Basic Electronics

Basic electronics and circuits

Edited by: Walid HAMDI

2 nd edition- September 2021

Edited by: instructor/ WALID HAMDI 1 Second edition September 2021

Preface

This book may be used as a complete set of course notes for students undertaking the study of

Electrical and Electronic Principles in the first year of Electronics and/or Electrical Diploma course. It also

provides coverage for some other courses, including foundation/ bridging courses which require the study

of Electrical and Electronic Engineering. The order of the chapters does not necessarily follow the order set

out in any syllabus, but rather follows a logical step-by-step sequence through the subject matter.

In order to encourage students to use other reference sources, all references related to all topics are

listed in the end of this book.

This edition included Supplementary Worked Examples and practical worksheets at the end of each

chapter.

All figures and schematics were made with Proteus program.


Introduction:

This student workbook provides a unit-by-unit outline of the basic in electronics training curriculum.

The following information is included together with space to take notes as you move through the

curriculum:

♦ The course objective.

♦ Course fundamentals.

♦ A list of new terms and words for the course.

♦ Equipment required for the practical experiment.

♦ The exercise objectives.

♦ Exercise discussion.

♦ Exercise notes.

♦The Appendix includes information.


A'SEEB VOCATIONAL COLLEGE

ELECTRONICS ENGINEERING DEPARTMENT

1 st Year – 1 st Semester Academic year: 2021-2022

Course name: Basic Electronics and Circuits

Course code: EECIM1102

Credit hours: 4 Contact hours: 11

Week/ Date

12/09/2021 – 16/09/2021

19/09/2021 – 23/09/2021

26/09/2021 – 30/09/2021

03/10/2021 – 07/10/2021

10/10/2021 – 14/10/2021

17/10/2021 – 21/10/2021

24/10/2021 – 28/10/2021

31/10/2021 – 04/11/2021

07/11/2021 – 11/11/2021

14/11/2021 – 18/11/2021

21/11/2021 – 25/11/2021

28/11/2021 – 02/12/2021

Topics

Work on semiconductor components and devices

Work on AC-DC/DC-AC power supplies.

Work on fundamental electronic circuits.

Troubleshoot and repair electronic circuits.


Course description

COURSE DETAILS

Specialization

Course title

Course code

Electronics

Basic electronics and circuits

EECIM1102

Credit hours 4

Year of study 1

Term 1

Total hours per term

Prerequisites

Practical 108

Theory 12

None

COURSE DESCRIPTION

Introduction to the fundamental laws, and circuits in electronics

Course Objectives:

Measuring the characteristics of semiconductor components,

Calculation of the main parameters of simple electrical circuits

and application

1. Calculate different electrical quantities.

Learning outcomes

Upon completion of

this

course, student

should

be able to:

Course content

2. Particularize electronic components and specify their application,

3. Measure the characteristics of electronic components and present

experimental results

4. Analyze electrical circuits and calculate the main parameters,

5. Develop, design and create simple electronic circuits,

6. Choose an engineering approach to solving problems, starting

from the acquired knowledge essential for the design of electronic

circuits.

Introduction. Passive components: diode, transistor (bipolar). Basic

transistor circuit: switch circuit.


COURSE TITLE: Basic electronics and circuits

SUBJECT: Semiconductors components and devices

COURSE CODE: EECIM1102

TITLE: Semiconductors components

After completing this competency, you should be able to:

Draw semiconductors schematic and symbol

Locate specifications from data-sheet

Implement semiconductors in different circuits

Test semiconductors components (diode, LED,)

Test bipolar transistor (NPN, PNP)

Analyze operating characteristic

I. ACTIVE COMPONENTS:

Active electronic components are those that can control the flow of electricity, and that’s because of the

material that is made from.

All active components are made from semi-conductor material.

INTRODUCTION TO SEMICONDUCTOR

Semiconductor materials such as silicon (Si), germanium (Ge) and gallium arsenide (GaAs), have

electrical properties somewhere in the middle, between those of a “conductor” and an “insulator”. They are

not good conductors not good insulators (hence their name “semi”-conductors).

The ability of semiconductors to conduct electricity can be greatly improved by replacing or adding certain

donor or acceptor atoms to this crystalline structure thereby, producing more free electrons than holes or

vice versa. That is by adding a small percentage of another element to the base material, either silicon or

germanium. Fig.27

Fig. 27

The process of adding donor or acceptor atoms to semiconductor atoms (the order of 1 impurity atom per

10 million (or more) atoms of the semiconductor) is called Doping. Thus doped silicon is no longer pure.

These donor and acceptor atoms are collectively referred to as “impurities”, and by doping a silicon material

with sufficient numbers of impurities, we can turn it into either an N-type or a P-type semiconductor

material.


a) N- type semiconductor:

In order for our silicon crystal to conduct electricity, we need

to introduce an impurity atom such as Arsenic, Antimony or

Phosphorus into the crystalline structure making it extrinsic

(impurities are added).

The resulting semiconductor basics material has an excess

of current-carrying electrons, each with a negative charge,

and is therefore referred to as an N-type.

b) P- type semiconductor:

If we go the other way, and introduce a “Trivalent” (3-electron)

impurity into the crystalline structure, such as Aluminum, Boron

or Indium, which have only three valence electrons available in

their outermost orbital, the fourth closed bond cannot be formed.

Therefore, a complete connection is not possible, giving the

semiconductor material an abundance of positively charged

carriers known as holes in the structure of the crystal where

electrons are effectively missing.

Then a semiconductor basics material is classed as P-type when its acceptor density is greater

than its donor density. Therefore, a P-type semiconductor has more holes than electrons.

c) PN junction:

A PN-junction is formed when an N-type material is fused together with a P-type material creating a

semiconductor diode.

In Figure below (a) the battery is arranged so that the negative terminal supplies electrons to the N-type

material. These electrons diffuse toward the junction. The positive terminal removes electrons from the P-

type semiconductor, creating holes that diffuse toward the junction. If the battery voltage is great enough to

overcome the junction potential (0.6V in Si), the N-type electrons and P-holes combine annihilating each

other. This frees up space within the lattice for more carriers to flow toward the junction. Thus, currents of

N-type and P-type majority carriers flow toward the junction. The recombination at the junction allows a

battery current to flow through the PN junction diode. Such a junction is said to be forward biased.


If the battery polarity is reversed as in Figure above (b) majority carriers are attracted away from the

junction toward the battery terminals. The positive battery terminal attracts N-type majority carriers,

electrons, away from the junction. The negative terminal attracts P-type majority carriers, holes, away from

the junction. This increases the thickness of the non-conducting depletion region. There is no

recombination of majority carriers; thus, no conduction. This arrangement of battery polarity is

called reverse bias.

1. DIODE

Signal Diodes are small two-terminal which conducts current when forward biased and blocks current flow

when reverse biased.

The forward voltage, V F, is a characteristic of the

semiconductor: 0.6v to 0.7v for silicon, 0.2 V for

germanium.

1.1 Signal Diode Parameters

Signal Diodes are manufactured in a range of voltage

and current ratings and care must be taken when

choosing a diode for a certain application.

a) Maximum Forward Current

The Maximum Forward Current ( I F(max) ) is as its name implies the maximum forward current allowed to

flow through the device. When the diode is conducting in the forward bias condition, it has a very small

“ON” resistance across the PN junction and therefore, power is dissipated across this junction (Ohm´s Law)

in the form of heat.

For example, our small 1N4148 signal diode has a maximum current rating of about 150mA with a power

dissipation of 500mW at 25°C. Then a resistor must be used in series with the diode to limit the forward

current, ( I F(max) ) through it to below this value.

b) Peak Inverse Voltage

The Peak Inverse Voltage (PIV) or Maximum Reverse Voltage ( V R(max) ), is the maximum

allowable Reverse operating voltage that can be applied across the diode without reverse breakdown and

damage occurring to the device.


The peak inverse voltage is an important parameter and is mainly used for rectifying diodes in AC rectifier

circuits with reference to the amplitude of the voltage were the sinusoidal waveform changes from a

positive to a negative value on each and every cycle.

c) Total Power Dissipation

Signal diodes have a Total Power Dissipation, (P D(max)) rating. This rating is the maximum possible power

dissipation of the diode when it is forward biased (conducting).

1.2 DIODE APPLICATIONS:

The application areas of diodes include communication systems as limiters, clippers, gates; computer

systems as logic gates, clampers; power supply systems as rectifiers and inverters; television systems as

phase detectors, limiters, clampers; radar circuits as gain control circuits, parameter amplifiers, etc.

a) Diode as a half wave rectifier

The most common and important application

of a diode is the rectification of AC power to

DC power. Below figure shows diode operation

in a rectifier:

The power diode in a half wave rectifier circuit

passes just one half of each complete sine wave

of the AC supply in order to convert it into a DC

supply. Then this type of circuit is called a “half-wave” rectifier because it passes only half of the incoming

AC power supply as shown below.

Equations:

V peak(out) = V peak(in) − 0. 7v

V avg = V peak

π

Example:

D1: 1N4001/1N4007

RL: 1KΩ


b) Diode as a full wave rectifier (Bridge rectifier)

The four diodes labeled D 1 to D 4 are arranged in “series pairs” with only two diodes conducting current

during each half cycle. During the positive half cycle of the supply, diodes D1 and D2 conduct in series

while diodes D3 and D4 are reverse biased and the current flows through the load.

c) IC bridge rectifier

The IC bridge rectifier, is based on 4 diodes, arranged

As shown in circuit (b).

The IC has 4 terminals:

2 terminals for AC inputs referred:

2 terminals for DC output referred: + -

Equations:

V peak(out) = V peak(in) − 2X0. 7v

V avg = 2.V peak

π


EECIM1102 Basic electronics and circuits Chapter 1: Semiconductors components

Worksheet N° 1 Diode testing - Experiment ...............

How to test diodes

Digital multimeter can test diodes using one of two methods:

1. Diode Test mode: almost always the best approach.

2. Resistance mode: typically used only if a multimeter is not equipped with a Diode Test mode.

Note: In some cases, it may be necessary to remove one end of the diode from the circuit in order to test

the diode.

Things to know about the Resistance mode when testing diodes:

Does not always indicate whether a diode is good or bad.

Should not be taken when a diode is connected in a circuit since it can produce a false reading.

CAN be used to verify a diode is bad in a specific application after a Diode Test indicates a diode is

bad.

A diode is best tested by measuring the voltage drop across the diode when it is forward-biased. A forwardbiased

diode acts as a closed switch, permitting current to flow.

A multimeter’s Diode Test mode produces a small voltage between test leads. The multimeter then displays

the voltage drop when the test leads are connected across a diode when forward-biased. The Diode Test

procedure is conducted as follows:

1. Make certain:

a) All power to the circuit is OFF

b) No voltage exists at the diode.

2. Turn the dial (rotary switch) to Diode Test mode ( ). It may share a space on the dial with

another function.

3. Connect the test leads to the diode. Record the measurement displayed.

4. Reverse the test leads. Record the measurement displayed.

Diode test analysis

A good forward-based diode displays a voltage drop ranging from 0.5 to 0.8 volts for the most

commonly used silicon diodes. Some germanium diodes have a voltage drop ranging from 0.2 to

0.3 V.

The multimeter displays OL when a good diode is reverse-biased. The OL reading indicates the

diode is functioning as an open switch.



Worksheet N° 2 Diode measurement - Experiment ................

A bad (opened) diode does not allow current to flow in either direction. A multimeter will display OL

in both directions when the diode is opened.

A shorted diode has the same voltage drop reading (approximately 0.4 V) in both directions.

A multimeter set to the Resistance mode (Ω) can be used as an additional diode test or, as mentioned

previously, if a multimeter does not include the Diode Test mode.

A diode is forward-biased when the positive (red) test lead is on the anode and the negative (black) test

lead is on the cathode.

The forward-biased resistance of a good diode should range from 1000 Ω to 10 MΩ.

The resistance measurement is high when the diode is forward-biased because current from the

multimeter flows through the diode, causing the high-resistance measurement required for testing.

A diode is reverse-biased when the positive (red) test lead is on the cathode and the negative (black) test

lead is on the anode.

The reverse-biased resistance of a good diode displays OL on a multimeter. The diode is bad if

readings are the same in both directions.

Tools and equipment:

Equipment Type/ reference

Diode

Multimeter

Resistor

Procedure:

Connect the circuit shown in (a). Here, the diode is forward-biased.

Set the DC power supply to output 4.5V.

Set the multimeter so that it reads 200mA DC full-scale.

Measure the current flowing and record it in the table.

Next, reverse the diode, as shown in (b). Now the diode is reverse biased.

Again measure and record the current flowing. (You need to change to a

more sensitive current range.)



Worksheet N° 2'' Diode measurement - Experiment ...............

Finally, for comparison purposes, make the same measurements on firstly a

short-circuit, as in (c), and an open-circuit, as in (d).

Again, record these measurements in the table below.

Equipment

(a) Forward bias

(b) Reverse bias

(c) Short-circuit

(d) Open-circuit

Measured current

Discussion and conclusion:

1. What do the results tell you about the resistance of the diode?

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

2. Does it conduct perfectly in the forward direction? Was it as good as a short-circuit?

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

3. Does the diode conduct at all in the reverse direction? Was it as good as an

open-circuit?

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

4. Write a summary of your findings.

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

Procedure

Forward characteristics

Build the circuit, to allow you to measure the forward

VF IF

characteristics of the diode (USE DIODE 1N4007).

0

Set the DC power supply for an output of 4.5V.

0.1

Set the voltmeter and the ammeter.

0.2

0.3

Use the ’pot’ to vary the voltage, V F, applied to the diode from 0.1V to 0.7V 0.4

in steps of 0.1V.

0.5

At each step, measure and record the forward current, I F, in the table.

0.6

Next invert the diode, and change the power supply voltage to 13.5V,

0.7

This allows you to measure the reverse characteristics of the diode

Change the ammeter to the 200µA DC range.

Once again, use the ’pot’ to vary the voltage applied to the diode, now called VR,

but this time you will only need to take current readings, IR, at 0V, 5V and 10V.

Record them in the table.

Use the axes provided to plot your results as a graph of applied voltage against current for both the forward

and reverse directions and for both diodes. Notice that the voltage and current scales are different for the

two directions.

Reverse characteristics

VR

0

5

10

IR



Homework Diode testing - Experiment Name: ....................................................................................

A- Diode characteristics

The following figure shows the diagram for a circuit to test the characteristics

of diode

RESULTS AND ANALYSIS

1. TABLE

PN forward biased

PN reverse biased

S.N

Voltmeter

Ammeter

Voltmeter

Ammeter

reading (Vf)

reading (If)

reading (Vr)

reading (Ir)

1 0.1 0 1 0

2 0.2 0 2 0

3 0.3 0.8 3 0

4 0.4 1.2 4 0.1

5 0.5 2 5 0.2

6 0.6 2.8 6 0.4

7 0.7 3.6 7 0.5

8 0.8 4.2 8 0.6

9 0.9 5.6 9 0.8

10 1 6.4 10 1


1. Plot the graph for ID1 versus VD and write down your observation from the graph.

If

Forward biased

Vf

Vr

Reverse biased

2. Observations:

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

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

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

Ir



Worksheet N° 3 Half-wave rectifier - Experiment ........

NOTE:

The diode is a ‘one-way valve’. It allows a current to flow through it in only one direction.

When it is forward-biased, a silicon diode conducts, with a voltage drop of about 0.7V across it.

When it is reverse-biased, it does not conduct (for low voltages, at any rate.)

INTRODUCTION:

A common use for a diode is to convert alternating current (AC) to direct current (DC) in a rectifier circuit,

exploiting the fact that a diode conducts only when the anode is more positive than the cathode.

PROCEDURE:

Build the circuit shown in Fig.28

Complete the table

Equipment REFERENCE SEPECIFICATIONS

Transformer

Diode

Resistor

Connect a dual trace oscilloscope*, using two ‘x10’ probes, so that Channel A displays at least two

complete cycles of the input waveform and Channel B displays the corresponding output. Connect

the oscilloscope ground terminals to the negative rail of the circuit.

Sketch the output waveform on the grid provided, and include labeled voltage and time axes.


Suggested oscilloscope (or equivalent) settings:

Time base - 10 ms/div Voltage range

Inputs A and B - ±5V DC with x10 probes

Discussion and conclusion:

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

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

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

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

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

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

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

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

*see worksheet: use of oscilloscope

PROCEDURE:

Build the circuit shown in Fig.28’’

Complete the table

Equipment REFERENCE SEPECIFICATIONS

Transformer

Diodes

Resistor

Connect an oscilloscope to display at least two

complete cycles of the output waveform (TP1).



Worksheet N° 4 Full-wave rectifier - Experiment ................

Sketch the output waveform on the upper grid, with labeled voltage and time axes.

Discussion

1. The Positive Half-cycle

During the positive half cycle of the supply,

diodes D1 and D2 conduct in series while diodes

D3 and D4 are reverse biased and the current flows

through the load as shown below.

2. The Negative Half-cycle

During the negative half cycle of the supply,

diodes D3 and D4 conduct in series, but diodes

D1 and D2 switch “OFF” as they are now reverse biased.

The current flowing through the load is the same direction as before.

Conclusion:

During each half cycle the current flows through two diodes instead of just one so the amplitude of the

output voltage is two voltage drops (2*0.7 = 1.4V) less than the input V MAX amplitude. The ripple frequency

is now twice the supply frequency (e.g. 100Hz for a 50Hz supply or 120Hz for a 60Hz supply.)

Measure the output voltage V DC = ........................................................

Measure the transformer output voltage V AB = .....................................

Comment

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2. LIGHT EMITTING DIODE: LED

The “Light Emitting Diode” or LED as it is more commonly called, is basically just a specialized type of

diode as they have very similar electrical characteristics to a PN junction diode. This means that an LED

will pass current in its forward direction but block the flow of current

in the reverse direction.

Light emitting diodes are made from a very thin layer of fairly

heavily doped semiconductor material and depending on the

semiconductor material used and the amount of doping, when

forward biased an LED will emit a coloured light at a particular

spectral wavelength.

When the diode is forward biased, electrons from the

semiconductors conduction band recombine with holes from the

valence band releasing sufficient energy to produce photons which

emit a monochromatic (single colour) of light. Because of this thin

layer a reasonable number of these photons can leave the junction

and radiate away producing a coloured light output.

Then we can say that when operated in a forward biased direction Light Emitting Diodes are

semiconductor devices that convert electrical energy into light energy.

Light Emitting Diode Colours

Both the forward operating voltage and forward current vary depending on the semiconductor material used

but the point where conduction begins and light is produced is about 1.2V for a standard red LED to about

3.6V for a blue LED.

The exact voltage drop will of course depend on the manufacturer because of the different dopant materials

and wavelengths used. The voltage drops across the LED at a particular current value, for example 20mA,


will also depend on the initial conduction V F point. As an LED is effectively a diode, its forward current to

voltage characteristics curves can be plotted for each diode colour as shown below.

Light Emitting Diodes I-V Characteristics.

LED Series Resistor. (Protective resistor)

The series resistor value R S is calculated by simply using Ohm´s Law, by knowing the required forward

current I F of the LED, the supply voltage V S across the combination and the expected forward voltage drop

of the LED, V F at the required current level, the current limiting resistor is calculated as:

Light Emitting Diode Example N o 1

An amber colored LED with a forward volt drop of 2 volts is to be connected to a 5.0v stabilized DC power

supply. Using the circuit above calculate the value of the series resistor required to limit the forward current

to less than 10mA.

Also calculate the current flowing through the LED if a 100Ω series resistor is

used instead of the calculated first.

1. Series resistor required at 10mA.

R S = V S − V F

= 5 − 2 = 300Ω

I −3 F 10 ∗ 10

2. With a 100Ω series resistor.

I F = V S − V F

= 5 − 2 = 0.03A = 30mA

R S 100


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Worksheet N° 5 LED - Experiment .................

INTRODUCTION: In this worksheet you will investigate a simple LED indicator circuit.

LEDs are not good at handling reverse bias. A reverse bias of more than 5V can permanently damage a

LED. Correct identification of the anode and cathode is therefore very important when using them in

practical circuits: The diagram shows the underside of a LED. The anode (A) is the longer

leg. The cathode (C) is shorter, and is next to the flat edge of the LED’s ‘skirt’.

PROCEDURE

Build the circuit shown opposite.

Set the DC power supply to provide an output of 7.5V.

Connect an ammeter, to measure the current through the

1kΩ

LED, and a voltmeter, to measure the voltage drop across the LED and series resistor.

Adjust the ’pot’ to vary the input voltage from 0V to 7.5V.

Measure and record the voltage and current at which the LED:

First becomes illuminated

Provides a reasonably bright output.

DISCUSSION AND CONCLUSION

1. List at least FOUR advantages of LED

LED state

producing visible light

providing bright light

LED + resistor

Voltage Current

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

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

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

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

2. LEDs are available in various colours. List some colours.

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

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


3. BIPOLAR JUNCTION TRANSISTOR (BJT)

3.1 Bipolar Transistor Basics

Simple diodes are made up from two pieces of semiconductor

material, either silicon or germanium to form a simple

PN-junction and we also learnt about their properties

and characteristics. If we now join together two individual

signal diodes back-to-back, this will give us two PN-junctions

connected together in series that share a common P or N

terminal. The fusion of these two diodes produces a three layer, two junctions, three terminal device

forming the basis of a Bipolar Junction Transistor, or BJT for short.

The word Transistor is an acronym, and is a combination of the words Transfer Varistor used to describe

their mode of operation way back in their early days of development. There are two basic types of bipolar

transistor construction, PNP and NPN, which basically describes the physical arrangement of the P-type

and N-type semiconductor materials from which they are made.

The Bipolar Transistor basic construction consists of two PN-junctions producing three connecting

terminals with each terminal being given a name to identify it from the other two. These three terminals are

known and labeled as the Emitter (E), the Base (B) and the Collector (C) respectively.

Bipolar Transistors are current regulating devices that control the amount of current flowing through them in

proportion to the amount of biasing voltage applied to their base terminal acting like a current-controlled

switch. The principle of operation of the two transistor types PNP and NPN, is exactly the same the only

difference being in their biasing and the polarity of the power supply for each type.


WARNING:

Transistors can be easily damaged if connected the wrong way in an electrical circuit. The above

package descriptions are only a few of the many types available. Do not take these pin

identifications for granted. Always look up your transistor type in the supplier’s catalogue. It will tell

you the case style used and there will be a page showing the pinout for all of the different

packages as shown below.

Common transistor pinouts

Pins

NPN

Pins

NPN

BC147

BC107

BC148

BC108

BC149

BC109

BC171

BC172

BC173

BC182

BC183

BC167

BC168

BC169

BC184

BC237

BC207

BC238

BC208

BC239

BC209

BC437

BC438

BC439

BC413

BC414

BC547

BC548

BC549

BC582

BC583

BC467

BC468

BC469

BC584

2N3903

2N3904

TIP3055

9013

9014

MJE

3055T

BD131

BD267A

BD139

TIP31A

BD263

TIP41A

2N3054

2N3055


3.2 BIPOLAR TRANSISTOR CONSTRUCTION

The construction and circuit symbols for both the PNP and NPN bipolar transistor are given above with the

arrow in the circuit symbol always showing the direction of "conventional current flow" between the base

terminal and its emitter terminal. The direction of the arrow always points from the positive P-type region to

the negative N-type region for both transistor types, exactly the same as for the standard diode symbol.

3.3 BIPOLAR TRANSISTOR CONFIGURATIONS

As the Bipolar Transistor is a three terminal device, there are basically three possible ways to connect it

within an electronic circuit with one terminal being common to both the input and output. Each method of

connection responding differently to its input signal within a circuit as the static characteristics of the

transistor vary with each circuit arrangement.

1. Common Base Configuration - has Voltage Gain but no Current Gain.

2. Common Emitter Configuration - has both Current and Voltage Gain.

3. Common Collector Configuration - has Current Gain but no Voltage Gain.

a) The Common Base (CB) Configuration

As its name suggests, in the Common Base or grounded base configuration, the BASE connection is

common to both the input signal AND the output signal with the input signal being applied between the

base and the emitter terminals. The corresponding output signal is taken from between the base and the

collector terminals as shown with the base terminal grounded or connected to a fixed reference voltage

point. The input current flowing into the emitter is quite large as its the sum of both the base current and

collector current respectively therefore, the collector current output is less than the emitter current input

resulting in a current gain for this type of circuit of "1" (unity) or less, in other words the common base

configuration "attenuates" the input signal.


The Common Base Transistor Circuit

This type of amplifier configuration is a non-inverting voltage amplifier circuit, in that the signal voltages Vin

and Vout are in-phase. This type of transistor arrangement is not very common due to its unusually high

voltage gain characteristics. Its output characteristics represent that of a forward biased diode while the

input characteristics represent that of an illuminated photo-diode. Also this type of bipolar transistor

configuration has a high ratio of output to input resistance or more importantly "load" resistance (RL) to

"input" resistance (Rin) giving it a value of "Resistance Gain". Then the voltage gain (Av) for a common

base configuration is therefore given as:

Common Base Voltage Gain:

A V = V out

V in

= I C ∗ R L

I E ∗ R IN

Where: Ic/Ie is the current gain: alpha (α) and RL/Rin is the resistance gain.

The common base circuit is generally only used in single stage amplifier circuits such as microphone preamplifier

or radio frequency (Rf) amplifiers due to its very good high frequency response.

b) The Common Emitter (CE) Configuration

In the Common Emitter or grounded emitter configuration, the input signal is applied between the base,

while the output is taken from between the collector and the emitter as shown. This type of configuration is

the most commonly used circuit for transistor based amplifiers and which represents the "normal" method

of bipolar transistor connection. The common emitter amplifier configuration produces the highest current

and power gain of all the three bipolar transistor configurations. This is mainly because the input

impedance is LOW as it is connected to a forward-biased PN-junction, while the output impedance is HIGH

as it is taken from a reverse-biased PN-junction.


The Common Emitter Amplifier Circuit

In this type of configuration, the current flowing out of the transistor must be equal to the currents flowing

into the transistor as the emitter current is given as Ie = Ic + Ib. Also, as the load resistance (RL) is

connected in series with the collector, the current gain of the common emitter transistor configuration is

quite large as it is the ratio of Ic/Ib and is given the Greek symbol of Beta, (β). As the emitter current for a

common emitter configuration is defined as Ie = Ic + Ib, the ratio of Ic/Ie is called Alpha, given the Greek

symbol of α. Note: that the value of Alpha will always be less than unity.

Since the electrical relationship between these three currents, Ib, Ic and Ie is determined by the physical

construction of the transistor itself, any small change in the base current (Ib), will result in a much larger

change in the collector current (Ic). Then, small changes in current flowing in the base will thus control the

current in the emitter-collector circuit. Typically, Beta has a value between 20 and 200 for most general

purpose transistors.

By combining the expressions for both Alpha, α and Beta, β the mathematical relationship between these

parameters and therefore the current gain of the transistor can be given as:

Alpha (α) = I C

I E

Beta (β) = I C

I B

I E = I C + I B

Where: "Ic" is the current flowing into the collector terminal, "Ib" is the current flowing into the base terminal

and "Ie" is the current flowing out of the emitter terminal.

Then to summarize, this type of bipolar transistor configuration has a greater input impedance, current and

power gain than that of the common base configuration but its voltage gain is much lower. The common

emitter configuration is an inverting amplifier circuit resulting in the output signal being 180 o out-of-phase

with the input voltage signal.


c) The Common Collector (CC) Configuration

In the Common Collector or grounded collector configuration, the collector is now common through the

supply. The input signal is connected directly to the base, while the output is taken from the emitter load as

shown. This type of configuration is commonly known as a Voltage Follower or Emitter Follower circuit.

The emitter follower configuration is very useful for impedance matching applications because of the very

high input impedance, in the region of hundreds of thousands of Ohms while having a relatively low output

impedance.

The Common Collector Transistor Circuit

The common emitter configuration has a current gain approximately equal to the β value of the transistor

itself. In the common collector configuration, the load resistance is situated in series with the emitter so its

current is equal to that of the emitter current. As the emitter current is the combination of the collector AND

the base current combined, the load resistance in this type of transistor configuration also has both the

collector current and the input current of the base flowing through it. Then the current gain of the circuit is

given as:

The Common Collector Current Gain

I E = I C + I B

A i = I E

= I C + I B

= I C

+ 1 = β + 1

I B I B I B

This type of bipolar transistor configuration is a non-inverting circuit in that the signal voltages of Vin and

Vout are in-phase. It has a voltage gain that is always less than "1" (unity). The load resistance of the

common collector transistor receives both the base and collector currents giving a large current gain (as

with the common emitter configuration) therefore, providing good current amplification with very little

voltage gain.


Bipolar Transistor Summary

The different transistor configurations given in the following table:

Characteristic

Common Common Common

Base

Emitter

Collector

Input Impedance Low Medium High

Output Impedance Very High High Low

Phase Angle 0 o 180 o 0 o

Voltage Gain High Medium Low

Current Gain Low Medium High

Power Gain Low Very High Medium


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Worksheet N° 6

CURRENT AND VOLTAGE CHARACTERISTICS

OF BJT - Experiment

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

OBJECTIVES

1. To practice how to test NPN and PNP transistors using multimeter.

2. To demonstrate the relationship between collector current (IC) and base current (IB).

3. To provide the opportunity for plotting the output characteristic curves for a transistor using measured

component values.

COMPONENT AND EQUIPMENT

PART A: Testing Transistor Diode Junction

A.1. 2N3904 NPN transistor

A.2. 2N3906 PNP transistor

A.3. Multimeter

A.4. Breadboard

PART B: Current and Voltage Characteristics of BJT

B.1. 100 kΩ resistor

B.2. 2N3904 NPN transistor

B.3. DC power supply (2 units)

B.4. Multimeter (2 units)

B.5. Breadboard

1. PROCEDURE PART A: Testing Transistor Diode Junction

1.1 Measuring Voltage across 2N3904 NPN transistor junction:

1.1.1 The schematic diagram and diode junction represented are shown in Fig. 29


1.1.2 Take your multimeter and select a low-resistance meter range.

1.1.3 Connect the meter’s positive lead to the transistor’s base lead, with the meter’s negative lead

connected to the transistor’s emitter lead. (NOTE: You have forward biased the transistor’s base-emitter

diodes junction.)

1.1.4 Record the display reading in Table 5.1.

1.1.5 Now, reverse the meter’s leads so that the positive lead is connected to the emitter and the negative

lead is connected to the base.

1.1.6 Record the display reading in Table 5.1.

1.1.7 Then, connect the meter’s positive lead to the base and the negative lead to the transistor’s collector

lead. 1.1.8 Record your result in Table 5.1.

1.1.9 After that, reverse the meter’s leads so that the positive lead is connected to the collector and the

negative lead is connected to the base.

1.1.10 Record the result in Table 5.1.

1.1.11 Now connect the meter’s positive lead to the collector the negative lead to the transistor’s emitter

lead. 1.1.12 Note and record this value in Table 5.1.

1.1.13 Then, reverse the meter’s leads so that the positive lead is connected to the emitter and the

negative lead is connected to the collector.

1.1.14 Record this value in Table 5.1. (NOTE: If transistor diode junction was forward biased, you should

have obtained a value between 0.5 and 0.8 and if the transistor diode junction was reverse biased, the

transistor would be in an “open-circuit” condition).

1.2 Measuring Voltage across 2N3906 PNP transistor junction:

1.2.1 The schematic diagram and diode junction represented are shown in Fig. 30

1.2.2 Take your multimeter and select a low-resistance meter range.

1.2.3 Now, connect the meter’s positive lead to the transistor’s base lead and the negative lead to the

transistor’s emitter lead.

1.2.4 Record your result in Table 5.2.

1.2.5 Then, reverse the meter’s leads so that the positive lead is connected to the emitter and negative lead

is connected to the base.

1.2.6 Note the meter reading, and record the result in Table 5.2.

1.2.7 Then connect the meter’s positive lead to the base and the negative lead to the transistor’s collector

lead. 1.2.8 Record your result in Table 5.2.

1.2.9 After that, reverse the meter’s leads so that the positive lead is connected to the collector and the

negative lead is connected to the base.

1.2.10 Record the result in Table 5.2.


1.2.11 Now, connect the meter’s positive lead to the collector and the negative lead to the transistor’s

emitter lead.

1.2.12 Note the meter reading, and record this result in Table 5.2.

1.2.13 After that, reverse the meter’s leads so that the positive lead is connected to the emitter and the

negative lead is connected to the collector.

1.2.14 Note the meter reading, and record this result in Table 5.2.

1.3 Compare the result of Table 5.1 and Table 5.2.Write your observations.

2. PART B: Current and Voltage Characteristics of BJT

2.1 Measure and record the actual resistance of your 100 KΩ resistor.

Fig. 31

2.2 Construct the circuit shown in Figure 31. Both supply voltages should initially be set to 0 VDC.

2.3 Calculate the value of V1 that is required to generate a current of 5 µA through R1.

2.4 Adjust VBB to obtain the value of V1 calculated in Step 2.3.

2.5 Adjust VCC so that VCE is 0.5 VDC.

2.6 Measure and record the value of IC in appropriate space in Table 5.3.

2.7 Increase VCC to provide a VCE of 1 VDC. Measure and record the corresponding value of IC in Table

5.3. 2.8 Repeat step 2.7 for each of the values of VCE listed in Table 5.3.

2.9 After completing the measurements in step 2.8 for IB = 5 µA, return VCE to 0 VDC. Calculate the value

of V1 that is required to generate a current of 10 µA through R1. Adjust VBB to provide this value of V1.

2.10 Repeat steps 2.5 through 8 for IB = 10 µA.

2.11 Repeat steps 2.5 through 2.10 until Table 5.3 is complete.

2.12 Using the data recorded in Table 5.3, plot the characteristic collector curves for the 2N3904 in graph

paper.


RESULTS:


PART B: Current and Voltage Characteristics of BJT

(1) For Step 2.1: The value of R1: _______________ Ω

(2) For Step 2.3: The value of V1: _______________ V


Wiring circuit

CALCULATION

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DISCUSSION


Again, a student used Digital Multimeter (DMM) to test the transistor as shown in Fig. 32.

From the testing, what type of the BJT transistor? ...................................................................

PART B:

Current and Voltage Characteristics of BJT Using the characteristic curves, predict the values of IC for each

of the IB and VCE combinations listed below.

IC = _________________ when IB = 25 µA and VCE = 10 Vdc

IC = _________________ when IB = 35 µA and VCE = 8 Vdc


CONCLUSION

Results for plotting characteristic curve (

resource: Rosenow, K. / University of Cleveland USA)

Ic (mA)

Comment:

VCE (V)

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SUBJECT: Semiconductors components and devices


TITLE: Transistor as a switch

Learn After completing this competency you should be able to:

Describe and analyze the operation and use NPN/PNP transistors in switching circuits

Use the following rules for transistor circuits:

• for V IN < 0.7 V, the transistor is off, V BE = V IN and V CE = the supply voltage

• for V IN ≥ 0.7 V, the transistor is on, V BE = 0.7 V and V CE = 0 V

• Select and apply I C = h FE I B until saturation is reached.

use data sheets to design switching circuits using NPN/PNP transistors

INTRODUCTION:

The simplest circuit that we can set up with a transistor

is shown below:

There are a few things to note about the way in which

the transistor is connected:

The emitter terminal is connected directly to the 0 V line.

A resistor has been added to the base terminal.

This is to limit the current flowing in the base circuit as only

a small current is needed to switch the transistor on.

The load (a lamp in this case) is connected into the collector circuit.

The flying lead shown in the circuit diagram can now either be connected to 0 V or to +6 V to demonstrate

the switching action of the transistor.

Case 1: Flying lead connected to 0 V

In this case, with the flying lead connected to 0 V there is

no difference in voltage between the base and the emitter

terminal, VIN = 0 V. Therefore, no current flows, the transistor

is switched off, and the lamp does not light.

Case 2: Flying lead connected to +6 V

In this case, with the flying lead connected to the +6 V line

there is a voltage difference between the base and emitter

terminals which causes current to flow from the base to the

emitter, V IN = 6 V. This switches the transistor on, which

allows a larger current to flow through the collector and

emitter, and the lamp lights.


This simple circuit provides a very good demonstration of the switching action of the transistor. The

flying lead could be connected to the output of a logic gate for example. Depending on whether the

output is at logic 0 or logic 1, will determine the state of the lamp.

There are two basic rules we have to remember about the transistor.

i. If V IN < 0.7 V, the transistor is off, V BE = V IN

ii. If

iii. V IN ≥ 0.7 V, the transistor is on, V BE = 0.7 V

Practical applications for transistor switching circuits

(a) Light activated switch

(i) The lamp comes on in darkness.

In bright light the resistance of the LDR (R2)

will be low. The voltage at point A is near to

0 V (VIN < 0.7 V) and the transistor will

be off as shown on the right.

In darkness the resistance of the LDR

will be very high. The voltage at point A is

now high (VIN > 0.7 V), and current flows

through the base circuit. The transistor

turns on and the lamp lights up.

(ii) The lamp comes on in bright light.

In darkness the resistance of the LDR will

be very high. The voltage at point A is near

to 0 V (VIN < 0.7 V) and the transistor will

be off as shown on the right.

In bright light the resistance of the LDR

decreases. The voltage at point A is

high and the transistor turns on

(VIN > 0.7 V), and the lamp lights up.

In both examples replacing R1 with a variable resistor would provide means of adjusting the sensitivity of

the systems.


EECIM1102 Basic electronics and circuits Chapter 4: Fundamental electronic circuit

Worksheet N° 7 Dark detector using LDR- Experiment ...........

1. Understand component function: the LDR

The LDR: Light Dependent Resistor, is a component that has a (variable) resistance that

changes with the light intensity that falls upon it. light sensing circuits.

2. Understand component pin out: transistor BC547

Fig. 34, represent the pin out of transistor Q1, model: BC547

Identify the type of transistor: NPN or PNP: ……………

3. Build the circuit Fig. 33

Adjust the potentiometer* (see Appendix A) and discuss its

function:

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What is the function of resistor R1?

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Measurement and testing:

LDR VB VCE LED

Light on ………. ………. ……….

Light off ………. ………. ……….

Conclusion and comment:

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2n2222

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The following graph shows the typical response that is obtained from a circuit like that shown above.

There are three key parts to this graph, which is known as the transistor voltage transfer characteristic.

(i)

(ii)

(iii)

Off region

This part of the characteristic when V IN is between 0 and 0.7 V shows that when the transistor is

completely switched off, no current flows through the base-emitter junction, no current flows

through the collector, and the voltage across the collector-emitter junction of the transistor (V OUT) is

equal to the supply voltage.

Linear region

When the voltage VIN increases above 0.7 V a base current starts to flow. The transistor behaves as

a current amplifier and the base current causes a larger amplified current to flow through the

collector and load. As VIN increases further, more current flows into the base and this allows

a further increase in the collector-emitter current.

Saturation

As VIN continues to increase, a point is reached where changes to VIN no longer cause any

change to VOUT, and we say that the transistor is saturated.

The saturation point is reached just before the voltage across the load reaches the full voltage of

the power supply and the voltage across the collector-emitter junction of the transistor VOUT is

about 0.2 V (i.e. nearly = 0 V).

Note:

We have referred to the voltage across the collector-emitter junction of the transistor as VOUT. It is

often referred to as VCE.


Transistor switching circuit

When the transistor is being used as a switch, we operate in the cut-off and saturation regions of

the characteristic, avoiding the linear region.

There are two reasons for avoiding the linear region when designing transistor switching circuits. Firstly,

the output device will not work correctly because the full supply voltage does not appear across the

load, as V CE will have a significant value. Secondly, because of this value of V CE and the current flowing

in the collector, power will be used up in the collector-emitter junction causing the transistor to overheat.

In this course we will only be considering switching circuits, and the following information will be important.

For V IN < 0.7 V: V BE = V IN and V CE = Supply voltage,

For V IN > 0.7 V: V BE = 0.7 V and V CE = 0 V;

Current gain (h FE)

In order to design circuits for transistors, there is also an important formula which needs to be considered.

This is the current gain formula for the transistor. We have mentioned several times that the transistor acts

as a current amplifier. Each transistor has a current gain called ‘h FE’ and this is defined by the following

current gain formula.

h FE = I c

I B

Where IC is the collector current, and IB is the base current.

For example:

(i) If IB = 10 mA and hFE = 120, what is IC?

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(ii) If IC = 800 mA and hFE = 250, what is IB?

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Always check after using the current gain formula that I B is smaller than I C.

Different types of transistor have different h FE values that can range from 10 to over 800 in value. You

will not be expected to remember the different values of h FE. You will either be told the h FE value for

the transistor, or you will be able to calculate it from values of I B and I C.


Selecting a suitable transistor

A brief look through an electronic component supplies catalogue or website, e.g. Rapid Electronics

or Maplin will reveal many pages dedicated to transistors. So how do you select the most appropriate

transistor for your application?

There are three key points to consider:

i) what is the maximum collector current that your load requires.

ii) what current gain, h FE , do you require.

iii) the cost.

The application will determine the most appropriate transistor to use. The investigations in this chapter will

suggest using one of two inexpensive transistors. The BC337 is a medium power transistor with a very

high h FE and a moderate collector current capability. The BD437 is a high power transistor with a moderate

h FE and a high current capability. In general, the higher the transistor collector current capability of a

transistor type the lower the value of h FE.

Note: Transistors can be easily damaged if connected the wrong way in an electrical circuit. Always look

up your chosen transistor type in the supplier’s catalogue or website. It will tell you the case style

from which you can work out the pin identification. If you look at the BC337 and the BD437 on the

previous page you will notice they have completely different pin connections.

Your teacher might give you different but equivalent transistors to use. You will need to know the

pin identification for those transistors.


Transistor switching circuit design calculations

We are now in a position to start looking at some design problems involving the transistor switching circuits.

So first a couple of examples:

Example 1:

The circuit below contains a transistor with a current gain, hFE = 125. The circuit switches a warning

lamp rated at 6 V, 200 mA.

a) Determine the collector current when the lamp is working at its rated voltage and current and

the transistor is just saturated.

Voltage across lamp = 6 V and VCE = 0 V therefore

IC = 200 mA

b) Calculate the base current.

Now that we know IC, we can find IB, by rearranging the current gain formula

I B =

h FE = I c

I B

I c 200. 10−3

= = 1.6 ∗ 10 −3 A = 1.6mA

h FE 125

c) Calculate the voltage across RB. Using Ohm’s Law

V B = R B . I B = 1 ∗ 10 3 ∗ 1.6 ∗ 10 −3 = 1.6v

d) Determine the value of V IN that will cause the transistor to just saturate.

To get the voltage for V IN , we just have to add on the voltage across the baseemitter

junction V BE , which is always 0.7 V, therefore

V IN = V RB + V BE

= 1.6 V + 0.7 V

e) Complete the following table to show:

= 2.3 V

i) the voltage V BE and V CE for the input voltages V IN given.

Input voltage, V IN V BE V CE Lamp On/Off?

ii) whether the buzzer will be On or Off.

0.2 V 0.2 V 6 V Off

For V IN < 0.7 V: V BE = V IN and V CE = Supply Voltage.

For V IN > 0.7 V: V BE = 0.7 V and V CE = 0 V.

2.5 V 0.7 V 0 V On


Example 2:

The temperature sensing circuit below contains a transistor

with a current gain, h FE = 400. The circuit switches a

warning buzzer on when the temperature in a greenhouse

gets too high. The resistance of the buzzer is 30 Ω.

a) Calculate the collector current when the transistor is just saturated.

When the transistor is just saturated, VOUT will be = 0 V. This means that the whole

voltage of the power supply must be across the buzzer.

I C = V supply

= 9 = 0.3A = 300mA

R Load 30

b) Calculate the base current.

I B =

h FE = I c

I B

I c

= 0.3

h FE 400 = 0.75mA

c) At a certain temperature the base current is 0.5 mA.

i) What is the new value of collector current?

ii) What is the new value of the voltage across the buzzer?

We know the transistor was just saturated with base current of 0.75 mA so now it is no longer

saturated.

i) I C = h FE ∗ I B = 400 ∗ 0.5 ∗ 10 −3 = 200mA

ii)

V Buzzer = I C ∗ R Buzzer = 200 ∗ 10 −3 ∗ 30 = 6V

d) When the base current was 0.5 mA it was found that the transistor became very hot and the

buzzer was quiet. Suggest a reason why this happened.

The transistor is not saturated, and is therefore operating in the linear region.

This results in the transistor overheating and can permanently damage the transistor.

The voltage across the buzzer is only 6 V so it sounds quiet.



SUBJECT: Work on AC-DC/DC-AC power supplies


TITLE: Power supplies

After completing this competency, you should be able to:

Draw schematic diagram of common types of power supply

Test components

Construct and build power supply circuit 12VDC

Analyze the function of each section

Draw waveform of each section

Diagnose abnormal function

Replace defective component

Repair defective connection

Check circuit worked before applying power

DESCRIPTION OF TASKS:

Power supplies in recent times have greatly improved in reliability but, because they have to handle

considerably higher voltages and currents than any or most of the circuitry they supply, they are often the

most susceptible to failure of any part of an electronic system.

Modern power supplies have also increased greatly in their complexity, and can supply very stable

output voltages controlled by feedback systems. Many power supply circuits also contain automatic

safety circuits to prevent dangerous over voltage or over current situations.

Warning

Power Supply Block Diagram

Safety Information

If you are considering building or repairing a power supply, especially one that is powered from mains

(line) voltages the power supply modules on this site will help you understand how many commonly

encountered circuits work. However you must realize that the voltages and currents present in many

power supplies are, at best dangerous, and can be present even when the power supply is switched off!

At worst, the high voltages present in power supplies can, and from time to time do KILL.

Edited by: Instructor/ WALID HAMDI 48 Second edition: September 2021

5V/9V/12V fixed power supply (Positive)

TP1

TP2

TP3

TP4

Circuit Diagram (Schematic Diagram)

L1 = Step down transformer with i/p of 230 AC 50 Hz and output of (XX ) - 0- (XX)) volts(rms).

XX = Required DC output voltage.

Here is the table for different voltages

Output voltage (DC Volts)

5 230: 5-0-5

9 230: 9-0-9

12 230:12-0-12

15 230: 15-0-15

Components

D1, D2 = Diodes 1N4003

D3 = Diode 1N4003/ 1N4001 (optional)

Transformer rating (rms Volts)

C1 = 1000µF aluminum electrolytic capacitor (For loads less than 100mA you can substitute with 220 µF

capacitor), Voltage rating = 2.5 times of Output Voltage.

C2 = 10 Micro Farad aluminum electrolytic capacitor

IC1 = 7805 for +5V DC output

=7809 for +9V DC output

=7812 for +12V DC output

=7815 for +15V DC outpu


Draw the waveform for each TP : Test Point

• Waveform on TP1 :



• Waveform on TP4


5V/9V/12V fixed power supply (Negative)

Circuit Diagram (Schematic Diagram)-2

L1 = Step down transformer with i/p of 230 AC 50 Hz and output of (XX ) - 0- (XX)) volts(rms).

XX = Required DC output voltages.

Output voltage (DC Volts)

5 230: 5-0-5

9 230:9-0-9

12 230:12-0-12

15 230:15-0-15

Transformer rating (rms Volts)

We use the same components value as previous circuit, only the regulator change

IC1 = 7905 for -5V DC output

=7909 for -9V DC output



Additional note: It's safer to put one heat sink to 78XXX and 79XX IC for safeguarding the IC from

overheating

In case you are using both the power supplies the ground connection of both positive and negative

power supplies can be shorted.


Voltage regulation.

For comparison of different types of power supplies, the following terms are commonly used :

Voltage regulation. The DC voltage available across the output terminals of a given power supply

depends upon load current. If the load current Idc is increased by decreasing RL (See Fig. 33 a), there is

greater voltage drop in the power supply and hence smaller d.c. output voltage will be available. Reverse

will happen if the load current decreases. The variation of output voltage w.r.t. the amount of load current

drawn from the power supply is known as voltage regulation and is expressed by the following relation :

In a well designed power supply, the full-load voltage is only slightly less than no-load voltage i.e. voltage

regulation approaches zero. Therefore, lower the voltage regulation, the lesser the difference between fullload

and no-load voltages and better is the power supply. Power supplies used in practice have a voltage

regulation of 1% i.e. full-load voltage is within 1% of the no-load voltage. (Fig. 33 b), shows the change of

d.c. output voltage with load current. This is known as voltage regulation curve.

Example 1. If the d.c. output voltage is 400V with no-load attached to power supply but decreases to 300V

at full-load, find the percentage voltage regulation.

Solution.

VNL = 400 V ; VFL = 300 V

% Voltage regulation = V NL−V FL

V FL

∗ 100 = 400−300

300

∗ 100 = 33.33 %

Example 2. A power supply has a voltage regulation of 1%. If the no-load voltage is 30V, what is the fullload

voltage?

Solution.

1 = V NL−V FL

V FL

VFL = 29.7v

∗ 100 = 30−V FL

V FL

∗ 100


Example 3. Two power supplies A and B are available in the market. Power supply A has no-load and fullload

voltages of 30V and 25V respectively whereas these values are 30V and 29V for power supply B.

Which is better power supply?

Solution.

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

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

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

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

Example 4. Fig. 34 shows the regulation curve of a power supply. Find (i) voltage regulation and (ii) minimum load

resistance.

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

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

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

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

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

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

Example 5.

Calculate the approximate DC output voltage of this power supply when it is not loaded:

Fig. 34

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

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

Example 6.

Identify the voltages that are supposed to appear between the listed test points:

1. VTP1−TP2 = ............................

2. VTP1−TP3 = ............................

3. VTP2−TP3 = ............................

4. VTP4−TP5 = ............................

5. VTP5−TP6 = ............................

6. VTP7−TP8 = ............................

7. VTP9−TP10 = .........................

Assume that the power transformer has a step-down ratio of 9.5:1.


Example 6a.

A technician is troubleshooting the power supply circuit in Example 6, with no DC output voltage. The output

voltage is supposed to be 15 volts DC:

The technician begins making voltage measurements between some of the test points (TP) on the circuit

board. he records these measurements:

1. VTP9−TP10 = 0 volts DC




5. VTP4−TP5 = 0 volts AC



Based on these measurements, what do you suspect has failed in this supply circuit? Explain your answer.

Also, critique this technician’s troubleshooting technique and make your own suggestions for a more efficient

pattern of steps.

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

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

Example 6b.

A technician is troubleshooting a power supply circuit with no DC output voltage. The output voltage is

supposed to be 15 volts DC:


board. What follows is a sequential record of her measurements:




4. VTP4−TP5 = 0.5 volts AC

5. VTP7−TP8 = 1.1 volts DC


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

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

Example 6c.

A technician is troubleshooting a power supply circuit with no DC output voltage. The output voltage is

supposed to be 15 volts DC:


board. What follows is a sequential record of his measurements:








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



SUBJECT: Work on fundamental electronic circuits


TITLE: Fundamental electronic circuit

After completing this competency you should be able to:

1. Read and interpret and draw a schematic diagram.

2. Read and interpret a datasheet.

3. Solve problem calculation.

I. INTRODUCTION:

1. COMMON ELECTRONICS SYMBOLS:


2. NAMES DESIGNATORS AND VALUES:

One of the biggest keys to being schematic-literate is being able to recognize which components are which.

The component symbols are important, but each symbol should be paired with both a name and value to

complete it.

b. Name and value:

Component names are usually a combination of one or two letters and a number. The letter part of the name

identifies the type of component: R's for resistors, C's for capacitors, U's for integrated circuits, etc. Each

component name on a schematic should be unique; if you have multiple resistors in a circuit, for example, they

should be named R 1, R 2, R 3, etc. Component names help us reference specific points in schematics.

The prefixes of names are pretty well standardized. For some components, like resistors, the prefix is just the

first letter of the component. Other name prefixes are not so literal; inductors, for example, are L's (because

current has already taken I). Here's a quick table of common components and their name prefixes:

Although these are the "standardized" names for component symbols, they're not universally followed. You

might see integrated circuits prefixed with IC instead of U, for example, or crystals labeled as XTAL's instead

of Y's. Use your best judgment in diagnosing which part is which. The symbol should usually convey enough

information.

c. Nets, Nodes and Labels:

Now all that remains is identifying how all of the symbols are connected together.

Schematic nets tell you how components are wired together in a circuit. Nets are represented as lines between

component terminals. Sometimes (but not always) they're a unique color, like the green lines in this schematic:


2.3 Junctions and Nodes

Wires can connect two terminals together, or they can connect dozens. When a wire splits into two directions,

it creates a junction. We represent junctions on schematics with nodes, little dots placed at the intersection of

the wires.

Nodes give us a way to say that "wires crossing this junction are connected". The absences of a node at a

junction means two separate wires are just passing by, not forming any sort of connection.


EECIM1102 Basic electronics and circuits Chapter 2: Fundamental electronic circuit

Worksheet N° 8 Schematic diagram ...........

1. Wiring diagram:

A wiring diagram is a simple visual representation

of the physical connections and physical layout of

an electrical system or circuit. It shows how the

electrical wires are interconnected and can also

show where fixtures and components may be

connected to the system.

2. Schematic diagram:

From the previous wiring diagram Fig. 35, complete the schematic diagram Fig.36

Fig. 35

3. Discussion:

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………


II. BASIC CIRCUIT CALCULATION:

1. Application for transistor: DC mode

Example 1: WORKED EXAMPLE

Find and in the transistor circuit of Fig. 37. Assume that the transistor operates in the active mode

(VBE=0.7) and that β=50.

Solution:

For the input loop, KVL gives:

−4 + I B . 20. 10 3 + V BE

Sine V BE =0.7 in the active mode,

I B = 4 − 0.7

20. 10 3 = 165μA

But : I C = β. I B =50*165μA = 8.25mA

For the output loop, KVL gives:

−V O − 100. I c + 6 = 0

So V O = 6 − 100. I c = 6 − 100 ∗ 8.25 ∗ 10 −3 = 5.175 V

Note that: Vo=VCE

Fig. 37

Example 2: WORKED EXAMPLE

For the transistor circuit in Fig.38, β=100 and VBE= 0.7V

Determine Vo and VCE

Solution:

For loop1, KVL gives:

5 − 10 ∗ 10 3 ∗ I B − V BE − V o = 0 (1)

knowing that I C ≅ I E , and I C = β. I B

and Ohm's law gives: V O = 200 ∗ I E

So (1) gives: 5 − 10 ∗ 10 3 ∗ I B − V BE − 200 ∗ β. I B = 0

So I B = 0.143mA

now V O = 200 ∗ I E = 2.86V

For the output loop (loop 2), KVL gives:

V O + V CE + 500 ∗ I C − 12 = 0

knowing that I C ≅ I E = 14.3mA

V CE = 12 − 500 ∗ 14.3 ∗ 10 −3 − 2.86 = 1.99V

Fig. 38


2. Problem solving

2.1 Example1: Calculate the following parameters for the circuit in Figure 39,

Questions

a. IB = .......................................................................................................

IC =........................................................................................................

b. VR = ......................................................................................................

c. VC = .......................................................................................................

Fig. 39

Calculation:

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

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

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

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

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

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

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

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

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

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

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


Appendix A: Use of oscilloscope

1. Front panel of a digital oscilloscope

No. Description No. Description

(1) Menu (11) Logic Analyzer

(2) LCD (12) Record

(3) Multi-Function Knob (13) Power Key

(4) Navigation Knob (14) USB Host

(5) HORIZONTAL (15) Digital Channel Input (MSO only)

(6) CLEAR (16) Function Menu Keys

(7) AUTO (17) Vertical

(8) RUN/STOP (18) Analog Channel Inputs

(9) SINGLE (19) Function Menu Keys

(10) Default & Print (20) Trigger


2. Basics of Oscilloscope

The main purpose of an oscilloscope is to graph an electrical signal as it varies over time. Most scopes

produce a two-dimensional graph with time on the x-axis and voltage on the y-axis.

Example1:

Note:

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………


Applications:

1. Determine the frequency and the peak voltage of this waveform, as displayed by an oscilloscope with a

vertical sensitivity of 2 volts per division and a time base of 0.5 milliseconds per division:

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………

2. Assuming the vertical sensitivity control is set to 2 volts per division, and the time base control is set to 10 μs

per division, calculate the amplitude of this “saw-tooth” wave (in volts peak and volts peak-to-peak) as well as

its frequency.

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………

3. Assuming the vertical sensitivity control is set to 0.5 volts per division, and the timebase control is set to 2.5

ms per division, calculate the amplitude of this sine wave (in volts peak, volts peak-to-peak, and volts RMS) as

well as its frequency.


……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………


Appendix B: The potentiometer

1.1 Introduction:

A potentiometer, also referred to as pot, may come in a wide variety of shapes and are used in many

applications in your daily life, for example to control the audio volume of the radio.

A pot is a manually adjustable variable resistor with three terminals. In the figure below you can see some

examples of potentiometers.

1.2 Symbols

In a circuit diagram, a potentiometer is represented by one of the two symbols below:

1.3 How Does a Potentiometer Work?

A potentiometer has 3 pins. Two terminals (1 and 3)

are connected to a resistive element and the third terminal (2)

is connected to an adjustable wiper.


1.4 Types of potentiometers

A wide variety of potentiometers exist. Manually adjustable potentiometers can be divided in rotary or linear

movement types. The tables below list the available types and their applications. Besides manually adjustable

pots, also electronically controlled potentiometers exist, often called digital potentiometers.

a. Rotary potentiometers

The most common type of potentiometer where the wiper moves along a circular path.

Type Description Applications

Single-turn pot

Multi-turn pot

Dual-gang pot

Concentric pot

Servo pot

Single rotation of approximately 270 degrees

or 3/4 of a full turn

Multiple rotations (mostly 5, 10 or 20), for

increased precision. They are constructed

either with a wiper that follows a spiral or helix

form, or by using a worm-gear.

Two potentiometer combined on the same

shaft, enabling the parallel setting of two

channels. Most common are single turn

potentiometers with equal resistance and taper.

More than two gangs are possible but not very

common.

Dual potmeter, where the two potentiometers

are individually adjusted by means of

concentric shafts. Enables the use of two

controls on one unit.

A motorized potmeter which can also be

automatically adjusted by a servo motor.

Most common pot, used in applications

where a single turn provides enough

control resolution.

Used where high precision and resolution

is required. The worm-gear multi turn pots

are often used as trimpots on PCB.

Used in for example stereo audio volume

control or other applications where 2

channels have to be adjusted in parallel.

Often encountered in (older) car radios,

where the volume and tone controls are

combined.

Used where manual and automatic

adjustment is required. Often seen in

audio equipment, where the remotecontrol

can turn the volume control knob.

Dual-gang potentiometer Concentric potentiometer Multi-turn potentiometer


b. Linear potentiometers

Potentiometers where the wiper moves along a linear path. Also known as slider, slide pot or fader.

Type Description Applications

Slide pot

Dual-slide pot

Multi-turn slide

Motorized fader

Single linear slider potentiometer, for

audio applications also known as a fader.

High quality faders are often constructed

from conductive plastic.

Dual slide potentiometer, single slider

controlling two potentiometers in parallel.

Constructed from a spindle which

actuates a linear potentiometer wiper.

Multiple rotations (mostly 5, 10 or 20), for

increased precision.

Fader which can be automatically

adjusted by a servo motor.

For single channel control or

measurement of distance.

Often used for stereo control in

professional audio or other

applications where dual parallel

channels are controlled.

Used where high precision and

resolution is required. The multi turn

linear pots are used as trimpots on

PCB, but not as common as the

worm-gear trimmer potentiometer.

Used where manual and automatic

adjustment is required. Common in

studio audio mixers, where the servo

faders can be automatically moved to

a saved configuration.

Slide potentiometer Motorized fader Multi-turn linear trimpot

c. Digital potentiometers

Digital potentiometers are potentiometers which are controlled electronically. In most cases they exist of an

array of small resistive components in series. Every resistive element is equipped with a switch which can

serve as the tap-off point or virtual wiper position. A digital potentiometer can be controlled by for example

up/down signals.

1.2 Standard values

Because potentiometers are variable there is no need for a wide range of values. While potentiometers can be

manufactured in every resistance value you can think of, most potentiometers have values in the following

range of multiples.

Common potentiometer values (multiples)

10 20 22 25 47 50


1.3 Applications of Potentiometers

A potentiometer essentially works as a voltage divider, however it is used in many industries and applications

too. Some of the applications are listed below, categorically:

1.3.1 Pots as Controllers:

Potentiometers can be used in user controlled input applications, where there is a requirement of

manual variation in the input. Like for example a throttle pedal is often a dual gang pot, used to increase the

redundancy of the system. Also, the joysticks that we use in machine control, is a classic example of pot used

as a user controlled input.

Another application where pots are used as controllers are in audio systems. The potentiometer

with logarithmic taper, is often used in audio volume control devices, this is so because our hearing has a

logarithmic response to sound pressure. A logarithmic taper pot will therefore naturally make the transition

from a loud to soft sound ( and vice versa), smoother to our ears. Mostly a motorized pot (with logarithmic

taper) is used for this application.

1.3.2 Pots as measuring devices:

Most common application of potentiometer is as voltage measuring devices. The name itself has that

implication. It was first manufactured for the purpose of measuring and controlling the voltage.

Since these devices convert the position of the wiper into an electrical output, they are used as

transducers to measure distance or angles.

1.3.3 Pots as tuners and calibrators:

Pots can be used in a circuit, to tune them to get the desired output. Also during the calibrations of a device, a

preset pot are often mounted on the circuit board. They are kept fixed for most of the time.

With this we have covered almost all the aspects so that now you know the basics of a potentiometer. Lets

have a quick recap of what we learnt:

Answer key page 37-38

Example 6



3. VTP2−TP3 = 0 volts





Example 6 a

The fuse is blown open.

Example 6 b

The transformer has an open winding.

Example 6 c

There is an ”open” fault between TP4 and TP6.


Appendix C: Symbols and formula


Basic Electrical Formula

1) Resistor in Series RT = R1 + R2 + R3 + …………. + Rn Ω

2) Resistor in Parallel 1/RT = 1/R1 + 1/R2 + ……….. + 1/Rn Ω

3) Inductor in Series LT = L1 + L2 + L3 + ………... + Ln H

4) Inductor in Parallel 1/LT = 1/L1 + 1/L2 + …………. + 1/Ln H

5) Capacitor in Series 1/CT = 1/C1 + 1/C2 + ………… + 1/Cn F

6) Capacitor in parallel CT = C1 + C2 + C3 + ………... + Cn F

7) Ohm’s Law V= I x R Volts(V) (Or) I = V / R Amps(A) (Or) R = V / I Ohms(Ω)

8) Power P = V x I Watts(W) (Or) P = I² x R Watts(W) (Or) P = V² / R Watts(W)

9) Energy E = P x T Joules(J)

Where P = Power in Watts (W) and T = Time in Seconds (S)

10) KVL VT = V1 + V2 + ……… + Vn Volts(V) in a closed circuit.

11) KCL IT = I1 + I2 + ……In Amps(A)

( Algebraic Sum of entering current = Algebraic Sum of leaving current)

12) Voltage Divider Rule V1 = (VT x R1)/ (R1+R2) Volts(V) , V2 = (VT x R2)/ (R1+R2) Volts(V)

13) Current Divider Rule I1 = ( IT x R2 ) / (R1 + R2) Amps(A), I2 = ( IT x R1 ) / (R1 + R2) Amps(A)

14) Vrms = Vp

= 0.707 ∗ Vp

√2

15) Vavg = 2 ∗ Vp = 0.637 ∗ Vp

π

16) Inductive Reactance XL = ωL = 2ᴫ.F.L Ohms (Ω)

17) Capacitive Reactance XC = 1/ (ωC) = 1/ (2π.F.C) Ohms (Ω)

18) Total Impedance Z = √(R 2 + X L 2 ) Ohms (Ω) (Or) Z = √(R 2 + X C 2 ) Ohms (Ω)


References

https://www.electronics-tutorials.ws/

https://www.makerspaces.com

https://www.instructables.com

www.allaboutcircuits.com

www.edutek.ltd.uk

https://www.ibiblio.org

www.learnabout-electronics.org

https://www.uen.org

www.gistrayagada.ac.in

https://www.docsity.com

https://www.cla.purdue.edu

https://www.researchgate.net

https://www.clemson.edu

www.engineering.nyu.edu

https://www.oreilly.com

https://www.abctlc.com

www.edutalks.org

www.keiabroad.org

https://lecturenotes.in

www.centecinc.com

https://www.open.edu

https://www.udemy.com

https://www.yamanelectronics.com

https://www.makerspaces.com


Annex: use of a breadboard.

+


-

-

Worksheet 6/ page 38

Results for plotting characteristic curve


You may notice electronic ICs have these pins Vcc, Vss, Vdd, Vee. Some ICs has Vcc and GND,

some have Vdd and Vss, some have Vcc and Vee. Now the question is what is the full form or

meaning of these pins and why different names are used for different ICs. You may also notice, in

electronic circuits these names are used with their power supply terminals. So, let's know why

different names are used.

Full Forms of Vcc, Vss, Vdd, Vee

Vcc = Voltage Collector Collector

Vdd = Voltage Drain Drain

Vss = Voltage Source Source

Vee = Voltage Emitter Emitter

GND = Ground

Now you may think why there is double terms are used such as Collector Collector, Drain Drain, etc.

These double terms refer that the particular pin connected to the many transistors, not to a single

transistor. For example, the Vcc not connected only to a single transistor, it connected with many

transistors.

What is Vcc, Vss, Vdd, Vee?

Vcc and Vdd are the positive supply voltage to an IC or circuit.

Vss and Vee are the negative supply voltage to an IC or electronic circuit.

Why different names are used?

We know that BJT(Bipolar Junction Transistor) has three terminal named as Emitter, Base, and

Collector.

FET(Field Effect Transistor) also has three terminals named as Gate, Drain, and Source.

So when an electronic circuit or IC is made using Bipolar Junction Transistors then the supply voltage

pins are denoted as Vcc and Vee

Vcc refers that the supply voltage pin is connected to the collector of the transistors.

Vee refers that the supply voltage pin is connected to the emitter of the transistors.


When an electronic circuit or IC is made using Filed Effect Transistors the supply voltage pins are

denoted as Vdd and Vss

Vdd refers that the supply voltage pin is connected to the drain of the transistor.

Vss refers that the supply voltage pin is connected to the source of the transistor.

Supply Voltage

BJT

(Bipolar Junction

Transistor)

FET

(Field Effect Transistor)

Positive Supply Voltage Vcc Vdd

Negative Supply Voltage Vee Vss


Basic Electronics

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