1 electronics use your mouse to move around the software. you can either click anywhere on the...

1

ElectronicsUse your mouse to move around the software. You can either click anywhere on the screen to get the next animation or click on a button if you can see one on the screen.

TPS 2002

Always move the mouse before you click it.

2

ElectronicsIntroduction

Ohm’s Law

Power Calculations

Resistors in Series and Parallel

Capacitors

Alternating Current

Waveforms

The Potential Divider

Transistor Circuits

Questions

3

Introduction

Basic Concepts

Simple Circuits

Questions

Main Menu

4

Basic Concepts

Return previous slide

Electric current is due to the flow of charge.

In a solid conductor, the charge is carried by electrons.

In a solid conductor, an electric current is due to the flow of electrons.

5

Basic Concepts


Conductors include:

copper

gold

silver

lead

All metals

And water (not distilled) which is why you should not use mains appliances in the presence of water.

6

Basic Concepts


Insulators include:

Rubber

Plastic

Most solid non metals

Glass

Glass, unless it is very hot, is one of the best insulators available.

7

Basic Concepts


Electric current (I) is measured in ampere (A) - I is the symbol used to indicate current.

The “amp” is a rather large unit for most electronic applications so we use the following sub-multiples:

1 mA = 0.001A that is 1 / 1 000 th of an ampere

You already know that 1mm is 1/1000th of an metre so there is nothing new here.

1 A = 0.000 001A that is 1 / 1 000 000 th of an ampere

8

Basic Concepts


Voltage is measured in volt (V)

1 mV = 0.001V that is 1 / 1 000 th of a volt

1 V = 0.000 001V that is 1 / 1 000 000 th of a volt

Common sub-multiples of the volt (less than a volt) include:

Common multiples of the volt (greater than a volt) include:

1 kV = 1 000 V

1 MV = 1 000 000 V

The kV and the MV are not common in electronics.

9

Basic Concepts


Resistance is measured in ohm ()

1 m = 0.001 that is 1 / 1 000 th of an ohm

1 = 0.000 001 that is 1 / 1 000 000 th of an ohm

This is pronounced micro ohm

Common sub-multiples of the ohm (less than an ohm) include:

Common multiples of the ohm (greater than an ohm) include:

1 k = 1 000

1 M = 1 000 000

The m and the are not common in electronics.

10

Basic Concepts


Capacitance is measured in farad (F)

1 m F = 0.001 F that is 1 / 1 000 th of a farad

1 F = 0.000 001 F that is 1 / 1 000 000 th of a farad

Common sub-multiples of the farad (less than a farad) include:

1 n F = 0.000 000 001 F that is 1 / 1 000 000 000 th of a farad

this is written in full as a nano farad

1 p F = 0.000 000 000 001 F that is 1 / 1 000 000 000 000 th of a farad

This is written in full as pico farad

11

Basic Concepts


Symbol Prefix Multiplication factorT tera 1012 1 000 000 000 000G giga 109 1 000 000 000M mega 106 1 000 000k kilo 103 1 000h hecto 102 100da deca 101 10d deci 10-1 0.1c centi 10-2 0.01m milli 10-3 0.001 micro 10-6 0.000 001n nano 10-9 0.000 000 001p pico 10-12 0.000 000 000 001f femto 10-15 0.000 000 000 000 001a atto 10-18 0.000 000 000 000 000 001

Here is a summary of many of the available multiples and sub-multiples

The most frequently used are in

bold.

12


Basic Concepts

Resistors are marked with a series of coloured rings to give us an idea of how big their resistance is.

The first band is orange. This means the first digit is 3.

The second band is blue. This means the second digit is 6.

The third band is red. This means that there are 2 zeros.

So the resistor is nominally 3 600 .or 3k6

This gold band indicates that the tolerance of the resistor is ±5% (plus or minus 5 percent). This means that its resistance is between 3 400 and 3 800 . We say it is nominally 3 600 .

13

Basic Concepts


The colours used for the first three bands and their meanings are as follows:

Colour Number Number of zerosBlack 0 noneBrown 1 0Red 2 00Orange 3 000Yellow 4 0 000Green 5 00 000Blue 6 000 000Violet 7 0 000 000Grey 8 00 000 000White 9 000 000 000

14

Basic Concepts


The colours used for the last band and their meanings are as follows:

Gold ± 5 %

Silver ± 10 %

No band ± 20 %

Resistors are manufactured in “preferred values”. That means that you can only buy certain values.

The preferred values for resistors with a tolerance of ±20% are: 10,15,22,33,47,68 and 100. These are just the first two significant figures. You can buy a 1500 but not a 2000 .

15

Basic Concepts


The preferred values for 10% resistors are:

10

12

15

18

22

27

33

39

47

56

58

82

100

&

16

Basic Concepts


The preferred values for 5% resistors are:

1011121315161820222427303336

39434751566268758291100

&

17

Basic Concepts

In summary:

Voltage is measured in

Current is measured in

Resistance is measured in

Capacitance is measured in

Volt (V)

Ampere (A)

ohm ()

farad (F)

3MV = 3 000 000 V

2kV = 2 000 V

5mV = 0.005 A

7A = 0.000 007 A

& 1nF = 0.000 000 001 F

1pF = 0.000 000 000 001 F

Home


18

Simple CircuitsSimple circuits have three main blocks of components in common that perform the same type of job. These are:

Input OutputProcessor

Return to menu slide

This is called a block diagram.

•The processor is the decision-making part of the system.

•The input is a sensor that transforms everyday phenomena such as temperature and heat to an electric signal that the processor can deal with.

•The output is a device that converts an electric signal from the processor into something that we want.

19



Return to previous slide

Examples of input devices include:

Pressure pads

LDRs

Thermistors

Reed switches

20




Examples of output devices include:

Lamps

LEDs

Motors

Solenoids

21




Examples of basic processors include:

Transistors

Operational amplifiers

22



What would the block diagram look like for a system that brought on a light when it got dark?

LDR LampProcessor


Home

23

Questions - Simple Circuits

1 Write down the units of voltage, capacitance, resistance and current. What is the symbol for current?

2 Write out 22mA and 420 pF in full.

3 Give three examples of input devices.

4 Write out a block circuit diagram for a device that could lift up a trap door when a beam of light was broken.

5 What is the nominal value and tolerance of this resistor?

6 What would the colours of a 47M resistor be?

ANSWER

ANSWER

ANSWER

ANSWER

ANSWER

ANSWER


24

Solutions - Simple Circuits

1 Write down the units of voltage, capacitance, resistance and current. What is the symbol for current?

Voltage volt (V)

Capacitance farad (F)

Resistance ohm ()

Current ampere (A)

The symbol for current is I

Return

25


2 Write out 22mA and 420 pF in full.

22mA = 22 x 0.001 A = 0 . 22 A

420 pF = 420 x 0.000 000 000 001 F = 0 . 000 000 000 42 F

Return

26

Return


3 Give three examples of input devices.

Pressure pads

LDRs

Thermistors

Reed switches

27


4 Write out a block circuit diagram for a device that could lift up a trap door when a beam of light was broken.

LDR MotorProcessor

or

LDR SolenoidProcessor

Note that a bulb is not an input device as it gives out light.

Return

28



The first band is brown. This means the first digit is 1.

The second band is black. This means the second digit is 0.

The third band is red. This means that there are 2 zeros.

So the resistor is nominally 1 200 .

This gold band indicates that the tolerance of the resistor is ±5% (plus or minus 5 percent).

Return

29


The first band is yellow. This means the first digit is 4.

The second band is violet. This means the second digit is 7.

The third band is blue. This means that there are 6 zeros.

The resistor is nominally 47 000 000 .


6 What would the colours of a 47M resistor be?

Return

30

Ohm’s Law

Ohm’s Law states that so long as the physical conditions remain constant, the current through a conductor is proportional to the voltage across it.

This gives us the formula:

Voltage = current x resistance

V = I R

We can rearrange this equation to give either

R = V / I

or

I = V / R


31

Ohm’s Law

What does it mean?

“Physical conditions remaining constant” - This really means as long as the temperature remains constant. Usually it does.

“The current through a conductor is proportional to the voltage across it” - this means that if you double the voltage, you get twice the current. Triple the voltage and you triple the current etc.

Voltage

CurrentLow resistance

High resistance


32

Ohm’s Law


Calculations using Ohm’s law fall into three types:

What is the resistance if ?

What is the current if ?

What is the voltage if ?(Use R = V / I) (Use V = I x R)

E.G.What resistance could you use with a 10V supply to limit the current to 15mA?

R = V / I = 10 / 0.015

667 so use 680

(Use I = V / R)E.G.A 430 resistor protects an LED in a 5V circuit. What is the current through the LED?I = V / R = 5 / 430= 12 mA

E.G.12mA runs through a prorctive resistor of resistance 820 . What is the voltage across the resistor ?V = IR = 0.012x820= 9.84 V

33

Ohm’s Law


The voltage across the diode is 0.7 V and the cell produces 1.5 V. What is the current through the resistor?

If you can’t see how to do it straight away, write the values given onto the diagram.

1.5V

0.7 V

Diode820

Voltage across the resistor = 1.5V (provided by the cell)

- 0.7V (lost across the diode) = 0.8V

Using I = V / R = 0.8 / 820 = 1mA

34

1 A power supply drives a current of 500mA through a bulb with a working resistance of 3. What voltage is the power supply?

2 A power supply provides 12 V to a bulb passing 3 A. What is the working resistance of the bulb?

3 A 47 k resistor has a pd of 9 V across it. What current passes through the resistor?

4 An 18V power supply is placed across a resistor of resistance 10k. What current will flow through the resistor?

5 The effective resistance of a small motor is 5 . What current passes through it if a cell of voltage 6 V is placed across it?

Ohm’s Law - Questions


Solution 1 Solution 2 Solution 3 Solution 4 Solution 5

Home

35

Ohm’s Law - Solutions


V = I x R

so V = 0.5 x 3

=1.5 volt

Return

36

2 A power supply provides 12 V to a bulb passing 3 A. What is the working resistance of the bulb?


V = I x R

So R = V / I

= 12 / 3

= 4 A

Return

37


3 A 47 k resistor has a pd of 9 V across it. What current passes through the resistor?

V = I x R

So I = V / R

= 9 / 47 000

= 0.000 191 A

= 191 A

Return

38


4 An 18V power supply is placed across a resistor of resistance 10k. What current will flow through the resistor?

V = I x R

so I = V / R

= 18 / 10 000

= 0.001 8 A

= 1.8 mA

Return

39

5 The effective resistance of a small motor is 5 . What current passes through it if a cell of voltage 6 V is placed across it?


V = I x R

So I = V / R

= 6 / 5

= 1.2 A

Return

40

Power Calculations

The detail from the bottom of an electrical appliance shown here gives a very useful, commonly used method of writing the power of the appliance.

20 VA is exactly the same as 20 W (20 watts).

The more powerful an appliance is, the greater the number will be. An electric fire might well be 2 or 3 kW (2 000 or 3 000 W).

1W is sometimes called 1VA because you can calculate the power by multiplying “the volts by the amps”!

Power = current x voltage or P = I VReturn main menu

41

Power Calculations

Remember that :

20VA means a power of 20W

and that

P = I x V


This device runs from a 230V mains supply. What can we learn from this information? Well, we know that P = IV; we also know the voltage and the power, so we can calculate the current I.

I = P / V

So I = 20 / 230

= 87 mA

42

Power Calculations


Can you match these typical power ratings with the device that they describe?

Ligh

t em

ittin

g di

ode

Halogen desk lamp

Power station

Electric lamp

Torch Bulb

43

Power Calculations


Light emitting diode

Halogen desk lamp

Power station

Electric lamp

Torch Bulb

44

Power Calculations - the formulae

Power = current x voltage

P = IV

so I = P / V and V = P / I

But from Ohm’s Law, V = I x R

So P = I x IR

P = I2R

And from Ohm’s Law, I = V / R

So P = (V/R) x V

P = V2/R


45

Power Calculations - Questions1 A diode has a voltage of 0.7V across it and a current of 100mA flowing through it. What is the power dissipated in the diode?

2 A wire carries a current of 5 mA and the power dissipated in the wire is 2.5 W. What is the voltage across the wire?

3 What is the current passing through a coil that dissipates 40mW when a voltage of 5V is applied across it?

4 A current of 10 mA passes through a 10 resistor. What is the power dissipated in the resistor?

5 A voltage of 9 V is applied across a 10 k resistor. What power is dissipated in the resistor?


Home

46

Power Calculations - Solutions1 A diode has a voltage of 0.7V across it and a current of 100mA flowing through it. What is the power dissipated in the diode?

P = IV

so P = 0.1 x 0.7

= 0.07

= 70 mW

47

Power Calculations - Solutions2 A wire carries a current of 5 mA and the power dissipated in the wire is 2.5 W. What is the voltage across the wire?

P = IV

So V = P / I

= 0.000 0025 / 0.005

= 0.000 5 W

= 0.5 mW

= 500 W

48

Power Calculations - Solutions3 What is the current passing through a coil that dissipates 40mW when a voltage of 5V is applied across it?

P = IV

So I = P / V

= 0.04 / 5

= 0.008 W

= 8mW

49

Power Calculations - Solutions4 A current of 10 mA passes through a 10 resistor. What is the power dissipated in the resistor?

P = I2R

= 0.012 x 10

= 0.001

=1mW

50

Power Calculations - Solutions5 A voltage of 9 V is applied across a 10 k resistor. What power is dissipated in the resistor?

P =V2/R

= 92 / 10 000

0.0081

= 8.1 mW

51


Resistors are said to be connected in series when the same current has to pass through each resistor i.e. the current does not have to split.

These three resistors are connected in series.

And so are these 5 resistors

Return to main menu

52


Resistors are said to be connected in parallel when the current has to split to pass through each resistor i.e. the current through each resistor might not be the same.

These three resistors are connected in parallel.

And so are these two.


53


These three resistors connected in series, could be replaced by one resistor of resistance 141M.

47 M 47 M 47 M

R1 R2 R3

That is 47 M + 47 M + 47 M = 141 M(This is not available so we might use a 150 M)

As a general formula we could write:

R total= R1 + R2 + R3 or R = R1 + R2 + R3


54

Resistors in Series and ParallelR1

R2

R3

The resistance of each of the resistors in this parallel network is 47 M. The effective resistance of three resistors is 15.7 M (use 16 M).

You would expect the resistance to be less than any of the individual resistances in the network as there are three possible routes for the electricity to take.The formula used to add the resistances is:

321

1111

RRRRtotal

or133221

321

RRRRRR

RRR

RTotal Return to previous slide

55

Resistors in Series and Parallel - Questions1 What is the combined resistance of a 39k and a 47k resistor connected in series?

2 What is the combined resistance of a 39k and a 47k resistor connected in parallel?

3 What is the combined resistance of a 10k , 20k and 47k resistor connected in series?

4 What is the combined resistance of a 10k , 20k and 47k resistor connected in parallel?

5 Suppose you need a 30k resistor but you only have a 27k and a handful of assorted resistors. What other resistor would you search for and how would you connect them together?


Home

56

Resistors in Series and Parallel - Answers1 What is the combined resistance of a 39k and a 47k resistor connected in series?


So R = 39 000 + 47 000 = 86 000

= 86k


57

Resistors in Series and Parallel - Answers2 What is the combined resistance of a 39k and a 47k resistor connected in parallel?

So 1/R = (1/39 000) +(1/47 000)

= 0.00002564 + 0.00002128 = 0.00004692

So R = 1/0.00004692 = 21 314

= 21.3k

= 22k


21

111

RRRTotal

58

Resistors in Series and Parallel - Answers3 What is the combined resistance of a 10 k, 20 k and 47 k resistor connected in series?


So R = 10 000 + 20 000 + 47 000

= 77 000

= 77 k


59

Resistors in Series and Parallel - Answers4 What is the combined resistance of a 10k, 20k and 47k resistor connected in parallel?

The numbers here could get very big so let us omit three zeros and give the answer in k as all the resistances are in k anyway.

R = (1/10) +(1/20) +(1/47)

= 0.1000 + 0.0500 + 0.0213 = 0.1713

=5.84 k

Use 5.8 kReturn to menu slide

321

1111

RRRRTotal

60

Resistors in Series and Parallel - Answers5 Suppose you need a 30 k resistor but you only have a 27 k and a handful of assorted resistors. What other resistor would you search for and how would you connect them together?

You are looking to increase the resistance by adding another resistor. This can only be done by adding a resistor in series.


We know that Rtotal is 30 k and R1 is 27 k

So 30 = 27 + R2

R2 = 30 - 27

= 3 k


61

CapacitorsCapacitors store charge. The greater the voltage that you apply to them, the greater the charge that they store. In fact the ratio of the charge stored to the voltage applied is called the capacitance.

Capacitance = Charge / Voltage

or C = Q / V

Capacitance is measured in farad (F) but the farad is a large unit of capacitance so you usually see microfarad F (millionth of a farad 10-6 or 0. 000 001 F), nanofarad nF (10-9) or picofarad pF (10-12).

The capacitor in the picture is a 470 F (0.000 47 F) polar (you must connect it the correct way around in the circuit) capacitor rated at 40V (the working voltage should not exceed 40V).

Return to main menu

62

Capacitors - Symbols

This is a non-polar capacitor - it does not matter which way around you place it in the circuit.

This is a polar capacitor. It is essential that the capacitor is connected into the circuit the correct way around.


63

Capacitors

Capacitor Characteristics

Time Constant Calculations

Capacitors in Series

Capacitors in Parallel

Questions


64


Closing the switch allows the capacitor to charge. As this happens, the voltage across the capacitor will rise in line with the fall of current through it as it becomes fully charged.

V

A

Voltage

Current

Time

Time


65


The circuit has now been adapted so that closing the switch allows the capacitor to discharge through the resistor. Note now that the current will fall as the voltage falls.

V

A

Voltage

Current

Time

Time


66


The circuit on the left allows us to investigate the charging and discharging of a capacitor simply.

Connecting the flying lead S to point X will charge the capacitor from the cell through the resistor R.

Connecting the flying lead S to Y will then discharge the capacitor through the resistor.

It has been found that increasing either the capacitance or the resistance will increase the time taken for the capacitor to charge.


A

V

X Y

C

R

S

67


When the flying lead S is connected to X, the capacitor will charge up through the resistor.At first there will be little or no charge in the capacitor so the current flows into the capacitor (via the resistor), quite rapidly.

The current through the resistor develops a voltage over the resistor. The voltage across the capacitor will be proportional to the charge in it. Since the charging has only just begun, it will be small but growing.

The capacitor begins to charge: It gets harder for more charge to flow into the capacitor so the current decreases. As the charge on the capacitor is increasing, the voltage across it increases too.


A

V

X Y

C

R

S

68

A

V

X Y

C

R

S


Eventually the capacitor will “fill”. This really means that it approaches the condition such that the voltage across it is equal to the supply voltage.

There will no longer be any current flowing.

The time taken to achieve this increases with increased capacitance and /or resistance.

Increasing the supply voltage makes no difference to the time taken for the voltage across the capacitor to approach the voltage across the supply.

The capacitor is said to be fully charged..


Home

69


The time constant is the time taken for:

the current or voltage to have fallen to 37% of its original value

or

the voltage to have risen to 63% of its original value

We can calculate the time constant for a circuit by multiplying the capacitance of the capacitor by the resistance of the resistor:

T = C x R

The units of the time constant are seconds if the resistance is in ohms and the capacitance in farads.

Return to menu

70


E.G. One

A 10M0 resistor is connected in series with a 470 pF capacitor. How long will it take to discharge the capacitor to 37V from an initial voltage of 100V?

Note that the voltage is falling to 37% of its initial value, so we are looking at one time constant.

Using T = C R

T = 0.000 000 000 470 x 10 000 000

= 0.004 7 s


71


E.G. Two

A 10M0 resistor is connected in series with a capacitor. If the time constant is 0.001s, what is the capacitance of the capacitor?

Using T = C R

0.001 = C x 10 000 000

C = 0.001 / 10 000 000

= 0.000 000 000 1 F

= 100 pF

Home


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Capacitors in SeriesC1 C2 C3

These three capacitors are connected in series. Their combined capacitance is given by:

321

1111

CCCCtotal

321

321

CCC

CCCCTotal

or

Return to menu

73

4 What is the combined capacitance of a 10 F, 20 F and 47 F capacitor connected in series?

The numbers here could get very small so let us omit 5 zeros and give the answer in F as all the capacitances are in F anyway.

C = (10 x 20 x 47) / (10 + 20 + 47)

= 9400 / 77

=122 F

321

321

CCC

CCCCTotal

Capacitors in Series

C1 C2 C3


Home

74


C3

C1

C2

These three capacitors are connected in parallel with each other.

Note that because they are in parallel, they must have the same voltage across each other.

The combined capacitance of the network of capacitances is given by:

Ctotal = C1 + C2 + C3

Return to menu

75


C3

C1

C2


Suppose that the capacitances are 10 F, 20 F and 47 F.

So C = 10 + 20 + 47

= 77 F


Home

76

Capacitors - Questions1 What is the combined capacitance of a 10pF and a 20pF capacitor connected in series?

2 What is the combined capacitance of a 10pF and a 20pF capacitor connected in parallel?

3 What is the time constant for a 1000 pF capacitor connected to a 100k resistance?

4 What capacitance would you need to combine with a 200 F capacitor to give the combinations a capacitance of 100 F? How would you connect them together to achieve this?

5 What resistor could you connect with a 47 F capacitor to give a time constant of 1.83 s ?


77

Capacitors - Answers

1 What is the combined capacitance of a 10pF and a 20pF capacitor connected in series?

The numbers here could get very small so let us omit 11 zeros and give the answer in pF as all the capacitances are in pF anyway.

C = (10 x 20) / (10 + 20)

= 200 / 30

=6.7 pF

321

321

CCC

CCCCTotal

C1 C2 C3

Return to menu

78



C3

C1

C2


So C = 10 + 20

= 30 pF

Return to menu

79


3 What is the time constant for a 1000 pF capacitor connected to a 100k resistance?

T = C x R

= (1000 x 0.000 000 000 001) x 100 000

= 0.000 1 s

Return to menu

80


4 What capacitance would you need to combine with a 200 F capacitor to give the combinations a capacitance of 100 F? How would you connect them together to achieve this?

In order to get a smaller capacitance, you need to connect them in series.

321

321

CCC

CCCCTotal

C1 C2 C3

So 100 = 200 x C2 / (200 + C2)100 (200 + C2) = 200C2

20 000 + 100C2 = 200C2

20 000 = 200C2 - 100C2

100 C2 = 20 000C2 = 200 F

Return to menu

81


5 What resistor could you connect with a 47 F capacitor to give a time constant of 1.83 s ?

T = C x R

1.83 = 0.000 047 x R

R = 1.83 / 0.000 047

= 38 936

Use 39 k

Return to menu

82

Alternating Current

Direct current (DC) is the current that comes from a cell or battery.

It is unidirectional. That is to say that the net drift of electrons is in one direction. This one direction will always be from positive to negative for electrons but negative to positive for conventional flow.

It is easier to convert voltages from one value to another if the direction of the current is rapidly changing.

This is called an alternating current (AC).

Return to main menu

83

Alternating Current

Alternating current has some strange properties:

•it can appear to pass through a capacitor

•it produces the discharges that you see in a plasma ball

•it can be stepped up (to higher voltages and lower current)

•it can be stepped down (to lower voltages and higher current)

Mains voltage is always due to an alternating current. It is used because it can be stepped up or down easily.


84

Alternating CurrentThroughout Europe, mains voltage is supplied at a frequency of 50 Hz.

You will remember that Hz is the abbreviation for hertz - the unit of frequency.This means that the electricity goes through one complete cycle 50 times every second.

This means that the voltage will:

•start from zero and build up in one direction until it reaches a maximum value (about 325 V).•Fall back to zero•Change direction and start to build up to a maximum value (about - 325 V)•Fall back to zero•50 times per second

Alternating voltages (and currents) can have extremely high frequencies. The current that produces radio waves can be many MHz (millions of hertz). Return to previous slide

85

Alternating Current - Questions

1 What is the frequency of the mains voltage in the UK and Europe?

2 What is the frequency of the mains voltage in the USA?

3 How long is a complete cycle of mains ac at 50 Hz?

4 How many times does the electricity in 50 Hz ac mains, change direction?

5 Write out 250 MHz in full.


Home

86

Alternating Current - Answers

1 What is the frequency of the mains voltage in the UK and Europe? 50Hz

2 What is the frequency of the mains voltage in the USA? 60Hz

3 How long is a complete cycle of mains ac at 50 Hz?

1/50th of a second = 0.02s

4 How many times does the electricity in 50 Hz ac mains, change direction?

Twice each cycle so 100 times

5 Write out 250 MHz in full

250 000 000 hertz

Return to menu

Home

87

Waveforms

As alternating currents and voltages vary with time, it is useful to have a graphical representation of them.

The electronic device that does this for us is called an oscilloscope.

The essence of an oscilloscope is its ability to plot a graph for us, showing the variation of voltage with time. It is particularly useful because it works extremely quickly and it barely interferes with the circuit from which you are taking the measurements.

When you first look at an oscilloscope, the number of knobs, levers and buttons can be bewildering but you will soon get used to it. In fact, you rarely need to use most of them.

Return to menu

88

WaveformsScreen showing the trace

On / off switch

x input

The trace shows a graph with voltage on the y axis (vertical) and time on the x axis (horizontal).


89

Waveforms

The waveform produced by mains voltage looks like this:

Voltage

time

This shape of waveform is called a sinusoidal wave.


90

Waveforms


Voltage

time

The arrows indicate a complete cycle. A cycle is the time that it takes to reach the next adjacent identical position on the waveform.


91


Voltage

time

This arrow indicates the peak voltage. For mains it is about 325V in the UK. The value of 230V that is quoted is the peak divided by the square root of 2. (It is the dc equivalent voltage that would produce the same heating effect as an ac with a peak voltage of 325V. Dividing by root 2 only works for sinusoidal waveforms). The peak voltage is the maximum voltage.

Waveforms


92


Voltage

time

This arrow indicates the peak to peak voltage. Notice that the peak to peak voltage is twice the peak voltage. For mains it is about 650V in the UK and Europe.

Waveforms


93

Waveforms

Another common waveform called the square wave looks like this:

Voltage

time

Waveforms



94

Waveforms


Voltage

time

Waveforms

This arrow indicates the peak voltage.


95

Waveforms


Voltage

time

Waveforms

This arrow indicates the peak to peak voltage. It will be twice the peak voltage


96

Waveforms

Another common waveform called the saw-tooth wave looks like this:

Voltage

time

Waveforms



97

Waveforms


Voltage

time

Waveforms

This arrow indicates the peak voltage.


98

Waveforms


Voltage

time

Waveforms

This arrow indicates the peak to peak voltage.


99

Waveforms - Questions

1 What is the name of the trace shown on the oscilloscope?

2 How many complete cycles can you see?

3 If the scale on the y-axis is 2V per division, estimate the peak voltage. What is the peak to peak voltage?

4 If the scale on the x-axis is 2ms per division, estimate the length of a cycle.

5 Using your answer to question 4, calculate the frequency of the wave.


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100

Waveforms - Answers

1 What is the name of the trace shown on the oscilloscope?

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101

Waveforms - Answers

2 How many complete cycles can you see?

ONE TWO THREE

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102

3 If the scale on the y-axis is 2V per division, estimate the peak voltage. What is the peak to peak voltage?

Waveforms - Answers

5 divisions

so 5 divisions x 2 V per division

= 10V

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103

Waveforms - Answers

4 If the scale on the x-axis is 2ms per division, estimate the length of a cycle.

Count as many complete cycles as you can to get as accurate an answer as possible.

3 cycles is about 35 divisions

1 cycle is about 11.7 divisions

11.7 divisions is 11.7 x 2 ms

= 23.4 ms

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104

Waveforms - Answers

5 Using your answer to question 4, calculate the frequency of the wave.

From question 4 one cycle is about = 23.4 ms23.4ms = 23.4 x 0.001s

= 0.0234sHow many of these can we get into 1 second?

= 1 / 0.0234= 42.7 Hz

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105


Voltage is sometimes referred to as potential difference. The potential divider simply divides up a potential or voltage.

In its simplest form it is two resistors placed across a power supply. The voltage across of each resistor is less than the supply voltage. Adding the voltage across each resistor will give the supply voltage. It is probably easiest to understand if you look at the diagram.

+9V

0 VHere the power supply is 9V. Note that both the resistances are the same.

The voltage from the supply will be split (divided) equally as 4.5V. Of course 4.5V + 4.5V = 9V.

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106

+9V

0 V

The Potential DividerIn this potential divider circuit, the resistances are not the same.

1 000 2 000

The bigger resistance here means that there will be a bigger voltage here.

The smaller resistance here means that there will be a smaller voltage here.

It is 2/3rds of the resistance.

It is 2/3rds of the resistance.

It is 1/3rd of the resistance.

It is 1/3rd of the resistance.

It develops 2/3rds of the voltage.

It develops 2/3rds of the voltage.

It develops 1/3rd of the voltage.

It develops 1/3rd of the voltage.

2/3 x 9V = 6V2/3 x 9V = 6V2/3 x 9V = 6V2/3 x 9V = 6V

1/3 x 9V = 3V1/3 x 9V = 3V1/3 x 9V = 3V1/3 x 9V = 3V

If you would like to work through this again, step back through the sequence using the left arrow key.


107

+V

0 V


R1

R2

V1

V2

You can calculate the voltage across each resistor using a formula too.

VV11 = V x R = V x R11 / (R / (R11 + R + R22))VV11 = V x R = V x R11 / (R / (R11 + R + R22))

V1 = 9 x 2 000 / (1 000 + 2000) = 9 x 2 / 3 = 6VV1 = 9 x 2 000 / (1 000 + 2000) = 9 x 2 / 3 = 6V


V1 = 9 x 1 000 / (1 000 + 2000) = 9 x 1 / 3 = 3VV1 = 9 x 1 000 / (1 000 + 2000) = 9 x 1 / 3 = 3V


108


The potential divider does not have to be made up of two fixed resistors. One of them could be variable, or even both.

R1

R2

V1

V2

As R1 increases so does V1 but V2 will fall.As R1 decreases so does V1 but V2 will rise.

It is just as if there is only one cake to go around (the voltage). If R1 increases then V1gets more cake so there is less left for V2!


109

The Potential DividerNow the variable resistor has been moved to the lower position in the network..

R1

R2

V1

V2

As R2 increases so does V2 but V1 will fall.As R2 decreases so does V2 but V1 will rise.

It can be handy to change the position of the variable resistor. Later you will see that it can change the action of a transistor circuit so make sure that you follow it.


110


+V

0V

With this potential divider, the “tap” in the middle is a “slider”. It probably moves along a track of carbon.

R1 and V1 will be the resistance and voltage “above” the slider.

R1V1

R2 and V2 will be the resistance and voltage “below” the slider.

R2 V2

As R1 increases, R2 decreases. This will result in V1 increasing and V2 decreasing.As R1 decreases, R2 increases. This will result in V1 decreasing and V2 increasing.


111



Here is a component that could be used as a potential divider.

The black ring is the carbon track.

You adjust it by putting a screwdriver in here and turning the outer metal ring.

112



You adjust it by putting a screwdriver in here and turning the outer metal ring.

Here is another:

113



It is possible to use many different components that vary their resistance in a potential divider circuit. Here are a few that you might find and the physical conditions that change their resistance.

Light Dependent Resistor (LDR) - decreases resistance with increased illumination.

Thermistor - decreases its resistance with increased temperature (negative temperature coefficient).

Microphone - changes resistance with sound.

Strain gauge - changes its resistance when stressed.

Photodiode - decreases resistance with increased illumination.

114

The Potential Divider - Questions

1 A 2k and a 3k resistor are used in potential divider using a 10V supply. Sketch a possible set up and label the resistors and the voltages across them.

2 Suggest two possible fixed resistors that could be used to obtain 3V from a 15V supply.

3 A 27k and a 62k resistor are used in potential divider using a 12V supply. Sketch a possible set up and label the resistors and the voltages across them.

4 A 15V supply is attached across a potential divider. If one of the resistors is a 390k and there is a voltage of 9V across the other, what is the second resistance?

5 A potential divider is created from a fixed resistor and an LDR. Explain how the network produces different voltages.


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Answer

Answer

Answer

Answer

Answer

115

The Potential Divider - Answers1 A 2k and a 3k resistor are used in potential divider using a 10V supply. Sketch a possible set up and label the resistors and the voltages across them.

+10V

0 V

R1 =

2 000

V1= 4V

V2 = 6V


V1 = 10 x 2 000 / (3 000 + 2000) = 10 x 2 / 5 = 4VV1 = 10 x 2 000 / (3 000 + 2000) = 10 x 2 / 5 = 4V


V1 = 10 x 3 000 / (3 000 + 2000) = 10 x 3 / 5 = 6VV1 = 10 x 3 000 / (3 000 + 2000) = 10 x 3 / 5 = 6V

R2 =

3 000

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116

The Potential Divider - Answers2 Suggest two possible fixed resistors that could be used to obtain 3V from a 15V supply.

VV11 = V x R = V x R11 / (R / (R11 + R + R22))VV11 = V x R = V x R11 / (R / (R11 + R + R22)) So 3 = 15 x R1 / (R1 + R2)

We could use more or less any combination. However, they should be quite high so that they do not drain a lot of current.

Let us choose R1 as being 100k.

3 = 15 x 100 / (100 +R2)

3(100 + R2) = 1 500

300 + 3R2 = 1 500

3R2 = 1 500 - 300 = 1 200

R2 = 1 200 / 3 = 400k.Return to menu

117

The Potential Divider - Answers3 A 27k and a 62k resistor are used in potential divider using a 12V supply. Sketch a possible set up and label the resistors and the voltages across them.

+12V

0 V

R1 =

27k V1= 3.64V

V2 = 8.36V


V1 = 12 x 27 / (27 + 62) = 12 x 27 / 89 = 3.64VV1 = 12 x 27 / (27 + 62) = 12 x 27 / 89 = 3.64V


V1 = 12 x 62 / (27 + 62) = 12 x 62 / 89 = 8.36VV1 = 12 x 62 / (27 + 62) = 12 x 62 / 89 = 8.36V

R2 =

62k

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118

The Potential Divider - Answers4 A 15V supply is attached across a potential divider. If one of the resistors is a 390k and there is a voltage of 9V across the other, what is the second resistance?


So V1 = 15 - 9 = 6V

Now 6 = 15 x R1 / (R1 +390)

6 (R1 + 390) = 15 R1

6 R1 + 2 340 = 15 R1

9R1 = 2 340

R1 = 2 340 / 9 = 260k so use 270k

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119

The Potential Divider - Answers5 A potential divider is created from a fixed resistor and an LDR. Explain how the network produces different voltages.

In bright illumination the LDR’s resistance falls. The voltage across the LDR will consequently fall while the voltage across the fixed resistance will rise.

As it goes dark, the LDR’s resistance will increase, increasing the voltage across the LDR. As this happens, the voltage across the resistor will fall.

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120

+0.7V

Transistor Circuits

base

emitter

collector

The transistor is a three connection component that can be used either as an amplifier or a switch.

Essentially the circuit is set up so as to try to force electrons through the emitter and out of the collector. This might be to light a bulb.

+V

0V

However, under normal circumstances, there is a very high resistance between the emitter and the collector so the bulb will not light.If we make the base go positive, the collector / emitter junction conducts and the bulb will light. Return to menu

121

Transistor Circuits

The clever part now is to control the base bias voltage that turns the transistor on.

+V

0V

The base bias voltage is the voltage between the base and the emitter. If it is anything much less that 0.7V, the transistor will be off.

The transistor switches on when it is 0.7V.

You should never allow the base bias voltage to get too high as this will overheat the base and burn out the transistor. For this reason you will frequently find a resistor connected to the base.

c

e

bRb

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122

Transistor Circuits


+V

0V

c

e

b

This can be achieved using a potential divider.

Rb

R1

R2

Correct selection of the two resistors R1 and R2 will take the base to 0.7V and turn the transistor on. Suppose R2 was much higher than R1. The voltage across R2 would be high so the transistor would switch on. Return

123

Transistor Circuits


+V

0V

c

e

bRb

R1

R2

Now suppose R2 was an LDR.In the bright light, its resistance would be low so the voltage across it would be low, the transistor switched off and the lamp off.

But suppose that it now goes dark!

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124

Transistor Circuits


+V

0V

c

e

bRb

R1

R2

It has just gone dark!

The resistance of the LDR rises.

The voltage across the LDR rises.

The base bias voltage reaches 0.7V

The transistor switches on. The bulb lights. Return

125

Transistor Circuits


+V

0V

c

e

bRb

R1

R2

Suppose that we now swap the positions of the resistor and the LDR.

The bulb will now come on in daylight! It might be useful as a warning light circuit in certain circumstances.

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126

Transistor Circuits


+V

0V

c

e

bRb

R1

R2

Now let us consider:•Ib the base current that flows into the transistor•Ie the emitter current that flows out of the transistor

Ib

Ie

•IC the collector current that flows into the transistor

Ie = Ib + Ic

Ic

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127

Transistor Circuits


+V

0V

c

e

bRb

R1

R2

Ib

Ie

Ie = Ib + Ic

IcThe base current will be very small as it has passed through R1 and Rb so it is almost true that Ie = Ic.

The ratio of Ic : Ib is important. It shows that the transistor is amplifying. It is often around about 100.

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128

Transistor Circuits


+V

0V

c

e

bRb

R1

R2

Ib

Ie

Ie = Ib + Ic

IcThat is to say that the collector current is a always a constant amount bigger than the base current. Feed a small current to the base and you get a big current in the collector.

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129

Transistor Circuits


+V

0V

c

e

bRb

R1

R2

Ib

Ie

Ie = Ib + Ic

IcThe ratio is called hfe.

hfe = Ic / Ib

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130

Click on a component to find out what it does.

Transistor Circuits

+V

0V

c

e

bRb

R1

R2

C1

Relay

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131


Capacitor

This stores charge. It acts as a time delay to any switching. If the transistor is on and tries to go off, it will act as a reservoir and keep the transistor on for a while longer.

Transistor Circuits

+V

0V

c

e

bRb

R1

R2

C1

Relay

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LDR Light Dependent Resistor

Its resistance decreases with increased illumination. In the dark, the resistance goes up turning the transistor on.

Transistor Circuits

+V

0V

c

e

bRb

R1

R2

C1

Relay

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133


Base Bias Resistor

This fixed resistor protects the base from too much current.

Transistor Circuits

+V

0V

c

e

bRb

R1

R2

C1

Relay

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Potential Divider

The LDR and R2 are a potential divider.

Transistor Circuits

+V

0V

c

e

bRb

R1

R2

C1

Relay

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135


Transistor

A small voltage at the base will allow current to flow through the emitter from the collector.

Transistor Circuits

+V

0V

c

e

bRb

R1

R2

C1

Relay

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136


The diode diverts the currents formed by this process.

Diode - this only allows current to flow in the direction of the arrow head. Rapid changes in the magnetic field of the relay can cause high voltage that would damage the transistor.

Transistor Circuits

+V

0V

c

e

bRb

R1

R2

C1

Relay

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137


Relay - current through the relay produces a magnetic field that throws a switch in another external circuit. The external circuit can be a much higher powered circuit.

Transistor Circuits

+V

0V

c

e

bRb

R1

R2

C1

Relay

CONTINUEReturn

138

Answers:

1 2 3 4 5

Transistor Circuits - Questions

1 The collector current in a circuit is 120mA when the base current is 3mA. What is hfe and the emitter current?

2 Why do we connect a resistor directly to the base of the transistor?

3 Sketch a circuit that will throw a relay in the dark that in turn will turn on a switch.

4 Why should a diode be connected across a relay in the collector circuit of a network?

5 Explain what happens when the resistance of the base bias resistor falls in a transistor circuit controlling a motor.


139

Transistor Circuits - Answers

1 The collector current in a circuit is 120mA when the base current is 3mA. What is hfe and the emitter current?

hfe = Ic / Ib

= 120 / 3

= 40

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140


2 Why do we connect a resistor directly to the base of the transistor?

The resistor limits the current entering the base. This stops the base from overheating due to excessive currents which would burn the transistor

out.

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141


3 Sketch a circuit that will throw a relay in the dark that in turn will turn on a switch.

+V

0V

c

e

b

R1

C1

Relay

LDR

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142


4 Why should a diode be connected across a relay in the collector circuit of a network?

The relay can create large voltages due to rapid changes in magnetic fields as they switch off. The diode provides a short circuit for the

current due to this voltage. As diodes only carry current in one direction, the diode connected to the relay will have no effect in normal operation

of the relay.

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143


5 Explain what happens when the resistance of the base bias resistor falls in a transistor circuit controlling a motor.

As the resistance falls, so will the voltage across it.

The voltage across the base will consequently fall.

This will turn off the transistor.

The collector current will fall to zero (or extremely close to zero).

The motor will switch off.

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144

Electronics Questions



3 What is the current passing through a heater that dissipates 40W when a voltage of 10V is applied across it?

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145

4 What is the combined resistance of a 30k , 20k and 47k resistor connected in parallel?

5 Suppose you need a 50k resistor but you only have a 47k and a handful of assorted resistors. What other resistor would you search for and how would you connect them together?



8 What is the time constant for a 1000 F capacitor connected to a 10M resistance?



146

9 If the scale on the y-axis is 4mV per division, estimate the peak voltage. What is the peak to peak voltage?

10 If the scale on the x-axis is 50 s per division, estimate the length of a cycle. Using your answer, calculate the frequency of the wave.



147


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148


The first band is violet. This means the first digit is 7.

The second band is green. This means the second digit is 5.

The third band is black. This means that there are no zeros.

So the resistor is nominally 75 .


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Electronics Answers

149


V = I x R

so V = 0.15 x 10

=1.5 V

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Electronics Answers

150

3 What is the current passing through a coil that dissipates 40W when a voltage of 10V is applied across it?

P = IV

So I = P / V

= 40 / 10

= 4 W

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Electronics Answers

151

4 What is the combined resistance of a 30k, 20k and 47k resistor connected in parallel?

321

321

RRR

RRRRTotal

The numbers here could get very big so let us omit three zeros and give the answer in k as all the resistances are in k anyway.

R = (30 x 20 x 47) / (30 + 20 + 47)

= 28 200 / 97

=291 k

Use 300 kReturn to menu

Electronics Answers

152

5 Suppose you need a 50 k resistor but you only have a 47 k and a handful of assorted resistors. What other resistor would you search for and how would you connect them together?

You are looking to increase the resistance by adding another resistor. This can only be done by adding a resistor in series.


We know that Rtotal is 50 k and R1 is 47 k

So 50 = 47 + R2

R2 = 50 - 47

= 3 k

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Electronics Answers

153


The numbers here could get very small so let us omit 11 zeros and give the answer in pF as all the capacitances are in pF anyway.

C = (47 x 20) / (47 + 20)

= 940 / 67

=14 pF

321

321

CCC

CCCCTotal

C1 C2 C3

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Electronics Answers

154


C3

C1

C2


So C = 47 + 20

= 67 pF

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Electronics Answers

155

8 What is the time constant for a 1000 F capacitor connected to a 10M resistance?

T = C x R

= (1000 x 0.000 001) x 10 000 000

= 10 000 s

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Electronics Answers

156

9 If the scale on the y-axis is 4mV per division, estimate the peak voltage. What is the peak to peak voltage?

5 divisions

so 5 divisions x 4mV per division

= 20mV

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Electronics Answers

157

10 If the scale on the x-axis is 50 s per division, estimate the length of a cycle.

Count as many complete cycles as you can to get as accurate an answer as possible.

3 cycles is about 35 divisions

1 cycle is about 11.7 divisions

11.7 divisions is 11.7 x 2 s

= 23.4 s

Electronics Answers

158

10 continued Using your previous answer, calculate the frequency of the wave.

From question 4 one cycle is about = 23.4 s23.4 s = 23.4 x 0.000 001s

= 0.000 023 4sHow many of these can we get into 1 second?

= 1 / 0.000 023 4s= 42.7 kHz

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Electronics Answers

159


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