traces follow each other continuously then there will be 200,000 scans in one second (200,000 x 5µs = 1s), the frequency is 200kHz. This concept is the one above all other that newcomers to electronics find most difficult to accept, the speed at which electronic devices can operate, far, far faster than our brains want to accept. The reciprocal of the time taken for one cycle of events is the frequency of that event. This is important and should be remembered. frequency =

1 time period

This allows us to make measurements of frequency on an oscilloscope by noting the time taken for one cycle and then calculating the reciprocal of that time.

6.4 div

Fig B.6

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For instance, in the example in Fig B.6 opposite, if the timebase setting is calibrated and switched to 0.2ms/div then the time taken for the cycle indicated is: 6.4 x 0.2 = 1.28ms and the frequency of the waveform represented will be: 1 . Hz 12 . 8 x 10 -3 = 78126 ●

Try the following example for yourself:

Assume that the timebase is correctly calibrated and switched to 20µs/div.

?

Fig B.7 ●

What frequency is represented in Fig B.7 if the two vertical lines represent one cycle of a waveform?

You should have arrived at about 5.95kHz. The reading of the time scale cannot be very accurate, certainly not to 5 parts in 600, so it might be better to call this 6kHz.

348



Y Amplifiers Turn your attention now to some of the controls on the lower panel, the Y amplifiers.

Y-POS. I

VOLTS/DIV.

1

.5

CH.I DC AC GD

.5 .1

20

5

30pF 400Vp-p max.

V

VAR. 2.5:1

CAL.

5 INV.I

CHI/II

TRIG.I/II

V

DC AC GD

20

5

DUAL

CH.II

50

10

mV

3

.1

2

10 20

Y-POS.I I

.2

1 50

2

10 1MΩ

VOLTS/DIV.

.2

1

10

ADD

1MΩ 400Vp-p max.

CHOP

HOR. INP.

20

5

2

mV

30pF

4

Fig B.8

1 This is the channel 1 (CH.1) Y amplifier shift or position control. It applies a direct voltage to the Y plates. ●

Try this now. Move the trace line up and down.

The effect is that you are applying a signal to the Y plates, only relatively very slowly. Electronics can do it much faster. Do not try to rotate the knob too quickly or you may damage the track of the control. ●

Set the timebase to minimum speed (200ms/div) and try moving the Y1 shift again. You can almost draw a sinewave, if you are careful, but of course it dies away very quickly.

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Dual Trace Operation ●

Set the timebase back to high speed at 0.2ms/div and position the trace in the top half of the screen.

●

2 Press the button marked DUAL to select both Y traces.

A second trace will now have appeared near to the center of the screen. ●

3 Move the new trace down to the lower half of the screen with the Y2 shift control.

●

Reduce the timebase speed again to 100ms/div.

You will see that the oscilloscope draws the Y1 and Y2 traces alternately. This is the simplest form of dual mode operation, but is not very satisfactory for low frequency signal inputs. You would have great difficulty in comparing waveforms on the two traces. ●

4 Press the button marked ALT/CHOP (or ADD/CHOP). Both traces are now drawn simultaneously.

What is happening is that the circuit chops between the two traces very many times during one scan, so quickly that you cannot see it doing it. This is the best mode of operation for timebase speeds below 2ms/div. You will see that operating the ALT/CHOP switch has little effect at timebase speeds of 2ms/div and above, but the difference is easily observed at 5ms/div and below.

350



Voltage Measurements

2

2 0V VOLTS/DIV. 0V

Y-POS.I

.5

1 CH.I

20 10

3

V

VAR. 2.5:1

CAL.

5

INV.I

CHI/II

TRIG.I/II

V

DC AC GD

20

5

DUAL

CH.II

50

10

mV

1

.1

2

10 20

Y-POS.I I

.2

1 50

5

30pF 400Vp-p max.

.5

0V.1

2

DC AC GD 1MΩ

VOLTS/DIV.

.2

1

10

ADD

1MΩ 400Vp-p max.

CHOP

HOR. INP.

20

mV

5

30pF

3

Fig B.9 ● ●

1 Set both channel input switches to AC, and 2 both

Y amplifier sensitivity switches to 0.1V/div.

●

3 Plug an oscilloscope probe lead into each of the input sockets.

●

Adjust the Y shift controls to locate the Y1 trace in the middle of the upper half of the screen and the Y2 in the lower. Locate the calibrator (CAL) terminal lug on the panel just below the screen and hook the CH.1 probe on. Note the amplitude given for this signal besides the terminal(s). If you have more than one voltage available, then select the one nearest to 0.2V. X-MAG.

CAL. 0.2V

COMPONENT

X10

2V

TESTER

Note: The ground clip is not needed since this internally.

is

completed

Fig B.10

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You will now have a square wave displayed on the upper trace. The vertical edges of the waveform are so fast that they do not have time to leave any evidence of their presence. It appears as though the change from negative to positive is instantaneous. Increasing the brightness to maximum may just show them very faintly. ●

Re-adjust for normal intensity.

The waveform should cover two divisions in the vertical direction (2 x 0.1V= 0.2V). ●

Clip the CH.2 probe on as well.

You now have waveforms displayed on both traces.

Y-POS. I

VOLTS/DIV. .5

CH.I DC AC

5

20

V

VAR. 2.5:1

CAL.

5

INV.I

.1

2

50

5

20 10

10 20

mV CHI/II

TRIG.I/II

Y-POS.I I

.2

1 50

10

30pF 400Vp-p max.

.5 .1

2

GD 1MΩ

VOLTS/DIV.

.2

1

V DUAL

10

CH.II DC AC GD

ADD

1MΩ 400Vp-p max.

CHOP

HOR. INP.

20

5

mV

30pF

Fig B.11

352

●

Press the INVERT 1 button and observe that the CH.1 display is inverted, the CH.2 trace remaining unaffected.

●

Increase the CH.1 Y amplifier sensitivity to 50mV/div and observe how many squares are now covered by the waveform.



AC/DC Operation ●

Return both amplifier input switches to the GD (ground) position.

The waveforms are removed. ●

of

Using the Y shift (position) controls centralize both traces across the middle the screen so that they are overlayed on top of each other.

You should now only be able to see one line. ●

Return the CH.1 Y amplifier input switch to AC and the waveform reappears at the center of the screen with the Y2 trace acting as a base (0V) line.

You are now looking at the AC component of the waveform. However, this waveform has a DC component equal in amplitude to the peak value of the AC signal. ●

Switch the CH.1 Y amplifier input switch to DC.

The waveform moves up to sit on the 0V base line provided by the Y2 trace. The DC component of the signal is now being passed to the display as well as the AC. In fact the waveform has two amplitude levels, 0V and 0.2V. This facility of being able to suppress the DC component if you wish can be very useful if a small AC component rides on top of a very large DC component. The AC can be inspected with the amplifier set to a very sensitive setting which would move the DC component well off the viewable screen area taking the AC component with it ! Generally speaking, it is better to retain the DC component of any waveform in the display if you can.

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Frequency Measurement Example You have already been introduced to this most important aspect of the oscilloscope's measurement capability. Let us now use it in practice. The calibration signal is only intended for checking the sensitivity of the Y amplifiers and probe compensation. The frequency of the signal is not precise, and therefore provides us with an excellent example for practice.

T = Time taken for one cycle.

T

Fig B.12 ●

Read off the number of divisions for one complete cycle - T as precisely as possible along the center line.

●

Multiply by the setting of the timebase selector to convert this into a time. 1

●

Use a calculator to take the inverse (reciprocal

x ) of this to give the

frequency. You should have found a frequency somewhere near 1kHz.

354



Trigger ●

Return the CH.1 Y amplifier input switch to GD and switch the CH.2 input to AC.

You can see the waveform, but it is not stable. This is because the trigger or synchronizing facility is automatically allocated to the CH.1 signal until you say otherwise.

Y-POS. I

VOLTS/DIV. .5

CH.I DC AC GD

.5 .1

20

5

30pF 400Vp-p max.

V

VAR. 2.5:1

CAL.

5

INV.I

.1

2

50

5

20 10

10 20

mV CHI/II

TRIG.I/II

Y-POS.I I

.2

1 50

2

10

1MΩ

VOLTS/DIV.

.2

1

V DUAL

10

CH.II DC AC GD

ADD

1MΩ 400Vp-p max.

CHOP

HOR. INP.

20

5

mV

30pF

Fig B.13 ●

Press the CHI/II TRIG.I/II button.

Trigger control is transferred to the CH.2 input waveform and the signal locks in. If you now reverse the settings to display the CH.1 waveform with CH.2 grounded, the waveform will be unstable again until you release the CHI/II TRIG.I/II button again. Automatic triggering is quite a complex operation and it is worth examining the theory of this a little more closely. ●

Switch trigger control back to CH.2 to unlock the display.

The displayed trace may be only marginally out of lock, giving a slowly moving waveform, or it may be considerably out, giving no readable waveform. Using the timebase fine tuning control (the one with the arrowhead) try to stop the trace from moving. You will find that this is very difficult, since the slightest thing will change the frequency enough to de-synchronize the waveform.

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You will probably find two different settings within the range of the control which will give you either one complete cycle or one and a half. ●

Switch control back to CH.1 to lock the trace again.

You find that there are very nearly two complete cycles when the control is in the properly calibrated position. As the fine timebase control is adjusted when the waveform is locked, all that happens is that the waveform is stretched or contracted to display more or less cycles. Note, however, that the trace always starts with the positive-going edge of the waveform.

Fig B.14

This is the trigger point, at the zero crossing of the test waveform (in a positivegoing direction). The timebase in the oscilloscope is held off until this point is reached and then allowed to run. In this way the displayed waveform always starts at the same point (crossing zero in a positive-going direction) so each successive trace overlays the previous one and the display appears stationary. There are several features on the timebase panel which affect the triggering.

356



Triggering

3 POWER

1

X-Y

on/off

X-POS.

TIME/DIV. SLOPE +/-

ms

2 .5 .2 .1

AT/NORM.

50

5

20

10

TR

µs 10

20

5

50

INTENS. TRIGGER SELECTOR

AC DC HF LF LINE

4

2

2

100

FOCUS

LEVEL

1

200

EXT.

100Vp-pmax.

µs

ms

5

TRIG. INP.

.5

CAL.

6

Fig B.15

1

AT/NORM. This means Automatic Trigger or NORMal operation. In

automatic triggering (button out) the action is as described above. With the button pressed the trigger point voltage level is adjustable by the LEVEL control 2 . The effect of this is to change the starting point voltage so that the display starts at any point you choose on the waveform. If you set the level higher or lower than the extremities of the test waveform then the timebase never triggers and there is no display, the screen remains blank. With the level button pointing vertically upwards the trigger point is the zero voltage crossing level. You cannot see the effect of this control if you only have the calibration waveform available. The square wave has only two levels, ON or OFF. However, if you have a signal source with sine or triangular waveform then connect this to one of the Y channel inputs, adjust for a good display using timebase (X) and sensitivity (Y) controls, then press the AT/NORM. button and adjust the level control. Observe the effect and then return the AT/NORM. button to the out position.

The +/- button 3 inverts the display by selecting the zero crossing trigger point when the waveform is negative going instead of positive.

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With any waveform displayed and locked, press the +/- button and observe. Return to the out position.

The displayed waveform can be very complex and contain components at many different frequencies. The automatic trigger circuits are periodic, i.e. they are sensitive to frequency. For some displays the trigger circuits may need a little help in the form of selecting the frequency. The calibration waveform is a middle frequency and any setting of the TRIGGER SELECTOR 4 except LINE will provide a stable display. The settings of this selector are: AC

The alternating component of the test waveform is passed to the trigger circuits. This will normally cover frequencies from DC to 10MHz.

DC

The DC component passes to the trigger circuits. To use this facility NORMal triggering must be selected.

HF

Frequencies above 10MHz.

LF

Frequencies below 1kHz. This would normally be used with a complex wave containing many frequency components where you wish to lock on to the low frequency component(s) rather than the high, such as an amplitude modulated carrier wave as used in radio communication.

LINE Many oscilloscopes are used for television servicing, so many are provided with line synchronizing pulse separators to lock onto these pulses which define the termination of each line of the picture.

picture information

line sync. pulse

Fig B.16 Television Waveform

358

line sync. pulse



This function will only lock on to short duration (5µs) negative-going pulses. It will sometimes be required to examine waveforms which are too weak to provide a satisfactory signal to the trigger circuits so that automatic triggering cannot be achieved. An alternative source of higher voltage waveform(s) at the same frequency will often be available. This alternative source can be fed in directly to the trigger as an "external" trigger source so that a weak but stable display can be achieved. The EXT. TRIG. button 5 selects this functi on, but at the same time switches off the intern al, automatic triggering. ●

Press the EXT. TRIG. button and note that the display is no longer locked.

●

Take the probe from the CH.2 input and plug it into the EXT. TRIG. input socket 6 . Couple this to the cal. signal.

Note that the display is again locked and that all of the other triggering functions can be selected with this input.

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Component Tester Many oscilloscopes are provided with this most valuable facility, which enables the instant display of the characteristics of many electronic components. An alternating voltage is applied to the component under test and also to the X plates of the oscilloscope. The current drawn flows in a series resistor mounted inside the oscilloscope, developing a volt drop across it which is proportional to the current drawn. This is applied to the Y plates. The instantaneous values of both voltage applied and current drawn are therefore plotted.

X-MAG.

X10

CAL. 0.2V

COMPONENT

2V TESTER

test component

Fig B.17 ●

●

Connect the component to be tested as in Fig B.17 above. Testmeter leads will be ideal for this purpose. Press the Component Tester button (arrowed).

The characteristic will immediately be displayed.

360



With no component connected the display will be the characteristic of an open circuit, no current, whatever the voltage. A lead connected between the two terminal sockets indicated will be a short circuit. Can you anticipate the display? Here are a few other samples:

Fig B.18

This facility is very useful when troubleshooting. By now you should feel more confident in the use of your oscilloscope. You will find it an invaluable instrument in future investigations of electronic circuits.

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Manual for D1750.pdf

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