Electronics LaboratoryManual

Prepared byAssoc. Prof. Dr. Haluk DENİZLİ

Assist. Prof. Dr. Hakan YETİŞRes. Assist. Dr. Atılgan ALTINKÖK

Res. Assist. Arzu ÖZRes. Assist. Ali YILMAZ

Department of PhysicsAbant Izzet Baysal University

Updated for Fall 2012

Edited by Ali Yılmaz, September 25, 2012

SAFETY WARNING

Before using this laboratory, read, understand and follow the Safety Precautions mentioned inside this manual.

This is an educational laboratory where high-voltage terminals and large current-carrying components and circuits are exposed for ease of measurements. Therefore, regardless of the voltage and current levels, these should be treated as high voltages and high currents, and the safety precautions mentioned in the manual must be followed.

SAFETY PRECAUTIONS

1. Why is safety important?

Attention and adherence to safety considerations is even more important in a power electronics laboratory than is required in any other undergraduate electrical engineering laboratories. Power electronic circuits can involve voltages of several hundred volts and currents of several tens of amperes. By comparison the voltages in many teaching laboratories rarely exceed 20V and the currents hardly ever exceed a few hundred milliamp.In order to minimize the potential hazards, we will use DC power supplies that never exceed voltages above 40-50V and will have maximum current ratings of 5A or less. However in spite of this precaution, power electronics circuits on which the student will work may involve substantially larger voltages (up to hundreds of volts) due to the presence of large inductances in the circuits and the rapid switching on and off of amperes of current in the inductances. For example a boost converter can have an output voltage that can theoretically go to infinite values if it is operating without load. Moreover the currents in portions of some converter circuits may be many times larger than the currents supplied by the DC supplies powering the converter circuits. A simple buck converter is an example of a power electronics circuit in which the output current may be much larger than the input DC supply current.

2. Potential problems presented by Power Electronic circuitsElectrical shock may take a life.Exploding components (especially electrolytic capacitors) and arcing circuits can cause blindness and severe burns.Burning components and arcing can lead to fire.

3. Safety precautions to minimize these hazards

3.1 General Precautions

Be calm and relaxed, while working in Lab.When working with voltages over 40V or with currents over 10A, there must be at least two people in the lab at all times.Keep the work area neat and clean.No paper lying on table or nearby circuits.Always wear safety glasses when working with other than signal-level power.Use rubber door mats to insulate yourself from ground, when working in the Lab.Be sure about the locations of fire extinguishers and first aid kits in lab.A switch should be included in each supply circuit so that when opened, these switches will de-energize the entire setup. Place these switches so that you can reach them quickly in case of emergency, and without reaching across hot or high voltage components.

3.2 Precautions to be taken when preparing a circuitUse only isolated power sources (either isolated power supplies or AC power through isolation power transformers). This helps in using a grounded oscilloscope. This reduces the possibility of risk of completing a circuit through your body. This also reduces the possibility of destroying the test equipment.

3.3 Precautions to be taken before powering the circuitCheck for all the connections of the circuit and scope connections before powering the circuit, to avoid shorting or any ground looping, that may lead to electrical shocks or damage of equipment.Check any connections for shorting two different voltage levels.Check if you have connected load at the output. This is very important in Boost and Buck-Boost Converters and converters based on them.

Double check your wiring and circuit connections. It is a good idea to use a point- to-point wiring diagram to review when making these checks.

3.4 Precautions while switching ON the circuit

Apply low voltages or low power to check proper functionality of circuits.Once functionality is proven, increase voltages or power, stopping at frequent levels to check for proper functioning of circuit or for any components is hot or for any electrical noise that can affect the circuit’s operation.

3.5 Precautions while switching on or shutting down the circuit

Reduce the voltage or power slowly till it comes to zero.Switch of all the power supplies and remove the power supply connections.Let the load be connected at the output for some time, so that it helps to discharge capacitor or inductor if any, completely.

3.6 Precautions while modifying the circuit

Switch on the circuit as per the steps in section 3.5.Modify the connections as per your requirement.Again check the circuit as per steps in section 3.3, and switch ON as per steps in section 3.4.

3.7 Other Precautions

No loose wires or metal pieces should be lying on table or near the circuit, to cause shorts and sparking.Avoid using long wires, that may get in your way while making adjustments or changing leads.Keep high voltage parts and connections out of the way from accidental touching and from any contacts to test equipment or any parts, connected to other voltage levels.When working with inductive circuits, reduce voltages or currents to near zero before switching open the circuits.BE AWARE of bracelets, rings, metal watch bands, and loose necklace (if you are

wearing any of them), they conduct electricity and can cause burns. Do not wear them near an energized circuit.Learn CPR and keep up to date. You can save a life.When working with energized circuits (while operating switches, adjusting controls, adjusting test equipment), use only one hand while keeping the rest of your body away from conducting surfaces.

PREFACE

GOAL

The purpose of the experiments described here is to acquaint the student with:

! (1) Analog & digital devices! (2) Design of circuits! (3) Instruments & procedures for electronic test & measurement.

The aim is to teach a practical skill that the student can use in the course of his or her own experimental research projects in physics, or another science.

At the end of this course, the student should be able to:

! (1) Design and build simple circuits of his or her own design.! (2) Use electronic test & measurement instruments such as oscilloscopes, timers, function generators, etc. in experimental research.

SCHEDULE

Exp. # ! Name of Exp.! ! ! Week

1. FAMILIARIZATION...................................................................October 4, 2012

2. THE SEMICONDUCTOR DIODE............................................October 11, 2012

3. HALF - WAVE & FULL - WAVE RECTIFIERS.......................October 18, 2012

4. THE ZENER DIODE..............................................................November 1, 2012

5. TRANSISTOR FAMILIARIZATION (BJT).............................November 8, 2012

6. THE SILICON CONTROLLED RECTIFIER (SCR).............November 15, 2012

7. Midterm I (Experiments 1, 2, 3, 4, 5 and 6)...........................November 22, 2012

8. THE DIAC & THE TRIAC................................................... November 29, 2012

9. THE FIELD EFFECT TRANSISTOR.....................................December 6, 2012

10. THE OPERATIONAL AMPLIFIER......................................December 13, 2012

11. Midterm II (Experiments 8, 9 and 10)..................................December 20, 2012

GRADING

QUIZ......................................................................................................................10 %

REPORT................................................................................................................15 %

MIDTERM I.........................................................................................................25 %

MIDTERM II .......................................................................................................20 %

FINAL …...............................................................................................................30 %

FAMILIARIZATION

OBJECTIVES:

1. To become familiar with the EEC740 mounting deck.2. Learn how to connect external supplies to the EEC470 deck.

EQUIPMENT REQUIRED:

Qty Apparatus1 Electricity & Electronics Constructor, EEC4701 Basic Electronics Kits 2, EEC4721 Power supply unit. External d.c and a.c supplies as appropriate (e.g.

Feedback Power Supply PS445)2 Multimeters or1 Milliammeter 100mA d.c and1 Voltmeter 20V d.c

PREREQUISITE ASSIGNMENTS:

None

KNOWLEDGE LEVEL:

Before working this assignment you should:• Know how to use an ammeter and a voltmeter.

Fig 1.1 EEC470 Deck Layouts

INTRODUCTION:

The Electricity & Electronics Constructor EEC470 is an unpowered mounting deck suitable for the construction of a wide variety of circuits which employ:Discrete components; Resistors, capacitors, transistors, etc.Integrated circuits; Analogue or digitalPower components; Thyristors and power transistors

The deck is used in conjunction with external power supplies and with various kits of components covering different areas of industrial electronics. Each kit is self- contained and is provided with a book of student assignments, of which this familiarization assignment is always the first.

THE DECK LAYOUT; The layout of the EEC470 Deck is shown in fig 1.1. The deck is supported by the following:Discrete Components; These are supplied in the various kits on two, three and four- pin carriers. Interlinking is done by “shorting” links on two-pin carriers or by flexible “patching” leads with 2mm pin terminations.Integrated Circuit Modules; These are supplied, where appropriate, in the standard module form used by Feedback in other digital and analogue constructional systems (series 341,346, etc). Many of these modules, particularly the digital types, obtain their power supply from the +5V and 0V bus-bars built in to this section. Other types will require power to be connected by patching leads.

Power Components; There is provision for inserting up to two semi-conductors mounted on heat sinks. The section is also useful for coupling to external equipment, using 2mm patch leads internally and 4mm leads externally.

Power Supplies; External supplies can be brought in on 4mm leads and patched internally using 2mm leads. The supplies listed on the panel are those available from the Feedback PS445 Power Supply unit designed for use with EEC470.The current ratings of the PS445 are as follows, and any similar power supplies may be used as available: -- Variable d.c V 0 to 20V regulated d.c at 350mA. -- Fixed d.c V +5V regulated d.c at 1A. 15V regulated d.c at 300mA.

-- A.C supplies 5,10,15,20 or 25V rms; 50 or 60Hz supplied at 300mA (isolated from other supplies).Connecting Leads; See table overleaf.

Q’t’y PlugDia.

Length Color Reference

9 4mm 450mm

1 each of: - brown, red,

orange, yellow, green, blue,

grey, white and black.

4 2/4mm 450mm 2 each of: - red and black.

3 2mm 300mm Red.

8 2mm 150mm Orange.

8 2mm 150mm Yellow.

5 2mm 100mm Green.

Patching Connections; Connections which are shown “- - - - - -“in the patching diagrams for any given assignment in the text.

PRACTICAL ACTIVITY:

To gain familiarity with the use of the deck, construct the simple circuit of fig 1.2 by patching the deck as shown in fig 1.1.

0 - 20V dc100 R

2200 R

A0 V

3300 R

V

Fig 1.2

Switch on the power supply.Set the power supply control to give 10VCopy the results table as shown in fig.1.6, reproduced at the end of this assignment, and record the current and voltage measurements for each resistor.Calculate the percentage error of your measurements.

COMPONENTS KITS:

The types and quantities of components supplied in each particular kit are shown in the separate section at the front of the relevant manual. The components maybe recognized either from the information in that section, by the marking on the carrier or by the component tray in which they are stored. Further information is contained in the assignments dealing with specific components.It is very helpful to be able to identify quickly, fixed resistors and capacitors. The following information will help.

Resistors; Most fixed resistors (except certain high wattage or high precision types) are coded by a set of colored bands as shown in fig 1.3.

Fig 1.3

The colors used and their meanings are shown in the fig of 1.4.

COLORBAND 1Value of 1st digit

BAND 2Value of 2nd digit

BAND 3Number of

Zeros

BAND 4Tolerance

BLACK 0 0 0 --

BROWN 1 1 1 --

RED 2 2 2 --

ORANGE 3 3 3 --

YELLOW 4 4 4 --

GREEEN 5 5 5 --

BLUE 6 6 6 --

VIOLET 7 7 7 --

GREY 8 8 8 --

WHITE 9 9 9 --

GOLD -- -- -- 5%

SILVER -- -- -- 0%

NONE -- -- -- 20%

Fig 1.4

As an example, if the bands are:1 Yellow2 Violet3 Red4 Gold…

…the resistor is of value 4700 (4.7 ) and has tolerance of 5%.

Capacitors:

Capacitors are sometimes colored-coded but more usually they have the value printed on the body.There is a good deal of variation between makers in exact form of the markings but most use the normal units of picofarad (pF), nanofarad (nF) or the microfarad ( ).

Often the F is omitted and sometimes, when there is no chance of confusion, the other letter also.Example;0.47 could appear as 0.47 or just 0.47. This could not be 0.47pF, as it would

be too low a value for any fixed capacitor. It could not be 0.47nF either because such a value would be expressed as 470pF.The tolerance and voltage rating of a capacitor are also often printed on the body, again with or without units. Thus you could find;

1.0 160V DC WKG or just;

1.0/20/160Large value capacitors (greater than about 1 ) are usually electrolytic

types, which MUST be correctly polarized. Usually one of the terminals is marked with a+ or a- and the capacitor must be inserted so as to apply the correct polarity.

Coding of Resistor and Capacitor Values: Resistor and capacitor values are usually expressed in the form of the nit symbol ( for resistance, or F for capacitance) together with one of the multiplier

prefixes from the table of fig 1.5.

Prefix Name Multiplying Power

T tera x 1012

G giga x 109

M mega x 106

k kilo x 103

m milli x 10-3

u or μ micro x 10-6

n nano x 10-9

prefix pico x 10-12

Fig 1.5

Thus, for example 4700 ohms can also be expressed as 4.7 . Also 0.1 μF can

be expressed as 100nF.In this form the decimal point frequently occurs and it is easy for a value to be misread if the point is badly printed or omitted entirely by error. To avoid this difficulty an alternative coding method is often used.For resistors the symbols R, K, M, G and T (all capitals) are used to represent 1,

and ohms respectively and are placed in the decimal point

position together with at least two figures.The following are a few examples:

Value Code

0.1 ohm OR1

0.22 ohm OR22

2.7 ohm 2R7

100 ohm 100R

1.5k ohms 1K5

2.2M ohms 2M2

1G ohms 1GO

1.5T ohms 1T5

For capacitors the symbols p, n, u (or μ) and m are used to represent and

farad respectively.

Some examples of these are as follows:

Value Code1.5 pF 1p522 pF 22p1 nF 1n0

0.1 μF 0u1 or 100n2.2 μF 2u2 or 2μ2

1500 μF 1m5

Theoretical ValuesTheoretical Values Experimental Values

Experimental Values

Percentage Errors

Percentage Errors

Resistor( )

Current( )

Voltage( )

Current( )

Voltage( )

Error inCurrent Value

Error inVoltage Value

1000

2200

3300

470

Fig 1.5

THE SEMICONDUCTOR DIODE

OBJECTIVES:

1. Ability to recognize diodes in various physical forms.2. Ability to determine the diode polarity and to understand the need for

correct connection.3. To obtain knowledge of the forward voltage/current characteristic and the

conduction voltage for germanium and silicon types.

EQUIPMENT REQUIRED:

Qty Apparatus1 Electricity & Electronic Constructor, EEC4701 Basic Electronics, Kit 2, EEC4721 Power supply unit 0 to 20V variable d.c, regulated (e.g. Feedback

Power Supply PS445)2 Multimeters, or1 5V d.c voltmeter and 1 High impedance (20 ) 0-3V d.c voltmeter.

PREREQUISITE ASSIGNMENTS:

Assignment 1

KNOWLEDGE LEVEL:

Before working this assignment you should:• Know the operation of series d.c circuits.

INTRODUCTION:

A Semiconductor Junction Diode (or just diode) is made from a piece of P-type and a piece of N-type semiconductor joined together. See fig2.1.

Fig 2.1 Function Diode

If a voltage (potential difference) is applied across the two terminals, the Diode will conduct electricity. The amount of current that flows depends upon the size and polarity of the applied voltage.The Diode is represented in circuits by the symbol shown in fig 2.2.

Cathode (K)N - type

Anode (A)P - type

Fig 2.2 Diode Symbol

Find and examine the two Diode supplied in the kit. They should appear as in fig 2.3.

Fig 2.3 Two Types of Diode

The Diode with the metal body (type 6F60) can handle larger currents and power than the other one.

EXPERIMENTAL PROCEDURE:

Determining Diode Polarity:

Construct the circuit of fig 2.4. Note that the resistor limits the current to safe value.

A

+ 10V dc

0V

V

AK

BYW54

10mAd.c.

4k7

Fig 2.4 Diode Test Circuit

Switch on the power supply.Set the power supply control to give 10V on the meter.Copy the result table as shown in fig.2.5, reproduced at the end of this assignment, and record the current measurement in the first row of the table.Now, switch off the power supply and reverse the BYW54 diode to give the circuit of fig 2.6.

A

+ 10V dc

0V

V

AK

BYW54

1mAd.c.

Fig.2.6 Test Circuit

Switch on the power supply and readjust the voltage to 10V.Read the new value of diode current and record it in the second row of your table.Study your results and answer the questions on the next page.

QUESTIONS:

1. Which side of a diode should be connected to the positive voltage supply to make it conduct current?

2. When the diode was connected the opposite way round was the current?

a) slightly smallerb) much smaller orc) too small to measure

Forward /Reverse Biased Connections:

When a diode is connected so as to conduct it is FORWARD BIASED. When a diode is connected so as NOT to conduct it is REVERSE BIASED.Fig 2.7 shows the two methods of connecting diodes.

AK

+

-

AK

+

- a) Forward Biased b) Reverse Biased

Fig 2.7 Diode Bias

Knowledge of the conduction characteristic when a diode is forward biased is very important and is the subject of the rest of this of this Assignment.

The Characteristics of Forward Biased Diodes:

Construct the circuit of fig 2.8. The 2.2 potentiometer will provide fine control

over the applied voltage.

If

V

V

2k2

+5V

0V

Vs

Vr

BYW54Vd

100R

Fig 2.8 Test Circuit

NOTE: And

Copy the results table as shown in fig 2.9, reproduced at the end of this Assignment, for your results.Turn the potentiometer to zero; fully clockwise.Switch on the power supply and adjust it to supply 5V.Adjust the potentiometer to give a voltage of 1V on the voltmeter showing .

Now use the power supply variable control to set to:

0, 0.1V, 0.2V, etc, up to 1.0V.

Note for each setting and enter it in your table.

Now, with the power supply variable control set to supply 20V, use the potentiometer to set to:

1.5V, 2.0V, 2.5V and 3.0V.Again enter the values of in your table.

Calculate and as shown in fig 2.9 and enter these also in the table.

Plot versus on the axes of the graph.

QUESTIONS:

3. At what approximate value of does the current begin to

rise noticeably?4. Does raise much above this value for larger values of ?

PRACTICAL CONSIDERATIONS AND APPLICATIONS:

Both BYW54 and 6F60 diodes are made of Silicon and the forward conduction voltage of about 0.6V is typical of silicon junctions. Also typical of silicon diodes is the very small reverse current. Some diodes are made of Germanium and these have a smaller conduction voltage of about 0.2V but they also pass greater reverse currents. The 6F60 diode passes a greater reverse current than BYW54. This is because 6F60 is designed for much larger forward currents – up to 6A average. At the low voltage used in this experiment the reverse amount will still be very small. Diodes can withstand high reverse voltage but will eventually break down at some voltage and may be irreparably damaged. Type 6F60 can take the higher voltage of 600V compared with 150V for BYW54. Diodes have very many applications at many different powers, voltage and current levels. A very important application is the production of direct voltage from alternating voltage and this is dealt with in Assignment 3 and 4 which cover Rectification.

SUMMARY: In this assignment you have learnt that:

1. A diode conducts when its anode is positive relative to its cathode, and does not conduct when the voltage is reversed.

2. Diodes have different shapes and sizes according to their voltage, current and power ratings.

3. Silicon diodes have a conduction voltage of about 0.6V whereas Germanium diodes have one of about 0.2V.

4. The forward characteristic of a diode is not a straight line trough zero but looks like fig.2.10.

Fig 2.10 Typical Silicon Diode – Forward Characteristic

EXERCISE:

Construct the circuit of 2.11 and apply the method used earlier in this Assignment to find and .

If

V

V

2k2

+5V

0V

Vs

Vr

BYW54Vd

100R

Fig 2.11 Diode Power Dissipation

Diode power dissipation=

Calculate the power dissipation of the diode, and check to see if the it becomes warm to the touch after a few minutes operation.

FURTHER READING:

If you wish to know more about the physics of semiconductor diode, read Appendix A. this is not, however, essential.!

Circuit Current(mA)

Fig. 2.4

Fig. 2.6

Fig 2.5

VS(V)

Vr(V)

Vd = Vr - Vr(V)

If = 10 Vr(mA)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.5

2.0

2.5

3.0

Fig 2.9

! ! ! ! HALF-WAVE RECTIFIERSOBJECTIVES:

Because of their ability to conduct current in one direction and block current in the other direction, diodes are used in circuits called rectifiers that convert ac voltage into dc voltage. Rectifiers are found in all dc power supplies that operate from an ac voltage source. A power supply is an essential part of each electronic system from the simplest to the most complex. In this experiment, you will study the most basic type of rectifier, the half-wave rectifier. After completing this experiment, you should be able to;

• Explain and analyze the operation of half-wave rectifiers• Describe a basic dc power supply and half-wave rectification• Determine the average value of a half-wave rectified voltage• Discuss the effect of barrier potential on a half-wave rectifier output• To understand the effect of a reservoir capacitor upon the rectified waveform and

its mean value.EQUIPMENT REQUIRED:

Qty Apparatus1 Electricity & Electronics Constructors EEC470

1 Basic Electronics Kit EEC4721 Power supply unit. A.C supply; 20Vrms; 50 or 60Hz. (Isolated from other supplies)

1 Multimeter or 50V d.c. voltmeter.1 Oscilloscope.

PREREQUISITE ASSIGNMENTS: Assignment 2.

KNOWLEDGE LEVEL:

Before working this assignment you should:• Know the operation of transformers• Know the operation of series and parallel ac circuits.• Know how to use an oscilloscope.

INTRODUCTION:

In assignment 2 you found that a diode conducts current in one direction (from anode to cathode) but not in the reverse direction. A widely used application of this feature is the conversion of alternating voltages to direct voltages (fig 3.1). This assignment studies the simplest circuits for achieving this conversion, Which is called RECTIFICATION or in some cases, DETECTION.

0 V

A Sinusoidal Alternating Voltage

V

0 V

A Direct Voltage

Fig.3.1

Average Value of the Half-Wave Output Voltage:

The average value of the half-wave rectified output voltage is the value you would measure on a dc voltmeter. Mathematically, it is determined by finding the area under the curve over a full cycle, as illustrated in fig 3.4, and then dividing by 2Pi, the number of radians in a full cycle. The result of this is expressed in Equation 3.1, where Vp is the peak value of the voltage. This equation shows that Vavg is approximately 31.8% of Vp for a half-wave rectified voltage.

Equation: 3.1 Vavg = Vp /

EXPERIMENTAL PROCEDURE SIMPLE HALF-WAVE RECTIFICATION:

As shown in the patching of fig 3.2, construct the circuit of 3.3.

V10 kΩ

BYW54

Vp (in)

Fig 3.3 Half-wave Rectification

Switch on the oscilloscope and the sinusoidal supply. With the oscilloscope d.c. coupled adjust the time base and the Y amplifier sensitivity to obtain a steady trace of about 4 cm vertical and 5 ms/cm horizontal. You should see a waveform as in fig 3.4.

T

Vpk

Fig 3.4 half-wave Rectified waveform

a) Observe the Half-wave rectified waveform together with input voltage signal and draw it to a graph paper.(for diode BYW54)

Measure and record time T and peak voltage Vpk.

Sketch the waveform and label it to show the periods when the diode is conducting and those when it is not. Time t depends upon the frequency of your power supply. For a 50Hz supply it should be 20 ms and for 60Hz it should be 17 ms. Confirm this. Vpk should be nearly equal to the peak voltage of the alternating supply:

b) By using a d.c. voltmeter determine the average voltage (Vav) and compare with Vp(in)/

QUESTIONS: 1. Why will Vpk not be exactly equal to this voltage? 2. How much will it differ? 3. The mean voltage you obtain is positive relative to zero. How could you obtain negative voltage? Confirm your answer by experiment.

The effect of a Reservoir Capacitor:

Very often when rectifying an alternating voltage, we wish to produce a steady direct voltage free from variations of the sort observed in fig 3.4. one way of doing this is to connect a capacitor in parallel with the load resistor as in fig 3.5

10 kΩ

BYW54

Vp (in) C

Fig: 3.5 Half-Wave Rectifiers with Reservoir Capacitor

0 V

C charges

Diode cuts off C discharges

Waveform with Capacitor

Waveform without Capacitor

Fig: 3.6 The Effect of a Reservoir Capacitor

The capacitor C (usually called the reservoir capacitor) becomes charged-up by the current trough the diode during the positive half-cycle. Then, when the supply voltage starts to reduce again, the capacitor keeps the output voltage high and the diode cuts off. Capacitor C then discharges trough R until the next positive half-cycle occurs.Now, add a capacitor of 1uF to your circuit.Observe the output waveform on the oscilloscope and note the value of the peak-to-peak variations in voltage. Note also the new mean voltage on the voltmeter.a) Observe the input and output voltage signals and draw them to graph paper for capacitor C=100nF, and C=1µF (Note that Vac far each capacitor)

For C=100nF, R=10kΩ;" " " " For C=1 µF, R=10kΩ;" " " " " " "

"

Vpp(in)=12V Vp(rec)=?f(Hz) Vdc500750

10001250150017502000

"

Vpp(in)=12V Vp(rec)=?f(Hz) Vdc4080

120160200240

QUESTION:

4. Is the new mean voltage greater or less than it was before?

Now replace the 1μF capacitor by a much larger value of 22 μF, making sure to connect the + side of the capacitor to the diode cathode(the capacitor is electrolytic and MUST be connected inn the correct polarity) and answer the following questions.

QUESTIONS:

5. The variations on the rectified waveform are called RIPPLE. Is the ripple now less or more than it was with the lower value capacitor? 6. Is the mean rectified voltage now greater or less?

PRACTICAL CONSIDERATIONS AND APPLICATIONS:

When rectification is used to provide a direct voltage power supply from an alternating source, the ripple is an undesirable feature. For a given capacitor value, a greater load current (smaller or resistor) discharges the capacitor more and so increases the ripple obtained fig 3.7 shows this.

0 V

Larger load current Small load current

Fig 3.7 the Effect of Load Current

Several methods are available to reduce ripple:1. Larger capacitors, the uses of which are limited due to cost and size, and

also because large capacitors can require very large charging currents to be supplied trough the diode.

2. Electronic stabilization this reduces ripple as well as keeping the output voltage steady when the load or input voltage changes.

3. Full-wave rectification. With this, the ripple is much reduced as every half-cycle of the input, instead of every other half-cycle, contributes to the rectified output.

In figure 3.8 it can be seen that capacitor charging occurs at half the previous interval and the amount of discharge for a given load current is therefore less.

0 V

Fig 3.8 Full-wave Rectification

Assignment 4 deals with methods of achieving full-wave rectification.When diodes are used for detection purposes in the reception of modulated radio signals, quite different considerations apply. These cannot be discussed in detail here but Feedback’s manual ACS2956, Analogue Communications Systems, will provide full information on this application.

SUMMARY:

In this assignment you learnt that:1. A simple diode circuit can convert an alternating voltage to a direct voltage.2. The mean value of the rectified voltage can be increased by using a reservoir

capacitor across the load.

3. A half-wave rectified voltage gives appreciable ripple which however, can be reduced by several means.

EXERCISE:

A half-wave rectifier, as in fig 3.9, produces a certain amplitude (from peak-to peak ) of ripple.

R

D

Vp (in) C V(out)

Fig 3.9 Exercise circuit

If the load resistor is reduced to half of its original value, what increase in capacitor value will restore the ripple to the same values as before?

Confirm your answer by practical experiment, starting with:

R= 10kΩ and C=22µF, D (6F60)

FULL - WAVE RECTIFICATION

OBJECTIVES:

1. Ability to recognize a full-wave rectified waveform, with and without a reservoir capacitor.

2. Understand the working of diode bridge circuit as a full-wave rectifier and its advantage over half – wave rectification.

3. Awareness of the two – diode method of obtaining full – wave rectifications.

EQUIPMENT REQUIRED:

Qty Apparatus 1 Electricity & Electronics Constructor EEC4701 Basic Electronic Kit EEC4721 Power supply unit. A.c. supply; 20V rms; 50 or 60Hz. (isolated from

other supplies). (e.g. Feedback Power Supply PS445)1 Multimeter or1 Voltmeter 50V d.c.1 Oscilloscope.

PREREQUISETIE ASSIGNMENT:

Assignment 2

KNOWLEDGE LEVEL:

Before working this assignment you should:• Know the operation of series and parallel a.c. circuits.• Know how to use an oscilloscope.

INTRODUCTION:

At the end of Assignment 3 we discussed ways of reducing the ripple or voltage variation on a rectified direct voltage. One of these was to use every half – cycle of the input voltage instead of every other half – cycle.

A circuit which allows us to do this is shown in fig 4.1, and is known as the DIODE BRIDGE.

Load

D1

D3D2

D4

A

B

Fig 4.1 A Diode Bridge Rectifier

During the positive half – cycle of the supply “A” is more positive than “B”. Diodes D1 and D2 therefore conduct while diodes D3 and D4 are reverse – biased. The current flows as shown in fig 4.2.

Fig 4.2 Positive Half – Cycle Fig 4.3 Negative Half –Cycle

Load

D1

D3D2

D4

+

-

D1

D3D2

D4

-

+

During the negative half – cycle the current flow is as represented by fig 4.3. ın each case the current in the load is in the same direction.

EXPERIMENTAL PROCEDURE:

A Bridge Rectifier with Resistive Load:

Select the Bridge Rectifier from the component kit. It appears as in fig 4.4a and fig 4.4b in the circuit, showing the rectifier terminals are labeled.

Fig 4.4 Bridge Rectifier

The terminals labeled + and – are so called because these are the polarities that will exist across the load.Construct the circuit of fig 4.5.

10kΩ V

OscilloscopeY input

Oscilloscopecommon

AC supply 20 V rms

Fig 4.5 Test Circuit

With the oscilloscope d.c. coupled, adjust the controls to obtain a steady trace of about 4cm vertical and 5ms/cm horizontal. You should observe a waveform as in fig 4.6. time “T” will be 10ms for 50Hz supply, and 8.5ms for 60Hz.

Vpk

0V T

Fig 4.6 Full –wave Rectified Waveform

Note the value of Vpk and also the mean value of output voltage indicated on the voltmeter. Compare these figures with those obtained in Assignment 3.

QUESTIONS:

1. Should Vpk be the same as it was for a half –wave rectifier? Does your observation confirm your answer?

2. How does the mean value compare with that found for half – wave rectification?

As the mean value of a half – cycle of sine wave is , and every half – cycle is

present, this should be the mean value measured. Confirm this from your readings.

The Effect of a Reservoir Capacitor:

Add a 1μF capacitor in parallel with the load resistor and note the new mean value and the peak – to – peak ripple amplitude of the rectified waveform. Compare these figures with those obtained in Assignment 3 for the same load and capacitor values.

PRACTICAL CONSIDERATIONS AND APPLICATIONS:

The alternating input voltage to a rectifier is usually obtained from the main supply trough a transformer, for two reasons:

1. To obtain the desired voltage by choice of the transformer ratio.2. To provide isolation from the main supply for safety reasons.

Fig 4.7 shows such an arrangement with a bridge rectifier.

OscilloscopeY input

Oscilloscopecommon

line

neutral

main supply

rectified output

Fig 4.7 Transformer – fed Bridge Rectifier

In this figure, although the load current is always in one direction, the current in the transformer secondary is alternating.

Fig 4.8 shows another method of full –wave rectification, using a centre – tapped transformer winding and two diodes.

Fig 4.8 Full – wave Rectifier using Two Diodes

The arrow show how current flows on alternate half – cycles. The value of the output waveform is exactly the same as that for a bridge circuit provided each half of the transformer windings has the same rms voltage as the whole of the winding in fig 4.7.

The circuit saves two diodes, but increases the cost of the transformer. In fig 4.8 each half – secondary winding must have the same voltage rating as the single secondary of fig 4.7. Suppose the half – secondaries were wound with wire of half the cross – sectional area, so as to fit the two into the same space as the one secondary of fig 4.7, and use the same amount of copper. Each half – secondary would then have twice the resistance.

The current flows in each half – secondary only on alternative half – cycles, but

would generate twice the loss in the active cycle.

Each half – secondary would thus develop as much heat as the single secondary of fig 4.7, i.e. twice as much for both. A larger transformer would therefore be required to avoid excessive heating. Its greater cost would usually outweigh the cost of the two diodes saved.In full – wave rectification the basic repetition rate of the ripple is twice that of the supply (e.g. 100Hz for a 50Hz supply). In half – wave the frequency is the same as the supply frequency. This is often useful as an indication that one half of a bridge or full –wave rectifier is faulty.

SUMMARY:

In this assignment you have learnt that:1. A bridge full – wave rectifier gives a greater mean value and

fewer ripples for a given load and reservoir capacitor than a half – wave rectifier.

2. The alternative full – wave circuit using a centre – tapped transformer and two diodes is less efficient than the bridge circuit because it requires a bigger transformer for a given output power.

EXERCISE:

Fig 4.9 shows the discharge curve for a reservoir capacitor in half – wave and full – wave rectification, for the same load and capacitor values.

Fig 4.9

A capacitor discharges into a resistor in an exponential fashion, which is with a rate of discharge that reduces as the discharge progresses.With this mind, would you expect the peak – to – peak ripple in full – wave to be?

a) That in half – wave

b) Less than

c) More than

d) Explain your answer and confirm it by reference to measurements made in Assignment 3 and 4 for similar load conditions.

THE ZENER DIODE

OBJECTIVES:

1 Ability to recognize zener diodes in various physical forms and to distinguish them from rectifying diodes.

2 Understand the constant-voltage characteristic of a reverse –biased zener diode.

3 Understand the use of a zener diode in a simple voltage regulator circuit.

EQUIPMENT REQUIRED: Qty Apparatus:

1 Electricity& Electronics constructor EEC470.1 Basic Electronics Kit EEC472.1 Power supply unit 0 to 20 V variable d.c, regulated. (e.g Feedback

Supply PS445)3 Multimeter, or 1 Voltmeter, 20V d.c. and 1 Ammeter, 100mA d.c and1 Ammeter, 1a d.c.

PREREQUISITE ASSINGMENTS: Assignment 2.

KNOWLEDGE LEVEL:

Before working this assignment you should:• Know what is meant by internal resistance and the effect it has on terminal

voltage.

INTRODUCTION:

In Assignment 2 you found that a reverse-biased diode passes neglect able current. You also learnt that it will eventually suffer breakdown and damage if the reverse voltage is made too high. See fig 5.1.

Fig 5.1: Reverse Breakdown of a Diode

Zener diodes are specially constructed to break down at controllable voltages and to do so without damage to this device. As we shall see, this feature can be put to good use.Two Zener diodes are contained in the EEC472 Kit. They are types BZY95C10 and BZY88C7V5 and are shown in fig 5.2 with the standard circuit symbol.

Fig 5.2 Zener Diodes and Symbol

Zener diodes look very similar to rectifier diodes and terminal names and identification methods are the same. The larger types, such as BZY95, have greater power and current capacities.The two types of diode can usually be distinguished only by their type numbers. For Zener diodes these often, but not always, contain the letter Z.

EXPERIMENTAL PROCEDURE:

The Zener Diode Reverse Characteristics:

Construct the circuit of fig 5.4.

Fig 5.4 Test Circuit

The method of obtaining the voltage-current characteristic is the similar to that of Assignment 2 but notice that the Zener diode is reverse-biased. Using the power supply variable control, set Vs to the values given in fig 5.5.For each value record Vr, then calculate:

and

Copy the results table as shown in fig 5.5, reproduced at the end of this assignment, and enter your results. Prepare a graph like fig 5.6 and plot against .

Fig 5.6 The Zener Diode Characteristic

QUESTIONS:

1. Describe the characteristic in your own words.2. The nominal voltage of the BZY887V5 is 7.5V. Does your graph

agree with this exactly? If not, can you suggest a reason for any difference?

3. Why is the series resistor in fig 5.4 necessary?Calculate the power dissipated in the diode for each value of and , and enter it

into the last column of your table. = x mW

Plot against on your graph.

4. The maximum allowable power dissipation of type BZY88 is 400mW. Does your maximum value of approach this limit?

A Simple Zener Diode Voltage Regulator:

The Zener diode has a region in its reverse characteristic of almost constant voltage regardless of the current trough the diode. This can be used to regulator or stabilize a voltage source against supply or load variations.

Fig 5.7 shows an unregulated voltage source supplying current to a variable load.

Fig 5.7 An Unregulated Power Source

If either Vs or changes, so will the voltage across the load .

One way of keeping this voltage more constant is to connect across the load a Zener diode whose breakdown voltage is the desired constant voltage. Fig 5.8 shows practical circuit of this kind.

Fig 5.8 A Simple Zener Diode Regulator

The object of this practical exercise is to discover:1. How much variation of Vs can be tolerated?2. How much variation of can be tolerated?

As shown in the patching diagram of fig 5.9, construct the circuit of fig 5.8.Copy the results table as shown in fig 5.10 reproduced at the end of this assignment, for your results.Now, remove the potentiometer temporarily to make =0, and then slowly increase

Vs until the diode just begins to conduct current say 1mA. Record Vs and =0 in the

first row of the table.Now set the potentiometer to maximum (clockwise) and replace it in the circuit.The extra current drawn by will reduce the diode current below 1 mA.

Increase Vs to 12V; the diode current will increase above 1mA. Then adjust until

the diode current just returns to approximately 1mA. Again record Vs and .

Repeat this for Vs=14, 16, 18 and 20V.Prepare a graph like fig5.11 and plot Vs versus .

Fig 5.11

Every point on your graph represents a condition where the Zener diode has only just reached its breakdown voltage.

Thus, for a given , lower value of Vs will take the diode out of breakdown and, for a

given Vs, a higher value of will the same.

Therefore the whole of one side of your graph is an area where the diode is not in breakdown and thus is not holding constant. Mark this area on your graph.

NOTE:If you cannot do this look at the fig 5.15 of this assignment for help.

You have now found what maximum load current and minimum supply voltage can be used without the load voltage falling below the Zener value.But what sets a limit to the minimum load current and the maximum supply voltage?The answer is the power dissipation allowable in the Zener diode.The maximum power dissipation allowed for BZY95C10 is 1.5 W. But = x

And = 10V… … (Almost constant)

Thus:

=1.5/10 A = 150mA

(Provided Id never goes higher than this the power limit will not be exceeded).

We use the same circuit but set the Id meter to the 1A d.c range. Copy the results table as shown in fig 5.12, reproduced at the end of this assignment, for your results.Now start by reducing to minimum (anti-clockwise) and then set Vs to 20V and

increase until the diode current Id reads 150mA. Read the load current and

record it against Vs=20V in the table. Reduce Vs in steps of 1V, each time resetting to give =150mA approximately and recording . Continue until it is no longer

possible to set Id to 150mA. Plot Vs versus on the same axes used for your

previous graph. Shade on your graph the area that now represents the useable range of Vs and .

PRACTICAL CONSIDERATIONS AND APPLICATIONS:

Zener diodes are widely used as voltage stabilizers and voltage references.They are manufactured with Zener voltage ratings of between about 2.7V and 200V, usually in ‘preferred’ voltage steps, e.g. 2.7,3.4,3.6,3.9,4.3,4.7,5.1 etc, just as 1 percent tolerance resistors are manufactured. The power dissipation rating of a Zener diode is an important parameter. Zener diodes are manufactured with power ratings between 400mW and in excess of 100 W. Although Zener Voltages are fairly insensitive to changes in diode current, they are however sensitive to temperature changes. Normally the Zener voltage is specified at a temperature of 25˚C, but diodes will have a temperature coefficient. Typical figures for this range from -5.0 mV/ for a 2.7 - volt device to +60 mV / for a 75 –volt device. The zero temperature coefficients is given around 5.6 volts Zener voltage. Zener diodes have many uses other than for providing stable or reference voltage sources (e.g. they can be used for clipping), thereby doing away with the need for a voltage source as the clipping reference. Zener diodes are often used for over-voltage protection, being connected across the load. The Zener voltage is chosen such that under normal operating conditions the diode is reverse-biased below the Zener voltage, so the device acts as an ordinary diode (i.e. non-conducting). If however the voltage rises above the Zener voltage, the diode will break down and pass a heavy current. The excess voltage may be dropped in a resistor, as in fig 5.13(a), or the fuse will blow, as in fig 5.13(b).

Fig 5.13 A Zener Diode used as Load Protection

SUMMARY: In this assignment you learnt that:

1. A Zener diode in its breakdown region has an almost constant voltage regardless of diode current.

2. This feature can be used to stabilize a varying voltage.3. There are limits on the variations of Vs and in a simple

stabilizer. These are: The need to keep the diode in the Zener region The need to keep the diode power dissipation below the

allowable maximum.EXERCISE:

Refer to your graph and decide which of the following combinations of Vs and (supply voltage and load current) are permissible in the circuit of fig 5.8.

VS(V)

IL (mA)

1 12 3002 19 2503 19 504 15 755 14 200

A SIMPLE ZENER DIODE VOLTAGE REGULATOR:

Fig 5.14 is the sort of graph that should be obtained.

Fig 5.14 Allowable Operating ConditionsEXERCISE:

Conditions 2 and 4 are permissible. Conditions 1 and 5 will give an output voltage below the Zener value. Conditions 3 will exceed the allowable Zener dissipation.

VS(V)

Vr(V)

Vd=VS-Vr (V)

Id = Vr (mA)

Pd = Vd x Id (mW)

0246

6.57

7.58

8.59

101520

Fig 5.5

VS(V)

IL (mA)

Id = 1 mA fixed

7.4 0

Id = 1 mA fixed10

Id = 1 mA fixed11 Id = 1 mA fixed

12

Id = 1 mA fixed

13

Id = 1 mA fixed

14

Id = 1 mA fixed

Fig 5.10

VS(V)

IL (mA)

All readings are for Idjust equal to 50 mA

16All readings are for Idjust equal to 50 mA

15 All readings are for Idjust equal to 50 mA14

All readings are for Idjust equal to 50 mA

13

All readings are for Idjust equal to 50 mA

12

All readings are for Idjust equal to 50 mA

11

All readings are for Idjust equal to 50 mA

Fig 5.12

TRANSISTOR FAMILIARIZATION

OBJECTIVES:

1. Ability to recognize transistors in various physical forms and to identify their terminals.

2. Understanding of the basic construction of PNP and NPN transistors.3. Understanding of junction biasing and the direction and magnitude of current

flows.

EQUIPMENT REQUIRED:

Qty Apparatus1 Electricity & Electronics Constructor EEC470.1 Basic Electronics kit EEC472.1 Power supply unit +5V and -15V variable d.c., regulated. (e.g. Feedback Power Supply PS445)2 Multimeter, or 1 Micro-ammeter, 100uA d.c. and

PREREQUISITE ASSIGNMENTS:! Assignment 2

KNOWLEDGE LEVEL:

Before working this assignment you should:

. Know the operation of parallel d.c.circuits.

INTRODUCTION:Transistors are three- terminal devices constructed in the form of two semiconductor junctions, rather like two junction diodes. Fig 6.1 shows the two types, NPN and PNP, governed by the physical arrangement of the P- and N- type semiconductor materials.

PNP TRANSISTOR NPN TRANSISTOR

Construction

Two-diodeanalogy

Symbol

PNP Transistor

E C

BNPN Transistor

B

CE

Fig 6.1 Two Types of Transistor

Each of the PN junctions in this diagram behaves individually like the simple diode you studied in Assignment 2, but when joined together in this way, the behavior is very different. In normal use the EMITTER-BASE diode is forward –biased and behaves almost exactly like an independent diode. The COLLECTOR-BASE diode, however, is reverse-biased and normally you would expect if to pass no current. But if the E-B diode is conducting forward current, this influences the reverse-biased C-B diode and causes it to pass almost as much reverse current

Fig 6.2 shows this for PNP and NPN types. The small difference current flows in the base circuit.

IE

IB IC

E

B

CNPN

EIE

IB IC

B

CPNP

B

CE E C

B

IB

IEIC IC

IB

IE

Is slightly less than ;

And is much less than or

Fig 6.2 Transistor current Flow

The ratio is usually called .

Because is almost as big as , is nearly 1.

= = nearly 1(e.g. 0.99)

The ratio is usually called

Thus

and

If =0.99,

= 0.99/0.01

= 99.

It is this large ratio between and that makes the transistor a useful amplifying

device when connected so that is derived from an input and provides an

output. In the Assignment we shall first identify some actual transistors and then confirm the

directions and magnitudes of currents, finding and in the process.

EXPERIMENTAL PROCEDURE:

Transistor Identification:

Select from the EEC472 the transistors BC107 and BCY70. Fig 6.3 illustrates these types.

Fig 6.3 Typical Low-power Transistor (as supplied)

Transistors are made in many other physical forms. Fig 6.4 shows some other types you could have to recognize.

Fig 6.4 Different Transistor Styles

Make sure you can accurately identify the terminals of the transistors in the kit.

MEASUREMENT OF TRANSISTOR CURRENT:

Part I: (NPN Transistor) Construct the circuit of fig 6.5.

Fig 6.5 Transistor Test Circuit

This uses the BC107-NPN transistor. The capacitor is provided to ensure that the circuit is

stable and has no effect on your measurements.Fill the table 6.9 according to NPN transistor (BC107)Turn the potentiometer to zero (clockwise) and switch on both power supplies.

Slowly increase by turning the potentiometer anti-clockwise until just begin to flow.

Connect the voltmeter temporarily between E and B on the transistor (3V d.c. range) and

note the value of in the table.

Remove the meter. And then continue increasing until approximately equals from

1mA, 2mA, 3mA,...,10mA and record the values of and in the fig. 6.9.

Plot versus graph and calculate using this graph

Now increase until approximately equals 10mA; again record and and plot

versus graph then calculate using this graph.

Part II: (PNP Transistor)

Construct the circuit of fig 6.6.

Fig 6.6

This uses the BCY70-PNP transistor. The capacitor is provided to ensure that the circuit is stable and has no effect on your measurements.

Fill the table 6.9 according to PNP transistor (BCY70)

QUESTIONS:

1. Did your reading of confirm that the forward-biased E—B

junction is acting like a simple diode? Explain.

Satisfy yourself that both and are flowing in the directions shown in fig 6.6.

For each set of readings calculate , and as follows:

; ;

2. Do your results show that and ,

a) Increase;b) Decrease , or;

c) Stay constant as increase?

PRACTICAL CONSIDERATIONS AND APPLICATIONS:

The measurements you made on the BC107 transistor used a circuit in which the E and C terminals were biased with voltages relative to the base B. for this reason this circuit is called

a COMMON BASE connection. It is also possible to bias the junctions with voltages relative to the EMITTER or COLLECTOR, giving COMMON EMITTER and COMMON COLLECTOR

connections as in fig 6.7.

Fig 6.7 Bias Arrangements for an NPN Transistor

In common-emitter, must be larger than to ensure that the C-B junction remains

reverse-biased. In common-collector, must be larger than to ensure that the E-B

junction remains forward-biased. These three connections have important differences in their responses to inputs. The common-emitter and common-collector circuits are the most

important connections since the common-base is used only in special circumstances. As with diodes, transistors can be made from Germanium instead of Silicon, but these are rarely

used.

SUMMARY: In this assignment you have learnt that:

1. Transistor has two basic forms; the PNP and the NPN.

2. A transistor is similar to two diode junctions, one forward and one reverse-biased.

3. The base current is much smaller than either the emitter or collector current, which they nearly equal are.

4. There are three basic bias connections for a transistor.

EXERCISE:

Fig 6.8 shows an incomplete circuit of PNP transistor in common-emitter connection.

Fig 6.8

Complete the circuit with a suitable collector bias voltage and show the direction and size of

the collector current .

Also find and

FURTHER READING:

If you wish to know more about the physics of transistors read Appendix A.

IC (mA)

IB(mA)

justmeasurable ----

1

2

3

4

5

6

7

8

9

10

Fig 6.9

THE SILICON CONTROLLED RECTIFIER

OBJECTIVES:

1. Recognition of SCR’s in different physical forms.2. Understanding of the two-transistor analogy and the different ways of triggering

an SCR.3. Knowledge of the terms ‘BREAKDOWN VOLTAGE’ and ‘HOLDING CURRENT’.4. Appreciation of the areas of application of SCR’s.

EQUIPMENT REQUIRED:

Qty Apparatus1 Electricity & Electronics Constructor EEC470.1 Basic Electronics Kit EEC472.1 Power supply unit 0 to 20V variable d.c regulated and +15V d.c. regulated.

A.c. supply; 20V rms 50 or 60 Hz (isolated from other supplies). (e.g. Feedback Power Supply PS445)

2 Multimeter or 1 Voltmeter 5V d.c. and 1 Millimeter 10/100mA ac/d.c.

PREREQUISITE ASSIGNMENTS:

Assignments 2 and 6.

KNOWLEDGE LEVEL:

Before working this assignment you should:• Know the operation of series and parallel a.c. circuits.

INTRODUCTION:

The Silicon Controlled rectifier, or SCR, is one of several semiconductor devices which are capable of acting as fast switches for large currents. The general name for these devices is THYRISTOR. Fig 8.1 shows the SCR symbol and some of the physical forms in which it is found.

Fig 8.1 Types of SCR and Graphical symbol

The SCR resembles a rectifier diode but if the anode is held positive relative to the cathode no current flows until a positive current is injected into the gate. The diode then switches on and will not switch off until the anode- cathode voltage is removed. Hence the name CONTROLLED RECTIFIER. Fig 8.2 shows the SCR as a three-junction device.

Fig 8.2 The Two-transistor Analogy of an SCR.

This can be regarded as two inter-connected two-junction transistors, one PNP and the other NPN. If A is made positive relative to K, and G is left unconnected, no current will flow because each transistor gets its base-emitter current from the other’s collector emitter current. So, until the one of the transistor is given some base current, nothing can happen.If now a current is injected into the base of transistor ‘2’, the resulting collector current flows in the base of ‘1’. This in turn causes a collector current in ‘1’ which increases the base current of ‘2’ and so on. Very rapidly the two transistors force each other to conduct to saturation; the current being limited only by resistance in the external circuit.

If the anode-cathode voltage is now reduced; the current also reduces until it goes below some critical value, and the transistors switch off again. Just as rapidly.

If a reverse voltage is applied to the SCR (anode negative to cathode) it behaves very much like an ordinary diode. No current passes until at, some high voltage, it breaks down completely.Firstly we shall measure two important quantities of a forward-biased SCR.

EXPERIMENTAL PROCEDURE SCR Measurements:

Construct the circuit of fig 8.3.

Fig 8.3 Test Circuit for Trigger Current.

Set the variable d.c. volts to 20V. Turn the potentiometer to zero (clockwise) and switch on the supplies. Slowly rotate the potentiometer, observing the gate current meter continuously, until the lamp suddenly lights. Record the gate current at which this occurs. Switch off the supplies and return the potentiometer to zero. Copy the results table as shown in fig 8.4, reproduced at the end of this assignment, for your results.Repeat the measurement several times. To ensure that you have the correct value. What you have found is the TRIGGER CURRENT (I ). Enter it into your table. Now

trigger the SCR on again and measure the voltage from anode to cathode. This is the SATURATION VOLTAGE V (sat). Enter this in the table.

QUESTION:

1- Look back at fig 8.2. do you expect the saturation voltage to be greater or less than 0.6V? Explain.

Finally connect the milliammeter (on range 100mA) in series with the lamp as in fig 8.5). See also fig 8.3).

Fig 8.5 Test Circuit for Holding Current.

Trigger the SCR on. The meter reads the lamp current for a 20V supply voltage. Temporarily disconnect the gate connection. Slowly reduce the supply voltage until the SCR current suddenly falls to zero. Note the value of the current at which this occurs. Repeat this procedure several times to ensure that you have the correct value. What you found is the HOLDING CURRENT (I ). Enter this also into your

table.

NOTE To switch an SCR ‘ON’ the GATE CURRENT must by at least I .

To switch an ‘OFF’ the ANODE CURRENT must be at most I .

QUESTION:

2- What do you think will happen in the circuit of fig 8.3 if you trigger the SCR on, and then reduce the gate current to zero again? Confirm your answer by experiment.

USE of an ALTERNATING ANODE SUPPLY:

We have found that the anode current must be reduced to below I to switch the

SCR off. This is the only way of switching off. You CANNOT do it by reducing the gate current. If the anode supply is an alternating voltage it will go negative every half-cycle, reducing the anode current to zero. Alter your circuit to that of fig8.6. Remember to use a.c meters.

Fig 8.6 test Circuit with an A.C Supply

QUESTIONS:

3- What do you observe now when you repeatedly increase and decrease the gate current?

Explain what you think is happening.4- Why does the lamp burn less brightly than it did with the 20V d.c supply?

PRACTICAL CONSIDERATIONS AND APPLICATIONS:SCR’s can be triggered unintentionally in several ways and we have to be aware of these as they could be the cause of wrong operation. Fig 8.7 illustrates this.

Fig 8.7 False Triggering Mechanisms

In (a) a high temperature increases the leakage current of the two ‘transistors’ in the SCR. This is the small collector-emitter current that flows when there is zero base current. If it becomes too great it will be enough to initiate the trigger action. If a very high forward voltage is applied, as in (b), the ‘transistors’ can break down and this too will initiate triggering.Fig 8.7(c) shows a very rapidly rising anode-cathode voltage. Every transistor has some capacitance from collector to emitter as shown in fig 8.8. A fast-rising anode-cathode supply causes small currents in these capacitors and can act to cause triggering.

Fig 8.8 Stray Capacitance Positions.

Steps must always be taken in practice to avoid each of these possible false trigger mechanisms.SCR’s are available to carry currents from less than 1A up to 1000A or more. They therefore find use in the switching of heavy electrical equipment, where they replace contractors. The following advantages should be obvious:

No moving partsNo contact arcingNo bad contacts due to corrosion or dirt

In addition to simply switching currents on and off, SCR’s can be made to control the mean value of a load current without dissipating large amounts of power. In this application they can replace bulky high wattage rheostats and save electrical energy at the same time. A good example of this is the control of theatre lighting. The introduction to Assignment 10 explains how this can be done.

SUMMARY:

In this assignment you have learnt that:

1.SCR’s can be triggered on by a gate current but triggered off only by reducing the anode current.

2.SCR’s can be regarded as two interconnected transistors.3.SCR’s can be triggered unintentionally by high temperature, over-voltage or a

rapidly rising anode voltage.

EXERCISE:

The C220E in fig 8.9 has a maximum steady anode current capability of 10A. Given that V (sat) is 1.5V at this current, find.

(a) The power dissipation in the SCR(b) The value of R(c) The power dissipation in R(d) The efficiency of the SCR as a switch for a 10A anode current.

Fig 8.9

TRIGGER CURRENT (I ) mA

SATURATION VOLTAGE (V ) V

HOLDING CURRENT (I ) mA

Fig 8.4

TRIGGER DEVICES – THE DIAC AND UJT

OBJECTIVES:

1. Ability to recognize the devices and their symbols.2. Understanding of the need for trigger devices with thyristor.3. Appreciation of the main features of the DIAC and UJT.

EQUIPMENT REQUIRED: Qty Apparatus

1 Electricity & Electronics Constructor EEC4701 Basic Electronics Kit EEC4721 Power supply unit 0 to 20V variable d.c. regulated +15V d.c. regulated.

a.c. supply; 15V rms 50 or 60 Hz (isolated from other supplies. e.g. Feedback Power Supply PS445)

1 2-channel oscilloscope 1 Multimeter or 1 3-30V d.c. voltmeter.

PREREQUISITE ASSIGNMENTS:

Assignment 8 or 9

KNOWLEDGE LEVEL:

Before working this assignment you should:• Know how to use an oscilloscope.

INTRODUCTION:

In assignment 8 and 9 two semiconductor switches (or thyristors) were studied; the SCR and TRIAC. In ON-OFF switching applications they could be triggered by simple

circuits producing steady gate currents. Fig 10.1 reminds us of such a circuit using an SCR.

Fig 10.1 An SCR Circuit.

A thyristor will switch off only when its supply voltage falls to zero.If we wish to control the mean value of a load current, rather than just switch it on and off, we have only one method available. This is illustrated in fig 10.2 for an SCR.

Fig 10.2 SCR Conduction for Different Delay Times.

A steady gate current would allow conduction over the full period of the positive half-cycle. If instead, a short pulse of gate current is applied at the TRIGGER POINTS, conduction occurs over part of the half-cycle only. This reduces the mean current.

The mean current can be varied by changing the delay time T between the start of the cycle and the trigger. This is known as PHASE CONTROL. Fig 10.3 explains why.

Fig 10.3 Phase Control of an SCR.

The supply wave A is delayed by the phase shift to give B. When B reaches a certain TRIGGER LEVEL the trigger circuit generates a gate pulse C for the SCR.

To achieve phase control, then, two things are needed:a) A variable phase shift circuit (usually passive components such as

resistors and capacitors).b) A trigger circuit that can produce a pulse when the delayed waveform

reaches a certain level.In this assignment we look at two devices which serve as trigger generators, the DIAC (Diode A.C. switch) and the UJT ( Uni-Junction Transistor).

EXPERIMENTAL PROCEDURE:

The DIAC;Find the DIAC-type ST2 in your EEC472 kit. It should appear as in fig 10.4, which also shows the symbol and the voltage-current characteristics.

Fig 10.4 The DIAC, its Graphical Symbol and Characteristics

The DIAC is made like a transistor but has no base connection. When a voltage greater than V is applied, breakdown occurs. In an ordinary diode the voltage

would then remain constant as the current increased. In the DIAC the transistor action causes the voltage to reduce as the current increases. This gives the characteristic a negative resistance, as shown in fig 10.4. The DIAC is symmetrical and therefore has the same characteristic for negative voltages. It is the negative resistance that makes the DIAC suitable as a trigger for an SCR or TRIAC.To test this, construct the circuit of fig 10.5 to the patching diagram of fig 10.6.

Fig 10.5 The DIAC Test Circuit

Set the variable d.c. supply to zero and switch on the supplies.

Slowly increase the variable d.c. voltage until the waveform at Y2 suddenly appears. That is, the DIAC ‘switches on’. Notice the very rapid rise of V , produced by the negative resistance. See fig 10.6.

Fig 10.6 DIAC Waveforms

Measure V and V from the oscilloscope.

V is the DIAC break over voltage.

V is the load voltage.

is the DIAC current immediately after switch-on.

V - V is the voltage across the DIAC immediately after switch on.

From these figures it is possible to construct an approximate characteristic for the DIAC as in fig 10.7. In fig 10.7 P represents the conditions just before switch-on, and Q the conditions just after. The values shown are not necessarily the correct ones. Construct the fig 10.13, and draw a graph like fig 10.7.

Fig 10.7 The DIAC Characteristic and Load Line

THE TRIACOBJECTIVES:

1. Recognition of a TRIAC device2. Understanding of the bidirectional nature of the TRIAC and its areas of application.3. Appreciation of the different behavior of the device in the four operating applications.

EQUIPMENT REQUIRED: Qty Apparatus

1 Electricity & Electronics constructor EEC4701 Basic Electronics Kit EEC4721 Power supply unit +5V d.c. regulated and +15V d.c. regulated.

A.c. supply; 15Vrms 50 or 60 Hz (isolated from other supplies. (e.g. Feedback Power Supply PS445)). 1 Multimeter or 1 Milliammeter 50mA d.c. PREREQUISITE ASSIGNMENTS:

Assignment 8

KNOWLEDGE LEVEL:

Before working this assignment you should:• Know how to use Cartesian axes involving four quadrants.

INTRODUCTION:

In assignment 8 we learnt that the SCR can be used to switch a unidirectional current but it will not conduct in reverse.

An alternating supply is often necessary to ensure that the SCR will switch off. It is rather inefficient, because it conducts only every other half-cycle (like a half-wave rectifier as described in Assignment 2).

Fig 9.1 shows the basic single-SCR circuit and also one way of using four SCR’s in a bridge to achieve controlled full-wave rectification.

Fig 9.1 Half-wave and Bridge SCR Circuits

Note that, in a bridge circuit, gates G1/G2 are triggered together on one half-wave and G3/G4 on the next.

A simpler and less expensive way of obtaining bidirectional conduction is to use a TRIAC. Fig 9.2 shows a typical device and its graphical symbol.

Fig 9.2 A TRIAC and its Graphical Symbol

A TRIAC, like an SCR, is a type of thyristor. Although it has the same basic four layers of semi-conductor materials, its detailed construction is too involved to be described here.

The behavior of the TRIAC is very similar to that of the SCR but it can be triggered into conduction by gate current for either polarity of the voltage between terminals T and T .

Fig 9.3 shows the four modes in which a TRIAC can be operated.

Fig 9.3 TRIAC Triggering Modes

With reference to Quadrant I, the TRIAC is usually triggered by a positive gate current (mode I +), but can be triggered by negative gate current, (mode I-). Similarly in Quadrant III negative I is usual (mode III-), but positive I is possible (mode

III+). Modes I- and III+ are, however, less sensitive than the usual modes, I+ and III-.

In the first Practical we shall confirm that a TRIAC can be triggered in any of the four modes.

EXPERIMENTAL PROCEDURE:

Triggering modes of the TRIAC;

Construct the circuit of fig 9.4.

Fig 9.4 Test Circuit for Trigger Current

Copy the results table as shown in fig 9.5, reproduced at the end of this assignment, for your results.

Set the potentiometer to zero (clockwise) and switch on the power supplies. Slowly increase the gate current until the lamp lights, noting the value of I when this occurs. Record this in

your table under I for mode I+.

Switch off and move link 1 to apply -15V to the gate circuit instead of +15V.

Repeat the measurement and record the value of I for mode I-, reversing the meter

connections if necessary.

Now move link 2 to apply -15V to the lamp. Repeat the measurement of I for mode III-.

Finally restore the gate supply to +15V and measure I for mode III+.

You should find that modes I+ and III- have similar values of I but that modes I- and III+

require comparably greater gate currents to cause triggering.

TRIAC’s, like SCR’s, require a minimum current, and called the HOLDING CURRENT, to keep them in conduction. You can confirm this if you wish for mode I+ by the same method used in Assignment 8.

A simple TRIAC Switch;

Construct the circuit of fig 9.6.

Fig 9.6 Simple TRIAC Switch Circuit.

If link A is not connected there is no gate current and the lamp does not light.If A is connected to point B, gate current flows at every half-cycle and the lamp lights.

QUESTIONS:

1 What do you expect to happen if A is connected to point C? Confirm by experiment and explain your answer.

2 Which mode or modes are in use for:‘A’ connected to ‘B’,‘A’ connected to ‘C’?

PRACTICAL CONSIDERATIONS AND APPLICATIONS:

TRIAC’s have the same breakdown voltage, temperature and rate of voltage rise limitations as SCR’s. They are available in a wide range of voltage and current ratings.

Apart from the bidirectional current capability and the need to consider the triggering modes, TRIAC’s are in most other respects similar to SCR’s in their range of applications.

SUMMARY:

In this assignment you have learnt that:

1 A TRIAC is a four-layer thyristor device similar to the SCR.2 The TRIAC can be triggered into conduction in either direction.3 There are four triggering modes, of which two are preferred.

EXERCISE:

Fig 9.7 shows a simple TRIAC switch circuit.

Fig 9.7 Simple TRIAC Switch Circuit.

Use your measured results for I and assume that the voltage from G to T will be about 1V.

Suggest a value for R which, when the switch is closed, will give:a) Half-wave operation of the lampb) Full-wave operation of the lamp

Confirm your answer experimentally.

MODE V I

I+I-III-III+

Fig 9.5

THE FIELD EFFECT TRANSISTOR

OBJECTIVES:

1. Understanding of the difference between bipolar and field-effect transistors.2. Ability to distinguish between JFET and MOSFET types and between N and P channel

construction.3. Be able to recognize the basic characteristics of a JFET.4. Know the principal advantages of FET’s and some applications.

EQUIPMENT REQUIRED:

Qty Apparatus1 Electricity & Electronics Constructor EEC4701 Basic Electronics Kit EEC4721 Power supply unit 0 to +20V variable d.c. and +- 15V d.c. regulated

(e.g. Feedback Power Supply PS445) 1 Oscilloscope 3 Multimeter or 1 Voltmeter, 3V d.c. and 1 Milliammeter, 50 A/10mA d.c. and

1 Voltmeter 50V d.c. 1 function Generator 2V peak to peak at 1 kHz (e.g. Feedback FG601)

PREREQUISITE ASSIGNMENTS:

Assignments 1, 6, 7

KNOWLEDGE LEVEL:

Before working this assignment you should:• Know how to use an oscilloscope.

INTRODUCTION:

Field-Effect Transistors (FET’s) are made in various forms. One type, the Junction FET (JFET) has a construction quite similar to the UJT (Assignment 10) but works in a different way.

Fig 11.1 JFETs-Constructions, Symbols, and Appearance.

As can be seen, the JFET has two forms; the N-channel and P-channel which are analogous to PNP and NPN in ordinary transistors. We shall look more closely at the N-channel type.Fig 11.2 shows an N-channel JFET and its bias voltages.

Fig 11.2 The Bias Arrangement for an N-channel JFET.

The CHANNEL is a resistive path trough which voltage V can drive a current I .

A voltage gradient is thus formed down the length of the channel, the voltage becoming less positive as we go from DRAIN to SOURCE. The PN junction thus has a high reverse bias at D and a lower reverse bias at S. This bias causes a ‘DEPLETION LAYER’, whose width increases with the bias.

Depletion means a reduction of available electrons to carry current. If V is made more negative,

the depletion layer increases in width at all points. The values of V and V both influence the

width of the depletion layer. This alters the effective channel resistance and hence I . Fig 11.3

shows this.

Fig 11.3 The Depletion Effect

As V increases negatively the channel is ‘squeezed’, reducing the current I . But the GATE-

CHANNEL junction is like a reverse-biased junction diode and thus carries only a very small current. I is controlled by V trough a ‘field effect’. Hence the name FET.

In the first Practical we shall see how V and V affect I .

EXPERIMENTAL PROCEDURE:

Characteristics of an N-Channel JFET:

Construct the circuit of fig 11.4.

Fig 11.4 Test Circuit for A Typical N-channel JFET.

Set the potentiometer anti-clockwise and the variable d.c. voltage to zero.Switch on the power supply. Now set V to the first value on the table.11.1, and read I for each value of V .

Then fill the table 11.1.

Plot I versus V graph for each value of V .

QUESTIONS: Study your graphs and answer the following questions.

1. Above which values of V is I almost unaffected by V when V = 0?

2. For a given value of V , (say 10V), do equal changes of V cause equal changes

of I ?

Understand what you answer implies.Now go back to your circuit and set V to 10V and V to -1.0V. Then alter the circuit to place the

ammeter in place of the link in the gate lead as in fig 11.5. And try to measure I .

Fig 11.5 Measuring Gate Current.

3. Can you now measure I or is it too small?

A JFET Amplifier:

An FET can be used to amplify signals in manner similar to a transistor in common-emitter connection. In this case we call it COMMON-SOURCE.To obtain an output voltage we insert a load resistance in the drain lead, the effects of this being represented on the characteristic by a load line.Fig 11.6 shows a practical amplifier circuit with a typical characteristic and load line.

Fig 11.6 Test Circuit and Characteristic.

Construct the circuit of fig 11.6Now apply an input of 2V peak-to-peak at 1000Hz from the generator and observe the output on the oscilloscope.And set the V = 12V, note the V value.

4. Is the output a good sine-wave?

Measure the peak-to-peak output voltage and calculate the voltage gain

Now draw load line for R= 1000 on the I versus V graph.

5. The input resistance of fig 11.6 can not be greater than the bias resistor R

PRACTICAL CONSIDERATIONS AND APPLICATIONS:

We have studied an N-channel JFET. A P-channel JFET is very similar in operation but uses reversed –bias voltages as in fig 11.7.

Fig 11.7 A P-channel JFET and Characteristic

Another form of FET exists whose gate is insulated from the channel (insulated gate FET). The most common insulation method is a metal-oxide layer and the type is called MOSFET. The gate on a JFET must not be biased in such a way as to forward-bias the PN junction however, in a MOSFET, no such limitation applies.It is therefore possible to bias the gate in either polarity. Fig 11.8 shows the usual characteristics of two types of N-Channel MOSFET.

Fig 11.8 N-channel MOSFET Characteristics and Graphical Symbols

The depletion-mode type is like a JFET with an extended gate bias range but the enhancement-mode is quite different since at V =0, no current flows at all. This is often a useful feature.

Two matching types of MOSFET exist with P-channel and have reverse polarities of bias. Their symbols are shown in fig 11.9.

Fig 11.9 P-channel MOSFET Symbols

We can put all these types together in a family tree as in fig 11.10.

Fig 11.10 An FET Family Tree.

MOSFETs have even higher input impedances than JFETsAll FETs are useful for amplification with minimum load on the source. Enhancement-type MOSFETs are especially valuable as electronic switches because with no bias they are normally non-conducting and the high gate resistance means that very little control current is needed. MOSFETs, due to their very high gate resistance, can easily accumulate large static charges and can become damaged unless carefully handled. Some types are fitted internally with protective zener diodes to prevent damage during handling.

SUMMARY:

In this assignment you have learnt that:

1 FETs can take various forms.2 All forms possess high gate resistance, but particularly the MOS types.3 As voltage amplifiers in a common-source circuit, FETs are not very linear

(i.e. they produce distortion if the output voltage is large). 4 FETs can be used as electronic switches.

EXERCISE:

An important parameter of an FET used as an amplifier is its ‘transconductance’. This is defined by:

Transconductance (g ) = (change in I ) / (change in V ) =

(Common source)

Study your graph, fig 11.7 and estimate the change in I for a 0.5V change in V when V =10V

and V =1.0V. Then find g .

The voltage gain (A) for a load resistor R is given by

A= where R is in ohms ( )

Use this expression to verify the voltage gain measured in the Assignment.

V V =

0V

V

=0.5V

V

=1.0V

V

=2.0V

V

=3.5V

V

=5.0V

V

=7.5V

V

=10V

V

=15V

V

=20.0V0.0-0.5-1.0-1.5-2.0-2.0

THE OPERATIONAL AMPLIFIER

OBJECTIVES:

1. Knowledge of its differential input and high gain.2. An understanding of its use as a summing amplifier.

EQUIPMENT REQUIRED:

Qty Apparatus1 Electricity & Electronics Constructor EEC4701 Amplifier Kit EEC4731 Power supply unit 0 to +20V variable d.c and 15V d.c regulated ( e.g.

Feedback Supply PS 445)2 Multimeter or2 Voltmeters 15V d.c

PREREQUISITE ASSIGNMNETS:

Assignment 7

KNOWLEDGE LEVEL:

Before working this assignment you should:• Know the operation of a.c coupled amplifier circuits with and without negative

feedback applied.

INTRODUCTION:

In the amplifiers we have looked at so far the inputs and outputs have been a.c coupled. This means that they are of no use for very low frequencies or for d.c amplification. The main problem in making an amplifier for d.c is ensuring that the mean d.c level of the output is the same as the mean d.c level at the input. In the single stage in the previous assignments this is obvious not so.

Operational Amplifiers:

The operational amplifier is a device that has all the properties required for d.c amplification. It contains several stages and circuitry for temperature drift compensation. Although it could be made using discrete transistors it is usually an integrated circuit (IC) with all the components on a single silicon chip. This makes operational amplifiers available in large quantities at very low cost.

Fig 16.1 Operational Amplifier

Fig 16.1 shows the circuit symbol for an operational amplifier (often abbreviated to “Op – amp”). We can see some important features:

1. It has positive and negative power connections. This so that the output can swing either side of zero volts.

2. It has a positive and a negative input. This means that signals applied to the positive input are not inverted while signals applied the negative input are inverted. Both inputs have the same total gain. This is called a differential input stage.

3. Although it is not shown on the symbol, the amplifier gain is very high. 100,000 would be a typical value.

Using the Operational Amplifier:

In a previous assignment we saw the benefits of negative feedback. We found that by sacrificing some gain the stage gain can be made independent of the device characteristics and be determined by circuit components that can be controlled very closely. As the gain in the op – amp is so high we can afford to lose some b using negative feedback thus producing a very stable amplifier system. We could feed some of the output back to the input --- but to make the feedback negative we must feed it back to the negative input. The negative input will now have a different signal on it than the actual input voltage (the sum of the input and the negative feedback). As we are not using the positive input we must connect it to ground. Combining these ideas we have the circuit in fig 16.2.

Fig 16.2 Op – amp feedback circuit

We will now apply some simple theory to this circuit.We know that the gain of the op – amp is very high, so will be vary small compared

with . If we assume it to be almost zero we can say:

(Very nearly)

But

This is very important as we have produced an amplifier the gain of which is the ratio of two resistors, and is independent of the actual gain of the op – amp. This is called “closed loop operation”. Notice that the amplifier inverts the signal.

Fig 16.3 Summing amplifier

If we add another input resistor equal to the value of as in fig 16.3, we can

modify the above equation thus:

We now have an amplifier that produces an output proportional to the SUM of the two input voltages. This is called a summing amplifier.If the two input resistors were not equal the equation would be modified thus

The operational amplifier we shall be using for our experiments is a type 741. It is very small, and comes in plastic pack 9mm x 6mm x 3mm. For convenience the device is soldered on a small board with pins to suit the EEC470 deck. There are a few practical problems. Although in theory when both inputs are zero the output should be zero, due to leakage currents and other causes there is a slight “offset” of the output. This can be removed by the offset null potentiometer mounted in the module. In the first activity we shall familiarize ourselves with the IC and try the offset null control. In the second activity we shall make a closed – loop feedback circuit.

EXPERIMENTAL PROCEDURE:

Dc Offset: The circuit we shall use is shown in fig 16.4. Set up it.

NOTE Operational amplifiers of the type used in the Kit require a fixed power supply of +15, 0, -15V d.c. This fixed supply is usually omitted from circuit diagrams showing operational amplifiers, in order to reduce the number of connections to pins 12 and 9 respectively of the 741 Op – Amp on fig 16.4 and subsequent diagrams.

Fig 10.4 Test circuit for d.c offset

Switch on the power supply. Turn the potentiometer, marked “zero”, in the op – amp module with a small screwdriver. Notice that the output changes between +15V and -15V very suddenly. It is not possible to set it to zero as the amplifier is operating in the open – loop mode with very high gain. The potentiometer is not fine enough in this condition. Now link A to B thus adding a 100 feedback resistor. The loop is now closed and the gain reduced.

QUESTIONS:

1. Can you calculate the gain?

Try to null the amplifier again.2. Can you set the output to zero? Why?

Op – Amp Performance:

Fig 16.5 Test circuit for Op – amp performance

Set up the circuit diagram in fig 16.5. The circuit is a summing amplifier. The variable voltage from the power supply is one input and the potentiometer is the other. Notice that

is different from . The output is related to the input by the ratio of the feedback

resistor to the input resistor for each input.

3. What is the relationship of the output to the input voltage?It will take the form:

Where x and y are resistance ratios.

Turn on the equipment and turn the input voltage controls up and down. Notice the polarity change in the output voltage .

Copy the results table as shown in fig 16.6, reproduced at the end of this assignment. Set up each input voltage condition in the table. Record the output voltage for each condition. Using the formula you found in Q3, calculate the expected output voltage each time and enter it in the table.

We have made an electric adding machine as the output voltage is the algebraic sum of the input voltages. It uses electric current as an analogue of the number. There are larger systems using this principle and they are called analogue computers.

PRACTICAL CONSIDERATIONS AND APPLICATIONS:

The operational amplifier has a wide variety of uses in control systems and instrumentation. By using other components around it the op – amp may be used to perform other mathematical operations, differentiations and integration, for example. (Hence the name operational amplifier). The thermal stability of an op – amp is excellent as all the components are mounted on the same silicon chip and the designer is able to counteract most of the drift problems.The 741 is a general-purpose op – amp; there are others which have higher input impedances and work up to a higher frequency. Although the 741 is quite insensitive to supply voltage variations it is normally used with a regulated supply.

SUMMARY:

In this assignment you have learnt that:1. The operational amplifier is a d.c amplifier which can amplify both positive and

negative signals and give positive or negative outputs.2. The amplifier has a differential input and very high gain.3. When operated in a closed in a closed loop the gain can be very closely controlled

by resistors.4. It can sum two separate inputs.

Input voltageInput voltage Output voltageOutput voltageCalculated Voltage

0.5 20.1 60.3 4-0.9 2-1.1 4-1.5 6

Fig 16.6